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How Do I Get My Dog to Drink More Water?
Written by Darwin's
Proper hydration is essential for your dog's overall health and well-being. However, some dogs may not drink enough water, leading to potential health issues. There are many effective and practical ways to encourage your dog to drink more water. By implementing these strategies, you can ensure that your furry companion stays adequately hydrated.
Understanding the Importance of Hydration: How Much Water Does A Dog Need?
Ensuring that your dog gets enough water is crucial because it helps maintain proper organ function, aids digestion, regulates body temperature, and promotes healthy skin and coat. Hydration is especially vital during hot weather to help prevent heat exhaustion or if your dog is active or nursing puppies.
Determining the appropriate amount of water for a dog to drink depends on several factors, including the dog's size, activity level, and environmental conditions. According to the American Kennel Club (AKC), dogs generally require 0.5 to 1 ounce of water per pound of body weight per day. For example, a 50-pound dog would need approximately 50 to 100 ounces (or about 6 to 12 cups) of water daily. However, it's important to note that this is a general guideline, and individual needs may vary. Factors such as exercise, weather conditions, and the presence of certain health conditions can increase a dog's water intake requirements. It's essential to monitor your dog's water consumption and ensure fresh, clean water is readily available throughout the day. Consulting with a veterinarian is advised for specific recommendations based on your dog's unique needs and circumstances. (Source: American Kennel Club, "How Much Water Should a Dog Drink?")
Signs and Symptoms of Dehydration in Dogs
To address your dog's water intake, it's crucial to be aware of the signs that indicate dehydration.
If you notice any of these signs of dehydration in your dog, it is crucial to take immediate action to rehydrate them. Offer fresh, clean water and encourage them to drink. If your dog continues to show signs of dehydration or if you are concerned about their health, it is advisable to consult your veterinarian for proper guidance and treatment.
Remember, prevention is key when it comes to dehydration. Always ensure that your dog has access to clean, fresh water throughout the day, especially during warm weather or after physical activity. Monitoring your dog's water intake and being attentive to any signs of dehydration will help maintain their overall health and well-being.
7 Ways to Get your Dog to Drink More Water
Make Fresh Water Easily Accessible
Ensure that fresh water is always available for your dog. Place multiple water bowls in easily accessible areas around your home and yard. Consider using spill-proof bowls, especially if you have an active or curious pup.
Choose the Right Water Bowl
The type of water bowl you use can influence your dog's willingness to drink. Opt for stainless steel or ceramic bowls, as they are less likely to retain odors or bacteria. Avoid plastic bowls, as they can leach chemicals and develop scratches that harbor bacteria.
Keep Water Fresh and Clean
Dogs prefer clean and fresh-tasting water. Change the water in your dog's bowls daily and wash them thoroughly. This will prevent the buildup of residue or contaminants that might discourage your dog from drinking.
Enhance Water Palatability
If your dog is hesitant to drink plain water, try adding flavor to make it more enticing. You can use low-sodium chicken or beef broth, or even a small amount of unsalted tuna water. However, avoid additives like sugar, salt, or artificial flavors.
Hydration through Food and Treats
Incorporate moisture-rich foods into your dog's diet to boost hydration. Wet dog food or adding water to dry kibble can help increase their overall water intake. Additionally, you can freeze low-sodium broth or water into ice cube trays and offer them as hydrating treats.
Consider a Dog Water Fountain
Some dogs are naturally drawn to running water. Investing in a dog water fountain can be an effective way to encourage them to drink more. The flowing water mimics a freshwater source, making it more appealing to your furry friend.
Set a Drinking Routine
Establishing a routine can help regulate your dog's water intake. Offer water at regular intervals throughout the day, especially after meals and physical activity. By creating consistency, you can encourage your dog to drink regularly.
Promoting Optimal Hydration for a Healthy and Happy Dog
Proper hydration is crucial for your dog's overall health and well-being. By implementing these simple tips, you can ensure that your dog stays hydrated and avoids potential health issues associated with dehydration. Remember to monitor their water intake and consult a veterinarian if you notice persistent signs of dehydration. With your care and attention, your beloved companion will thrive with a healthy and hydrated lifestyle.
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187501 | https://en.wikipedia.org/wiki/Censure_in_the_United_States | Jump to content
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Contents
1 Presidential censures
1.1 Adopted resolutions
1.2 Other censure attempts
2 Senatorial censures
3 House censures
4 Cabinet censures
5 Censure at other levels of government
6 Chronology of censures
7 References
8 Further reading
Censure in the United States
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From Wikipedia, the free encyclopedia
Formal statement of disapproval towards a U.S. government figure
Not to be confused with Censorship in the United States.
Censure is a formal, public, group condemnation of an individual, often a group member, whose actions run counter to the group's acceptable standards for individual behavior. In the United States, governmental censure is done when a body's members wish to publicly reprimand the president of the United States, a member of Congress, a judge or a cabinet member. It is a formal statement of disapproval. It relies on the target's sense of shame or their constituents' subsequent disapproval, without which it has little practical effect when done on members of Congress and no practical effect when done on the president.
The United States Constitution specifically grants impeachment and conviction powers, respectively, to the House of Representatives and Senate. It also grants both congressional bodies the power to expel their own members, though it does not mention censure. Each body adopts rules allowing censure, which is "stronger than a simple rebuke, but not as strong as expulsion." In general, each house of Congress is responsible for invoking censure against its own members; censure against other government officials is not common. Because censure is not specifically mentioned as the accepted form of reprimand, many censure actions against members of Congress may be listed officially as rebuke, condemnation, or denouncement.
Like a reprimand, a censure does not remove a member from their office so they retain their title, stature, and power to vote. There are also no legal consequences that come with a reprimand or censure. The main difference is that a reprimand is "considered a slap on the wrist and can be given in private and even in a letter," while a censure is "a form of public shaming in which the politician must stand before their peers to listen to the censure resolution."
Presidential censures
[edit]
Adopted resolutions
[edit]
There have been four cases in U.S. history where the House of Representatives or the Senate adopted a resolution that, in its original form, would censure the president.
The 1834 censure of President Andrew Jackson "remains the clearest case of presidential censure by resolution." In 1834, while under Whig control, the Senate censured Jackson, a member of the Democratic Party, for withholding documents relating to his actions in defunding the Bank of the United States. During the waning months of Jackson's term, his Democratic allies succeeded in expunging the censure.
In 1860, the House of Representatives adopted a resolution admonishing both President James Buchanan and Secretary of the Navy Isaac Toucey for allegedly awarding contracts on the basis of "party relations." The House may have intended this resolution as a lesser reprimand than a formal censure.
In two other cases, the Senate adopted a resolution that was originally introduced to censure the president, but that, in its final form, did not overtly censure the president.
In 1864, during the American Civil War, Senator Garrett Davis introduced a resolution to censure President Abraham Lincoln for allowing two individuals to resume their service as generals after winning election to Congress. The final resolution adopted by the Senate required generals to be "re-appointed in the manner provided by the Constitution," but did not overtly censure Lincoln.
In 1912, Senator Joseph Weldon Bailey introduced a resolution censuring President William Howard Taft for allegedly interfering with a disputed Senate election. The final Senate resolution did not specifically refer to Taft, but stated that presidential interference in a disputed Senate race would warrant censure.
Other censure attempts
[edit]
Several other presidents have been subject to censure attempts in which no formal resolution was adopted by either the House or the Senate. In 1800, Representative Edward Livingston of New York introduced a censure motion against President John Adams. In 1842, Whigs attempted to impeach President John Tyler following a long period of hostility with the president. When that action could not get through Congress, a select Senate committee dominated by Whigs censured Tyler instead. In 1848, Congressman George Ashmun led an effort to censure President James K. Polk, on the grounds that the Mexican–American War had been "unnecessarily and unconstitutionally begun by the President." The House of Representatives voted to add Ashmun's censure as an amendment to a resolution under consideration by the House, but the resolution itself was never adopted by the House. In 1871, Senator Charles Sumner introduced an unsuccessful resolution to censure President Ulysses S. Grant for deploying ships to the Dominican Republic without the approval of Congress. In 1952, Congressman Burr Powell Harrison introduced a resolution censuring President Harry S. Truman for seizing control of steel mills during the 1952 steel strike. The resolution ultimately did not receive a vote.
President Richard M. Nixon was the subject of several censure resolutions introduced in the House of Representatives; most of the resolutions were related to the Watergate scandal. In 1972, a resolution censuring Nixon for his handling of the Vietnam War was introduced. A separate series of censure resolutions were introduced after the "Saturday Night Massacre" in October 1973. Another series of resolutions were introduced in July 1974. None of the resolutions were adopted, but Nixon resigned from office in August 1974.
In 1998, resolutions to censure President Bill Clinton for his role in the Clinton–Lewinsky scandal were introduced and failed. The activist group MoveOn.org originated in 1998, after the group's founders began a petition urging the Republican-controlled Congress to "censure President Clinton and move on"—i.e., to drop impeachment proceedings, pass a censure of Clinton, and focus on other matters. From 2005 to 2007, members of Congress introduced several resolutions to censure President George W. Bush and other members of the Bush administration. Most of the resolutions focused on Bush's handling of the Iraq War, but one resolution concerned the administration's "unlawful authorization of wiretaps of Americans" and two others alleged that Bush and Attorney General Alberto Gonzales had violated "statutes, treaties, and the Constitution." From 2013 to 2016, members of Congress introduced several resolutions to censure President Barack Obama. These resolutions charged that Obama had usurped the "legislative power of Congress” or had acted unlawfully. None of the resolutions to censure Bush or Obama were adopted.
On August 18, 2017, a resolution was introduced in the House to censure President Donald Trump for his comments "that 'both sides' were to blame for the violence in" the Unite the Right rally. On January 18, 2018, another motion to censure Trump was introduced in the House of Representatives by Rep. Cedric Richmond (D), who at the time was the Chairman of the Congressional Black Caucus, for Trump's remark, alleged by people in the room, stating "Why do we want all these people from 'shithole countries' coming here?" According to people in the room at the time, Trump was referring to people from Haiti and African nations coming to the United States. The censure motion failed to reach any legislative action. This comment was alleged to have been made on January 11, 2018, in an Oval Office meeting with lawmakers regarding immigration.
Senatorial censures
[edit]
See also: List of United States senators expelled or censured
The U.S. Senate has developed procedures for taking disciplinary action against senators through such measures as formal censure or actual expulsion from the Senate. The Senate has two basic forms of punishment available to it: expulsion, which requires a two-thirds vote; or censure, which requires a majority vote. Censure is a formal statement of disapproval. While censure (sometimes referred to as condemnation or denouncement) is less severe than expulsion in that it does not remove a senator from office, it is nevertheless a formal statement of disapproval that can have a powerful psychological effect on a member and on that member's relationships in the Senate.
In the history of the Senate, 10 U.S. Senators have been censured, the most famous being Joseph McCarthy. Their transgressions have ranged from breach of confidentiality to fighting in the Senate chamber and more generally for "conduct that tends to bring the Senate into dishonor and disrepute".
House censures
[edit]
See also: List of United States representatives expelled, censured, or reprimanded
The House of Representatives is authorized to censure its own members by the scope of United States Constitution (Article I, Section 5, clause 2). In the House of Representatives, censure is essentially a form of public humiliation carried out on the House floor. As the Speaker of the House reads out a resolution rebuking a member for a specified misconduct, that member must stand in the House well and listen to it. This process has been described as a morality play in miniature.
Most cases arose during the 19th century. Censure has been successful 26 times. In the modern history of the United States House Committee on Standards of Official Conduct (since 1966), censure has been successful nine times.
Cabinet censures
[edit]
The first attempted use of censure in the United States was directed at George Washington's treasury secretary Alexander Hamilton, who was accused of misadministration of two Congressionally authorized loans under the Funding Act of 1790 by William Giles.
Augustus Hill Garland, Attorney General in Grover Cleveland's administration, was censured in 1886 for failing to provide documents about the firing of a federal prosecutor.
Censure at other levels of government
[edit]
In Houston Community College System v. Wilson (2022) the Supreme Court of the United States held that the First Amendment to the United States Constitution does not prevent local government bodies from censuring their own members.
Chronology of censures
[edit]
To date, Andrew Jackson is the only sitting President of the United States to be successfully censured, although his censure was subsequently expunged from official records. Between 2017 and 2020, several Members of Congress introduced motions to censure President Donald Trump for various controversies, including as a possible substitute for impeachment during the Trump-Ukraine scandal, but none were successful.
On December 2, 1954, Republican Senator Joseph McCarthy from Wisconsin was censured by the United States Senate for failing to cooperate with the subcommittee that was investigating him, and for insulting the committee that was recommending his censure.
On June 10, 1980, Democratic Representative Charles H. Wilson from California was censured by the House of Representatives for "financial misconduct", as a result of the "Koreagate" scandal of 1976. "Koreagate" was an American political scandal involving South Koreans seeking influence with members of Congress. An immediate goal seems to have been reversing President Richard Nixon's decision to withdraw troops from South Korea. It involved the KCIA (now the National Intelligence Service) funneling bribes and favors through Korean businessman Tongsun Park in an attempt to gain favor and influence. Some 115 members of Congress were implicated.
On July 20, 1983, Representatives Dan Crane, a Republican from Illinois, and Gerry Studds, a Democrat from Massachusetts, were censured by the House of Representatives for their involvement in the 1983 Congressional page sex scandal.
On July 12, 1999, the U.S. House of Representatives censured (in a 355-to-0 vote) a scientific publication titled "A Meta-analytic Examination of Assumed Properties of Child Sexual Abuse Using College Samples", by Bruce Rind, Philip Tromovich, and Robert Bauserman; (see Rind et al. controversy) which was published in the American Psychological Association's "Psychological Bulletin (July 1998).
On July 31, 2007, retired Army General Philip Kensinger was censured by the United States Army for misleading investigators of the Pat Tillman death in 2004.
On July 6, 2009, South Carolina Republican Governor Mark Sanford was censured by the South Carolina Republican Party executive committee for traveling overseas on taxpayer funds to visit his mistress.
On October 13, 2009, the mayor of Sheboygan, Wisconsin, Bob Ryan, was censured due to a YouTube video that showed him making sexually vulgar comments about his sister-in-law taken at a bar on a cell phone. The censure was voted 15-0 by the Sheboygan Common Council. His powers were also quickly reduced by the Common Council, and he was ultimately removed from office two and a half years later in a recall election for continued improprieties in office.
In November 2009, members of the Charleston County Republican Party censured Republican Senator Lindsey Graham of South Carolina in response to his voting to bail out banks and other Wall Street firms, and for his views on immigration reform and cap-and-trade climate change legislation.
On December 2, 2010, Democratic Rep. Charlie Rangel from the State of New York was censured after an ethics panel found he violated House rules, specifically failing to pay taxes on a villa in the Dominican Republic, improperly soliciting charitable donations, and running a campaign office out of a rent-stabilized apartment meant for residential use.
On January 4, 2010, members of the Lexington County Republican Party censured Senator Lindsey Graham of South Carolina for his support of government intervention in the private financial sector and for “debasing” longstanding Republican beliefs in economic competition.
On January 22, 2013, the Arizona Republican Party censured longtime Sen. John McCain for his record of occasionally voting with Democrats on some issues.
On February 6, 2021, the Wyoming Republican Party censured Rep. Liz Cheney, the House Republican Conference Chair and third highest-ranking member of the House Republican leadership, for her vote to impeach former President Donald Trump during his second impeachment.
On February 13, 2021, the Louisiana Republican Party censured Senator Bill Cassidy, the senior U.S. senator from Louisiana, for his vote to convict former President Donald Trump during his second impeachment trial.
On February 15, 2021, the North Carolina Republican Party's central committee voted to censure U.S. Senator Richard Burr for his vote to convict former president Donald Trump during his second impeachment trial.
On March 16, 2021, the Alaska Republican Party censured U.S. Senator Lisa Murkowski for her vote to convict former president Donald Trump during his second impeachment trial.
On November 17, 2021, the Democratic-controlled U.S. House of Representatives passed a measure to censure Republican Rep. Paul Gosar for posting an anime video of him killing fellow Representative Alexandria Ocasio-Cortez and attacking President Joe Biden.
On January 22, 2022, the Arizona Democratic Party censured U.S. Senator Kyrsten Sinema for blocking voting rights.
On February 4, 2022, the Republican National Committee voted to formally censure Rep. Liz Cheney and Rep. Adam Kinzinger for their participation in the United States House Select Committee on the January 6 Attack.
On June 21, 2023, the Republican-controlled U.S. House of Representatives passed a measure to censure Democratic Rep. Adam B. Schiff for pressing allegations that Donald Trump's 2016 presidential campaign colluded with Russia, a week after a first attempt to censure Schiff was blocked.
On November 7, 2023, in a 234–188 vote the U.S. House of Representatives censured Rep. Rashida Tlaib (D-Mich.) for her remarks related to the Israel–Hamas war. This marked the second attempt to censure Tlaib, who was accused of "promoting false narratives" and "calling for the destruction of the state of Israel". Tlaib had shared a video on social media that used the phrase "from the river to the sea". The censure was supported by 22 Democrats and drew attention as a symbolic move, given Tlaib's status as the only Palestinian American in Congress. Despite criticism from members of both parties, most Democrats opposed the censure, emphasizing freedom of speech.
On December 8, 2023, the U.S. House of Representatives voted to censure Rep. Jamaal Bowman (D-NY) in a 214-191 vote for pulling a fire alarm to stall a House vote on September 26, 2023. He was seen on video attempting to open a door, and then pulling the fire alarm. He claimed he thought that the alarm would open the door. On October 30, Rep. Bowman pled guilty to a misdemeanor count and was fined $1000.
On March 6, 2025, the U.S. House of Representatives voted to censure Rep. Alexander N. Green (D-TX) in a 224–198–2 vote, with 10 Democrats joining most Republican in voting in favor of the censure. The vote was held after Green disrupted President Donald Trump's March 4, 2025 address to a joint session of Congress by pointing his cane at the dais and shouting, "You have no mandate to cut Medicaid." Green was escorted out of the House Chamber by the sergeant-at-arms after repeatedly interrupting the address.
References
[edit]
^ a b c Lehman, Jeffrey; Phelps, Shirelle (2005). West's Encyclopedia of American Law, Vol. 2 (2 ed.). Detroit: Thomson/Gale. p. 291. ISBN 9780314201546.
^ "U.S. Senate Reference". Retrieved November 13, 2015.
^ "U.S. Senate: Censure".
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^ "Discipline & Punishment | US House of Representatives: History, Art & Archives". history.house.gov. Retrieved January 28, 2021.
^ "U.S. Senate: About Censure". www.senate.gov. Retrieved January 28, 2021.
^ "Charles Rangel censured on House floor - what does censure mean?". Christian Science Monitor. December 2, 2010.
^ "Resolutions to Censure the President: Procedure and History" (PDF). Retrieved January 5, 2024.
^ a b Hudiburg, Jane A.; Davis, Christopher M. (February 1, 2018). "Resolutions to Censure the President: Procedure and History" (PDF). Congressional Research Service. pp. 4–5.
^ "U.S. Senate: Art & History Home > Historical Minutes > 1801–1850 > Senate Censures President". Retrieved November 13, 2015.
^ Whitelaw, Nancy. Andrew Jackson Frontier President.[full citation needed]
^ a b Hudiburg & Davis (2018), pp. 5–6.
^ Hudiburg & Davis (2018), pp. 6–7.
^ Hudiburg & Davis (2018), p. 7.
^ "American President: John Tyler: Domestic Affairs". Millercenter.org. Archived from the original on November 27, 2010. Retrieved November 28, 2010.
^ Hudiburg & Davis (2018), pp. 8–9.
^ Hudiburg & Davis (2018), p. 9.
^ "S.Res. 44". Thomas.loc.gov. February 12, 1999. Archived from the original on July 18, 2012. Retrieved November 19, 2010.
^ "H.J.Res. 139". Thomas.loc.gov. Archived from the original on July 14, 2012. Retrieved November 19, 2010.
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^ "H.J.Res. 140". Thomas.loc.gov. December 17, 1998. Archived from the original on July 23, 2012. Retrieved November 19, 2010.
^ Benen, Steve (December 2, 2014). "Censure and move on?". MSNBC.
^ Weiner, Rachel (March 15, 2013). "MoveOn.org moving to petition-driven model". The Washington Post.
^ Hudiburg & Davis (2018), pp. 10–11.
^ "Reps. Nadler, Watson Coleman, and Jayapal Announce Censure Resolution Against President Trump for Blaming 'Both Sides' for Violence in Charlottesville, VA and Excusing Behavior of 'Unite the Right' Participants" (Press release). Office of Congressman Jerrold Nadler. August 16, 2017. Retrieved August 17, 2017.
^ Marcos, Cristina (August 18, 2017). "Pelosi endorses push to censure Trump". The Hill. Retrieved August 23, 2018.
^ "Trump decries immigrants from 'shithole countries' coming to US". CNN. January 11, 2018.
^ "Cedric Richmond is leading an effort to censure Donald Trump". BayouBrief.com. January 12, 2018.
^ a b "U.S. Senate: Reference Home > United States Senate Election, Expulsion and Censure Cases". Senate.gov. March 26, 2009. Retrieved November 28, 2010.
^ "U.S. Senate: Art & History Home > Origins & Development > Powers & Procedures > Expulsion and Censure". Senate.gov. Retrieved November 28, 2010.
^ "U.S. Senate:Home > Art & History Home > Origins & Development > Powers & Procedures > Expulsion and Censure". Retrieved August 6, 2007.
^ 83rd U.S. Congress (July 30, 1954). "Senate Resolution 301: Censure of Senator Joseph McCarthy". U.S. National Archives and Records Administration. Retrieved October 30, 2013.{{cite web}}: CS1 maint: numeric names: authors list (link)
^ Maskell, Jack. "Expulsion, Censure, Reprimand, and Fine: Legislative Discipline in the House of Representatives" (PDF). Congressional Research Service. The House of Representatives - in the same manner as the United States Senate - is expressly authorized within the United States Constitution (Article I, Section 5, clause 2) to discipline or "punish" its own Members ... to protect the institutional integrity of the House of Representatives, its proceedings, and its reputation.
^ a b c Bresnahan, John (November 18, 2010). "Charlie Rangel to face censure vote". Politico.
^ a b "Punishment in the House". The New York Times. November 18, 2010.
^ "A Lonely Guilty Verdict for Charlie Rangel". Politics. U.S. News & World Report. November 24, 2010. Retrieved November 28, 2010.
^ Kleinfield, N. R. (December 3, 2010). "Amid Routine Business, History and Humiliation". The New York Times. p. A28.
^ Sonmez, Felicia (November 17, 2021). "House censures Rep. Gosar, ejects him from committees over violent video depicting slaying of Rep. Ocasio-Cortez". The Washington Post. Retrieved November 17, 2021.
^ Sheridan, Eugene R. (1992). "Thomas Jefferson and the Giles Resolutions". The William and Mary Quarterly. 49 (4): 589–608. doi:10.2307/2947173. ISSN 0043-5597. JSTOR 2947173.
^ Liptak, Adam (March 25, 2022). "Censure of Politician Did Not Violate First Amendment, Supreme Court Rules". The New York Times. p. A16. Retrieved October 29, 2023.
^ "U.S. Senate: Art & History Home > Historical Minutes > 1801-1850 > Senate Censures President". Retrieved April 1, 2006.
^ "House Democrats Intro First Motion to Censure Trump". The Daily Beast. August 18, 2017. Retrieved May 4, 2020.
^ "Small group of Democrats floats censure instead of impeachment". Politico. December 10, 2019. Retrieved May 4, 2020.
^ Petre, Linda (February 5, 2020). "Senate GOP drives stake through talk of Trump censure". The Hill. Retrieved May 4, 2020.
^ "Committee on Standards of Official Conduct". Archived from the original on March 29, 2008.
^ Berry, Kenneth K.; Berry, Jason (January–February 2000). "The Congressional Censure of a Research Paper: Return of the Inquisition?". Skeptical Inquirer. Vol. 24, no. 1. pp. 20–21. Archived from the original on September 28, 2008. Retrieved June 21, 2008.
^ Lewis, Neil A. (August 1, 2007). "Retired General is Censured for Role in Tillman Case". The New York Times. Retrieved April 8, 2008.
^ Hamby, Peter (July 7, 2009). "South Carolina GOP votes to censure Sanford". CNN. Retrieved December 15, 2009.
^ "No available copy of article exists". Retrieved October 14, 2009.[dead link]
^ a b Phillips, Kate (January 5, 2010). "Senator Graham Censured Again". The New York Times.
^ "Arizona GOP rebukes McCain for not being conservative enough". CNN. January 26, 2014. Archived from the original on January 28, 2014.
^ "Wyoming GOP censures Rep. Liz Cheney over impeachment vote". AP News. Associated Press. January 7, 2021.
^ Williams, Jordan (February 13, 2021). "Louisiana GOP votes to censure Cassidy over impeachment vote". The Hill. Retrieved February 14, 2021.
^ Ward, Myah (February 15, 2021). "GOP Sen. Burr censured by North Carolina GOP after Trump conviction vote". Politico.
^ "Lisa Murkowski censured by Alaska Republicans for voting to convict Trump". The Guardian. March 16, 2021. Retrieved March 16, 2021.
^ Sonmez, Felicia (November 17, 2021). "House censures Rep. Gosar, ejects him from committees over violent video depicting slaying of Rep. Ocasio-Cortez". The Washington Post. Archived from the original on November 17, 2021. Retrieved November 17, 2021.
^ Shahey, Maeve (January 22, 2022). "Arizona Democratic Party censures Sinema over voting rights stance". Politico.
^ Orr, Gabby (February 4, 2022). "In censure of Cheney and Kinzinger, RNC calls events of January 6 'legitimate political discourse'". CNN.
^ Wang, Amy B.; Alfaro, Mariana (June 21, 2023). "House passes measure to censure Adam Schiff". The Washington Post. ISSN 0190-8286. Retrieved July 17, 2023.
^ Zhou, Li (November 9, 2023). "The House censure of Rashida Tlaib, explained". Vox.
^ Grisales, Claudia (November 7, 2023). "House votes to censure Rep. Rashida Tlaib for Israel-Hamas war comments". NPR. Retrieved November 7, 2023.
^ "House votes to censure Rep. Jamaal Bowman for pulling fire alarm". Retrieved January 5, 2024.
^ "Rep. Jamaal Bowman pleads guilty to a misdemeanor for pulling a fire alarm in House office building". AP News. October 26, 2023. Retrieved January 5, 2024.
^ Schnell, Mychael (March 6, 2025). "House votes to censure Al Green for disrupting Trump speech". The Hill. Retrieved March 6, 2025.
^ "Democratic Rep. Al Green removed from chamber after outburst during Trump address". ABC News. March 5, 2025. Retrieved March 5, 2025.
^ "Democratic Rep. Al Green removed after disrupting Trump's speech". NBC News. March 5, 2025. Retrieved March 5, 2025.
Further reading
[edit]
Butler, Anne M.; Wolff, Wendy (1995). United States Senate Election, Expulsion and Censure Cases, 1793–1900. Washington, DC: Government Printing Office.
"Enforcement of Ethical Standards in Congress". Final Report of the Joint Committee on the Organization of Congress. December 1993. Archived from the original on July 25, 2007.
"Resolutions Censuring the President: History and Context, 1st-114th Congresses" (PDF). Congressional Research Service.
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187502 | https://math.stackexchange.com/questions/71157/proof-of-dividing-fractions-fraca-bc-d-fracadbc | algebra precalculus - Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ - Mathematics Stack Exchange
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Proof of dividing fractions a/b c/d=a d b c a/b c/d=a d b c
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For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
a b÷c d=a b⋅d c=a d b c?a b÷c d=a b⋅d c=a d b c?
algebra-precalculus
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edited Oct 23, 2022 at 9:41
Jam
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asked Oct 9, 2011 at 16:44
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Write a b÷c d a b÷c d as
a b c d.a b c d.
Suppose you wanted to clear the denominator of this compound fraction. You could try multiplication by d c d c, but you'll have to multiply the top and the bottom of the fraction to avoid changing it. So, you end up with
a b c d=a b c d⋅d c d c=a b⋅d c c d⋅d c=a d b c 1=a d b c.a b c d=a b c d⋅d c d c=a b⋅d c c d⋅d c=a d b c 1=a d b c.
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edited Oct 9, 2011 at 19:44
Arturo Magidin
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answered Oct 9, 2011 at 17:45
Austin MohrAustin Mohr
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There is something assumed about the order of operations in the notation a/b÷c/d a/b÷c/d. I see this as a problem in pedagogy when fractions are first introduced in schools. Somehow we have to assume that the slash bars are to be done before the division symbol, even though each of those symbols stands for "divide". If I were you, I'd always stay in the habit of writing fractions as a b a b rather than a/b a/b.
end rant.
Traditionally, a b÷c d a b÷c d means: Beginning with the quantity a b a b, divide this by the quantity c d c d. So let's think of a small example not dealing with fractions first:
8÷2.8÷2.
Now, we all know the answer is 4 4. But I want to draw your attention to the fact that this expression is the same as:
8⋅1 2.8⋅1 2.
What happened? Well division by 2 2 is the same as multiplication by 1 2 1 2, because 1 2 1 2 is the multiplicative inverse, or reciprocal of 2 2. By defintion, the reciprocal of a number x x is a number y y such that x⋅y=1 x⋅y=1. Every nonzero number has a reciprocal. If x≠0 x≠0, then its reciprocal is 1 x 1 x (because x⋅1 x=1 x⋅1 x=1). So, for x≠0 x≠0:
8÷x=8⋅1 x.8÷x=8⋅1 x.
Now, if x x is a fraction, such as c d c d, then the most natural way to write the reciprocal of x x is by "flipping" the fraction. The reciprocal of c d c d is d c d c (well... because c d⋅d c=c d c d=1 c d⋅d c=c d c d=1). Thus:
8÷c d=8⋅d c.8÷c d=8⋅d c.
Finally, it does not matter so much what the first factor of the expression is. It could be another fraction:
a b÷c d=a b⋅d c=a d b c.a b÷c d=a b⋅d c=a d b c.
Hope this helps!
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answered Oct 9, 2011 at 17:15
Shaun AultShaun Ault
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The fraction x=a b x=a b is the solution to b⋅x=a b⋅x=a. Similarly y=c d y=c d solves d⋅y=c d⋅y=c. Now, multiply the left-hand-side with b⋅x b⋅x and the right hand-size with a a which preserves the equality since a=b⋅x a=b⋅x. We get
(b⋅x)⋅(d⋅y)(b⋅d)⋅(x⋅y)==c⋅a(c⋅a)(b⋅x)⋅(d⋅y)=c⋅a(b⋅d)⋅(x⋅y)=(c⋅a)
Thus (x⋅y)(x⋅y) is the fraction a⋅c b⋅d a⋅c b⋅d.
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answered Oct 9, 2011 at 16:55
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Division is normally the inverse operation to multiplication, so - as I have explained to one of my daughters this week - dividing by n n is the same as multiplying by 1/n 1/n. In one sense this is a totally trivial observation, but in another it involves a much deeper understanding of what is going on.
It follows that division, where it makes sense, can be replaced by multiplication by the reciprocal (or inverse), which is the precise case my daughter wanted to understand. Getting hold of this opens up a world of algebraic possibilities, so it is well worth doing.
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answered Oct 9, 2011 at 18:23
Mark BennetMark Bennet
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HINTx/z=y⟺x=y z.x/z=y⟺x=y z. Thus a b/c d=a b d c⟺a b=a b d c c d a b/c d=a b d c⟺a b=a b d c c d
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answered Oct 9, 2011 at 17:12
Bill DubuqueBill Dubuque
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Suppose a b a b and c d c d are fractions. That is, a a, b b, c c, d d are whole numbers and b≠0 b≠0, d≠0 d≠0. In addition we require c≠0 c≠0.
Let a b÷c d=A a b÷c d=A. Then by definition of division of fractions , A A is a unique fraction so that A×c d=a b A×c d=a b.
However, (a b×d c)×c d=a b×(d c×c d)=a b×(d c c d)=a b×(d c d c)=a b(a b×d c)×c d=a b×(d c×c d)=a b×(d c c d)=a b×(d c d c)=a b.
Then by uniqueness (of A A), A=a b×d c A=a b×d c.
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edited Oct 14, 2011 at 11:02
answered Oct 12, 2011 at 21:43
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187503 | https://en.wikipedia.org/wiki/Sequential_pattern_mining | Jump to content
Sequential pattern mining
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From Wikipedia, the free encyclopedia
Data mining technique
Sequential pattern mining is a topic of data mining concerned with finding statistically relevant patterns between data examples where the values are delivered in a sequence. It is usually presumed that the values are discrete, and thus time series mining is closely related, but usually considered a different activity. Sequential pattern mining is a special case of structured data mining.
There are several key traditional computational problems addressed within this field. These include building efficient databases and indexes for sequence information, extracting the frequently occurring patterns, comparing sequences for similarity, and recovering missing sequence members. In general, sequence mining problems can be classified as string mining which is typically based on string processing algorithms and itemset mining which is typically based on association rule learning. Local process models extend sequential pattern mining to more complex patterns that can include (exclusive) choices, loops, and concurrency constructs in addition to the sequential ordering construct.
String mining
[edit]
String mining typically deals with a limited alphabet for items that appear in a sequence, but the sequence itself may be typically very long. Examples of an alphabet can be those in the ASCII character set used in natural language text, nucleotide bases 'A', 'G', 'C' and 'T' in DNA sequences, or amino acids for protein sequences. In biology applications analysis of the arrangement of the alphabet in strings can be used to examine gene and protein sequences to determine their properties. Knowing the sequence of letters of a DNA or a protein is not an ultimate goal in itself. Rather, the major task is to understand the sequence, in terms of its structure and biological function. This is typically achieved first by identifying individual regions or structural units within each sequence and then assigning a function to each structural unit. In many cases this requires comparing a given sequence with previously studied ones. The comparison between the strings becomes complicated when insertions, deletions and mutations occur in a string.
A survey and taxonomy of the key algorithms for sequence comparison for bioinformatics is presented by Abouelhoda & Ghanem (2010), which include:
Repeat-related problems: that deal with operations on single sequences and can be based on exact string matching or approximate string matching methods for finding dispersed fixed length and maximal length repeats, finding tandem repeats, and finding unique subsequences and missing (un-spelled) subsequences.
Alignment problems: that deal with comparison between strings by first aligning one or more sequences; examples of popular methods include BLAST for comparing a single sequence with multiple sequences in a database, and ClustalW for multiple alignments. Alignment algorithms can be based on either exact or approximate methods, and can also be classified as global alignments, semi-global alignments and local alignment. See sequence alignment.
Itemset mining
[edit]
Some problems in sequence mining lend themselves to discovering frequent itemsets and the order they appear, for example, one is seeking rules of the form "if a {customer buys a car}, he or she is likely to {buy insurance} within 1 week", or in the context of stock prices, "if {Nokia up and Ericsson up}, it is likely that {Motorola up and Samsung up} within 2 days". Traditionally, itemset mining is used in marketing applications for discovering regularities between frequently co-occurring items in large transactions. For example, by analysing transactions of customer shopping baskets in a supermarket, one can produce a rule which reads "if a customer buys onions and potatoes together, he or she is likely to also buy hamburger meat in the same transaction".
A survey and taxonomy of the key algorithms for item set mining is presented by Han et al. (2007).
The two common techniques that are applied to sequence databases for frequent itemset mining are the influential apriori algorithm and the more-recent FP-growth technique.
Applications
[edit]
With a great variation of products and user buying behaviors, shelf on which products are being displayed is one of the most important resources in retail environment. Retailers can not only increase their profit but, also decrease cost by proper management of shelf space allocation and products display. To solve this problem, George and Binu (2013) have proposed an approach to mine user buying patterns using PrefixSpan algorithm and place the products on shelves based on the order of mined purchasing patterns.
Algorithms
[edit]
Commonly used algorithms include:
GSP algorithm
Sequential Pattern Discovery using Equivalence classes (SPADE)
FreeSpan
PrefixSpan
MAPres
Seq2Pat (for constraint-based sequential pattern mining)
See also
[edit]
Collocation extraction – Computational technique to find word sequences
Process mining – Data mining technique using event logs
Sequence analysis – Identification and study of genomic sequences
Sequence analysis in social sciences – Analysis of sets of categorical sequences
Sequence clustering
Sequence labeling
References
[edit]
^ Mabroukeh, N. R.; Ezeife, C. I. (2010). "A taxonomy of sequential pattern mining algorithms". ACM Computing Surveys. 43: 1–41. CiteSeerX 10.1.1.332.4745. doi:10.1145/1824795.1824798. S2CID 207180619.
^ Bechini, A.; Bondielli, A.; Dell'Oglio, P.; Marcellonii, F. (2023). "From basic approaches to novel challenges and applications in Sequential Pattern Mining". Applied Computing and Intelligence. 3 (1): 44–78. doi:10.3934/aci.2023004. hdl:11568/1172332.
^ Tax, N.; Sidorova, N.; Haakma, R.; van der Aalst, Wil M. P. (2016). "Mining Local Process Models". Journal of Innovation in Digital Ecosystems. 3 (2): 183–196. arXiv:1606.06066. doi:10.1016/j.jides.2016.11.001. S2CID 10872379.
^ Abouelhoda, M.; Ghanem, M. (2010). "String Mining in Bioinformatics". In Gaber, M. M. (ed.). Scientific Data Mining and Knowledge Discovery. Springer. doi:10.1007/978-3-642-02788-8_9. ISBN 978-3-642-02787-1.
^ Han, J.; Cheng, H.; Xin, D.; Yan, X. (2007). "Frequent pattern mining: current status and future directions". Data Mining and Knowledge Discovery. 15 (1): 55–86. doi:10.1007/s10618-006-0059-1.
^ George, A.; Binu, D. (2013). "An Approach to Products Placement in Supermarkets Using PrefixSpan Algorithm". Journal of King Saud University-Computer and Information Sciences. 25 (1): 77–87. doi:10.1016/j.jksuci.2012.07.001.
^ Ahmad, Ishtiaq; Qazi, Wajahat M.; Khurshid, Ahmed; Ahmad, Munir; Hoessli, Daniel C.; Khawaja, Iffat; Choudhary, M. Iqbal; Shakoori, Abdul R.; Nasir-ud-Din (1 May 2008). "MAPRes: Mining association patterns among preferred amino acid residues in the vicinity of amino acids targeted for post-translational modifications". Proteomics. 8 (10): 1954–1958. doi:10.1002/pmic.200700657. PMID 18491291. S2CID 22362167.
^ Hosseininasab A, van Hoeve WJ, Cire AA (2019). "Constraint-Based Sequential Pattern Mining with Decision Diagrams". Proceedings of the AAAI Conference on Artificial Intelligence. 33: 1495–1502. arXiv:1811.06086. doi:10.1609/aaai.v33i01.33011495. S2CID 53427299.
^ "Seq2Pat: Sequence-to-Pattern Generation Library". GitHub. 9 April 2022.
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Proving a cubic polynomial has no rational roots [duplicate]
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Monic f(x)∈Z[x]f(x)∈Z[x] has no rational root if f(0)&f(1)f(0)&f(1) odd [Parity Root Test, Modular Root Test] (2 answers)
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This is an exercise in polynomials/algebra/discrete mathematics I have just met:
For odd integers a,b∈Z a,b∈Z we are asked to show the polynomial x 3+a x+b x 3+a x+b has no rational roots.
The first thing that came to my mind is the rational root test which says that for a rational root x=p q x=p q written as a reduced fraction where p,q∈Z p,q∈Z with q≠0 q≠0, then q q is a factor of 1 and p p is a factor of b b which means that a rational root must actually be an odd integer (positive or negative) factor of b b, but I am stuck here. All help appreciated.
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edited Apr 24, 2023 at 19:59
Bill Dubuque
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asked Aug 24, 2020 at 20:30
Croc2AlphaCroc2Alpha
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You've got the right idea. Your denominator q q should be ±1±1, so your rational root must actually be an integer. As a result, you just need to show that
x 3+a x+b≠0 x 3+a x+b≠0
for all integers x,a,b x,a,b where a a and b b are odd. Since you're given that a a and b b are odd, it makes sense to look at this polynomial modulo 2 2; what do you see?
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answered Aug 24, 2020 at 20:34
Carl SchildkrautCarl Schildkraut
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actually since b b is odd and x x is an integer factor of it can we also deduce that x x must be odd?Croc2Alpha –Croc2Alpha 2020-08-24 20:37:27 +00:00 Commented Aug 24, 2020 at 20:37
1 @kroner Yes, exactly, although that isn't strictly necessary.Carl Schildkraut –Carl Schildkraut 2020-08-24 20:44:41 +00:00 Commented Aug 24, 2020 at 20:44
thank you very much Croc2Alpha –Croc2Alpha 2020-08-24 20:45:45 +00:00 Commented Aug 24, 2020 at 20:45
Please search to help avoid adding more dupe Q&A's to dupes of FAQs.Bill Dubuque –Bill Dubuque 2020-08-24 20:56:28 +00:00 Commented Aug 24, 2020 at 20:56
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If the root r r is odd, r 3+a r+b r 3+a r+b is an odd number, a contradiction.
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answered Aug 24, 2020 at 20:40
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6Monic f(x)∈Z[x]f(x)∈Z[x] has no rational root if f(0)&f(1)f(0)&f(1) odd [Parity Root Test, Modular Root Test]
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187505 | https://artofproblemsolving.com/wiki/index.php/Rearrangement_Inequality?srsltid=AfmBOopwUtz49lWgLaotSlJcAyeRhmFaP83kTfGLvxpAuZlNnDi6haeZ | Art of Problem Solving
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Rearrangement Inequality
The Rearrangement Inequality states that, if is a permutation of a finiteset (in fact, multiset) of real numbers and is a permutation of another finite set of real numbers, the quantity is maximized when and are similarly sorted (that is, if is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of ). Conversely, is minimized when and are oppositely sorted (that is, if is less than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members ).
Contents
[hide]
1 Introductory
2 Intermediate
2.1 Proof of the Rearrangement Inequality
3 Uses
4 See Also
5 External Links
Introductory
Consider the following simple application: suppose you are involved in the hold-up of a convenience store. You note, as you are emptying the register, that there are different numbers of each denomination (penny, nickel, dime, quarter, dollar bill, five dollar bill, ten dollar bill and twenty dollar bill) in the register. When would your take be maximized? Certainly, you would hope that there would be the largest number of twenty dollar bills, then the next largest number of tens, etc. Meanwhile, you would find yourself very disappointed if there were more pennies than nickels, more nickels than dimes, and so on. This is a simple application of the rearrangement inequality. It is also an application of the greedy algorithm, so one possible interpretation of the rearrangement inequality is that sometimes, the greedy algorithm works.
Intermediate
Proof of the Rearrangement Inequality
The proof of the Rearrangment Inequality can be handled with proof by contradiction. Only the maximization form is proved here; the minimization proof is virtually identical.
Before we begin the proof properly, it is useful to consider the case where . Without loss of generality, sort and so that and . By hypothesis, . Expanding and taking some terms to the other side of the inequality, we get , as desired.
Now for the general case. Again, without loss of generality, set and ; and suppose the grouping that maximizes the desired sum of products is not the one that pairs with , with , and so on. This means that there is at least one instance where is paired with while is paired with , where and . However, using the technique seen above to prove the inequality for , we can see that the sum of products can only increase if we instead pair with and with (unless both a's or both b's are equal, in which case either we can choose another pair of products or note that the current arrangement is actually identical to the desired one), which contradicts our assumption that the arrangement we had was already the largest one.
Note: The minimization equality can be very easily proved by noting that if we have the set , ordered in decreasing order and the set , ordered in increasing order, then the maximum sum is just . Thus, by negating all values the inequality follows.
Uses
The Rearrangement Inequality has a wide range of uses, from MathCounts level optimization problems to Olympiad level inequality problems. A relatively simple example of its use in solving higher-level problems is found in the proof of Chebyshev's Inequality. It is particularly useful in that it does not require any terms of either sequence to be positive or negative, unlike the power-mean family of inequalities.
See Also
Chebyshev's Inequality
Power Mean Inequality
External Links
The Rearrangement Inequality by Dragos Hrimiuc
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187506 | https://www.ruf.rice.edu/~kekule/SignificantFigureRules1.pdf | Following are Summaries from Two Chemistry Education Web Sites Concerning Significant Figure Rules From There are three rules on determining how many significant figures are in a number: 1. Non-zero digits are always significant. 2. Any zeros between two significant digits are significant. 3. A final zero or trailing zeros in the decimal portion ONLY are significant. Focus on these rules and learn them well. They will be used extensively throughout the remainder of this course. You would be well advised to do as many problems as needed to nail the concept of significant figures down tight and then do some more, just to be sure. Please remember that, in science, all numbers are based upon measurements (except for a very few that are defined). Since all measurements are uncertain, we must only use those numbers that are meaningful. A common ruler cannot measure something to be 22.4072643 cm long. Not all of the digits have meaning (significance) and, therefore, should not be written down. In science, only the numbers that have significance (derived from measurement) are written. If you're not convinced significant figures are important, you may want to read the Significant Figure Fable that follows. Rule 1: Non-zero digits are always significant. Hopefully, this rule seems rather obvious. If you measure something and the device you use (ruler, thermometer, triple-beam balance, etc.) returns a number to you, then you have made a measurement decision and that ACT of measuring gives significance to that particular numeral (or digit) in the overall value you obtain. Hence a number like 26.38 would have four significant figures and 7.94 would have three. The problem comes with numbers like 0.00980 or 28.09. Rule 2: Any zeros between two significant digits are significant. Suppose you had a number like 406. By the first rule, the 4 and the 6 are significant. However, to make a measurement decision on the 4 (in the hundred's place) and the 6 (in the unit's place), you HAD to have made a decision on the ten's place. The measurement scale for this number would have hundreds and tens marked with an estimation made in the unit's place. Like this: 1 Rule 3: A final zero or trailing zeros in the decimal portion ONLY are significant. This rule causes the most difficulty with ChemTeam students. Here are two examples of this rule with the zeros this rule affects in boldface: 0.00500 0.03040 Here are two more examples where the significant zeros are in boldface: 2.30 x 10¯5 4.500 x 1012 What Zeros are Not Discussed Above Zero Type #1: Space holding zeros on numbers less than one. Here are the first two numbers from just above with the digits that are NOT significant in boldface: 0.00500 0.03040 These zeros serve only as space holders. They are there to put the decimal point in its correct location. They DO NOT involve measurement decisions. Upon writing the numbers in scientific notation (5.00 x 10¯3 and 3.040 x 10¯2), the non-significant zeros disappear. Zero Type #2: the zero to the left of the decimal point on numbers less than one. When a number like 0.00500 is written, the very first zero (to the left of the decimal point) is put there by convention. Its sole function is to communicate unambiguously that the decimal point is a decimal point. If the number were written like this, .00500, there is a possibility that the decimal point might be mistaken for a period. Many students omit that zero. They should not. Zero Type #3: trailing zeros in a whole number. 200 is considered to have only ONE significant figure while 25,000 has two. This is based on the way each number is written. When whole number are written as above, the zeros, BY DEFINITION, did not require a measurement decision, thus they are not significant. However, it is entirely possible that 200 really does have two or three significant figures. If it does, it will be written in a different manner than 200. 2 Typically, scientific notation is used for this purpose. If 200 has two significant figures, then 2.0 x 102 is used. If it has three, then 2.00 x 102 is used. If it had four, then 200.0 is sufficient. See rule #2 above. How will you know how many significant figures are in a number like 200? In a problem like below, divorced of all scientific context, you will be told. If you were doing an experiment, the context of the experiment and its measuring devices would tell you how many significant figures to report to people who read the report of your work. Zero Type #4: leading zeros in a whole number. 00250 has two significant figures. 005.00 x 10¯4 has three. Exact Numbers Exact numbers, such as the number of people in a room, have an infinite number of significant figures. Exact numbers are counting up how many of something are present, they are not measurements made with instruments. Another example of this are defined numbers, such as 1 foot = 12 inches. There are exactly 12 inches in one foot. Therefore, if a number is exact, it DOES NOT affect the accuracy of a calculation nor the precision of the expression. Some more examples: There are 100 years in a century. 2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water. There are 500 sheets of paper in one ream. Interestingly, the speed of light is now a defined quantity. By definition, the value is 299,792,458 meters per second. Are Significant Figures Important? A Fable A student once needed a cube of metal that had to have a mass of 83 grams. He knew the density of this metal was 8.67 g/mL, which told him the cube's volume. Believing significant figures were invented just to make life difficult for chemistry students and had no practical use in the real world, he calculated the volume of the cube as 9.573 mL. He thus determined that the edge of the cube had to be 2.097 cm. He took his plans to the machine shop where his friend had the same type of work done the previous year. The shop foreman said, "Yes, we can make this according to your specifications - but it will be expensive." "That's OK," replied the student. "It's important." He knew his friend has paid $35, and he had been given $50 out of the school's research budget to get the job done. He returned the next day, expecting the job to be done. "Sorry," said the foreman. "We're still working on it. Try next week." Finally the day came, and our friend got his cube. It looked very, 3 very smooth and shiny and beautiful in its velvet case. Seeing it, our hero had a premonition of disaster and became a bit nervous. But he summoned up enough courage to ask for the bill. "$500, and cheap at the price. We had a terrific job getting it right -- had to make three before we got one right." "But--but--my friend paid only $35 for the same thing!" "No. He wanted a cube 2.1 cm on an edge, and your specifications called for 2.097. We had yours roughed out to 2.1 that very afternoon, but it was the precision grinding and lapping to get it down to 2.097 which took so long and cost the big money. The first one we made was 2.089 on one edge when we got finshed, so we had to scrap it. The second was closer, but still not what you specified. That's why the three tries." "Oh!" Rules for Rounding Off Now that "everyone" has a calculator that will give a result to six or eight (or more) figures, it is important that we know how to round the answer off correctly. The typical rule taught is that you round up with five or more and round down with four or less. THIS RULE IS WRONG! However, please do not rush off to your elementary school teacher and read 'em the riot act! The problem lies in rounding "up" (increasing) the number that is followed by a 5. For example, numbers like 3.65 or 3.75, where you are to round off to the nearest tenth. OK, let's see if I can explain this. When you round off, you change the value of the number, except if you round off a zero. Following the old rules, you can round a number down in value four times (rounding with one, two, three, four) compared to rounding it upwards five times (five, six, seven, eight, nine). Remember that "rounding off" a zero does not change the value of the number being rounded off. Suppose you had a very large sample of numbers to round off. On average you would be changing values in the sample downwards 4/9ths of the time, compared to changing values in the sample upward 5/9ths of the time. This means the average of the values AFTER rounding off would be greater than the average of the values BEFORE rounding. This is not acceptable. We can correct for this problem by rounding "off" (keeping the number the same) in fifty percent of the roundings-even numbers followed by a 5. Then, on average, the roundings "off" will cancel out the roundings "up." 4 The following rules dictate the manner in which numbers are to be rounded to the number of figures indicated. The first two rules are more-or-less the old ones. Rule three is the change in the old way. When rounding, examine the figure following (i.e., to the right of) the figure that is to be last. This figure you are examining is the first figure to be dropped. 1. If it is less than 5, drop it and all the figures to the right of it. 2. If it is more than 5, increase by 1 the number to be rounded, that is, the preceeding figure. 3. If it is 5, round the number so that it will be even. Keep in mind that zero is considered to be even when rounding off. Example #1 - Suppose you wish to round 62.5347 to four significant figures. Look at the fifth figure. It is a 4, a number less than 5. Therefore, you will simply drop every figure after the fourth, and the original number rounds off to 62.53. Example #2 - Round 3.78721 to three significant figures. Look at the fourth figure. It is 7, a number greater than 5, so you round the original number up to 3.79. Example #3 - Round 726.835 to five significant figures. Look at the sixth figure. It is a 5, so now you must look at the fifth figure also. That is a 3, which is an odd number, so you round the original number up to 726.84. Example #4 - Round 24.8514 to three significant figures. Look at the fourth figure. It is a 5, so now you must also look at the third figure. It is 8, an even number, so you simply drop the 5 and the figures that follow it. The original number becomes 24.8. When the value you intend to round off is a five, you MUST look at the previous value ALSO. If it is even, you round down. If it is odd, you round up. A common question is "Is zero considered odd or even?" The answer is even. Here are some more examples of the "five rule." Round off at the five. 3.075 3.85 22.73541 0.00565 2.0495 This last one is tricky (at least for high schoolers being exposed to this stuff for the first time!). The nine rounds off to a ten (not a zero), so the correct answer is 2.050, NOT 2.05. 5 Would your teacher be so mean as to include problems like this one on a test? In the ChemTeam classroom, the sufferers (oops, I mean students) have learned to shout "YES" in unison to such easy questions. Lastly, before we get to the problems. Students, when they learn this rule, like to apply it across the board. For example, in 2.0495, let's say we want to round off to the nearest 0.01. Many times, a student will answer 2.04. When asked to explain, the rule concerning five will be cited. However, the important number in this problem is the nine, so the rule is to round up and the correct answer is 2.05. Math With Significant Figures Addition and Subtraction In mathematical operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. Let's state that another way: a chain is no stronger than its weakest link. An answer is no more precise that the least precise number used to get the answer. Let's do it one more time: imagine a team race where you and your team must finish together. Who dictates the speed of the team? Of course, the slowest member of the team. Your answer cannot be MORE precise than the least precise measurement. For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the numbers ONLY. Here is what to do: 1) Count the number of significant figures in the decimal portion of each number in the problem. (The digits to the left of the decimal place are not used to determine the number of decimal places in the final answer.) 2) Add or subtract in the normal fashion. 3) Round the answer to the LEAST number of places in the decimal portion of any number in the problem. WARNING: the rules for add/subtract are different from multiply/divide. A very common student error is to swap the two sets of rules. Another common error is to use just one rule for both types of operations. Multiplication and Division The following rule applies for multiplication and division: The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer. This means you MUST know how to recognize significant figures in order to use this rule. 6 Example #1: 2.5 x 3.42. The answer to this problem would be 8.6 (which was rounded from the calculator reading of 8.55). Why? 2.5 has two significant figures while 3.42 has three. Two significant figures is less precise than three, so the answer has two significant figures. Example #2: How many significant figures will the answer to 3.10 x 4.520 have? You may have said two. This is too few. A common error is for the student to look at a number like 3.10 and think it has two significant figures. The zero in the hundedth's place is not recognized as significant when, in fact, it is. 3.10 has three significant figures. Three is the correct answer. 14.0 has three significant figures. Note that the zero in the tenth's place is considered significant. All trailing zeros in the decimal portion are considered significant. Another common error is for the student to think that 14 and 14.0 are the same thing. THEY ARE NOT. 14.0 is ten times more precise than 14. The two numbers have the same value, but they convey different meanings about how trustworthy they are. Four is also an incorrect answer given by some ChemTeam students. It is too many significant figures. One possible reason for this answer lies in the number 4.520. This number has four significant figures while 3.10 has three. Somehow, the student (YOU!) maybe got the idea that it is the GREATEST number of significant figures in the problem that dictates the answer. It is the LEAST. Sometimes student will answer this with five. Most likely you responded with this answer because it says 14.012 on your calculator. This answer would have been correct in your math class because mathematics does not have the significant figure concept. Example #3: 2.33 x 6.085 x 2.1. How many significant figures in the answer? Answer - two. Which number decides this? Answer - the 2.1. Why? It has the least number of significant figures in the problem. It is, therefore, the least precise measurement. Example #4: (4.52 x 10¯4) ÷ (3.980 x 10¯6). 7 How many significant figures in the answer? Answer - three. Which number decides this? Answer - the 4.52 x 10¯4. Why? It has the least number of significant figures in the problem. It is, therefore, the least precise measurement. Notice it is the 4.52 portion that plays the role of determining significant figures; the exponential portion plays no role. For practice with significant figures, go to: From It is important to be honest when reporting a measurement, so that it does not appear to be more accurate than the equipment used to make the measurement allows. We can achieve this by controlling the number of digits, or significant figures, used to report the measurement. Determining the Number of Significant Figures The number of significant figures in a measurement, such as 2.531, is equal to the number of digits that are known with some degree of confidence (2, 5, and 3) plus the last digit (1), which is an estimate or approximation. As we improve the sensitivity of the equipment used to make a measurement, the number of significant figures increases. Postage Scale 3 ±1 g 1 significant figure Two-pan balance 2.53 ±0.01 g 3 significant figures Analytical balance 2.531 ±0.001 g 4 significant figures Rules for counting significant figures are summarized below. Zeros within a number are always significant. Both 4308 and 40.05 contain four significant figures. 8 Zeros that do nothing but set the decimal point are not significant. Thus, 470,000 has two significant figures. Trailing zeros that aren't needed to hold the decimal point are significant. For example, 4.00 has three significant figures. If you are not sure whether a digit is significant, assume that it isn't. For example, if the directions for an experiment read: "Add the sample to 400 mL of water," assume the volume of water is known to one significant figure. Addition and Subtraction with Significant Figures When combining measurements with different degrees of accuracy and precision, the accuracy of the final answer can be no greater than the least accurate measurement. This principle can be translated into a simple rule for addition and subtraction: When measurements are added or subtracted, the answer can contain no more decimal places than the least accurate measurement. 150.0 g H2O (using significant figures) + 0.507 g salt 150.5 g solution Multiplication and Division With Significant Figures The same principle governs the use of significant figures in multiplication and division: the final result can be no more accurate than the least accurate measurement. In this case, however, we count the significant figures in each measurement, not the number of decimal places: When measurements are multiplied or divided, the answer can contain no more significant figures than the least accurate measurement. Example: To illustrate this rule, let's calculate the cost of the copper in an old penny that is pure copper. Let's assume that the penny has a mass of 2.531 grams, that it is essentially pure copper, and that the price of copper is 67 cents per pound. We can start by from grams to pounds. We then use the price of a pound of copper to calculate the cost of the copper metal. 9 10 There are four significant figures in both the mass of the penny (2.531) and the number of grams in a pound (453.6). But there are only two significant figures in the price of copper, so the final answer can only have two significant figures. Rounding Off When the answer to a calculation contains too many significant figures, it must be rounded off. There are 10 digits that can occur in the last decimal place in a calculation. One way of rounding off involves underestimating the answer for five of these digits (0, 1, 2, 3, and 4) and overestimating the answer for the other five (5, 6, 7, 8, and 9). This approach to rounding off is summarized as follows. If the digit is smaller than 5, drop this digit and leave the remaining number unchanged. Thus, 1.684 becomes 1.68. If the digit is 5 or larger, drop this digit and add 1 to the preceding digit. Thus, 1.247 becomes 1.25. |
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HomePor aplicaciónAplicaciones Analitica en LaboratorioAplicaciones de índice de refracciónÍndice de refracción: todo lo que es preciso saber Snell's Law: The Law of Refraction
Snell's Law: The Law of Refraction
Formula, Definition, Questions and More about Snells Law
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Snell's law, also known as the Snell-Descartes law, relates the incidence and refraction angles to the refraction indices of the media involved. As shown in the figure, the law determines that the product of the sine of the angle formed between the ray of light (1), the normal straight line (2) and the refractive index of the media (n 1 and n 2) must be constant.
n 1 sinα = n 2 sinβ
The figure shows how a light beam (1, blue arrow) is deflected when it crosses at a certain angle from an optically less dense (n 1) to an optically more dense medium (n 2), e.g. from air to water.
But when the beam crosses from one medium to another perpendicular to the dotted line, no directional change takes place (green arrow).
Snell's law states that the ratio of the refractive index of each of the media is proportional to the ratio of the incident and refractive angle of the light beam. Therefore:
Total Internal Reflection and the Critical Angle
Total internal reflection is defined as a process whereby all light traveling from an optically more dense medium towards an optically less dense one is reflected back into the optically more dense medium. Here is a detailed explanation of this phenomenon using the figure at left.
The dark blue line: If a light beam passes from an optically more dense (n 2) to an optically less dense medium (n 1), it is deflected.
From the dark blue to the green line: If the angle of incidence α is increased, then it can reach a critical value (1) at which the light beam no longer enters the less dense medium (n 1) and is instead refracted exactly along the interface of the two media. The angle (1) is known as the critical angle of total internal reflection. Note that the refractive angle β = 90°.
From the green to the light blue line: When the critical angle is exceeded, all light is reflected back into the optically more dense medium (n 2). This is known as total internal reflection (2).
The refractive index n1 is calculated from this critical angle α:
β = 90° —> sin β = 1
Remember: 1 (green) is the critical angle, and 2 (light blue) is the total internal reflection.
The Refractometer Principle Based on Snell's Law
Using Snell’s Law, scientists engineered refractometers capable of accurately measuring the refractive index of both liquids and semi-solid samples. The accompanying schematic shows this mechanism in action.
When measuring refractive index:
A light-emitting diode or LED serves as the light source and emits a light beam. The LED beam that is emitted then passes through:
The polarization filter,
The interference filter, and
The focal lens. After filtering, the LED beam then reaches:
The sample held by the sapphire prism.
Because the angle of incidence is greater than the critical angle:
Light is reflected and deflected via a lens in the sapphire prism to
The optical CCD sensor, which measures the critical angle.
Additionally, a modern digital refractometer controls the temperature at the prism/sample boundary automatically, further enhancing measurement accuracy.
Want to Know More About Refractive Index?
Learn more about the definition, applications, how to measure it and much more.
Índice de refracción: todo lo que es preciso saber
Tips and Hints to Achieve the Best Results
Did you know that insufficient cleaning of the prism can falsify your next measurement?
Do you know which samples can be measured with a refractometer?
This guide explains what precautions should be taken to avoid errors when measuring the refractive index, BRIX, HFCS or concentration of liquids. Download our free guide.
Guía de medición del índice de refracción
FAQ
What is Snell’s law of refraction?
Snell's law relates the incidence and refraction angles to the refraction indices of the media involved.
What is Snell’s law formula?
The formula of Snell’s law is: n1 sin α = n2 sin β.
In the above equality, if we consider that n2 > n1, then sin β< sen α and β<α. We can then conclude that when light passes from a less refractive medium to a more refractive medium, the speed of light decreases and the light beam (1) approaches the normal straight line (2). That is the angle that the light beam forms with the normal straight line decreases.
What is "n" in Snell’s law?
The “n” in Snell’s law refers to the refractive index value of the media. More specifically, the term n1 = the index of refraction of the incident medium, while n2 = index of refraction of the refractive medium.
What is the history of Snell’s law? Who discovered Snells law?
This law was discovered in 1621 by the Dutch mathematician and astronomer Willebrord Snel van Royen (also known as Willebrord Snellius). In 1637, René Descartes was the first to publish this law that hold's the name Snell's Law today. Despite this, Willebrord Snellius generally holds the credit for the discovery of Snell's Law.
Therefore, Snells law may also be known as the “Snell-Descartes Law”.
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187508 | https://blogs.cornell.edu/info2040/2014/11/07/human-error-in-the-urn-game/ | Human Error in the Urn Game : Networks Course blog for INFO 2040/CS 2850/Econ 2040/SOC 2090
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Human Error in the Urn Game
In class, one of the applications for Bayes’ Rule we discussed was to calculate probabilities in the urn game. That is, say we have an urn with red and blue marbles inside, two of one color and one of the other color, with a 1/2 chance of either color being the majority. If we draw a red marble, it is not very obvious what the probability is that the urn is majority red GIVEN that we drew a red marble. However, it is clear what the probability is that the urn is majority red (1/2), what the probability is that we draw a red marble given a majority red urn (2/3), and what the probability is of drawing a red marble at all (1/2). Bayes’ rule gives us a relationship between these 4 probabilities, so that we can find the conditional probability we want in terms of the other 3 probabilities (2/3 in this case). This method extends to multiple rounds of the urn game, where players draw marbles in turn and announce their guess of what the majority color of the urn is, so that each player knows the guesses of all players that went prior (but not necessarily the actual color of their draw).
However, there is one major assumption we made in our analysis of the urn game. In our calculations, we assumed all players were perfectly reasoning individuals who would always guess the more probable majority color from the information they were given (and would guess the color of their draw if they found themselves with an equal probability of either outcome). Unfortunately, the real world is not so ideal. People make mistakes in their guesses, which can alter the outcome of the game. In this study, the researchers played the urn game the same way that we describe it, but they used a slightly different formula in their probability calculations to account for human error. In particular, their formula had a precision parameter, λ, that indicated the amount of error in the players’ decisions. When λ=0, their probability expression for the likelihood of a player making the right choice simplifies to 1/2, which means that the player is essentially picking randomly – they have a 1/2 chance of picking the right (i.e. higher probability) answer and a 1/2 chance of picking the wrong answer. When λ=∞, their probability for the player making the right choice simplifies to 1, which means the player will always make the right choice. Within this framework that accounts for error, they then ran the test under different conditions – in one case, the subjects got a reward just for participating, while in the other cases, subjects got rewards only when they guessed correctly. What they found was that error in guesses was much higher in the participation-reward scenario. This makes some sense since people are more likely to make mistakes, guess randomly, or even guess the wrong answer when the payoff for guessing correctly is the same as if they guess incorrectly. Another interesting result they found was that cascade decisions were far less likely in the participation-reward scenario, about 40% compared to about 75% in the scenarios rewarding correct guesses. Again, this phenomenon can probably be explained by the fact that people are more likely to guess whatever they want when there is no incentive for guessing correctly, so that a cascade is less likely to happen since the participants may guess differently than the cascading answer (since they have nothing to lose). Meanwhile, in the case where there is a reward for correct guesses, cascades and cascade answers are much more likely, since people are more likely to join the cascade and guess the same as everyone else (under the assumption that they will be more likely to guess correctly if they do so).
The urn game is an interesting situation to examine when considering conditional probabilities and the occurrence of information cascades or herd mentality scenarios. However, it is important to remember that human nature dictates that we are not perfect and are liable to make mistakes or incorrect decisions. Therefore, while it is interesting to examine the outcomes of the urn game under various ideal conditions, one must also take into account human error when looking to model more realistic situations.
November 7, 2014 | category: Uncategorized
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187509 | https://www.nationalfamilyplanning.org/file/2024-national-conference/DRH_Nguyen_SL_NFPRHA-2024---clean.pdf | CDC Contraception Guidance: U.S. Medical Eligibility for Contraceptive Use (US MEC) U.S. Selected Practice Recommendations for Contraceptive Use (US SPR) Antoinette T. Nguyen, MD, MPH, FACOG Senior Medical Officer Division of Reproductive Health 21 May 2024 2024 NFPRHA National Conference National Center for Chronic Disease Prevention and Health Promotion The findings and conclusions in this presentation are those of the author and do not necessarily represent the official position of the Centers for Disease Control and Prevention.
I have no conflicts of interest to disclose.
Disclosures CDC Contraception Guidance • Evidence-based clinical guidance for health care providers • Remove unnecessary medical barriers to accessing and using contraception • Support the provision of person-centered contraceptive counseling and services - Reproductive autonomy - Shared decision making Improving Contraception Access: Clinical Practice Guidelines • World Health Organization (WHO) global recommendations - First MEC published in 1996, with CDC technical assistance - First SPR published in 2001, with CDC technical assistance - Currently, MEC 5th edition (2015) and SPR 3rd edition (2016) • 2010: US MEC first adapted from WHO • 2013: US SPR first adapted from WHO • 2016: US MEC and US SPR update • Interim updates - 2017/2020: update of US MEC recommendation for women at high risk of HIV - 2021: new US SPR recommendation on self-administration of subcutaneous depot medroxyprogesterone acetate (DMPA-SC) History • Clinical guidance for safe use of contraceptive methods by medical conditions and characteristics • > 1800 recommendations for > 60 conditions and characteristics, e.g. - Is it safe for a patient with hypertension to use combined oral contraceptives?
- Is it safe for an adolescent to use an intrauterine device (IUD)?
U.S. Medical Eligibility Criteria for Contraceptive Use (US MEC) US MEC Categories US MEC, 2016 US MEC, 2016 • Clinical guidance that address provision of contraception, management of side effects, and issues related to contraceptive method use, e.g.
- How to be reasonably certain that a person is not pregnant - When to start a specific method - What exams and tests are needed - What follow-up is needed - How to manage bleeding irregularities and other problems - How many pill packs to provide U.S. Selected Practice Recommendations for Contraceptive Use (US SPR) US SPR: How to be reasonably certain a person is not pregnant US SPR, 2016 US SPR: When to start using specific contraceptive methods US SPR, 2016 • Title X family planning guidelines • Endorsed by professional organizations • Used by service delivery organizations • Training • US MEC/SPR app: downloaded over 440,000 times • Champions How the guidance is used US MEC/SPR Update • Determine the scope of the update - Public input through Federal Register Notice • Convene scoping meeting with subject matter experts (SMEs) to gather individual input on potential updates • Update existing and conduct new systematic reviews - Plan to publish in peer-reviewed journals • Conduct patient engagement listening sessions • Convene meeting with SMEs to gather individual input on the evidence and recommendations • CDC determines the final recommendations - Publication of updated US MEC and US SPR in Morbidity and Mortality Weekly Reports (MMWR) US MEC/SPR update process (2022-2024) • Increased emphasis on person-centered contraceptive counseling and provision • Use of gender-inclusive language • Updated terminology for certain conditions, e.g.
- Thrombophilia and hematologic conditions - Subcategories for cirrhosis and solid organ transplantation US MEC/SPR: General revisions • Addition of chronic kidney disease - Nephrotic syndrome - Hemodialysis - Peritoneal dialysis • Inclusion of additional contraceptive methods - New formulations of combined pills, patches and vaginal rings - New formulations of progestin only pills - New dose of progestin intrauterine device (IUD) - Vaginal pH modulator US MEC: New recommendations • Postpartum • Post-abortion: first trimester medication abortionwith mifepristone • DVT/PE: on anticoagulation therapy • Systemic lupus erythematous (SLE): positive or unknown antibodies • Cirrhosis • Liver tumors: hepatocellular adenoma • Sickle cell disease • Solid organ transplant • High risk for HIV • Additional conditions with increased risk of thrombosis (e.g., major surgery with prolonged immobilization, thrombophilia, superficial venous thrombosis, valvular heart disease, peripartum cardiomyopathy) US MEC: Updated recommendations • US MEC can help providers remove unnecessary medical barriers to accessing and using contraception • Most people can safely use most contraceptive methods • Contraceptive counseling and services should be offered in a non-coercive manner that honors a person’s values, goals, and reproductive autonomy through a shared decision-making process with providers • When applying US MEC classifications, providers should discuss the risks of a particular contraceptive method as well as the health risks associated with pregnancy US MEC: Take-home messages • New recommendations - Testosterone use and risk of pregnancy among transgender, gender diverse, and non-binary persons with a uterus - Self-administration of subcutaneous injectable contraception • Updated recommendations - Provision of medications for IUD placement - Bleeding irregularities during implant use • Changes to align with updates to US MEC 2024 US SPR: Updates • US SPR can help providers decrease medical barriers to initiating and using contraception • Most people can start most contraceptive methods at any time • Few, if any, exams or tests are needed • Routine follow-up generally not required • Recommendations for person-centered counseling and management of potential issues with contraceptive initiation and continuation US SPR: Take-home messages US MEC/SPR App • US MEC summary table (English, Spanish) • US SPR quick reference charts - When to start contraceptive methods and routine follow up - What to do if late, missed or delayed CHC or POP - Management of IUD when PID is found - Management of bleeding irregularities while using contraception Additional US MEC/SPR provider tools • Online access 2024 US MEC and US SPR will be published in late summer • Broad and diverse set of partners • Update and disseminate provider tools and app • Meeting and conference presentations • Publications and other outreach Dissemination and implementation For more information, contact CDC 1-800-CDC-INFO (232-4636) TTY: 1-888-232-6348 cdc.gov The findings and conclusions in this report are those of the authors and do not necessarily represent the official position of the U.S. Centers for Disease Control and Prevention.
Antoinette Nguyen, MD, MPH oms6@cdc.gov CDC, Division of Reproductive Health CDC Contraceptive Guidance for Health Care Providers | Reproductive Health | CDC |
187510 | https://www.nagwa.com/en/explainers/757135794890/ | Lesson Explainer: Distance and Displacement | Nagwa
Lesson Explainer: Distance and Displacement | Nagwa
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Lesson Explainer: Distance and Displacement Physics • First Year of Secondary School
In this explainer, we will learn how to define distance as the length of a path between two positions and displacement as the straight-line distance between two positions.
Let us first discuss distance.
When an object moves from one point to another, it moves along a path that connects those points. The path that the object moves along has a length. This length equals the distance that the object moves.
A path between two points can be a straight line between them. The following figure shows a straight path that an object moves along.
A path between two points can also be curved, as shown in the following figure.
For both straight lines and curves, it makes no difference to the distance traveled which point the object travels from and which it travels to, as the length of the line is the same either way. A distance does not have a direction, only a magnitude.
Quantities that have magnitudes but not directions are scalar quantities; hence, distance is a scalar quantity.
When an object moves, the motion can be between more than two points.
Suppose that an object travels from a point A to a point B, and then from point B to a point C, as shown in the following figure.
The movement of the object can be split into the movement from A to B and the movement from B to C.
The distance, , that the object travels is given by
Let us look at an example in which the distance traveled along a path that changes direction is determined.
Example 1: Determining Distance Traveled along a Path That Changes Direction
What is the total distance walked by someone along the lines shown in the diagram?
Answer
The person walks along three straight lines. The distance that they move is the sum of the lengths of these lines. The distance moved, , is given by
We can see that distance always increases as an object moves. The least distance that an object can move is zero, when it remains at rest.
Let us now look at another example in which the distance traveled along a path that changes direction is determined.
Example 2: Determining Distance Traveled along a Path That Changes Direction
What is the total distance covered by someone who walks along the lines shown in the diagram?
Answer
The person walks along three straight lines. The distance that they move is the sum of the lengths of these lines. The distance moved, , is given by
Suppose that the object also travels from point C back to point A, as shown in the following figure.
The distance, , that the object travels is now given by
We suppose that the object repeatedly travels from A to B, from B to C, and from C to A, making the journey several times. We can call the number of times that the object repeats this journey .
We can now call the distance that the object travels , which is given by
Let us look at an example in which the distance traveled along a closed path is determined.
Example 3: Determining Distance Traveled along a Path That Changes Direction
What is the total distance covered by someone who walks along the lines shown in the diagram, not walking on any line more than once?
Answer
The person walks along three straight lines. No line is walked more than once and no line is not walked, so each line is walked once.
The distance that the person moves is the sum of the lengths of these lines. The distance moved, , is given by
What has been shown in these examples for distances moved in straight lines also applies to distances moved along curved paths.
Suppose that an object travels along a circular path, as shown in the following figure.
Let us suppose also that the object travels once around the circular path, returning to its starting point and not reversing direction. The distance that the object moves equals the circumference of the circle.
Suppose instead that an object travels along the path shown in the following figure that takes the object from A to B, then from B to C, and finally from C to A.
The distance that the object moves is the sum of the lengths of the curved paths between the points.
Distance has now been explained.
Let us now discuss displacement.
When an object changes position, as well as moving a distance, it also has a displacement.
Displacement is also a quantity that describes the separation of points from each other, but it is not the same thing as distance.
The reason for displacement being different from distance is that displacement has a direction. Quantities that have a direction as well as a magnitude are vector quantities, so displacement is a vector quantity. Displacement is often represented by the symbol .
Consider the line connecting the points shown in the following figure.
An object could move from A to B or from B to A. The displacement of the object traveling from A to B is in the opposite direction to the displacement of the object traveling from B to A.
Let us suppose the distance from A to B is 1 metre. This is the same as the distance from B to A.
The displacement of an object that moves from A to B is 1 metre, but the displacement of an object that moves from B to A is metre, as shown in the following figure.
We can see from this that the distance between A and B equals the magnitude of the displacement of an object moving from A to B and it equals the magnitude of the displacement of an object moving from B to A. The direction of the displacement is shown by the positive or negative sign of the displacement.
The direction that is positive is from A to B in this example. Which direction is considered positive can be freely chosen. Whichever direction is considered positive, the opposite direction must be considered negative.
A displacement has a direction, so a displacement between two points must be a straight line between the points. A curved path changes direction along its length, so it does not have one specific direction.
Let us now look at an example in which displacements of points from other points are determined.
Example 4: Determining the Displacements between Positions
A speedboat passes by markers at the points A, B, and C, as shown in the diagram. Positive displacement is considered to be away from A, toward C.
What is the boat’s displacement from A when it is at B?
What is the boat’s displacement from C when it is at B?
What is the boat’s displacement from A when it is at C?
What is the boat’s displacement from C when it is at A?
Answer
The positive direction for displacement is stated in the question to be from A toward C. This is true whatever point the question asks for the displacement to be taken from.
When the boat is at B, the displacement from A to B is in the same direction as from A toward C, so it is in the positive direction, as shown in the following figure.
The distance from A to B is given by the distance from A to C minus the distance from B to C, so the displacement from A to B is given by
When the boat is at B, the displacement from C to B is in the opposite direction to the direction from A toward C, so it is in the negative direction, as shown in the following figure.
The distance from C to B is 180 m, so the displacement from C to B is given by
When the boat is at C, the displacement from A to C is in the positive direction, as shown in the following figure.
The distance from A to C is 250 m, so the displacement from A to C is given by
When the boat is at A, the displacement from C to A is in the negative direction, as shown in the following figure.
The distance from C to A is 250 m, so the displacement from C to B is given by
An object can return to its starting point by moving some distance along a line and then reversing the same distance along that line. The following figure shows points A and B connected by a straight line.
If an object travels from A to B and back to A, it has zero displacement. The distance moved by the object will not be zero, however, but will be twice the distance from A to B.
Let us now look at an example in which the distance and displacement due to the motion of an object that reverses direction are compared.
Example 5: Determining the Net Displacement of an Object That Changes Direction
A leaf is blown by the wind. The leaf moves 5 m forward and then 3 m backward.
What is the distance moved by the leaf?
What is the leaf’s net forward displacement?
Answer
The leaf moves in a straight line forward a distance of 5 m and then moves in a straight line backward a distance of 3 m. The distance that the leaf moves is the sum of the lengths of these paths. The distance moved, , is given by
The question asks for the net forward displacement of the leaf, so we should consider the forward motion of the leaf to be positive and must therefore consider the backward motion of the leaf as negative. The net forward displacement of the leaf is given by
If the motion of an object includes a change of direction that is not a complete reversal of that direction, then the object does not move along one line. The object can then be considered as having displacement in the -direction and in the -direction, as shown in the following figure.
The object travels equal distances in the -direction and in the -direction. The object has two displacements, each in a different direction.
Let us now look at an example in which the displacements in the and directions of an object that moves are determined.
Example 6: Determining the Net Displacement of an Object in Perpendicular Directions
A person walks from point A to point B, as shown in the diagram.
What is the displacement of point B from point A in the -direction?
What is the displacement of point B from point A in the -direction?
Answer
The diagram shows that the positive -direction is to the right. The object moves 4 m to the right and also moves 1 m to the left. The displacement in the -direction is given by
The diagram shows that the positive -direction is upward. The object moves 3 m upward and also moves 5 m downward. The displacement in the -direction is given by
An object can return to its starting point by moving along a closed path that changes direction. The path that the object takes to return to its starting position can consist of straight lines, curves, or both straight lines and curves, as shown in the following figure.
In the closed paths shown in the preceding figure, only the straight lines can represent displacements. Only displacements are vectors, so only the straight lines have arrows showing a direction.
Let us look at an example involving the displacements in the - and -directions of objects that move along closed paths.
Example 7: Determining the Net Displacement of an Object along a Closed Path
Two people walk along triangular lines that connect the points A, B, and C shown in the diagram. The first person walks from point A along a triangular path that returns them to point A. When the first person returns to point A, they stop. The second person walks from point B along a triangular path that returns them to point B. When the second person returns to point B, they stop.
What is the displacement of the first person from point A in the -direction when they stop?
What is the displacement of the first person from point A in the -direction when they stop?
What is the displacement of the second person from point B in the -direction when they stop?
What is the displacement of the second person from point B in the -direction when they stop?
Answer
The first person starts at point A and walks a triangular path back to point A, where they stop. Point A is the point at which the motion of the first person starts and the point at which it ends. The displacement of the person is, therefore, zero. A displacement of zero is zero in any direction, so the displacement in the -direction is zero and the displacement in the -direction is zero.
The motion of the second person is almost exactly the same as that of the first person, the only difference being that the second person starts at point B rather than point A.
The different starting positions of the two people make no difference to their displacements, as each person returns to their starting position and so both have zero displacement.
Let us now summarize what has been learned in these examples.
Key Points
A distance is the length of a path between two points.
The path between points can be a straight line or a curve.
The direction that an object moves between two points has no effect on the distance that the object moves. Distance has magnitude but no direction, so it is a scalar quantity.
The total distance moved by an object that moves between multiple points is the sum of the distances that it moves between those points.
A displacement is a straight-line distance from one point to another point.
A displacement has a direction as well as a magnitude, so it is a vector quantity.
For motion along a line, a direction must be chosen from one end of the line to the other for which the displacement is taken as positive. For the opposite direction, the displacement is taken as negative.
The magnitude of the displacement along a straight-line path between two points is the distance between those points along that path.
The motion of an object that travels from a point back to that same point produces zero displacement
For the motion of an object including changes of direction that are not a complete reversal of direction, the object will have displacements along more than one line.
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187511 | https://math.stackexchange.com/questions/4427999/bayesian-updating-with-binomial-data-and-a-uniform0-theta-prior | probability - Bayesian updating with Binomial data and a Uniform([0,$\theta$]) prior - Mathematics Stack Exchange
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Bayesian updating with Binomial data and a Uniform([0,θ θ]) prior
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Assume p p has a Uniform([0,θ])Uniform([0,θ]) prior, where θ∈(0,1)θ∈(0,1) is a known constant. Let X∼Binomial(n,p)X∼Binomial(n,p) for some n∈N n∈N. What is posterior μ(p|X)μ(p|X)?
(In the θ=1 θ=1 case it's well known that p|X=x∼Beta(1+x,1+n−x)p|X=x∼Beta(1+x,1+n−x). What happens when θ<1 θ<1?)
Edit: Assume for simplicity that this prior is correctly specified. (i.e. p p is "actually" drawn according to Uniform([0,θ])Uniform([0,θ]), and then X X is drawn as above given p p). However, if you know the answer for the case where this prior is misspecified (prior is Uniform([0,θ])Uniform([0,θ]), but p p is actually drawn according to Uniform([0,1])Uniform([0,1])) then even better!
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edited Apr 15, 2022 at 2:35
cluelessmathematiciancluelessmathematician
asked Apr 15, 2022 at 1:59
cluelessmathematiciancluelessmathematician
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If X X is the number of successes out of n n tries, it is either 0, 1 or above. If θ∈(0,1)θ∈(0,1) then the uniform prior's support never includes 1 or any number above it. Has your prior ruled out every possible observation except zero? I don't think Bayes theorem behaves right when you observe something to which you've ascribed zero prior probability.Retracted –Retracted 2022-04-15 02:06:07 +00:00 Commented Apr 15, 2022 at 2:06
Very good question! For a fixed θ∈(0,1)θ∈(0,1), the Uniform([0,θ])Uniform([0,θ]) prior indeed means that p∈(θ,1]p∈(θ,1] are "ruled out" ex-ante. I should have explicitly stated in my question that I assume this prior is correctly specified (i.e. p p is actually drawn according to the prior Uniform([0,θ])Uniform([0,θ]))cluelessmathematician –cluelessmathematician 2022-04-15 02:28:35 +00:00 Commented Apr 15, 2022 at 2:28
P.S. This question is actually a stepping stone for figuring out the case where this prior is misspecified (i.e. the prior is Uniform([0,θ])Uniform([0,θ]), but p p is "objectively" drawn according to Uniform([0,θ])Uniform([0,θ])). (A seminal paper about this kind of learning is Berk (1966))cluelessmathematician –cluelessmathematician 2022-04-15 02:31:33 +00:00 Commented Apr 15, 2022 at 2:31
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Why not just do the calculation:
p∼Uniform(0,θ),f(p)=1 θ 1(0≤p≤θ),p∼Uniform(0,θ),f(p)=1 θ 1(0≤p≤θ),
and
X∣p∼Binomial(n,p),X∣p∼Binomial(n,p),
so that
f(p∣x)∝Pr[X=x∣p]f(p)=p x(1−p)n−x 1(0≤p≤θ).f(p∣x)∝Pr[X=x∣p]f(p)=p x(1−p)n−x 1(0≤p≤θ).
This is a truncated beta; to show this, we integrate with respect to p p to find the normalization constant that makes the posterior likelihood a proper density:
∫θ p=0 f(p∣x)d p=∫θ p=0 p x(1−p)n−x d p∫p=0 θ f(p∣x)d p=∫p=0 θ p x(1−p)n−x d p
which is an incomplete beta function. For integers satisfying 0≤x≤n 0≤x≤n, there is a closed polynomial form; e.g., if n=7 n=7 and x=3 x=3, we get θ 4 280(70−224 θ+280 θ 2−160 θ 3+35 θ 4)θ 4 280(70−224 θ+280 θ 2−160 θ 3+35 θ 4), and then the posterior density is p 3(1−p)4 p 3(1−p)4 divided by this amount. It is not difficult to write this polynomial in terms of a finite sum; e.g.,
∫θ p=0 p x(1−p)n−x d p=∫θ p=0 p x∑k=0 n−x(n−x k)(−p)k d p=∑k=0 n−x(n−x k)(−1)k∫θ p=0 p x+k d p=∑k=0 n−x(n−x k)(−1)k θ x+k+1 x+k+1.∫p=0 θ p x(1−p)n−x d p=∫p=0 θ p x∑k=0 n−x(n−x k)(−p)k d p=∑k=0 n−x(n−x k)(−1)k∫p=0 θ p x+k d p=∑k=0 n−x(n−x k)(−1)k θ x+k+1 x+k+1.
So an explicit answer for your posterior density would be:
f(p∣x)=p x(1−p)n−x 1(0≤p≤θ)∑n−x k=0(n−x k)(−1)k θ x+k+1 x+k+1.f(p∣x)=p x(1−p)n−x 1(0≤p≤θ)∑k=0 n−x(n−x k)(−1)k θ x+k+1 x+k+1.
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answered Apr 15, 2022 at 2:58
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187512 | http://www.evlm.stuba.sk/~velichova/xmathkluc/diff/practise/examples/ex13.xml | EXAMPLE 1
Find the derivative of the function
f(x) = (3x - 2x2)(5 + 4x).
SOLUTION
Following the Product rule we have
f ' (x) = (3x - 2x2)(0 + 4) + (5 + 4x) (3 - 4x)=
= 12x - 8x2 + 15 + 12x - 20x - 16x2=
= - 24x2 +4x + 15
EXAMPLE 2
Find the derivative of the function
f(x) = 2x(x2 + 3x).
SOLUTION
Following the Product rule we have
f ' (x) = 2x(2x + 3) + 2(x2 + 3x) =
= 4x2 + 6x + 2x2 + 6x =
= 6x2 + 12x
Simplifying the function f(x) we have
f(x) = 2x3 +6x2
and following the Constant, Power and Sum rules we receive the same result
f ' (x) = 6x2 + 12x
EXAMPLE 3
Find the derivative of the function
f(x) =
(
1
x
+
1
)
(
x
-
1
)
.
SOLUTION
We begin with rewriting the function as
f(x) = (x -1 + 1)(x - 1)
and following the product rule we have
f ' (x) = (x -1 + 1)(1) + (x - 1) (-x -2)
= x -1 + 1 - x -1 + x -2 =
= 1 + x -2 =
x
2
+
1
x
2
EXAMPLE 4
Find the derivative of the function
f(x) = 3(x4 - 1)( 5x2 + 2x).
SOLUTION
Following the Constant rule and the Product rule we can write
f ' (x) = 3[(x4 - 1)(10x + 2) + ( 5x2 + 2x)(4x3)] =
= 3(10x5 + 2x4 - 10x - 2 + 20x5 + 8x4) =
= 3(30x5 + 10x4 - 10x -2) |
187513 | https://franknielsen.github.io/blog/MahalanobisThalesTheorem/index.html | Thales circle theorem extended to Mahalanobis geometry
Frank Nielsen Frank.Nielsen@acm.org
October 20, 2017
This column is also available in pdf: filename
1 Thales’ theorem in Euclidean geometry
In planar Euclidean geometry, Thales’ theorem states that any triangle pqr circumscribing a circle with one pair (p,q) of antipodal points is necessarily a right triangle. A pair (p,q) of antipodal points of a smooth convex object is such that the tangent lines at p and q are parallel to each other. See Figure 1 for an illustration, and for a historical account (Thales of Miletus, 624–546 BC).
Theorem 1 (Thales’ circle theorem) Any triangle circumscribed by a circle with one side being a diameter is right-angle.
Figure 1: Thales’ circle theorem: Triangle pqr is right-angle at r where [pq] is a diameter. (p,q) is a pair of antipodal points.
2 Thales’ theorem in Mahanalobis geometry
Let DA(p,q) denote the Mahalanobis distance between two points p and q, for a positive definite matrix A ≻ 0:
When A = I is the 2 × 2 identity matrix, the Mahalanobis distance amounts to the Euclidean distance DE(p,q) = ∥p - q∥ = .
A Mahalanobis circle CA(c,r) of center c and radius r is defined as follows:
A Mahalanobis circle has an ellipsoid (Euclidean) shape.
Let us generalize Thales’ theorem as follows:
Theorem 2 (Thales’ Mahalanobis circle theorem) Any triangle circumscribed by a Mahalanobis circle with one pair of points being antipodal is right-angle.
Proof: Consider the Cholesky decomposition of A: A = LL⊤ = U⊤U with L (U = L⊤) a lower triangular matrix (an upper triangular matrix, respectively) with positive diagonal elements. The Mahalanobis distance amounts to calculate an ordinary Euclidean distance on affinely transformed points x′ = L⊤x = Ux:
Thus a Mahalanobis circles CA transforms affinely to a Euclidean circle CE = CI, and antipodal pairs of points on CA remain antipodal in CE.
Two vectors u and v are perpendicular in the Mahalanobis geometry if and only if u⊤Av = 0. That is, if u⊤Av = u⊤LL⊤v = (L⊤u)⊤L⊤v = u′⊤v′ = 0.
A triangle pqr circumscribing the Mahalanobis circle CA with (p,q) an antipodal pair in Mahalanobis geometry transforms into a triangle p′q′r′ circumscribing the Euclidean circle C′ = {L⊤x : x ∈ CA(c,r)} with (p′,q′) an antipodal pair. Therefore p′q′r′ is a right-angle triangle in Euclidean geometry, and:
Therefore, pqr is a right-angle triangle in Mahalanobis geometry. □
Figure 2: Thales’ circle theorem in Mahalanobis geometry: Triangle pqr is right-angle at r where [pq] is an antipodal pair of points. Notice that Mahalanobis geometry is generally not conformal so that a Mahalanobis right-angle does not visualize as a Euclidean right-angle.
Note that Mahalanobis geometry is not conformal when A ⁄= λI (for λ > 0), the scaled identity matrix. Therefore angles are not preserved in Mahalanobis geometry: That is, a Mahalanobis right-angle cannot be visualized as a Euclidean right-angle in general.
Squared Mahalanobis distances are the only symmetric Bregman divergences . But Thales’ theorem do not extend to other (asymmetric) Bregman divergences.
References
Jean-Daniel Boissonnat, Frank Nielsen, and Richard Nock. Bregman Voronoi diagrams. Discrete & Computational Geometry, 44(2):281–307, 2010.
Frank Nielsen and Richard Nock. On the smallest enclosing information disk. Information Processing Letters, 105(3):93–97, 2008.
Christoph J Scriba and Peter Schreiber. Geometry in the Greek-Hellenistic era and late antiquity. In 5000 Years of Geometry, pages 27–116. Springer, 2015. |
187514 | https://nfnetwork.org/data/uploads/resources/resource_files/the_network_edge_volume_4.pdf | The Network Edge: Volume 4 - August 2013 The Network Edge brings you quarterly updates on the latest neurofibromatosis (NF) research and clinical advances from recent scientific publications. The Network Edge is organized into ‘bite sized’ sections by NF topic area, so you can focus in on the information that is of most interest for you. The Network Edge Features… - The Bottom Line: Each section starts with a summary sentence highlighting the ‘take home’ points from that section. - Highlighting Federally Funded Research: All research identified as being either fully or partly funded by the Congressionally Directed Medical Research Neurofibromatosis Research Program (CDMRP NFRP) or the National Institutes of Health (NIH) is tagged CDMRP or NIH after the author name. - A Global NF Picture: To keep you abreast of all NF research advances, The Network Edge includes publications from the United States and around the world. Country of origin of the research study is indicated after the author name. - The Network Edge Archive: At the end of this Volume of The Network Edge there is a table showing topics covered by past Volumes. This should help if you wish to search for further information in The Network Edge archive. Highlights from Volume 4 of The Network Edge: New biological markers to better predict development of MPNSTs Imaging brain structure: a new approach to measuring NF1 vision loss in NF1? A potentially helpful role for radiotherapy following NF2 meningioma surgery Café-au-lait marks may carry secrets of NF1 biology Children with NF1 have reduced brain blood flow New NF2 drug therapy shows promise Growth factor TGF-beta contributes to NF1 bone abnormalities Special considerations for surgical management of very large NF1 plexiform tumors Sleep disturbances in NF1 A new mouse model for studying NF1 learning disabilities Early screening for BRCA mutations in women with NF1 to detect breast cancer risk Multiple-spinal-menigioma disease, a new genetic condition defined NF Network 2 The Network Edge: Volume 4 - Contents 1. NF2 Clinical Management 2. NF1 Clinical Management 3. NF1 Bony Abnormalities 4. NF1 and the Eye: Optic Pathway Gliomas and Other Ocular Features 5. Heart and Blood Vessel Abnormalities in NF1 6. NF1 Malignant Peripheral Nerve Sheath Tumors 7. NF1 Learning Disabilities 8. Altered Brain Function in NF1 9. Other Clinical Features of NF1 10. NF Genetics Update 11. What’s New in Schwannomatosis? 12. NF1 and Breast Cancer: An Update 13. What’s New in NF2 Biology? 14. What’s New in NF1 Biology? Disclaimer: The Network Edge is a quarterly lay summary and synthesis of published scientific articles related to neurofibromatosis. Every effort has been made to ensure that the information provided accurately reflects and interprets the original articles. The Network Edge is not intended as a substitute for the medical advice of physicians. The reader should regularly consult a physician in matters relating to his/her health and particularly with respect to any symptoms that may require diagnosis or medical attention. The author and the Neurofibromatosis Network hereby disclaim liability to any party for loss, damage, or disruption caused by errors or omissions. The Network Edge © Neurofibromatosis Network, 2013 NF Network 3 1. NF2 Clinical Management The Bottom Line: The NF2 radiosurgery debate: weighing the benefits of cochlear implants. Radiotherapy/Radiosurgery Massager et al. (Belgium) analyzed data from eighteen persons with NF2 who had received Gamma-knife ‘model C’ stereotactic radiosurgery for treatment of vestibular schwannoma tumors between 2001 and 2010. The average follow up time on these persons after treatment was around four and a half years. Sixteen vestibular schwannoma tumors in twelve persons were studied, and tumor growth was found to be controlled in 80% of tumors. No one receiving radiosurgery developed facial weakness, and over three-quarters of the group had preservation of hearing. This study suggests positive effects of radiosurgery in NF2. Lustgarten (Venezuela) presented a review of publications centered on the use of radiosurgery in NF2, highlighting the fact that NF2 vestibular schwannomas are surgically challenging. These tumors are lobular, don’t have an extensive network of blood vessels like other tumors and grow faster than sporadic schwannomas. They also grow unpredictably, are typically diagnosed in younger persons than are sporadic schwannomas, and tend to engulf and even invade the surrounding cochlear and facial nerves (sporadic schwannomas instead compress and displace the nerves, making them easier to remove surgically without damaging the nerves). In short, NF2 vestibular schwannomas are more challenging to manage with traditional surgery. From a review of the literature, Lustgarten concluded that radiosurgery may help control NF2 tumor growth, but that hearing preservation outcomes are worse for those with NF2 receiving surgery than they are for those with sporadic schwannomas. He concludes with the following viewpoint: “There is positive and negative evidence as to whether radiosurgery might have detrimental effects if used in NF2, but that when all options and risks are presented to the person with NF2 they will choose the option that best suits their needs, whether this decision is based on time commitment, cost or invasiveness of the procedure”. Cochlear Implants Pai et al. (United Kingdom) reviewed the outcomes of cochlear implant (CI) surgeries in five persons with NF2 and two persons with a single vestibular schwannoma, conducted between 2000 and 2012. All seven persons had profound hearing loss at the time of surgery. The study monitored speech perception and quality of life after CI surgery. Three of the five participants with NF2 group achieved satisfactory speech perception, and one of the five achieved improved environmental sound and better communication ability. Importantly, however, a comparison showed that the participants with just one vestibular schwannoma gained more benefit from the CI than those with two. NF Network 4 2. NF1 Clinical Management The Bottom Line: Tackling surgery for very large plexiform tumors; monitoring for tumor re-growth after surgery, malignancy or infection; face transplant success update; transitioning from pediatric to adult NF1 care. Plexiform Neurofibroma It can be difficult for surgeons to remove an entire plexiform neurofibroma, since these often grow intimately with internal organs and not all tumor tissue can easily be removed. It is important to monitor the tumor site after surgery for tumor regrowth or malignant transformation. Nguyen et al. (United States and Germany) examined the records of fifty-two persons with NF1, who between them had 56 plexiform tumors, and who had undergone plexiform surgery and monitoring for up to 6 years following surgery. 8% of the group (6 children and 2 adults) had residual tumor left behind after surgery that progressed and required a second surgery. Overall tumor progression rates were higher in children than adults. The most successful surgeries were done on small plexiform neurofibromas where the majority of tumor could be removed at the outset. Very large plexiforms tumors are called giant neurofibromas, and require special management both during and after surgery. Vélez et al. (Spain) describe the surgical care and clinical management of a twenty-two-year-old woman with a giant neurofibroma in the lower back and buttock. The tumor has previously been de-bulked, but that surgery had led to excessive bleeding. The tumor had then continued to grow and required further surgery. The medical team imaged the tumor by MRI prior to surgery and saw that it contained an extensive network of blood vessels which placed the tumor at risk of excessive bleeding during surgery. In the days leading up to surgery, some of the blood vessels feeding the tumor were artificially blocked by the injection of small synthetic particles (called embolization) to help limit bleeding during surgery. A surgical stapler was used to manage bleeding during surgery. These treatments significantly controlled bleeding during surgery for what proved to be a sixteen kilogram (thirty-two pound) tumor. The patient required further surgery due to wound infection, but recovered fully and within two years was able to walk without support. The authors propose that this approach of carefully managed embolization could be a valuable tool for the treatment of giant neurofibromas. They also propose that the clinical definition of giant neurofibroma be established as one constituting 20% or more of a person’s body weight. Face Transplantation The French surgical team led by Dr. Laurent Lantieri pioneered a face transplantation approach and to date have done seventeen of these procedures including two persons with NF1. Sedaghati-Nia et al. (France) review the team’s progress to date with this technique and the challenges it still presents. Face transplantation is a lengthy surgery of fifteen to twenty eight hours. Regulating anesthesia and fluid levels during surgery is a major challenge. Multiple blood transfusions may be required due to hemorrhaging. The NF1 surgeries were particularly challenging because the entire face was being grafted, whereas for other conditions, such as burns, only part of the face needed to be grafted. In the period following surgery, the transplant recipients face challenges including opportunistic bacterial infections due to being immune suppressed. Nevertheless this pioneering technique offers promise to many including those who are most greatly affected by facial neurofibromas. NF Network 5 Transitioning from Pediatric to Adult Care NF1 is a lifelong condition requiring ongoing medical care, and transitioning from pediatrician to adult doctor can be challenging, as highlighted by Van Lierde et al. (Italy). To place this in context, NF1 is a rare disease (i.e. affecting fewer than 200,000 Americans); and there are an estimated 5,000 and 8,000 rare diseases, many with childhood onset. Like NF1, many rare diseases are complex conditions and can include a learning disability, presenting challenges in securing effective lifetime services. Adult doctors are often less familiar with rare conditions than are pediatricians. In Italy, as in many countries, there are no national guidelines for NF1 pediatric-to-adult transition, and no resources to support this. This report presented a ‘call to action’ to advocate for our governments to address this gap. This would actually be to the government’s benefit, since ignoring this issue will ultimately lead to higher costs for healthcare services in the long run. 3. NF1 Bony Abnormalities The Bottom Line: Altered bone metabolism in children with NF1; identifying clinical predictors of NF1-related scoliosis; the link between pain and spine stability in NF1; high levels of growth factor TGF-beta-1 contribute to bone abnormalities in NF1; highlighting potential orthodontic impacts of NF1 Altered Bone Metabolism Bone mineralization is reduced in people with NF1, but it is not clear why or what its impact is. Armstrong et al. (United States and Canada) study eighteen children with NF1 but no visible bone abnormalities, and compare them to their brothers and sisters who don’t have NF1 but have a similar diet and lifestyle. All of these children were compared to a larger group of ‘control’ children with no family history of NF1. The children with NF1 had reduced bone mineral content in the lumbar spine and femur (leg bone), than did their siblings or the control children. The children with NF1 had reduced density of trabecular (porous) bone in the tibia (leg bone) as well as reduced bone strength and reduced resistance to bending and stress. Children with NF1 had higher levels of blood calcium (suggesting more turnover of bone) than their brothers and sisters or the controls, though this needs further investigation. However, all of the children had the same incidence of arm and leg fractures and the same Vitamin D levels. The study concluded that children with NF1 might have a specific deficit in trabecular bone metabolism, and that this might be due to a delay in puberty and production of sex hormones. Scoliosis Scoliosis, or curvature in the backbone, is the most common bone abnormality in NF1 seen in about 30% of cases. Many people will require surgery for scoliosis at some point in life, but there are no reliable predictors for when or whether this will be needed. Lykissas et al. (United States) searched for such predictors by examining the past medical records of fifty six individuals with NF1. 63% of the group had three or more abnormal features of the spine, and these were the individuals in greatest need of surgery. The features most likely to lead to surgery were vertebral scalloping (concave regions appear on the back of the vertebrae) and dural ectasia (ballooning of the membrane around the spinal cord). The group proposes that studying the extent of these features could potentially predict the likely need of surgery in scoliosis. Bone Abnormalities as a Source of Pain Dural ectasia mentioned above is actually quite rare in NF1. It can cause significant pain but it is not clear why. A study of four persons with NF1 and dural ectasia by Khoo Bao et al. (United Kingdom) NF Network 6 showed that around the area of dural ectasia there are stress fractures in the bone, sometimes including fractures in the pedicles, which are small bony connector structures within the backbone. It is controversial whether the dural ectasia causes these fractures or vice versa. This new study suggests that these could occur simultaneously, agitating each other. Unfortunately this is a difficult aspect of NF1 to detect and study. When persons with NF1 present with pain, the first thought by the health professional is usually that the cause is malignancy, which is evaluated through MRI or PET-CT. MRI is not optimal for evaluating bone fractures, but CT is effective, and therefore this should be kept in mind during CT imaging in a person with NF1 presenting with pain. Detecting spinal fractures early is important as these can be managed by spinal stabilization, which would help avoid future deterioration. Facial and Jaw Bone Abnormalities NF1 can affect the teeth, mouth and jaw, either during development of these structures, or later because of tumors pressing on these structures. These can affect quality of life, and present a challenge for physicians once the malformations are established. Kopczynski et al. (Poland) report on a nine-year old girl with NF1 related tumors in the face and salivary gland on one side. The tumors had distorted the chin toward the unaffected side of the face and affected the angle of the mouth. CAT scanning revealed that the development of the bony structures had been affected by the growing tumors. Unfortunately, the tumor continued to grow after removal, and efforts to use clinical devices to correct the abnormality were not successful. The authors highlight this case for clinicians to be aware from a very early age that children with NF1 can develop abnormalities in the face and jaw, affecting dental development as well as physical development, and suggest orthodontic check-ups be an integral part of care from the time of NF1 diagnosis so that any problems may be identified and treated as early in life as possible. Bone Biology Rhodes et al. CDMRP (United States) propose that hyperactivity in the cell signal transforming growth factor beta-1 (TGF-beta-1) contributes to bony abnormalities in NF1. They show that mice with NF1 bony abnormalities have blood levels of TGF-beta-1 five to six times higher than normal. High levels of TGF-beta-1 were also seen in blood taken from people with NF1 bone abnormalities. Further study showed that in the bone itself, TGF-beta-1 is made by osteoblasts, the cells that give rise to new bone; and it is these cells that over express TGF-beta-1 in the mouse bone. When ‘functional’ neurofibromin (NF1 protein) is re-introduced into defective mouse osteoblasts, TGF-beta-1 signaling is normalized; and when an inhibitor of TGF-beta-1 is given to the mouse model, fractures in these mice are better able to heal. These findings support a role for TGF-beta-1 in mediating the occurrence of bony abnormalities in NF1 and suggest this growth factor pathway could be a future focus for targeting drug therapies. 4. NF1 and the Eye: Optic Pathway Gliomas and Other Ocular Features The Bottom Line: Examining the brain structure to predict potential vision loss in NF1; new clinical eye features in NF1; evaluating optic nerve tortuosity; optic pathway glioma and diencephalic syndrome. Clinical Eye Measurements in NF1 Cassiman et al. (Belgium) discuss the importance and controversies of the ophthalmic exams done in the course of NF1 clinical diagnosis and management. These include diagnostic identification of Lisch nodules and monitoring for optic pathway glioma, sphenoid dysplasia of the bony eye socket, and plexiform neurofibromas around the eye orbit. There is some controversy around how screening and follow up for the different features is done, and this group highlights a need to improve these by better NF Network 7 understanding the natural history of these features. For optic pathway glioma in particular, vision is principally used as a measure of improvement in response to drug therapy, but there is no single widely used and accepted visual acuity assessment protocol. One of the challenges is that measuring visual acuity in children can be difficult because it depends on the full cooperation of the child, who is often quite young. Seeking an alternative measure of visual acuity, de Blank et al. NIH (United States) measure brain tissue microstructure, specifically in the optic radiations, the nerve fibers that carry images from the eye to the visual cortex, the part of the brain that ‘interprets’ these were examined by diffusion tensor imaging (DTI) in the ophthalmology records of children with optic pathway glioma. The study showed that loss of visual acuity correlated with progressive structural changes in the optic radiations including changes in nerve fiber integrity. This brain measurement could be further refined and used to predict vision loss, so that children at greatest risk could receive appropriate clinical management. A Japanese group recently published three papers exploring other effects in the eye that may be associated with NF1 and may serve as clinical markers. Makino and Tampo (ref. 1, ref. 2) (Japan) present a report on a thirty-five-year-old man and a twenty-year-old man, both with NF1, who came to their clinic for an eye exam but did not report vision issues. Infrared fundus autofluorescence (IR-FAF) showed these men to have multiple, bright patchy lesions in the choroid of the eye which lies behind the retina. The team suggests that these choroidal abnormalities be explored as a diagnostic tool for NF1 to use in addition to looking for Lisch nodules. The same team, Makino et al. (Japan), report an unusual feature in the eye of a 30-year-old woman with NF1: abnormalities in the blood vessels of the retina. This is a rare condition in the general population, and the group believes this is the first case reported in NF1. This woman also had the choroidal abnormalities mentioned above. What was intriguing was that in monitoring this woman over a number of years, the abnormal blood vessels seemed to undergo dynamic changes. In contrast, when abnormalities in the blood vessels of the retina are seen in the general population, these remain quite static. The group will pursue this finding. Optic Nerve Tortuosity Children with NF1 often have tortuous (‘squiggly’) optic nerves; in the general population, the optic nerve is straighter. It is not really clear what a tortuous optic nerve means in NF1, and there has been no way of effectively measuring this. Ji et al. NIH (United States) took on this question with a goal of developing a ‘tortuosity index’ that could be measured from MRI images to create a universal scale. The index was developed taking into account the overall dimensions of the optic nerve. This study examined retrospective data from children captured over a period of eight years from children with NF1 who had undergone MRI imaging for tumor exams, and children without NF1 who had undergone MRI for headache evaluation. Tortuosity was greater in the NF1 group (both in those with and without a diagnosed optic pathway glioma) than those without NF1. The tortuosity index proved to be more reliable than prior methods such as visually analyzing the optic nerves and making a subjective assessment. In this study, 84% of children with NF1 were classified with tortuosity. The next step is to correlate tortuosity with vision status, optic pathway glioma status, etc. to see if tortuosity might be a predictor of other NF features. Diencephalic Syndrome Diencephalic syndrome can occur in infants and causes profound emaciation in the child. Affected children lack fat tissue under the skin, have difficulty gaining weight even when taking in normal calorie amounts, are hyperactive and may develop hypothalamic brain tumors. Cavicchiolo et al. (Italy) describe a case of diencephalic syndrome in a three-year-old with NF1 and an optic pathway NF Network 8 glioma that was initially stable but began to grow one year after diagnosis. It is believed this tumor growth impinged on the hypothalamus and resulted in diencephalic syndrome. Once the child received treatment to stabilize the tumor growth, the diencephalic syndrome subsided. The authors suggest that children with optic pathway gliomas should be monitored for symptoms of diencephalic syndrome. 5. Heart and Blood Vessel Abnormalities in NF1 The Bottom Line: Brain blood flow is reduced in children with NF1; link to glaucoma in NF1; potential blood vessel rupture in NF1. Blood Brain Flow Heart and blood vessel abnormalities can occur in NF1 and affect any region of the body. As discussed in Volume 3 of The Network Edge, this includes the brain where abnormal growth of brain blood vessels in children, called Moyamoya, can lead to stroke. However the full spectrum of abnormalities in brain blood flow in NF1 is not known. Yeom et al. (United States) evaluated retrospective data from children with and without NF1 who had been examined between 2010 and 2012 using a specialized imaging technique called arterial spin-labeled perfusion (ASL). This provides a highly sensitive measure of brain blood flow. ASL had been used in children with NF1 to examine brain structure; and in children without NF1 it had been used for a variety of clinical investigations such as headaches or dizziness. The study included a very select subpopulation of fourteen children with NF1 aged twenty-two months to eighteen years old, and age matched children without NF1. The study excluded those with diagnosed Moyamoya, those with psychiatric problems, and those who had been exposed to chemotherapy or radiation, all of which might independently affect blood brain flow. The study found significantly reduced brain blood flow in various regions of the brain in children with NF1, not seen in children without NF1. Brain blood flow is tightly linked to the body’s metabolism and the group proposes that an underlying metabolic difference in the NF1 children is undermining brain blood flow. The authors looked for a connection with learning disabilities but found none. They acknowledged this is a very small and early stage study but will explore this further. Structural Blood Vessel Abnormalities Blood vessel abnormalities in NF1 can extend to the tiniest blood vessels such as those in the eye. Pichi et al. (Italy) report on the case of a thirteen-year-old boy with NF1 who came to their clinic reporting pain and blurred vision in one eye, and was found to have high pressure in the eye. This had previously been described and treated as hypertensive uveitis. Imaging the boy’s eye revealed tortuous blood vessels with poor flow; this was subsequently confirmed as retinal ischemia. This was finally diagnosed as neovascular glaucoma caused by the occlusion of abnormal blood vessels. This case highlights the potential for glaucoma in NF1 as a side effect of abnormal blood vessels. Puvanesarajah et al. (United States) report on a thirty-two-year-old man with NF1 presenting with upper back pain, muscle weakness, numbness in the extremities and difficulty urinating. Imaging revealed a neurofibroma in the spine, and subsequent surgery showed there had been a blood vessel rupture. This turned out to be an intercostal arterial aneurysm (weakness in the wall of an artery located between the ribs). Following surgery to remove the tumor and repair the aneurysm, the man made a full recovery. This case highlights the potential for blood vessel rupture in NF1. NF Network 9 6. NF1 Malignant Peripheral Nerve Sheath Tumors The Bottom Line: Identifying new molecular ‘predictors’ of MPNST; progress in searching for new MPNST drug targets; utilizing MRI and PET together in MPNST diagnosis. Molecular Markers of MPNST People with NF1 have a one hundred times greater risk of developing malignant peripheral nerve sheath tumor (MPNST) than do the general population; half of all cases of MPNST are seen in NF1. Early detection of MPNSTs is critical, as these are difficult to manage. In the past couple of Volumes of The Network Edge we have highlighted research focused on identifying markers for malignant peripheral nerve sheath tumors. The goal is to be able to predict whether a person is at risk of developing MPNSTs so they can receive the appropriate clinical management. Park, Sawitzki et al. (Germany and Korea) reviewed what has been published about this field. They identified fifty-six possible blood markers for MPNST and used these to screen blood samples from one hundred and four persons with NF1, with and without MPNST, and from forty-one persons without NF1. From this study, four markers of particular interest were identified: epidermal growth factor, interferon-gamma, interleukin-6, and tumor necrosis factor-alpha. All four of these had an altered expression in NF1. Another two markers, IGFBP1 and RANTES, were elevated specifically in persons with NF1 and MPNST. These six markers were further analyzed to see whether marker level correlated with NF1 tumor load (the volume of tumors identified in an individuals’ body by MRI scan). IGFBP1 was found to correlate with tumor load. The two markers IGFBP1 and RANTES may have future applications as predictors of MPNST occurrence in NF1. Two other recent studies look at a role for miRNAs as MPNST predictors. miRNAs are small cellular elements which play a role in many normal cell functions. Certain miRNAs are elevated to unusually high blood levels in cancer. Volume 3 of The Network Edge reported on the work of Weng et al. suggesting that miRNAs could predict whether someone with NF1 is at risk of developing MPNSTs. Weng et al. (China) have expanded this study and examined the genetics of two hundred persons with NF1 and one hundred and fifty six persons with NF1 and MPNST to look within eleven different miRNA signaling pathways for the presence of 53 mutations called ‘single nucleotide polymorphisms (SNPs). This study confirmed that there are specific miRNA mutations which individually, or in combination, might promote the risk of MPNSTs. Masliah-Planchon et al. (France) examined the activity level of three hundred and seventy seven well know miRNAs in a series of NF1 related dermal neurofibromas, plexiform neurofibromas and MPNST tumor samples and cells. The study found a variety of fluctuations in miRNA levels. Most significantly this included fluctuations in miRNAs that regulate the tumor suppressor gene PTEN, resulting in a drop in PTEN activity, which is typically associated with cancer. PTEN activity was at higher levels in dermal neurofibromas and plexiform neurofibromas than in MPNSTs. Identification of these miRNAs could help inform future research strategies toward predicting and treating MPNSTs. Rahrmann et al. NIH (United States) have developed a unique approach to studying the genetic mutations underlying NF1. The group uses the ‘Sleeping Beauty’ transposon system which introduces small pieces of DNA randomly into the genome of Schwann cells and their precursor cells in mice. Many of these mice go on to form tumors. Studying these helps to identify which mutations may be driving MPNST development. The group studied two hundred and sixty nine neurofibromas and one hundred and six MPNSTs that grew in mice following the ‘Sleeping Beauty’ treatment. These revealed six NF Network 10 hundred and ninety five new genetic sites in the neurofibromas, and eighty seven new genetic sites in the MPNSTs, that seemed to be significant. Comparing this to human data, the group identified a number of new candidate gene mutations that could have a role in MPNSTs. In addition a new gene of unknown function, Foxr2, was identified. Foxr2 turned out to be a proto-oncogene (i.e. being cancer-associated) and is increased in expression in MPNST cells compared to normal Schwann cells. When Foxr2 is introduced into normal cells it promotes them to make tumors. Foxr2 may have a role in maintaining growth of MPNSTs. Overall this study has turned up a number of interesting leads for future studies to identify and develop treatments for MPNSTs. Clinical Management of MPNSTs Finally, moving to the clinical imaging setting, Ahlawat et al. (United States) report a case of a fifty-three-year-old man with NF1 reporting progressive pain in his left leg. PET imaging revealed two masses in the thigh that were suspect MPNSTs. Anatomical and functional MRI analysis followed, and the functional MRI analysis, which examines tumor metabolism, actually suggested that the two suspect tumors had different features. Subsequently both tumors were surgically excised. One was indeed found to be an MPNST, while the other was found to be a benign schwannoma which had essentially ‘mimicked’ an MPNST. From this study the authors highlight the importance of using anatomical and functional MRI as well as PET when seeking to diagnose an MPNST. 7. NF1 Learning Disabilities The Bottom Line: A new mouse model for studying NF1 learning disabilities shows promise. Wozniak et al. CDMRP, NIH (United States) created a genetically-engineered mouse that is prone to develop optic pathway gliomas, and also has NF1-related learning disabilities. These mice, termed Nf1-OPG, have previously been shown to have low levels of the nerve signal dopamine in an area of the brain called the striatum. This was believed to be contributing to learning disabilities in these animals, since this behavior can be corrected by the drug L-Dopa, which compensates for reduced dopamine. In this study, the Nf1-OPG mice were subjected to maze tests and intellectual challenges. The Nf1-OPG mice showed a reduced wish to explore their environment, and generally moved around less, particularly when the environment was “less safe” (e.g. in a maze, when the arm of maze was ‘open ended’ versus closed). The Nf1-OPG mice also showed less interest in sniffing smells (“olfactory investigations”) and were less willing to stand on their hind limbs (”rear”) to reach a suspended ball. When these mice received the drug L-Dopa, these behaviors became more normalized. Future studies by the group will look at changes in signaling pathways within the brain to see if these correlate with the behavioral changes seen in the Nf1-OPG mice. The model certainly seems to be useful for further study of NF1-related learning disabilities. NF Network 11 8. Altered Brain Function in NF1 The Bottom Line: Poor sleep quality for those with NF1; interpreting UBOs/T2-hyperintensities; special considerations for epilepsy in NF1. Sleep Disturbances NF1-related sleep disturbances are not often discussed, but Leschziner et al. (United Kingdom) focus on sleep patterns in a study of one hundred and fourteen individuals with NF1. The study considered factors that might influence sleep, such as drugs prescribed, physical complications and employment status. Overall, those with NF1 had a poorer sleep quality and more sleep disturbances than did the general population. The study suggests this to be influenced by a number of factors including pain, anxiety, depression and learning disabilities. UBOs/T2-hyperintensities A challenge for physicians in imaging and monitoring tumors in the brain of children with NF1 is the appearance of T2-hyperintensities – also known as ‘bright spots’ or ‘unidentified bright objects’ (UBOs) - that are visible on scans. These hyperintensities usually disappear with age and are not cause for concern in kids, but they can confuse the picture of what is being visualized in the brain. The reason for UBOs appearing and disappearing is not well understood; they are not formally recognized by the NIH guidelines for NF1 diagnosis. Khan et al. (United Kingdom) endeavor to better characterize UBOs, and the parts of the brain where they develop, with a view to considering whether they could be used as formal diagnostic clues for NF1. A retrospective review of twenty-two childrens’ MRI scans was included. 81% of the children had UBOs, primarily in two regions of the brain: the globus pallidus, which regulates voluntary movement, and the cerebellum, which regulates balance. To a lesser extent, UBOs were found in the cerebral hemispheres and brainstem. UBOs were infrequent in children under four, but more common in children between four and twelve. UBOs were found to be twice as common in males, than in females. The UBOs did not seem to progress with age even when NF1 symptoms progressed. 9. Other Clinical Features of NF1 The Bottom Line: A biological basis for kidney abnormalities in NF1; lung lesions and NF1; a rare cerebellar tumor; screening for high blood calcium may detect hyperthyroidism in NF1. Kidney Abnormalities in NF1 Kidney abnormalities in NF1 that are unrelated to tumor growth are rare. Afshinnia et al. (United States) report on a case of a forty-two-year-old woman with NF1 (and a strong family history of NF1) presenting with migraine. Clinical analysis revealed that she had proteinuria (protein in the urine) and focal segmental glomerulosclerosis (FSGS), both of which suggest damaged kidney structure and function. This is quite unusual in NF1 and the group believes it may be only the third such case reported. The woman had multiple tumors in the chest but no tumor growth around the kidney. There was no family history of kidney disease. Fortunately the woman responded well to treatment by the drug lisinopril. The investigators dig beyond the clinical and into the biology of this case and propose the kidney abnormalities seen may be rooted in organ development and are driven by abnormalities in mTOR signaling and MAP kinase signaling – both of which are known also to have roles in tumor growth. NF Network 12 Brain Tumors Benign brain tumors such as pilocytic astrocytomas are frequent in NF1. Rarely, a malignant brain tumor – such as glioblastoma or anaplastic astrocytoma - will be identified. Any type of NF1-related brain tumor in a region of the brain called the cerebellum is very rare, so a malignant tumor of the cerebellum in NF1 is extremely rare. Brokinkel et al. (Germany) describe a very rare case of an NF1 related cerebellar anaplastic astrocytoma in a fifty-four-year-old-man. The tumor was partially resected by surgery then treated with chemotherapy. Though the man made recovery and treatment initially seemed to stem tumor growth, he deteriorated over the next sixteen months until receiving the drug levodopa. The authors believe this was a highly unique case and in scanning the literature could find only one other reported NF1-related cerebellar anaplastic astrocytoma in a nine-year-old girl. Finally, in a continuing medical education report, Antonio et al. (Brazil) present a helpful chronology of the major advances made in NF1 research and clinical care from its historical identification until today, including some of the challenges that remain. (This paper is very readable and available free by looking up the reference listed below on www.pubmed.gov). 10. NF Genetics Update The Bottom Line: Special considerations for spinal NF1; examining NF1 gene mutations in international populations; role for the NF1 gene in pheochromocytoma in the general population; defining a new SMARCE1 related genetic condition, multiple-spinal-meningioma disease. Spinal NF1 Spinal NF1 is a form of NF1 where the person has multiple spinal tumors on both sides of the spinal cord, on the nerve roots of the spine; tumors may affect all nerve roots. It has traditionally been thought to have only a minimal presence of the other tumor types and features of NF1 (e.g. bone abnormalities, learning disabilities). Spinal NF1 typically has a later life onset than ‘full blown’ NF1. Risk of malignancy or severe health issues is linked to the extent to which tumors develop around the spinal cord. A few recent publications examine this form of NF1. As with NF1 in general, spinal neurofibromatosis can be inherited or can occur spontaneously. Ruggieri et al. (Italy) report on a pair of female twins aged sixteen who are monozygotic (i.e. they are identical twins arising from the same egg), as well as a fourteen-year-old boy, all of whom developed spinal neurofibromatosis. Neither the twins nor the boy had a parent with NF1. The twin girls developed spinal neurofibromatosis at age four. All three children were diagnosed because of the presence of large, symmetrical spinal neurofibromas and accompanying neuropathy (nerve degeneration and pain). From a review of past publications, the authors believe this study has yielded some new findings. They believe this is the first reported case of monozygotic twins with spinal neurofibromatosis, and they believe age 4 to be the youngest reported diagnosis of spinal neurofibromatosis. Genetic analysis showed all three to have missense mutations in a region of the NF1 gene called exon 39. Finally, all three had MRI-detected signal abnormalities in a region of the brain called the basal ganglia, which the authors also believe to be a unique finding. Burkitt Wright et al. (United Kingdom) ask whether spinal neurofibromatosis can really be accurately diagnosed based on clinical features – as is largely currently the case - when no genetic analysis is done. An accurate diagnosis of spinal NF1 is an important issue for families since this will help NF Network 13 dictate clinical care plans and will hopefully mean a less severe outcome for the person than ‘full blown’ NF1. It is also important for family planning and genetic counseling purposes, since spinal neurofibromatosis tends to occur through multiple generations in families, but its diagnosis in later life often means someone is diagnosed after having children. The research group examined five families with spinal NF1 identified through the same neurofibromatosis clinic. These persons were found to have a broad spectrum of NF1 genetic mutations indicating that spinal neurofibromatosis can occur due to a number of different mutations. The family members also presented quite a variety of additional NF1 features: café-au-lait spots were found to be quite common and there were occasional occurrences of other tumor types such as optic pathway glioma. When these families were analyzed for genetic mutations, splice and missense mutations were found to be associated with ‘strictly spinal’ NF1 (and were seen in four of the five families studied) though frameshift and nonsense mutations may rarely be associated. From their studies the group proposes that if anyone is suspected to have spinal NF1 they should be subject as early as possible to a genetic analysis so that the right clinical monitoring plan and, if necessary, genetic counseling can be provided. International NF1 Genetics Studies Two recent publications looking at different international populations with NF1 review the range of mutations seen in the NF1 gene. Van Minkelen et al. (The Netherlands) presented a review of 18 years of data from almost two thousand people diagnosed clinically or genetically with NF1. The individuals studied had over one thousand different individual mutations, although definitive mutations were not identified for everyone even when there was a clear clinical diagnosis of NF1. The mutations were cross-referenced with the physical features of NF1 in each person to see if ‘genotype-phenotype’ links could be identified. The strongest link seen was that the presence of gene microdeletions was linked to a six-fold increase in children needing special education. The group commented that emerging genetic technologies should allow screening for novel mutations. Nemethova et al. (Slovakia) report on a study of the range of gene mutations in one hundred and eight persons in the Slovakia population diagnosed with NF1. The population exhibited a range of physical features of NF1, and half had family members with NF1. Frameshift mutations were the most common type of mutation, in around 40% of the population; missense and splicing mutations were rarer, each seen only in about 10% of the population. Gene deletions and duplications made up about 6% of cases. A remaining five cases had nontypical splicing variants. Overall this represents a similar repertoire of mutations seen in other countries, except that there was a high representation of small deletions/insertions and a decreased proportion of nonsense mutations, than seen elsewhere. Hopefully this eastern European population can add to what has been seen in other countries. The NF1 Gene and Pheochromocytoma Volume 3 of The Network Edge reported that in a small number of persons with NF1, pheochromocytoma (an adrenal tumor type) was seen in conjunction with gastrointestinal stromal tumors (GIST, a potentially malignant tumor of the gut). Both pheochromocytoma and GIST are quite rare NF1-associated tumors but were found to co-exist in a small number of patients. In a new study, Welander et al. (Sweden) report from a search of public gene databases that NF1 gene mutations may actually be an important player not only in NF1-related pheochromocytomas, but also in person in the general population who develop pheochromocytoma. This finding could help increase a focus on pheochromocytoma and the role of the NF1 gene in it occurrence. SMARCE1 Gene and Multiple-Spinal-Meningioma Disease NF Network 14 Smith et al. (United Kingdom) have identified a new genetic condition, multiple-spinal-meningioma disease, due to a mutation in the SMARCE1 gene on Chromosome 17. Like the SMARCB1 gene, which is associated with a proportion of cases of schwannomatosis, SMARCE1 is a chromatin-remodeling complex gene that also has a tumor suppressor function. Meningiomas are benign tumors that can occur in NF2 (50-75% likelihood over lifetime), and NF2 gene mutations are frequently seen in meningiomas in the general population too. However meningiomas are not typically associated with either NF1 or schwannomatosis. Three unrelated individuals with a family history of spinal schwannomas were clinically diagnosed with spinal meningiomas but with no NF2 or SMARCB1 mutations. These persons were found to have previously unidentified SMARCE1 mutations. This important study, published in Nature Genetics therefore puts a new genetic disease (and potential gene of interest in NF as well) on the map. Is it NF1? Balikcioglu et al. (United States) present a case of three African-American brothers with adrenal hypoplasia congenita, a rare inherited condition that in boys causes adrenal failure in infancy or childhood, and in later life, infertility due to gonadotropin deficiency and defective spermatogenesis. It is caused by mutations in the NR0B1 gene, which has sex-linked functions. In the family presented, the authors report the first ever case where boys with this condition also have features of NF1 or possibly Legius Syndrome. These included café au lait spots, skeletal issues, large head circumference, and freckling. However, no mutations in NF1 gene or the SPRED1 (Legius Syndrome) gene were found. This is an interesting case as it highlights the possibility of syndromes that may appear to be NF1 or Legius Syndrome but are not verified by genetic analysis. 11. What’s New In Schwannomatosis? The Bottom Line: Highlighting the importance of educating radiologists about schwannomatosis. Koontz et al. (United States) provide a summary of the advances and challenges in the field of schwannomatosis. Though much has been learned about this rare form of neurofibromatosis, it remains unfamiliar to many radiologists. It is important that radiologists become educated about schwannomatosis, and how to differentiate it from NF2, because imaging plays such an important role in schwannomatosis diagnosis. 12. NF1 and Breast Cancer: An Update The Bottom Line: Early screening for BRCA mutations in women with NF1 may help with early detection of breast cancer risk. As we have reported in past Volumes of The Network Edge, there is a growing body of evidence that women with NF1 are at increased risk of developing breast cancer compared to the general population. Campos et al. (Spain) report on two women from the same family affected by both NF1 and early-onset breast cancer (diagnosed at ages forty and thirty-five). Mutation analyses of these women identified a type of genetic mutation called a nonsense mutation in the NF1 gene; and a type of genetic NF Network 15 mutation called a frameshift mutation in BRCA1, a gene which is closely linked to increased risk of breast cancer and which is located on the same section of Chromosome 17 as the NF1 gene. The sisters had a family history of NF1 including a brother with an NF1 mutation but no BRCA1 mutation. This study suggests there may be value in early screening for BRCA gene mutations in women with NF1 so that any added risk of developing breast cancer can be fully evaluated and managed. 13. What’s New in NF2 Biology? The Bottom Line: New tumor drug therapy shows promise. Tanaka et al. (United States) report on the candidate NF2 therapy NXD30001. This drug targets HSP90, a ‘chaperone protein’ in the cell whose normal role is to stabilize a range of signaling molecules involved in cell division and growth. The HSP90-regulated signals are hyperactive in NF2 tumors and HSP90 itself is expressed at higher than normal levels in a variety of tumors and tends to correlate with a poor prognosis. Treatment with the drug NXD30001 breaks down HSP90 and reduces signaling in these pathways with a goal of slowing tumor growth. NXD30001 was tested in cells lacking NF2 gene function, as well as in primary cells taken directly from human schwannoma and meningioma tumors. The drug suppressed division of these cells, and genetic analysis of the cells showed that an array of molecular targets had been inhibited. The drug also inhibited tumor growth in animal models of NF2. NXD30001 therefore represents a promising candidate therapy for NF2 tumors. The drug has been shown to effectively cross the blood-brain barrier, which bodes well for its development into a treatment. 14. What’s New in NF1 Biology? The Bottom Line: Café au lait marks yield new biology leads to understanding NF1; progress in pilocytic astrocytoma biology. Biology of Café au Lait Marks A hallmark physical feature of NF1 is café au lait marks on the skin. These look like very large freckles and can be quite numerous. The tan color in café au lait spots comes from pigment cells within the skin called melanocytes. Though these markings do not present any health concerns, their biological development is of interest to scientists, because understanding how café au lait spots form might tell us a bit more about the genetic causes of NF1. Arun et al. (Canada) examined this question using normal human melanocytes. They found that in normal cells, the NF1 gene product, neurofibromin, interacts inside the cell with microtubules (structural elements), specifically a component of the microtubules called the ‘dynein heavy chain 1’ (DHC). This neurofibromin-DHC interaction normally regulates the activity of pigmentation particles called melanosomes within the melanocyte. When the NF1 gene was inactivated in these cells (as it is in people with NF1), and no neurofibromin was made, the pigmentation particles become unregulated and oversized within the cell. This contributes to the appearance of café au lait marks. The loss of neurofibromin’s regulatory activities on microtubules may well contribute to other physical features on NF1, and the authors plan to pursue this. NF Network 16 Biology of Pilocytic Astrocytomas Pilocytic astrocytomas are the most common glial cell tumor occurring in children, and 15-20% of these cases are in NF1. These tumors are low grade, i.e. not aggressively malignant. Though they can usually be effectively removed by surgery, some of the brain regions where these tumors grow - the optic pathway, brainstem and cerebellum – can be tricky to operate on without resulting in loss of vision or another function. Chen and Gutmann NIH (United States) present a review of these tumors to better understand their cellular origins, biology and how they might be treated by drug therapies. The first molecular signal to be identified as contributing to pilocytic astrocytomas was the mutation of the NF1 gene. Non-NF1 associated pilocytic astrocytomas have an abnormal signal called fusion-BRAF (f-BRAF). F-BRAF and abnormal NF1 signaling both cause over-activity of the cell signal ‘mammalian target of rapamycin’ (mTOR). As a result, drugs targeting mTOR are being assessed as clinical therapies for pilocytic astrocytomas. There are good mouse models of pilocytic astrocytomas which have helped these studies tremendously. For example, various candidate genes are inactivated in different brain cell populations to see if tumors arise. The local cellular environment around where tumors form is being explored for potential ‘drivers’ of tumor growth, which might include the influence of another cell type called microglia. (The microenvironment concept is also seen in plexiform neurofibromas where non-tumor cells called mast cells are believed to promote the growth of the tumor; this interaction has been effectively targeted in the clinic with the drug Gleevec). It may also be the case that children developing pilocytic astrocytoma have additional differences in their genome that further amplify the likelihood of these tumors developing. This summary provides a comprehensive view of progress made to date in understanding and treating the unique tumors that are pilocytic astrocytomas. NF Network 17 References Afshinnia F, Vega-Warner V, Killen P. (2013) Focal segmental glomerulosclerosis in association with neurofibromatosis type 1: a case report and proposed molecular pathways. Clin Kidney J. Apr;6(2):208-210. Ahlawat S, Blakeley J, Montgomery E, Subramaniam RM, Belzberg A, Fayad LM. (2013) Schwannoma in neurofibromatosis type 1: a pitfall for detecting malignancy by metabolic imaging. Skeletal Radiol. May 7. [Epub ahead of print] Antônio JR, Goloni-Bertollo EM, Trídico LA. (2013) Neurofibromatosis: chronological history and current issues. An Bras Dermatol. Jun;88(3). pii: S0365-05962013000300329. doi: 10.1590/abd1806-4841.20132125. Armstrong L, Jett K, Birch P, Kendler DL, McKay H, Tsang E, Stevenson DA, Hanley DA, Egeli D, Burrows M, Friedman JM. (2013) The Generalized Bone Phenotype in Children With Neurofibromatosis 1: A Sibling Matched Case-Control Study. Am J Med Genet A. May 25. doi: 10.1002/ajmg.a.36001. [Epub ahead of print] Arun V, Worrell L, Wiley JC, Kaplan DR, Guha A. (2013) Neurofibromin interacts with the cytoplasmic Dynein Heavy Chain 1 in melanosomes of human melanocytes. FEBS Lett. Apr 10. pii: S0014-5793(13)00264-0. doi: 10.1016/j.febslet.2013.03.035. [Epub ahead of print] Balikcioglu PG, Gómez R, Vargas A, Aradhya S, Messiaen LM, Lacassie Y. (2013) Adrenal hypoplasia congenita with phenotypic features suggestive of neurofibromatosis type 1 among three African-American brothers. Am J Med Genet A. Jul 4. doi: 10.1002/ajmg.a.36031. [Epub ahead of print] Brokinkel B, Schober O, Ewelt C, Heindel W, Hargus G, Stummer W, Holling M, Wölfer J. (2013) Cerebellar Anaplastic Astrocytoma in an Adult with Neurofibromatosis Type 1: Case Report and Review of Literature. J Neurol Surg A Cent Eur Neurosurg. May 13. [Epub ahead of print] Burkitt Wright EM, Sach E, Sharif S, Quarrell O, Carroll T, Whitehouse RW, Upadhyaya M, Huson SM, Evans DG. (2013) Can the diagnosis of NF1 be excluded clinically? A lack of pigmentary findings in families with spinal neurofibromatosis demonstrates a limitation of clinical diagnosis. J Med Genet. Jun 28. [Epub ahead of print] Campos B, Balmaña J, Gardenyes J, Valenzuela I, Abad O, Fàbregas P, Volpini V, Díez O. (2013) Germline mutations in NF1 and BRCA1 in a family with neurofibromatosis type 1 and early-onset breast cancer. Breast Cancer Res Treat. Apr 27. [Epub ahead of print] Cassiman C, Legius E, Spileers W, Casteels I. (2013) Ophthalmological assessment of children with neurofibromatosis type 1. Eur J Pediatr. May 25. [Epub ahead of print] NF Network 18 Cavicchiolo ME, Opocher E, Daverio M, Bendini M, Viscardi E, Bisogno G, Perilongo G, Da Dalt L. (2013) Diencephalic syndrome as sign of tumor progression in a child with neurofibromatosis type 1 and optic pathway glioma: a case report. Childs Nerv Syst. Apr 25. [Epub ahead of print] de Blank PM, Berman JI, Liu GT, Roberts TP, Fisher MJ. (2013) Fractional anisotropy of the optic radiations is associated with visual acuity loss in optic pathway gliomas of neurofibromatosis type 1. Neuro Oncol. 2013 May 7. [Epub ahead of print] Chen YH, Gutmann DH. (2013) The molecular and cell biology of pediatric low-grade gliomas. Oncogene Apr 29. doi: 10.1038/onc.2013.148. [Epub ahead of print] Ji J, Shimony J, Gao F, McKinstry RC, Gutmann DH. (2013) Optic nerve tortuosity in children with neurofibromatosis type 1. Pediatr Radiol. May 2. [Epub ahead of print] Khan A, Beri S, Baheerathan A, Balki A, Hussain N, Gosalakkal J. (2013) Globus pallidus high-signal lesions: A predominant MRI finding in children with neurofibromatosis type 1. Ann Indian Acad Neurol. 2013 Jan;16(1):53-6. doi: 10.4103/0972-2327.107702. Khoo Bao JN, Ogunwale B, Huson SM, Ealing J, Whitehouse RW. (2013) Spinal bone defects in neurofibromatosis type I with dural ectasia: stress fractures or dysplastic? A case series. Eur Radiol. Jun 28. [Epub ahead of print] Koontz NA, Wiens AL, Agarwal A, Hingtgen CM, Emerson RE, Mosier KM. (2013) Schwannomatosis: the overlooked neurofibromatosis? AJR Am J Roentgenol. Jun;200(6):W646-53. doi: 10.2214/AJR.12.8577. Kopczyński P, Flieger R, Matthews-Brzozowska T. (2012) Changes in the masticatory organ in patients with Recklinghausen's disease - a case report. Contemp Oncol (Pozn). 16(5):453-5. doi: 10.5114/wo.2012.31780. Epub 2012 Nov 20. Leschziner GD, Golding JF, Ferner RE. (2013) Sleep Disturbance as Part of the Neurofibromatosis Type 1 Phenotype in Adults. Am J Med Genet A. May 1:0. doi: 10.1002/ajmg.a.35915. [Epub ahead of print] Lustgarten L. (2013) The impact of stereotactic radiosurgery in the management of neurofibromatosis type 2-related vestibular schwannomas. Surg Neurol Int. Apr 17;4(Suppl 3):S151-5. doi: 10.4103/2152-7806.110663. Print 2013. Lykissas MG, Schorry EK, Crawford AH, Gaines S, Rieley M, Jain V. (2013) Does the Presence of Dystrophic Features in Neurofibromatosis Type 1 Patients With Spinal Deformities Increase the Risk of Surgery? Spine (Phila Pa 1976). May 15. [Epub ahead of print] Makino S, Endoh K, Tampo H. (2013) Retinal microvascular abnormalities in neurofibromatosis type 1 associated with congenital retinal macrovessels. Case Rep Ophthalmol Med. 2013:604191. doi: 10.1155/2013/604191. Epub 2013 May 23. Makino S, Tampo H. (ref. 1)(2013) Choroidal abnormalities in a patient with neurofibromatosis type 1. Intern Med. 52(12):1445-6. NF Network 19 Makino S, Tampo H. (ref. 2) (2013) Optical coherence tomography imaging of choroidal abnormalities in neurofibromatosis type 1. Case Rep Ophthalmol Med. 2013:292981. doi: 10.1155/2013/292981. Epub 2013 Apr 22. Masliah-Planchon J, Pasmant E, Luscan A, Laurendeau I, Ortonne N, Hivelin M, Varin J, Valeyrie-Allanore L, Dumaine V, Lantieri L, Leroy K, Parfait B, Wolkenstein P, Vidaud M, Vidaud D, Bièche I. (2013) MicroRNAome profiling in benign and malignant neurofibromatosis type 1-associated nerve sheath tumors: evidences of PTEN pathway alterations in early NF1 tumorigenesis. BMC Genomics. Jul 13;14(1):473. [Epub ahead of print] Massager N, Delbrouck C, Masudi J, De Smedt F, Devriendt D. (2013) Hearing preservation and tumour control after radiosurgery for NF2-related vestibular schwannomas. B-ENT. 2013;9(1):29-36. Nemethova M, Bolcekova A, Ilencikova D, Durovcikova D, Hlinkova K, Hlavata A, Kovacs L, Kadasi L, Zatkova A. (2013) Thirty-Nine Novel Neurofibromatosis 1 (NF1) Gene Mutations Identified in Slovak Patients. Ann Hum Genet. Jun 12. doi: 10.1111/ahg.12026. [Epub ahead of print] Nguyen R, Ibrahim C, Friedrich RE, Westphal M, Schuhmann M, Mautner VF. (2013) Growth behavior of plexiform neurofibromas after surgery. Genet Med. 2013 Apr 18. doi: 10.1038/gim.2013.30. [Epub ahead of print] Pai I, Dhar V, Kelleher C, Nunn T, Connor S, Jiang D, O'Connor AF. (2013) Cochlear implantation in patients with vestibular schwannoma: A single United Kingdom center experience. Laryngoscope. Apr 24. doi: 10.1002/lary.24056. [Epub ahead of print] Park SJ, Sawitzki B, Kluwe L, Mautner VF, Holtkamp N, Kurtz A. (2013) Serum biomarkers for neurofibromatosis type 1 and early detection of malignant peripheral nerve-sheath tumors. BMC Med. Apr 23;11(1):109. [Epub ahead of print] Pichi F, Morara M, Lembo A, Ciardella AP, Meduri A, Nucci P. (2013) Neovascular glaucoma induced by peripheral retinal ischemia in neurofibromatosis type 1: management and imaging features. Case Rep Ophthalmol. Apr 13;4(1):69-73. doi: 10.1159/000350956. Print 2013 Jan. Puvanesarajah V, Lina IA, Liauw JA, Coon AL, Witham TF. (2013) Intercostal aneurysm causing spinal cord compression in an NF1 patient. Eur Spine J. Apr 19. [Epub ahead of print] Rahrmann EP, Watson AL, Keng VW, Choi K, Moriarity BS, Beckmann DA, Wolf NK, Sarver A, Collins MH, Moertel CL, Wallace MR, Gel B, Serra E, Ratner N, Largaespada DA. (2013) Forward genetic screen for malignant peripheral nerve sheath tumor formation identifies new genes and pathways driving tumorigenesis. Nat Genet. May 19. doi: 10.1038/ng.2641. [Epub ahead of print] Rhodes SD, Wu X, He Y, Chen S, Yang H, Staser KW, Wang J, Zhang P, Jiang C, Yokota H, Dong R, Peng X, Yang X, Murthy S, Azhar M, Mohammad KS, Xu M, Guise TA, Yang FC. (2013) Hyperactive transforming growth factor-β1 signaling potentiates skeletal defects in a neurofibromatosis type 1 mouse model. J Bone Miner Res. May 23. doi: 10.1002/jbmr.1992. [Epub ahead of print] NF Network 20 Ruggieri M, Polizzi A, Salpietro V, Incorpora G, Nicita F, Pavone P, Falsaperla R, Nucifora C, Granata F, Distefano A, Padua L, Caltabiano R, Lanzafame S, Gabriele AL, Ortensi A, D'Orazi V, Panunzi A, Milone P, Mankad K, Platania N, Albanese V, Pavone V. (2013) Spinal Neurofibromatosis with Central Nervous System Involvement in a Set of Twin Girls and a Boy: Further Expansion of the Phenotype. Neuropediatrics. Jun 18. [Epub ahead of print] Sedaghati-Nia A, Gilton A, Liger C, Binhas M, Cook F, Ait-Mammar B, Scherrer E, Hivelin M, Lantieri L, Marty J, Plaud B. (2013) Anaesthesia and intensive care management of face transplantation. Br J Anaesth. May 23. [Epub ahead of print] Smith MJ, O’Sullivan J, Bhaskar SS, Hadfield KD, Poke G, Caird J, Sharif S, Eccles D, Fitzpatrick D, Rawluk D, du Plessis D, Newman WG, Evans DG. (2013) Loss-of-function mutations in SMARCE1 cause an inherited disorder of multiple spinal meningioma. Nature Genetics, 45(3): 295-299. Tanaka K, Eskin A, Chareyre F, Jessen WJ, Manent J, Niwa-Kawakita M, Chen R, White CH, Vitte J, Jaffer ZM, Nelson SF, Rubenstein AE., Giovannini M (2013)Therapeutic Potential of HSP90 Inhibition for Neurofibromatosis type 2.Clin Cancer Res. May 28. [Epub ahead of print] Van Lierde A, Menni F, Bedeschi MF, Natacci F, Guez S, Vizziello P, Costantino MA, Lalatta F, Esposito S. (2013) Healthcare transition in patients with rare genetic disorders with and without developmental disability: Neurofibromatosis 1 and williams-beuren syndrome. Am J Med Genet A. May 21. doi: 10.1002/ajmg.a.35982. [Epub ahead of print] van Minkelen R, van Bever Y, Kromosoeto JN, Withagen-Hermans CJ, Nieuwlaat A, Halley DJ, van den Ouweland AM. (2013) A clinical and genetic overview of 18 years Neurofibromatosis type 1 molecular diagnostics in the Netherlands. Clin Genet. May 8. doi: 10.1111/cge.12187. [Epub ahead of print] Vélez R, Barrera-Ochoa S, Barastegui D, Pérez-Lafuente M, Romagosa C, Pérez M. (2013) Multidisciplinary management of a giant plexiform neurofibroma by double sequential preoperative embolization and surgical resection. Case Rep Neurol Med. 2013;2013:987623. doi: 10.1155/2013/987623. Epub Mar 28. Welander J, Söderkvist P, Gimm O. (2013) The NF1 gene: a frequent mutational target in sporadic pheochromocytomas and beyond. Endocr Relat Cancer. Jul 4;20(4):C13-C17. Print 2013. Weng Y, Chen Y, Chen J, Liu Y, Bao T. (2013) Common genetic variants in the microRNA biogenesis pathway are associated with malignant peripheral nerve sheath tumor risk in a Chinese population. Cancer Epidemiol. Jun 10. pii: S1877-7821(13)00079-9. doi: 10.1016/j.canep.2013.05.003. [Epub ahead of print] Wozniak DF, Diggs-Andrews KA, Conyers S, Yuede CM, Dearborn JT, Brown JA, Tokuda K, Izumi Y, Zorumski CF, Gutmann DH. (2013) Motivational Disturbances and Effects of L-dopa Administration in Neurofibromatosis-1 Model Mice. PLoS One. Jun 10;8(6):e66024. doi: 10.1371/journal.pone.0066024. Print 2013. Yeom KW, Lober RM, Barnes PD, Campen CJ. (2013) Reduced Cerebral Arterial Spin-Labeled Perfusion in Children with Neurofibromatosis Type 1. AJNR Am J Neuroradiol. Jun 13. [Epub ahead of print] NF Network 21 The Network Edge Archive This table indicates which topics have been covered in past Volumes of The Network Edge. Past Volumes may be accessed online at: CONTENTS Volume 1 - Fall 2012 Volume 2 - Winter 2013 Volume 3 - Spring 2013 Volume 4 - Summer 2013 CDMRP NFRP Updates X X NF1 Clinical Trials X X NF1 Clinical Management X X X X NF1 Learning Disabilities X X X X NF1 Bony Abnormalities X X X X NF1 Malignant Peripheral Nerve Sheath Tumors X X Heart and Blood Vessel Abnormalities in NF1 X X X Increased Breast Cancer Risk in NF1 X X Other Clinical Features of NF1 X X X What’s New in NF1 Biology? X X X X NF2 Clinical Trials X X NF2 Clinical Management X X X X What’s New in NF2 Biology? X X X X Schwannomatosis Update X X X Legius Syndrome Update X X The Evolving Link Between NF and Cancer X Altered Brain Function in NF1 X NF1 and the Eye: Optic Pathway Gliomas and Other Features X NF Genetics Update X Pheochromocytoma in NF1 X 213 S. Wheaton Avenue, Wheaton, IL 60187 Phone 630-510-1115 www.nfnetwork.org admin@nfnetwork.org |
187515 | https://www.youtube.com/watch?v=QTw4XlvfSNA | Find the Missing Numerator or Denominator | Solve Fraction Equations | Eat Pi
Eat Pi
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Posted: 11 Jun 2022
In this video, I teach you how to find the missing numerator or denominator of a fraction equation. I go over a couple methods including cross multiplication.
0:00 - x/12 + 8/12 = 11/12
1:20 - x/4 + 2/3 = 11/12
4:45 - 1/4 + 2/x = 11/12
If you have any questions, please leave them in the comment section below! Also, if you find the videos helpful, please like, share, and subscribe!
25 comments
Transcript:
x/12 + 8/12 = 11/12 what's up you friggin geniuses so in this video i'm going to teach you how to find the missing numerator or denominator of a fraction equation all right and we're specifically going to go over some addition problems here okay so let's start with this problem right here so here we have a missing numerator right so we have x over 12 plus 8 over 12 is equal to 11 over 12 right so whenever you're solving a problem like this the first thing you want to do is always make sure your denominators match up okay that's the very first step on all these problems so as you can see here we have all matching denominators right they're all 12. so in that case that means we just have to worry about the numerators the numbers on top right so all we're trying to solve is what's on top so here we have x plus eight is equal to eleven right let's write it out x plus eight is equal to eleven right now to solve for x right here we just need to get rid of this plus twelve so we can just sorry 8 so we just have to subtract 8 and what you do to one side of the equation you have to do to the other right so we have to subtract 8 on this side also right so then here these 8's cancel out so then we're just left with this x right here so we have x is equal to 11 minus 8 which is equal to 3. all right so then your answer right here would be x is equal to 3. all right not too bad right let's do a x/4 + 2/3 = 11/12 couple more all right so here's a similar problem so again we're missing the numerator right so we have x over 4 this time plus two-thirds is equal to 11 over 12. so again you want to make sure all the denominators match in this case they don't right so that's the first thing that we have to deal with all right so how can we make all these denominators match right well first of all we have to figure out what's the smallest number that all of these denominators fit into okay and in this case it would be 12 right because 12 fits into itself one time 3 fits into 12 evenly right 4 times and 4 fits into 12 evenly three times okay so we want to change all these bottom numbers to 12 right so first of all to change 4 into 12 all we have to do is multiply by 3 right because 4 times 3 is equal to 12 right but since we're multiplying by 3 on the bottom we also have to multiply by 3 on the top okay we can't just change fractions willy-nilly all right we have to be consistent all right so if we're going to do something to the bottom we got to do it to the top right so then on this one over here to change 3 to 12 all we have to do is multiply by 4 right because 3 times 4 is equal to 12. but again stay consistent so if we're going to multiply by 4 on the bottom we also have to multiply by 4 on top okay and then here we already have 12 right so we don't have to do anything we have to change anything here or in the numerator all right so let's simplify some things so first of all here on top x times 3 that's simply equal to 3x right so we can just write it as 3 x and then that's going to be over the bottom over here 4 times 3 that's equal to 12. and then we're going to add this other fraction right here right so on top we have 2 times 4 which is equal to 8 and that's going to go over 3 times 4 which again is 12 and then that's equal to 11 over 12. right 11 over 12. all right now you can see we have all the same denominators right so now that we have all the same denominators now like in the last problem all we have to do is worry about the top over here right so we have 3x plus 8 is equal to 11 right so 3 x plus 8 is equal to 11 right so we're trying to solve for x right here so we want to isolate this by itself on one side of this equal sign right so the first thing we can do is get rid of this plus eight and just like in the last problem we can do that by subtracting eight and what we do to one side of the equation again you do that to the other side right so then on this side again these cancel out so then we're just left with 3 x on this side right and then that's equal to 11 minus 8 which is equal to 3. okay now in order to get rid of this 3 since we're multiplying here in order to undo that multiplication we just have to divide and you just divide by whatever you're trying to get rid of which is three right so we're going to divide the side by three and again what you do to one sided equation you do to the other so divide the side by 3 also all right so then on this side these 3's cancel out so then we're just left with this x right here so then we get x is equal to 3 divided by 3 which is equal to 1. boom all right so then your answer over here is x is equal to uno to 1. 1/4 + 2/x = 11/12 all right last problem and here we're trying to figure out what the denominator is right so here we have 1 4 plus 2 over x is equal to 11 over 12 right now again the first step was always trying to figure out what's the smallest number that all your denominators fit into right but in this case we can't do that because we don't know what this denominator is right it's just a variable it's just an x okay so what we're going to have to do instead is like in a regular linear equation is just try and get this x by itself on one side of the equal sign okay so the first thing that we can do is get rid of this fraction right here the one-fourth all right so this positive one-fourth over here i'm going to subtract one-fourth okay and what we do to one side of an equation again we do to the other right so we're going to subtract one-fourth from this side also okay so then on this side these 4 cancel out so then we're just left with this 2 over x right positive 2 over x on this side so we have i'll write it over here 2 over x and then that's equal to 11 12 minus 1 4 okay so in order to add or subtract fractions again they have to have the same denominators right and in this case they don't but they both fit into this number 12 right here right because 12 fits into itself one time evenly and four fits into twelve evenly right three times so again we're going to multiply the bottom by three and multiply the top by three okay so then simplifying this we're going to have 11 over 12 minus and then on top over here we're going to have 1 minus or sorry 1 times 3 1 times 3 is equal to 3 and on the bottom we have 4 times 3 which is equal to 12. okay so then now we have the same denominators so now we can just solve straight across right so eleven minus three that's equal to eight okay so here this simplifies to eight over twelve right so this thing up here again simplifies to just eight over twelve eight over 12. okay so now here we're just left with a fraction is equal to another fraction right so in this case we can use cross multiplication all right if you're not sure if you don't remember what cross multiplication is i'll link a video to that in the card above but basically the way you solve that is just by multiplying diagonally or crossing okay so for instance we're going to multiply these two together and then we're going to set that equal to this other cross right here when we multiply those two together okay so again we're going to multiply these two together this diagonal first so we have 2 and 12 right so 2 times 12 that's equal to 24 and then we're going to set that equal to our other cross over here in orange the x and the eight so eight times x is just equal to eight x right so we're just going to write that as eight x okay so here we have 24 is equal to 8x right so again the goal is just to solve for x right here right so we need to get rid of this 8 so we can do that by dividing by 8 right and again what you do to one side you do to the other okay so then here these eights cancel out so then on this side we're just left with this x so x is equal to 24 divided by eight which is equal to three all right so then your answer right here for your denominator would be x is equal to 3. so if you found the video helpful definitely leave a thumbs up down below and if you have any other questions or want to see any other examples just let me know in the comment section below |
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Basics
Learn more about activity, concentration, potential and voltage here.
Back to Principles of Electrochemical Potential Measurement
Basic Terms
Activity and Concentration – both terms are used in electrochemistry. But what is the difference?
The concentration of substance c indicates the amount of substance n of a dissolved species in a defined volume and thus the unit is mol/L. Furthermore, the designation molarity M is commonly used in chemistry for the concentration of substance in mol/L.
The concentration of substance is dependent on the temperature because the volume is temperature-dependent. You can also relate the concentration of substance of a dissolved species to the weight of the solvents in kg. Then, you get the molality b in mol/kg. The molality is temperature-independent. The molality is especially found in older tables.
Moreover, we can find the term of the activity of a species. The activity of a substance is a thermodynamic value, a kind of a corrective concentration.
In a solution the charged ions are surrounded by solvation shells (hydrate shells in water as solvent) and thus sealed off each other. The ions are less effective. The activity therefore corresponds to the remaining effective concentration, which is usually lower than the concentration because of the interactions between the ions in the solution. Only in very diluted solutions are the activity and the concentration almost equal.
In highly concentrated solutions, however, the activities can reach values higher than the concentration because there are not enough solvate molecules to completely solvate all ions.
The deviation between activity and concentration is reflected in a correction factor, the so-called activity coefficient y, γ or f (depending on the kind in which the concentration of substance is stated).
a = y ∗ c c = concentration of substance in mol/L (molarity)
a = γ ∗ c c = concentration of substance in mol/kg (molality)
a = f ∗ c c = content of the dissolved substance as mole fraction
Universal definition of the activity (does not correspond to the physicochemical definition) according to Jander, Jahr; Maßanalyse; 15. Auflage, 1989, de Gruyter
Potential – Voltage – Electromotive Force
All three terms refer to a voltage difference between two systems. If the reference point for the measurement is known, it is called potential; otherwise, it is called voltage.
Electromotive force refers to the potential difference of a cell, which is measured if the cell works reversibly and no current flows.
According to the convention of the IUPAC 1, a standard potential (or standard voltage) E is calculated by the equation
E = E(right) – E(left) (standard electrode potential of the right cell minus standard electrode potential of the left cell).
The IUPAC² also determines: “The standard potential of an electrochemical reaction, abbreviated as standard potential, is defined as the standard potential of a hypothetical cell, in which the electrode (half-cell) at the left of the cell diagram is the standard hydrogen electrode (SHE) and the electrode at the right is the electrode in question.“
If you always consider the standard hydrogen electrode as the “left” cell, the standard potentials of the electrochemical series automatically arise, indicated as standard reduction potential. In old tables one often finds standard oxidation potentials, in which case the algebraic signs must be reversed. Whether oxidation or reduction potentials are indicated in a table can be seen from the algebraic signs for the reaction Cu/Cu 2+, because then the standard potential is +0.34 V.
In doing so, the question arises as to which electrode has to be connected to which input of the measurement instrument, because the algebraic sign of the measured potential changes depending on the connection of the electrodes. The values of the electrochemical series are the result of the measurements if you connect the standard hydrogen electrode to the minus input (COM). If you measure against another reference electrode, you connect it to the minus input (COM). Thus, the measurement electrode is always connected to the positive pole:
E = E(measurement electrode) – E(reference electrode).
The algebraic sign of the measured potentials shows which of the electrode is more positive, in which direction the current flows and which of the reaction runs spontaneously, since the potential is directly related to the equilibrium constant of the reactants. E > 0 for E(right) > E(left) means that the right electrode is positively charged towards the left electrode. There is therefore a shortage of electrons at the right electrode caused by the reduction of particles. This electrode is also called the cathode, where the reduction takes place. Therefore, the electrones flow from left to right. Thus, the electric current, unfortunately defined as a flow of positive particles, flows in the opposite direction.
There is an excess of electrons at the left electrode, because particles are being oxidised there. This electrode is called the anode, where the oxidation takes places. The reaction therefore tends to runs spontaneously from left (oxidation) to right (reduction). E < 0 for E(right) < E(left) by implication means that the reaction runs spontaneously from right to left.
Conversion of the potentials of common reference electrodes (png)
1: Cohen et al: Quantities, units and symbols in physical chemistry, IUPAC Green Book, 3rd edn., 2nd printing, IUPAC & RSC Publishing, Cambridge, 2008, p. 71
2: Cohen et al: Quantities, units and symbols in physical chemistry, IUPAC Green Book, 3rd edn., 2nd printing, IUPAC & RSC Publishing, Cambridge, 2008, p. 74
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187518 | https://goodfaithmedia.org/global-christian-population-projected-to-reach-3-3-billion-by-2050/ | Global Christian Population Projected to Reach 3.3 Billion by 2050
by Zach Dawes Jr | Feb 13, 2023 | News
There will be over 2.6 billion Christians worldwide by the middle of 2023 and around 3.3 billion by 2050, according to a report published in early January by the Center for the Study of Global Christianity at Gordon Conwell Theological Seminary.
The 2,604,381,000 estimate for mid-year 2023 Christian population total is a 44,506,000-person increase from the mid-year 2022 total.
The new estimates of the worldwide Christian population for both 2025 and 2050 are higher than the 2022 report’s estimates. The 2023 report projects a Christian population of 2,662,979,000 in 2025 (up 25.8 million from the 2022 estimate) and a population of 3,342,878,000 in 2050 (up 8.8 million).
The number of Christians as a percentage of the world population declined from 34.5% to 32.3% from 1900 to 2000, before rising 0.1% from 2000 to 2022. If the report’s estimates prove accurate, Christians would comprise 34.4% of the global population in 2050.
From 2000 to 2022, the global Christian population increased by 1.18% annually, with Pentecostals / Charismatics seeing the highest annual growth rate (1.88%), followed by evangelicals (1.79%), independents (1.65%), Protestants (1.58%), Roman Catholics (0.93%), unaffiliated Christians (0.82%) and Orthodox Christians (0.53%).
In 2050, Roman Catholics are projected to remain the most populous Christian tradition (1.5 billion), followed by Pentecostals / Charismatics (1 billion), Protestants (881 million), evangelicals (620 million), independents (616 million), Orthodox (307 million) and unaffiliated (131 million).
Annual Christian population growth rates from 2000 to the present have been highest in Africa (2.76%), followed by Asia (1.62%), Latin American (1.03%), Oceana (0.92%), Northern America (0.29%) and Europe (0.04%). Overall, the Global South’s Christian population has increased by an annual percentage rate of 1.8% during this time period, compared to a rate of 0.12% for the Global North.
From 2000 to the present, the global Muslim population increased at the highest annual rate of all faith traditions at 1.87%, increasing by 697.8 million adherents to 2,007,352,000 (mid-year 2023 projection).
The global Sikh population increased at the next highest annual rate (1.56%), followed by Hindus (1.2%), Christians (1.18%), ethnoreligionists (1.16%), Buddhists (0.77%), Jews (0.7%), Agnostics (0.61%), nonreligionists (0.56%), Chinese folk-religionists (0.34%), atheists (0.29%) and new religionists (0.26%).
The full report is available here.
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Home › Resources › Accounting › Net Profit Margin
Table of Contents
What is Net Profit Margin?
Net Profit Margin Formula
Download CFI's Free Net Profit Margin Template
Video Explanation of Net Profit Margin
Understanding the Ratio
Limitations of Net Profit Margin Ratio
Financial Analysis
Additional Resources
Net Profit Margin
Net income divided by total revenue, expressed as a percentage
Written byCFI Team
Read Time 6 minutes
Over 2.8 million + professionals use CFI to learn accounting, financial analysis, modeling and more. Unlock the essentials of corporate finance with our free resources and get an exclusive sneak peek at the first module of each course. Start Free Start Free
What is Net Profit Margin?
Net Profit Margin (also known as “Profit Margin” or “Net Profit Margin Ratio”) is a financial ratio used to calculate the percentage of profit a company produces from its total revenue. It measures the amount of net profit a company obtains per dollar of revenue gained. The net profit margin is equal to net profit (also known as net income) divided by total revenue, expressed as a percentage.
The typical profit margin ratio of a company can be different depending on which industry the company is in. As a financial analyst, this is important in day-to-day financial analysis.
Source: CFI Financial Analysis Fundamentals Course.
Net Profit Margin Formula
Net Profit Margin = Net Profit ⁄ Total Revenue x 100
Net profit is calculated by deducting all company expenses from its total revenue. The result of the profit margin calculation is a percentage – for example, a 10% profit margin means for each $1 of revenue the company earns $0.10 in net profit. Revenue represents the total sales of the company in a period.
Calculation Example #1
Company XYZ and ABC both operate in the same industry. Which company has a higher net profit margin?
Step 1: Write out the formula
Net Profit Margin = Net Profit/Revenue
Step 2:Calculate the net profit margin for each company
Company XYZ:
Net Profit Margin = Net Profit/Revenue = $30/$100 = 30%
Company ABC:
Net Profit Margin = Net Profit/Revenue = $80/$225 = 35.56%
Company ABC has a higher net profit margin.
Calculation Example #2
Company A and company B have net profit margins of 12% and 15% respectively. Both companies earned $150 in revenue. How much net profit did each company make?
Step 1:Write out formula
Net Profit Margin = Net Profit/Revenue
Net Profit = Net Margin Revenue
Step 2:Calculate net profit for each company
Company A:
Net Profit = Net Margin Revenue = 12% $150 = $18
Company B:
Net Profit = Net Margin Revenue = 15% $150 = $22.50
Calculation Example #3
Company A and B earned $83.50 and $67.22 in net profit respectively. Both companies have a net profit margin of 18.22%. How much revenue did each company earn?
Step 1:Write out formula
Net Profit Margin = Net Profit/Revenue
Revenue = Net Profit/Net Profit Margin
Step 2:Calculate revenue for each company
Company A:
Revenue = $83.50/18.22% = $458.29
Company B:
Revenue = $67.22/18.22% = $368.94
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Video Explanation of Net Profit Margin
Below is a video explanation from CFI’s Financial Analysis Fundamentals Course of how net profit margin is calculated and what it means when analyzing a company’s performance.
Get more video tutorials in CFI’s Financial Analyst Training Program.
Understanding the Ratio
The net profit margin ratio is used to describe a company’s ability to produce profit and to consider several scenarios, such as an increase in expenses which is deemed ineffective. It is used extensively in financial modeling and company valuation.
Net profit margin is a strong indicator of a firm’s overall success and is usually stated as a percentage. However, keep in mind that a single number in a company report is rarely adequate to point out overall company performance. An increase in revenue might translate to a loss if followed by an increase in expenses. On the other hand, a decrease in revenue, followed by tight control over expenses, might put the company further in profit.
Other common financial metrics are EBITDA and Gross Profit.
A high net profit margin means that a company is able to effectively control its costs and/or provide goods or services at a price significantly higher than its costs. Therefore, a high ratio can result from:
Efficient management
Low costs (expenses)
Strong pricing strategies
A low net profit margin means that a company uses an ineffective cost structure and/or poor pricing strategies. Therefore, a low ratio can result from:
Inefficient management
High costs (expenses)
Weak pricing strategies
Investors need to take numbers from the profit margin ratio as an overall indicator of company profitability performance and initiate deeper research into the cause of an increase or decrease in the profitability as needed.
Limitations of Net Profit Margin Ratio
When calculating the net profit margin ratio, analysts commonly compare the figure to different companies to determine which business performs the best.
While this is common practice, the net profit margin ratio can greatly differ between companies in different industries. For example, companies in the automotive industry may report a high profit margin ratio but lower revenue as compared to a company in the food industry. A company in the food industry may show a lower profit margin ratio, but higher revenue.
It is recommended to compare only companies in the same sector with similar business models.
Other limitations include the possibility of misinterpreting the profit margin ratio and cash flow figures.A low net profit margin does not always indicate a poorly performing company. Also, a high net profit margin does not necessarily translate to high cash flows.
Limitations Example #1 – Comparing Companies
A jewelry company that sells a few expensive products may have a much higher profit margin as compared to a grocery store that sells many cheap products.
It’d be inappropriate to compare the margins for these two companies, as their operations are completely different.
Limitations Example #2 – Companies with Debt
If a company has higher financial leverage than another, then the firm with more debt financing may have a smaller net profit margin due to the higher interest expenses. This negatively affects net profit, lowering the net profit margin for the company.
Limitations Example #3 – Depreciation Expense
Companies with high property plant & equipment (PP&E) assets will be affected by higher depreciation expenses, lowering the firm’s net profit margin. This may be misleading because the company could have significant cash flow but may seem inferior due to their lower profit margin.
Limitations Example #4 – Manipulation of Profit
Management may reduce long-term expenses (such as research and development) to increase their profit in the short-term. This can mislead investors looking at net margin, as a company can boost their margin temporarily.
Financial Analysis
Calculating the net margin of a business is a routine part of financial analysis. It is part of a type of analysis known as vertical analysis, which takes every line item on the income statement and divides it into revenue. To compare the margin for a company on a year-over-year (YoY) basis, a horizontal analysis is performed. To learn more, read CFI’s free guide to analyzing financial statements.
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187521 | https://www.math.cmu.edu/~mlavrov/arml/15-16/number-theory-10-04-15-solutions.pdf | Number Theory Misha Lavrov Diophantine equations Western PA ARML Practice October 4, 2015 2.1 Warm-up 1. (ARML 1993) There are several values for a prime p with the property that any five-digit multiple of p remains a multiple of p if you “rotate the digits”. One such value is 41 (for example, since 50635 is a multiple of 41, so are 55603, 35506, 63550, and 6355); another such value is 3. Compute the value of p that is greater than 41.
Rotating the digits takes a number 10a + b to 104b + a. (For example, 50635 = 5063 · 10 + 5 becomes 55603 = 5 · 104 + 5063.) If ( 10a + b ≡0 (mod p) a + 104b ≡0 (mod p) then a ≡−104b (mod p), so 0 ≡10a + b ≡(−105b) + b = (1 −105)b (mod p), which means p | (1 −105)b.
The only way to guarantee this is to have p | 105 −1 = 99999. Since 99999 = 32 · 41 · 271, the solution we’re looking for is p = 271.
2.2 Exponential Diophantine equations 1. Solve over the integers: (a) 2x −1 = 3y.
The only solutions are 21 −1 = 30 and 22 −1 = 31.
Take the equation modulo 8. We have 32 ≡1 (mod 8), so 3y is either 1 or 3 modulo 8. On the other hand, 2x −1 ≡7 (mod 8), assuming x ≥3, and 7 is neither 1 nor 3.
Therefore x ≤2. Now we try x = 0, x = 1, and x = 2: x = 0 doesn’t work (giving 20 −1 = 0, not a power of 3) but x = 1 and x = 2 both produce solutions.
(b) 7x + 4 = 3y.
There are no solutions.
Take the equation modulo 3. On the left-hand side, we have 7x+4 ≡1x+4 ≡2 (mod 3).
On the right-hand side, we have 3y ≡0 (mod 3), unless y = 0, in which case 3y ≡1 (mod 3), which is still not 2.
(c) 3x + 2 = 5y.
The only solution is 31 + 2 = 51.
Take the equation modulo 9. The powers of 5 modulo 9 are 1, 5, 7, 8, 4, 2, 1, . . . . Assuming x ≥2, the right-hand side is 2 modulo 9, so we must have y ≡5 (mod 6).
Now take the equation modulo 7, chosen because 56 ≡1 (mod 7). This means 5y = 55 · (56)k ≡3 (mod 7), so 3x ≡3 −2 = 1 (mod 7). The powers of 3 modulo 7 are 1, 3, 2, 6, 4, 5, 1, . . . , so we must have x ≡0 (mod 6).
In particular, x is even, so 3x = 729x/6. Since 728 = 23 · 7 · 13, we take the equation modulo 13. On the left-hand side, we get 729x/6 + 2 ≡1x/6 + 2 = 3 (mod 13). On the right-hand side, since 56 ≡−1 (mod 13), we have 5y ≡±55 ≡±5 (mod 13), which is either 5 or 8.
This is a contradiction, so we must have x < 2. Trying x = 0 and x = 1, we find the only solution.
(d) 2x + 1 = 3y.
The only solutions are 21 + 1 = 31 and 23 + 1 = 32.
Assume y ≥2 and take the equation modulo 9. Then we have 2x ≡−1 (mod 9). The powers of 2 modulo 9 are 1, 2, 4, −1, −2, −4, 1, . . . , repeating every 6 steps, so x ≡3 (mod 6). In particular, x is divisible by 3.
Then we have 2x + 1 = (2x/3)3 + 1 = (2x/3 + 1)(22x/3 −2x/3 + 1). This is equal to 3y, so both factors must be powers of 3. In particular, 2x/3 + 1 is a power of 3, so if (x, y) is a solution to the Diophantine equation and y ≥2, there is another solution with x/3 in place of x. We can keep dividing x by 3 until we descend to a solution with y < 2.
When y = 0 there is no solution, and when y = 1 we get the solution (1, 1). Therefore all solutions must descend to the (1, 1) solution. This gives us the x = 3 solution found above, but 29 + 1 is not a power of 3, so we have exhausted all solutions.
(e) 3x + 4y = 5z.
The only solutions are 30 + 41 = 51 and 32 + 42 = 52.
Take the equation modulo 3. We’ll deal with the x = 0 case later; if x > 0, we get 3x + 4y ≡0 + 1y (mod 3) on the right, and 5z ≡(−1)z on the left. This tells us that z is even.
Now we have 3x = 25z/2 −4y = (5z/2 +2y)(5z/2 −2y), so both 5z/2 +2y and 5z/2 −2y are powers of 3. But their sum is 2 · 5z/2, which is not divisible by 3, so one of the powers of 3 (the smaller one) must be 30 = 1, and we are left with the equations ( 5z/2 + 2y = 3x, 5z/2 −2y = 1.
Taking the difference, we get 3x −1 = 2y+1. This is the equation in part (d), so we must have x = y = 2 or y = 0 and x = 1. The first option gives us the (2, 2, 2) solution, and the second option can’t find a value of z.
It remains to consider the x = 0 case, where we get 4y +1 = 5z. The y = 1 solution we’ve already found, so assume y ≥2 and take the equation mod 8. Since 52 ≡1 (mod 8), z must be even, so we have a difference of squares once again: (5z/2 + 2y)(5z/2 −2y) = 1.
But this is impossible to satisfy, since the factors can’t both be 1 or both -1, so there are no further solutions to be found.
2. Find all positive integers x and y such that 2x + 3y is a perfect square.
The only solutions are 20 + 31 = 4, 23 + 30 = 9, and 24 + 32 = 25.
Try y = 0. Then 2x +1 = k2 for some k, so 2x = k2 −1 = (k +1)(k −1). This is only possible when k −1 = 2 and k + 1 = 4, giving us one of the solutions.
Otherwise, y > 0, so we have (−1)x + 0 ≡k2 (mod 3). But k2 can only be 0 or 1 modulo 3, so x must be even. Then we have a difference of squares: 3y = (k + 2x/2)(k −2x/2).
So both k + 2x/2 and k −2x/2 are powers of 3. But their difference is 2x/2+1, which is not divisible by 3. Therefore k −2x/2 = 30 = 1. Solving for k, we get k = 2x/2 + 1, so 2x + 3y = (2x/2 + 1)2 = 2x + 2x/2+1 + 1.
This means that 3y = 2x/2+1 + 1, which has only two solutions, by problem 1(d). We can have x/2 + 1 = y = 1, giving 20 + 31 = 4, or x/2 + 1 = 3 and y = 2, giving us 24 + 32 = 25.
3. (BMO 1981) Find the smallest positive value of |12m −5n|, where m, n are positive integers.
Clearly, |121 −51| = 7 is achievable. Is any smaller value possible? We have 12m −5n ≡ 0m −1n ≡1 (mod 2), 12m −5n ≡0m −(−1)n ̸≡0 (mod 3), and 12m −5n ≡2m −0n ̸≡0 (mod 5), which rules out 2, 3, 4, 5, and 6. So it remains to check if there are any solutions to 12m −5n = ±1.
Taking the equation modulo 4, we get 0m −1n ≡±1 (mod 4), so the 1 must be negative, and we have 5n −12m = 1.
Taking the equation modulo 3, we ge (−1)n −0m ≡1 (mod 3), so n must be even.
Taking the equation modulo 5, we get 0n −2m ≡1 (mod 5), which is possible only for m ≡2 (mod 4). So m must be even as well.
But now we have the difference of squares (5n/2)2 −(12m/2)2 = 1, which factors as (5n/2 − 12m/2)(5n/2 + 12m/2) = 1. So both factors must be 1 or else both -1, which is impossible as 12m/2 > 0.
So an absolute difference of 1 is rulled out, and the smallest achievable value is 7.
2.3 Other Diophantine equations 1. Show that there are no integer solutions to x3 + y3 + z3 = 400.
Take the equation modulo 9. It’s easy to check that all perfect cubes are 0, 1, or -1 modulo 9, so the remainder modulo 9 of x3 + y3 + z3 can be any of {−3, −2, −1, 0, 1, 2, 3}. However, 400 ≡4 (mod 9).
2. (PUMaC 2009) Find all prime numbers p which can be written as p = a4 + b4 + c4 −3 for some primes a, b, and c (not necessarily distinct).
Write the right-hand side as (a4 −1) + (b4 −1) + (c4 −1). We have x4 −1 ≡0 (mod 2) unless x is even, x4 −1 ≡0 (mod 3) unless x ≡0 (mod 3), and x4 −1 ≡0 (mod 5) unless x ≡0 (mod 5).
Therefore a4 + b4 + c4 −3 ≡0 (mod 2) unless a, b, or c is 2; it is divisible by 3 unless a, b, or c is 3; and it is divisible by 5 unless a, b, or c is 5. We can check that p = 2, p = 3, and p = 5 are too small to be a solution, so the only possibility is p = 24 + 34 + 54 −3 = 719, which is indeed prime.
3. (USAMO 1979) Determine all non-negative integer solutions, apart from permutations, of the equation n4 1 + n4 2 + · · · + n4 14 = 1599.
Modulo 16, any perfect fourth power is either 0 or 1, so the sum on the right-hand side can be anything from 0 to 14 modulo 16. But 1599 = 1600 −1, so it is 15 modulo 16, and there are no solutions.
4. Find all integer solutions to x2 + 2x = y2.
The only solutions are 02 + 20 = 12 and 62 + 26 = 102.
We have 2x = y2 −x2 = (y + x)(y −x), so both y + x and y −x are powers of 2. Write y + x = 2i and y −x = 2j; we have x = 2i−1 −2j−1. The equation 2x = y2 −x2 becomes 22i−1−2j−1 = 2i+j so 2i−1 −2j−1 = i + j.
Since i > j, we have 2i−1 −2j−1 ≥2i−2, while i + j < 2i. Thus, 2i−2 < 2i, which means i > 2i−3. This is true only for i ≤5: the right-hand side grows much faster than the left.
Checking all values 0 ≤j ≤i ≤5, we only find the two solutions above.
5. Show that for any integers x, y ≥2, 222...2 | {z } x −333...3 | {z } y ≥11.
Almost any modulus will work. Modulo 100, the power tower of 2’s will eventually stabilize at 36, and the power tower of 3’s at 87, giving a lower bound of 49. It then suffices to check that no small values of x and y do better than 33 −222 = 11. |
187522 | https://en.wikipedia.org/wiki/OR_gate | Jump to content
OR gate
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From Wikipedia, the free encyclopedia
Digital logic gate type
This article is about OR in the sense of an electronic
logic gate (e.g. CMOS 4071). For OR in the purely logical sense, see Logical disjunction.
| | |
--- |
| | This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "OR gate" – news · newspapers · books · scholar · JSTOR (September 2012) (Learn how and when to remove this message) |
The OR gate is a digital logic gate that implements logical disjunction. The OR gate outputs "true" if any of its inputs is "true"; otherwise it outputs "false". The input and output states are normally represented by different voltage levels.
Description
[edit]
| OR gate truth table | | |
---
| Input | | Output |
| A | B | A OR B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Any OR gate can be constructed with two or more inputs. It outputs a 1 if any of these inputs are 1, or outputs a 0 only if all inputs are 0. The inputs and outputs are binary digits ("bits") which have two possible logical states. In addition to 1 and 0, these states may be called true and false, high and low, active and inactive, or other such pairs of symbols.
Thus it performs a logical disjunction (∨) from mathematical logic. The gate can be represented with the plus sign (+) because it can be used for logical addition. Equivalently, an OR gate finds the maximum between two binary digits, just as the AND gate finds the minimum.
Together with the AND gate and the NOT gate, the OR gate is one of three basic logic gates from which any Boolean circuit may be constructed. All other logic gates may be made from these three gates; any function in binary mathematics may be implemented with them.
It is sometimes called the inclusive OR gate to distinguish it from XOR, the exclusive OR gate. The behavior of OR is the same as XOR except in the case of a 1 for both inputs. In situations where this never arises (for example, in a full-adder) the two types of gates are interchangeable. This substitution is convenient when a circuit is being implemented using simple integrated circuit chips which contain only one gate type per chip.
Symbols
[edit]
There are two logic gate symbols currently representing the OR gate: the American (ANSI or 'military') symbol and the IEC ('European' or 'rectangular') symbol. The DIN symbol is deprecated.
The "≥1" on the IEC symbol indicates that the output is activated by at least one active input.
MIL/ANSI Symbol
IEC Symbol
DIN Symbol
As of Unicode 16.0.0, the OR gate is also encoded in the Symbols for Legacy Computing Supplement block as U+1CC15 LOGIC GATE OR.
Hardware description and pinout
[edit]
OR gates are basic logic gates, and are available in TTL and CMOS ICs logic families. The standard 4000 series CMOS IC is the 4071, which includes four independent two-input OR gates. The TTL device is the 7432. There are many offshoots of the original 7432 OR gate, all having the same pinout but different internal architecture, allowing them to operate in different voltage ranges and/or at higher speeds. In addition to the standard 2-input OR gate, 3- and 4-input OR gates are also available. In the CMOS series, these are:
4075: triple 3-input OR gate
4072: dual 4-input OR gate
Variations include:
74LS32: quad 2-input OR gate (low power Schottky version)
74HC32: quad 2-input OR gate (high speed CMOS version) - has lower current consumption/wider voltage range
74AC32: quad 2-input OR gate (advanced CMOS version) - similar to 74HC32, but with significantly faster switching speeds and stronger drive
74LVC32: low voltage CMOS version of the same.
Implementations
[edit]
CMOS OR gate (NOT gate is visible on the right)
NMOS OR gate
BJT OR gate
OR gate using diodes
Analytical representation
[edit]
is the analytical representation of OR gate:
OR gates with many inputs
[edit]
OR gates with multiple inputs are designated with the same symbol, with more lines leading in. While direct implementations with more than three inputs are possible in logic families like CMOS, these are inefficient. More efficient implementations use a cascade of NOR and NAND gates, as shown in the picture below.
12-input OR gate realized via a cascade of NOR and NAND gates.
Alternatives
[edit]
Further information: NAND logic and NOR logic
If no specific OR gates are available, one can be made from NAND or NOR gates in the configuration shown in the image below. Any logic gate can be made from a combination of NAND or NOR gates.
| Desired gate | NAND construction | NOR construction |
---
| | | |
Wired-OR
[edit]
With active low open collector logic outputs, as used for control signals in many circuits, an OR function can be produced by wiring together several outputs. This arrangement is called a wired OR. This implementation of an OR function typically is also found in integrated circuits of N or P-type only transistor processes.
See also
[edit]
Wikimedia Commons has media related to OR gates.
AND gate
NOT gate
NAND gate
NOR gate
XOR gate
XNOR gate
Boolean algebra
Logic gate
References
[edit]
^ "Logic OR Gate Tutorial". Electronics Tutorials. 20 August 2013.
^ "OR Gate". Hyperphysics.phy-astr.gsu.edu. Retrieved 2012-09-24.
^ Broesch, James D. (2012). Practical Programmable Circuits: A Guide to PLDs, State Machines, and Microcontrollers. Elsevier Science. p. 19. ISBN 978-0323139267.
^ U.S. Department of Defense (February 26, 1962). Graphical Symbols for Logic Diagrams (Report). MIL-STD-806.
^ Harris, David Harris, Sarah (2007). Digital design and computer architecture (1st ed.). San Francisco, Calif.: Morgan Kaufmann. p. 21. ISBN 9780123704979.{{cite book}}: CS1 maint: multiple names: authors list (link)
^ Brumbach, Michael E. (January 2010). Industrial electricity (8th ed.). Clifton Park, N.Y.: Delmar. p. 546. ISBN 9781435483743.
^ Semiconductor Group. Overview of IEEE Standard 91-1984: Explanation of Logic Symbols (PDF) (Report). Texas Instruments. p. 4. SDYZ001A.
^ "Multiple-input Gates". All About Circuits. Retrieved 2024-02-04.
| Common logical connectives | | |
---
| Tautology/True | | |
| Alternative denial (NAND gate) Converse implication Implication (IMPLY gate) Disjunction (OR gate) | |
| Negation (NOT gate) Exclusive or (XOR gate) Biconditional (XNOR gate) Statement (Digital buffer) | |
| Joint denial (NOR gate) Nonimplication (NIMPLY gate) Converse nonimplication Conjunction (AND gate) | |
| Contradiction/False | |
| Philosophy portal | | |
Retrieved from "
Categories:
Logic gates
Boolean algebra
Digital electronics
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| 目录[隐藏] 1 什么是分层最佳抽样 2 分层最佳抽样的公式 3 分层最佳抽样举例 |
[编辑]
什么是分层最佳抽样
| 市场调查方法 |
| | | | A | | 案头调研 | | 案例研究法 | | B | | 不重复抽样 | | C | | 抽样调查 | | 重置抽样 | | 抽签法 | | 产品留置测试 | | D | | 多维尺度法 | | 定量研究方法 | | 定性研究方法 | | 典型调查法 | | 电话调查 | | 多阶段抽样 | | 等距抽样 | | 独立控制配额抽样 | | 等距量表 | | 等比量表 | | E | | 二手资料调研 | | 二路焦点小组 | | F | | 非概率抽样 | | 分层抽样 | | 分层比例抽样 | | 分层最佳抽样 | | G | | 观察法 | | 概率抽样 | | 拐点调研 | | 滚雪球抽样 | | H | | 会议调查 | | J | | 焦点访谈法 | | 经验判断法 | | 随机抽样 | | 家庭日记法 | | 经销商访谈 | | K | | 可行性研究 | | 控制实验法 | | L | | 联合分析法 | | 留置调查 | | 垃圾调研法 | | 类别量表 | | M | | 面谈访问法 | | 盲测 | | 描述性调研 | | 媒介调查法 | | P | | PPS | | 判断抽样 | | 配额抽样 | | 平衡量表法 | | 评价量表 | | 配对比较量表 | | Q | | Q分类法 | | R | | 任意抽样 | | S | | 容量测定法 | | SEM模型 | | 深层访谈法 | | 双重抽样 | | 实验调查法 | | 实地调研 | | 数值分配量表 | | 随机号码表法 | | 顺序量表 | | T | | 投影技法 | | 推销估计法 | | 投射研究 | | 探索性调研 | | W | | 文献调查法 | | 问卷调查法 | | 网络调研 | | 文案调查法 | | 无准备访问 | | 网上调查 | | X | | 询问法 | | 辛迪加调研 | | 行踪分析 | | 相互控制配额抽样 | | Y | | 邮寄调查 | | 因果性调研 | | Z | | 主观概率法 | | 整群抽样 | | 重点调查 | | 逐户寻找法 | |
| [编辑] | | |
分层最佳抽样又称“非比例抽样”，是根据各层基本单位标准差的大小，来确定各层样本数目的抽样方法。
在各层基本单位之间的差异过分悬殊、某些层的重要性大于其他层的情况下，采取非比例抽样时，在这些层抽取的样本数就多；反之，抽取的样本数就少。如果采取同时兼顾层的大小和层内差异程度的大小来抽样，则有利于提高综合样本对总体全貌的代表性，并可以提高样本的可信程度。
[编辑]
分层最佳抽样的公式
采用分层最佳抽样法，确定各样本数目的计算公式如下：
式中：
ni：第i层应抽出的样本数目；
n：样本总数目；
Ni：第i层的调查单位数；
Si：第i层调查单位的样本标准差。
[编辑]
分层最佳抽样举例
某地有居民20000户，其中高、中、低收入户分别为4000户、12000户、4000户。又已知高收入户的标准差为300元，中收入户的标准差为200元， 低收入户的标准差为100元。现要抽选200户做样本，进行购买力的调查，用分层最佳抽样法分配各层的样本数目。
本题中，已知各层居民收入标准差，即：高收入层（n1）=300、中收入层（n2）=200、低收入层（n3）=100。为了便于计算，见列表：
| | | | |
--- --- |
| 各层次（不同经济收入） | 各层的调查单位数（户）Ni | 各层的样本标准差（元）Si | 乘积NiSi |
| 高 中 低 | 4000 12000 4000 | 300 200 100 | 1200000 2400000 400000 |
| | 20000 | | 4000000 |
按公式计算，各层的样本数目为：
高收入层样本数目：（户）
中收入层样本数目：（户）
低收入层样本数目：（户）
应用分层最佳抽样方法计算出的各层样本抽取数同分层比例抽样法抽出的样本数相比较，可以看出，因各层标准差大小不同，家庭收入高的分层样本增加了20个（从40个变为60个），家庭收入中等的分层样本数，仍然为120个，而家庭收入低的分层样本数减少了20个（从40个变为20个）。高收入户和低收入户在调查总体中单位数都是4000户，为什么从高收入户中产生样本数目是60户，从低收入户中产生样本数目只有20户。这是因为，高收入户的标准差大（300元），从中抽取样本数目就要多一些。低收入户的标准差小（100元），从中抽取的样本数可以少一些。这样抽选到的综合样本比原先仅考虑分层比例抽样得的综合样本更具有对调查总体的代表性，其抽样调查推断的总体结果准确性程度会有所提。
从理论上说，各层中的标准差估计值，反映的是各层的单位特征值和各层平均值之间的差异。如果某层中各单位特征值比较接近，差异较小，那么从理论上说，标准差就小。因此，少抽取一些数目的样本，仍然可以代表、反映该层的大致情况。如果某层内各单位差异较大，那么标准差就较大，因而要适当多选一些样本才更合理。
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187524 | https://askfilo.com/user-question-answers-chemistry/how-many-sigma-and-pi-bonds-are-in-35373937353335 | Question asked by Filo student
How many sigma and pi bonds are in CO2?
Views: 5,643 students
Updated on: Oct 22, 2023
Text SolutionText solutionverified iconVerified
To solve the question, we need to identify the number of sigma (σ) and pi (π) bonds in the compound CO2.
Step 1. Recall the definition of a sigma bond and a pi bond.
A sigma bond (σ) is a type of covalent bond where the electron density is concentrated along the axis connecting the nuclei of the bonded atoms. A pi bond (π) is a covalent bond formed by the side-by-side overlap of two p-orbitals.
Step 2. Count the number of sigma bonds in CO2.
We find that CO2 consists of two double bonds between the carbon and two oxygen atoms. Each double bond consists of one sigma bond and one pi bond. Therefore, there are 2 sigma bonds in CO2.
Step 3. Count the number of pi bonds in CO2.
Since there are two double bonds in CO2, there are 2 pi bonds.
Therefore, the number of sigma and pi bonds in CO2 are 2 and 2, respectively.
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| Question Text | How many sigma and pi bonds are in CO2? |
| Updated On | Oct 22, 2023 |
| Topic | All topics |
| Subject | Chemistry |
| Class | High School |
| Answer Type | Text solution:1 |
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187525 | https://www.sciencedirect.com/science/article/pii/S2950236524000173 | Skip to main contentSkip to article
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2. Material & Methods
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187526 | https://www.3erp.com/blog/brass-machining-grades/ | By Ronan Ye in Company News | Apr 13, 2023
Brass Grades Suitable for CNC Machining: 18 Options to Choose From
Ronan Ye
Rapid Prototyping & Rapid Manufacturing Expert
Specialize in CNC machining, 3D printing, urethane casting, rapid tooling, injection molding, metal casting, sheet metal and extrusion
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Contents
1. Defining CNC Machining and Brass Grades2. What are The Different Types of Machinable Brass?3. Selecting the Right Brass Grade for Your Project4. The Versatility of Brass in CNC Machining
CNC machining has revolutionized the world of manufacturing, offering unparalleled precision and efficiency. One material that has stood the test of time in this field is brass, an alloy of copper and zinc.
With its excellent machinability and unique properties, it’s no wonder that brass is a popular choice for numerous applications. In this comprehensive guide, we will explore the various brass grades suitable for CNC machining and dive into their unique characteristics and applications.
Defining CNC Machining and Brass Grades
Brass machining is a manufacturing process that involves the use of computer numerical control (CNC) machines to cut, shape, and manipulate brass, which is a metal alloy primarily composed of copper and zinc.
In brass CNC machining, a design is first created using computer-aided design (CAD) software. The design is then converted into numerical code or G-code, which the CNC machine can understand. The CNC machine interprets this code to perform the necessary operations, such as drilling, milling, turning, or cutting, to create the desired brass component.
What are The Different Types of Machinable Brass?
Various brass grades have been developed to cater to different applications and requirements in CNC machining. These grades are identified by their copper and zinc content, as well as other alloying elements present.
C200 – C20000 Brass Grades
The C200 series, also known as the alpha brasses, contain a high copper content ranging from 95% to 99%. These alloys are known for their excellent corrosion resistance, making them a popular choice for plumbing fixtures and fittings, as well as electrical and electronic components.
Advantages:
Corrosion Resistance: C200 brass grades exhibit exceptional corrosion resistance due to their high copper content. This characteristic makes them ideal for use in environments where they are exposed to moisture, chemicals, or other corrosive substances, ensuring durability and a long service life.
Electrical Conductivity: The high copper content in C200 brass grades also results in excellent electrical conductivity. As a result, these alloys are often used in electrical and electronic applications, such as connectors, terminals, and wiring components.
Ductility and Formability: C200 brass grades possess good ductility and formability, allowing them to be easily shaped and formed into various components. This property is particularly beneficial for applications that require intricate designs and complex geometries.
Common Applications:
Plumbing fittings and fixtures: C200 brass grades are often used in the production of faucets, valves, and other plumbing components due to their corrosion resistance and durability.
Electrical and electronic components: The excellent electrical conductivity of C200 brass grades makes them suitable for use in connectors, terminals, and other electrical components.
Radiator cores: C200 brass grades are commonly used in the manufacturing of radiator cores for automotive and industrial applications, owing to their good heat transfer properties and corrosion resistance.
Coins and medals: The attractive appearance and resistance to wear make C200 brass grades an ideal choice for minting coins and crafting medals.
Decorative and ornamental applications, including plaques, medals, and sculptures
Architectural applications, such as trim, moldings, cornices, window frames and hardware
Fire sprinkler systems, Heat exchanger tubes and components
Main Disadvantages:
Machinability: The high copper content in C200 brass grades can make them more difficult to machine compared to other brass alloys with lower copper content. This may limit their use in certain applications where intricate shapes and complex geometries are required.
Cost: Due to the high copper content, C200 brass grades can be more expensive than other brass alloys with lower copper content
C210
C210 Brass Grade, also known as Gilding Metal or C21000 Brass, is a copper-zinc alloy known for its deep red color, ductility, and ease of cold working. This alloy consists primarily of copper (95%) and zinc (5%), which makes it an ideal choice for applications that require excellent formability and a rich, reddish hue.
Chemical Properties:
Copper (Cu): 95%
Zinc (Zn): 5%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.90 g/cm³
Melting Point: 925-960°C (1700-1760°F)
Tensile Strength: 280-455 MPa (40,600-66,000 psi)
Yield Strength: 70-340 MPa (10,150-49,300 psi)
Elongation: 30-60%
Hardness (Brinell): 60-130 BHN
Electrical Conductivity: 37% IACS (International Annealed Copper Standard)
C220
C220 Brass Grade, also known as Commercial Bronze or C22000 Brass, is a copper-zinc alloy characterized by its excellent ductility, strength, and corrosion resistance. This alloy is primarily composed of copper (90%) and zinc (10%), making it an ideal choice for various applications requiring a combination of formability and moderate strength.
Chemical Properties:
Copper (Cu): 90%
Zinc (Zn): 10%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.80 g/cm³
Melting Point: 920-950°C (1690-1740°F)
Tensile Strength: 315-520 MPa (45,700-75,400 psi)
Yield Strength: 100-345 MPa (14,500-50,000 psi)
Elongation: 25-55%
Hardness (Brinell): 70-150 BHN
Electrical Conductivity: 34% IACS (International Annealed Copper Standard)
C230
Brass Grade, also known as Red Brass or C23000 Brass, is a copper-zinc alloy known for its high corrosion resistance and excellent formability. This alloy primarily consists of copper (85%) and zinc (15%), which provides a unique combination of strength, ductility, and resistance to various types of corrosion.
Chemical Properties:
Copper (Cu): 85%
Zinc (Zn): 15%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.71 g/cm³
Melting Point: 915-940°C (1680-1725°F)
Tensile Strength: 300-550 MPa (43,500-79,800 psi)
Yield Strength: 70-310 MPa (10,150-45,000 psi)
Elongation: 20-60%
Hardness (Brinell): 60-140 BHN
Electrical Conductivity: 28% IACS (International Annealed Copper Standard)
C260
C260 Brass Grade, also known as Cartridge Brass or C26000 Brass, is a copper-zinc alloy known for its excellent ductility, strength, and corrosion resistance. This alloy primarily consists of copper (70%) and zinc (30%), offering a balanced combination of formability, strength, and a golden yellow color.
Chemical Properties:
Copper (Cu): 70%
Zinc (Zn): 30%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.53 g/cm³
Melting Point: 930-965°C (1710-1770°F)
Tensile Strength: 310-620 MPa (45,000-90,000 psi)
Yield Strength: 90-370 MPa (13,000-53,700 psi)
Elongation: 15-45%
Hardness (Brinell): 80-160 BHN
Electrical Conductivity: 28% IACS (International Annealed Copper Standard)
C272
C272 Brass Grade, also known as Yellow Brass or C27200 Brass, is a copper-zinc alloy recognized for its good ductility, strength, and corrosion resistance. This alloy mainly consists of copper (63%) and zinc (37%), which provides a desirable combination of formability, strength, and a yellowish-gold color.
Chemical Properties:
Copper (Cu): 63%
Zinc (Zn): 37%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.48 g/cm³
Melting Point: 905-935°C (1660-1715°F)
Tensile Strength: 315-585 MPa (45,700-84,800 psi)
Yield Strength: 100-360 MPa (14,500-52,200 psi)
Elongation: 15-45%
Hardness (Brinell): 90-165 BHN
Electrical Conductivity: 28% IACS (International Annealed Copper Standard)
C274
C274 Brass Grade, also known as Yellow Brass or C27400 Brass, is a copper-zinc alloy appreciated for its excellent ductility, moderate strength, and corrosion resistance. This alloy primarily consists of copper (62%) and zinc (38%), offering a balanced combination of formability, strength, and a golden-yellow color.
Chemical Properties:
Copper (Cu): 62%
Zinc (Zn): 38%
Other trace elements: Lead (Pb), Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.47 g/cm³
Melting Point: 900-940°C (1650-1725°F)
Tensile Strength: 310-575 MPa (45,000-83,400 psi)
Yield Strength: 90-355 MPa (13,050-51,500 psi)
Elongation: 20-45%
Hardness (Brinell): 85-160 BHN
Electrical Conductivity: 27% IACS (International Annealed Copper Standard)
C300 – C30000 Brass Grades
The C300 to C30000 brass grades, also known as intermediate brasses or alpha-beta brasses, are characterized by a balance of copper and zinc content, providing them with a unique set of attributes that cater to various applications across different industries.
Advantages:
Strength and Ductility: C300 to C30000 brass grades offer a balance between strength and ductility, making them suitable for applications that require both robustness and the ability to be easily shaped and formed. These alloys provide a compromise between the strength of beta brasses and the ductility of alpha brasses.
Corrosion Resistance: Similar to alpha and beta brasses, C300 brass grades exhibit good corrosion resistance due to their copper content. This characteristic makes them suitable for use in environments where they are exposed to moisture, chemicals, or other corrosive substances, ensuring durability and a long service life.
Machinability: Intermediate brasses generally offer good machinability. This property enables them to be easily shaped, cut, and formed into various components, making them a popular choice for machining applications.
Common Applications:
Fasteners: screws, nuts, bolts, and washers
Plumbing fittings: valves, faucets, and pipe fittings
Automotive components: radiators, fuel tanks, and transmission parts
Electrical connectors and components: terminals, switches, and plugs
Marine hardware: propellers, boat fittings, and pump components
Ammunitions: casings and cartridge components
Gears and bearings: for machinery and equipment
Architectural elements: decorative hardware, door handles, and window fittings
Musical instruments: bells, cymbals, and brass instrument components
Locks and padlocks: for security and durability
Industrial machinery components: bushings, gears, and wear plates
Consumer goods: zippers, buttons, and belt buckles
Art and sculpture: castings and decorative elements
Molding and extrusion: for complex shapes and profiles
Coinage: due to its corrosion resistance and durability
Disadvantages
Cost: The increased zinc content in C300 to C30000 brass grades may lead to higher production costs compared to alpha brasses with lower zinc content.
Not as strong as beta brasses: While C300 grades provide a balance between strength and ductility, they may not be as strong as beta brasses, which could be a disadvantage in some applications where higher strength is required.
C360
C360 Brass Grade, also known as Free-Machining Brass or Free-Cutting Brass, is a high-grade copper-zinc alloy with excellent machinability. This alloy is composed primarily of copper (61%), zinc (35%), and lead (3%). The addition of lead results in improved machinability, which makes it ideal for producing intricate and complex parts through machining processes.
Chemical Properties:
Copper (Cu): 61%
Zinc (Zn): 35%
Lead (Pb): 3%
Other trace elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.47 g/cm³
Melting Point: 930-935°C (1710-1715°F)
Tensile Strength: 370-470 MPa (53,700-68,100 psi)
Yield Strength: 210-320 MPa (30,500-46,400 psi)
Elongation: 12-25%
Hardness (Brinell): 100-160 BHN
Electrical Conductivity: 26% IACS (International Annealed Copper Standard)
C314
C314 Brass Grade, commonly referred to as Leaded Commercial Bronze or C31400 Brass, is an alloy of copper, zinc, and lead. Renowned for its remarkable machinability, this alloy exhibits a blend of copper (89%), zinc (9%), and lead (2%), making it suitable for a range of applications that require easy machining, fair strength, and corrosion resistance.
Chemical Properties:
Copper (Cu): 89%
Zinc (Zn): 9%
Lead (Pb): 2%
Minor traces of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.77 g/cm³
Melting Point Range: 910-945°C (1670-1735°F)
Tensile Strength: 295-515 MPa (42,800-74,700 psi)
Yield Strength: 80-325 MPa (11,600-47,100 psi)
Elongation: 20-60%
Hardness (Brinell): 65-140 BHN
Electrical Conductivity: 33% IACS (International Annealed Copper Standard)
C330
C330 Brass Grade, often referred to as Low-Leaded Brass or C33000 Brass, is an alloy composed of copper, zinc, and a small amount of lead. This alloy is valued for its good machinability, ductility, and moderate strength. The primary constituents of C330 are copper (65%), zinc (34%), and lead (1%), which make it suitable for various applications that demand easy machining and fair strength.
Chemical Properties:
Copper (Cu): 65%
Zinc (Zn): 34%
Lead (Pb): 1%
Minute traces of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.49 g/cm³
Melting Point Range: 900-950°C (1650-1740°F)
Tensile Strength: 300-570 MPa (43,500-82,700 psi)
Yield Strength: 90-350 MPa (13,050-50,750 psi)
Elongation: 10-50%
Hardness (Brinell): 75-160 BHN
Electrical Conductivity: 27% IACS (International Annealed Copper Standard)
C335
C335 Brass Grade, commonly known as Free-Machining Brass or C33500 Brass, is an alloy consisting of copper, zinc, and a small quantity of lead. This alloy is notable for its excellent machinability, fair ductility, and moderate strength. The primary components of C335 are copper (62.5%), zinc (36.5%), and lead (1%), making it a suitable material for a range of applications that require easy machining and reasonable strength.
Chemical Properties:
Copper (Cu): 62.5%
Zinc (Zn): 36.5%
Lead (Pb): 1%
Trace amounts of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.45 g/cm³
Melting Point Range: 895-955°C (1640-1750°F)
Tensile Strength: 290-560 MPa (42,000-81,200 psi)
Yield Strength: 85-345 MPa (12,300-50,000 psi)
Elongation: 12-45%
Hardness (Brinell): 70-155 BHN
Electrical Conductivity: 26% IACS (International Annealed Copper Standard)
C353
C353 Brass Grade, often referred to as High-Leaded Brass or C35300 Brass, is an alloy composed of copper, zinc, and a significant amount of lead. This alloy is valued for its exceptional machinability, adequate ductility, and fair strength. The primary constituents of C353 are copper (62%), zinc (35.5%), and lead (2.5%), making it an appropriate material for various applications requiring efficient machining and reasonable strength.
Chemical Properties:
Copper (Cu): 62%
Zinc (Zn): 35.5%
Lead (Pb): 2.5%
Minute quantities of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.44 g/cm³
Melting Point Range: 890-950°C (1635-1740°F)
Tensile Strength: 285-550 MPa (41,300-79,800 psi)
Yield Strength: 80-340 MPa (11,600-49,300 psi)
Elongation: 10-40%
Hardness (Brinell): 65-150 BHN
Electrical Conductivity: 25% IACS (International Annealed Copper Standard)
C365
C365 Brass Grade, also known as Leaded Muntz Metal or C36500 Brass, is an alloy made up of copper, zinc, and a considerable amount of lead. This alloy is recognized for its excellent machinability, reasonable ductility, and satisfactory strength. The primary components of C365 are copper (59%), zinc (39%), and lead (2%), making it a suitable choice for a range of applications that require efficient machining and acceptable strength.
Chemical Properties
Copper (Cu): 59%
Zinc (Zn): 39%
Lead (Pb): 2%
Trace elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties
Density: 8.41 g/cm³
Melting Point Range: 885-955°C (1625-1750°F)
Tensile Strength: 280-540 MPa (40,600-78,300 psi)
Yield Strength: 75-335 MPa (10,900-48,600 psi)
Elongation: 8-35%
Hardness (Brinell): 60-145 BHN
Electrical Conductivity: 24% IACS (International Annealed Copper Standard)
C377
C377 Brass Grade, often referred to as Forging Brass or C37700 Brass, is an alloy composed of copper, zinc, and a trace amount of lead. This alloy is known for its excellent hot forging capabilities, good ductility, and moderate strength. The primary components of C377 are copper (58-61%), zinc (37-40%), and lead (1-2.5%), making it an ideal material for a variety of applications that require ease of hot forging and decent strength.
Chemical Properties:
Copper (Cu): 58-61%
Zinc (Zn): 37-40%
Lead (Pb): 1-2.5%
Trace amounts of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.44 g/cm³
Melting Point Range: 890-935°C (1635-1715°F)
Tensile Strength: 280-450 MPa (40,600-65,250 psi)
Yield Strength: 90-260 MPa (13,050-37,700 psi)
Elongation: 20-40%
Hardness (Brinell): 65-110 BHN
Electrical Conductivity: 27% IACS (International Annealed Copper Standard)
C385
C385 Brass Grade, commonly known as Architectural Brass or C38500 Brass, is an alloy consisting of copper, zinc, and a significant amount of lead. This alloy is valued for its outstanding machinability, good ductility, and moderate strength. The primary constituents of C385 are copper (57%), zinc (40%), and lead (3%), making it an appropriate material for a variety of applications that demand efficient machining and fair strength.
Chemical Properties:
Copper (Cu): 57%
Zinc (Zn): 40%
Lead (Pb): 3%
Trace amounts of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.40 g/cm³
Melting Point Range: 880-955°C (1615-1750°F)
Tensile Strength: 275-530 MPa (39,900-76,900 psi)
Yield Strength: 70-330 MPa (10,150-47,900 psi)
Elongation: 6-30%
Hardness (Brinell): 55-140 BHN
Electrical Conductivity: 23% IACS (International Annealed Copper Standard)
C400 – C40000 Brass Grades
The C400 to C40000 brass grades, also known as beta brasses, are characterized by a higher zinc content compared to alpha brasses, which bestows them with distinct attributes that cater to specific applications across various industries.
Advantages
Strength and Hardness: C400 to C40000 brass grades are known for their impressive strength and hardness, making them suitable for applications requiring robust materials capable of withstanding significant wear and tear. These alloys are often used in load-bearing components and parts that experience high levels of friction.
Corrosion Resistance: Similar to alpha brasses, C400 to C40000 brass grades exhibit good corrosion resistance due to their copper content. However, the higher zinc content can result in reduced corrosion resistance compared to their alpha counterparts. Nonetheless, these alloys are still suitable for many applications where corrosion resistance is required.
Machinability: Beta brasses generally offer good machinability. This property enables them to be easily shaped, cut, and formed into various components, making them a popular choice for machining applications.
Common Applications:
Marine equipment: underwater fastenings, hull components, and portholes
Pump and valve components: impellers, housings, and shafts
Condenser tubes: for heat exchangers in power plants and industrial processes
Turned parts: precision components for machines and instruments
Telecommunication equipment: connectors, contact pins, and sockets
Electrical relay parts: offering low electrical resistance and good conductivity
Ordnance components: specialized parts for military and defense applications
Jewelry and accessories: cufflinks, bracelets, and brooches
Antifriction applications: wear-resistant components in machines and equipment
Boiler components: tubes, fittings, and heat exchanger elements
Pinions and worm drives: for power transmission in machinery
Sprinkler systems: fittings, nozzles, and valves for fire protection
Water meters and fittings: for accurate measurement and control of water flow
Reproduction of antique hardware: authentic-looking parts for restoration projects
Marine-grade electrical components: switches, receptacles, and wiring devices resistant to saltwater environments
C443
C443 Brass Grade, often referred to as Admiralty Brass or C44300 Brass, is an alloy composed of copper, zinc, and a small amount of arsenic. This alloy is known for its excellent corrosion resistance, good ductility, and moderate strength. The primary components of C443 are copper (70%), zinc (29%), and arsenic (0.02-0.06%), making it an appropriate material for various applications that require resistance to corrosion and a fair level of strength.
Chemical Properties:
Copper (Cu): 70%
Zinc (Zn): 29%
Arsenic (As): 0.02-0.06%
Trace amounts of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.53 g/cm³
Melting Point Range: 900-945°C (1650-1735°F)
Tensile Strength: 310-480 MPa (45,000-69,600 psi)
Yield Strength: 105-290 MPa (15,200-42,000 psi)
Elongation: 25-55%
Hardness (Brinell): 80-120 BHN
Electrical Conductivity: 28% IACS (International Annealed Copper Standard)
C464
C464 Brass Grade, commonly known as Naval Brass or C46400 Brass, is an alloy consisting of copper, zinc, and a trace amount of tin. This alloy is appreciated for its exceptional corrosion resistance, particularly in seawater environments, as well as its good ductility and moderate strength. The primary constituents of C464 are copper (60%), zinc (39.25%), and tin (0.75%), making it an ideal material for various applications that necessitate corrosion resistance and decent strength.
Chemical Properties:
Copper (Cu): 60%
Zinc (Zn): 39.25%
Tin (Sn): 0.75%
Trace elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.41 g/cm³
Melting Point Range: 905-940°C (1660-1725°F)
Tensile Strength: 300-450 MPa (43,500-65,250 psi)
Yield Strength: 100-280 MPa (14,500-40,600 psi)
Elongation: 20-45%
Hardness (Brinell): 75-115 BHN
Electrical Conductivity: 26% IACS (International Annealed Copper Standard)
C485
C48500 Brass Grade, also known as Leaded Naval Brass or C485 Brass, is an alloy composed of copper, zinc, tin, and a small amount of lead. This alloy is recognized for its impressive corrosion resistance, especially in marine environments, in addition to its good ductility and moderate strength. The primary components of C48500 are copper (58.5%), zinc (39%), tin (1%), and lead (1.5%), making it a suitable choice for a variety of applications that require corrosion resistance and decent strength.
Chemical Properties:
Copper (Cu): 58.5%
Zinc (Zn): 39%
Tin (Sn): 1%
Lead (Pb): 1.5%
Trace elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Nickel (Ni)
Physical Properties:
Density: 8.39 g/cm³
Melting Point Range: 900-935°C (1650-1715°F)
Tensile Strength: 290-430 MPa (42,100-62,350 psi)
Yield Strength: 95-270 MPa (13,800-39,150 psi)
Elongation: 18-40%
Hardness (Brinell): 70-110 BHN
Electrical Conductivity: 23% IACS (International Annealed Copper Standard)
C770
C770 Brass Grade, commonly known as Nickel Silver or C77000 Brass, is an alloy composed of copper, nickel, and zinc. This alloy is appreciated for its excellent corrosion resistance, good strength, and attractive appearance. The primary components of C770 are copper (55-65%), nickel (10-30%), and zinc (18-27%), making it a suitable material for various applications that require corrosion resistance and appealing aesthetics.
Chemical Properties:
Copper (Cu): 55-65%
Nickel (Ni): 10-30%
Zinc (Zn): 18-27%
Trace amounts of elements: Iron (Fe), Aluminum (Al), Silicon (Si), and Lead (Pb)
Physical Properties:
Density: 8.80 g/cm³
Melting Point Range: 1000-1100°C (1832-2012°F)
Tensile Strength: 450-800 MPa (65,250-116,000 psi)
Yield Strength: 140-700 MPa (20,300-101,500 psi)
Elongation: 2-40%
Hardness (Brinell): 130-240 BHN
Electrical Conductivity: 7% IACS (International Annealed Copper Standard)
Applications:
Decorative and ornamental elements in architectural projects, such as door and window fittings, railings, and trim
Musical instrument parts, including flutes, saxophones, and horns, due to its good strength and acoustic properties
High-quality cutlery, flatware, and utensils due to its appealing appearance and corrosion resistance
Watch and clock components due to its durability and resistance to tarnish and corrosion
Jewelry making, including chains, pendants, and earrings, due to its attractive appearance and resistance to tarnish and corrosion
Disadvantages:
Low electrical conductivity compared to other copper alloys, which may limit its use in specific electrical applications
Moderate ductility and formability may not be suitable for specific formability applications
Higher cost compared to other brass alloys, like C260 or C272, due to the presence of nickel
Here is a table for each grade and its main advantages:
| Brass Grade | Main Advantages |
--- |
| C360 | High machinability, excellent corrosion resistance, and good strength |
| C230 | Good corrosion resistance, excellent hot-working characteristics, and good strength |
| C210 | Excellent cold-working characteristics, good corrosion resistance, and good strength |
| C220 | Good machinability, excellent cold-working characteristics, and good strength |
| C260 | Good machinability, excellent cold-working characteristics, and good strength |
| C272 | Excellent machinability, good corrosion resistance, and good strength |
| C274 | Excellent machinability, good corrosion resistance, and good strength |
| C314 | Excellent machinability, good corrosion resistance, and good strength |
| C330 | Good hot-working characteristics, excellent corrosion resistance, and good strength |
| C335 | Good hot-working characteristics, excellent corrosion resistance, and good strength |
| C353 | Excellent cold-working characteristics, good corrosion resistance, and good strength |
| C365 | Good machinability, excellent cold-working characteristics, and good strength |
| C385 | Good machinability, excellent cold-working characteristics, and good strength |
| C443 | Excellent corrosion resistance in saltwater environments, good strength, and excellent heat-transfer properties |
| C464 | Excellent corrosion resistance in seawater and marine environments, good strength, and excellent hot-working characteristics |
| C485 | Excellent corrosion resistance in seawater and marine environments, good strength, and excellent hot-working characteristics |
| C770 | Excellent wear resistance, good corrosion resistance, and good strength |
Selecting the Right Brass Grade for Your Project
When choosing the appropriate brass grade for your CNC machining project, it’s essential to consider the specific requirements of your application. Factors such as machinability, strength, corrosion resistance, and cosmetic appearance should all be taken into account when making your decision. To aid in your selection process, consider the following guidelines:
Prioritize Machinability
If your project requires high-speed machining and tight tolerances, opt for a brass grade with excellent machinability, such as C36000 Free Cutting Brass. This grade is specifically designed for complex parts that demand precise machining.
Consider Corrosion Resistance
For components exposed to corrosive environments or harsh conditions, prioritize a brass grade with superior corrosion resistance. C46400 Naval Brass and C48500 Leaded Naval Brass are both excellent choices for marine and other corrosive applications.
Balance Strength and Machinability
In applications that require both strength and machinability, consider a brass grade that offers a good balance of these properties. C37700 Forging Brass is an excellent choice for parts requiring forging and machining, while C48500 Leaded Naval Brass combines strength with enhanced machinability.
Evaluate Aesthetic Requirements
If the appearance of your machined part is a crucial factor, choose a brass grade that offers a visually appealing finish. Most brass grades exhibit an attractive golden hue, but some, such as C36000 Free Cutting Brass, may have a slightly more yellow appearance due to their higher zinc content.
Make Sure that Your Machining Shop Supports the Brass Grade
At 3ERP, we take pride in offering top-of-the-line CNC brass machining services that bring your vision to life with unparalleled accuracy and precision. Our expertise in working with a variety of brass alloys ensures that your custom parts and components boast remarkable quality and performance.
Our cutting-edge CNC machine centers, combined with our team of highly experienced technicians, enable us to produce custom brass parts that meet even the most exacting specifications.
The Versatility of Brass in CNC Machining
Brass is a versatile and reliable material for CNC machining, offering a range of grades to suit various applications and requirements. By understanding the unique properties of each brass grade, you can make an informed decision about which is best suited to your project.
Whether you need excellent machinability, superior corrosion resistance, or a combination of strength and appearance, there is a brass grade that will meet your needs. Armed with this comprehensive guide, you can confidently navigate the world of brass grades and find the perfect material for your CNC machining project.
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187527 | https://www.acs.org/content/dam/acsorg/education/k-8/inquiry-in-action/fifth-grade/g5-l1.4-bkgd.pdf | 5th Grade - Lesson 1.4 The Water Cycle Teacher Background The water cycle relies on the processes of evaporation and condensation. Evaporation The evaporation of water happens over a wide range of temperatures. At any temperature, the molecules of a substance are moving at a variety of speeds (kinetic energies). Evaporation happens when the molecules at the surface of a liquid move fast enough to break away from other molecules in the liquid and become a gas. Evaporation happens in room-temperature water and even in cold water because at those temperatures, a portion of water molecules have enough energy to break away from other water molecules (evaporate). When a faster-moving molecule is at the surface, it can break away from other molecules even though most of the other molecules are moving more slowly. On cold days, water evaporates, but it evaporates more slowly than it would on a warmer day. Adding energy (heating) increases the rate of evaporation Although water can evaporate at low temperatures, the rate of evaporation increases as the temperature increases. This makes sense because at higher temperatures, more molecules are moving faster; therefore, it is more likely for a molecule to have enough energy to break away from the liquid to become a gas. Condensation The flip side of evaporation is condensation. For condensation to occur, molecules of water vapor in the air need to be moving slow enough so that when they collide with other molecules of water vapor, they attract to become liquid water. Removing energy (cooling) increases the rate of condensation As the temperature decreases, the rate of condensation increases. This is because a lower temperature means that more molecules are moving more slowly. If molecules move slower, they are more likely to attract and change their state from a gas to a liquid. Evaporation Increased Rate of Evaporation Condensation Increased Rate of Condensation Grade 5 - Lesson 1.4 The Water Cycle 1 www.acs.org/inquiryinaction ©American Chemical Society 2019 Evaporation and Condensation Happen at the Same Time At any temperature, evaporation and condensation are actually occurring at the same time. Faster molecules from the liquid evaporate while slower molecules from the gas condense. Depending on the conditions, one process will happen at a faster rate than the other resulting in net evaporation or net condensation. Other factors affect evaporation and condensation There’s more that influences evaporation and condensation than just temperature. The amount of water vapor in the air is also a big factor. When the air is dry, water evaporates faster than it condenses so there is a net high rate of evaporation. But if the air is very humid, the rate of condensation would be high so even if water evaporated, the net rate of evaporation would not be as high as on a dry day. Relative humidity Since water vapor in air condenses to liquid water as the air cools, cooler air tends to contain less water vapor than warmer air. This is the basis for the measurement of relative humidity which measures the amount of water vapor in the air relative to the maximum amount of water vapor the air can hold at that temperature. Since warm air can hold more water vapor than colder air, a given amount of water vapor in warm air will have a lower relative humidity than the same amount of water vapor in cool air. Since cold air can’t hold as much water vapor as warmer, a given amount of water vapor in cold air will have a higher relative humidity than the same amount of water vapor in warm air. Fog Depending on the amount of water vapor and the temperature of the air and ground, condensation causes a lot of the different types of moisture that we see. With the right combination of water vapor and temperature, condensed water vapor can form a mist that is visible and close to the ground, called fog. Dew Condensation is also the cause of dew. Dew is liquid water that has condensed from water vapor and is often found on grass and your car on a cold morning. Frost When it is really cold outside, frost can form on surfaces such as glass and on plants. Some frost forms when water vapor in the air condenses to liquid water and then freezes to form ice. Frost can also form in a different way when the humidity and temperature are just right. In these cases, the water vapor in the air changes directly to ice crystals on a cold surface without passing through a liquid phase. Grade 5 - Lesson 1.4 The Water Cycle 2 www.acs.org/inquiryinaction ©American Chemical Society 2019 |
187528 | https://math.stackexchange.com/questions/4515395/are-opinions-considered-propositions | logic - Are opinions considered propositions? - Mathematics Stack Exchange
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Are opinions considered propositions?
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I was wondering the thought. My textbook says:
A proposition is still a proposition whether its truth value is known to be true, known to be false, unknown, or a matter of opinion.
"It is a nice day" - This is not a proposition
"All Politicians are dishonest" - This is a proposition
"The movie was funny" -This is a proposition.
Then wouldn't statement 1 be considered a proposition, especially since it says that being a matter of opinion doesn't change the fact is a proposition? Or maybe this doesn't matter at all?
logic
logic-translation
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edited Aug 20, 2022 at 7:46
Mauro ALLEGRANZA
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asked Aug 20, 2022 at 3:34
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9 This is more suited for philosophy stack exchange. To give you a cursory answer however, a proposition is a truth-bearer- an object that is capable of having a truth value.emesupap –emesupap 2022-08-20 03:41:17 +00:00 Commented Aug 20, 2022 at 3:41
4 What textbook is it?Suzu Hirose –Suzu Hirose 2022-08-20 03:45:51 +00:00 Commented Aug 20, 2022 at 3:45
I personally don't know if I agree if a statement can still be a proposition in a mathematical sense if the truth value is a matter of opinion. But I'm sure someone on here knows more than I do about this.blakedylanmusic –blakedylanmusic 2022-08-20 03:48:15 +00:00 Commented Aug 20, 2022 at 3:48
1 i would throw this book into the garbage.lola –lola 2022-08-20 14:05:42 +00:00 Commented Aug 20, 2022 at 14:05
in particular, you are right that if you adopt the (quite useless) definition of proposition given in the book, the number 1 must be a proposition as much as 3 is.lola –lola 2022-08-20 14:07:33 +00:00 Commented Aug 20, 2022 at 14:07
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"It is a nice day" - This is not a proposition
"All Politicians are dishonest" - This is a proposition
"The movie was funny" -This is a proposition.
I think your textbook is translating the above as
N(x)N(x)
∀x[P(x)→¬H(x)]∀x[P(x)→¬H(x)]
F(c).F(c).
Since propositions do not contain any free variable, (1) is disqualified from being a proposition. On the other hand, c c is a constant and (3) is a proposition.
A proposition is still a proposition whether its truth value is known to be true, known to be false, unknown, or a matter of opinion.
Here, your textbook is not defining a proposition, but just trying to say that a subjective proposition's truth value alternates across contexts; for example, the truth value of the proposition “Jan 1 2020 was a nice day” varies according to the particular location and the definition of “nice day”.
Incidentally, this is why the widespread “either true or false but not both” characterisation of a proposition is misleading: “for each x,x 2 x,x 2 is not a negative number” is a proposition, and is true in real analysis and false in complex analysis (though it does have a definite truth value under each interpretation).
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edited Sep 19, 2024 at 6:00
answered Aug 20, 2022 at 5:14
ryangryang
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The day might be "nice" to one person but not "nice" to another person. "It is a nice day" is NOT either true or false to all.George Ivey –George Ivey 2022-08-20 12:40:43 +00:00 Commented Aug 20, 2022 at 12:40
1 @GeorgeIvey What's with the caps lock? And your comment is entirely consistent with my Answer.ryang –ryang 2022-08-20 12:55:56 +00:00 Commented Aug 20, 2022 at 12:55
“it is a nice day” has “it”, which in the standard meaning as it’s used in natural language has the same usage as “the” in “the movie was funny”, i.e. it is of type N(c)N(c) where c = the present day.lola –lola 2022-08-20 14:05:11 +00:00 Commented Aug 20, 2022 at 14:05
1 i understood that you were just attempting at making excuses for the book. i wanted to remark that the book’s example is quite debatable and that the best thing would be to find another book.lola –lola 2022-08-20 14:43:42 +00:00 Commented Aug 20, 2022 at 14:43
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Product Rule
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The Product Rule is a method of taking the derivative of the product of two functions. It states that the derivative of .
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187530 | https://www.acog.org/clinical/clinical-guidance/committee-opinion/articles/2011/07/vitamin-d-screening-and-supplementation-during-pregnancy | Vitamin D: Screening and Supplementation During Pregnancy | ACOG
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Vitamin D: Screening and Supplementation During Pregnancy
Vitamin D: Screening and Supplementation During Pregnancy
Committee Opinion CO
Number 495
July 2011
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Number 495
Committee on Obstetric Practice
This document reflects emerging clinical and scientific advances as of the date issued and is subject to change. The information should not be construed as dictating an exclusive course of treatment or procedure to be followed.
ABSTRACT: During pregnancy, severe maternal vitamin D deficiency has been associated with biochemical evidence of disordered skeletal homeostasis, congenital rickets, and fractures in the newborn. At this time, there is insufficient evidence to support a recommendation for screening all pregnant women for vitamin D deficiency. For pregnant women thought to be at increased risk of vitamin D deficiency, maternal serum 25-hydroxyvitamin D levels can be considered and should be interpreted in the context of the individual clinical circumstance. When vitamin D deficiency is identified during pregnancy, most experts agree that 1,000–2,000 international units per day of vitamin D is safe. Higher dose regimens used for treatment of vitamin D deficiency have not been studied during pregnancy. Recommendations concerning routine vitamin D supplementation during pregnancy beyond that contained in a prenatal vitamin should await the completion of ongoing randomized clinical trials.
Vitamin D is a fat-soluble vitamin obtained largely from consuming fortified milk or juice, fish oils, and dietary supplements. It also is produced endogenously in the skin with exposure to sunlight. Vitamin D that is ingested or produced in the skin must undergo hydroxylation in the liver to 25-hydroxyvitamin D (25-OH-D), then further hydroxylation primarily in the kidney to the physiologically active 1,25-dihydroxyvitamin D. This active form is essential to promote absorption of calcium from the gut and enables normal bone mineralization and growth. During pregnancy, severe maternal vitamin D deficiency has been associated with biochemical evidence of disordered skeletal homeostasis, congenital rickets, and fractures in the newborn 12.
Recent evidence suggests that vitamin D deficiency is common during pregnancy especially among high-risk groups, including vegetarians, women with limited sun exposure (eg, those who live in cold climates, reside in northern latitudes, or wear sun and winter protective clothing) and ethnic minorities, especially those with darker skin 345. Newborn vitamin D levels are largely dependent on maternal vitamin D status. Consequently, infants of mothers with or at high risk of vitamin D deficiency are also at risk of vitamin D deficiency 56.
For the individual pregnant woman thought to be at increased risk of vitamin D deficiency, the serum concentration of 25-OH-D can be used as an indicator of nutritional vitamin D status. Although there is no consensus on an optimal level to maintain overall health, most agree that a serum level of at least 20 ng/mL (50 nmol/L) is needed to avoid bone problems ref078910. Based on observations of biomarkers of vitamin D activity, such as parathyroid hormone, calcium absorption, and bone mineral density, some experts have suggested that vitamin D deficiency should be defined as circulating 25-OH-D levels less than 32 ng/mL (80 nmol/L) 11. An optimal serum level during pregnancy has not been determined and remains an area of active research.
In 2010, the Food and Nutrition Board at the Institute of Medicine of the National Academies established that an adequate intake of vitamin D during pregnancy and lactation was 600 international units per day 12. Most prenatal vitamins typically contain 400 international units of vitamin D per tablet. Summarizing recent observational and interventional studies, the authors of a recent clinical report from the Committee on Nutrition of the American Academy of Pediatrics suggested that a daily intake higher than that recommended by the Food and Nutrition Board may be needed to maintain maternal vitamin D sufficiency 13. Although data on the safety of higher doses are lacking, most experts agree that supplemental vitamin D is safe in dosages up to 4,000 international units per day during pregnancy or lactation 12.
At this time there is insufficient evidence to support a recommendation for screening all pregnant women for vitamin D deficiency. For pregnant women thought to be at increased risk of vitamin D deficiency, maternal serum 25-OH-D levels can be considered and should be interpreted in the context of the individual clinical circumstance. When vitamin D deficiency is identified during pregnancy, most experts agree that 1,000–2,000 international units per day of vitamin D is safe. Higher dose regimens used for the treatment of vitamin D deficiency have not been studied during pregnancy. Recommendations concerning routine vitamin D supplementation during pregnancy beyond that contained in a prenatal vitamin should await the completion of ongoing randomized clinical trials. At this time, there is insufficient evidence to recommend vitamin D supplementation for the prevention of preterm birth or preeclampsia.
References
Pawley N, Bishop NJ. Prenatal and infant predictors of bone health: the influence of vitamin D. Am J Clin Nutr 2004;80:1748S–51S.
Article Locations:Article Location
2. Gale CR, Robinson SM, Harvey NC, Javaid MK, Jiang B, Martyn CN, et al. Maternal vitamin D status during pregnancy and child outcomes. Princess Anne Hospital Study Group. Eur J Clin Nutr 2008;62:68–77.
Article Locations:Article Location
3. Hollis BW, Wagner CL. Assessment of dietary vitamin D requirements during pregnancy and lactation. Am J Clin Nutr 2004;79:717–26.
Article Locations:Article Location
4. Lee JM, Smith JR, Philipp BL, Chen TC, Mathieu J, Holick MF. Vitamin D deficiency in a healthy group of mothers and newborn infants. Clin Pediatr (Phila) 2007;46:42–4.
Article Locations:Article Location
5. Bodnar LM, Simhan HN, Powers RW, Frank MP, Cooperstein E, Roberts JM. High prevalence of vitamin D insufficiency in black and white pregnant women residing in the northern United States and their neonates. J Nutr 2007;137:447–52.
Article Locations:Article LocationArticle Location
6. Dijkstra SH, van Beek A, Janssen JW, de Vleeschouwer LH, Huysman WA, van den Akker EL. High prevalence of vitamin D deficiency in newborn infants of high-risk mothers [published erratum appears in Arch Dis Child 2007;92:1049]. Arch Dis Child 2007;92:750–3.
Article Locations:Article Location
7. Holick MF. Vitamin D deficiency. N Engl J Med 2007;357:266–81.
Bouillon R, Norman AW, Lips P. Vitamin D deficiency. N Engl J Med 2007;357:1980–1; author reply 1981–2.
Article Locations:Article Location
9. Vitamin D supplementation: Recommendations for Canadian mothers and infants. Paediatr Child Health 2007;12:583–98.
Article Locations:Article Location
10. National Institutes of Health, Office of Dietary Supplements. Vitamin D. Available at: Retrieved December 16, 2010.
Article Locations:Article Location
11. Hollis BW, Wagner CL. Normal serum vitamin D levels. N Engl J Med 2005;352:515–6; author reply 515–6.
Article Locations:Article Location
12. Institute of Medicine of the National Academies (US). Dietary reference intakes for calcium and vitamin D . Washington, DC: National Academy Press; 2010.
Article Locations:Article LocationArticle Location
13. Wagner CL, Greer FR. Prevention of rickets and vitamin D deficiency in infants, children, and adolescents. American Academy of Pediatrics Section on Breastfeeding; American Academy of Pediatrics Committee on Nutrition [published erratum appears in Pediatrics 2009;123:197]. Pediatrics 2008;122:1142–52.
Article Locations:Article Location
Copyright July 2011 by the American College of Obstetricians and Gynecologists, 409 12th Street, SW, PO Box 96920, Washington, DC 20090-6920. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, posted on the Internet, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from the publisher. Requests for authorization to make photocopies should be directed to: Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400.
ISSN 1074-861X
Vitamin D: screening and supplementation during pregnancy. Committee Opinion No. 495. American College of Obstetricians and Gynecologists. Obstet Gynecol 2011;118:197–8.
Topics
Delivery of health careDietary guidelinesDietary supplementsHealth personnelMass screeningNutritionNutrition policyPregnancyPregnancy complicationsVitamin deficiencyVitamins
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De Moivre's Theorem
De Moivre's Theorem is a very useful theorem in the mathematical fields of complex numbers. It allows complex numbers in polar form to be easily raised to certain powers. It states that for and , .
Proof
This is one proof of de Moivre's theorem by induction.
If :
If , the formula holds true because Assume the formula is true for . Now, consider :
Therefore, the result is true for all nonnegative integers .
If , one must consider when is a positive integer.
And thus, the formula proves true for all integral values of .
Generalization
Note that from the functional equation where , we see that behaves like an exponential function. Indeed, Euler's identity states that . This extends de Moivre's theorem to all .
See Also
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187532 | https://www.youtube.com/watch?v=_rK02neOF18 | Trusses Method of Joints | Mechanics Statics | Learn to Solve Questions
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Learn how to solve for forces in trusses step by step with multiple examples solved using the method of joints. We talk about determining force directions, what compression and tension is and much more.
🔹Breaking forces into components:
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Intro(00:00)
Determine the force in each member of the truss. (02:29)
Determine the force in each member of the truss and state (05:49)
The maximum allowable tensile force in the members (08:08)
Find more at
Book used: R. C. Hibbeler and K. B. Yap, Engineering Mechanics Statics.
Hoboken: Pearson, 2017.
624 comments
Transcript:
Intro Let’s talk about trusses and how to solve for
unknown forces. This is a truss. You see them on bridges, roofs and loads of other places. Each
of these individual pieces is called a member and they are usually connected with a pin, and we call
that a joint. Each member can either be in tension or compression. If the forces are pulling on the
ends of the member, it’s in tension and if its pushing at the at the 2 ends, it’s in compression.
Today, we are going to look at a method called the method of joints. The way to think about these
problems is to realize that if the whole truss is in equilibrium, then each and every member and
joint is also in equilibrium. So in simple terms, if the whole truss isn’t moving, then neither
are any of the parts. That means we can write our equilibrium equations and solve for the forces
at each member. Now let’s say we have a truss like this. The way we solve it is to start at a
location where we have at least one known force and a maximum of 2 unknowns. So in this truss,
this is the best spot to start. The next step is to isolate this joint, and then assume the
direction of the force. Is the force going to come towards the pin, or away from it? It’s just an
assumption so it doesn’t really matter which you pick, though over time, you will notice that you
can make a very good guess. Now if your assumption is wrong, you will get a negative value, so then
you know it’s the other way around. Once you find the force in the member, you have to determine
whether the member is in tension or compression. Now assume we figured out that at this joint,
we have the forces like this. You might assume, well this force is coming towards the pin, so
it looks like it’s pulling on the member, which means its in tension. This is not the case. What’s
really happening is that we are finding the force applied to the pin, which means the pin exerts
an equal but opposite force onto the member. So if a force is coming towards the pin, then that
member is in compression. If a force is going away from the pin, that member is in tension. Lastly,
if we have a force coming towards the pin, at the opposite end of the member, the
force is going towards the other pin, so its always in opposite directions. The same
is true if a force is going away from the pin, in that case, at the other end, the
force is going away from that pin. To solve the problems you face, it’s really
important that you know how to break forces into x and y components. If you need a refresh,
or you forgot, please check the description. Now let’s go through some examples and see
how we can actually solve for unknown forces. Let’s take a look at this problem where we need
to find the force in each member of the truss Determine the force in each member of the truss. and whether they are in tension or compression.
The best place to start our analysis is a location where we know at least one force and 2 unknowns,
which is right here. So let’s draw point D separately. I am going to assume that
both forces are going away from the pin. Now any force not lying on the x and y axis
has to be broken into x and y components. Next, we write our equations of equilibrium. First,
for y-axis forces because we can easily solve for the force in member DC. We will pick up to be
positive. Solving gives us the force in member DC. Now since we assumed it to be going away from the
pin, and we got a positive value, that means our assumption was right. That also means this member
is in tension since anytime a force is leading away from the pin, the member is in tension. Next,
x-axis forces, we will assume left to be positive. Don’t forget to plug in the value for member DC we
just found. Let’s solve. Notice we got a negative value, that means our assumption was wrong and
that the force is actually coming towards the pin. Since this force is coming towards the
pin, this member is in compression. Now we need to pick another point to write our
equations of equilibrium. Point E has too many unknowns since we have 3 unknown members, so the
next best spot is point C. We know the force of member DC, which we found in the previous
step. Since at D it was leaving the pin, that means at C, it’s also leaving the pin,
in other words, going towards D. We also have forces CE and CB. I am going to assume force CE
is coming towards pin C and force CB is going away from pin C. Now let’s write our equations
of equilibrium starting with the y axis forces. We got a positive value, so our assumption
is right, and its coming towards the pin so it’ll be in compression. Next x axis
forces. Let’s solve. We got a positive value, so our assumption was right and since the force
is going away from pin C, it’s in tension. Now we will pick point B. Let’s draw the forces.
So we know the force in member BC, which was going away from point C, which means at point B, it’s
going towards point C, in other words, leaving B. For force BE, I will assume it comes towards
the pin and force BA goes away from the pin. Now for our equations. First, y axis
forces. If we divide all terms by sin 60, we can see that the force in member BE is
actually the same as the force in member BA. Next, x axis forces. Let’s solve the 2 equations.
Lastly, we can look at point E. Since it’s a roller, we’d have a single force straight
upwards. We know force DE, CE, and BE. Note the directions since all we are doing
is flipping them. So for example, at pin D, we found that force DE was coming towards pin D,
which means the same force is going towards pin E, so it’s just flipped. For force EA, I will assume
it comes towards the pin. Now for our equations. First y axis forces. This gives us the reaction at
the roller. Next x axis forces. Since force EA is coming towards the pin, it’s in compression.
Now we found all the forces in each member. Let’s take a look at this problem where
we need to figure out the force in each Determine the force in each member of the truss and state member and whether they are in tension or
compression. Where is the best place to start? That’s point D since we know a force already
being applied and we would only have 2 unknowns. We have force DE and force DC. We
also need the angle at the top, and that can be found using trigonometry since
this whole system is a right-angle triangle. We can use inverse of tan, which is opposite
over adjacent, and solve for theta. Now that we have the angle, I am going to assume force
DE comes towards the pin and force DC is going away from the pin. Let’s write our equations. We
will start with an equation for x axis forces. Let’s solve. Since this force is going towards pin
D, it’ll be in compression. Next, y axis forces, don’t forget to use the force in member DE
we just found. Let’s solve. This force is heading away from point D, so it’ll be in tension.
Next we will look at point C. So we have the 900 N force, then we have the force of member DC, which
is going away from point C. I will assume force CE is coming towards point C, and force CB to be
going away from pin C. Let’s write our equations. First, x axis forces. So this is in compression.
Next, y axis forces. This member is in tension. Now we move on to point E. Before we do
anything else, we need to figure out the angles. Now there is probably a bunch of ways to get
to this answer, but one way to do it without knowing these 2 angles are equal, is to find the
length of member EC. We can do that by using tan. Then we see another right-angle triangle, so
this angle can be found using inverse of tan. This angle up here is simply 53.13 degrees.
Now the bottom angle is also 53.13 because we just have a big right-angle triangle.
That means this angle is also 53.13 degrees. Now we have these forces at E. For the unknown
force EB, I will assume it goes away from the pin and for force EA, I will assume it comes
towards the pin. Now let’s write our equations. First x axis forces. Let’s simplify. Now y
axis forces. We can now solve the 2 equations. And those are our answers, we didn’t even need
to figure out the reactions at the supports. Let’s take a look at one last question. In
the question, we are told that the maximum The maximum allowable tensile force in the members tensile force a member can handle is 5 kN and the
maximum compression a member can handle is 3 kN. We need to figure out the maximum force P
that can be applied to the truss without going above the limits. Before we do anything,
let’s figure out the angles inside this truss. Since this is an isosceles triangle, this angle is
30 degrees, which means this angle is 120 degrees. The same is true for the top triangle.
Now the best place to start is at point C. We will solve this like we know the value of P. I
am going to assume that forces CD and CB is coming towards pin C. Let’s write our equations starting
with y axis forces. Let’s simplify. Next, x axis forces. Let’s simplify. So the goal is to just
write all the forces in each member in terms of P. Now we move onto point D. I will assume force DA
comes towards the pin and force DB leaves the pin. We also have the force we just found and
another force P. Let’s write our equations. Now we will simplify these equations so
we can write each force in terms of P. We will now move to point A, which has
a roller for a support. That means we will have one vertical reaction upwards, along
with forces AD and AB. I will assume force AB comes towards pin A. Now let’s write our
equations. First, y axis forces. Let’s simplify. Now, x axis forces. Let’s simplify. We now have
written all the forces in each member in terms of force P. Let’s look at all the members that are
in compression. We have 4 such members, CB, CD, DA and AB. From these 4, we see that member DA would
have the largest value because it’s multiplied by the largest number. We also know that the maximum
allowable compressive force for each member is 3 KN. So let’s solve for P by assuming that member
DA has a force of 3 kN. We get a value of 1.3 kN. Now we need to check something, which is to see
if this value is okay for our member in tension. The maximum allowable tensile force is 5 kN and
the only member in tension is member DB. Let’s substitute 1.3 kN to see if member DB would carry
a force larger than 5 kN. We see that it won’t, which means our P value is within range. So the
maximum value of P that can be applied is 1.3 kN. That should cover the types of problems you will
face when it comes to trusses and using the method of joints. I hope this video helped and thanks
for watching. Best of luck with your studies! |
187533 | https://www.scientia-education.com/manuals/AI_HL/splitted/TI84_for_AI_HL_1_1_scientific_notation_and_rounding.pdf | TI-84 manual for the IB www.scientia-education.com 1.1 Scientific notation and rounding 1.1.1 Switch the TI-84 to scientific mode Recall that any number can be written in scientific notation (or normal form): ±[number between 1 and 9.999...] × 10integer.
For example, 142857 = 1.42857 × 105.
The TI-84 uses the symbol “E” for scientific notation. It means ×10.... For example, if you compute 4.51 ∗1011 the calculator should display 4.51E11.
Note that for the IB, you must write numbers using ×10... notation, and not E notation By default, any number between 0.001 and 999′999′999 will not be written in scientific notation.
You can change that so any number is written in scientific notation. Press and select SCI as below (2nd line): Press . Typing 142857 should thus display 1.42857E5.
1.1.2 Rounding up automatically You cannot round up to n significant figures directly with the calculator (it basically doesn’t know what significant figures are), but it can round up to any decimals between 0 and 9.
10 Copyright © 2021 Scientia-Education Learning and Consulting.
All rights reserved.
TI-84 manual for the IB www.scientia-education.com If you are struggling with rounding, you can set-up the calculator to display any amount of decimals (between 0 and 9).
Suppose you want to display a number up to 3 decimals. To do so, press and select 3 (3rd line): Press .
If you type 0.1234 the calculator should display 0.123. If you want to switch back to normal mode, press and select FLOAT(3rd line): Press .
11 Copyright © 2021 Scientia-Education Learning and Consulting.
All rights reserved. |
187534 | https://www.encyclopedie-environnement.org/en/zoom/deciphering-benson-bassham-calvin-cycle/ | | Focus
1/4 | The path of carbon in photosynthesis
Deciphering the Benson-Bassham-Calvin Cycle
PDF
1. A technological revolution for deciphering the cycle
Around 1945, the american chemists A.A. Benson, J.A. Bassham and M. Calvin (Figure 1) tackled the task of identifying the first carbon-containing product of photosynthesis that had escaped all scientific investigation. After the World War II, these researchers had just acquired two new important technological allies in Berkeley, California:
a radioactive tracer, 14C, which had just been isolated by S. Ruben and M. Kamen in 1941 ,
new techniques for the chromatographic separation of the carbon-containing compounds from metabolism, developed a few years earlier in England by R. Martin and A. Synge.
These two tools will allow Dr. Calvin and his co-workers to identify the path taken by carbon during the various biochemical reactions for assimilation of the photosynthetic carbon.
These first experiments were not carried out on leaves but on unicellular algae (Chlorella or Senedesmus), which were easier to use. The experimental protocol is as follows:
The unicellular algae are placed in a suspension medium that flows at a constant rate from a tank through a transparent illuminated coil.
When photosynthesis is stationary, under illumination at a given temperature, sodium bicarbonate (Na14CO3) is introduced by a syringe into the suspension medium.
Incubation in the presence of 14C is limited to the time the suspension continues to descend through the coil after injection (a few seconds).
At the end, the plant material is fixed in boiling methanol (Figure 2).
The constituents present in the extract are then separated by two-dimensional chromatography.
The chromatogram is then X-rayed in the dark. The black spots on the radio-chromatogram indicate the presence of the radioactive compounds that were formed during the 14C labelling of the algae.
In order to identify these compounds, radiochromatograms are compared with chromatograms coloured by different chemical reagents specific to organic acids, amino acids, sugars and phosphorylated compounds (Figure 3).
.
2. The Benson-Bassham-Calvin Cycle
All the molecules identified with the methods described above were organized in a cycle comprising 3 phases. Figure 4 schematically depicts the Benson-Bassham-Calvin Cycle.
Phase 1 – Carbon Dioxide (CO2) fixation
3 RuBP (C5 molecule) + 3 CO2 (C1 molecule) → 6 3-PGA (C3 molecule) (catalyzed by the Rubisco)
Phase 2 – Reduction of phosphoglyceric acid to triose-phosphate
The 3-phosphoglyceric acid (3-PGA) is reduced in two steps to glyceraldehyde 3-phosphate (G3P). A molecule with 3 carbons and a phosphate group, G3P is a triose phosphate. Both reactions consume NADPH2 and part of the ATP produced by thylacoids during the photochemical reactions of photosynthesis (see Shedding light on Photosynthesis):
6 3-PGA (C3 molecule) + 6 ATP ⇌ 6 1,3-bisPGA (C3 molecule) + 6 ADP
6 1,3-bisPGA (C3 molecule) + 6 NADPH + 6 H+ ⇌ 6 G3P (C3 molecule) + 6 NADP+ + 6 Pi
Phase 3 – Regeneration of the CO2 acceptor
The role of the cycle is to regenerate the RuBP. The set of reactions in phase 3 consists of reconverting molecules with 3 carbon atoms into molecules with 5 carbon atoms in order to be able to reset the cycle. This phase consumes very large quantities of glyceraldehyde 3-phosphate (5 molecules out of 6 formed). Phosphorylated C6, C4, C7 and finally C5 backbones are made (see Figure 4), as can be represented schematically as follows:
C3P + C3P → C6-diP → C6-P + Pi
C6-P + C3P → C4P + C5P
C4P + C3P → C7-diP→ C7-P + Pi
C7P + C3P → 2 C5P
Only the last stage of this phase consumes ATP. The overall balance sheet can be summarized as follows:
5 G3P (C3 molecule) → 3 ribulose-5-P (C5 molecule) + 2 Pi
3 ribulose-5-P (C5 molecule) + 3 ATP ⇌ 3 RuBP (C5 molecule) + 3 ADP
To summarize, only one of the 6 triose phosphate molecules produced by this cycle will be used for the production of sugars and, more broadly, other metabolites that are the basis of biomass creation (Figure 4). It will be used for the biosynthesis of starch, amino acids or lipids in the chloroplast or exported out of the chloroplast and transformed into sucrose by the enzymes of the cytoplasm (see Focus Sucrose or Starch?).
Notes and References
Cover image. Structures of somes molecules of the Benson-Bassham-Calvin Cycle (Public domain)
Andrew Alm Benson (1917-2015), James Alan Bassham (1922-2012) & Melvin Calvin (1911-1997), are American chemists whose contributions to the discovery of the carbon cycle in plants are considerable. Benson, a specialist in carbon compounds and sugars, is credited in particular with the discovery of ribulose 1,5-bisphosphate on which the carbon of CO2 is fixed. Calvin was awarded the 1961 Nobel Prize in Chemistry for his work on the assimilation of carbon dioxide by plants.
Martin Kamen (1913-2002) and Samuel Ruben (1913-1943) are two American chemists, co-discoverers of carbon-14 (14C). With a half-life of 5730 years, 14C can be very easily used in metabolic experiments. It is one of the three naturally abundant isotopes (along with 12C and 13C), but the only radioactive one.
Richard Laurence Millington Synge (1914-1994), and Archer John Porter Martin (1910-2002) are both English chemists. They were co-winners of the 1952 Nobel Prize in Chemistry for the invention of “partition chromatography”.
One of the devices -offered by Andrew Benson to Roland Douce- is kept at the Musée des Confluences in Lyon (France).
More:
Farineau J. & Morot-Gaudry F., 2011, La Photosynthèse, Quae, ISBN 978-2-7592-0903-3
Morot-Gaudry F., Moreau F., Prat R., Maurel C. & Sentenac H. (2017) Biologie végétale : Nutrition et métabolisme – 3e édition, Dunod
Video: A conversation between Andrew Benson and Professor Bob Buchanan on the discovery of the Benson-Bassham-Calvin Cycle |
187535 | https://www.khanacademy.org/math/geometry-home/triangle-properties/altitudes/v/proof-triangle-altitudes-are-concurrent-orthocenter | Use of cookies
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187536 | https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.15%3A_Hess's_Law_of_Heat_Summation | 17.15: Hess's Law of Heat Summation - Chemistry LibreTexts
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17: Thermochemistry
Introductory Chemistry (CK-12)
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17.15: Hess's Law of Heat Summation
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[ "article:topic", "energy", "enthalpy", "thermochemistry", "Hess\'s law", "combustion", "showtoc:no", "reaction", "hydrocarbons", "license:ck12", "authorname:ck12", "source@ "welding", "acetylene", "heat summation" ]
[ "article:topic", "energy", "enthalpy", "thermochemistry", "Hess\'s law", "combustion", "showtoc:no", "reaction", "hydrocarbons", "license:ck12", "authorname:ck12", "source@ "welding", "acetylene", "heat summation" ]
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Contents
1. Home
2. Bookshelves
3. Introductory, Conceptual, and GOB Chemistry
4. Introductory Chemistry (CK-12)
5. 17: Thermochemistry
6. 17.15: Hess's Law of Heat Summation
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Introductory Chemistry (CK-12)
Front Matter
1: Introduction to Chemistry
2: Matter and Change
3: Measurements
4: Atomic Structure
5: Electrons in Atoms
6: The Periodic Table
7: Chemical Nomenclature
8: Ionic and Metallic Bonding
9: Covalent Bonding
10: The Mole
11: Chemical Reactions
12: Stoichiometry
13: States of Matter
14: The Behavior of Gases
15: Water
16: Solutions
17: Thermochemistry
18: Kinetics
19: Equilibrium
20: Entropy and Free Energy
21: Acids and Bases
22: Oxidation-Reduction Reactions
23: Electrochemistry
24: Nuclear Chemistry
25: Organic Chemistry
26: Biochemistry
Back Matter
17.15: Hess's Law of Heat Summation
Last updated Mar 14, 2025
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17.14: Heat of Combustion
17.16: Standard Heat of Formation
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Table of contents
1. Adding Heats of Reaction
2. Summary
Calculating the
energy
involved in the operation of an acetylene torch is no simple
matter
. Since there is a complex series of reactions taking place, simple methods for determining the
heat
of reaction will not work. We need to develop new approaches to these calculations.
Adding Heats of Reaction
It is sometimes very difficult or even impossible to measure the
enthalpy change
for a reaction directly in the laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because multiple side reactions may occur at the same time. Fortunately, it is possible to measure the
enthalpy change
for a reaction with an indirect method. Hess's
law
of
heat
summation states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a
heat
of reaction for the final equation.
An example will illustrate how Hess's
law
can be used. Acetylene (CA 2HA 2) is a gas that burns at an extremely high temperature (3300 oC) and is used in welding. On paper, acetylene gas can be produced by the reaction of solid carbon (graphite) with hydrogen gas.
2C(s,graphite)+HA 2(g)→CA 2HA 2(g)ΔH=?
Unfortunately, this reaction would be virtually impossible to perform in the laboratory because carbon would react with hydrogen to form many different hydrocarbon
products
simultaneously. There is no way to create conditions under which only acetylene would be produced.
However, enthalpy changes for combustion reactions are relatively easy to measure. The heats of combustion for carbon, hydrogen, and acetylene are shown below, along with each
balanced
equation.
C(s,graphite)+OA 2(g)→COA 2(g)ΔH=−393.5 kJ HA 2(g)+1 2OA 2(g)→HA 2O(l)ΔH=−285.8 kJ CA 2HA 2(g)+5 2OA 2(g)→2COA 2(g)+HA 2O(l)ΔH=−1301.1 kJ
To use Hess's
law
, we need to determine how the three equations above can be manipulated so that they can be added together to result in the desired equation (the formation of acetylene from carbon and hydrogen).
In order to do this, we will go through the desired equation, one
substance
at a time—choosing the
combustion reaction
from the equations above that contains that
substance
. It may be necessary to either reverse a
combustion reaction
, or multiply it by some factor in order to make it "fit" to the desired equation. The first reactant is carbon and in the equation for the desired reaction, the
coefficient
of the carbon is a 2. So, we will write the first
combustion reaction
, doubling all of the coefficients and the ΔH:
2C(s,graphite)+2OA 2(g)→2COA 2(g)ΔH=2(−393.5)=−787.0 kJ
The second reactant is hydrogen and its
coefficient
is a 1, as it is in the second
combustion reaction
. Therefore, that reaction will be used as written.
HA 2(g)+1 2OA 2(g)→HA 2O(l)ΔH=−285.8 kJ
The product of the reaction is CA 2HA 2 and its
coefficient
is also a 1. In
combustion reaction
#3, the acetylene is a reactant. Therefore, we will reverse reaction 3, changing the sign of the ΔH:
2COA 2(g)+HA 2O(l)→CA 2HA 2(g)+5 2OA 2(g)ΔH=1301.1 kJ
Now, these three reactions can be summed together. Any
substance
that appears in equal quantities as a reactant in one equation and a product in another equation cancels out algebraically. The values for the enthalpy changes are likewise added.
2C(s,graphite)+2OA 2(g)→2COA 2(g)ΔH=−787.0 kJ HA 2(g)+1 2OA 2(g)→HA 2O(l)ΔH=−285.8 kJ 2COA 2(g)+HA 2O(l)→CA 2HA 2(g)+5 2OA 2(g)ΔH=1301.1 kJ 2C(s,graphite)+HA 2(g)→CA 2HA 2(g)ΔH=228.3 kJ
So, the
heat
of reaction for the combination of carbon with hydrogen to produce acetylene is 228.3 kJ. When one
mole
of acetylene is produced, 228.3 kJ of
heat
is absorbed, making the reaction endothermic.
Summary
Hess's law of heat summation states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a heat of reaction for the final equation.
Hess's law is used to calculate the heat of reaction for processes that cannot be measured directly.
This page titled 17.15: Hess's Law of Heat Summation is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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187537 | http://www2.math.ou.edu/~jjackson/teaching/1408printable.pdf | 14.8: Lagrange Multipliers Julia Jackson Department of Mathematics The University of Oklahoma Fall 2021 Overview In the previous section, in two distinct contexts we wanted to find the absolute maximum and minimum values of a two- or three-variable function subject to some constraint.
The first was when calculating the absolute maximum and minimum values of a two-variable function on a closed, bounded subset of its domain. In these problems, each boundary component provided a distinct constraint, and we sought the absolute maximum and minimum values of our function on each on our way to solving the overall problem.
Overview, cont.
The second was in solving optimization problems. In these, we wanted to calculate the absolute maximum or minimum value of a three- (or sometimes two-) variable function subject to a directly stated constraint.
In this section, we explore a second method for solving such problems.
This new technique, called the method of Lagrange multipliers, is a handy alternative to the method we learned in the previous section. Each is useful in certain contexts, and mastery of both methods will maximize your flexibility in solving the problems of the previous section.
Table of Contents The Method of Lagrange Multipliers Exercises The Method To find the absolute maximum and minimum values of a function f (x, y, z) (resp. f (x, y)) subject to the constraint g(x, y, z) = k (resp.
g(x, y) = k) for k a constant, perform the following: 1. Find all constants λ and ordered triples (x, y, z) (resp. ordered pairs (x, y)) such that ∇f = λ∇g and g = k are simultaneously true.
2. Evaluate f at all the ordered triples (resp. ordered pairs) from the previous step. The largest value of f is the absolute maximum value of f subject to the constraint, and the smallest value of f is the absolute minimum value of f subject to the constraint.
Example Find all the extreme values of the function f (x, y) = x2 + 2y 2 on the circle x2 + y 2 = 1.
Here we want to find all extreme values of the function f (x, y) = x2 + 2y 2 subject to the constraint x2 + y 2 = 1. We already know a method for solving this problem, thanks to our work in the previous section, but let’s see how the method of Lagrange multipliers handles it.
We begin by introducing a function name to the non-constant part of the constraint: g(x, y) := x2 + y 2 Example, cont.
This done, our next goal is to find all constants λ and ordered pairs (x, y) satisfying ∇f (x, y) = ⟨2x, 4y⟩= λ ⟨2x, 2y⟩= λ∇g(x, y) and x2 + y 2 = 1 By comparing the components of the vectors (more precisely, vector fields) in the first equation, we arrive at the more useful set of equations: 2x = λ2x (1) 4y = λ2y (2) x2 + y 2 = 1 (3) Thus, our goal, according to the method of Lagrange multipliers, is to find all constants λ and ordered pairs (x, y) satisfying all three of these equations at once.
Example, cont.
There are many ways to solve this system, but here’s one that stands out to me: First, note that: (1) ⇒2x(1 −λ) = 0 ⇒x = 0 or λ = 1 Therefore, for (1) to be true, we must have either x = 0 or λ = 1. We now investigate each case separately.
Example, cont.
First, suppose that x = 0. Note that: (3) and x = 0 ⇒y 2 = 1 ⇒y = ±1 Therefore, if x = 0, (1) and (3) can only be true at the same time if y = ±1. But then note that: (2), x = 0, and y = 1 ⇒4 = 2λ ⇒λ = 2 and (2), x = 0, and y = −1 ⇒−4 = −2λ ⇒λ = 2 Example, cont.
Therefore, if x = 0, then all three equations are true at the same time: at the point (0, 1) with λ = 2 or at the point (0, −1) with λ = 2 We will hang onto the points (0, 1) and (0, −1) for later.
Example, cont.
Now we investigate the case where λ = 1. Note that: (2) and λ = 1 ⇒4y = 2y ⇒2y = 0 ⇒y = 0 Therefore, if λ = 1, (1) and (2) can only be true at the same time if y = 0. But then note that: (3), λ = 1, and y = 0 ⇒x2 = 1 ⇒x = ±1 Example, cont.
Therefore, if λ = 1, then all three equations are true at the same time: at the point (1, 0) with λ = 1 or at the point (−1, 0) with λ = 1 Example, cont.
At this point, we’ve found all possible solutions to this system of equations, as we’ve exhausted all the cases we discovered. Indeed, we found that for the first equation to be true, it must be that either x = 0 or λ = 1, and then we investigated which possible values of y and λ; and x and y, respectively, can satisfy the rest of the system in each case.
You may find it somewhat challenging to keep track of which case you’re working on at a given time, especially if the number of cases grows large; and you may find it especially challenging to know when you’ve finished investigating all of the cases you found. To that end, you may find a tree like the one on the following slide helpful.
Example, cont.
Cases x = 0 y = 1 λ = 2 y = −1 λ = 2 λ = 1 y = 0 x = 1 x = −1 This is a chart that you can construct as you go through solving the system of equations. For example, in this problem we would proceed as follows: the first equation tells us that for all equations in the system to be true simultaneously, we must have that either x = 0 or λ = 1. So, we add a node below “Cases” for each of these. Then, when we investigate the x = 0 case, we see that y must either be 1 or −1 when x = 0, so we add these nodes below x = 0, etc. When all nodes terminate with a value of x, y, and λ, you’re finished!
Example, cont.
Let’s finish up the problem. In solving the system of equations above, we obtained the following points: (0, 1), (0, −1), (1, 0), and (−1, 0) We now plug all of these into f (x, y). The method of Lagrange multipliers tells us that the largest value we get is the absolute maximum value of f (x, y) on the circle x2 + y 2 = 1; and the smallest value we get is the absolute minimum value of f (x, y) on the circle x2 + y 2 = 1.
Example, cont.
We have: f (0, 1) = 02 + 2 · 12 = 2 f (0, −1) = 2 f (1, 0) = 1 f (−1, 0) = 1 Therefore, the absolute maximum value of f (x, y) on the unit circle is f (0, 1) = f (0, −1) = 2, and the absolute minimum value of f (x, y) on the unit circle is f (1, 0) = f (−1, 0) = 1.
Typical Challenges As we got a peek at in the previous example, the most complicated portion of using the method of Lagrange multipliers is typically solving the system of (generally non-linear) equations which arises from the equations ∇f = λ∇g and g = k. In particular, the challenge often lies in keeping track of a number of cases.
Below we examine another example, and give a handy technique that can be used to work one’s way through a number of the typical systems of equations that arise in these problems.
Example Find the maximum volume of a rectangular box with no lid constructed from 12m2 of cardboard.
You may remember this problem from §14.7. Now we will use the method of Lagrange multipliers to obtain the same solution we did there.
We begin, again, by sketching and labelling the box.
Example Once again, we wish to calculate the maximum volume of this box. Given our labels in the diagram, this means that we wish to find the maximum value of the three-variable function: V (x, y, z) := xyz subject to the constraint that the box must constructed from exactly 12m2 of cardboard. That is, the surface area of the box (which, again, has no top) must be exactly 12m2. Given our diagram labels on the previous slide, this means that x, y, and z must satisfy the constraint: xy + 2xz + 2yz = 12 Ah! We see from this setup that the method of Lagrange multipliers could be used to solve this problem. So... let’s give it a try!
Example, cont.
We begin by giving a name to the non-constant portion of the constraint equation. We’ll stick with g(x, y, z) for our name, since that’s the one used in the statement of the method of Lagrange multipliers. That is, we declare: g(x, y, z) := xy + 2xz + 2yz With this squared away, the method of Lagrange multipliers says that to calculate the absolute maximum and minimum values of V (x, y, z) subject to the constraint g(x, y, z) = 12, we must first find all ordered triples (x, y, z) and constants λ such that the equations: ∇V (x, y, z) = ⟨yz, xz, xy⟩= λ ⟨y + 2z, x + 2z, 2x + 2y⟩= λ∇g(x, y, z) and xy + 2xz + 2yz = 12 are true at the same time.
Example, cont.
By comparing the components of the vectors (more precisely, vector fields) of the first equation on the previous slide, we arrive at a more workable set of equations: yz = λ(y + 2z) (1) xz = λ(x + 2z) (2) xy = λ(2x + 2y) (3) xy + 2xz + 2yz = 12 (4) Thus, our goal is to find all constants λ and all ordered triples (x, y, z) that satisfy all three of these equations at once.
Take a few minutes, and see if you can solve this system of equations on your own before moving on.
Example, cont.
There are many ways to solve any system of equations, but here’s one technique that you may not have seen before, and that I find particularly useful in this case (and in many others).
Note that the left-hand sides of the first three equations are all very nearly the same, with one variable missing in each. So, we’ll start by multiplying each of these by its “missing” variable to obtain the following modified set of equations: xyz = λx(y + 2z) (1′) xyz = λy(x + 2z) (2′) xyz = λz(2x + 2y) (3′) xy + 2xz + 2yz = 12 (4) The list of equations on the previous slide are all true at the same time precisely when this list of equations is. Therefore, we will work with this modified list.
Example, cont.
Now would be an excellent time to remind ourselves of the values that x, y, and z can take. All three variables represent edge lengths, and therefore can take only positive values in order to remain physically meaningful. There is, of course, another major restriction on all three (the constraint equation), but at the very least it’s good to know that any cases we come across, or any ordered triples we find that involve negative values of x, y, or z may be discarded immediately.
Now let’s proceed in solving the system on the previous slide.
Example, cont.
The left-hand side of the first three equations being the same provides a major advantage in how we may proceed, as this means that the right-hand sides of these equations must also be the same. In particular: (1′) and (2′) ⇒λx(y + 2z) = λy(x + 2z) ⇒λxy + 2λxz = λxy + 2λyz ⇒2λz(x −y) = 0 Therefore, if (1′) and (2′) are true at the same time, we must have at least one of: λ = 0, z = 0, or x = y Example, cont.
Let’s examine each case.
First, since z must be a positive number (as, again, it’s the length of one of the edges of the box), we can immediately discard the case z = 0.
Similarly, consider the case λ = 0. Plugging λ = 0 into, say, (1′) gives: xyz = 0 which in turns implies that at least one of x, y, and z is zero. So, again, we can discard this case, as all three of these numbers must be positive.
Thus, the only way that (1′) and (2′) can be true at the same time is if x = y.
Example, cont.
With this new understanding in mind, let’s proceed in a similar way with equations (1′) and (3′): (1′), (3′), and x = y ⇒λx(x + 2z) = λz(4x) ⇒λx2 + 2λxz = 4λxz ⇒λx2 = 2λxz ⇒xλ(x −2z) = 0 Therefore, if (1′), (2′), and (3′) are true at the same time, we must have that: x = 0 or x = 2z Example, cont.
Just as before, we can immediately discard the case x = 0, since x must be a positive number.
Therefore, combining all of our above work, the only way that (1′), (2′), and (3′) can be true at the same time is if: x = y and x = 2z We’re now in a great position to find all of the ordered triples (x, y, z) and values of λ that satisfy all four equations at the same time.
Example, cont.
First, notice that: x = y, x = 2z, and (4) ⇒x2 + x2 + x2 = 12 ⇒x2 = 4 ⇒x = ±2 Since x cannot be negative, this means that the only x-value that can satisfy all four equations at once is x = 2. For this value of x, we also have y = 2 and z = 1. Further: x = 2, y = 2, z = 1, and (1′) ⇒4 = 8λ ⇒λ = 1 2 Thus, the only way that all four equations in our system can be true at the same time is if (x, y, z) = (2, 2, 1) and λ = 1 2 Example, cont.
To be certain that we’ve covered all possible cases, we might make a tree like the following as we go: Cases λ = 0 z = 0 x = y x = 0 x = 2z x = 2 y = 2 z = 1 λ = 1 2 Example, cont.
Therefore, by the method of Lagrange multipliers, the absolute maximum volume of a rectangular box with no lid, constructed from 12m2 of material is V (2, 2, 1) = 4m3 Comments The previous examples demonstrate both the utility and the pitfalls of the method of Lagrange multipliers. Indeed, the first example was fairly efficient to solve using this method whereas some cleverness is required to solve it using the methods of the previous section (see the first exercise at the end of the section). On the other hand, the method of the previous section fairly efficiently solved our optimization problem, whereas the method of Lagrange multipliers required a bit of cleverness, as well as careful tracking of a number of cases, to solve a system of four equations.
This is a general feature of these two methods. The method of Lagrange multipliers trades the potential complexity of solving a slightly longer system of equations for the potential complexity of computing a list of partial derivatives and setting them equal to zero simultaneously.
Comments, cont.
All of this is to say that both methods have their utility, and you should bear both in mind when trying to find absolute extrema on a closed, bounded set; or when trying to solve optimization problems. Thoughtful practice and experience will help you to decide which will prove more useful for a given problem.
Table of Contents The Method of Lagrange Multipliers Exercises Exercises 1. Use the methods from the previous section to solve the first example, above. Which do you find easier in this case: those methods, or the method of Lagrange multipliers?
2. Find the extreme values of the function f (x, y) = 3x + y on the circle x2 + y 2 = 10 two ways: first, by using Lagrange multipliers, and second, by using the methods of the previous section. Which do you prefer?
3. Find the extreme values of h(x, y) = x2 + 2y 2 on the unit disk x2 + y 2 ≤1. At which step in the solution can the method of Lagrange multipliers be used?
4. Find the points on the sphere x2 + y 2 + z2 = 4 that are closest to and farthest from the point (3, 1, −1).
Solutions 2. The absolute minimum value of f (x, y) on the circle x2 + y 2 = 10 is f (−3, −1) = −10; and the absolute maximum value of f (x, y) on this same circle is f (3, 1) = 10.
3. The absolute minimum value of h(x, y) on the unit disk is h(0, 0) = 0; and the absolute maximum value of h(x, y) on the unit disk is h(0, 1) = h(0, −1) = 2. The method of Lagrange multipliers can only help us find the absolute maximum and minimum values of h(x, y) on the boundary of the disk, i.e. on the unit circle.
4. The point on the sphere x2 + y 2 + z2 = 4 which is closest to the point (3, 1, −1) is 6 √ 11, 2 √ 11, −2 √ 11 ; and the point on this same sphere farthest from the point (3, 1, −1) is −6 √ 11, −2 √ 11, 2 √ 11 . |
187538 | https://asphalt.fandom.com/wiki/Urn_problem | Urn problem | Asphalt Wiki | Fandom
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In probability theory and statistics, an urn problem is an idealized thought experiment in which some objects of real interest (such as atoms, people, cars, etc.) are represented as colored balls in an urn or other container. One pretends to remove one or more balls from the urn; the goal is to determine the probability of drawing one color or another, or some other properties. A number of important variations are described below.
An urn model is either a set of probabilities that describe events within an urn problem, or it is a probability distribution, or a family of such distributions, of random variables associated with urn problems.
Basic urn model[]
In this basic urn model in probability theory, the urn contains x white and y black balls, well-mixed together. One ball is drawn randomly from the urn and its color observed; it is then placed back in the urn (or not), and the selection process is repeated.
Possible questions that can be answered in this model are:
Can I infer the proportion of white and black balls from n observations? With what degree of confidence?
Knowing x and y, what is the probability of drawing a specific sequence (e.g. one white followed by one black)?
If I only observe n balls, how sure can I be that there are no black balls? (A variation on the first question)
Example[]
One example of an urn problem is the binomial distribution: the distribution of the number of successful draws (trials), i. e. extraction of white balls, given n draws. This is of great importance for players because basically all in-game random processes involving random containers like Pro Kit Boxes or Card Packs are realizations of the urn model: Instead of balls, cards are drawn from the container. Obtaining a desired card (e.g. a blueprint or an engine card) is a success, not obtaining it is a failure.
See also[]
Experiment
Binomial distribution
Bernoulli process
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Probability theory
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187539 | https://en.wikipedia.org/wiki/K-outerplanar_graph | Jump to content
k-outerplanar graph
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From Wikipedia, the free encyclopedia
In graph theory, a k-outerplanar graph is a planar graph that has a planar embedding in which the vertices belong to at most concentric layers. The outerplanarity index of a planar graph is the minimum value of for which it is -outerplanar.
Definition
[edit]
An outerplanar graph (or 1-outerplanar graph) has all of its vertices on the unbounded (outside) face of the graph. A 2-outerplanar graph is a planar graph with the property that, when the vertices on the unbounded face are removed, the remaining vertices all lie on the newly formed unbounded face. And so on.
More formally, a graph is -outerplanar if it has a planar embedding such that, for every vertex, there is an alternating sequence of at most faces and vertices of the embedding, starting with the unbounded face and ending with the vertex, in which each consecutive face and vertex are incident to each other.
Properties and applications
[edit]
The -outerplanar graphs have treewidth at most . However, some bounded-treewidth planar graphs such as the nested triangles graph may be -outerplanar only for very large , linear in the number of vertices.
Baker's technique covers a planar graph with a constant number of -outerplanar graphs and uses their low treewidth in order to quickly approximate several hard graph optimization problems.
In connection with the GNRS conjecture on metric embedding of minor-closed graph families, the -outerplanar graphs are one of the most general classes of graphs for which the conjecture has been proved.
A conjectured converse of Courcelle's theorem, according to which every graph property recognizable on graphs of bounded treewidth by finite state tree automata is definable in the monadic second-order logic of graphs, has been proven for the -outerplanar graphs.
Recognition
[edit]
The smallest value of for which a given graph is -outerplanar (its outerplanarity index) can be computed in quadratic time.
References
[edit]
^ Bodlaender, Hans L. (1998), "A partial -arboretum of graphs with bounded treewidth", Theoretical Computer Science, 209 (1–2): 1–45, doi:10.1016/S0304-3975(97)00228-4, hdl:1874/18312, MR 1647486
^ Baker, B. (1994), "Approximation algorithms for NP-complete problems on planar graphs", Journal of the ACM, 41 (1): 153–180, doi:10.1145/174644.174650, S2CID 9706753.
^ Chekuri, Chandra; Gupta, Anupam; Newman, Ilan; Rabinovich, Yuri; Sinclair, Alistair (2006), "Embedding -outerplanar graphs into ", SIAM Journal on Discrete Mathematics, 20 (1): 119–136, doi:10.1137/S0895480102417379, MR 2257250, S2CID 13925350
^ Jaffke, Lars; Bodlaender, Hans L.; Heggernes, Pinar; Telle, Jan Arne (2017), "Definability equals recognizability for -outerplanar graphs and -chordal partial -trees" (PDF), European Journal of Combinatorics, 66: 191–234, doi:10.1016/j.ejc.2017.06.025, MR 3692146
^ Kammer, Frank (2007), "Determining the smallest such that is -outerplanar", in Arge, Lars; Hoffmann, Michael; Welzl, Emo (eds.), Algorithms: ESA 2007, 15th Annual European Symposium, Eilat, Israel, October 8-10, 2007, Proceedings, Lecture Notes in Computer Science, vol. 4698, Springer, pp. 359–370, doi:10.1007/978-3-540-75520-3_33
Retrieved from "
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187540 | https://math.stackexchange.com/questions/4472169/given-that-fx-xk-where-k-0-what-is-the-range-of-fx-on-the-interv | functions - Given that $f(x) = x^k$ where $k < 0$, what is the range of $f(x)$ on the interval $1, \infty)$? - Mathematics Stack Exchange
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Given that f(x)=x k f(x)=x k where k<0 k<0, what is the range of f(x)f(x) on the interval [1,∞)1,∞)?
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Not sure how to solve this question. I set f(x)f(x) as y y and tried rewriting x x as y√k y k – then break it up into cases where x>0 x>0 and x<0 x<0. But that didn't exactly work out since it got a little confusing... A bit help? Also, edit: When they say, [1,∞)[1,∞), do they mean the domain or range or something else?
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edited Jun 14, 2022 at 18:06
user1043968
user1043968
asked Jun 14, 2022 at 5:30
user1043968 user1043968
8
1 How can x x be negative when it is given to be in [1,∞)[1,∞)?Kavi Rama Murthy –Kavi Rama Murthy 2022-06-14 05:33:35 +00:00 Commented Jun 14, 2022 at 5:33
You can ignore my first attempt – I was pretty sure it was wrong anyways, sorry about that.user1043968 –user1043968 2022-06-14 05:34:29 +00:00 Commented Jun 14, 2022 at 5:34
The answer is (0,1, x=1 y√k 1 y k. S i n c e x∈[1,+∞)S i n c e x∈[1,+∞) we obtain y>0 y≤1 y>0 y≤1 and hence Range =(0,1].
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answered Jun 14, 2022 at 7:26
user1054388 user1054388
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Just plug it in: Since k<0,if you plug a 1 in, f(x) = 1. If x is super large,the output approaches 0 but never reaches it. So the answer is (0, 1]
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answered Nov 28, 2022 at 2:31
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2.9
Rotation Symmetry
Written by:CK-12 | Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 01, 2025
Lesson
Rotation Symmetry
A shape has symmetry if it can be indistinguishable from its transformed image.
A shape has rotation symmetry if there exists a rotation less than that carries the shape onto itself.
If you can rotate a shape less than about some point and the shape looks like it never moved, it has rotation symmetry.
There are 3 ways to name rotation symmetry.
Order: The order of rotation symmetry of a geometric figure is the number of times you can rotate the geometric figure so that it looks exactly the same as the original figure before you get back to where you started.
In the lefthand image below, the shape has two positions that are indistinguishable, so it has rotation symmetry of order 2.
On the right, the shape has the positions that are indistinguishable, so it has rotation symmetry of order 3.
Fraction of a turn:If you look at how far you turn the shape to get it to look the same, you can think of that as a fraction of how far you go to get all the way around.
On the left, you turn way around, so it has turn symmetry.
On the right, it turns of the way around to look like itself again, so it has turn symmetry.
Angle: The angle of rotation symmetry is the smallest angle the figure can be rotated to coincide with itself. So the half turn becomes half of and the one-third turn becomes a third of
The shape on the left has rotation symmetry.
The shape on the right has rotation symmetry.
Identifying Rotation Symmetry
A rectangle is an example of a shape with rotation symmetry.
A rectangle can be rotated about its center and it will look exactly the same and be in the same location. The only difference is the location of the named points.
A rectangle has half-turn symmetry, and therefore is order 2.
Does a square have rotation symmetry?
Yes, a square can be rotated counterclockwise (or clockwise) about its center and the image will be indistinguishable from the original square.
Identifying Angles of Rotation
How many angles of rotation cause a square to be carried onto itself?
Rotations of andin either direction will all cause the square to be carried onto itself.
A square has quarter-turn rotation symmetry, and so has an order of 4.
Rotation Symmetry of a Trapezoid
Consider rotation symmetry in trapezoids.
In a generic trapezoid as well as isosceles trapezoid, there is no rotation symmetry.
A trapezoid must be rotated a full to again appear in its original position.
Rotation Symmetry of a Circle
Rotate a circle about its center through any angle and it fits onto itself.
A circle has rotation symmetry around the center for every angle.
A circle has an unlimited number of angles of symmetry and the order of its rotation is infinite.
Rotation Symmetry
Drag the cursor to rotate the polygon and observe the angle of rotation.
Examples
Example 1
What happens when you rotate the regular pentagon below clockwise about its center? Why is special in this case?
When you rotate the regular pentagon about its center, it will look exactly the same. This is because the regular pentagon has rotation symmetry, and is the minimum number of degrees you can rotate the pentagon in order to carry it onto itself.
Example 2
Does each capital letter below have rotation symmetry? If so, state the angles of rotation that carry the letter onto itself.
Capital letter N
Yes, it does have rotation symmetry. It can be rotated
Capital letter S
Yes, it does have rotation symmetry. It can be rotated
Capital letter H
Yes, it does have rotation symmetry. It can be rotated
Capital letter B
No, it does not have rotation symmetry.
CK-12 PLIX Interactive
| |
| Summary |
| Rotation symmetry occurs when a shape can be rotated less than 360° and still look the same. There are three ways to name rotation symmetry: order, fraction of a turn, and angle. A circle has infinite rotation symmetry, as it can be rotated around its center through any angle and still fit onto itself. |
Review
What does it mean for a shape to have symmetry?
What does it mean for a shape to have rotation symmetry?
Why does the stipulation of less than exist in the definition of rotation symmetry?
For each of the following shapes, state whether or not it has rotation symmetry. If it does, state the number of degrees you can rotate the shape to carry it onto itself.
Equilateral triangle
Isosceles triangle
Scalene triangle
Parallelogram
Rhombus
Regular pentagon
Regular hexagon
Regular 12-gon
Regular -gon
Circle
Kite
Where will the center of rotation always be located for shapes with rotation symmetry?
Does every polygon that has rotation symmetry also have reflection symmetry? Why or why not?
Does every polygon that has reflection symmetry also have rotation symmetry? Why or why not?
Does a line have rotation symmetry? How about reflection symmetry? Explain.
Does an angle have rotation symmetry? How about reflections symmetry? Explain?
Harold argues that a semi-circle has rotation symmetry, but Jay disagrees. Who is correct and why?
Tina argues that our reflections in the mirror show that humans have rotational symmetry, the image in a mirror is a rotation of the original. Brenda disagrees. Who is correct and why?
Vanessa wants a house with a floor plan that has reflection symmetry, while Jessica prefers one that has rotation symmetry. Draw sketches of each. Which do you prefer and why?
Many animals, plants, or other objects in the world have reflection or rotation symmetry. Give examples and describe the symmetry in as much detail as possible.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
Rotation symmetry occurs when a shape can be rotated less than 360° and still look the same.
There are three ways to name rotation symmetry: order, fraction of a turn, and angle.
A circle has infinite rotation symmetry, as it can be rotated around its center through any angle and still fit onto itself.
Vocabulary
Rotation
Point
Angle
rectangle
center
square
Trapezoid
isosceles trapezoid
circle
Polygon
Pentagon
Test Your Knowledge
Question 1
Which picture has rotational symmetry?
a
b
c
With rotational symmetry, a shape or image can be rotated and it still looks the same.
Neither the star or arrow has rotational symmetry, only double-curve does.
Question 2
Find the angle of rotation for a regular pentagon.
a
60 degrees
b
108 degrees
c
120 degrees
d
72 degrees
A regular pentagon has rotational symmetry of 5.
The angle of rotation is @$\begin{align}\frac{360^{\circ}}{n}\end{align}@$.
@$$\begin{align}\frac{360^{\circ}}{n} = \frac{360^{\circ}}{5} = 72^{\circ}\end{align}@$$
The angle of rotation for a regular pentagon is 72 degrees.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Rotation Symmetry Principles - Basic
Rotation Symmetry Examples - Basic
This End Up
Shapes for Rotational Symmetry
| Image | Reference | Attributions |
---
| | | Credit: CK 12 Source: CK 12 License: CC BY-NC 3.0 |
| | | License: CC BY-NC |
| | | Credit: Melissa Sanders |
| | | License: CC BY-NC |
| | | License: CC BY-NC |
| | | Credit: Brady Holt Source: License: CC BY-NC |
| | | License: CC BY-NC |
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Square – Definition, Properties, Examples, Facts, FAQs
Home » Math Vocabulary » Square – Definition, Properties, Examples, Facts, FAQs
What is a Square in Math?
Properties of a Square
Solved Examples On Square
Practice Problems On Square
Frequently Asked Questions On Square
Square – Introduction
Take a look at the images given below. You might have come across objects like a photo frame, or a craft paper in day-to-day life. Can you identify what is common in them?
All of them have a square shape.
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Definition of Square in Math
A square is a regular polygon having four equal sides and equal angles that measure 90° each.
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What is a Square in Math?
A square is a two-dimensional closed shape with 4 equal sides and 4 vertices. Its opposite sides are parallel to each other. We can also think of a square as a rectangle with equal length and breadth.
Looking around, you can find many things that resemble the square shape. Common examples of this shape include a chessboard, craft papers, bread slice, photo frame, pizza box, a wall clock, etc.
Properties of a Square
It has 4 sides and 4 vertices.
Its sides are equal in length.
All interior angles are equal and right angles, which means that each angle measures 90°.
The sum of all the interior angles is 360°.
Its two diagonals bisect each other at right angles.
Area and Perimeter of a Square
Area represents space occupied by a shape or figure whereas perimeter is the length of the outer boundary of the shape. Let’s discuss the formula for finding the area and perimeter of a square.
Area
The area of a two-dimensional shape is defined as the amount of space covered by the shape if we were to keep it on a flat table.
For a square of side length “s” units, the area is given by the formula:
Area $= \text{side} \times \text{side} = \text{S}^2$
The area is expressed in square units, such as $\text{cm}^2$, $\text{cm}^2$, etc.
Perimeter
The perimeter of a two-dimensional shape is defined as the total length of its boundary.
For a square of side length “s” units, the perimeter is given by the formula:
Perimeter $= \text{side} + \text{side} + \text{side} + \text{side} = 4$ $\text{x}$ $\text{s}$
The perimeter is expressed in linear units, such as cm, inches, m, etc.
Solved Examples On Square
Example 1: The side of a square paper is 12 feet. Find the area of the paper.
Solution:
We know that the area of a square is given by $\text{s}^2$, where $\text{s} =$ length of the side.
For the given square, s $= 12$ feet
Therefore, the area of the square paper is given by:
Area $= \text{s}^2 = 12 \times 12 = 144$ sq. ft.
Example 2: If the perimeter of a square measures 68 cm, what is the measure of its side?
Solution:
We know that the perimeter of a square is given by 4 x side.
It is given that the perimeter is 68 cm.
Therefore, 4 x side $= 68$
Which means, side $= \frac{68}{4} = 17$ cm
Example 3: What is the perimeter of a square that has a side of 15 meters?
Solution: We know that the perimeter of a square is given by 4 x s, where s represents the length of each side.
It is given that the side s $= 15$ meters.
Therefore, perimeter $= 4 \times 15 = 60$ meters.
Practice Problems On Square
Square
Attend this quiz & Test your knowledge.
1
If the side of a square chess board is 20 cm in length, find the board's perimeter.
80 cm
170 cm
200 cm
150 cm
CorrectIncorrect
Correct answer is: 80 cmWe know that the perimeter of a square is given by 4 x s, where s represents the length of each side. It is given that the side s $= 20$ cm. Therefore, the perimeter $= 4 x 20 = 80$ cm.
2
What is the measure of an interior angle of a square?
60 degrees
360 degrees
90 degrees
120 degrees
CorrectIncorrect
Correct answer is: 90 degreesSince all the interior angles of a square are right angles, the measure of each one of them is 90 degrees.
3
What is the area of a square with a side length of 25 cm?
625 sq cm
500 sq cm
125 sq cm
None of the above
CorrectIncorrect
Correct answer is: 625 sq cmWe know that the area of a square is given by $\text{s}^2$, where $\text{s} =$ length of the side. It is given that the side $\text{s} = 25$ cm. Therefore, Area $= \text{s}^2 = 25 \times 25 = 625$ sq. cm.
Frequently Asked Questions On Square
How to identify a polygon as a square?
A polygon that is made of four equal sides, with all the interior angles measuring 90 degrees, is a square.
Is the side of a square and its diagonal the same length?
No, the side of a square and its diagonal aren’t of the same length. The diagonal of a square is greater in length than its side.
If two squares have the same perimeter, will they have the same area?
Yes, if two squares have the same perimeter, it means that they also have sides of the same length. This, in turn, implies that they also have the same area.
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187544 | https://medlineplus.gov/ency/article/001438.htm | Skip navigation
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Polyarteritis nodosa
URL of this page: //medlineplus.gov/ency/article/001438.htm
Polyarteritis nodosa
Polyarteritis nodosa is a serious inflammatory blood vessel disease. The small and medium-sized arteries become swollen and damaged.
Causes
Arteries are the blood vessels that carry oxygen-rich blood to organs and tissues. The cause of polyarteritis nodosa is unknown. The condition occurs when certain immune cells attack the affected arteries. The tissues that are fed by the affected arteries do not get the oxygen and nourishment they need. Damage occurs as a result.
More adults than children get this disease.
People with active hepatitis B or hepatitis C may develop this disease.
Symptoms
Symptoms are caused by damage to affected organs. The skin, joints, muscles, gastrointestinal tract, heart, kidneys, and nervous system are often affected.
Symptoms include:
Abdominal pain
Decreased appetite
Fatigue
Fever
Joint aches
Muscle aches
Unintentional weight loss
Weakness
If nerves are affected, you may have numbness, pain, burning, and weakness. Damage to the nervous system may cause strokes or seizures.
Exams and Tests
No specific lab tests are available to diagnose polyarteritis nodosa. There are a number of disorders that have features similar to polyarteritis nodosa. These are known as "mimics."
You will have a complete physical exam.
Lab tests that can help make the diagnosis and check for disease mimics include:
Complete blood count (CBC) with differential, creatinine, tests for hepatitis B and C, and urinalysis
Erythrocyte sedimentation rate (ESR) or C-reactive protein (CRP)
Serum protein electrophoresis
Serum complement levels
HIV test
Hepatitis B and hepatitis C tests
Cryoglobulins
Anti-phospholipid antibodies
Blood cultures
Other blood tests will be done to check for similar conditions, such as systemic lupus erythematosus (ANA) or granulomatosis with polyangiitis (ANCA)
Arteriogram
Tissue biopsy
Treatment
Treatment involves medicines to suppress inflammation and the immune system. These may include steroids, such as prednisone. Similar medicines, such as azathioprine, methotrexate or mycophenolate that allow for reducing the dose of steroids are often used as well. Cyclophosphamide is used in severe cases.
For polyarteritis nodosa related to hepatitis, treatment may involve plasmapheresis and antiviral medicines.
Outlook (Prognosis)
Current treatments with steroids and other medicines that suppress the immune system (such as azathioprine or cyclophosphamide) can improve symptoms and the chance of long-term survival.
The most serious complications most often involve the kidneys and gastrointestinal tract.
Without treatment, the outlook is poor.
Possible Complications
Complications may include:
Heart attack
Intestinal necrosis and perforation
Kidney failure
Stroke
When to Contact a Medical Professional
Contact your health care provider if you develop symptoms of this disorder. Early diagnosis and treatment may improve the chance of a good outcome.
Prevention
There is no known prevention. However, early treatment can prevent some damage and symptoms.
Alternative Names
Periarteritis nodosa; PAN; Systemic necrotizing vasculitis
Images
Artery cut section
Microscopic polyarteritis 2
Circulatory system
References
Luqmani R, Awisat A. Polyarteritis nodosa and related disorders. In: Firestein GS, Budd RC, Gabriel SE, Koretzky GA, McInnes IB, O'Dell JR, eds. Firestein & Kelley's Textbook of Rheumatology. 11th ed. Philadelphia, PA: Elsevier; 2021:chap 95.
Shanmugam VK. Vasculitis and other uncommon arteriopathies. In: Sidawy AN, Perler BA, eds. Rutherford's Vascular Surgery and Endovascular Therapy. 10th ed. Philadelphia, PA: Elsevier; 2023:chap 138.
Stone JH. The systemic vasculitides. In: Goldman L, Cooney KA, eds. Goldman-Cecil Medicine. 27th ed. Philadelphia, PA: Elsevier; 2024:chap 249.
Review Date 5/9/2024
Updated by: Neil J. Gonter, MD, Assistant Professor of Medicine, Columbia University, NY and private practice specializing in Rheumatology at Rheumatology Associates of North Jersey, Teaneck, NJ. Review provided by VeriMed Healthcare Network. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.
Related MedlinePlus Health Topics |
187545 | https://www.mathbootcamps.com/calculating-the-mean/ | Calculating the mean - MathBootCamps
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Calculating the mean
The mean is a way of measuring the center of a data set. That is, it is a way of trying to describe the typical data value. For symmetric data sets, it does a good job of this. But, for skewed data sets or data sets with outliers it can be a bit misleading. Before we see how it is calculated, let’s talk about notation.
[adsenseWide]
Notation
The mean is represented in two different ways, depending on whether or not it represents the mean of a sample or the mean of a population. Both are calculated the same, and the difference between the two is only important in some settings.
| Population Mean | Sample Mean |
--- |
| (This is the Greek letter Mu. Read this as “mew”.) | (Read this as “x-bar”.) |
For this guide, we will assume that we are working with sample data, so we will use .
Calculation Using the Formula
When you learned how to find the average, you were likely taught the arithmetic average. This is where you add up all the values and then divide by however many values were in the data set. The mean is calculated in the exact same way.
Example
Find the mean of the data set below.
10 12 11 8 14 12
Using the idea above:
So, the mean for this data set is approximately 11.2. Looking at the original data, this does a good job of describing the center of this data set.
Calculation Using a TI83 or TI84 Graphing Calculator
The mean can easily be found using the function 1varstats on Ti83/84 graphing calculators. Here, we will go through the steps one by one. If you want to see a video of how to do this, scroll down to the bottom (or click here)!
Step 1: Enter your data in L1
To enter data in your calculator, press [STAT] and then go to 1: Edit by pressing [ENTER] or .
Now to enter the data, type each number and press enter. Note that if you already have data in your list, highlight the very top where it says L1 and press [CLEAR] followed by [ENTER].
Step 2: Calculate 1-var-stats
Once your data is in the list, press [STAT] again and then go to the CALC menu. From here, choose 1: 1varstats.
Now press enter twice and you will get a list of summary statistics. The first value that comes up is the mean!
Note: if you have a newer calculator, the menu looks a little different now. Instead of pressing enter twice, you will press enter and then have a 1-var stats menu come up. Just click CALCULATE at this menu and you will have the same information come up as you see above.
As you can see we get the exact same value as we did above. This is very nice for working with larger data sets.
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187546 | https://simple.wikipedia.org/wiki/Carboxylic_acid | Carboxylic acid - Simple English Wikipedia, the free encyclopedia
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From Simple English Wikipedia, the free encyclopedia
General structure of a carboxylic acid
A carboxylic acid is any molecule that has a COOH group. This group contains a carbonyl (C=O double bond) together with an hydroxyl group (OH) on the same carbon atom. Because it is easy to remove the proton with even a weak base, these compounds are called acids. An example of a carboxylic acid is acetic acid, which is also known as vinegar.
Carboxylic acids are found a lot in food. Many types of fat molecules are actually carboxylic acids. For example, chocolate and coconuts have these acids. They are also used a lot in soaps and detergents.
The smaller carboxylic acids dissolve in water. The bigger ones dissolve only in organic solvents.
List
[change | change source]
This section lists the first few carboxylic acids.
Straight-chain, saturated carboxylic acids | Carbon atoms | Common name | IUPAC name | Chemical formula | Common location or use |
--- ---
| 1 | Carbonic acid | Carbonic acid | OHCOOH | Blood and tissues (bicarbonate buffer system) |
| 1 | Formic acid | Methanoic acid | HCOOH | Insect stings |
| 2 | Acetic acid | Ethanoic acid | CH 3 COOH | Vinegar |
| 3 | Propionic acid | Propanoic acid | CH 3 CH 2 COOH | Preservative for stored grains, body odour |
| 4 | Butyric acid | Butanoic acid | CH 3(CH 2)2 COOH | Butter |
| 5 | Valeric acid | Pentanoic acid | CH 3(CH 2)3 COOH | Valerian |
| 6 | Caproic acid | Hexanoic acid | CH 3(CH 2)4 COOH | Goat fat |
| 7 | Enanthic acid | Heptanoic acid | CH 3(CH 2)5 COOH | |
| 8 | Caprylic acid | Octanoic acid | CH 3(CH 2)6 COOH | Coconuts |
| 9 | Pelargonic acid | Nonanoic acid | CH 3(CH 2)7 COOH | Pelargonium |
| 10 | Capric acid | Decanoic acid | CH 3(CH 2)8 COOH | Coconut and Palm kernel oil |
| 11 | Undecylic acid | Undecanoic acid | CH 3(CH 2)9 COOH | |
| 12 | Lauric acid | Dodecanoic acid | CH 3(CH 2)10 COOH | Coconut oil and hand wash soaps |
| 13 | Tridecylic acid | Tridecanoic acid | CH 3(CH 2)11 COOH | |
| 14 | Myristic acid | Tetradecanoic acid | CH 3(CH 2)12 COOH | Nutmeg |
| 15 | Pentadecylic acid | Pentadecanoic acid | CH 3(CH 2)13 COOH | |
| 16 | Palmitic acid | Hexadecanoic acid | CH 3(CH 2)14 COOH | Palm oil |
| 17 | Margaric acid | Heptadecanoic acid | CH 3(CH 2)15 COOH | |
| 18 | Stearic acid | Octadecanoic acid | CH 3(CH 2)16 COOH | Chocolate, waxes, soaps, and oils |
| 19 | Nonadecylic acid | Nonadecanoic acid | CH 3(CH 2)17 COOH | Fats, vegetable oils, pheromone |
| 20 | Arachidic acid | Icosanoic acid | CH 3(CH 2)18 COOH | Peanut oil |
This short article about chemistry can be made longer. You can help Wikipedia by adding to it.
| Expand v t e Functional groups |
| Hydrocarbons (only C and H) | Alkene Vinyl Allyl 1-Propenyl Crotyl Alkyl Methyl Ethyl Propyl Butyl Pentyl Alkyne Carbene Methine Allene Benzyl Cumulene Methylene bridge Methylene group Phenyl |
| Only carbon, hydrogen, and oxygen (only C, H and O) | | R-O-R | Acetal Alcohol Alkoxy Methoxy Ether Enol ether Epoxide Peroxy Hydroperoxy Dioxiranes Ethylenedioxy Methylenedioxy | | carbonyl | Aldehyde Ketone Acyl Acetyl Acryloyl Benzoyl Ynone | | carboxy | Carboxyl Acetoxy Carboxylic anhydride Ester Ortho ester | |
| Only one element, not being carbon, hydrogen, or oxygen (one element, not C,H or O) | | Nitrogen | Amine Hydrazone Nitrate Nitrile Nitro Azo Carbamate Cyanate Imide Imine Isocyanate Isonitrile Nitrene Nitroso Nitrosooxy Amide Oxime | | Phosphorus | Phosphonate Phosphonous | | Sulfur | Sulfone Sulfoxide Thial Thioester Thioketone Thiol Disulfide Persulfide Sulfo Sulfonic acid Thionoester Sulfide Sulfino Sulfinyl Sulfonyl Thionyl Thiosulfinate Thiosulfonate Thioxanthate Xanthate | | Selenium | Selenol Selenonic acid Seleninic acid Selenenic acid Selone | | Tellurium | Tellurol Telluroketone | | halo | Fluoroethyl | |
| Other | Sulfonamide Isothiocyanate Phosphoramides Sulfenyl chloride Thiocyanate |
Retrieved from "
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Carboxylic acid
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187547 | https://www.youtube.com/watch?v=l_w8XT61wPo | ¿Cómo calcular la velocidad inicial y la aceleración? | MRUV
AulaNerd
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Description
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Posted: 22 Jul 2020
En este video te explico cómo resolver ejercicios utilizando un sistema de ecuaciones de dos incógnitas.
Ejemplo
Un caballo con aceleración constante cubre la distancia de 80 m entre dos puntos en 7 s. Su rapidez al pasar por el segundo punto es 15 m/s.
a) ¿Qué rapidez tenía en el primer punto?
b) ¿Qué aceleración tiene el caballo ?
34 comments
Transcript:
[Música] bien hola chicos el día de hoy vamos a trabajar un ejemplo más de movimiento rectilíneo uniformemente variable el ejemplo dice así un caballo con la aceleración constante cubre la distancia de 80 metros entre dos puntos en siete segundos su rapidez al pasar por el segundo punto es de 15 metros sobre segundo qué rapidez tenía en el primer punto y que aceleración tiene lo que observamos en este problema es lo siguiente observamos un caballo que va a recorrer una distancia de 80 metros comprendidos entre el punto y el punto b esto lo va a realizar en un tiempo de 7 segundos y la velocidad que se registra en el punto b es de 15 metros sobre segundo nos preguntan entonces qué rapidez tenía en el primer punto y qué aceleración tiene el caballo nuestros datos serán entonces la velocidad final la registrada en el segundo punto es de 15 metros sobre segundo el tiempo que es de 7 segundos la distancia que desde 80 metros y nos preguntan la aceleración del caballo y la velocidad inicial del caballo o bien la velocidad en el primer punto lo primero que vamos a realizar será observar las ecuaciones y determinar cuál de estas nos servirá para solucionar el problema entonces la primera pregunta es qué rapidez tenía en el primer punto entonces buscamos la velocidad inicial observamos que en la primera actuación tenemos la velocidad inicial sin embargo también tenemos la aceleración y si nos damos cuenta la aceleración también es una variable en este problema por lo tanto tenemos dos incógnitas no es factible utilizar veamos la segunda tenemos también la velocidad inicial y la aceleración y en la tercera de igual forma tenemos la aceleración y la velocidad inicial por lo tanto no es factible utilizar una sola de estas ecuaciones que realizaremos entonces utilizaremos dos ecuaciones por comodidad utilizaremos la primera y la tercera entonces tendríamos la primera ecuación y la segunda ecuación tenemos dos ecuaciones con dos incógnitas vamos a utilizar el método de igualación para resolver este problema por comodidad igualar las aceleraciones porque digo que es por comodidad simple y sencillamente porque en esta ecuación de acá la aceleración ya se encuentra despejado ahora lo único que tengo que hacer es en esta ecuación despejar a aceleración como realizar esto pues observó que el 2 y la distancia se encuentran multiplicando por lo tanto los pasaré al otro lado a dividir y tendría que la aceleración es igual a la velocidad final al cuadrado menos la velocidad inicial al cuadrado dividido dos veces la distancia entonces ya puedo igualar la aceleración de la primera ecuación con la aceleración de la segunda ecuación donde esta aceleración marca en celeste será esta información y la aceleración que se encuentra marcada en rojo será la información que se encuentra en rojo por lo tanto yo tendré lo siguiente la velocidad final al cuadrado menos la velocidad inicial al cuadrado dividido dos veces la distancia es igual a la velocidad final menos la velocidad inicial dividido el tiempo de esta ecuación que acabo de encontrar voy a buscar la velocidad inicial por lo tanto debo de realizar un despeje el 2 y la distancia se encuentran acá dividiendo los voy a pasar del otro lado del signo igual a multiplicar el tiempo de igual forma se encuentra dividiendo lo pasaré del otro lado del signo igual a multiplicar entonces tendría lo siguiente el tiempo por la velocidad final al cuadrado menos la velocidad inicial al cuadrado es igual a 2 veces la distancia por la velocidad final menos la velocidad inicial acá voy a utilizar la propiedad distributiva del producto para multiplicar el tiempo por la velocidad final al cuadrado y el tiempo por la velocidad inicial al cuadrado de igual forma en el segundo miembro de la ecuación voy a utilizar la propiedad distributiva del producto para multiplicar dos veces la distancia por la velocidad final y dos veces la distancia por la velocidad inicial tendría entonces lo siguiente el tiempo por la velocidad final al cuadrado menos el tiempo por la velocidad inicial al cuadrado es igual a 2 veces la distancia por la velocidad final menos 2 veces la distancia por la velocidad inicial ahora voy a sustituir datos el tiempo era de 7 segundos la velocidad final es de 15 metros sobre segundo y la distancia es de 80 metros ahora bien tendría la siguiente ecuación voy a realizar los productos que se encuentran indicados 7 por 15 al cuadrado es 1575 menos 7 por la velocidad inicial al cuadrado es igual a 2 por 80 por 15 es igual a 2400 - 2 por 80 de 160 por la velocidad inicial como podemos observar esta es una ecuación de segundo grado por lo tanto debo de igualar la a cero para poder solucionarlo como puede observar 1575 se encuentra positivo lo pasaré al otro lado del signo igual como negativo y menos 7 velocidad inicial al cuadrado es tan negativo lo pasaré del otro lado del signo igual como positivo tendría entonces lo siguiente ahora como puede observar este 1575 negativo y este 2400 son términos semejantes por lo tanto voy a hacer la respectiva reducción entonces tendría que 7 por la velocidad inicial al cuadrado menos 160 por la velocidad inicial más 825 este 825 salió de la reducción de estos dos términos semejantes bien como puedo observar esta es una ecuación de segundo grado que obedece a la siguiente estructura a x al cuadrado más bx más c es igual a 0 por lo tanto podríamos solucionar la que sea con nuestra calculadora por factorización o bien como lo vamos a hacer en este ejemplo con esta fórmula recordando que la variable x es la velocidad inicial y que a es 7 veces menos 160 y c es 825 entonces vamos a reescribir esta ecuación en la respectiva fórmula diríamos entonces que la velocidad inicial va a ser igual a menos ve que es menos 160 más menos más menos la raíz debe al cuadrado la raíz de menos 160 al cuadrado menos 4 por a que es 7 por c que es 825 dividido dos veces a que es dos veces siete entonces realizando las respectivas operaciones tendríamos que la velocidad inicial es 160 más menos la raíz de 2.500 dividido 14 entonces por ser una ecuación de segundo grado nos quedarán dos soluciones la primera solución para la velocidad nos dice que es de 15 como podemos observar 15 es la velocidad final por lo tanto no es esta la que estamos buscando pero si la velocidad número 2 que es de 7.85 metros sobre sí podríamos nosotros entonces dar una respuesta a esta pregunta la velocidad del caballo al pasar por el primer punto es de 7.85 metros sobre segundo encontrada a la velocidad inicial será mucho más fácil encontrar la aceleración ya que la ecuación número uno nos dice que la aceleración es la velocidad final menos la velocidad inicial divido el tiempo por lo tanto sustituyendo datos y realizando estas operaciones tendríamos que la aceleración es de 1.02 metros sobre segundo al cuadrado y daríamos así entonces la respuesta para la segunda pregunta la aceleración del caballo es de 1.02 metros sobre segundo al cuadrado bien chicos eso es todo sean buenos [Música] |
187548 | https://www.geeksforgeeks.org/maths/difference-between-an-arithmetic-sequence-and-a-geometric-sequence/ | Difference between an Arithmetic Sequence and a Geometric Sequence - GeeksforGeeks
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Difference between an Arithmetic Sequence and a Geometric Sequence
Last Updated : 18 Feb, 2024
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Arithmetic is a mathematical operation that deals with numerical systems and related operations. It's used to get a single, definite value. The word "Arithmetic" comes from the Greek word "arithmos," which meaning "numbers." It is a field of mathematics that focuses on the study of numbers and the properties of common operations such as addition, subtraction, multiplication, and division.
A sequence is a collection of items in a specific order (typically numbers). Arithmetic and geometric sequences are the two most popular types of mathematical sequences. Each consecutive pair of terms in an arithmetic sequence has a constant difference. A geometric sequence, on the other hand, has a fixed ratio between each pair of consecutive terms.
Arithmetic Sequence
If the difference between any two consecutive terms is always the same, a sequence of integers is termed an Arithmetic Sequence. Simply put, it indicates that the next number in the series is calculated by multiplying the preceding number by a set integer. Further, an Arithmetic Sequence can be written as,
a, a + d, a + 2d, a + 3d, a + 4d
where a = the first term
d = common difference between terms.
For example, in the following sequence: 5, 11, 17, 23, 29, 35, ..., the constant difference is 6.
Geometric Sequence
If the ratio of any two consecutive terms is always the same, a sequence of numbers is called a Geometric Sequence. Simply put, it means that the next number in the series is calculated by multiplying a set number by the preceding number. Further, a Geometric Sequence can be expressed as:
a, ar, ar2, ar3, ar4…
where a = first term
d = common difference between terms.
For instance, 2, 6, 18, 54, 162,... The constant multiplier is 3 in this case.
How can you tell the difference between an Arithmetic sequence and a Geometric sequence?
To tell the difference between arithmetic and geometric sequence, the following points are important,
An arithmetic Sequence is a set of numbers in which each new phrase differs from the previous term by a fixed amount. Geometric Sequence is a series of integers in which each element after the first is obtained by multiplying the preceding number by a constant factor.
When there is a common difference between subsequent terms, represented as ‘d,' a series can be arithmetic. The sequence is said to be geometric when there is a common ratio between succeeding terms, indicated by ‘r.'
The new term in an arithmetic sequence is obtained by adding or subtracting a fixed value from the previous term. In contrast to geometric sequence, the new term is found by multiplying or dividing a fixed value from the previous term.
The variation between the members of an arithmetic sequence is linear. In contrast, the variation in the sequence's elements is exponential.
Infinite arithmetic sequences diverge, while infinite geometric sequences converge or diverge, depending on the situation.
Difference between an arithmetic sequence and a geometric sequence
| S.No. | Arithmetic sequence | Geometric sequence |
---
| 1 | Arithmetic Sequence is a set of numbers in which each new phrase differs from the previous term by a fixed amount. | A geometric sequence is a collection of integers in which each subsequent element is created by multiplying the previous number by a constant factor. |
| 2 | Between successive words, there is a common difference. | Between successive words, they have the same common ratio. |
| 3 | Subtraction or addition are used to get terms. | Division or Multiplication are used to get terms. |
| 4 | Example: 5, 11, 17, 23, 29, 35,... | Example: 2, 6, 18, 54, 162,... |
Sample Problems
Question 1: What is a Geometric Sequence, and why is it called that?
Answer:
Because the numbers go from one to another by diving or multiplying by a similar value, it's called a geometric sequence.
Question 2: Is it possible for an Arithmetic Sequence to also be Geometric?
Answer:
In mathematics, an arithmetic sequence is defined as a sequence in which the common difference, or variance between subsequent numbers, remains constant. The geometric sequence, on the other hand, is characterized by a stable common ratio between subsequent values. As a result, a sequence cannot be both geometric and arithmetic at the same time.
Question 3: In an arithmetic sequence, what is 'a'?
Answer:
An arithmetic sequence is a set of terms in which the difference between two succeeding members of the series is a constant term, 'a' is the first term of an in the arithmetic sequence.
Question 4: What is the procedure for determining the nthterm of an arithmetic sequence?
Answer:
a n = 2n + 1 is the formula for finding the n th term of an arithmetic sequence or the n th term could be written as a + (n - 1) d.
Where 'a' is the first term and 'd' is common difference of an arithmetic sequence.
Question 5: What is the procedure for determining the nthterm of a geometric sequence?
Answer:
a n = ar n − 1 is the formula for finding the n th term of a geometric sequence where 'a' is the first term and 'd' is the common ratio of a geometric sequence.
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187549 | https://languageposters.com/pages/spanish-verbs-recibir-conjugation?srsltid=AfmBOor1FLCsNlqHPJhB5ccIq90LDpJsJHMztC1WkvdNeKPv30jrZ6Ca | Recibir Conjugation - Conjugate Recibir in Spanish – LanguagePosters.com
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Spanish Verbs
Recibir Conjugation
Recibir conjugation
Recibir is a Spanish regular IR verb meaning "to receive". Verbs are considered regular if they follow a predictable pattern when conjugated.
Recibir is ranked on the 100 Most Used Spanish Verbs Poster as the #7 most used regular verb.
Present Tense
Recibir Conjugation
Recibir Participio
Recibir Gerundio
Recibir Present Continuous
Past Tenses
Recibir Past Tense
Recibir Preterite
Recibir Imperfect
Recibir Present Perfect
Recibir Past Perfect
Future Tenses
Recibir Future Tense
Recibir Conditional Tense
Subjunctive Tenses
Recibir Subjunctive
Further Reading
Spanish Regular Verbs
Spanish Conjugation Chart
100 Most Used Spanish Verbs Poster w/ Study Guide
A selection of the most used irregular and regular Spanish verb conjugations. A must have for anyone who wants to learn Spanish!
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Recibir Conjugation
| Pronoun | Conjugation | Translation |
---
| Yo | recibo | I receive |
| Tú | recibes | You receive |
| Él/Ella/Ud. | recibe | He/She receives |
| Nosotros | recibimos | We receive |
| Vosotros | recibís | You (plural) receive |
| Ellos/Ellas/Uds. | reciben | They receive |
Irregular forms are highlighted in red
Recibir Participio
The Participio or Past Participle of Recibir is recibido. This is used to form the Recibir Present Perfect and the Recibir Past Perfect.
Recibir Gerundio
The Gerundio or Present Participle of Recibir is recibiendo. This is used to form the Recibir Present Continuous.
Recibir Present Continuous
The Present Continuous (presente progresivo or presente continuo) is used to describe actions that are happening right now or around now. To form the present continuous tense, combine the Estar conjugation with the Recibir Gerundio (or Present Participle).
Example: Él está siendo muy egoísta últimamente. (He is being very selfish lately.)
| Pronoun | Conjugation | Translation |
---
| Yo | estoy recibiendo | I am receiving |
| Tú | estás recibiendo | You are receiving |
| Él/Ella/Ud. | está recibiendo | He/She is receiving |
| Nosotros | estamos recibiendo | We are receiving |
| Vosotros | estáis recibiendo | You (plural) are receiving |
| Ellos/Ellas/Uds. | están recibiendo | They are receiving |
Irregular forms are highlighted in red
Back to top
Recibir Past Tense
Spanish has several past tenses, each serving a specific purpose to express actions in the past.
Additionally, the use of Spanish past tenses can vary significantly across regions due to cultural and linguistic differences.
These are some of the most commonly used Spanish past tenses:
Recibir Preterite
The Preterite Tense (Pretérito or Indefinido) is used for completed actions in the past that have a clear beginning or end.
Example: Ayer comí pizza. (Yesterday, I ate pizza).
| Pronoun | Conjugation | Translation |
---
| Yo | recibí | I received |
| Tú | recibiste | You received |
| Él/Ella/Ud. | recibió | He/She received |
| Nosotros | recibimos | We received |
| Vosotros | recibisteis | You (plural) received |
| Ellos/Ellas/Uds. | recibieron | They received |
Irregular forms are highlighted in red
Back to top
Recibir Imperfect
The Imperfect Tense (Imperfecto) describes ongoing, habitual, or background actions in the past, often without a defined start or end.
Example: Cuando era niño, jugaba fútbol todos los días. (When I was a child, I used to play soccer every day).
| Pronoun | Conjugation | Translation |
---
| Yo | recibía | I used to receive |
| Tú | recibías | You used to receive |
| Él/Ella/Ud. | recibía | He/She used to receive |
| Nosotros | recibíamos | We used to receive |
| Vosotros | recibíais | You (plural) used to receive |
| Ellos/Ellas/Uds. | recibían | They used to receive |
Irregular forms are highlighted in red
Back to top
Recibir Present Perfect
The Present Perfect Tense (Presente Perfecto) is used to describe actions that have been completed recently or have relevance to the present moment. The Present Perfect Tense is formed by combining the Present Tense of the auxiliary verb Haber Conjugation with the Recibir Participio.
Example: He leído ese libro. (I have read that book).
| Pronoun | Conjugation | Translation |
---
| Yo | he recibido | I have received |
| Tú | has recibido | You have received |
| Él/Ella/Ud. | ha recibido | He/She has received |
| Nosotros | hemos recibido | We have received |
| Vosotros | habéis recibido | You (plural) have received |
| Ellos/Ellas/Uds. | han recibido | They have received |
Irregular forms are highlighted in red
Back to top
Recibir Past Perfect
The Past Perfect Tense (Pluscuamperfecto) refers to actions completed before another action in the past. The Past Perfect Tense is formed by combining the Imperfect Tense of the auxiliary verb Haber Conjugation with the Recibir Participio.
Example: Había comido antes de salir. (I had eaten before leaving).
| Pronoun | Conjugation | Translation |
---
| Yo | había recibido | I had received |
| Tú | habías recibido | You had received |
| Él/Ella/Ud. | había recibido | He/She had received |
| Nosotros | habíamos recibido | We had received |
| Vosotros | habíais recibido | You (plural) had received |
| Ellos/Ellas/Uds. | habían recibido | They had received |
Irregular forms are highlighted in red
Back to top
Recibir Future Tense
The Future Tense (Futuro) is used to describe actions or events that will happen in the future. It often expresses certainty or plans about what is yet to come.
Example: Mañana comeré pizza. (Tomorrow, I will eat pizza).
| Pronoun | Conjugation | Translation |
---
| Yo | recibiré | I will receive |
| Tú | recibirás | You will receive |
| Él/Ella/Ud. | recibirá | He/She will receive |
| Nosotros | recibiremos | We will receive |
| Vosotros | recibiréis | You (plural) will receive |
| Ellos/Ellas/Uds. | recibirán | They will receive |
Irregular forms are highlighted in red
Back to top
Recibir Conditional Tense
The Conditional Tense (Condicional) is used to express hypothetical situations, polite requests, or actions that depend on other conditions. It is often used in combination with the Future Tense.
Example: Me gustaría un café, por favor.(I would like a coffee, please).
| Pronoun | Conjugation | Translation |
---
| Yo | recibiría | I would receive |
| Tú | recibirías | You would receive |
| Él/Ella/Ud. | recibiría | He/She would receive |
| Nosotros | recibiríamos | We would receive |
| Vosotros | recibiríais | You (plural) would receive |
| Ellos/Ellas/Uds. | recibirían | They would receive |
Irregular forms are highlighted in red
Back to top
Recibir Subjunctive
The Subjunctive Tense (Subjuntivo) is used to express doubt, uncertainty, desire, or emotion. It is often used in dependent clauses introduced by specific conjunctions or verbs.
Example: Espero que seas feliz. (I hope you are happy).
| Pronoun | Conjugation | Translation |
---
| Yo | reciba | I receive |
| Tú | recibas | You receive |
| Él/Ella/Ud. | reciba | He/She receives |
| Nosotros | recibamos | We receive |
| Vosotros | recibáis | You (plural) receive |
| Ellos/Ellas/Uds. | reciban | They receive |
Irregular forms are highlighted in red
Back to top
Spanish Regular Verbs
A verb is called a regular verb when its conjugation follows a consistent and predictable pattern. In contrast, a verb that does not follow these standard patterns is called an irregular verb. In Spanish, the three regular conjugation patterns are based on the verb endings: -AR, -ER, and -IR.
Back to top
Spanish Conjugation Chart
Looking for more Spanish verbs like Recibir? Check out our Spanish Conjugation Chart, the 100 Most Used Spanish Verbs Poster!
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187550 | https://physics.stackexchange.com/questions/506702/is-the-mean-life-being-1-lambda-a-coincidence-or-a-definition | Stack Exchange Network
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Is the mean life being 1/$\lambda$ a coincidence or a definition?
We were studying radioactivity and after we worked out the half life to be $ln2/\lambda$ the professor then said, this is the average time it takes for half the nuclei to decay to find the average time for 1 nuclei to decay its just $1/\lambda$.
Is there some statistical proof to this or is this just a definition.
1 Answer 1
By definition, suppose the cumulative distribution function of the time for a single nucleus to decay is $P(t)$, then the half life $T_{1/2}$ will satisfy $P(T_{1/2})=1/2$, and the average time for a nucleus to decay is $T_0=\int_0^\infty t(1-P(t))dt$ .
In the case of radioactive decay, or any other exponential decay, we know that $P(t)$ has the form $P(t)=1-e^{-\lambda t}$, based on the observation $-dN/dt\propto N$. Calculation shows that $T_0=1/\lambda$ and $T_{1/2}=\ln2/\lambda$.
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187551 | https://en.wiktionary.org/wiki/reveal | reveal - Wiktionary, the free dictionary
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1 EnglishToggle English subsection
1.1 Etymology
1.2 Pronunciation
1.3 Noun
1.3.1 Derived terms
1.3.2 Related terms
1.3.3 See also
1.4 Verb
1.4.1 Derived terms
1.4.2 Translations
1.5 Anagrams
reveal
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From Wiktionary, the free dictionary
English
[edit]
English Wikipedia has an article on:
reveal
Wikipedia
Etymology
[edit]
From Middle Englishrevēlen, from Old Frenchreveler, from Latinrevēlāre(“to reveal, uncover”), from re-(“back, again”) + vēlāre(“to cover”), from vēlum(“a cloth, covering, curtain, veil, awning, sail”) (whence also Englishveil, Englishunveil, Englishvoile, Russianвуаль(vualʹ), Russianзавуали́ровать(zavualírovatʹ, “to veil”) (compare typologically)).
Also compare typologically English uncloak, unmask, Russian разобла́чать(razobláčatʹ) (akin to облаче́ние(oblačénije), срыва́ть покро́вы(sryvátʹ pokróvy) (покро́в(pokróv, “cover, covering”)).
Pronunciation
[edit]
(Received Pronunciation)IPA(key): /ɹɪˈviːl/
(General American)IPA(key): /ɹɪˈvil/
Audio (California):Duration: 2 seconds.0:02(file)
(General Australian)IPA(key): /ɹɪˈviːl/, [ɹɪˈvɪil]
Audio (Queensland):Duration: 2 seconds.0:02(file)
Hyphenation: re‧veal
Rhymes: -iːl
Noun
[edit]
reveal (pluralreveals)
The outer side of a window or doorframe. synonyms▲quotations▼Synonyms:jamb, revel
2010, Carter B. Horsley, The Upper East Side Book:The building has a one-story rusticated limestone base and a canopied entrance with a doorman beneath an attractive, rusticated limestone window reveal on the second floor and a very impressive and ornate limestone window reveal on the third floor flanked by female figures.
(cinematography,narratology,comedy, usually informal) A revelation; an uncovering of what was hidden in the scene or story. quotations▼The comedian had been telling us about his sleep being disturbed by noise. Then came the reveal: he was sleeping on a bed in a department store.
2002, Blain Brown, Cinematography, →ISBN:A simple dolly or crane move can be used for an effective reveal. A subject fills the frame, then with a move, something else is revealed.
2017 February 23, Katie Rife, “The Girl With All The Gifts tries to put a fresh spin on overripe zombie clichés”, in The Onion AV Club:Once you find out what’s going on—the girl is a “hungry,” this film’s term for zombies—it’s still interesting enough, if not quite as powerful. That’s basically what you’re in for with this British postapocalyptic survival horror tale, which starts off strong but dilutes its impact with every consecutive reveal.
2019, Douglas Rushkoff, “Survival of the Richest”, in Extinction Rebellion, editor, This Is Not A Drill, London: Penguin, →ISBN:Even Westworld—based on a science-fiction novel where robots run amok—ended its second season with the ultimate reveal: human beings are simpler and more predictable than the artificial intelligences we create.
For more quotations using this term, see Citations:reveal.
Derived terms
[edit]
face reveal
gender reveal
sex reveal
Related terms
[edit]
unveil
veil
velum
voile
See also
[edit]
dénouement
plot twist
Verb
[edit]
reveal (third-person singular simple presentreveals, present participlerevealing, simple past and past participlerevealed)
(transitive) To uncover; to show and display that which was hidden. synonyms▲quotations▼Synonyms:uncover, unfold, unveil; see alsoThesaurus:reveal
c. 1625, Edmund Waller, Of the Danger His Majesty (being Prince) Escaped in the Road at St Andero Light was the wound, the prince's care unknown, / She might not, would not, yet reveal her own.
2013 June 7, Gary Younge, “Hypocrisy lies at heart of Manning prosecution”, in The Guardian Weekly, volume 188, number 26, page 18:The dispatches revealed details of corruption and kleptocracy that many Tunisians suspected, but could not prove, and would cite as they took to the streets.
(transitive) To communicate that which could not be known or discovered without divine or supernaturalinstruction. synonyms▲Synonyms:disclose, divulge; see alsoThesaurus:divulge
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revealed religion
revelation
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show ▼±to uncover
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Arabic: كَشَفَ(kašafa) Aragonese: revelar Azerbaijani: üzə çıxarmaq Bulgarian: разкривам(bg)(razkrivam) Catalan: revelar(ca) Chinese: Mandarin: 透露(zh)(tòulù), 顯示/ 显示(zh)(xiǎnshì) Czech: odhalit(cs) Danish: afsløre Dutch: onthullen(nl), zich ontpoppen Egyptian: (prj ẖr) Esperanto: malkaŝi Estonian: paljastama Finnish: paljastaa(fi) French: révéler(fr) Galician: revelar(gl) German: enthüllen(de) Greek: αποκαλύπτω(el)(apokalýpto)Ancient: φαίνω(phaínō), ἐκφαίνω(ekphaínō), μηνύω(mēnúō) Hebrew: גילה(gilá), חשף(he)(khasáf) Hungarian: felfed(hu) Icelandic: afhjúpa Ido: revelar(io) Irish: foilsighOld Irish: foilsigidir Italian: rivelare(it), gettare la maschera, uscire allo scoperto, mostrarese stesso, svelare(it) Japanese: 現す(ja)(あらわす, arawasu), 表す(ja)(あらわす, arawasu), 表わす(ja)(あらわす, arawasu) Korean: 나타내다(ko)(natanaeda), 드러내다(ko)(deureonaeda) Kurdish: Central Kurdish: دەرخستن(derxistin) Latin: acclārō, exhibeō(la), patefaciō, revēlō Macedonian: открива(otkriva) Malay: dedah(ms) Ngazidja Comorian: upvenua Norwegian: avsløre Old Church Slavonic: авити(aviti) Old English: ætīewan Ottoman Turkish: كشف ایتمك(keşf etmek) Persian: مکشوف ساختن(makšuf sâxtan) Polish: odkrywać(pl)impf, odkryć(pl)pf, odsłaniać(pl)impf, odsłonić(pl)impf, ujawniać(pl)impf, ujawnić(pl)pf Portuguese: revelar(pt) Russian: выявля́ть(ru)impf(vyjavljátʹ), раскрыва́ть(ru)impf(raskryvátʹ), пока́зывать(ru)impf(pokázyvatʹ) Scots: kithe Serbo-Croatian: открити, otkriti(sh) Spanish: revelar(es), propalar(es) Swahili: -toboa, -dhihirisha(sw) Swedish: uppenbara(sv), avslöja(sv), avtäcka Telugu: బయటపెట్టు(te)(bayaṭapeṭṭu), వెల్లడించు(te)(vellaḍiñcu) Thai: เปิดเผย(th)(bpə̀ət-pə̌əi) Turkish: açığa vurmak(tr) Ugaritic: 𐎁𐎙𐎊(bġy) Ukrainian: розкривати(rozkryvaty), виявляти(uk)(vyjavljaty), показувати(uk)(pokazuvaty), з'ясовувати(zʺjasovuvaty) Vietnamese: để lộ, tiết lộ(vi) Welsh: datguddio(cy) Yiddish: אַנטפּלעקן(antplekn)
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[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
show ▼±to communicate that which could not be known or discovered without divine or supernatural instruction
[Select preferred languages] [Clear all]
Arabic: أَوْحَى(ʔawḥā), أَنْزَلَ(ʔanzala)(literally “to send down”), نَزَّلَ(nazzala)(literally “to send down”)
Azerbaijani: nazil etmək
Catalan: divulgar(ca)
French: révéler(fr), laisser voir(fr)
Galician: revelar(gl)
German: offenbaren(de)
Italian: rivelare(it)
Latin: dēclārō, ēnūntiō
Ngazidja Comorian: uɓulia, upvenulia
Plautdietsch: openboaren
Portuguese: divulgar(pt)
Spanish: revelar(es)
Swahili: kufunulia
Vietnamese: mặc khải
Welsh: datguddio(cy)
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[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
Anagrams
[edit]
Leaver, laveer, leaver, vealer
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187552 | https://www.arthroscopytechniques.org/article/S2212-6287(24)00445-6/fulltext | The Remplissage Technique as a Treatment for Shoulder Instability - Arthroscopy Techniques
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Technical Note | ShoulderVolume 14, Issue 4103292 April 2025 Open access
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The Remplissage Technique as a Treatment for Shoulder Instability
Robert A.Cecere, B.S.
Robert A.Cecere, B.S.0009-0006-3533-6083
Correspondence
Address correspondence to Robert A. Cecere, B.S., Sports Medicine Institute, Hospital for Special Surgery, 535 E 70th St, New York, NY 10021, U.S.A.
cecerer@hss.edu
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Sports Medicine Institute, Hospital for Special Surgery, New York, New York, U.S.A.
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cecerer@hss.edu ∙ Anna B.Williams, B.A.
Anna B.Williams, B.A.0000-0001-5385-1456
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Sports Medicine Institute, Hospital for Special Surgery, New York, New York, U.S.A.
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∙ Lawrence V.Gulotta, M.D.
Lawrence V.Gulotta, M.D.
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Sports Medicine Institute, Hospital for Special Surgery, New York, New York, U.S.A.
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Affiliations & Notes Article Info
Sports Medicine Institute, Hospital for Special Surgery, New York, New York, U.S.A.
Publication History:
Received June 5, 2024; Accepted September 6, 2024; Published online November 8, 2024
DOI: 10.1016/j.eats.2024.103292 External LinkAlso available on ScienceDirect External Link
Copyright: Published by Elsevier Inc. on behalf of the Arthroscopy Association of North America.
User License: Creative Commons Attribution – NonCommercial – NoDerivs (CC BY-NC-ND 4.0) | Elsevier's open access license policy
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Abstract
Patients with a history of shoulder instability commonly experience recurrent shoulder dislocations. Correcting their shoulder instability is vital to preventing recurrence and debilitating injury. Patients presenting with an engaging Hill-Sachs lesion are at a particularly high risk of future dislocations. The remplissage procedure, designed as a supplement to the Bankart repair, prevents the Hill-Sachs lesion from engaging with the glenoid, decreasing the risk of future dislocations when compared with arthroscopic Bankart repair alone. One barrier to its use has been the technical difficulty of the procedure. We present the senior author’s technique for performing remplissage.
Technique Video
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Technique Video
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Recurrent shoulder dislocation is a common outcome for patients recovering from a history of shoulder instability. Managing these patients by correcting their shoulder instability is vital to preventing recurrence and debilitating injury. The remplissage procedure was introduced as a supplement to the Bankart repair to decrease the failure rate in patients with a Hill-Sachs lesion.1
1.
Purchase, R.J. ∙ Wolf, E.M. ∙ Hobgood, E.R. ...
Hill-Sachs “remplissage”: An arthroscopic solution for the engaging Hill-Sachs lesion
Arthroscopy. 2008; 24:723-726
Full Text
Full Text (PDF)
Scopus (339)
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As a supplement to the Bankart repair, the remplissage procedure can be performed through the portals from the Bankart repair, without the need for new incisions.2
2.
Provencher, M.T. ∙ Frank, R.M. ∙ Leclere, L.E. ...
The Hill-Sachs lesion: Diagnosis, classification, and management
J Am Acad Orthop Surg. 2012; 20:242-252
Crossref
Scopus (248)
PubMed
Google Scholar
The procedure prevents the Hill-Sachs lesion from engaging with the glenoid, preventing the lesion from continuing to be “off track,” specifically when the patient abducts or externally rotates the shoulder.3
3.
Hurley, E.T. ∙ Toale, J.P. ∙ Davey, M.S. ...
Remplissage for anterior shoulder instability with Hill-Sachs lesions: A systematic review and meta-analysis
J Shoulder Elbow Surg. 2020; 29:2487-2494
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The procedure should be considered when the patient presents with an off-track Hill-Sachs lesion and subcritical glenoid bone loss of less than 15%.4
4.
Polio, W. ∙ Brolin, T.J.
Remplissage for anterior shoulder instability: History, indications, and outcomes
Orthop Clin North Am. 2022; 53:327-338
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Surgical Technique
Imaging
The computed tomography (CT) scan is the superior imaging modality for the detection of bone loss in possible candidates for the remplissage procedure.2
2.
Provencher, M.T. ∙ Frank, R.M. ∙ Leclere, L.E. ...
The Hill-Sachs lesion: Diagnosis, classification, and management
J Am Acad Orthop Surg. 2012; 20:242-252
Crossref
Scopus (248)
PubMed
Google Scholar
To mitigate the risk of underestimating the size of the Hill-Sachs lesion, 3-dimensional CT reconstruction is considered the current gold standard; 2-dimensional CT scans present a risk of underestimating the size of the Hill-Sachs lesion.5
5.
Miniaci, A. ∙ Gish, M.W.
Management of anterior glenohumeral instability associated with large Hill-Sachs defects
Tech Shoulder Elbow Surg. 2004; 5:170-175
Crossref
Scopus (140)
Google Scholar
Three-dimensional CT scans provide the easiest visualization of the glenoid track for evaluation and provide an estimate of the degree of osseous deficiency of the glenoid when evaluating a patient preoperatively for remplissage consideration.6
6.
Vopat, M.L. ∙ Peebles, L.A. ∙ McBride, T. ...
Accuracy and reliability of imaging modalities for the diagnosis and quantification of Hill-Sachs lesions: A systematic review
Arthroscopy. 2021; 37:391-401
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Scopus (17)
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There is considerable agreement that identification of a medium- to large-sized, off-track Hill-Sachs lesion with less than substantial glenoid bone loss (defined as <25% osseous deficiency of the glenoid) validates considering a patient for remplissage. The estimation of glenoid bone loss should be confirmed arthroscopically before continuing with the procedure.2
2.
Provencher, M.T. ∙ Frank, R.M. ∙ Leclere, L.E. ...
The Hill-Sachs lesion: Diagnosis, classification, and management
J Am Acad Orthop Surg. 2012; 20:242-252
Crossref
Scopus (248)
PubMed
Google Scholar
,5
5.
Miniaci, A. ∙ Gish, M.W.
Management of anterior glenohumeral instability associated with large Hill-Sachs defects
Tech Shoulder Elbow Surg. 2004; 5:170-175
Crossref
Scopus (140)
Google Scholar
Technique for Remplissage
The patient is placed in the beach-chair position. After the shoulder and upper extremity are prepared and draped in the usual sterile fashion, an articulating arm holder is used to help hold and position the arm throughout the operation. A standard posterior portal is established, and a diagnostic arthroscopy of the shoulder joint is performed. Once a tear of the anterior labrum, as well as an engaging Hill-Sachs lesion, is confirmed, attention is turned toward surgical repair. A standard anterior portal is made just superior to the subscapularis in the rotator interval. An anterolateral portal is established just over the biceps tendon, anterior to the leading edge of the supraspinatus (Fig 1). The arthroscope is then inserted into the anterolateral portal. This facilitates excellent visualization of the glenohumeral joint as well as the labrum circumferentially.
Figure viewer
Fig 1 After a standard diagnostic arthroscopy from the posterior portal, an anterolateral portal is made over the biceps tendon, anterior to the leading edge of the supraspinatus. The arthroscope is then inserted into the portal, allowing for visualization of the glenohumeral joint and labrum (as shown by the arrow).
While viewing through the anterolateral portal, the Hill-Sachs lesion is identified (Fig 2). With a combination of electrocautery and a 4.5-mm shaver, the Hill-Sachs lesion is prepared by removing the pseudo-membrane, which is typically over its surface. This can be facilitated by slightly externally rotating the arm in the articulating arm holder. Anchors are then placed through the posterior portal (Fig 3). Depending on the size of the Hill-Sachs lesion, either one 5.5-mm triple-loaded anchor or two 4.5-mm double-loaded anchors (Arthrex, Naples, FL) are placed in the usual fashion. The anchors are placed in the middle of the Hill-Sachs lesion. Care is taken not to place them adjacent to the articular surface for fear of overconstraining the shoulder joint. The anchors are typically placed 1 cm from the insertion of the rotator cuff.
Figure viewer
Fig 2 Viewing from the anterolateral portal, the Hill-Sachs lesion (identified by the arrow) is identified and is prepared by removing the pseudo-membrane with electrocautery and a 4.5-mm shaver through the posterior portal.
Figure viewer
Fig 3 Viewing from the anterolateral portal, once the Hill-Sachs lesion has been prepared by removing the pseudo-membrane, anchors (Arthrex) are placed through the posterior portal.
At this point, a posterolateral portal is established. This is achieved with spinal needle localization (Fig 4). After the incision for this portal is made, an 8.5-mm-diameter hard plastic cannula is placed under the deltoid and onto the bursal surface of the rotator cuff (Fig 5). The tip of the cannula can be swept superior, inferior, and side to side to ensure that it is free of the deep deltoid fascia and is in the bursal space. Through this cannula, the sutures from the previously placed anchors are then retrieved using a tissue penetrator in a mattress stitch configuration (Fig 6). Care is taken to ensure that the tissue penetrator is placed through the tendinous portion of the teres minor and infraspinatus and not through the muscular portion.
Figure viewer
Fig 4 After Hill-Sachs lesion preparation and anchor placement, a spinal needle is used to establish the posterolateral portal (as shown by the arrow).
Figure viewer
Fig 5 After spinal needle placement, an incision for the posterolateral portal is made. Next, an 8.5-mm-diameter hard plastic cannula is placed in the portal, under the deltoid, and onto the bursal surface of the rotator cuff (as shown by the arrow).
Figure viewer
Fig 6 Viewing from the anterolaterla portal, through the cannula in the posterolateral portal, the sutures from the previously placed anchors are retrieved using a tissue penetrator in a mattress stitch configuration. This is viewed from the anterolateral portal. After the sutures are passed through the tendon of the posterior rotator cuff, attention is turned toward the Bankart repair.
After the sutures are passed through the tendon of the posterior rotator cuff, attention is turned toward the Bankart repair. The Bankart repair is performed according to the surgeon’s preference.
After repair of the Bankart lesion, attention is turned toward completing the remplissage. The previously placed sutures, which are still in the cannula in the posterolateral portal, are pulled so that the cannula is firmly placed against the bursal side of the rotator cuff. The sutures are then secured using a snap, which in turn pins the cannula against the bursal side of the rotator cuff and protects the sutures within the cannula (Fig 7).
Figure viewer
Fig 7 After repair of the Bankart lesion, attention is turned toward completing the remplissage. The sutures located in the posterolateral portal are pulled(as shown by the arrow) so that the cannula is placed against the bursal side of the rotator cuff, protecting the sutures within the cannula.
The arthroscopic trocar is then placed through the anterolateral portal into the subacromial space. By use of a sweeping motion, the subacromial space is cleared and the trocar is used to find the cannula in that space. Once the trocar is against the cannula, the arthroscope is placed. From the posterior portal, a 4.5-mm shaver is then introduced. The 4.5-mm shaver is used to perform a limited bursectomy with the confidence of knowing that the sutures are protected inside the cannula, which is pinned to the bursal side of the rotator cuff with the snap holding it in place (Fig 8). Once adequate visualization of the cannula and the sutures is achieved, the snap is removed. The sutures are then tied using standard arthroscopic knot-tying techniques (Fig 9). The arthroscope is inserted into the cannula and the anterior portal so that the finalized remplissage procedure can be viewed (Fig 10, Video 1).
Figure viewer
Fig 8 Once the sutures have been protected by the cannula, the arthroscopic trocar is placed through the anterolateral portal into the subacromial space, and the subacromial space is cleared. Once the cannula has been located, the arthroscope is placed into the anterolateral portal. Now, a 4.5-mm shaver, inserted through the posterior portal, is used to perform a limited bursectomy with the confidence of knowing that the sutures are protected inside the cannula.
Figure viewer
Fig 9 After the bursectomy, adequate visualization of the cannula and the sutures can be achieved from the anterolateral portal, and the snap is then removed. The sutures are tied using standard arthroscopic knot-tying techniques.
Figure viewer
Fig 10 The arthroscope is inserted into the cannula and the anterolateral portal. The finalized remplissage procedure can be viewed.
Postoperatively, the patient is placed in a shoulder immobilizer for 4 weeks. During this time, the patient is allowed to perform hand, wrist, and elbow range of motion; however, pendulum exercises are not performed. At 4 weeks, the patient is allowed to remove the sling, at which time he or she begins formal physical therapy to work on both passive and active range of motion. However, strengthening is limited during this time. The goal is for the patient to have near full range of motion by 12 weeks. If this is achieved, then a strengthening program is initiated. The goal is to return to sports or full activity by 5 to 6 months postoperatively (Table 1).
Weeks 1-4
Patient placed in shoulder immobilizer
Hand, wrist, and elbow ROM
No pendulum exercises
Remove sling at 4 wk; begin formal PT, working on PROM and AROM
Strengthening limited
Week 7
135° PROM with forward elevation
50° PROM with external rotation and internal rotation in scapular plane
45° PROM with external rotation at 90° of abduction
115° AROM with forward elevation
Minimal pain
Week 8
May begin strengthening exercises
Week 12
Full PROM and AROM without pain
Initiate strengthening program
5-6 mo
Resume athletic activities
Table 1
Remplissage Postoperative Benchmarks
AROM, active range of motion; PROM, passive range of motion; PT, physical therapy; ROM, range of motion.
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Discussion
Overall, the results of patients with medium to large Hill-Sachs lesions and mild glenoid bone loss who have undergone Bankart repair with the addition of remplissage are favorable.1
1.
Purchase, R.J. ∙ Wolf, E.M. ∙ Hobgood, E.R. ...
Hill-Sachs “remplissage”: An arthroscopic solution for the engaging Hill-Sachs lesion
Arthroscopy. 2008; 24:723-726
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Comparison of outcome measures and scores preoperatively to postoperatively almost universally shows a dramatic favorable change in shoulder stability. Compared with Bankart repair alone, Bankart repair with remplissage has shown a statistically significantly lower dislocation recurrence rate and lower overall rate of instability (defined as recurrent dislocation or subluxation).3
3.
Hurley, E.T. ∙ Toale, J.P. ∙ Davey, M.S. ...
Remplissage for anterior shoulder instability with Hill-Sachs lesions: A systematic review and meta-analysis
J Shoulder Elbow Surg. 2020; 29:2487-2494
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PubMed
Google Scholar
Cadaveric models have shown decreased instability after remplissage, with shoulders undergoing Bankart repair with remplissage showing significantly reduced anterior instability and restoration of native glenohumeral stability.7
7.
Wu, C. ∙ Ye, Z. ∙ Lu, S. ...
Biomechanical analysis reveals shoulder instability with bipolar bone loss is best treated with dynamic anterior stabilization for on-track lesions and with remplissage for off-track lesions
Arthroscopy. 2024; 40:1982-1993
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Additionally, when compared with an open Latarjet procedure, Bankart repair with remplissage has a statistically significantly lower rate of complications and morbidities and has the same rate of recurrence.1
1.
Purchase, R.J. ∙ Wolf, E.M. ∙ Hobgood, E.R. ...
Hill-Sachs “remplissage”: An arthroscopic solution for the engaging Hill-Sachs lesion
Arthroscopy. 2008; 24:723-726
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Scopus (339)
PubMed
Google Scholar
Studies have reported percentages as high as 91% for patients returning to postoperative athletic activities at a similar competitive level to the preoperative level after Bankart repair with remplissage.8
8.
Gouveia, K. ∙ Harbour, E. ∙ Athwal, G.S. ...
Return to sport after arthroscopic Bankart repair with remplissage: A systematic review
Arthroscopy. 2023; 39:1046-1059
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Although various meta-analyses and systematic reviews have disproved the theory that patients experience substantial loss in external rotation after Bankart repair with remplissage, studies analyzing the outcomes with respect to specific sports have substantiated that even small postoperative decreases in external rotation can have significant effects on patients and their postoperative activities. In particular, overhead-throwing patients have reported return-to-play rates as low as 50% that result from limits to their range of motion after Bankart repair with remplissage (Table 2).8
8.
Gouveia, K. ∙ Harbour, E. ∙ Athwal, G.S. ...
Return to sport after arthroscopic Bankart repair with remplissage: A systematic review
Arthroscopy. 2023; 39:1046-1059
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Scopus (10)
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Pros
Decreases risk of further dislocation
High RTS rate
Uses pre-existing portals from Bankart repair
Cons
Can lead to small limitations on external rotation, leading to lower RTS rates for overhead athletes
Indications
Large, off-track Hill-Sachs lesion
Subcritical glenoid bone loss
Contraindications
Critical glenoid bone loss
Osseous deficiency of glenoid > 25%
Table 2
Summary of Advantages and Disadvantages of Bankart Repair With Remplissage
RTS, return to sport.
Open table in a new tab
Disclosures
The authors declare the following financial interests/personal relationships which may be considered as potential competing interests: L.V.G. has ownership interest in Responsive Arthroscopy and Imagen; serves on the speakers bureau of Smith & Nephew and Zimmer Biomet; is a consultant for Zimmer Biomet; and serves on the editorial or governing board of HSS Journal. Both other authors (R.A.C., A.B.W.) declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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eyJraWQiOiI4ZjUxYWNhY2IzYjhiNjNlNzFlYmIzYWFmYTU5NmZmYyIsImFsZyI6IlJTMjU2In0.eyJzdWIiOiI0Yzg4MDJhNzNmNzZjNGNiODViZTAzN2QxOGU2YjAxZCIsImV4cCI6MTc1OTE2Njc1MCwia2lkIjoiOGY1MWFjYWNiM2I4YjYzZTcxZWJiM2FhZmE1OTZmZmMifQ.kUrJzYntk7vK-iJpn27dcPUr2PxN344w9NMAG0liAp99lkGfkvLP8NUeDvcQS6iBZT_W3XP60-1Gtd1JZmGLdoXq0G-omcPYDsE8I7TlvfollxZWYpEzlr6vQXLyzbRAK7xSoEcjFBB_ENuiNoli-CCKNWnwWofQgNGb11x8B3CmGCBYW9z-qHEEcWRPEaEppmf0DgOp0_BrIkwNbhDvQ5wmfov0X6AIBYbbNEsS2z2zifkxjSvFL401RYztRZJ-tuHj-QfLsobEXFuUA8btoXnrY4p4DGiXR1zZJs7Cm8pePbJgdlvm3fXiQHACOj-sjA2dOiCXU_QI0NA7rRjC0A
Video (71.09 MB)
Video 1
After a standard diagnostic arthroscopy from the posterior portal, the arthroscope is placed into the anterolateral portal and the Hill-Sachs lesion is visualized. The lesion is prepared and anchors are placed through the posterior portal. At this time, attention can shift to the Bankart lesion. After Bankart lesion repair, the remplissage procedure begins by pulling and securing the previously placed sutures in the posterolateral portal. The cannula is positioned against the bursal side of the rotator cuff to protect the sutures. An arthroscopic trocar is inserted through the anterolateral portal to clear the subacromial space and locate the cannula. A 4.5-mm shaver is used from the posterior portal for a limited bursectomy. Once the cannula and sutures are visible, the sutures are tied using standard arthroscopic techniques, completing the remplissage procedure.
PDF (48.68 KB)
ABW disclosure for Rempl
PDF (70.46 KB)
COI Robert Cecere
PDF (63.08 KB)
LVG Disclosure for Rempl
References
1.
Purchase, R.J. ∙ Wolf, E.M. ∙ Hobgood, E.R. ...
Hill-Sachs “remplissage”: An arthroscopic solution for the engaging Hill-Sachs lesion
Arthroscopy. 2008; 24:723-726
Full Text
Full Text (PDF)
Scopus (339)
PubMed
Google Scholar
2.
Provencher, M.T. ∙ Frank, R.M. ∙ Leclere, L.E. ...
The Hill-Sachs lesion: Diagnosis, classification, and management
J Am Acad Orthop Surg. 2012; 20:242-252
Crossref
Scopus (248)
PubMed
Google Scholar
3.
Hurley, E.T. ∙ Toale, J.P. ∙ Davey, M.S. ...
Remplissage for anterior shoulder instability with Hill-Sachs lesions: A systematic review and meta-analysis
J Shoulder Elbow Surg. 2020; 29:2487-2494
Full Text
Full Text (PDF)
Scopus (40)
PubMed
Google Scholar
4.
Polio, W. ∙ Brolin, T.J.
Remplissage for anterior shoulder instability: History, indications, and outcomes
Orthop Clin North Am. 2022; 53:327-338
Full Text
Full Text (PDF)
Scopus (8)
PubMed
Google Scholar
5.
Miniaci, A. ∙ Gish, M.W.
Management of anterior glenohumeral instability associated with large Hill-Sachs defects
Tech Shoulder Elbow Surg. 2004; 5:170-175
Crossref
Scopus (140)
Google Scholar
6.
Vopat, M.L. ∙ Peebles, L.A. ∙ McBride, T. ...
Accuracy and reliability of imaging modalities for the diagnosis and quantification of Hill-Sachs lesions: A systematic review
Arthroscopy. 2021; 37:391-401
Full Text
Full Text (PDF)
Scopus (17)
PubMed
Google Scholar
7.
Wu, C. ∙ Ye, Z. ∙ Lu, S. ...
Biomechanical analysis reveals shoulder instability with bipolar bone loss is best treated with dynamic anterior stabilization for on-track lesions and with remplissage for off-track lesions
Arthroscopy. 2024; 40:1982-1993
Full Text
Full Text (PDF)
Scopus (1)
PubMed
Google Scholar
8.
Gouveia, K. ∙ Harbour, E. ∙ Athwal, G.S. ...
Return to sport after arthroscopic Bankart repair with remplissage: A systematic review
Arthroscopy. 2023; 39:1046-1059
Full Text
Full Text (PDF)
Scopus (10)
PubMed
Google Scholar
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187553 | https://www.youtube.com/watch?v=TpIBLnRAslI | Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy
Khan Academy
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Posted: 12 Oct 2011
What the angle bisector theorem is and its proof
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131 comments
Transcript:
What I want to do
first is just show you what the angle
bisector theorem is and then we'll actually
prove it for ourselves. So I just have an
arbitrary triangle right over here, triangle ABC. And what I'm going
to do is I'm going to draw an angle bisector
for this angle up here. And we could have done it
with any of the three angles, but I'll just do this one. I'll make our proof
a little bit easier. So I'm just going to bisect
this angle, angle ABC. So let's just say that's the
angle bisector of angle ABC, and so this angle
right over here is equal to this
angle right over here. And let me call this point down
here-- let me call it point D. The angle bisector
theorem tells us that the ratio between
the sides that aren't this bisector-- so when I put
this angle bisector here, it created two smaller triangles
out of that larger one. The angle bisector
theorem tells us the ratios between the other
sides of these two triangles that we've now created
are going to be the same. So it tells us that
the ratio of AB to AD is going to be equal to the
ratio of BC to, you could say, CD. So the ratio of--
I'll color code it. The ratio of that,
which is this, to this is going to be equal to
the ratio of this, which is that, to this right
over here-- to CD, which is that over here. So once you see the
ratio of that to that, it's going to be the same as
the ratio of that to that. So that's kind of a cool
result, but you can't just accept it on faith because
it's a cool result. You want to prove
it to ourselves. And so you can imagine
right over here, we have some ratios set up. So we're going to prove it
using similar triangles. And unfortunate for us, these
two triangles right here aren't necessarily similar. We know that these two angles
are congruent to each other, but we don't know
whether this angle is equal to that angle
or that angle. We don't know. We can't make any
statements like that. So in order to actually set
up this type of a statement, we'll have to construct
maybe another triangle that will be similar to one
of these right over here. And one way to do it would
be to draw another line. And this proof
wasn't obvious to me the first time that
I thought about it, so don't worry if it's
not obvious to you. What happens is if we can
continue this bisector-- this angle bisector
right over here, so let's just continue it. It just keeps going
on and on and on. And let's also-- maybe we can
construct a similar triangle to this triangle
over here if we draw a line that's parallel
to AB down here. So let's try to do that. So I'm just going to say,
well, if C is not on AB, you could always find
a point or a line that goes through C that
is parallel to AB. So by definition, let's
just create another line right over here. And let's call this
point right over here F and let's just pick
this line in such a way that FC is parallel to AB. So this is parallel to
that right over there. And we could just
construct it that way. And now we have some
interesting things. And we did it that
way so that we can make these two triangles
be similar to each other. So let's see that. Let's see what happens. So before we even
think about similarity, let's think about what we know
about some of the angles here. We know that we have
alternate interior angles-- so just think about these
two parallel lines. So I could imagine AB
keeps going like that. FC keeps going like that. And line BD right
here is a transversal. Then whatever this
angle is, this angle is going to be as well, from
alternate interior angles, which we've talked a lot
about when we first talked about angles with
transversals and all of that. So these two angles are
going to be the same. But this angle and
this angle are also going to be the same, because
this angle and that angle are the same. This is a bisector. Because this is a
bisector, we know that angle ABD is the
same as angle DBC. So whatever this angle
is, that angle is. And so is this angle. And that gives us kind
of an interesting result, because here we have
a situation where if you look at this
larger triangle BFC, we have two base angles
that are the same, which means this must be an
isosceles triangle. So BC must be the same as FC. So that was kind of cool. We just used the transversal and
the alternate interior angles to show that these
are isosceles, and that BC and FC
are the same thing. And that could be
useful, because we have a feeling that this
triangle and this triangle are going to be similar. We haven't proven it yet. But how will that help us get
something about BC up here? But we just showed that BC
and FC are the same thing. So this is going to
be the same thing. If we want to
prove it, if we can prove that the ratio of
AB to AD is the same thing as the ratio of FC
to CD, we're going to be there because BC, we
just showed, is equal to FC. But let's not start
with the theorem. Let's actually get
to the theorem. So FC is parallel
to AB, [? able ?] to set up this one
isosceles triangle, so these sides are congruent. Now, let's look at some
of the other angles here and make ourselves
feel good about it. Well, we have this. If we look at triangle ABD, so
this triangle right over here, and triangle FDC, we
already established that they have one set of
angles that are the same. And then, and then
they also both-- ABD has this angle right
over here, which is a vertical angle
with this one over here, so they're congruent. And we know if two triangles
have two angles that are the same, actually
the third one's going to be the same as well. Or you could say by the
angle-angle similarity postulate, these two
triangles are similar. So let me write that down. You want to make sure you get
the corresponding sides right. We now know by
angle-angle-- and I'm going to start at
the green angle-- that triangle B-- and
then the blue angle-- BDA is similar to triangle--
so then once again, let's start with the
green angle, F. Then, you go to the blue angle, FDC. And here, we want to eventually
get to the angle bisector theorem, so we want to look at
the ratio between AB and AD. Similar triangles,
either you could find the ratio between
corresponding sides are going to be
similar triangles, or you could find
the ratio between two sides of a similar triangle
and compare them to the ratio the same two corresponding sides
on the other similar triangle, and they should be the same. So by similar triangles,
we know that the ratio of AB-- and this, by the way,
was by angle-angle similarity. Want to write that down. So now that we know
they're similar, we know the ratio of AB to
AD is going to be equal to-- and we could even look here
for the corresponding sides. The ratio of AB, the
corresponding side is going to be CF-- is
going to equal CF over AD. AD is the same thing
as CD-- over CD. And so we know the ratio of AB
to AD is equal to CF over CD. But we just proved to
ourselves, because this is an isosceles triangle, that
CF is the same thing as BC right over here. And we're done. We've just proven AB over
AD is equal to BC over CD. So there's two things we
had to do here is one, construct this other
triangle, that, assuming this was parallel, that gave
us two things, that gave us another angle to show
that they're similar and also allowed us
to establish-- sorry, I have something
stuck in my throat. Just coughed off camera. So I should go get a
drink of water after this. So constructing
this triangle here, we were able to both
show it's similar and to construct this
larger isosceles triangle to show, look, if we can
find the ratio of this side to this side is the same
as a ratio of this side to this side, that's
analogous to showing that the ratio of this side
to this side is the same as BC to CD. And we are done. |
187554 | https://www.youtube.com/watch?v=-xixOm1IvzE | MATHCOUNTS Mini #80 - Circles and Triangles Revisited
MATHCOUNTS Foundation
40200 subscribers
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Description
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Posted: 2 Apr 2018
April 2018
This video explores solving problems using properties of circles and triangles.
Download the Activity Sheet at
Download the Solutions at
Video by Art of Problem Solving's Richard Rusczyk, a MATHCOUNTS alum. Visit Art of Problem Solving for many more educational resources.
24 comments
Transcript:
this month's lesson is routine Puerto Rico now it's geometry so I got my man Harvey or he insisted on coming I think because he has a lot of fans in Puerto Rico come on man admit it anyway anyway I'm not gonna need you to smut our because I've got this I've seen lots of problems like this here we have a coin it's wedged in the corner of a drawer then we have this rectangular box out here it's got a two and a half inch edge so this right here is two and a half now one corner of this box is one and a half inches from the corner of the drawer so this right here is one and a half and then the other corner appears two inches in the corner of the drawer there so it's drawing definitely not to scale and we're looking for the diameter of the coin it's sitting in here I'm a circle here it's tangent to a bunch of things out here when I have a circle tangent to some lines I like to draw in the radii so our radius right there throwing the radii to those points of tangency because that gives me right angles there right angle there right angle there right there and of course this is a corner of my drawer there's a right angle right here as well this little box there that puppy's a square so these out here both are as well now these tangents or two tangents from this point they're equal so I'm go ahead and throw in some variables here I'm gonna call this X that means this is X as well there's another really powerful thing that's all I'm radii two points of tangency just seems real appear gonna call that Y so this one is y as well so now I can build some equations that's why we throw variables onto our diagrams we try to build equations X plus y is 2.5 and then y plus r is 2 then R plus X is 1.5 these are really pretty equations nice that's some nice pretty symmetry going on over here on the Left we have every combination of two of the three variables when I see equations like this oh I'm awfully tempted to just add all three of them up we had all three of these we get 2x is too wise to ours over here so I have 2 times the sum of our plus X plus y that equals the sum of these 3 over here which is 6 divided by 2 I see that the sum of our plus X plus y is 3 I want the diameter of the coin so I'm going to target our X plus y is sitting right up here is 2.5 so the X plus y there's 2.5 that means that R has to be 0.5 I have to be really careful here word problem long word problem last step is read the question we want the diameter not the radius double the radius the diameter is 1 inch sorry Harv on to the next problem didn't need Harvey at all on that one here we go figure shows right triangle ABC bear and side lengths are 5 12 13 and we've got squares drawn on each side and then we connect these points here to build on these triangles and we want the area of the whole hexagon I got it all broken up into pieces already let's find the areas of those pieces well this piece right here this is 5 this is 12 5 times 12 is 60 have that gives us 30 is the area of that triangle this is a square with side length 5 this is 25 this is a square with side length 12 so it's area is 144 and then up here we have a square with side length 13 so it's area is 169 right triangle over here this side is 5 this side is 12 its areas 30 as well and then this triangle this side length is 5 but not right triangle this triangle this side length is 12 but the side length is 13 this is their team but those aren't altitudes [Music] yes hmm okay Harvey uh I was wrong could use your help man what you want me to beg all right dude man please help me I'm stuck here I just need a little guidance here the all the triangles are the same area it's obvious no you got to help me out I mean these two are the same that's obvious but these two it's obvious he's no help at all no seriously you're no help at all all right just a little bit all right we're going to do this ourselves cuz Harvey is not being helpful today here we go all right I got this banks right here this is five I needed altitude so I'm gonna go ahead and draw in the altitude I'm have to extend this side a bit there and I have a base over here this is 12 I'm gonna want to do the same thing over here I'm going to find the altitude to that side length of 12 Sun dive extend that side as well the second these two triangles look the same and these two triangles look like this triangle go ahead and continue this here and this out here that all these triangles they look the same can we prove that we've got a rectangle right here and then well this angle and this angle here they're complements this angle is complementary that so these two angles are the same this triangle is congruent to this triangle so we just took this rectangle split it in half now what's going on up here this angle complimentary that's this angle equals this these two angles are complementary because this in here the a/c is a right angle so this angle has to be the same as that one of course this side da the same as this here AC so these two right triangles they are congruent of course they're congruent to this one as well so this triangle congruent to this triangle has area 30 I want the area of this triangle I want the area of that triangle all right but this triangle is congruent to this one so this length right here this altitude it's the same length as BC this altitude is 12 and this base of course this is 5 so the area of this triangle 5 times 12 divided by 2 yeah yeah yeah I know you're right Harvey the areas are the same it's obvious you saw that right away no you saw a faster way whatever let's get this other triangle we get the same drill going on here this is 5 because this triangle right here is congruent to this triangle so a B has the same length as this altitude over here so this altitude is 5 our base down here CH has length 12 12 times 5 60 divided by 2 this is indeed area 30 so now we're ready finally to finish finish the problem we want the area the whole hexagon so we have to add all of these up let's see I've got one two three four four 30s that's 120 and then my 25 and 144 is gonna give me 169 I Sarah vagary and theorem right there 169 and then I had another 169 here let's just add these up this gives me 8 carry 1 5 and then this is a for our area is 458 all right Harvey show us the obvious way what's the obvious way for us to see that the areas of all these triangles are the same you're not gonna tell me come on excuse a little hint we just rotate da e and then everything lines up we rotate triangle da e it's all you're gonna tell me you're not gonna explain it it's obvious he says it's obvious I don't see it well maybe you do maybe you could explain it to me |
187555 | https://quizlet.com/603594912/amino-acids-at-ph-7-flash-cards/ | Amino acids at pH 7 Flashcards | Quizlet
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187556 | https://www.koppglass.com/blog/glass-thermal-properties-and-their-role-product-design | Skip to main content
Glass Thermal Properties and Their Role in Product Design
This is the first article in a three-part series that reviews the thermal, optical, and mechanicalproperties of glass. We will define common glass properties and explain their application and importance in component design.
It is critical to have a thorough understanding of glass thermal properties when designing with a glass lens or filter. When exposed to sudden or even gradual changes in temperature, improperly designed glass lenses will perform poorly and can even occasionally fail. Their thermal properties determine how they will perform in different operating conditions; this information will help you select a glass composition that will perform best for your application and environment.
Common values for the thermal properties of borosilicate glass are listed in the table below. In this article, we will discuss these properties as well as important processing temperatures.
| Thermal Property | Common Values of Borosilicate Glass |
--- |
| Linear Thermal Expansion | α = 30 - 60 x 10-7/°C |
| Thermal Conductivity | K = 1 Watt/m°C |
| Specific Heat | C= 800 J/kg°C |
Coefficient of Linear Thermal Expansion
The coefficient of thermal expansion (CTE) is a measure of how much volume changes as a material is heated or cooled. It is defined by
where V and T are volume and temperature, and its units are 1/°C. For glasses, the linear thermal expansion is often discussed. For isotropic amorphous materials such as glass that have small thermal expansions, the linear coefficient is accurately described by
Application:
If a temperature is unevenly applied to a glass, different areas of the glass will expand by varying amounts and internal stresses will develop. This could result in glass fracture or failure.
In applications where glasses are closely installed with other materials, their thermal expansions need to match. Ceramic enamels are often applied to glass lenses to block unwanted light. The CTE for the enamel must be similar to that of the glass or the enamel will crack and chip. Another example that demonstrates the importance of CTE occurs when a glass lens is fitted tightly into a metal fixture, such as in stage lighting. If the expansions of the materials are not taken into consideration, and adequate space is not provided, then the glass could crack and fail due to applied stress from the fixture.
Thermal Shock Resistance
The thermal shock resistance of a glass indicates how likely it is to break when its temperature suddenly changes. It is defined as the maximum change in temperature (ΔT) that a glass can withstand upon rapid heating or cooling. It can be related to other glass properties by
where σ is the internal stress necessary to cause cracking or failure, ν is Poisson’s ratio, E is Young’s modulus and α is the coefficient of linear thermal expansion of the glass.
Application:
Thermal shock resistance is often tested by taking heated glass lenses and rapidly cooling them through methods such as immersion in an ice bath. This type of testing can indicate the ability for glass lenses to withstand large changes in temperature when installed in application. For example, glass lenses used with high power lighting may get hot during application and experience rapid cooling when exposed to rain, snow, or other environmental factors. In these dynamic environments, it is critical to select the correct type of glass to ensure the lens’ ability to withstand thermal shock.
Thermal Conductivity
Thermal conductivity represents how well a glass conducts or transfers heat. It is defined as
where q is the heat flow measured in watts (or J/s), A is the cross-sectional area of the glass, and dT/dx is the temperature gradient applied to the glass. Good thermal conductors will allow heat to travel through the material very quickly, much like good electrical conductors will allow for faster charge movement.
Application:
It is often desirable for glasses to have low thermal conductivity in applications and act as a thermal insulator. LEDs, for instance, perform better in cooler temperatures and will output more light, according to research performed by the Lighting Research Center. If a temperature controlled LED fixture were to operate in a hot environment, then using a glass lens with a low thermal conductivity in that fixture would reduce the heat flow through the glass to the LED and improve its energy efficiency.
Specific Heat
The specific heat of a glass is the heat needed to raise the temperature of the glass by 1°C per unit weight:
where Q is heat, m is mass and T is temperature. If the thermal conductivity shows how much heat will flow through a material, the specific heat shows how quickly heat will raise the temperature of a glass.
Application:
The specific heat of a glass part could be an important consideration for applications where the glass operates at high temperatures. Consider a lighting fixture that uses a glass lens with a quartz halogen bulb; these bulbs often operate at high temperatures, outputting a large amount of heat. If the lens is designed with a lower specific heat, it will reach its equilibrium temperature faster and reduce the warm-up time of the system.
Important Glass Temperatures
There are typically five important temperatures that are often discussed in the production and design of glasses.
Melting point is the temperature at which the raw materials melt to a liquid state.
Working point is the temperature at which the glass melt is shaped or molded.
Softening dilatometric point is the temperature at which the glass begins to deform during heating when measured in a dilatometer.
Annealing point is the temperature at which the residual stresses in a glass are reduced over a matter of minutes.
Strain point is the temperature at which the residual stresses in a glass are reduced over a matter of hours.
These values are typically given as a range of temperatures rather than as a single point.
Application:
A sound knowledge of these temperature points is very important for glass manufacturers; it helps ensure production efficiency as well as high quality products. But it is also important for application design so that the right glass is chosen for a specific job. If a glass lens is going to be used in a high temperature environment, like the lens for a spotlight, its softening point has to be higher than the operating temperature of the light or the glass could lose its desired shape. These temperatures are also critical for setting parameters for the annealing, tempering, or heat-strengthening of glasses.
Thermal Dependence of Other Glass Properties
Many other glass properties may be affected by a change in temperature. For example, the chromaticity or color of a glass is often dependent on its thermal history. Reds and yellow colors in glass are usually developed during manufacturing by a process called striking, where the glass is reheated and cooled to promote specific colors through oxidation, reduction, or precipitation reactions of the colorants. In some cases, the glasses may even change color under normal operating conditions as the glasses equilibrate with the light source temperature.
As discussed above, glasses that are heated expand by an amount proportional to their coefficient of thermal expansion. This change in volume may also affect both the density and the refractive index of the glass. Typically, the density will decrease as the spacing between the ions in the glass increases. The refractive index, however, can either increase or decrease with temperature depending on both the change in ion spacing as well as the change in the electron cloud surrounding the ions.
Just as it is important to understand the thermal nature of a glass for temperature sensitive applications, it is often necessary to take the transmission, chromaticity, and refractive index of a glass into account during the design of a lens. The next article in this series will discuss the optical properties of glassand how these properties affect the suitability of a composition for different applications.
Learn More About Glass
To help you design better-performing glasses lenses, we created a comprehensive eBook that includes more than 40 pages of information on the thermal, optical, and mechanical properties of glass.
If you want to learn how to design glass lenses and components that are optimized for both your performance requirements and operating environment, download our free eBook.
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About the author: Kopp Glass
Kopp Glass, Inc. manufactures high-performance glass for mission critical applications. Grounded in Pittsburgh, PA’s industrious history, Kopp employs applications engineering expertise to develop solutions that meet demanding specifications. Kopp’s material science mastery supports an expansive composition portfolio including over 200 glasses that transmit wavelengths from UV to visible to IR. For nearly a century, Kopp has helped customers effectively navigate emerging technologies and develop innovative products that enable a safer and more productive world.
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187557 | https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-7-for-ccss/section/4.5/related/lesson/prices-involving-discounts-msm6/ | Prices Involving Discounts | CK-12 Foundation
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Back To Solving Markup and Markdown Problems with PercentsBack
4.5
Prices Involving Discounts
Written by:Jen Kershaw, M.ed |Catherine Kwok
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
[Figure 1]
Penny wants a new television. She sees that the one she wants is on sale at two different stores. The original price is $399. First, Pick is having a 25% off sale on all TVs. Halmarket is selling the same TV for $299. Which store has the better deal?
In this concept, you will learn how to find prices involving discounts.
Finding Prices Involving Discounts
A discount is the difference between the original price and the reduced price. It is described as a number of dollars or as a percent.
Find the reduced price in two steps:
First, find the dollar amount of the discount. Multiply the original price by the discount.
Next, subtract the dollar value of the discount from the original price. The total is the reduced price of the item.
Let’s look at an example.
Tracy went shopping for a new pair of sneakers. She found a blue pair that was $58. The sign said that they were 15% off the original price. How much is the discount? How much did Tracy end up paying for the sneakers?
First, find the discount amount. Find 15% of $58. 15% is also 0.15. Multiply the original price by the discount.
58×0.15=8.70
Next, subtract the discount from the original price.
58−8.70=49.30
The discount was $8.70. Tracy paid $49.30 for the sneakers.
You can also find the reduced price by multiplying the original price by the percent of the price you will pay. The percent you are paying is 100% minus the discount %.
Let’s look at Tracy’s sneakers again.
The sneakers are 15% off. Use mental math to calculate the percent Tracy will be paying. Think, “100 minus 15 is 85.”
Next, multiply the original price by the percent she will be paying. 85% is also 0.85.
58×0.85=49.30
This also gives the reduced price of the sneakers, $49.30.
Examples
Example 1
Earlier, you were given a problem about Penny and the TVs.
Pick has it on sale for 25% off and Halmarket has it on sale for $299. Originally the price was $399. To compare the prices, Penny needs to find the sale price of the TV at First Pick.
First, find the discount amount. Find 25% of $399. 25% is also 0.25. Multiply the original price by the discount.
399×0.25=99.75
Next, subtract the discount from the original price.
399−99.75=299.25
The TV at First Pick is on sale for $299.25. Compare $299.25 to the price of the TV at Hal market, $299.
The TV is cheaper at Hal market, but only by $0.25.
Example 2
Find the discount and the reduced price.
Kara bought a new dress. The original price of the dress was $65.50. There was a 15% discount on the dress.
What was the amount of the discount? How much did Kara pay for the dress?
First, find the discount amount. Find 15% of $65.50. 15% is also 0.15. Multiply the original price by the discount.
65.50×0.15=9.83
(Note: Dollars are only measured to the second decimal digit. Round 9.825 to the nearest hundredths place.)
Next, subtract the discount from the original price.
65.50−9.83=55.67
The discount was $9.83. Kara spent $55.67 on the dress.
Example 3
If a $50.00 shirt is 25% off, how much would you pay for the shirt?
First, find the discount amount. Find 25% of $50. 25% is also 0.25. Multiply the original price by the discount.
50×0.25=12.50
Next, subtract the discount from the original price.
50−12.50=37.50
You would pay $37.50.
Example 4
If a video game that usually costs $45.50 is 30% off, how much would you pay for the game?
First, use mental math to find the percentage you would end up paying for the video game. Think, “If the discount is 30%, I would pay 70% of the price.” 70% is also 0.7.
Then, multiply the original price by percent you would pay.
45.50×0.7=31.85
You would pay $31.85.
Example 5
If a backpack was reduced to $30 and the original price was $40, what percent was the discount?
First, use the given information to find the reduced amount. The reduced amount is the difference between the original price and the sale price.
40−30=10
Next, write an equation to find the discount percent. Remember, the original price multiplied by the discount percent is the reduced amount. Use x to represent the unknown discount percent.
40 x=10
Then, solve for x to find the discount percent.
40 x 40=10 40 x=0.25
Finally, convert the decimal to a percent.
0.25=25%
The backpack was discounted 25%.
Review
Find the reduced price. Round your answer to the nearest whole cent.
Original price: $19.95, discount 15%
Original price: $20.00, discount 50%
Original price: $35.50, discount 10%
Original price: $50.00, discount 30%
Original price: $100.00, discount 20%
Original price: $75.00, discount 30%
Original price: $29.95, discount 20%
Original price: $18.00, discount 10%
Original price: $47.50, discount 10%
Original price: $75.00, discount 30%
Original price: $125.00, discount 20%
Original price: $225.50, discount 10%
Original price: $456.00, discount 25%
Original price: $530.00, discount 30%
Original price: $750.00, discount 12%
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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Back to Prices Involving Discounts
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187558 | https://aasnova.org/2019/04/19/compact-objects-charging-toward-merger/ | Published Time: 2019-04-19T16:00:54+00:00
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Compact Objects Charging Toward Merger
6
By Susanna Kohleron 19 April 2019 Features
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Artist’s illustration showing two inspiralling neutron stars shortly before they merge. Could electric charge play a role in the radiation we see from compact-binary mergers? [Goddard Media Studios/NASA]
When two compact objects — neutron stars or black holes — merge, will they emit light? A recent study looks at a neglected factor that could affect the answer: electric charge.
Dark or Light?
Artist’s impression of two merging neutron stars producing a gamma-ray burst. [National Science Foundation/LIGO/Sonoma State University/A. Simonnet]
Most theories agree that a compact binary containing a neutron star can emit light when it merges. This is because these systems contain lots of neutron-rich matter that can then radiate in the final stages of merger, in the form of gamma-ray bursts, kilonovae, and afterglows.
But what about compact binaries containing two black holes? Or so-called “plunging” black-hole–neutron-star mergers in which the neutron star plunges directly into the black hole before it can be disrupted? Are these mergers all doomed to darkness?
Possible Charge
Not according to Bing Zhang, a scientist at University of Nevada Las Vegas. Recently, Zhang proposed that black holes might carry electric charge in a surrounding magnetosphere. As charged black holes spiral around and around each other during a merger, they could generate electromagnetic radiation: a characteristic signal that rises sharply just before merger.
Now Zhang is back with a generalized model for the merger of charged compact objects, which also explores possible signatures from electrically charged neutron stars. In a new study, he works out the details and reports on where we might be able to detect these signals.
Searching for a Signal
All compact binaries containing a neutron star should emit radiation from electric charge, since neutron stars are definitely charged — they’re essentially spinning magnets. But for most systems containing a neutron star, Zhang demonstrates, the radiation associated with the object’s charge will be non-detectable, since it’s so much dimmer than other electromagnetic signatures from merger (like a gamma-ray burst).
The Crab pulsar is a highly magnetized, spinning neutron star that powers the Crab nebula seen in this composite image. [X-ray: NASA/CXC/SAO/F.Seward; Optical: NASA/ESA/ASU/J.Hester & A.Loll; Infrared: NASA/JPL-Caltech/Univ. Minn./R.Gehrz]
There’s hope, though, in the scenario of a plunging neutron-star–black-hole merger. If the neutron star is less than 20% the size of the black hole, it can be consumed whole, preventing any of the typical electromagnetic signatures from occurring. In this case, the radiation from the charged, inspiralling neutron star is the only electromagnetic signal present.
If the neutron star in such a system has a magnetic field similar to that of the Crab pulsar — possible in young star clusters — the charge signal can reach detectable levels, according to Zhang’s calculations. In fact, it’s possible that we could observe such a signal as a fast radio burst, the mysterious millisecond radio bursts that we’ve seen originating from beyond our galaxy.
Looking Ahead
Many unknowns are still present in this picture. How is the electric radiation converted into observable emission? How commonly do we expect plunging neutron-star–black-hole mergers to occur as described? Will we be able to link radiation from charged mergers to a gravitational-wave chirp?
One thing is for certain: if we can, indeed, observe the light from charge in a compact-binary merger, this would provide an exciting new opportunity to further probe these distant, exotic systems.
Citation
“Charged Compact Binary Coalescence Signal and Electromagnetic Counterpart of Plunging Black Hole–Neutron Star Mergers,” Bing Zhang 2019 ApJL873 L9. doi:10.3847/2041-8213/ab0ae8
black holescompact binariesfast radio burstsmergersneutron stars
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Redox titrations
Redox Titrations: Mastering Analytical Chemistry Techniques Dive into the world of redox titrations and unlock the power of precise chemical analysis. Learn to quantify oxidizing and reducing agents, and apply this crucial skill to real-world scenarios in various scientific fields.
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Now Playing:Redox titrations – Example 0a
Intros
Using titration for redox reactions.
Using titration for redox reactions.
Recap of titration.
3. Using titration for redox reactions.
Titration for redox reactions.
4. Redox titration: Worked example
5. Winkler method to find BOD.
6. Worked example: Winkler method.
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Examples
Find the full equation and concentration of substances in a redox titration. A solution containing Co2+ ions of unknown concentration is made. 25mL of this Co2+ solution was measured and was titrated by 0.25M MnO4- solution until equivalence point was reached. An average titre of 16.20 mL MnO4- solution was required. The reaction produces Mn2+ and Co3+ ions. Write the full redox equation for this reaction and find the concentration of the aqueous Co2+ solution.
Find the full equation and concentration of substances in a redox titration. A solution containing I- ions of unknown concentration is made. 25mL of this solution is measured precisely and is titrated by 0.18M MnO4- solution until the equivalence point is reached. This is repeated, to find an average titre of 19.55 mL MnO4- solution being needed to completely react the I- ions. Write the full redox equation for this reaction and find the concentration of the aqueous I- solution.
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Redox titrations
Jump to:NotesConceptFAQsPrerequisites
Notes
In this lesson, we will learn:
To recall the practical use of titration experiments
How titration applies to redox reactions.
How to calculate chemical quantities required in redox reactions.
How to use the Winkler titration method to find the biochemical oxygen demand (BOD).
Notes:
We learned the basics of a titration with its use in acid-base chemistry in Acid-base-titration. Just like acid-base titrations are used to find the concentration of acids and bases, a redox titration can be done to find the unknown concentration of a chemical in a redox process. The working out and calculations are detailed in Acid-base-titration and is summarized in the image below. Chemical A and chemical B in a redox titration would simply be the two chemicals in the redox (the reducing and oxidizing agent):
Redox titrations will involve a reducing and oxidizing agent reacting together, but indicator is normally not used like it is in acid-base titrations. This means that one of the reactants used has to be one with a color difference between its reduced and oxidized form. There are two good options:
Potassium permanganate (KMnO4) is an oxidizing agent that is purple in solution, but turns colorless when reduced to Mn2+ ions.
Potassium iodide (KI) in solution gives I- ions that get oxidized (lots of chemicals can be used for this part) into brown-colored I2 in solution. Then in a redox titration, I2 can be reduced back to colorless I- ions. Starch can be added (it acts like an indicator for I2) to this, which is blue-black when I2 is present, the color fading when I2 becomes I- again.
WORKED EXAMPLE: A solution containing Co2+ ions of unknown concentration is made. 25mL of this Co2+ solution was measured and was titrated by 0.2M MnO4- solution until equivalence point was reached. 19.40 mL of the MnO4- solution was required. The first thing that needs doing is the finding out of the two half-reactions:
Manganese in MnO4- will be reduced to Mn2+ ions as shown in the half-equation: MnO4- + 8H+ + 5e-→Mn2+ + 4H2O
Co2+ ions can be oxidized to Co3+ according to the half-equation: Co2+→Co3+ + e- The method for working out half-equations in redox was covered in Half equations.Next, the combining of the two half-reactions will give us the overall equation 1 x [ MnO4- + 8H+ + 5e-→Mn2+ + 4H2O ] 5 x [ Co2+→Co3+ + e- ] These balance for electrons and give the overall equation: MnO4- + 8H+ + 5Co2+→Mn2+ + 4H2O + 5Co3+ This reaction has the cobalt solution as the unknown, so MnO4- with known concentration is being added by burette. MnO4- is purple and as it is added to the cobalt solution, the purple color will disappear as Co2+ reacts it away. When equivalence point is reached the purple color will no longer be removed as there will be no more Co2+ to remove the MnO4- and the purple color that it causes. Therefore the equivalence point is shown by the appearance of the purple color of the MnO4- thats now in excess. The number of moles of MnO4- can be calculated using the information in the question: Mol MnO4- = 19.40 mL 1000mL1L∗1L0.2molMnO4− = 3.88 10-3 mol MnO4- Looking at the equation, we can see a 1:5 MnO4- to Co2+ ratio. The equivalence point will have five times as many moles of cobalt as manganese, then. Mol Co2+ = 3.88 10-3 MnO4- 1molMnO4−5molCo2+ = 0.0194 mol Co2+ With the moles of Co2+ ions now found in 25 mL volume of the sample used, we can calculate the concentration. [Co2+] = 0.025L0.0194molCo2+ = 0.776 M Co2+
The Winkler method (or Winkler titration) is a way of finding the concentration of oxygen in water systems using a redox titration reaction. This is important for knowing the biochemical oxygen demand (BOD) of a water system, which is a good indicator of water purity. A lot of decaying matter like dead organisms and sewage in the water will cause an upsurge in bacteria, which demand oxygen in their respiratory processes. The Winkler method makes oxygen the limiting reagent in a redox titration process. A common redox reaction to do this with is the titration of iodine by thiosulfate ions. The limited supply of oxygen initially oxidises iodide to produce the iodine:
Mn2+ (aq) + O2 (aq) + 4 OH- (aq) → 2 MnO2 (aq) + 2 H2O (l) MnO2 (aq) + 4 H+ (aq) + 2 I- (aq) → Mn2+ (aq) + I2 + 2 H2O (l) The iodine then reacts with the thiosulfate: I2 (aq) + 2 S2O32- (aq) → S4O62- (aq) + 2 I- (aq) The amount of iodine that reacts in the final step is entirely dependent on how much oxygen was available to turn it from iodide into iodine. This is what the Winkler method is trying to find out. The concentration of oxygen is an indicator of how much is available and not being used up by microorganisms in the water. At any temperature, there is a general solubility of oxygen in water. With a typical atmosphere of 21% O2 gas, O2 solubility at 298K and 100 kPa (room temperature and pressure) is around 8.210-3 g dm-3. Any gap between this value and oxygen concentration in your sample indicates the biochemical oxygen demand (BOD). Worked example: Using the Winkler method of titration to find BOD.
A 750 mL sample of water from a lake is saturated with oxygen and left for a week. Afterward, the titration of iodine by thiosulfate ions is run, with the iodine being prepared in solution by reacting with oxygen in the following complete reaction scheme: Mn2+ (aq) + O2 (aq) + 4 OH- (aq) → 2 MnO2 (aq) + 2 H2O (l) MnO2 (aq) + 4 H+(aq) + 2 I- (aq) → Mn2+ (aq) + I2 + 2 H2O (l) I2 (aq) + 2 S2O32- (aq) → S4O62- (aq) + 2 I- (aq) 5.7 mL of a 0.05 M thiosulfate (S2O32-) solution completely reacted the I2 present in the water sample. Calculate how many moles of O2 were present in the water. Assuming 8.210-3 g dm-3 solubility of oxygen in this water, determine the BOD of this water sample. This question needs to be answered like a regular titration question first. We need to know how much oxygen is there. We can get this information by the amounts of thiosulfate and the molar ratios involved.
There is a 1:2 ratio of iodine to thiosulfate, so
mol S2O32- = 0.05 M S2O32- 10005.7mLS2O32− =2.85 10-4 mol I2 = 22.85×10−4 = 1.425 × 10-4 + There is a 1:1 ratio with iodine and MnO2, which itself is in a 2:1 ratio with O2. So we need to cut this amount in half again:
mol O2 = 21.425×10−4 = 7.125 × 10-5 You could say overall there is a 4 : 1 thiosulfate : O2 ratio in this reaction sequence, so cutting the thiosulfate moles by 4 will give us moles of O2.+ This second step needs us to find quantities in grams per decimeter cube. We are currently in moles, so we need to convert to grams and divide by volume in dm3 (where 1 dm3 = 1 L)
g O2 = 7.125 × 10-5 mol × 32molg =2.28 × 10-3g O2 This is the amount found in a 750 mL sample. This is 3/4 of a litre (or dm3), so this value needs to be divided by 0.75 to find the value for 1 dm-3. 0.75dm32.28×10−3gO2 = 3.04 × 10-3 g dm-3 This concentration is the amount of oxygen still available in the water sample. Assuming from above that a saturated sample at room temperature and pressure has a solubility of 8.2 × 10-3 g dm-3, the gap between our concentration and this concentration is the biochemical oxygen demand (BOD): BOD=8.2 × 10-3 - 3.04 × 10-3 = 5.16 × 10-3 g dm-3 This is the concentration of oxygen being used up by (micro)organisms in the water system, assuming the level is at equilibrium now.
Concept
Introduction to Redox Titrations
Redox titrations are a fundamental analytical technique in chemistry, essential for determining the concentration of oxidizing or reducing agents in a solution. Our introduction video provides a comprehensive overview of this crucial concept, serving as an excellent starting point for students and researchers alike. Similar to acid-base titrations, redox titrations involve the gradual addition of one solution to another until the reaction reaches completion. However, instead of neutralizing acids and bases, redox titrations focus on the transfer of electrons between oxidizing and reducing agents. This process allows for precise quantification of these agents in various samples. Understanding redox titrations is vital for applications in environmental analysis, industrial quality control, and pharmaceutical research. By mastering this technique, chemists can accurately measure the concentration of important substances like vitamin C, chlorine in water, or iron in ore samples. The principles of redox titrations form the foundation for more advanced electrochemical methods and are indispensable in modern analytical chemistry.
FAQs
What is meant by redox titration?
Redox titration is an analytical technique used to determine the concentration of an oxidizing or reducing agent in a solution. It involves the transfer of electrons between the analyte and a standardized titrant, resulting in a change in oxidation states. The endpoint is typically detected through a color change or using electrochemical methods.
2. What is the difference between a redox titration and an acid-base titration?
The main difference lies in the type of reaction occurring. Redox titrations involve electron transfer and changes in oxidation states, while acid-base titrations involve proton transfer. Redox titrations can analyze a wider range of substances and often use different indicators or endpoint detection methods compared to acid-base titrations.
3. Why is acid needed in redox titration?
Acid is often added in redox titrations to create favorable conditions for the reaction. It can help prevent side reactions, adjust the pH to an optimal range for the redox reaction, and in some cases, it's necessary for the formation of reactive species. For example, in permanganate titrations, acid is required to form the reactive Mn2+ species.
4. What is being oxidized in a redox titration?
In a redox titration, either the analyte or the titrant is being oxidized, while the other is being reduced. The species being oxidized loses electrons and increases its oxidation state. For example, in the titration of Fe2+ with KMnO4, the Fe2+ is being oxidized to Fe3+, while the MnO4- is being reduced to Mn2+.
5. How to solve redox titrations?
To solve redox titrations, follow these steps: 1) Balance the redox equation, 2) Determine the mole ratio between the analyte and titrant, 3) Calculate the moles of titrant used, 4) Use the mole ratio to find the moles of analyte, and 5) Calculate the concentration or mass of the analyte. Always consider the stoichiometry of the reaction and any dilutions made during the process.
Prerequisites
Understanding the foundation of redox titrations is crucial for mastering this important analytical technique in chemistry. One of the key prerequisite topics that students should grasp is calculating cell potential (voltaic cells). This fundamental concept plays a vital role in comprehending the principles behind redox titrations and their applications in various chemical analyses.
Redox titrations are a type of volumetric analysis used to determine the concentration of an analyte in a solution by utilizing oxidation-reduction reactions. To fully appreciate the intricacies of redox titrations, it's essential to have a solid understanding of cell potentials and how they relate to the spontaneity of redox reactions. The ability to calculate and interpret cell potentials provides valuable insights into the direction and extent of electron transfer in redox reactions, which is the cornerstone of redox titrations.
When performing redox titrations, students often encounter color changes that indicate the endpoint of the reaction. These color changes are directly related to the redox processes occurring in the solution. By understanding cell potentials and predicting redox reactions, students can better interpret these visual cues and accurately determine the endpoint of the titration.
Moreover, the concept of cell potentials is crucial for selecting appropriate indicators in redox titrations. Many indicators used in these titrations are themselves redox-active species, and their color changes are dependent on the oxidation state of the indicator molecules. A thorough understanding of cell potentials helps in choosing the right indicator for a specific redox titration, ensuring accurate and reliable results.
In addition, knowledge of cell potentials aids in understanding the limitations and potential sources of error in redox titrations. For instance, competing side reactions or incomplete reactions can affect the accuracy of the titration. By applying their understanding of cell potentials, students can predict these potential issues and take appropriate measures to minimize their impact on the analysis.
Furthermore, the ability to calculate and interpret cell potentials is invaluable when working with more complex redox systems, such as those involving multiple redox couples or in non-standard conditions. This knowledge allows students to adapt their titration techniques and calculations to a wide range of analytical scenarios, making them more versatile and proficient in chemical analysis.
In conclusion, a solid grasp of calculating cell potentials and predicting redox reactions is indispensable for students approaching the study of redox titrations. This prerequisite knowledge not only enhances their understanding of the underlying principles but also improves their practical skills in performing and interpreting redox titrations accurately. By mastering these fundamental concepts, students will be well-equipped to tackle more advanced topics in analytical chemistry and electrochemistry.
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187560 | https://www.news-medical.net/news/20230313/The-2022-NICE-guidelines-for-the-diagnosis-and-management-of-gout.aspx | The 2022 NICE guidelines for the diagnosis and management of gout
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The 2022 NICE guidelines for the diagnosis and management of gout
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By Suchandrima BhowmikReviewed by Benedette Cuffari, M.Sc.Mar 13 2023
A new editorial study on Gout, Urate, and Crystal Deposition Disease discusses the 2022 National Institute for Health and Care Excellence (NICE) guidelines for the management of gout.
Study:Crystal Clear? The 2022 NICE Guideline for the Diagnosis and Management of Gout.Image Credit: mapo_japan / Shutterstock.com
What is gout?
Gout is one of the most common forms of inflammatory arthritis that lead to pain and disability. Although effective treatments are available for gout, care for this condition is largely considered suboptimal.
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Only a few people with gout receive the definitive ‘curative’ urate-lowering therapy (ULT). Of these individuals, some achieve target serum urate levels required for the dissolution of monosodium urate crystal and clinical remission.
Current guidelines to treat gout
Clinical practice guidelines are statements that comprise recommendations for the optimization of patient care.
Several clinical guidelines have been published by the American College of Rheumatology (ACR), American College of Physicians (ACP), British Society for Rheumatology (BSR), and European Alliance of Associations for Rheumatology (EULAR) to optimize the diagnosis and management of gout. However, different recommendations can be made due to differences in methodologies, the types of permitted evidence to be included in the recommendation, and which multidisciplinary stakeholders are involved in the development of the guidelines.
The highest profile disagreement between guideline recommendations concerning gout has been caused by the ACP ‘treat-to-avoid-symptoms’ approach to ULT. This approach differs from other national and international guidelines that include a ‘treat-to-target’ strategy, in which titration of the ULT dose is used to achieve target serum urate levels.
NICE gout guidelines
NICE published new guidelines in June 2022 for the diagnosis and management of gout. These guidelines were established following the systematic search and review of relevant literature, combined with the consensus view, expert opinion, and patient perspectives.
The multidisciplinary guideline committee consisted of one orthopedic surgeon, one dietician, one pharmacist, two lay representatives, an independent chair, as well as methodological experts from the NICE National Guideline Center. These recommendations cover the diagnosis, assessment, and management of gout flares, lifestyle and diet, information and support, long-term gout management, and referral to specialist services.
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As compared to previously published guidelines regarding gout management, NICE recommends allopurinol or febuxostat as first-line ULT in people with gout without a history of major cardiovascular disease, while other guidelines recommended febuxostat only as a second-line ULT. NICE guidelines also concluded that no difference exists in cost-effectiveness or clinical efficacy between febuxostat and allopurinol.
The NICE guideline was in agreement with the ACR, BSR, and EULAR guidelines in recommending a ‘treat-to-target’ approach to ULT, which is in contrast to ACP guidelines.
Furthermore, NICE recommends a target serum urate level of less than 360 µmol/L as compared to a universal target serum urate level of less than 300 µmol/L recommended by BSR guidelines.
NICE also recommends follow-up and disease monitoring, which was not included in previous guidelines. These new guidelines also recommend annual monitoring of patient serum urate levels, even after they have achieved the target level.
Limitations
ULT recommendations by NICE were limited to xanthine oxidase inhibitors, with uricosuric drugs not considered. This made the guideline ineffective for people with gout who have contra-indications, are intolerant of, and do not respond to febuxostat and allopurinol. However, recommendations for treatment with uricosurics could be found in the BSR, EULAR, and ACR guidelines.
Conclusions
The 2022 NICE guidelines provide clarity on the diagnosis and management of gout. Including febuxostat as a first-line ULT can expand access to therapeutics for gout patients, while disease monitoring and follow-ups can provide healthcare professionals with an organized framework to manage the condition and ultimately lead to better health outcomes. Thus, the NICE guidelines have the potential to improve the management, prevention, and awareness of gout.
Journal reference:
Dahanayake, C., Jordan, K. M., & Roddy, E. (2023). Crystal Clear? The 2022 NICE Guideline for the Diagnosis and Management of Gout. Gout, Urate, and Crystal Deposition Disease. doi:10.3390/gucdd1010002.
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Suchandrima Bhowmik
Suchandrima has a Bachelor of Science (B.Sc.) degree in Microbiology and a Master of Science (M.Sc.) degree in Microbiology from the University of Calcutta, India. The study of health and diseases was always very important to her. In addition to Microbiology, she also gained extensive knowledge in Biochemistry, Immunology, Medical Microbiology, Metabolism, and Biotechnology as part of her master's degree.
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187561 | https://www.dictionary.com/e/word-of-the-day/2016/01/25/cupidity | Word of the Day - cupidity | Dictionary.com
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Word of the Day
Apr 10 putative
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vibrissa
[ vahy-bris-uh ]
noun
one of the stiff, bristly hairs growing about the mouth of certain animals, as a whisker of a cat.
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Vibrissa, “one of the stiff hairs growing about the mouth of an animal, such as a cat’s whisker,” is restricted pretty much to (human) anatomy, ornithology, and zoology. Vibrissa is the singular of the Late Latin plural noun vibrissae, a word that occurs only once, in a work by Sextus Pompeius Festus, a Roman grammarian and lexicographer who flourished in the late 2nd century a.d. Festus defines vibrissae as “the nose hairs of a human being, so called because when they are pulled out, the head shakes (caput vibrātur)” (vibrissae does in fact derive from the Latin verb vibrāre “to shake”). This “human” sense is the original meaning in English in the late 17th century, but it is no longer common; the more general zoological and ornithological meaning arose in the first half of the 19th century. The singular form vibrissa first appears in English in the first quarter of the 19th century.
how is vibrissa used?
I stroked his splendid vibrissae, the stiff, sensitive whiskers that a walrus uses to search for bivalves through the seabed’s dark murk, and that feel like slender tubes of bamboo.
Natalie Angier, "The Walrus: Smart, sophisticated and ever closer to the edge," _New York Times_, May 20, 2008
Whiskers – technically called vibrissae in mammals – are an important part of my cats’ sensory arrays. When Margarita abruptly tears across the apartment for reasons I can only speculate about, her whiskers can tell her if she’s cutting to[o] close to a wall so that she doesn’t run headlong into the doorway.
Riley Black, "Dinosaur Whiskers?" _National Geographic_, March 27, 2015
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putative
[ pyoo-t uh-tiv ]
adjective
commonly regarded as such; reputed; supposed.
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Putative, “supposed, so called, commonly regarded,” ultimately comes from Late Latin putātīvus “considered, reckoned, presumptive,” a derivative of the Latin verb putāre “to think, consider,” originally a farming or country word meaning “to trim, prune (trees), scour or clean (wool); purify, refine (gold).” In Latin putāre is not much used in its original senses, but it is very common in its developed senses, “to go over in the mind, ponder; to go over in words, discuss; estimate, deem, consider.” Putative entered English in the 15th century.
how is putative used?
Theputative black hole would have to be feeding at one-millionth of its potential rate if it were there at all, Dr. Gultekin said.
Dennis Overbye, "Missing: One Black Hole With 10 Billion Solar Masses," _New York Times_, January 19, 2021
Jules had to remember: Oh right, Ibsen, the putative reason Ash had gone to Oslo. Isben’s Ghosts.
Meg Wolitzer, _The Interestings_, 2013
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behoove
[ bih-hoov ]
verb (used with object)
to be worthwhile to, as for personal profit or advantage.
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Behoove, also spelled behove in British English, nowadays is an impersonal verb meaning “it is necessary or proper (for someone to do something).” Behoove comes from Middle English bihoven “to need, be constrained; to be needed or required.” Bihofen, already mostly used as an impersonal verb in Middle English, comes from Old English behōfian, bihōfian “to need, require,” used both personally and impersonally. Behoove entered English before the end of the 9th century.
how is behoove used?
The current pandemic, which has curtailed normal interaction, throws into dramatic relief the central importance of teaching not only for our students’ learning, but also for their overall well-being. It behooves us all, after COVID-19, to build a more resilient system that includes rewards and support that encourage collaboration toward our common educational goal.
Lisa M. Di Bartolomeo and Pablo García Loaeza, "Teaching and Tenure: Part 1," _Inside Higher Ed_, March 29, 2021
In this troll-saturated context, it’s hard to care about street-level trolls and their movie boycotts. In fact, it would probably behoove us to stop caring about “trolls” at all.
Emma Grey Ellis, "Trolls Are Boring Now," _Wired_, March 13, 2019
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cupidity
[ kyoo-pid-i-tee ]
noun
eager or excessive desire, especially to possess something; greed.
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Cupidity “excessive desire; greed” comes from Old French cupidité, from Latin cupiditās (inflectional stem cupiditāt-) “passionate desire, yearning, longing; greed; lust,” a derivative of the adjective cupidus, which has the same meanings. Cupidus is in turn derivative of the verb cupere “to wish, wish for, desire,” which (unfortunately) has no reliable etymology. Cupidity entered English in the 15th century.
how is cupidity used?
Their enemies are not man. They are intolerance, fanaticism, dictatorship, cupidity, hatred and discrimination which lie within the heart of man.
Thich Nhat Hanh to Rev. Martin Luther King, June 1, 1965, in _Dialogue_, 1965
He rushed with ravenous eagerness at every bait which was offered to his cupidity.
Thomas Babington Macaulay, _The History of England from the Accession of James II_, Vol. 5,
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Apr 06 adminicle
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megillah
[ m uh-gil-uh; Sephardic Hebrew m uh-gee-lah ]
noun
a lengthy, detailed explanation or account.
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Megillah, a slang term usually meaning “a lengthy, detailed, complicated story, especially a tedious one” comes from Yiddish megile. Megile is part of the Yiddish phrase di gantse megile “the whole (tedious) story.” The Yiddish noun comes from Biblical Hebrew məgillāh “scroll, roll, volume,” a collective noun generally referring to any of the five Biblical books assigned for public recitation in synagogues on certain Jewish feast days, but specifically to the recitation of the Book of Esther during Purim “(the Feast of) Lots,” celebrated in late winter or early spring. Məgillāh is a derivative of the verb gālal “to roll.” Megillah entered English in its liturgical sense in the mid-17th century; its slang sense dates from the early 20th century.
how is megillah used?
It was Bella’s daughter, Liz Abzug, who suggested that Mr. Fierstein create a play about her mother. …(She actually hoped he would write a musical, but that’s a whole other megillah.)
Katherine Rosman, "Being Bella? Harvey Fierstein Doesn't Need a Dress to Try," _New York Times_, October 10, 2019
It’s long, which is a given when you consider the authorship — clocking in at a shade over 16 hours, this eight-episode megillah’s running time falls somewhere in between Burns’ look at WWII (The War) and his recent exploration of the conflict in Vietnam (The Vietnam War).
David Fear, "Country Music' Review: Ken Burns' Epic, Essential Look at an American Artform,"_Rolling Stone_, September 14, 2019
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adminicle
[ ad-min-i-k uh l ]
noun
an aid; auxiliary.
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Adminicle “an aid; auxiliary” comes ultimately from Latin adminiculum “prop (for vines), a stake or pole for support”; in Roman legal usage adminiculum means “an argument supporting a claim.” Adminiculum is a compound beginning with the Latin preposition and prefix ad, ad– “to, toward, at,” and ending with the diminutive suffix –culum, which is the source of the English suffixes –cule (as in molecule and ridicule) and, via Old French, –cle (as in article and canticle). The midsection mini– of adminiculum is problematic, but it is probably related to moenia “defensive walls of a town.” Adminiculum entered English in the mid-16th century.
how is adminicle used?
In fact it is very evident that to Dr. Osgood Classical Mythology is an adminicle to the study of Milton and not a study in itself.
"Brief Mention", _American Journal of Philology_, Vol. 21 No. 82, 1900
His routine of labor, while so burdened with woe, would have crushed him, were it not for the memory of his love, which was an adminicle to his strength.
Anson D. Eby, _Showers of Blessing_, 1908
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bricolage
[ bree-k uh-lahzh, brik-uh- ]
noun
a construction made of whatever materials are at hand; something created from a variety of available things.
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The noun bricolage in French means “do it yourself,” formed from the verb bricoler “to do odd jobs, do small chores; make improvised repairs,” from Middle French bricoler “to zigzag, bounce off,” ultimately a derivative of the Old French noun bricole “a trifle.” The French suffix –age, completely naturalized in English –age, as in carriage, marriage, passage, voyage, comes from –āticum, a noun suffix from the neuter of the Latin adjective suffix –āticus. Bricolage entered English in the second half of the 20th century.
how is bricolage used?
Indeed, if we scratch beneath the surface, English is a veritable bricolage of these ‘borrowed’ words.
Tim Lomas, "The Magic of 'Untranslatable' Words," _Scientific American_, July 12, 2016
So, for now, with my basket in one hand and my daughter’s little palm in the other, we’ll continue to walk the world in search of people, spaces and moments that move our soul and gather them into a living piece of art, a bricolage of memories called home.
Stevie Trujillo, "The Wager of Raising a Child Abroad," _New York Times_, February 16, 2018
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Apr 03 cackleberry
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leporine
[ lep-uh-rahyn, -rin ]
adjective
of, relating to, or resembling a rabbit or hare.
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Leporine, “pertaining to or resembling a rabbit or hare,” a technical term in zoology, comes straight from the Latin adjective leporīnus, a derivative of the noun lepus (inflectional stem lepor-) “hare.” The etymology of lepus is obscure, but it may be related to Greek dialect léporis (Sicily) and lebērís (Marseille). Leporine entered English in the mid-17th century.
how is leporine used?
Of course, the Easter Bunny isn’t our only leporine hero. There is a general fascination with hares, bunnies, and rabbits in children’s literature and other aspects of popular and folk culture around the world.
Ellen C. Caldwell, "The Easter Bunny, or, Why We Love Rabbits," JSTOR Daily, March 25, 2016
His face looked naked, his teeth big and leporine.
Karen Joy Fowler, _We Are All Completely Beside Ourselves_, 2013
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cackleberry
[ kak-uh l-ber-ee ]
noun
a hen's egg used for food.
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Cackleberry, “an egg, a hen’s egg,” is a piece of facetious American slang. The word is a compound of the verb cackle “to utter a shrill, broken cry such as a hen makes” and the common noun berry “small fruit without a pit,” also used often in compounds such as strawberry or gooseberry.
how is cackleberry used?
“Cackleberries,” said Gately, picking up one of the eggs and examining it as though it were an emerald. “A genuine cackleberry.”
Beirne Lay, Jr. and Sy Bartlett, _Twelve O'Clock High!_ 1948
Klock had played swell ball all week, scampering around station one like a hare—the March variety, of course—but he wasn’t hitting hard enough to imperil the shell of a cackleberry.
James W. Egan, "Cuckoo Klock," _Munsey's Magazine_, Vol. 73, June to September, 1921
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187562 | https://www.sciencedirect.com/topics/neuroscience/ventral-posteromedial-nucleus | Skip to Main content
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Ventral Posteromedial Nucleus
In subject area:Neuroscience
The Ventral Posteromedial Nucleus (VPM) is a part of the ventral posterior nucleus in the thalamus that receives input from the trigeminothalamic tracts and is involved in processing tactile sensation, particularly for the head region.
AI generated definition based on: Fundamental Neuroscience for Basic and Clinical Applications (Fifth Edition), 2018
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The Somatosensory System I
2018, Fundamental Neuroscience for Basic and Clinical Applications (Fifth Edition)S. Warren, ... R.P. Yezierski
Ventral Posterior Nucleus
The ventral posterior nucleus, sometimes called the ventrobasal complex, is a wedge-shaped cell group located caudally in the thalamus. Its lateral border abuts the internal capsule, and ventrally it borders on the external medullary lamina. The ventral posterior nucleus is composed of the laterally located VPL and the medially located ventral posteromedial nucleus (VPM). Although these nuclei have also been termed the ventralis caudalis externus and ventralis caudalis internus in humans, the more widely used and recognized terms VPL and VPM are used in this book. The VPL is separated from the VPM by fibers of the arcuate lamina. The ventral posterior nucleus (VPM and VPL) is supplied by thalamogeniculate branches of the posterior cerebral artery, and compromise of these vessels can result in loss of all tactile sensation over the contralateral body and head (Fig. 17.10).
The VPL receives ascending input from the medial lemniscus, and input to the VPM is from the trigeminothalamic tracts. Within the VPL, medial lemniscal fibers from the contralateral cuneate nucleus terminate medial to those from the gracile nucleus. As a result, the representation of the lower extremity is lateral, and that of the upper extremity is medial in the VPL (Fig. 17.10). The representation of an individual body part is organized as a C-shaped lamina. Tactile signals are also represented in other thalamic nuclei receiving lemniscal input, including the ventral posterior inferior nucleus and the pulvinar and lateral posterior group.
In addition to their somatotopic organization, the medial lemniscal fibers that terminate in the ventral posterior nucleus are segregated on the basis of their functional properties. Rapidly and slowly adapting inputs terminate on different cell groups within the core region of the VPL. Pacinian inputs and inputs arising from joints and muscles are confined to a shell region on the posterior, rostral, and anterior edges of the nucleus. Individual lemniscal axons arborize in the sagittal plane to terminate on longitudinal cell clusters, called rods, in the VPL. This arrangement of inputs and target cells creates representations consisting of neurons with similar receptive fields and submodalities arranged along a rostrocaudal axis.
The VPL for the trunk and extremities (and VPM for the head) contains two populations of identified neurons. The first consists of large-diameter multipolar cells that give rise to axons that traverse the posterior limb of the internal capsule and terminate mainly in the primary (SI) and secondary (SII) somatosensory cortices. These thalamocortical cells and fibers are the third-order neurons in the PCMLS that provide excitatory (glutaminergic) input to the cortex. The second population consists of inhibitory (γ-aminobutyric acid [GABA]ergic) local circuit interneurons, which receive excitatory corticothalamic inputs and influence the firing rates of thalamocortical cells. In addition, these thalamocortical cells are also influenced by GABAergic input from the thalamic reticular nucleus and by excitatory (glutaminergic) corticothalamic fibers that arise in layer VI of the primary and secondary somatosensory cortices.
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Book2018, Fundamental Neuroscience for Basic and Clinical Applications (Fifth Edition)S. Warren, ... R.P. Yezierski
Chapter
Thalamus
2004, The Rat Nervous System (Third Edition)Henk J. Groenewegen, Menno P. Witter
Posterior Nucleus
The posterior thalamic nucleus (Po; also indicated as posteror complex) is situated in the caudal part of the thalamus, bordered caudally by the pretectal nuclei. In its caudal aspect, the Po is situated medial to the posterior intralaminar (PIL) and the medial geniculate (MGM) nuclei, more rostrally this position is taken by the ventral posterior complex, in particular the ventral posteromedial nucleus (Figures 30–42 in Paxinos and Watson, 1998). Dorsally, the Po is bordered by the lateral dorsal and lateral posterior nuclei; medial to the Po the intralaminar nuclei are situated. The posterior nucleus in rats is a heterogeneous area, in particular in its caudal aspects. Within the region generally considered to belong to the Po, several subnuclei have been identified on the basis of differential staining patterns, e.g. the ethmoid, scaphoid, and retroethmoid nuclei (Paxinos et al., 1999). The Po appears as a relatively cell-sparse area in Nissl-stained sections, standing out against the cell-dense ventral posteromedial nucleus medially. In acetylcholineseterase-stained sections, the Po shows moderate activity for this enzyme and also on the basis of this staining the posterior nucleus can be reasonably well-delineated from its neighboring nuclei (Plates 31, 33, 35, and 37 in Paxinos and Watson, 1998). With respect to calcium-binding proteins, the Po contains a population of calbindin D28K-positive neurons, while calretinin and parvalbumin are present only in a light plexus of fibers (Arai et al., 1994; e.g., Figs. 195, 201, 202, 208, and 209 in Paxinos et al., 1999).
Afferent and efferent connections
Like the ventral posterior complex, the posterior complex receives a significant input from the spinal cord and brain stem trigeminal complex (Cliffer et al., 1991; Chiaia et al., 1991a, 1991b; see also Tracey, Chapter 25, and Waite, Chapter 26, this volume). However, most of these ascending projections are less dense and they terminate in a more diffuse way in the Po than in the adjacent ventral posterior nuclei (e.g., Chiaia et al., 1991a; Villanueva et al., 1998). Moreover, projections to the posterior nucleus from the spinothalamic tract and the spinal trigeminal nucleus appear to be relatively more dense than those from the dorsal column nuclei and the principal trigeminal nucleus (Chiaia et al., 1991a; McAllister and Wells, 1981). Further inputs to the Po come from the inferior colliculus (LeDoux et al., 1987) as well as from the vestibular nuclei (Shiroyama et al., 1999). Reciprocal connections have been described between the posteromedial part of the Po and the region of the nucleus ruber (Roger and Cadusseau, 1987).
Recent studies have demonstrated that the trigeminal complex gives rise to several types of fibers ascending to the thalamus which differentially distribute over the ventral posteromedial nucleus and the Po, as well as over other extrathalamic nuclei. While most fibers originating from the principal trigeminal nucleus terminate in a very specific manner in the barreloids in the ventral posteromedial nucleus, a smaller contingent of principal trigeminal fibers ascends to the Po. In addition, these fibers give off collaterals to the tectum, the zona incerta, the ventral part of ventral posteromedial nucleus, and the medial part of medial geniculate nucleus. The neurons in the principal trigeminal nucleus giving rise to the projections to the Po are sensitive to stimulation of multiple whiskers, in contrast to those that project to the barreloids in the ventral posteromedial nucleus which transmit signals from single whiskers only (Veinante and Deschenes, 1999). Thalamic projection fibers originating in the spinal trigeminal complex can also be categorized in several types: one type of fiber projects primarily to the ventral part of ventral posteromedial nucleus; the other type of fiber projects predominantly to the caudal part of the Po while giving off collaterals to, among others, the tectum, the zona incerta, and the ventral lateral thalamic nucleus (Veinante et al., 2000a).
The Po, like the ventral posterior complex, projects to the primary and secondary sensory cortices, largely following a somatotopical organization (Fabri and Burton, 1991). Thalamocortical fibers from the posterior nucleus terminate in the upper part of layer VI as well as in layer I of the cortex (Zhang and Deschenes, 1998). In the barrel cortex, the projection fibers from the Po do not terminate within the barrels, but they preferentially target the interbarrel areas (Lin and Lu, 1993). Cortical projections to the Po primarily arise from the somatosensory areas S1 and S2, but additional projections originate from motor, premotor (frontal eye field), and insular cortices (Veinante et al., 2000b; Guandalini, 2001). While S1 projects predominantly to dorsal parts of the posterior complex, S2 targets more ventral and medial parts, although there exist substantial areas of overlap (Shi and Cassell, 1998b; Veinante et al., 2000b). The corticothalamic projections to Po arise predominantly from layer V pyramidal neurons as collaterals from axons that project to the striatum and descend into the brain stem (Fig. 4) (Levesque et al., 1996; Veinante et al., 2000b). The fibers originating in layer V of the barrel cortex, possibly predominantly originating from neurons in interbarrel cortical areas, exhibit giant terminals of the RL-type in the Po and they may be considered drivers in this nucleus (see section “Some General Aspects of Thalamic Organization”; Hoogland et al., 1987, 1991; Bourassa et al., 1995; Wright et al., 2000; Veinante et al., 2000b). The other cortical areas mentioned above, as well as the deep layer VI neurons in the cortical barrel field, project fibers with much smaller boutons in the Po and these may be considered modulators (Welker et al., 1988; Hoogland et al., 1991; Bourassa et al., 1995).
Functional aspects
It may be clear that the Po receives ascending inputs from various different modalities (see above), i.e., somatosensory, auditory, visual, and vestibular. Although there is probably convergence within the posterior nucleus of fibers carrying information about these different modalities, the somatosensory modality strongly dominates. Whereas the Po might be considered a primary relay thalamic nucleus, the driving input from layer V of the sensory cortex places this nucleus also in the category of higher order thalamic nuclei (see section “Some General Aspects of Thalamic Organization”). As may be clear from the discussions of the afferents and efferents of both the ventral posteromedial nucleus and the Po, the whisker system in rats occupies a large part of the somatosensory system, and the functional anatomy of this system has been analyzed in great detail (e.g., Moore et al., 1999; Ahissar and Zackenhouse, 2001). Therefore, the most concrete suggestions for a functional role of the Po have been made with respect to the behavioral functions of the whisker system. The whisker system not only provides information about passive movements of individual or groups of whiskers but the whiskers are also actively moved in certain frequencies to “explore” the surroundings. The fact that the Po receives its main cortical input from layer V neurons probably indicates that this concerns an “efference copy” of decending (motor) cortical signals to the brain stem. This has led Veinante et al. (2000b) to the suggestion that the Po plays a role at the interface between the sensory and motor aspects of the whisker system.
More specifically, the posterior complex receives the motor signals for active whisker movements as well as the sensory feedback. The Po, together with parts of the trigeminal complex, might thus be involved in monitoring the dynamics of the self-initiated whisker movements, a mechanism that may be necessary for a sensory organ such as the whiskers that lack proprioceptors (Veinante et al., 2000b). Whether the Po has other functions at the interface between the sensory and motor systems remains to be established. The reciprocal connections between the posterior nucleus and the red nucleus provide an indication in that direction (Roger and Cadusseau, 1987; Cadusseau and Roger, 1990; Arnault et al., 1994).
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Book2004, The Rat Nervous System (Third Edition)Henk J. Groenewegen, Menno P. Witter
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Taste
2017, Reference Module in Neuroscience and Biobehavioral PsychologyM.E. Frank, M.A. Barry
Parvicellular Ventral Posteromedial Thalamic Nucleus
The gustatory pathway to the thalamus is largely ipsilateral; VPMpc thalamic neurons respond only to stimulation of the tongue on the same side in most species (Fig. 3). However, the situation in humans is not as clear. In the few available clinical cases, ipsilateral gustatory losses usually follow pontine or midbrain lesions, but there are some contradictory examples and some experimental studies suggest that projections from the NST to the thalamus are bilateral. VPMpc lies immediately medial to the VPM and is contiguous with the facial–oral somatosensory representation in VPM. Neurons in VPMpc are smaller than in the VPM. VPMpc neurons respond to taste stimuli, but unlike the taste part of the NST, neurons that respond to somatosensory stimuli or somatosensory and taste stimuli are also present. General viscerosensory information may also be processed in VPMpc. Thus, unlike somatosensory, visual, and auditory thalamic nuclei, VPMpc, the thalamic nucleus that processes taste information, also processes information from other modalities.
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Reference work2017, Reference Module in Neuroscience and Biobehavioral PsychologyM.E. Frank, M.A. Barry
Chapter
Thalamus
2004, The Rat Nervous System (Third Edition)Henk J. Groenewegen, Menno P. Witter
Sensory Nuclei
Lateral Geniculate Nucleus
The lateral geniculate nucleus in rats is a relatively flattened, more or less oval-shaped nucleus on the dorsolateral surface of the caudal thalamus. It can be subdivided into a dorsal lateral geniculate nucleus (DLG) and a ventral lateral geniculate nucleus (VLG). Between the DLG and the VLG a third component is recognized, the intergeniculate leaflet (IGL). The DLG constitutes the main thalamic relay of visual information to the primary visual cortex, the VLG has a number of characteristics in common with the thalamic reticular nucleus, and only its lateral part receives direct retinal input. The IGL constitutes a relay between the retina and the hypothalamus and is involved in circadian functions. For a comprehensive treatment of the anatomical and functional aspects of the lateral geniculate complex, including the DLG, VLG, and IGL, see Sefton et al. (Chapter 32, this volume).
Dorsal lateral geniculate nucleus (DLG)
The dorsal lateral geniculate nucleus can be readily identified in Nissl-stained sections (Plate 37 in Paxinos and Watson, 1998) as well in acetylcholinesterase-stained sections in which the DLG shows moderate activity (Plate 39 in Paxinos and Watson, 1998). The cytoarchitecture of the DLG of rats is rather homogeneous. The majority of dorsal lateral geniculate neurons are thalamocortical projection cells. Unlike most other principal thalamic nuclei in the rat, the DLG contains several types of interneurons, namely, GABAergic, most of which coexpress NADPH diaphorase, and solely GABAergic or NADPH diaphorase-containing neurons (Ohara et al., 1983; Jones, 1985; Gabbott and Bacon, 1994). In contrast to many other mammalian species, the rat DLG is not clearly laminated, although fiber bundles running in a ventrolateral to dorsomedial orientation, parallel to the optic tract, impose a certain orientation on the neurons in the dorsal lateral geniculate nucleus. However, the largely segregated terminations of the optic fibers from the contralateral and ipsilateral eyes reveal a form of “hidden lamination” in the rat DLG (Reese, 1988; also Jones, 1985; Price, 1995). The lateral lamina directly adjacent to the optic tract, also called the “outer shell,” receives input from the contralateral eye. The medial part of the DLG, called the “inner core,” consists of two regions, the most medial region receives input from the contralateral eye and the lateral region is innervated by the ipsilateral eye (Reese, 1988; Jones, 1985). This organization refers to the caudolateral part of the DLG in which there is a binocular representation of the visual field; in the rostroventral part of the DLG retinal fibers from only the contralateral eye terminate, representing a large part of the temporal visual field (Reese, 1988). Calcium-binding proteins are differentially distributed in fibers and neurons in the dorsal lateral geniculate nucleus. Whereas calretinin and parvalbumin are present only in fibers, calbindin D28K is also expressed in (inter)neurons, in particular in the outer shell (Luth et al., 1993; Figure 230 in Paxinos et al., 1999). The plexus of calretinin fibers is most dense also in the outer shell (Figure 230 in Paxinos et al., 1999). Parvalbumin fibers are most probably derived from the retina and the reticular thalamic nucleus, and calretinin fibers from the retina (Luth et al., 1993; Arai et al., 1992). Calbindin D28K-containing fibers may be derived from the superior colliculus (Lane et al., 1997).
Affernt and efferent Projections
The DLG forms the main relay between the retina and the primary visual cortex (area 17 or V1), the optic terminations in the dorsal lateral geniculate nucleus being retinotopically organized (see above; Jones, 1985; Reese, 1988). The inner core receives, in its two ocular laminae, inputs from the contralateral nasal and the ipsilateral temporal retina, mapping the contralateral visual hemifield. The outer shell receives a projection from the complete contralateral visual hemifield only. These retinotopic maps in the contralateral and ipsilateral laminae of the DLG are in complete register. Lines of projection are oriented rostroventromedially from the optic tract at the thalamic surface through the different laminae of the DLG (Reese, 1988). Retinal inputs terminate as RL-type boutons, indicating that the retinal fibers are the “drivers” of the DLG (Jones, 1985; Sherman and Guillery, 2001; also II). The retinal afferents contact the dendrites of both the thalamocortical neurons and the interneurons, and they take part in glomerulus-like formations separated from the rest of the neuropil by glial elements.
Cortical inputs to the dorsal lateral geniculate nucleus are primarily derived from the primary visual cortex (area 17). The different cortical layers of the primary visual cortex have differential projection patterns to the visual thalamus (Bourassa and Deschennes, 1995). Fibers originating in the upper part of layer VI project to the DLG and they terminate in rostrocaudally oriented bands or “rods” that run parallel to the lines of projection of retinal afferents. Neurons in the deeper part of layer VI project to the lateral part of the lateral posterior thalamic nucleus and these fibers give off collaterals to the DLG where they take part in the formation of the rods. Neurons in layer V of the visual cortex do not target the DLG, but their main axon descends to the brain stem while giving off collaterals to the ventral lateral geniculate nucleus and to the lateral posterior and lateral dorsal thalamic nuclei (Bourassa and Deschennes, 1995). Layer VI corticothalamic fibers issue collaterals to the reticular thalamic nucleus while layer V axons do not. The terminating fibers in the DLG, originating from layer VI have small “en passant” varicosities which show at the ultrastrutural level the characteristics of the RS-type boutons and may be considered as “modulators” (Jones, 1985; Price, 1995; Bourassa and Deschennes, 1995; Sherman and Guillery, 2001)
Subcortical inputs to the DLG arise in the ventral lateral geniculate nucleus, reticular thalamic nucleus, superior colliculus, and several brain stem nuclei (Reese, 1988; Coleman and Mitrofanis, 1996; Moore et al., 2000). The inputs from the superior colliculus terminate in the peripheral zone of the outer shell and are probably associated with the calbindin DD28K-positive fiber plexus (Reese, 1988; Lane et al., 1997). Brain stem inputs include those from retinal input receiving nuclei like the nucleus of the optic tract, the olivary pretectal nucleus, and the parabigeminal nuclei (Schmidt et al., 1995). Further inputs are derived from the locus coeruleus (noradrenaline), the dorsal raphe nucleus (serotonin), and the laterodorsal tegmental nucleus (acetylcholine) (e.g., Papadopoulos and Parnavelas, 1990).
The output of the DLG is primarily directed at the primary visual cortex (area 17), terminating in layer IV, while there are lesser inputs to layers I and VI (Ribak and Peters, 1975; Jones, 1985). The peristriate area 18 also receives a weak input from the DLG (Sanderson et al., 1991). The geniculocortical pathway uses glutamate as neurotransmitter (Kharazia and Weinberg, 1994; Saez et al., 1998).
Ventral lateral geniculate nucleus
The VLG, like the reticular thalamic nucleus, is embryologically derived from the ventral thalamus. The ventral lateral geniculate nucleus could be viewed as a caudodorsal extension of the reticular thalamic nucleus. Cytoarchitectonically, the VLG can be subdivided into a lateral, magnocellular part (VLGMC) and a somewhat smaller, medial parvicellular part (VLGPC). The two parts are separated by a fiber-rich, cell-free zone (Jones, 1985). Neurons in the magnocellular part contain nitric oxide synthase and enkephalin; those in the parvicellular part substance P and calretinin (Meng et al., 1998; for review see Harrington, 1997). Enkephalinergic neurons have also been identified in the VLG (Hermanson et al., 1995).
Afferent and efferent projections
Afferents to the ventral lateral geniculate nucleus make a clear distinction between the medial VLGPC and the lateral VLGMC. The VLGPC receives extensive inputs from the brain stem, in particular from the reticular formation, the deep layers of the superior colliculus, the periaqueductal gray matter, peribrachial regions, the laterodorsal tegmental nucleus, the locus coeruleus, the substantia nigra pars reticulata, and deep cerebellar nuclei (Kolmac and Mitrofanis, 2000; Vaudano and Legg, 1992). The VLGMC receives strong projections mainly from the retina and layer V of the visual cortex (Hickey and Spear, 1976; Takahashi, 1985; Bourassa and Deschennes, 1995) and very few fibers from the brain stem (Kolmac and Mitrofanis, 2000). Unlike the dorsal lateral geniculate nucleus, there are no projections from the VLG to the cerebral cortex. Instead, the VLG has rather extensive projections to the dorsal thalamus, comparable with the intergeniculate leaflet (see below). Thus, the medial, parvicellular part of the VLG projects to the parafascicular and lateral dorsal thalamic nuclei as well as to the reuniens and rhomboid nuclei in the midline. The lateral, magnocellular VLG sends fibers to the dorsal lateral geniculate and lateral posterior thalamic nuclei (Moore et al., 2000; Kolmac et al., 2000). Hypothalamic projections from the VLG reach the lateral and posterior hypothalamus and the perifornical area. The ventral lateral geniculate nucleus further projects to the zona incerta, the pretectal nuclei, the deep and intermediate layers of the superior colliculus, the dorsal and medial terminal nuclei of the accessory optic system, the periaqueductal gray, the peripeduncular region, and the accessory inferior olive (Moore et al., 2000).
Intergeniculate leaflet
The IGL is a distinct, dorsoventrally narrow region between the dorsal and ventral lateral geniculate nuclei, which extends virtually over the entire rostrocaudal length of the lateral geniculate complex (Hickey and Spear, 1976; Moore and Card, 1994). Like the ventral lateral geniculate nucleus, it is a derivative of the ventral thalamus. Precise borders of the IGL are difficult to establish in Nissl-stained sections but can more readily be identified with the staining for glial fibrillary acidic protein (GFAP) and several peptides like neuropeptide Y, substance P, and enkephalin, as well as the neurokinin-1 receptor. The IGL contains several types of small- to medium-sized neurons most of which have their dendritic arborizations within the nucleus (Moore and Card, 1994; Piggins et al., 2001).
Afferent and efferent connections
The main sources of input to the intergeniculate leaflet are fibers from the retina and from the contralateral IGL. Retinal fibers terminate as RL-type terminals and may be considered the driving afferents of the IGL (Mikkelsen, 1992; Moore and Card, 1994). The neurons that give rise to the commissural connections contain enkephalin (Card and Moore, 1989). Further inputs to the IGL originate in the suprachiasmatic nucleus, posterior hypothalamic area, superior colliculus, and several brain stem nuclei, among them the locus coeruleus, the raphe, and the laterodorsal tegmental area (Moore et al., 2000). There is a dense substance P-immuoreactive plexus in the intergeniculate leaflet, the origin of which is unknown, however (Piggins et al., 2001). The efferents of the IGL primarily reach the hypothalamus, in particular the suprachiasmatic nucleus and anterior hypothalamic regions, forming the so-called geniculohypothalamic tract. Neurons in the IGL projecting to the suprachiasmatic nucleus contain neuropeptide Y (Card and Moore, 1989; Harrington, 1997). Mikkelsen (1994) showed direct projections from the intergeniculate leaflet to the subcommissural organ and the pineal gland. Further outputs of the IGL reach the midline thalamic nuclei, in particular, the paraventricular, but also the reuniens and rhomboid nuclei, as well as the dorsal and lateral hypothalamus and the zona incerta. Brain stem targets include the superficial layers of the superior colliculus, the periaqueductal gray, and several accessory optic nuclei (Moore et al., 2000).
Functional aspects of the lateral geniculate complex
The DLG must be considered as the main thalamic gateway for visual information from the retina to reach the cerebral cortex. This nucleus rather faithfully maps the external visual field onto the primary visual cortex (area 17, V1). The functional aspects of the DLG have been the subject of a vast body of literature and have recently been elegantly reviewed by others (e.g., Sherman and Guillery, 2001). For a further elaboration on this subject see Sefton et al. (Chapter 32, this volume).
The functions of the VLG have been less well-established. In view of its origin from the ventral thalamus, as well as some of its connectional characteristics, i.e., the rather extensive projections to the dorsal thalamus (see above), the functions of the VLG might, at least in part, be compared with the those of the reticular thalamic nucleus (see also section “Reticular Nucleus”; Sefton et al., Chapter 32, this volume). However, the functional aspects of the VLG are probably much more differentiated. As a result of their differential sets of afferent and efferent fibers, the medial and lateral parts of the VLG (VLGPC and VLGMC, respectively) are bound to have different, although possibly related, functions. Whereas the lateral VLGMC is more intimately related to the dorsal lateral geniculate nucleus and visual cortices, the medial VLGPC is more closely associated with the hypothalamus, in particular, the suprachiasmatic nucleus. Yet, these two parts of the VLG also share a number of afferents and efferents, as well as with the IGL (see below). Therefore, in a recent review, Harrington (1997) has argued that this entire ventral lateral geniculate complex may be regarded to fulfill distinct, yet interrelated, functions in controlling visuo-motor responses and circadian rhythms. Details of these functions, however, remain to be established. Functional aspects of the IGL seem to have been studied in slightly more detail (see also Sefton et al., Chapter 32, this volume). The retinal input to the IGL originates from a specific set of retinal ganglion cells that convey luminance information (Moore et al., 1995). Furthermore, the strong reciprocal connections of IGL with the suprachiasmatic nucleus as well as its connections with midline thalamic nuclei and the pineal gland fit in with an important role for the IGL in circadian functions (Moore and Card, 1994; Harrington, 1997). While the IGL projections to the suprachiasmatic nucleus provide an indirect way to influence autonomic and neuroendocrine circadian rhythms, there are also direct connections of IGL fibers with neuroendocrine neurons in other parts of the hypothalamus (Horvath, 1998).
Ventral Posterior Complex
The ventral posterior complex (VP) in rats occupies an extensive ventrolateral thalamic area, appearing immediately rostromedial to the medial geniculate complex in the caudal third of the thalamus and extending into the rostral third of the thalamus. The VP is bordered ventrally and laterally by the medial lemniscus and the reticular thalamic nucleus and dorsomedially by the posterior complex (see page 417). At rostral levels the posterior complex is gradually replaced by the ventrolateral nucleus, which is situated medial to the VP (see Figs. 28–38 in Paxinos and Watson, 1998). The ventral posterior complex is the main relay for sensory inputs to reach the cerebral cortex and it can be divided into at least three main parts: the ventral posterolateral nucleus (VPL) receiving spinal somatosensory inputs, the ventral posteromedial nucleus (VPM), receiving trigeminal somatosensory inputs, and the parvicellular medial parts of both the VPL and VPM, i.e., the VPPC (see page 419), which forms the main thalamic relay for gustatory and visceral ascending pathways.
The VPL and VPM are distinguishable not only based upon their connectivity (see below) but also on the basis of cyto- and chemoarchitectonics. The VPM stands out as a more densely packed nucleus in Nissl-stained sections as compared to the VPL and the posterior complex (e.g., Plates 33 and 37 in Paxinos and Watson, 1998). The VPL and the posterior complex demonstrate a moderate level of activity for acetylcholinesterase (AChE), while the activity is very low in the VPM (e.g., Plates 31 and 35 in Paxinos and Watson, 1998). Most neurons in the VPM and VPL are medium-sized thalamocortical neurons. In contrast to other species, only very few GABAergic interneurons are present in the ventral posterior complex of rats (Harris and Hendrickson, 1987; Price, 1995). The neurons in the VPL are arranged in rostrocaudal and dorsoventral rows that are roughly parallel with the external medullary lamina and these rows curve partially around the rostral pole of the VPM (McAllister and Wells, 1981). In both the VPL and the VPM a dense plexus of parvalbumin fibers is present, while only few calretinin- or calbindin D28K-positive fibers occur. In subregions of both the VPL and VPM, calbindin D28K-positive neurons are present (Arai et al., 1994; e.g., Figs. 188, 194, 202, and 208 in Paxinos et al., 1999).
The VPM is for the most part organized in so-called barreloids, first described in mice and later also identified in rats (Van der Loos, 1976) (Fig. 2). Barreloids are the representations of individual whiskers at the level of the thalamus (see below) formed by cellular aggregates; barreloids can best be visualized using mitochondrial markers such as cytochrome oxidase (Land and Simons, 1985; Haidarliu and Ahissar, 2001). These microstructures in the VPM are most apparent in young rats, but can also be demonstrated in adults. Whereas thalamic barreloids are considered to convey the information from single whiskers, the dendrites of neurons in these barreloids may cross the boundaries into neighboring barreloids providing a neural substrate for crosstalk between barreloids (Desilets-Roy et al., 2002).
Afferent and efferent connections
The ventral posterolateral and ventral posteromedial nuclei receive their main ascending inputs from somatosensory afferents originating in the spinal cord, dorsal column nuclei, and trigeminal complex (see also Tracey, Chapters 7 and 25, and Waite, Chapter 26, this volume). These somatosensory projections are somatotopically organized such that afferents from the trunk and limbs terminate in the VPL and those from the head terminate in the VPM. Spinal and trigeminal fibers not only reach the VPL and VPM but also distribute over much wider areas of the thalamus, including the posterior nucleus (see page 417) and the intralaminar nuclei (see section “Midline and Intralaminar Thalamic Nuclei”). Spinothalamic fibers, transferring among others nociceptive signals, originate from different laminae of the dorsal horn as well as the central gray of the spinal cord and terminate as large boutons in the ventral posterolateral nucleus (e.g., McAllister and Wells, 1981; Burstein et al., 1990; Dado et al., 1994; Katter et al., 1996; Kobayashi, 1998). Nociceptive information may reach the VPL also indirectly via the caudal medullary reticular formation (medullary dorsal reticular nucleus) in addition to the direct spinothalamic pathway (Villanueva et al., 1998). Dorsal column nuclear afferents likewise terminate with large boutons in the VPL (McAllister and Wells, 1981; Villanueva et al., 1998). Spinal and lemniscal fibers, in part, converge on individual thalamocortical neurons, the lemniscal fibers on the soma, and more proximal parts of dendrites of the neurons than the spinothalamic fibers (Peschanski and Ralston, 1985). Lemniscal, but also spinothalamic, fibers probably use glutamate as neurotransmitter (De Biasi et al., 1994). Spinothalamic fibers have also been shown to contain substance P (Battaglia et al., 1992; Nishiyama et al., 1995).
The spinal and principal trigeminal nuclei both project to the ventral posteromedial nucleus as well as, less densely, to the posterior complex (cf. also Waite, Chapter 26, this volume). Projections to the VPM are more focussed, those to the medial part of posterior nucleus more diffuse (Chiaia et al., 1991a). The whisker-sensitive parts of the spinal and principal trigeminal nuclei appear to give rise to two trigeminothalamic pathways that reach different parts of the barreloids. These pathways possibly play different functional roles in the relay of information from the whiskers to the somatosensory cortex (Williams et al., 1994; Veinante and Deschenes, 1999; Veinante et al., 2000a; Pierret et al., 2000).
From the principal trigeminal nucleus two types of thalamic projection fibers originate. Fibers of the first type, arising from small neurons confined to the so-called barrelettes in the principal trigeminal nucleus which each have a receptive fields for a single whisker, project densely to the “core compartments” of single barreloids in the dorsomedial two-thirds of the VPM. The second type of fiber originates from larger neurons with receptive fields for multiple whiskers. These fibers terminate in much wider thalamic regions, including the ventral part of the VPM containing the so-called “tail compartments” of the barreloids (Pierret et al., 2000; Veinante and Deschenes, 1999). Specific parts of the spinal trigeminal complex, in particular its interpolar division but also the rostral part of the caudal division, reach the ventral one-third of the VPM, as well as in the interbarreloid areas in the dorsal VPM (Williams et al., 1994; Veinante and Deschenes, 1999). The fibers from the interpolar trigeminal division are thin and form small-sized bushy arbors within the ventral posteromedial nucleus.
The VPL and VPM receive afferents from various other subcortical areas, among them a serotonergic input from the dorsal raphe nucleus and a GABAergic input from the thalamic reticular nucleus (see section “Reticular Nucleus”) (Cox et al., 1996; Kirifides et al., 2001). Reticular thalamic afferents originate from different types of neurons which have different termination patterns in the ventral posterior complex: some reticular fibers terminate as clusters; others have a wide and more diffuse termination pattern. This indicates that the inhibitory influences of the reticular thalamic nuceus serve different roles in the ventrobasal complex (Cox et al., 1996). The barreloids in the VPM are the thalamic modules in the pathways that lead from the whiskers via the barrelettes in the trigeminal complex to the barrels in the somatosensory cortex (S1). There is an almost exclusive one-to-one relationship from the individual peripheral whiskers to the corresponding cortical barrels. Thus, the thalamocortical projections from the barreloids in the VPM to the barrel cortex, like their corticothalamic counterparts, are very strictly topographically organized (Lu and Lin, 1993; Land et al., 1995; see below). In contrast, thalamocortical fibers from the posterior complex terminate only in the interbarrel areas (Lu and Lin, 1993). Thalamocortical axons from the VPL and VPM terminate predominantly in layer IV of the primary sensory cortex and they most likely use glutamate as neurotransmitter (Kharazia and Weinberg, 1994). Apart from the main terminations in layer IV, thalamocortical axons from the VPM issue also terminal branches in layers I and V/VI (Lu and Lin, 1993; Zhang and Deschennes, 1998).
The corticothalamic projections related to the VPL and VPM are organized in a complex way and reflect the hypothesized way in which the afferents of the thalamus are organized as drivers and modulators (see section “Some General Aspects of Thalamic Organization). Corticothalamic projections from the barrel cortex originating from pyramidal neurons in the upper part of layer VI terminate exclusively in the VPM where they arborize in long rostrocaudally oriented bands or “rods.” A rod originating from a single barrel in the cortex makes contact with a series of barreloids that together represent an arc of whiskers (Hoogland et al., 1987; Bourassa et al., 1995). Neurons in the deeper parts of layer VI of the barrel cortex primarily project to the medial part of posterior thalamic nucleus but, in addition, they give off collaterals to the VPM where they participate in the formation of rods (Fig. 3). Neurons in layer VI located in the interbarrel areas project exclusively to the posterior nucleus (Wright et al., 2000). All corticothalamic fibers originating in layer VI give off collaterals to the reticular nucleus and they have long branches that have numerous en passant boutons (Bourassa et al., 1995; Levesque et al., 1996; Wright et al., 2000). Corticothalamic fibers originating from pyramidal neurons in layer V neurons in the barrel field have main axons descending to the brain stem. They give off collaterals to the posterior and intralaminar thalamic nuclei, but there are no collaterals to the reticular nucleus and VPM. These layer V fibers terminate in a few clusters with large boutons in the posterior thalamic nucleus (Fig. 4) (Hoogland et al., 1991; Bourassa et al., 1995; Veinante et al., 2000b).
Functional aspects
The VPL and VPM are the primary thalamic relays for somatic sensory, i.e., nociceptive and tactile/kinestetic, information from the body and the head, respectively. For a comprehensive treatment of the pain system and the role of the thalamus in transmitting and modulating pain signals, see Chapter 27 by Willis et al. in this volume.
In the rat VPM, the barreloid area representing the whisker field occupies a large part of the nucleus, signifying the importance of the rodent whisker system for navigation and exploration of the environment. This is in part an active process during which the whiskers are being moved in exploratory activities. It has been hypothesized that one of the two functionally different “channels” that reach the VPM, i.e., the pathway that originates in the spinal trigeminal complex, may play a role in this active “whiskering,” while the second channel originating in the principal trigeminal nucleus conveys information about passive deflections (Pierret et al., 2000).
As indicated above, there exists an almost exclusive one-to-one relationship of individual whiskers with single barreloids in the ventral posteromedial nucleus. However, for the whisker system to function as an integrated system, which is of course necessary in order to interprete the detailed environmental information, integration of information from the different whiskers is taking place at several levels along the ascending pathway from the periphery to the barrel cortex. The ventral posteromedial nucleus is thought to play an important role in this process. This is signified by the complex interrelationships that exist between the VPM and the barrel cortex, as well as by the differentiated position of the reticular thalamic nucleus. For example, as indicated above, whereas individual barreloids project to single cortical barrels, return projections from the cortex take the form of rods in the VPM, “interconnecting” barreloids representing several or all whiskers in an arc (Hoogland et al., 1987; Bourassa et al., 1995). In the reticular and posterior thalamic nuclei efferents from a row of cortical barrels converge to a common termination site (Welker et al., 1988). The reticular nucleus, sending at least two types of fibers with distinct distribution patterns to the VPM, plays a significant and differentiated role in the modulation of information at the level of the VPM (Cox et al., 1996).
There is an extensive body of literature about the electrophysiological and functional aspects of the thalamic relay of somatic sensory information in rats, in particular in relation to the whisker system, but further elaboration on this subject is beyond the scope of this chapter. For a more comprehensive view on this subject, the reader is referred to recent reviews (e.g., Moore et al., 1999; Ahissar and Zackenhouse, 2001).
Posterior Nucleus
The posterior thalamic nucleus (Po; also indicated as posteror complex) is situated in the caudal part of the thalamus, bordered caudally by the pretectal nuclei. In its caudal aspect, the Po is situated medial to the posterior intralaminar (PIL) and the medial geniculate (MGM) nuclei, more rostrally this position is taken by the ventral posterior complex, in particular the ventral posteromedial nucleus (Figures 30–42 in Paxinos and Watson, 1998). Dorsally, the Po is bordered by the lateral dorsal and lateral posterior nuclei; medial to the Po the intralaminar nuclei are situated. The posterior nucleus in rats is a heterogeneous area, in particular in its caudal aspects. Within the region generally considered to belong to the Po, several subnuclei have been identified on the basis of differential staining patterns, e.g. the ethmoid, scaphoid, and retroethmoid nuclei (Paxinos et al., 1999). The Po appears as a relatively cell-sparse area in Nissl-stained sections, standing out against the cell-dense ventral posteromedial nucleus medially. In acetylcholineseterase-stained sections, the Po shows moderate activity for this enzyme and also on the basis of this staining the posterior nucleus can be reasonably well-delineated from its neighboring nuclei (Plates 31, 33, 35, and 37 in Paxinos and Watson, 1998). With respect to calcium-binding proteins, the Po contains a population of calbindin D28K-positive neurons, while calretinin and parvalbumin are present only in a light plexus of fibers (Arai et al., 1994; e.g., Figs. 195, 201, 202, 208, and 209 in Paxinos et al., 1999).
Afferent and efferent connections
Like the ventral posterior complex, the posterior complex receives a significant input from the spinal cord and brain stem trigeminal complex (Cliffer et al., 1991; Chiaia et al., 1991a, 1991b; see also Tracey, Chapter 25, and Waite, Chapter 26, this volume). However, most of these ascending projections are less dense and they terminate in a more diffuse way in the Po than in the adjacent ventral posterior nuclei (e.g., Chiaia et al., 1991a; Villanueva et al., 1998). Moreover, projections to the posterior nucleus from the spinothalamic tract and the spinal trigeminal nucleus appear to be relatively more dense than those from the dorsal column nuclei and the principal trigeminal nucleus (Chiaia et al., 1991a; McAllister and Wells, 1981). Further inputs to the Po come from the inferior colliculus (LeDoux et al., 1987) as well as from the vestibular nuclei (Shiroyama et al., 1999). Reciprocal connections have been described between the posteromedial part of the Po and the region of the nucleus ruber (Roger and Cadusseau, 1987).
Recent studies have demonstrated that the trigeminal complex gives rise to several types of fibers ascending to the thalamus which differentially distribute over the ventral posteromedial nucleus and the Po, as well as over other extrathalamic nuclei. While most fibers originating from the principal trigeminal nucleus terminate in a very specific manner in the barreloids in the ventral posteromedial nucleus, a smaller contingent of principal trigeminal fibers ascends to the Po. In addition, these fibers give off collaterals to the tectum, the zona incerta, the ventral part of ventral posteromedial nucleus, and the medial part of medial geniculate nucleus. The neurons in the principal trigeminal nucleus giving rise to the projections to the Po are sensitive to stimulation of multiple whiskers, in contrast to those that project to the barreloids in the ventral posteromedial nucleus which transmit signals from single whiskers only (Veinante and Deschenes, 1999). Thalamic projection fibers originating in the spinal trigeminal complex can also be categorized in several types: one type of fiber projects primarily to the ventral part of ventral posteromedial nucleus; the other type of fiber projects predominantly to the caudal part of the Po while giving off collaterals to, among others, the tectum, the zona incerta, and the ventral lateral thalamic nucleus (Veinante et al., 2000a).
The Po, like the ventral posterior complex, projects to the primary and secondary sensory cortices, largely following a somatotopical organization (Fabri and Burton, 1991). Thalamocortical fibers from the posterior nucleus terminate in the upper part of layer VI as well as in layer I of the cortex (Zhang and Deschenes, 1998). In the barrel cortex, the projection fibers from the Po do not terminate within the barrels, but they preferentially target the interbarrel areas (Lin and Lu, 1993). Cortical projections to the Po primarily arise from the somatosensory areas S1 and S2, but additional projections originate from motor, premotor (frontal eye field), and insular cortices (Veinante et al., 2000b; Guandalini, 2001). While S1 projects predominantly to dorsal parts of the posterior complex, S2 targets more ventral and medial parts, although there exist substantial areas of overlap (Shi and Cassell, 1998b; Veinante et al., 2000b). The corticothalamic projections to Po arise predominantly from layer V pyramidal neurons as collaterals from axons that project to the striatum and descend into the brain stem (Fig. 4) (Levesque et al., 1996; Veinante et al., 2000b). The fibers originating in layer V of the barrel cortex, possibly predominantly originating from neurons in interbarrel cortical areas, exhibit giant terminals of the RL-type in the Po and they may be considered drivers in this nucleus (see section “Some General Aspects of Thalamic Organization”; Hoogland et al., 1987, 1991; Bourassa et al., 1995; Wright et al., 2000; Veinante et al., 2000b). The other cortical areas mentioned above, as well as the deep layer VI neurons in the cortical barrel field, project fibers with much smaller boutons in the Po and these may be considered modulators (Welker et al., 1988; Hoogland et al., 1991; Bourassa et al., 1995).
Functional aspects
It may be clear that the Po receives ascending inputs from various different modalities (see above), i.e., somatosensory, auditory, visual, and vestibular. Although there is probably convergence within the posterior nucleus of fibers carrying information about these different modalities, the somatosensory modality strongly dominates. Whereas the Po might be considered a primary relay thalamic nucleus, the driving input from layer V of the sensory cortex places this nucleus also in the category of higher order thalamic nuclei (see section “Some General Aspects of Thalamic Organization”). As may be clear from the discussions of the afferents and efferents of both the ventral posteromedial nucleus and the Po, the whisker system in rats occupies a large part of the somatosensory system, and the functional anatomy of this system has been analyzed in great detail (e.g., Moore et al., 1999; Ahissar and Zackenhouse, 2001). Therefore, the most concrete suggestions for a functional role of the Po have been made with respect to the behavioral functions of the whisker system. The whisker system not only provides information about passive movements of individual or groups of whiskers but the whiskers are also actively moved in certain frequencies to “explore” the surroundings. The fact that the Po receives its main cortical input from layer V neurons probably indicates that this concerns an “efference copy” of decending (motor) cortical signals to the brain stem. This has led Veinante et al. (2000b) to the suggestion that the Po plays a role at the interface between the sensory and motor aspects of the whisker system.
More specifically, the posterior complex receives the motor signals for active whisker movements as well as the sensory feedback. The Po, together with parts of the trigeminal complex, might thus be involved in monitoring the dynamics of the self-initiated whisker movements, a mechanism that may be necessary for a sensory organ such as the whiskers that lack proprioceptors (Veinante et al., 2000b). Whether the Po has other functions at the interface between the sensory and motor systems remains to be established. The reciprocal connections between the posterior nucleus and the red nucleus provide an indication in that direction (Roger and Cadusseau, 1987; Cadusseau and Roger, 1990; Arnault et al., 1994).
Gustatory and Visceral Nuclei
The ventromedial part of the ventral posterior complex represents the relay for gustatory and visceral information from the periphery to the insular cortex (Norgren and Leonard, 1973; Cechetto and Saper, 1987). The medial part of the ventral posterior complex contains neurons that are smaller than those in the other parts of the VP, and it is therefore referred to as the parvicellular ventral posterior nucleus (VPPC). The VPPC is ventrolaterally bordered by the medial lemniscus and extends medially just ventral to the parafascicular thalamic nucleus and, more rostrally, ventral to the paracentral and central medial nuclei. The VPPC is rostrally “replaced” by the ventromedial nucleus (see Plates 34–37 in Paxinos and Watson, 1998). The VPPC is a rather thin sheet of neurons and it is generally divided into medial and lateral parts, i.e., the parvicellular part of the ventral posteromedial nucleus and the parvicellular part of the ventral posterolateral nucleus (e.g., Cechetto and Saper, 1987; Shi and Cassell, 1998a). Since the medial part of VPPC relays gustatory information (Cechetto and Saper, 1987; Lundy and Norgren, Chapter 28, this volume), this part of the nucleus is indicated by some authors as the gustatory nucleus (e.g., Shi and Cassell, 1998a). Apart from a moderate number of calbindin D28K-positive neurons, the VPPC only contains moderate to low amounts of calcium-binding proteins in fibers (Arai et al., 1994). The medial part of the VPPC contains a dense plexus of calcitonin gene-related peptide-containing fibers (Yasui et al., 1989).
Afferent and efferent connections
Major ascending inputs to the VPPC are derived from the parabrachial nucleus (Norgren and Leonard, 1973; Cechetto and Saper, 1987; Bester et al., 1999; Krout and Loewy, 2000a). Other brain stem inputs to the VPPC arise from specific parts of the spinal and principal trigeminal nuclei, as well as from the laterodorsal tegmental nucleus, locus coeruleus, nucleus of the tractus solitarius, the A5 region, and the cuneate nucleus (Krout et al., 2002). The VPPC, in turn, projects to the lateral and central nuclei of the amygdala, the amygdalostriatal transition area directly dorsal to the central amygdaloid nucleus, and the more rostrally located ventral parts of the caudate–putamen (fundus striati). Cortical projections are primarily directed at the granular and dysgranular insular areas of both the posterior insular and parietal insular cortices, while the anterior insular cortices receive only a minor projection from the VPPC (Kosar et al., 1986a, 1986b; Cechetto and Saper, 1987; Turner and Herkenham, 1991; Nakashima et al., 2000). In most of these efferent VPPC projections to cortical and subcortical targets a mediolateral topographical arrangement exists. Cortical afferents to the VPPC are primarily derived from the insular cortical areas to which it projects. Thus, the granular and dysgranular posterior insular cortices project predominantly to the medial VPPC, while the granular and dysgranular parietal insular cortices send fibers primarily to the lateral part of the VPPC (Shi and Cassell, 1998a, 1998b). The rostrally located dysgranular and agranular insular cortices have only very few projections to the VPPC (Shi and Cassell, 1998a).
Functional aspects
The main function of the VPPC is the transfer of gustatory and visceral information to the granular and dysgranular insular cortices. These insular areas are situated between olfactory cortical areas more ventrally, i.e., around the rhinal sulcus, and the sensorimotor cortices involved with mouth/tongue, head, and forelimbs in the frontoparietal areas just dorsal to the insular cortices. Combined anatomical and physiological studies have demonstrated that the gustatory area is situated more medially in the VPPC, and the visceral area is located more laterally (Kosar et al., 1986a, 1986b; Cechetto and Saper, 1987). In a recent, detailed tracing study, it has been shown that the parabrachial projections to the medial part of the VPPC originate from the two main gustatory relay nuclei in the parabrachial complex. Based on light microscopic observations, the parabrachial fibers in the VPPC have large terminals, suggesting primary sensory relay of information (Bester et al., 1999). The subdivision of the VPPC into a medial gustatory and a lateral visceral part is in line with the functional topography that has been established in the granular and dysgranular posterior insular cortices (Kosar et al., 1986a, 1986b; Cechetto and Saper, 1987; Shi and Cassell, 1998a, 1998b).
Behavioral studies, using specific lesions of the gustatory part of the VPPC, suggest that this thalamic region is not essential for taste discrimination per se, but rather that it is important for more complicated behavioral functions in which the performance is dependent upon “manipulation” of gustatory information (Reilly and Pritchard, 1997). This could be in line with anatomical observations that parabrachial projections bypass the thalamus to reach directly the amygdala and insular cortices (Shipley and Sanders, 1982; Yasui et al., 1989).
For a more comprehensive review of the rat gustatory system and the role of the thalamus in this system, see Chapter 28 by Lundy and Norgren in this volume.
Medial Geniculate Nucleus
The medial geniculate nucleus (MG) forms the most caudal extension of the thalamus and its caudal half is situated laterally alongside the mesencephalon. For a detailed discussion of the auditory thalamus see Malmierca and Merchán (Chapter 31, this volume). The rostral part of the MG is located ventromedially to the lateral geniculate complex. The MG is the principal auditory relay nucleus of the thalamus and consists of several subnuclei that all have different functions within the auditory system. The medial geniculate nucleus can be subdivided into medial (MGM), ventral (MGV), and dorsal (MGD) subnuclei (Jones, 1985; Clerici and Coleman, 1990). The “auditory” thalamus further consists of a number of smaller nuclei that are positioned medially and ventromedially to the main MG complex, i.e., the suprageniculate nucleus (SG), the posterior limitans thalamic nucleus (PLi), the posterior intralaminar thalamic nucleus (PIL), and the lateral part of the parvicellular subparafascicular nucleus (SPF; see section “Midline and Intralaminar Thalamic Nuclei”). A marginal zone (MZMG) “covers” the dorsal, lateral, and ventral aspects of the MGD and MGV.
The MGD and MGV are separated by the so-called midgeniculate bundle which is mainly derived from the inferior colliculus. The MGV can be further subdivided into a ventral and an ovoid part based on fiber architectonics and cytoarchitectonics (Clerici and Coleman, 1990). The main cell type in MGV is small- to medium-sized and has bushy tufted dendrites forming fibrodendritic laminae that are regularly oriented in association with the fiber bundles from the inferior colliculus in a dorsolateral to ventromedial direction (Winer et al., 1999a). In the oval part of the MGV, the orientation of cells and colliculular afferent fiber bundles is more spiral-like (Clerici and Coleman, 1990). The MGD is rather heterogeneous in its neuronal composition and can be subdivided into several subnuclei. Neurons in the MGD have radiating, tufty dendrites (Winer et al., 1999a). Also in the MGM, the neuronal population is rather heterogeneous, ranging from small to magnocellular. Few interneurons (about 1%) have been identified in the MG complex, most of them being GABAergic (Winer and Larue, 1988; Winer et al., 1999a).
Afferent and efferent projections
Ascending afferents to the MG primarily originate in the inferior colliculus. While the MGV receives its afferents mainly from the central nucleus of the inferior colliculus, the MG complex as a whole collects fibers from the cortices of the inferior colliculus. Like the MGV, the MGM receives, in addition, collicular fibers from the central nucleus (Jones, 1985; LeDoux et al., 1987). The central nucleus of the inferior colliculus is strictly tonotopically organized, and the MGV transfers this highly organized information to the primary auditory cortex Te1 in the temporal lobe (Winer et al., 1999b). As such, the MGV, in particular the ovoid subnucleus, forms the main thalamic input for area Te1. The projections from the MGV terminate primarily in layers III and IV of Te1, but there are also projections to the junction of layers V and VI and to superficial layer I (Romanski and LeDoux, 1993; Cetas et al., 1999; Winer et al., 1999b). The projections from the MGV in Te1 are strongly convergent, highly topographic, and spatially focal, and they originate from only one type of neuron in the MGV. Its caudal part excepted, the MGV does not project outside area Te1. The cortical projections from the MGM and the MGD are more divergent, not only directed toward Te1 but, in a much denser pattern, also to surrounding cortical areas Te2 and Te3. These thalamocortical projections arise from several types of geniculate neurons and mostly terminate in the middle cortical layers (Roger and Arnault, 1989; Arnault and Roger, 1990; Winer et al., 1999b).
Corticothalamic fibers from the temporal cortical areas Te1 (primary auditory cortex) and Te2 and Te 3 (together indicated as the “auditory belt” cortex) terminate in different parts of the medial geniculate complex (Shi and Cassell, 1997). Cortical area Te1 projects primarily to the MGV and more moderately to the ventral part of the MGD. Area Te3 sends its most dense projections to the dorsal part of the MGD, directly adjacent to the lateral posterior nucleus which also receives a dense input from Te2 (Shi and Cassell, 1997). Strikingly, the ovoid subnucleus of MGV is largely avoided by these corticothalamic projections, indicating a lack of reciprocity in the cortical connections of this part of the MGV. At the ultrastructural level, corticothalamic projections arising from the primary auditory cortex terminate as small and large (“giant”) terminals in the medial geniculate complex. The smaller, most numerous corticothalamic terminals are present throughout the MG complex, the larger terminals have been observed in the ventral part of the MGD (Rouiller and Welker, 1991) as well as in the dorsal part of the marginal division (MZMG; Bartlett et al., 2000). Excitatory collicular terminals are rather variable in size and include small and large profiles, the larger ones being predominant in the MGV (Bartlett et al., 2000). This heterogeneity in the morphology of the ascending, colliculogeniculate projections indicates a higher complexity in the auditory pathways than in the ascending projections to other sensory thalamic nuclei. Furthermore, part of the collicular afferents of the MGV and MGD appear to be inhibitory (Bartlett and Smith, 1999). Yet, the general organization of ascending driving inputs and descending modulatory cortical inputs seems to hold also for the MGV. Driving cortical inputs from the primary auditory cortex may reach parts of the medial geniculate complex (MGD and MZMG) other than that from which it receives its primary thalamic input, i.e., the MGV (see section “Some General Aspects of Thalamic Organization”).
All medial geniculate nuclei, except the MGV, project to other temporal association cortices, like the perirhinal cortex, as well as to the amygdala. A common feature of the medially located MGM, SG, and PIL nuclei is that they target the upper part of layer I of the temporal association cortices, in addition to more specific projections to deeper layers of these cortices (Linke and Schwegler, 2000). The SG, in addition, projects to the medial agranular frontal cortex (Kurokawa and Saito, 1995). Projections to the amygdala, primarily terminating in the lateral nucleus and, to a lesser degree, the basal nuclei, arise from the medially located subnuclei MGM, SG, and PIL and the lateral part of the parvicellular subparafascicular nucleus (LeDoux et al., 1985, 1990; Turner and Herkenham, 1991; Namura et al., 1997; Doron and LeDoux, 1999, 2000). Apart from projections to the amygdala, the medial subnuclei of the medial geniculate complex also have reciprocal connections with basal ganglia structures. In particular the caudal parts of the striatum, including the amygdalostriatal transition zone, and the caudal globus pallidus participate in these connections with the MGM, SG, PIL, and lateral subparafascicular nucleus (Moriizumi and Hattori, 1992; Shammah-Lagnado et al., 1996).
The MGM, as well as the SG and PIL, not only receives ascending auditory information from the inferior colliculus but is also reached by fibers from the superior colliculus (Linke, 1999). In addition, the MGM receives direct projections from the dorsal cochlear nucleus, bypassing the inferior colliculus. In view of the fact that both the superior colliculus and the dorsal cochlear nucleus transfer multimodal information, it seems likely that the medially located subnuclei of the MG complex do not only process auditory information (Malmierca et al., 2002).
Functional aspects of the medial geniculate complex
It is evident from anatomical and physiological data that the MGV is the main thalamic relay to the primary auditory cortex, subserving the specific tonal analysis of sounds (e.g., LeDoux et al., 1987; Romanski and LeDoux, 1993; Bordi and LeDoux, 1994a; Winer et al., 1999b). The functions of the MGD are less clear, but are most probably concerned with nontonal aspects of sounds and the integration with other sensory modalities. The projections from the inferior colliculus stem from a part of the colliculus that is not tonotopically organized and in the MGD the projections to the temporal cortices are also less strictly organized, terminating primarily in the auditory association cortices (LeDoux et al., 1987; Clerici and Coleman, 1990). The superficial, dorsal region of the MGD may even be a visual-recipient rather than an auditory-recipient area of the medial geniculate complex (Sun et al., 1996; Shi and Cassell, 1997). For a more detailed discussion of the functional aspects of the auditory thalamus, see Chapter 31 by Malmierca and Merchan in this volume.
The functions of the MGM, as well as those of the PIL, SG, and SPF nuclei, must be interpreted in the context of their connections with the temporal association cortices, the amygdala, and the basal ganglia (see above). In particular the relationships of these MG nuclei with the amygdala have been the focus of electrophysiological and behavioral studies. It is clear from such studies that these pathways, which connect the auditory system primarily with the limbic association areas of the brain, are associated with the emotional and mnemonic aspects of sounds. The extensive and elegant functional anatomical studies of LeDoux and colleagues have demonstrated that the MGM, the PIL, and the lateral part of the parvicellular subparafascicular nucleus play an important role in the association of emotionally negative, e.g., noxious, stimuli and the context in which these stimuli occur (fear conditioning; LeDoux, 1993, 2000). The convergence of auditory and somatosensory inputs, but possibly also of visual inputs via the superior colliculus (Linke et al., 1999), in these nuclei of the MG complex appear to form the neuronal basis for such associations that are behaviorally expressed via the amygdala (LeDoux et al., 1987; Bordi and LeDoux, 1994a, 1994b; Linke et al., 1999; LeDoux, 2000). In the context of fear conditioning, plastic changes have been demonstrated in both the MG complex and the amygdala (LeDoux, 2000; Maren et al., 2001; cf. also Komura et al., 2001). The MGM and the PIL have also been shown to form a crucial link in pathways that lead to the neuroendocrine expression of audiogenic stress via the amygdala and, ultimately, the hypothalamic paraventricular–hypophysis system (Campeau et al., 1997; Campeau and Watson, 2000). Thus, the medially located nuclei of the medial geniculate complex are important for the association of auditory and other sensory inputs and the “translation” of this information, via the amygdala and temporal association cortices, in behavior and emotional reactions.
Finally, it is important to note that several nuclei of the medial geniculate complex not only project to the cerebral cortex or subcortical forebrain structures, but that in particular the medially located MGM, PIL, SPF, and SG also have descending projections to the inferior colliculus (Senatorov and Hu, 2002; Winer et al., 2002). Thus, parallel to descending cortical projections to brain stem auditory centers, these medially located MG nuclei are also in a position to provide feedback to the early relays in the ascending auditory pathways.
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Book2004, The Rat Nervous System (Third Edition)Henk J. Groenewegen, Menno P. Witter
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Thalamus
2012, The Human Nervous System (Third Edition)Jürgen K. Mai, F. Forutan
Ventromedial Posterior Nucleus (VMpo)
VMpo may be considered a satellite of the VP complex (Nieuwenhuys et al., 2008). It represents the entry of the spinothalamic (s-th) and spinal trigemino-thalamic (V th) fibers (Craig et al., 1994; Blomqvist et al., 2000). Considering the funneling of several fiber systems into this area, the designations of Ncl. ventro-caudalis portae (V.c.por) and Ncl. limitans portae (Li.por) by Hassler (1959, 1982) appear appropriate. The area corresponds to PO of the “Michigan school” and to the Ncl. basalis medialis/Regio basalis of Percheron (2004).
VMpo shows characteristic neighborhood relations: It has a border to the midbrain, is located adjacent to limitans nucleus and to the medial geniculate body. VMpo lies ventral to ventrocaudal to VPL, ventrolateral to VPM and directly adjacent to the anterior and medial pulvinar nucleus (Figure 19.24). Due to its complex neighborhood relationship and the blurred boundaries VMpo impresses as a poorly outlined area rather than a circumscribed nucleus, a nucleus that is left over after other things have been carved out. Its architecture is dominated by the spreading fibers of the medial lemniscus and the spino- and trigeminothalamic tract. This is reflected in the cell and fiber staining with an inhomogeneous patchy pattern and without clear limitation. The cells are arranged in typical cell swarms; they are medium-sized or round with medium intensity in the Nissl staining (Figure 19.17H).
The inhomogeneous appearance corresponds to the focally increased calbindin (Cb), calretinin (CR), substance P, CART- and CD15-IR (Figure 19.25). VMpo cells signal pain and temperature (Craig et al., 1994; Blomqvist at al., 2000; cf. Willis and others 2001; Graziano and Jones 2004; Lenz et al., 2010).
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The Diencephalon
2018, Fundamental Neuroscience for Basic and Clinical Applications (Fifth Edition)G.A. Mihailoff, D.E. Haines
Lateral Thalamic Nuclei
This large collection of thalamic neurons is grouped into dorsal and ventral tiers. The relatively small group of dorsal tier nuclei includes the lateral dorsal and lateral posterior nuclei along with the much larger pulvinar nucleus (pulvinar) (Figs. 15.7B-D, 15.8D, 15.9, and 15.10). The connections of the lateral dorsal and lateral posterior nuclei are formed with the cingulate gyrus and parietal lobe, respectively (Fig. 15.10). The large pulvinar nucleus consists of anterior, medial, lateral, and inferior subdivisions. The inferior division receives input from the superior colliculus and projects to the visual association cortex. Other portions of the pulvinar project to areas of the temporal, parietal, and frontal lobes that are especially concerned with visual function and eye movements (Fig. 15.11).
The large ventral tier of the lateral group consists of three separate nuclei (Figs. 15.7, 15.8B, C, and 15.9). The ventral anterior (VA) nucleus and the slightly more caudal ventral lateral (VL) nucleus are important motor-related nuclei; the ventral posterior nucleus, consisting of ventral posterolateral (VPL) and ventral posteromedial (VPM) nuclei, convey somatosensory information to the cerebral cortex.
The VA is composed of a large parvocellular portion and a small magnocellular part. The former receives input from the medial segment of the globus pallidus, and the latter receives afferents from the reticular portion of substantia nigra. The efferent projections from the VA are diffuse and appear to include selected parts of the frontal lobe (Fig. 15.11).
The VL (Figs. 15.7B and 15.10) is also composed of three subdivisions: a pars oralis, a pars medialis, and a pars caudalis. The largest of these, the pars oralis, receives a dense projection from the internal segment of the ipsilateral globus pallidus; some of these afferents enter the caudal subdivision. In contrast, the pars caudalis subdivision of the VL receives its main input from the contralateral cerebellar nuclei. Consequently, pallidal and cerebellar projections are largely segregated within this nucleus. The output of the VL reflects its segregated input in that the oral and caudal parts project to largely separate areas of the frontal lobe (Fig. 15.11).
The larger and more laterally located VPL nucleus and the comparatively smaller and more medially located VPM nucleus both receive somatosensory input from the contralateral side of the body (Figs. 15.7C and 15.10). The medial lemniscus and spinothalamic fibers terminate in a somatotopic manner (cervical fibers medial, sacral fibers lateral) within the VPL, whereas trigeminothalamic fibers from the spinal trigeminal nucleus and the principal trigeminal sensory nucleus terminate in the VPM. Both the VPL and VPM project to the somatosensory cortex of the parietal lobe (Fig. 15.11).
A small group of cells called the ventral posterior inferior nucleus is situated ventrally between the VPL and VPM. These cells process vestibular input and project to lateral areas of the postcentral gyrus that are located in the depths of the central sulcus. Similarly, a small group of cells forming the rostral (oral) portion of the VPL receives cerebellar input and projects to the precentral gyrus of the frontal lobe; this nucleus probably represents a few cells that have been displaced from the slightly more rostrally located VL. This cell group is also called the ventral intermediate nucleus because of its location between the VL and VPL.
The medial (MGB) and lateral (LGB) geniculate nuclei (body) are considered parts of the lateral thalamic nuclear group (Figs. 15.7D, 15.8D, and 15.10). The MGB receives ascending auditory input via the brachium of the inferior colliculus and projects to the primary auditory cortex in the temporal lobe. The LGB receives visual input from the retina via the optic tract and in turn projects to the primary visual cortex on the medial surface of the occipital lobe (Fig. 15.11).
Located in the posterior thalamus at about the level of the pulvinar and geniculate nuclei is a cluster of cell groups collectively called the posterior nuclear complex. This complex consists of the suprageniculate nucleus, the nucleus limitans, and the posterior nucleus. These nuclei are positioned superior to the medial geniculate and medial to the rostral pulvinar. The posterior nuclear complex receives and sends to the cortex nociceptive cutaneous input that is transmitted over somatosensory pathways.
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Book2018, Fundamental Neuroscience for Basic and Clinical Applications (Fifth Edition)G.A. Mihailoff, D.E. Haines
Chapter
Thalamus
2012, The Human Nervous System (Third Edition)Jürgen K. Mai, F. Forutan
Lateral Region 634
: Motor Thalamus 636
: Ventral Anterior Nucleus (VA) 637
: Ventrolateral Nucleus (VL) 642
: Sensory Thalamus 644
: Ventroposterior Nucleus or Complex (Ncl. ventralis posterior, VP) 644
: Ventral Posterolateral Nucleus (VPL) 645
: Ventral Posteromedial Nucleus (VPM) 645
: Superior Ventroposterior Nucleus, VPS (Kaas) (Oral Ventroposterior Nucleus, VPO) 646
: Ventral Posterior Nucleus, Parvicellular Part (VPPC) (Nucleus ventroposterior medialis, pars parvocellularis, VPMpc) 646
: VM, VPI and VMpo – Entry Zones of Afferents to the Lateral Region in Humans 647
: Ventromedial Posterior Nucleus (VMpo) 649
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Book2012, The Human Nervous System (Third Edition)Jürgen K. Mai, F. Forutan
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Thalamus
2012, The Human Nervous System (Third Edition)Jürgen K. Mai, F. Forutan
Ventrolateral Nucleus (VL)
VL as defined here corresponds to the major cerebellar territory (Table 19.1). It is placed intermediate between the territory for the basal ganglia and the sensory afferents. Because of this location the designation of “n. intermédiare” (Vogt, 1909) has been adopted by many other researchers (Crouch, 1934; Vogt and Vogt, 1941; Percheron, 1977). Jones (2007), Mai et al. (2008) and Morel et al. (1997) referred to this territory as VLp. For reasons already described, the nomenclature of Ilinsky and Kultas-Ilinsky (1987) is used in this text. In the human this area can be divided into a medial oral (VLM) and a lateral caudal division (VLL) on the basis of cell size (Figure 19.20). Hassler (1959) has fragmented the cerebellar territory in several compartments (Table 19.1). The entry zone of the cerebellar fibers is regarded as separate nucleus (VPI, VLb).
The ventrolateral nucleus (VLp) exhibits an oblique rostromedial to caudolateral orientation with respect to the axial stereotactic plane (Figure 19.16). The nucleus begins rostrally behind VAM (VAmc), medially and caudally to the VAL (VLa) and ends in front of VP. Medially it is bordered by the internal medullary lamina; caudolaterally it reaches the external medullary lamina. Dorsally the superior region is located and ventrally the basal ventrolateral nucleus, VLb (VM; Figure 19.24). Like in the monkey (Kuo and Carpenter, 1973; Percheron, 1996; de las Heras, 1998; McFarland and Haber, 2000) the human VL is relatively poor in cells. These consist of two classes in respect to cell size (30–40 μm2 and 20 μm2) (Van Buren and Borke, 1972; cit Ohye, 1990). The large cells, particularly those in the ventral part, are mostly polygonal, stain deeply, have many wide processes and are often arranged in rows. They are significantly larger than those of VAL and VPM; compared with the caudally adjacent ventral posterolateral thalamic nucleus (anterior part, VPLa) with cells of similar size, they are distributed largely homogeneously and are less densely arranged. The lateral and basal regions are characterized by heavily stained fiber bundles (Figure 19.24). Nissl sections show a clear contrast if compared with the dense islands of small cells in the pallidal region on the one hand and the still smaller, more densely packed cells of the posterior somesthetic region. In the border areas, the demarcation is difficult, despite the significant cytological differences, since the cross-sectional area has complicated neighborhood relations and the outer contour may be tortuous.
Against the pallidal VAL the cerebellar VL is distinguished by its lack of calbindin- (CB) and the presence of moderate calretinin- (CR) IR (Münkle et al., 2000; Figure 19.18F,H). Parvalbumin-IR has been described as a discriminative marker for the cerebellar area (Percheron, 2004, p. 627; Münkle et al., 1999). Fang et al. (2006) described the VLp (VL) in a prosimian monkey distinctly labeled by AChE and parvalbumin but also in CO preparations. Lenz et al. (2010) described much weaker AChE activity within VL (VLp) than in VAL (VLa).
Thanks to the large cells in VL (VLp) the transition to the anterior and anterodorsal pulvinar nucleus (APul) is relatively clear (Figure 19.20). Immune staining against neurofilaments (SMI311) allows the delineation because the dendrites of the cells in the anterodorsal pulvinar nucleus are not distinctly visible, as is the case in the other subnuclei of the pulvinar. VL (VLp) is negative for CB-IR and moderately positive for CR-IR; it is therefore easily distinguishable from VAL (VLa).
The pathways of the monkey and human cerebello-thalamic fibers are reviewed by Voogd in the Chapter 15. Anterograde tracing methods in monkeys located the terminal area of the cerebellar fibers in a rather distinct cytoarchitectonic region. According to Percheron (1977) the afferents form a continuous band that occupies the entire ventrodorsal extent of the VL without overlap with afferents from the basal ganglia and the spinal cord. It appears that this region can be characterized in the human thalamus by its intensive parvalbumin-IR (Morel et al., 1996). Individual cerebellar afferents form either long, complex, widely distributed branches (in different VL locations) with divergent axon collaterals and distant terminal fields or end as single individual adjacent terminal fields (Mason et al., 2000). The latter arrangement appears as if the terminals are lined up, forming stacks that could potentially represent clusters of neurons related to cooperating muscle groups. This arrangement, similar to the “onion skin” model of Vitek et al. (1995, 1996) could be the basis of the suggested somatotopic organization as reported by several anatomical and physiological studies in monkey (see Mason et al., 2000).
In spite of the rather distinctive architecture of the cerebellar territory also in humans, the translation to human neurophysiology has produced some confusion, particularly in the field of neurosurgery (see Ohye, 1982, 1990; Tasker et al., 1982; Krack et al., 2002; Percheron, 2004). The cerebellothalamic zone that corresponds most probably to Hassler’s Ncl. ventralis oralis, V.o (VL of Walker) (Hassler, 1949b, 1950, 1982) comprises several components which could not definitely be related to specific modalities and sources. Hassler distinguished in the anterior part (V.o.a) the terminal area for the fibers from the inner pallidum and in the posterior part (V.o.p) that for the cerebellar fibers (with an inner segment for the isthmo-thalamic fibers) (Figure 19.16B; Table 19.1). The correspondence of his delineations with the actual pallidal and cerebellar territories has been a matter of debate and has influenced the discussions about the optimal location for stereotactic interventions. Many investigators believe that V.o.p is associated with the pallidothalamic pathway (Ohye et al., 1990; Percheron et al., 1996; Percheron, 2004). The interpretation of V.o.p as either a pallidal or cerebellar associated field has influenced the target location in stereotaxy and may still be incorrect (Bakay, 2009).
The morphology of the areas that were designated as V.o.a and V.o.p by Hassler appears quite uniform. Both areas contain cells of comparable size (V.o.a: 300–450 μm2, V.o.p: 200–300 μm2 with a mean of 250 μm2) with a very similar cell density (80–150/mm2/50 microns average thickness) (Hirai and co-workers, cited in Ohye et al., 1990); the fiber density is somewhat higher in the more anterior part; the difference is, however, not suited for a demarcation of the two components. Morphological features together with physiological and clinical findings make it likely that the areas V.o.a and V.o.p represent a common territory for the pallidal fibers (Molnar et al., 2005) and consequently the termination of cerebellar fibers thus “shifted” to the more caudal level of V.im. This does not exclude the possibility that a part of V.o.p receives cerebellar input (see Krack et al., 2002). This assignment would correspond also to the very different cell types in the pallidal and cerebellar territories. This interpretation seems also supported by the results of stereotactic surgery, because it turned out that it is not V.o.a that is the best target for the treatment of the so-called essential tremor, but the more caudal Ncl. ventralis intermedius of the thalamus (VIM) (see below).
V.im is according to Hassler et al. (1979) located between V.o and V.c (Figure 19.16). The external division (V.im.e) is regarded as the most effective stereotactic target for the treatment of tremor (Percheron, 2004). In this place the axons of the brachium conjunctivum are concentrated and so-called “tremoro-synchronized” neurons (“tremor cells”) are found there (Guiot et al., 1962; see Tasker et al., 1982). Because these neurons respond to kinesthetic stimuli (Ohye and Narabayashi, 1979) it was suggested that this area (VPO; Kaas: VPS) receives lemniscal proprioceptive afferents.
Percheron (2004) has transferred patient data into atlas templates, in which VPS (VPO) was registered. He could demonstrate that the “tremoro-synchronized” neurons are located in front of the representation of the surface sensitivity (VPC), within the cerebellar territory (Vim-VL) and – to a lesser extent – within the pallidal territorry. According to these calculations the most efficient target for lesioning or stimulation in cases of tremor is the kinesthetic zone in VLL (V.im.e/VImL), a region that responds to kinesthetic/passive movements about a joint and that is known to project to the motor cortex (Molnar, 2005).
In addition to the cerebellar afferents fibers from other sources were demonstrated in experimental animals and postulated in humans. These include fibers from the tectum (Percheron, 2004), from the vestibular nuclei, and from the interstitial nucleus of Cajal (see Ohye, 1990) which were described to terminate in the rostral division of VL (VLp; V.o.i). The organization of the efferents from different VL areas to the cortex and to the striatum is discussed in Chapters 15 and 29Chapter 15Chapter 29.
The widespread distribution in VL with multiple terminal fields has been described by Mason et al. (2000) as specific structural feature of the dentatothalamic connection. The assembly of terminals in supposed specializations of stacks could be correlated to the representation of one or multiple topographic maps of body parts. The coordinated activation could recruit muscle groups for complex motor activities (multiple muscle synergies). The overall effect of the afferents from the cerebellum pertains to balance and fine motor skills.
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Book2012, The Human Nervous System (Third Edition)Jürgen K. Mai, F. Forutan
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Update on Emerging Treatments for Migraine
2020, Progress in Brain ResearchShu-Ting Chen, Jr-Wei Wu
3.4The central processing of migraine pain
The ventral posteromedial nucleus of the thalamus is the principal structure for conveying nociceptive information from trigeminovascular neurons to higher cortical regions (Noseda et al., 2011). Also, studies showed other thalamic nuclei also involved in the pain signal processing of dura mater, such as ventroposteromedial (VPM), posterior and lateral posterior/dorsal thalamic nuclei (Noseda et al., 2010; Zagami and Lambert, 1991). In addition, thalamic nuclei modulate non-headache symptoms of migraine, such as photophobia or allodynia (Noseda et al., 2016; Rossi and Recober, 2015). Higher cortical areas, including the primary (S1) and secondary (S2) somatosensory areas, the insula (Ins), the anterior cingulate cortex (ACC), and the prefrontal cortex, comprise the ‘pain matrix’ responsible for pain processing (Goadsby et al., 2017a). Prior research has shown that the amygdala also plays a role in migraine pain through modulation of the emotional aspect of pain perception. For example, altered amygdala functional connectivity is linked to migraine chronification (Chen et al., 2017).
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Book series2020, Progress in Brain ResearchShu-Ting Chen, Jr-Wei Wu
Chapter
Trigeminal Nerve
2014, Clinical Anatomy of the Cranial NervesPaul Rea
The Nuclei
| Nucleus | Information Carried | Location | Projection |
--- --- |
| Spinal trigeminal nucleus | Pain, temperature, and light touch | Medulla | Ventral posteromedial nucleus of the dorsal thalamus |
| Pontine trigeminal nucleus | Discriminative and light touch, proprioception of the jaw | Pons | Ventral posteromedial nucleus of the thalamus |
| Mesencephalic trigeminal nucleus | Proprioception of the face (lower jaw) | Pons and midbrain | Motor nuclei of the trigeminal nerve |
| Trigeminal motor nucleus | Motor information | Pons | Muscles of mastication and tensor veli palatini, tensor tympani, anterior belly of digastric and mylohyoid |
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Book2014, Clinical Anatomy of the Cranial NervesPaul Rea
Related terms:
Sensation of Taste
Solitary Tract
Barrel Cortex
Ventral Posterolateral Nucleus
Thalamic Reticular Nucleus
2,5-Dimethoxy-4-iodoamphetamine
Migraine
Brainstem
Thalamus
Somatosensory Cortex
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187563 | https://www.nerc.com/pa/comp/CE/Enforcement%20Actions%20DL/NOC-2660%20Full%20PDF_PUBLIC_Redacted_Part%202_Final.pdf | Attachment 4 Record documents for the violation of CIP-003-3 R6 4.a The Entity’s Self-Report (RFC2017017568); 4.b The Entity’s Mitigation Plan designated as RFCMIT012980 submitted ; 4.c The Entity’s Certification of Mitigation Plan Completion dated ; 4.d ReliabilityFirst’s Verification of Mitigation Plan Completion dated ; 4.e The Entity’s Self-Report (RFC2017018261); 4.f The Entity’s Mitigation Plan designated as RFCMIT013213-1 submitted ; 4.g The Entity’s Certification of Mitigation Plan Completion dated ; 4.h ReliabilityFirst’s Verification of Mitigation Plan Completion dated ; 4.i The Entity’s Self-Report (RFC2017018760); 4.j The Entity’s Mitigation Plan designated as RFCMIT013443 submitted ; 4.k The Entity’s Certification of Mitigation Plan Completion dated ; 4.l ReliabilityFirst’s Verification of Mitigation Plan Completion dated NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Self Report Page 3 of 3 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 7 Page 2 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Plan Details 1. Milestone 1 will provide evidence showing access has been revoked for 1 and 2.
2. Milestone 2 will show that a disciplinary action has been taken to correct the employee behavior.
3. Milestone 3: Update procedures to indicate that the leader must approve reassignment of 4. Milestone 4: Update job aid to identify as NERC-CIP asset when assigning to the owner.
Identify and describe the action plan, including specific tasks and actions that your organization is proposing to undertake, or which it undertook if this Mitigation Plan has been completed, to correct the violation(s) identified above in Section C.1 of this form: Provide the timetable for completion of the Mitigation Plan, including the completion date by which the Mitigation Plan will be fully implemented and the violations associated with this Mitigation Plan are corrected: July 28, 2017 Proposed Completion date of Mitigation Plan: Milestone Activities, with completion dates, that your organization is proposing for this Mitigation Plan: Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 05/03/2017 Revoke Access Evidence showing access has been revoked for 1 and 2.
05/03/2017 No 06/14/2017 Disciplinary Action Disciplinary Action 06/14/2017 No 07/28/2017 Job aid Update Update job aid to identify as NERC-CIP asset when assigning to the owner.
No 07/28/2017 Procedure update Update procedures to indicate that the leader must approve reassignment of No Additional Relevant Information 7 Page 5 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts.
By implementing the mitigation plan proposed in section D, will minimize similar issues. The disciplinary action is designed to correct the employee's behavior.
Prevention By completion of the mitigation plan will minimize similar issues. The disciplinary action is designed to correct the employee's behavior.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 7 Page 6 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Verification of Mitigation Plan Completion Milestone 1: Revoke Access. File 1, “RFC2017017568 Certification Package”, 1 and 2 access removal as evidence showing access has been revoked for 1 and 2. Milestone # 1 Completion verified. Milestone 2: File 1, “RFC2017017568 Certification Package”, Disciplinary Action Email as evidence of an email by Employee Relation stating a corrective action was taken June 14, 2017, for the employee. Milestone # 2 Completion verified. Milestone 3: Job aid Update. File 1, “RFC2017017568 Certification Package”, as evidence that shows an update has been made for a job aid stating that will be identified as a NERC-CIP asset when assigning to the owner. also provided an email communication as evidence an email was sent to subject matter experts that the job aid has been modified/updated. Milestone # 3 Completion verified. Milestone 4: Procedure update. File 1, “RFC2017017568 Certification Package”, Leader Approve of as evidence that shows an update has been made for a job aid stating that the leader must approve reassignment of the It was noted in the mitigation plan that a procedure will be updated. However, made the update to the job aid. also provided an email communication as evidence NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION an email was sent to subject matter experts that the aid has been modified/updated. Milestone # 4 Completion verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst August 22, 2017 Self Report access qualifications and the 1 removal of access for the employee that no longer required access.
Additional Entity Comments: Comment From User Name Additional Comments No Comments Document Name Description Size in Bytes From Additional Documents Description of Entitlements.docx This file contains description of entitlements noted in this violation.
13,052 Entity Page 4 of 4 08/22/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst October 17, 2017 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 10 Page 2 of 10/17/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst October 17, 2017 Relevant information regarding the identification of the violation(s): This potential non-compliance was identified by during the Q2 On 03/01/2017, prior to the identification of this potential non-compliance, documented and updated its process to address process gaps when bulk loading employees access records. 10 Page 6 of 10/17/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst October 17, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending missing qualification 08/01/2017 Update the Request for access SWI Update the Request for access SWI to require validation that users are loaded into prior to performing bulk load 08/01/2017 No 08/30/2017 Conduct a of access Conduct a for the employees with missing authorization records and remove any inappropriate access 08/04/2017 No Additional Relevant Information 10 Page 8 of 10/17/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst October 17, 2017 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts.
The potential impact to the BES is Lower because users gained access without the proper qualifications. The actual impact is low due to compensating controls of the that worked as designed to catch any inaccurate access. Additionally, users maintained all qualification required to have access to the BCSI. removed access to the users without proper qualification along with conducting an off-cycle to validate if access was needed for each employee identified in this potential non-compliance.
Prevention The completion of the Mitigation Plan as outlined and implemented will help to ensure that during user access bulk uploads all users access is provisioned properly and that all users access has a corresponding authorization record.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 10 Page 9 of 10/17/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION credentials to have the provisioned access. After identification, the subject matter expert proposed 2 definitive paths in order to revoke access. Milestone # 3 Completion verified Milestone 4: Remove access. Proposed Completion Date: July 21, 2017 Actual Completion Date: July 21, 2017 File 1, “RFC2017018261 Certification Package”, Remove access- RFC2017018261, Pages 1 through 4, shows an email from the subject matter expert with included screen grabs showing that the access had been removed due to the lack of required qualifications. Milestone # 4 Completion verified Milestone 5: Update the Request for access SWI. Proposed Completion Date: August 1, 2017 Actual Completion Date: August 1, 2017 File 1, “RFC2017018261 Certification Package”, SWI-Template, Pages 1 through 9, show the updated template (8-1-2017) and the email communication that went out to affected staff per this change of process/ procedure per this milestone. Milestone # 5 Completion verified Milestone 6: Conduct a Review of access. Proposed Completion Date: August 30, 2017 Actual Completion Date: July 21, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION File 1, “RFC2017018261 Certification Package”, Remove Access-RFC2017018261, Pages 1 through 4, shows the entity SME response to removing access with the users access that needs to be removed. Milestone # 6 Completion verified The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: November 28, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Self Report order to ensure that the individuals with active electronic access or unescorted physical access have authorization records. This process supports CIP004 R4 Part 4.2 Incident description 2017 Q3 process identified that users had access to share drive that holds the BCIS but the authorization record were not found in Further investigation showed that these as well, were marked for removal, and the supervisors were communicated to remove the access. The Supervisors for users had requested a revocation in based on the 2017 Q1 results. While system records indicates that the requests were completed and access was revoked, the he users still had access to the shared drives for which access was requested to be revoked. This implies that workflow was closed without the process confirming that the access to the shared drive was revoked.
In addition to the users from Q1, the results for 2017 Q3, indicated users as provisioned in different from what the results expected. These were treated as a regular outcome of the process that required follow-up and correction. The revocation process was immediately followed and corrected by the Supervisors.
What is the problem?
2017 Qtr3 revealed that users identified to be removed in 2017Q1 process continued to have access to the shared drives (designated storage locations).
As ked as designed but authorization records for such access did not exist, this has been recorded as a violation of CIP-004- R4 P4.1 - "Process to authorize based on need, as determined by the Responsible Entity,except for CIP Exceptional Circumstances: 4.1.3. Access to designated storage locations, whether physical or electronic, for BES Cyber System Information.".
Root Cause of Possible Violation: As per the & RCA performed on 10/31/2017, the root cause was identified to be Lack of validation of removal of access when identified during the process.
How was the violation discovered?
On 09/29/2017, while concluding 2017 Q3 determined that 6 users discrepancies from Q1 process remained unresolved in Q3.
Explain how is it determined that the Noncompliance is related to documentation, performance, or both.
On examining the root causes listed above, it was determined that noncompliance is related to a gap in the currently defined Process.
The Supervisor closed the revocation of access ticket without checking for evidence of revocation. The process SWI needs to be updated and recirculated to the users involved with access management.
Timeline: 09/29/2017 - 2017 Q3 completed.
09/29/2017 - Access management analyst reported users with access to shared drives and no valid record. showed access revoked.
These users had been repeatedly been an issue since Q1.
10/31/2017 - RCA was performed to understand the sequence of events in the process and it was identified that there was a gap in the process. The process did not require evidence of access revocation prior to closing an tickets.
Page 2 of 3 12/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 01, 2017 Self Report Mitigating Activities: Description of Mitigating Activities and Preventative Measure: Corrective Actions: Access for these users have been revoked since identification of this issue.
Mitigating and Preventive measures: Update the process to include validation of the removal of access.
Date Mitigating Activities Completed: Impact and Risk Assessment: Description of Potential and Actual Impact to BPS: Potential Impact Potential impact of users having access to these could be severe as the share drive includes the BCSI relevant to process, procedures, and programs.
Actual Impact The users with access to shared drives that needed to be revoked, were personal who had background clearance and were trained in NERC CIP standards prior to being granted access via The revocation process did not get concluded for these users since the Supervisor closed the ticket to revoke access without evidence.
Minimal Severe Actual Impact to BPS: Potential Impact to BPS: Risk Assessment of Impact to BPS: The risk was measured as low because all of the users with access to the share drive had a valid PRA and had a valid training record.
Additional Entity Comments: Comment From User Name Additional Comments No Comments No Documents Additional Documents Size in Bytes From Document Name Description Page 3 of 3 12/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 9 Page 2 of 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Results of the RCA: (What is the root cause?) As per the & RCA performed 10/31/2017, the root cause was identified to be Lack of validation of removal of access when identified during the process.
Relevant information regarding the identification of the violation(s): On 09/29/2017, while concluding 2017 Q3 determined that users discrepancies from Q1 process remained unresolved in Q3. 9 Page 5 of 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Plan Details Milestone 1 - Access for these users to be revoked for the shared drives to correct current access requirements Access for these users have been revoked since identification of this issue. The Supervisors were contacted and requested to remove access (outside of and monitor that the action was taken, including emailing an evidence to the account management team for each discrepancy resolved.
Milestone 2 - Update the process to include validation of the removal of access Review and update process process for Access removal request. process will include validation of removal actions. The team that executes quarterly process is team of members. This team has incorporated the validation of removal in current practice (in order to complete Q4 process). However, it will be formalized by end of December.
Milestone 3 - Communicate the process to include validation of the removal of access Communicate the update to the process to the set of Supervisors responsible for managing the access authorizations.
Identify and describe the action plan, including specific tasks and actions that your organization is proposing to undertake, or which it undertook if this Mitigation Plan has been completed, to correct the violation(s) identified above in Section C.1 of this form: Provide the timetable for completion of the Mitigation Plan, including the completion date by which the Mitigation Plan will be fully implemented and the violations associated with this Mitigation Plan are corrected: January 10, 2018 Proposed Completion date of Mitigation Plan: Milestone Activities, with completion dates, that your organization is proposing for this Mitigation Plan: Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 11/29/2017 1. Access for these users to be revoked for the shared drives Contact the Supervisors of all users to revoke access to the shared drive per the results.
11/29/2017 No 01/10/2018 2. Update the process to include validation of the removal of access Review and update process process for Access removal request. process will include validation of removal actions. The team that executes quarterly process is team of members. This team No 9 Page 6 of 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Confidential Non-Public Information December 14, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending has incorporated the validation of removal in current practice (in order to complete Q4 process).
However, it will be formalized by end of December.
01/10/2018 3. Communicate the updated process Communicate the update to the team responsible for managing the access authorizations.
No Additional Relevant Information 9 Page 7 of 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts. has not identified any risk to the BES. The risk was measured as low because all of the users with access to the share drive had a valid HR clearance as employees and had a valid NERC CIP training record.
Validation of removal has been incorporate in current practice and will be tested while performing the for Q4.
Potential impact of users having access to these could be severe as the share drive includes the BCSI relevant to process, procedures, and programs.
These include revocation of access for 6 users whose roles have undergone a change and hence access is no longer needed to the shared drive, and an update of the process to ensure revocation was successfully performed before tickets are closed in the future.
Prevention will prevent such occurrences in the future by updating the gap in the process itself. Currently, an ticket can be closed by the Supervisor without reviewing access revocation evidence. In the updated process, an evidence would need to be provided as a backup to close an ticket. This would prevent an occurrence whereby shows no access, while the user continued to have access in the shared drive.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 9 Page 8 of 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: January 18, 2018 File 1, “RFC2017018760 Certification Package”, Milestone 3- Submit, Page 2 shows an email with an attached process which was sent to 6 email addresses. File 2, “RFC2017018760 Milestone 3 Submit”, Page 2, shows the email that was sent in regards to the updated process. Milestone # 3 Completion verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: March 12, 2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Attachment 5 Record documents for the violation of CIP-004-6 R5 5.a The Entity’s Self-Report (RFC2017017152) submitted ; 5.b The Entity’s Self-Report (RFC2017017152) submitted ; 5.c The Entity’s Mitigation Plan designated as RFCMIT012807-1 submitted ; 5.d The Entity’s Certification of Mitigation Plan Completion dated ; 5.e ReliabilityFirst’s Verification of Mitigation Plan Completion dated NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst February 27, 2017 Self Report Risk Assessment of Impact to BPS: Electric System assets as a result of this potential violation.
Additional Entity Comments: Comment From User Name Additional Comments No Comments No Documents Additional Documents Size in Bytes From Document Name Description Page 3 of 3 02/27/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst March 01, 2017 Self Report No Documents Page 3 of 3 03/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst April 12, 2017 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 9 Page 2 of 04/12/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst April 12, 2017 Plan Details Milestones below show the detailed actions is undertaking to mitigate the violation.
Identify and describe the action plan, including specific tasks and actions that your organization is proposing to undertake, or which it undertook if this Mitigation Plan has been completed, to correct the violation(s) identified above in Section C.1 of this form: Provide the timetable for completion of the Mitigation Plan, including the completion date by which the Mitigation Plan will be fully implemented and the violations associated with this Mitigation Plan are corrected: May 23, 2017 Proposed Completion date of Mitigation Plan: Milestone Activities, with completion dates, that your organization is proposing for this Mitigation Plan: Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 01/11/2017 Instance 1- Hold Security / Vendor Meeting Go through all leader's employees managed to determine if all employees still working on account 01/11/2017 No 01/11/2017 Instance 1- Revoke access Revoke NERC Access 01/10/2017 No 01/11/2017 Instance 1- Training Retrain leaders on non-EE deactivation process 01/11/2017 No 03/08/2017 Instance 2- Milestone 1 will update the System Access Control Procedures shared account inventory to reflect the current shared account inventory.
03/16/2017 No 03/08/2017 Instance 2- Milestone 2 A procedure will be developed by to include a check list for transitioning SMEs between the roles.
03/23/2017 No 05/23/2017 Instance 2- Milestone 3 Will perform quality check across all BCAs to see if there are other similar occurrences (Extent No 9 Page 6 of 04/12/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst April 12, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending of condition) Additional Relevant Information 9 Page 7 of 04/12/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst April 12, 2017 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts.
The risk to the BES is determined to be minimum because both the terminations were volunteer separations.
Review of logs shows no activity from respective IDs after the separation date.
Prevention By completion of the mitigation plan will minimize similar issues by updating the System Access Control Procedures shared account inventory to reflect the current shared account inventory and the procedure will be developed by to include a check list for transitioning SMEs between roles.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 9 Page 8 of 04/12/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone #1 Completion Verified. Milestone 2: Instance 1- Revoke access. File 1, “RFC2017017152 Certification Package- Submit (ZIP File Folder)”, Instance 1 –Milestone 2- Submit(File Name), Pages 2 through 7, show how CIP environment is labeled in order to provide evidence and explanation into the requests to revoke access according to Milestone 2 and to demonstrate that no access/access events occurred during the time of this potential noncompliance. Pages 8 through 11 provide tickets for access revocation as required by Milestone 2. Milestone #2 Completion Verified. Milestone 3: Instance 1- Training. File 1, “RFC2017017152 Certification Package- Submit (ZIP File Folder)”, Instance 1 –Milestone 3 –Submit (File Name), Page 2 of 14, provides an email with attachments that was delivered to affected employees in order to reinforce the revocation process. Page 3 of 14, shows the process workflow for the revocation of access while Pages 5 through 12, walk a user through the steps of how to formally request and/or remove access via their enterprise business system. Milestone # 3 Completion Verified. Milestone 4: Instance 2- Milestone 1 File 1, “RFC2017017152 Certification Package- Submit (ZIP File Folder)”, Instance 2 –Milestone 1 –Submit (File Name), Pages 7 and 8, illustrate the update to the System Access Control Procedures shared account inventory as specified by this milestone. Page 12 shows the revision history of this update from January 30, 2017, which specifies these changes and their location within the document. Milestone #4 Completion Verified. NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone 5: Instance 2- Milestone 2 File 1, “RFC2017017152 Certification Package- Submit (ZIP File Folder)”, Instance 2 –Milestone 2 –Submit (File Name), Pages 2 through 5, show the checklist for transitioning SMEs required by this milestone. Milestone # 5 Completion Verified. Milestone 6: Instance 2 -Milestone 3. File 2, “Instance 2- Milestone 3- Submit-New (Zip File Folder)”, Instance 2- Milestone 3 – Submit –New (File Name), Pages 15 through 265, show the extent of condition analysis performed in regard to access revocation when employees have changed roles, or left the company. Milestone # 6 Completion Verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: June 22, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Attachment 6 Record documents for the violation of CIP-005-5 R2 6.a The Entity’s Self-Report (RFC2018019570); 6.b The Entity’s Mitigation Plan designated as RFCMIT013868 submitted ; 6.c The Entity’s Certification of Mitigation Plan Completion dated ; 6.d ReliabilityFirst’s Verification of Mitigation Plan Completion dated NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst June 13, 2018 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 14 Page 2 of 06/13/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst June 13, 2018 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 08/15/2018 No Additional Relevant Information 14 Page 12 of 06/13/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone 6: Proposed Completion Date: December 6, 2017 Actual Completion Date: November 20, 2017 File 2, “ Certification Package Updated,” Milestone 6 – Submit at Pages 2 and 3, contains a signed attestation, which includes a statement explaining that Milestone # 6 Completion verified. Milestone 7: Proposed Completion Date: January 30, 2018 Actual Completion Date: January 30, 2018 File 2, Certification Package Updated,” Milestone 7 – Submit at Pages 2 through 12, shows that the entity Milestone # 7 Completion verified. Milestone 8: . Proposed Completion Date: June 25, 2018 Actual Completion Date: June 25, 2018 File 2, “ Certification Package Updated,” Milestone 9 – Submit at Pages 2 through 8, contains: (a) pre- and post- and Milestone # 8 Completion verified. Milestone 9: Proposed Completion Date: June 28, 2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: June 22, 2018 File 2, “ Certification Package Updated,” Milestone 10 – Submit at Pages 2 through 11, shows the updated program, which includes The same file at Pages 12 and 13 shows the email sent out regarding the updates along with contact information if staff had questions or concerns. Milestone # 9 Completion verified. Milestone 10: Create a process for . Proposed Completion Date: July 3, 2018 Actual Completion Date: August 3, 2018 File 2, “ Certification Package Updated,” Milestone 11 – Submit at Pages 2 through 10, contains a process diagram, a standard work instruction, and emails communicating the diagram and standard work instruction. File 2, Certification Package Updated,” Milestone 10 – Submit at Pages 2 through 14, shows an updated program, which includes Milestone # 10 Completion verified. Milestone 11: Proposed Completion Date: July 30, 2018 Actual Completion Date: July 5, 2018 File 2, “ Certification Package Updated,” Milestone 11 – Submit at Pages 2 through 15, shows the standard work instruction that was updated to reflect revocation of access after 120 days of consecutive non-use. Milestone # 11 Completion verified. Milestone 12: Perform risk assessment on . NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Proposed Completion Date: August 6, 2018 Actual Completion Date: July 20, 218 File 2, “ Certification Package Updated,” Milestone 13 – Submit at Pages 2 through 5, contains documents evidencing the entity’s risk assessment of Milestone # 12 Completion verified. Milestone 13: Deploy Proposed Completion Date: August 13, 2018 Actual Completion Date: July 23, 2018 File 2, “ Certification Package Updated,” Milestone 14 – Submit at Pages 2 through 37, shows the approved change order requests for the Milestone # 13 Completion verified. Milestone 14: Communicate updated and newly created process(es). Completion Date: August 15, 2018 Actual Completion Date: August 3, 2018 File 2, “ Certification Package Updated”, Milestone 7 – Submit, Milestone 10 – Submit, Milestone 11 – Submit, and Milestone 12 – Completion Report, contain the relevant communications and/or records of communications. Milestone # 14 Completion verified. The Mitigation Plan is hereby verified complete. NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Anthony Jablonski Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: February 7, 2019 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Attachment 7 Record documents for the violations of CIP-006-6 R1 7.a The Entity’s Self-Report (RFC2017017304); 7.b The Entity’s Mitigation Plan designated as RFCMIT012854 submitted ; 7.c The Entity’s Certification of Mitigation Plan Completion dated ; 7.d ReliabilityFirst’s Verification of Mitigation Plan Completion dated ; 7.e The Entity’s Self-Report (RFC2017017547); 7.f The Entity’s Mitigation Plan designated as RFCMIT012890 submitted ; 7.g The Entity’s Certification of Mitigation Plan Completion dated ; 7.h ReliabilityFirst’s Verification of Mitigation Plan Completion dated ; 7.i The Entity’s Self-Report (RFC2017018166); 7.j The Entity’s Mitigation Plan designated as RFCMIT013214 submitted ; 7.k The Entity’s Certification of Mitigation Plan Completion dated ; 7.l ReliabilityFirst’s Verification of Mitigation Plan Completion dated ; 7.m The Entity’s Self-Report (RFC2017018857); 7.n The Entity’s Mitigation Plan designated as RFCMIT013482 submitted ; 7.o The Entity’s Certification of Mitigation Plan Completion dated ; 7.p ReliabilityFirst’s Verification of Mitigation Plan Completion dated NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst March 17, 2017 Self Report Additional Entity Comments: Comment From User Name Additional Comments No Comments No Documents Additional Documents Size in Bytes From Document Name Description Page 4 of 4 03/17/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 12 Page 2 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Plan Details Corrective Actions: A PSP Door and Alarm Study (Milestone 1) was conducted on 02/10/2017 covering the PSP doors at and benchmarking other industrial sites to identify common equipment and human performance issues and resolutions. The study included an inventory of all PSP door hardware, examination of all maintenance performed and re-examination of past violations regarding access control. The results of the study led to a two phased approach to implementing PSP door security operations, maintenance and testing for the PSP doors: • Phase 1: most problematic doors • Phase 2: Remaining doors Four PSP doors providing alternate access to PSPs (ie. Not the primary entrance) were identified as high failure PSP doors during the Door and Alarm Study. These alternate access doors were temporarily roped off on 02/13/2017 with "Emergency Use Only" signs posted in an effort to reduce the high alarm volumes that consume security resources (Milestone 2).
Preventive Actions: • Define, document and communicate PSP Program roles and responsibilities to include Physical Security Program Owner accountability and business unit/vendor responsibilities as they relate to PSP operations and maintenance (Milestone 3).
• Develop Functional Requirements Document (FRD) to address business, functional, non-functional and stakeholder requirements for PSP doors and door hardware located in industrial security environments (Milestone 4).
• Develop a detailed pre-specification for PSP single door and double door design types. Pre-specifications will cover the door and associated door hardware for PSP doors located in industrial security environments (Milestone 5).
• Develop and execute a Pilot security operations and maintenance test plan for phase one PSP doors based on functional requirements and industrial design pre-specifications. Pilot Test Plan encompasses two standards: A PSP Single Door standard and PSP Double Door standard (Milestone 6).
• Conduct Test of One on PSP Single Door and PSP Double Door (Milestone 7) with the expected outcome of a 'Go-No Go' determination for implementing the Phase One PSP Door Replacement Plan for the PSP doors and/or door hardware at (Milestone 8) and Phase Two PSP Door Replacement Plan for the remaining PSP doors and/or door hardware at (Milestone 9).
Identify and describe the action plan, including specific tasks and actions that your organization is proposing to undertake, or which it undertook if this Mitigation Plan has been completed, to correct the violation(s) identified above in Section C.1 of this form: Provide the timetable for completion of the Mitigation Plan, including the completion date by which the Mitigation Plan will be fully implemented and the violations associated with this Mitigation Plan are corrected: July 31, 2017 Proposed Completion date of Mitigation Plan: Milestone Activities, with completion dates, that your organization is proposing for this Mitigation Plan: Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 02/10/2017 Milestone 1 Conduct PSP Door and Alarm Study.
Study was conducted 02/10/2017 No 12 Page 6 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending on all PSP doors at and included an inventory of all PSP door hardware, examination of all maintenance performed and re-examination of past violations regarding access control. Other industrial sites were benchmarked as part of the study to identify common equipment and human performance issues and resolutions. A two phased approach will be taken to provide a comprehensive industrial security approach for PSP doors: - Phase 1: most problematic doors - Phase 2: Remaining doors 02/13/2017 Milestone 2 Temporarily Block Off High Failure PSP Doors. PSP doors providing alternate access to PSPs (ie. Not the primary entrance) were identified as high failure PSP doors in the Door and Alarm Study.
These alternate access doors have been temporarily roped off and "Emergency Use 02/13/2017 No 12 Page 7 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending Only" signs were posted in an effort to reduce the high alarm volumes that consume security resources.
04/05/2017 Milestone 4 Develop Functional Requirements Document (FRD).
FRD will address business, functional, non-functional and stakeholder requirements for PSP doors and door hardware located in industrial security environments.
04/05/2017 No 04/09/2017 Milestone 6 Develop Pilot Test Plan.
Develop and execute a Pilot security operations and maintenance test plan for phase one PSP doors based on functional requirements and industrial design pre-specifications. Pilot Test Plan encompasses two standards: A PSP Single Door standard and PSP Double Door standard.
04/10/2017 No 04/10/2017 Milestone 5 Develop PSP Door Pre-specifications.
Develop a detailed pre-specification for PSP single door and double door design 04/08/2017 No 12 Page 8 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending types. Pre-specifications will cover the door and associated door hardware for PSP doors located in industrial security environments.
04/26/2017 Milestone 7 Conduct Test of One on PSP Single Door and PSP Double Door.
Expected outcome is a 'Go-No Go' determination for implementing Phase One PSP Door Replacement Plan and following Phase Two PSP Door Replacement Plan.
No 05/31/2017 Milestone 3 Define and Document PSP Program Roles and Responsibilities.
Define, document and communicate PSP Program roles and responsibilities to include Physical Security Program Owner accountability and business unit/vendor responsibilities as they relate to PSP operations and maintenance.
No 06/30/2017 Milestone 8 Implement Phase One PSP Door Replacement Plan.
Implement phase one replacement No 12 Page 9 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending plan for the PSP doors and/or door hardware at 07/30/2017 Milestone 9 Implement Phase Two Rollout.
Implement phase replacement plan for the remaining PSP doors and/or door hardware at No Additional Relevant Information 12 Page 10 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst May 01, 2017 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts. PSP doors providing alternate access to PSPs (ie. Not the primary entrance) were identified as high failure PSP doors during the Door and Alarm Study. These alternate access doors were temporarily roped off on 02/13/2017 with "Emergency Use Only" signs posted in an effort to reduce the high alarm volumes that consume security resources (Milestone 2).
Prevention Successful completion of the Mitigation Plan as laid out in Section D will minimize the probability of incurring further access control failures associated with the PSP doors and door hardware located at Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 12 Page 11 of 05/01/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone 2: Temporarily Block Off High Failure PSP Doors. PSP doors providing alternate access to PSPs (i.e. Not the primary entrance) were identified as high failure PSP doors in the Door and Alarm Study. These alternate access doors have been temporarily roped off and "Emergency Use Only" signs were posted in an effort to reduce the high alarm volumes that consume security resources. Proposed Completion Date: February 13, 2017 Actual Completion Date: February 13, 2017 File 1, “RFC2017017304 Submission” Milestone 2 Submit, Page, 2, shows a positive result of the entities’ mitigation plan and actions taken after incident discovery. This document shows the number of nuisance alarms in which were generated prior to containment and countermeasures were effectively implemented. It shows the numbers in the thousands dropping to double digits almost immediately and then to the single digits within roughly 60 days’ time. Milestone # 2 Completion verified Milestone 3: Define and Document PSP Program Roles and Responsibilities. Define, document and communicate PSP Program roles and responsibilities to include Physical Security Program Owner accountability and business unit/vendor responsibilities as they relate to PSP operations and maintenance. Proposed Completion Date: May 31, 2017 Actual Completion Date: June 8, 2017 File 1, “RFC2017017304 Submission”, Milestone3 Submit, Pages 2 through 6, illustrate the RACI model in which the entity created in order to identify responsibilities related to PSP Programs. Milestone # 3 Completion verified Milestone 4: Develop Functional Requirements Document (FRD). FRD will address business, functional, non-functional and stakeholder requirements for PSP doors and door hardware located in industrial security environments. Proposed Completion Date: April 5, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: April 5, 2017 File 1, “RFC2017017304 Submission”, Milestone 4 Submit, Page 2, shows the functional requirements document which addresses stakeholder requirements residing within industrial security environments. Milestone # 4 Completion verified Milestone 5: Develop PSP Door Pre-specifications Develop a detailed pre-specification for PSP single door and double door design types. Pre-specifications will cover the door and associated door hardware for PSP doors located in industrial security environments. Proposed Completion Date: April 10, 2017 Actual Completion Date: October 13, 2017 File 1, “RFC2017017304 Submission”, Milestone 5 Submit, Pages 2 through 7, are an updated version of the entity Access control hardware specifications. This document sets forth the entity standard based upon door classification as to what door specifications need to be followed and/ or addressed according to company policy and procedure. This Latest revision is effective October 13, 2017. Milestone # 5 Completion verified Milestone 6: Develop Pilot Test Plan. Develop and execute a Pilot security operations and maintenance test plan for phase one PSP doors based on functional requirements and industrial design pre-specifications. Pilot Test Plan encompasses two standards: A PSP Single Door standard and PSP Double Door standard. Proposed Completion Date: April 10, 2017 Actual Completion Date: April 8, 2017 File 1, “RFC2017017304 Submission”, Milestone 6 Submit, Pages 6 through 10, illustrate the pilot door test plan after the implementation of new hardware. This test plan proposed a specified and expected outcome to determine if the door passed/ failed. Based on Pages 7 through 10, it shows the expected outcome vs. the actual outcome and the remediation’s if expected outcome was not obtained and/or failed. NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone # 6 Completion verified Milestone 7: Conduct Test of One on PSP Single Door and PSP Double Door. Expected outcome is a 'Go-No-Go' determination for implementing Phase One PSP Door Replacement Plan and Following Phase Two PSP Door Replacement Plan. Proposed Completion Date: April 26, 2017 Actual Completion Date: April 26, 2017 File 1, “RFC2017017304 Submission”, Milestone 7 Submit, Pages 2 through 6, illustrate the actual door test plan after the implementation of new hardware. This test plan proposed a specified and expected outcome to determine if the door passed/ failed. Based on Pages 7 through 10, it shows the expected outcome vs. the actual outcome and the remediation’s if expected outcome was not obtained and/or failed. Milestone # 7 Completion verified Milestone 8: Implement Phase One PSP Door Replacement Plan. Implement Phase on replacement plan for the PSP doors and/or door hardware at Proposed Completion Date: June 30, 2017 Actual Completion Date: June 30, 2017 File 1, “RFC2017017304 Submission”, Milestone 8 Submit, Pages 2 through 6, shows the first phase of the door replacement plan in regards to this milestone. Milestone # 8 Completion verified Milestone 9: Implement Phase Two Rollout. Implement phase replacement plan for the remaining PSP doors and/or door hardware at Proposed Completion Date: October 13, 2017 Actual Completion Date: September 29, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION File 1, “RFC2017017304 Submission”, Milestone 9 Submit, Page 4, shows the phase 2 replacement implementation including start and completion dates. As indicated by this milestone. Milestone # 9 Completion verified The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: December 5, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 10 Page 2 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone # 1 Completion verified. Milestone 2: Verify alarming functionality. File 1, “RFC2017017547 Certification Package”, Milestone 2 Submit, Pages 2 through 7, show the testing results of the door functionality after repair and prior to placing the door back into service. Milestone # 2 Completion verified. Milestone 3: Discipline two contract employees. File 1, “RFC2017017547 Certification Package”, Milestone 3 Submit, Page 1, is a description of the disciplinary action that was taken while Page 2 is a signed attestation stating that disciplinary action was taken. Milestone # 3 Completion verified. Milestone 4: Perform Alternate Measures. File 4, “RFC2017017547 Updated Milestone 4, 5 and 6”, Milestone 4 Submit update Pages 1 through 9, and information provided via teleconference show the alternate measures log along with a description of how/ what these alternate measures are and how they were carried out. The blanks within the evidence are also part of this mitigation plan and the entity did take immediate corrective actions for security officers who did not complete the logs as required. Milestone #4 Completion verified. Milestone 5: Conduct Physical walk down of PSP. File 4, “RFC2017017547 Updated Milestone 4, 5 and 6”, Milestone 5 submit update, Pages 2 through 27, show the updated tampering verification log which was discussed in detail via a teleconference with entity staff in regards to ensuring that they are/ were checking for signs of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION physical and electronic tampering. Since the teleconference the entity has changed its procedures and policies in order to reflect the checks for electronic and physical checks when in regards to tampering. Milestone # 5 Completion verified. Milestone 6: Review all barrier logs for March 2017. File 4, “RFC2017017547 Updated Milestone 4, 5 and 6”, Milestone 6 submit update, Pages 2 through 45, provide Q&A responses from RF to entity in regards to this mitigation plan as well as the testing log with additional callouts that were discussed via teleconference with the entity. The entity has provided additional callouts as to the blank areas in the logs and have since made corrections to their documentation in order to account for items that are not applicable (N/A) instead of leaving them blank and incomplete. This evidence provides account for their door testing sequence and the previous teleconference provided insight into how this testing is conducted. Milestone # 6 Completion verified. Milestone 7: Reinforcement of procedures. File 1, “RFC2017017547 Certification Package”, Milestone 7 Submit, Pages 2 and 3, show the training topics and the attendees of the training required by milestone 7. Milestone # 7 Completion verified. The Mitigation Plan is hereby verified complete. Date: NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst August 07, 2017 Self Report Additional Entity Comments: Comment From User Name Additional Comments No Comments No Documents Additional Documents Size in Bytes From Document Name Description Page 3 of 3 08/07/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst September 08, 2017 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 8 Page 2 of 09/08/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst September 08, 2017 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending tampering of the cyber asset hardware occurred since the exposure on May 25 07/27/2017 Update procedures Update and disseminate procedures to address NERC CIP requirements for cyber assets 07/27/2017 No 08/18/2017 Verify no tampering of the software Determine baselines and review (" logs to verify that no tampering of the cyber asset software occurred during the period May 25, 2017 through June 19, 2017 08/18/2017 No 11/17/2017 Update records documents The records documents, which entail plans and schematics, will be updated to include to prevent access.
No Additional Relevant Information 8 Page 6 of 09/08/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst September 08, 2017 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts.
The risk is minimal since the area that was exposed has been fixed by blocking the as of the end of the day on June 19, 2017. The access to this PSP is controlled by badge access. Therefore, compensating controls are in place that would have prevented a greater impact to the BES.
Prevention By updating and disseminating procedures to address NERC CIP requirements for cyber asset, including informing the initial Agenda has been updated to include a checklist item "Potential NERC/CIP Impact (Y/N). (If yes, complete Construction Checklist " The addition of this item now brings attention to this area and any new construction/remodeling projects are now aware of the requirement and projects will contain contingencies if NERC CIP requirements for cyber assets are involved directly or indirectly.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 8 Page 7 of 09/08/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: May 25, 2017 File 1, “RFC2017018166 Certification Package”, Milestone 1- Submit, Pages 2 through 6, show the results of the physical walk-down to verify that there was no tampering of assets physically. Milestone # 1 Completion verified. Milestone 2: Update procedures. Proposed Completion Date: July 27, 2017 Actual Completion Date: July 27, 2017 File 1, “RFC2017018166 Certification Package”, Milestone 2- Submit, Pages 2 through 20, provide documentation particularly a checklist (Page 3 section 7) in regards to contacting and a cyber SME in the event that a PSP and or adjacent location will be affected. In addition, new instruction for notifying construction managers was sent out to affected parties to notify them of procedural changes in this regard. Milestone # 2 Completion verified. Milestone 3: Verify no tampering of the software. Proposed Completion Date: August 18, 2017 Actual Completion Date: August 18, 2017 File 1, “RFC2017018166 Certification Package”, Files 3- Submit, Page 5 through 116, shows the results of the affected assets showing that no deviation from the existing baseline existed. Milestone # 3 Completion verified. Milestone 4: Update records documents. Proposed Completion Date: November 17, 2017 Actual Completion Date: November 6, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION File 1, “RFC2017018166 Certification Package”, Milestone 4- Submit, Page 2, shows the architectural drawing showing in order to prevent access as determined by this milestone in regards to updating the records and documents. Milestone # 4 Completion verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: November 28, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Self Report preventive maintenance on the door immediately following the incident, including degreasing the lock that was preventing it from falling into correct locking mechanism and ensuring the latch was functioning as required following the incident. such doors exist at the To check extent of condition, reviewed the alarm logs for all doors since last inspection of the doors to ensure that no forced entry alarm had been recorded from any of the doors. It was confirmed this was the only incident of such a nature.
What is the problem?
A malfunctioning door at opened despite an invalid access by the card read for door # The user had authorized unescorted physical access, since the user entry into the PSP was not logged, this has been recorded as a violation of CIP-006- R1 P1.8 - "Log (through automated means or by personnel who control entry) entry of each individual with authorized unescorted physical access into each Physical Security Perimeter, with information to identify the individual and date and time of entry." Root Cause of Possible Violation: As per the & RCA performed on 12/08/2017, the door locking mechanism malfunctioned and did not secure the lock in the desired place due to lack of a maintenance program How was the violation discovered?
On 11/28/2017, the ( at the received an Invalid Access Attempt and an immediate second alarm for Forced entry from card reader # Further investigation by the physical security team, determined valid unescorted access for the employee but a malfunctioning door that did not lock into place allowing for forced entry alarm.
Explain how is it determined that the Noncompliance is related to documentation, performance, or both.
On examining the root causes listed above, it was determined that noncompliance is related to the door malfunctioning i.e. technical problem, due to lack of a maintenance program at the Timeline: 1. 28 November 2016 - the ( at the received an Invalid Access Attempt and an immediate second alarm for Forced entry from card reader # 2. 28 December 2016 - A member of the Security team, reviewed the 2 alarms on video and then reached on-site to discuss incident with employee who had triggered the alarm to confirm authorized unescorted access card of the employee , and an invalid Fob swiped by the employee instead of the valid card in error.
3. 28 December 2016 - This member of the security team, examined the door locking magnetic bars and noticed they were not latching as required. He performed a degreasing of the lock and ensured the locking mechanism worked securely before leaving the site.
Mitigating Activities: Description of Mitigating Activities and Preventative Measure: Corrective (Immediate) Activities: An immediate corrective maintenance on door with card reader # was performed following the incident to ensure the door's locking mechanism was working as required. It was checked and corrected to secure entry.
Mitigating Activities: To Counter the invalid access alarms as a false occurrence, due to multiple cards (such as key fobs and access cards together), a broadcast Page 2 of 4 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Self Report communication will be circulated to all Employees by 12/15/2017 to ensure they carry their door access cards separate from other readable electronic chips that might cause an Invalid Access Attempt alarms.
Alarm log for forced-in and forced-out instances for all doors was reviewed for similar forced entry alarms to ensure no malfunctioning of the door mechanism could have caused a potential insecure unlogged entry into the PSP. The logs were reviewed since the last reported inspection performed at the door. No such incident was identified in this review.
Preventative Measures: To ensure that the doors functions as required, a maintenance contract with a service provider for all doors located in the will be finalized by 12/31/2017. This maintenance would include maintenance and repair to avoid malfunctioning of the doors, including the locking mechanism.
Once the maintenance contract is finalized, the details of the contract would be entered into a WO to monitor the Project Management of the execution of the contract for all doors.
A SWI for the maintenance program of the doors will be created.
Date Mitigating Activities Completed: Impact and Risk Assessment: Description of Potential and Actual Impact to BPS: Actual Impact Since the user had authorized unescorted access granted to the PSP, his entry into the door did not cause any risk. The door logs were also reviewed since last inspection to check for similar forced alarm entries. Post review, it was confirmed that there are no alarms since last inspection.
Potential Impact Potential impact of a forced entry alarm to a PSP was low since the user had authorized unescorted access and was a daily worker in the Minimal Minimal Actual Impact to BPS: Potential Impact to BPS: Risk Assessment of Impact to BPS: The risk assessed to the BES is low based on access level of the employee who triggered the alarm and a review of all alarm logs at to determine no such incident that been recorded for door malfunctioning causing a forced door alarm since last inspection performed on the doors in December 6, 2017.
Additional Entity Comments: Comment From User Name Additional Comments No Comments Additional Documents Page 3 of 4 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst December 14, 2017 Self Report No Documents Size in Bytes From Document Name Description Page 4 of 4 12/14/2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Confidential Non-Public Information January 09, 2018 Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 9 Page 2 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst January 09, 2018 Violation(s) This Mitigation Plan is associated with the following violation(s) of the reliability standard listed below: Requirement Violation ID Date of Violation Requirement Description RFC2017018857 11/28/2017 CIP-006-6 R1.
Each Responsible Entity shall implement one or more documented physical security plan(s) that collectively include all of the applicable requirement parts in CIP-006-6 Table R1 – Physical Security Plan.
Brief summary including the cause of the violation(s) and mechanism in which it was identified: On November 28, 2017 at approximately, 4:41 PM, the ( at the received an "Invalid Access Attempt" and an immediate second alarm for "Forced entry" from card reader # (Door of PSP). immediately responded to these alarms and when the physical security team reached onsite they confirmed that employee ID that had triggered the invalid access attempt was a employee with authorized unescorted access to the area. However, the employee was carrying multiple cards along with his door access card, the card reader read FOB for his gym access. The card reader denied access and alerted an Invalid access to the This indicates that the card reader was working as required and the Invalid access attempt alert was appropriate. Despite denied access, the door could be pulled open by the employee, who did not realize the access was denied. This triggered the second alarm at for Forced entry.
On investigation by the security personnel, the door's locking mechanism was malfunctioning and was not locking into place. The employee was therefore able to pull open the door despite a denied access with the Key FOB.
The security personnel who attended to the alarm had seen a video preview of the employee to confirm the incidents causing the alarm. He performed a preventive maintenance on the door immediately following the incident, including degreasing the lock that was preventing it from falling into correct locking mechanism and ensuring the latch was functioning as required following the incident. such doors exist at the To check extent of condition, reviewed the alarm logs for all doors since last inspection of the doors to ensure that no forced entry alarm had been recorded from any of the doors. It was confirmed this was the only incident of such a nature.
What is the problem?
A malfunctioning door at opened despite an invalid access by the card read for door # The user had authorized unescorted physical access, since the user entry into the PSP was not logged, this has been recorded as a violation of CIP-006- R1 P1.8 - "Log (through automated means or by personnel who control entry) entry of each individual with authorized unescorted physical access into each Physical Security Perimeter, with information to identify the individual and date and time of entry." Root Cause of Possible Violation: As per the & RCA performed on 12/08/2017, the door locking mechanism malfunctioned and did not secure the lock in the desired place due to lack of a maintenance program Explain how is it determined that the Noncompliance is related to documentation, performance, or both.
On examining the root causes listed above, it was determined that noncompliance is related to the door malfunctioning i.e. technical problem, due to lack of a maintenance program at the Timeline: 1. 28 November 2016 - the ( at the received an Invalid Access Attempt and an immediate second alarm for Forced entry from card reader # 2. 28 December 2016 - A member of the Security team, reviewed the 2 alarms on video and then reached on-site to discuss incident with employee who had triggered the alarm to confirm authorized unescorted access card of the employee , and an invalid Fob swiped by the employee instead of the valid card in error.
3. 28 December 2016 - This member of the security team, examined the door locking magnetic bars and noticed they were not latching as required. He performed a degreasing of the lock and ensured the locking mechanism worked securely before leaving the site. 9 Page 4 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst January 09, 2018 Relevant information regarding the identification of the violation(s): On 11/28/2017, the ( at the received an Invalid Access Attempt and an immediate second alarm for Forced entry from card reader # Further investigation by the physical security team, determined valid unescorted access for the employee but a malfunctioning door that did not lock into place allowing for forced entry alarm. 9 Page 5 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst January 09, 2018 Plan Details Mitigating Activity: Milestone 1: Review alarm log for forced-in and forced-out instances for all doors. Purpose of this milestone is to verify that the similar condition did not exist on any other door. The review was performed for the period of last inspection to current date. All the doors were found functioning without any issues. This review was performed by analyst in The evidence will be review of logs for all the PSP doors in Preventative activities: The milestones below directly address the root cause and help reduce the risk of such occurrences in future.
Milestone 2: A SWI for the maintenance program of the doors will be created. Purpose of this SWI is to help the assignee of the WO to perform the work.
Milestone 3: The WO is a required artifact to perform any and all the jobs in In order to the mitigation, will create a recurring Work Order (WO) in for monthly maintenance of all PSP doors. Purpose of this milestone is to ensure that every month a WO is assigned by the system (no human interaction) without a failure. The assignee will perform the work and will have to close the WO otherwise will create an automatic escalation.
Identify and describe the action plan, including specific tasks and actions that your organization is proposing to undertake, or which it undertook if this Mitigation Plan has been completed, to correct the violation(s) identified above in Section C.1 of this form: Provide the timetable for completion of the Mitigation Plan, including the completion date by which the Mitigation Plan will be fully implemented and the violations associated with this Mitigation Plan are corrected: January 12, 2018 Proposed Completion date of Mitigation Plan: Milestone Activities, with completion dates, that your organization is proposing for this Mitigation Plan: Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 12/13/2017 Review alarm log for forced-in and forced-out instances for all doors.
Purpose: Ensure that a similar condition does not exist on any other door.
Evidence: An excel sheet with review from 12/13/2017 No 12/21/2017 Develop SWI for the maintenance program of the doors Purpose: Help the assignee of the WO to perform the work.
Evidence: A newly created SWI 12/21/2017 No 9 Page 6 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst January 09, 2018 Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 01/12/2018 Create a recurring Work Order (WO) in for monthly maintenance of all PSP doors Purpose: Ensure that WO is issued and Preventive Maintenance is performed. The SWI noted in last milestone is reviewed/provided as Pre-Specification with monthly WO.
Evidence: A recurring Work Order No Additional Relevant Information 9 Page 7 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst January 09, 2018 Reliability Risk Reliability Risk While the Mitigation Plan is being implemented, the reliability of the bulk Power System may remain at higher Risk or be otherwise negatively impacted until the plan is successfully completed. To the extent they are known or anticipated : (i) Identify any such risks or impacts, and; (ii) discuss any actions planned or proposed to address these risks or impacts.
Based on the review of logs of all the doors does not see any risk or a negative impact as all other doors functioned without a failure. All the processes and procedures for alarming, alerting and performing an immediate preventive maintenance worked as designed.
Prevention A monthly preventive maintenance will reduce the risk of occurrence of similar issues in future.
Describe how successful completion of this plan will prevent or minimize the probability further violations of the same or similar reliability standards requirements will occur This activity was completed on 12/08/2017.
Communicate to all the employees in the advising them to carry the door access key separate from other access cards or key fobs. Purpose of this milestone to raise the awareness that the non "electronically readable" cards/fobs may create noise when used along with door access card.
Describe any action that may be taken or planned beyond that listed in the mitigation plan, to prevent or minimize the probability of incurring further violations of the same or similar standards requirements 9 Page 8 of 01/09/2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone 3: Create a recurring Work Order (WO) in for monthly maintenance of all PSP doors. Proposed Completion Date: January 12, 2018 Actual Completion Date: December 16, 2017 File 3, “RFC2017018857 Additional Data on request”, Milestone 3- Submit Page 3, shows the tasks created until April in order to maintain the doors as required by this milestone. Page 2 shows the configuration of that work order showing its frequency and frequency units. Milestone # 3 Completion verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: March 10, 2018 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Attachment 8 Record documents for the violations of CIP-007-3a R3 8.a The Entity’s Self-Report (RFC2016016341); 8.b ReliabilityFirst’s Verification of Mitigating Activities Completion dated ; 8.c The Entity’s Self-Report (RFC2016016342); 8.d The Entity’s Mitigation Plan designated as RFCMIT012397-1 submitted ; 8.e The Entity’s Certification of Mitigation Plan Completion dated ; 8.f ReliabilityFirst’s Verification of Mitigation Plan Completion dated NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Milestone 4: Patching compliance- were conducted by Business Unit that included all employees involved in patching by October 5, 2016. File 2, “Mitigating Activities”, MA 3a) Patch Management Program Compliance Stand- FINAL, Pages 1 through 3, show the topics covered by the entity in their meetings. File 2, “Mitigating Activities”, MA 3b) – PMP Meeting Sign-in sheet, Page 1 shows the departments sign in sheet for the previously mentioned stand-down. File 2, “Mitigating Activities”, MA 3c) 20161006064118581, Page 1, shows the facility sign in sheet for the previously mentioned File 2, “Mitigating Activities”, MA 3d) Network Eng – Oct5 NERC CIP patching sign in sheet, Page 1, shows the sign in sheet for the previously mentioned stand-down. File 2, “Mitigating Activities”, MA 3e) patching Page 1, shows the department sign in sheet for the previously mentioned Milestone #4 Completion verified. The Mitigating Activities is hereby verified complete. Kristen Senk Senior Counsel ReliabilityFirst Corporation Date: NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Compliance Notices Section 6.2 of the NERC CMEP sets forth the information that must be included in a Mitigation Plan. The Mitigation Plan must include: (1) The Registered Entity's point of contact for the Mitigation Plan, who shall be a person (i) responsible for filing the Mitigation Plan, (ii) technically knowledgeable regarding the Mitigation Plan, and (iii) authorized and competent to respond to questions regarding the status of the Mitigation Plan. This person may be the Registered Entity's point of contact described in Section B. (2) The Alleged or Confirmed Violation(s) of Reliability Standard(s) the Mitigation Plan will correct. (3) The cause of the Alleged or Confirmed Violation(s). (4) The Registered Entity's action plan to correct the Alleged or Confirmed Violation(s). (5) The Registered Entity's action plan to prevent recurrence of the Alleged or Confirmed violation(s). (6) The anticipated impact of the Mitigation Plan on the bulk power system reliability and an action plan to mitigate any increased risk to the reliability of the bulk power-system while the Mitigation Plan is being implemented. (7) A timetable for completion of the Mitigation Plan including the completion date by which the Mitigation Plan will be fully implemented and the Alleged or Confirmed Violation(s) corrected. (8) Implementation milestones no more than three (3) months apart for Mitigation Plans with expected completion dates more than three (3) months from the date of submission. Additional violations could be determined or recommended to the applicable governmental authorities for not completing work associated with accepted milestones. (9) Any other information deemed necessary or appropriate. (10) The Mitigation Plan shall be signed by an officer, employee, attorney or other authorized representative of the Registered Entity, which if applicable, shall be the person that signed the Self Certification or Self Reporting submittals. (11) This submittal form may be used to provide a required Mitigation Plan for review and approval by regional entity(ies) and NERC.
• The Mitigation Plan shall be submitted to the regional entity(ies) and NERC as confidential information in accordance with Section 1500 of the NERC Rules of Procedure.
• This Mitigation Plan form may be used to address one or more related alleged or confirmed violations of one Reliability Standard. A separate mitigation plan is required to address alleged or confirmed violations with respect to each additional Reliability Standard, as applicable.
• If the Mitigation Plan is accepted by regional entity(ies) and approved by NERC, a copy of this Mitigation Plan will be provided to the Federal Energy Regulatory Commission or filed with the applicable governmental authorities for approval in Canada.
• Regional Entity(ies) or NERC may reject Mitigation Plans that they determine to be incomplete or inadequate.
• Remedial action directives also may be issued as necessary to ensure reliability of the bulk power system.
• The user has read and accepts the conditions set forth in these Compliance Notices. 10 Page 2 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst ( process ( NERC-CIP asset change control process for new assets).
Table 1 Timeline Date Event May 2008 System Installation May 2008 - September 2015 Administrative Software packages installed 10/20/15 go live 7/1/16 v5 goes live 7/16/16 Identification of issue 7/16/16 Started review of Patch Source and applicable patches 7/16 - 7/29/16 All other packages and Operating System were patched during monthly patch cycle.
9/29/16 Completed review of Patch Source and applicable patches 10/10/16 Installed patches 10/24/16 Knowledge share 12/07/2016 Effectiveness Review 3/07/2017 Extent of Condition review of patch sources Relevant information regarding the identification of the violation(s): During a Mock Audit, the SME was discussing their patch sources. They were in the process of removing software that was not essential to their system. team members were in the audience listening to this discussion when they realized that they have some of the same software and that it was not covered by 10 Page 5 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION ReliabilityFirst Milestone Activity Proposed Completion Date (Shall not be greater than 3 months apart) Actual Completion Date Description Entity Comment on Milestone Completion Extension Request Pending 12/07/2016 Effectiveness Review Group review of the taken mitigation activities.
12/07/2016 No 03/07/2017 Extent of Condition review of patch sources review of current patch sources 02/28/2017 No 03/31/2017 Software Disposition Split the software packages, Operating Systems, etc. into two categories No 06/30/2017 Software Verification Verify that the remaining software is needed No 07/14/2017 Remove software that is not needed Remove the unneeded software No 07/31/2017 Determine patch source and update as necessary For software with a business reason not covered by identify a new patch source and update (or mitigate) as necessary.
No Additional Relevant Information Milestone Activity Completion Date Revise patch source 09/09/2016 Evaluate patch sources 09/29/2016 Install required patches 10/11/2016 Knowledge share 10/24/2016 Effectiveness Review 12/07/2016 Extent of Condition review of patch sources 3/07/2017 Software Disposition 3/31/2017 Software Verification 6/30/2017 Remove software that is not needed 7/14/2017 Determine patch source and update as necessary 7/31/2017 10 Page 8 of NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Verification of Mitigation Plan Completion Milestone 1: Revise patch source template. File 1, “RFC2016016342 Certification Package”, Milestone 1 Evidence, Pages 3 through 33, show the affected patch source documents with the applicable changes highlighted and bookmarked. Page 2 summarizes the work order used to make the changes, showing completion on September 9, 2016. Proposed Completion Date: September 9, 2016 Actual Completion Date: September 9, 2016 Milestone # 1 Completion verified. Milestone 2: Evaluate patch new sources. File 1, “RFC2016016342 Certification Package”, Milestone 2 Evidence, Pages 3 through 78, show the identification and assessment of patches for the newly identified applications. The date of assessment is shown as September 29, 2016. Proposed Completion Date: September 29, 2016 Actual Completion Date: September 29, 2016 Milestone # 2 Completion verified. Milestone 3: Install required patches. File 1, “RFC2016016342 Certification Package”, Milestone 3 Evidence, Page 5, shows completion of patching on October 2, 2016. Pages 6 and 7, show details of the changes made to the installed software. Proposed Completion Date: October 11, 2016 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: October 2, 2016 Milestone # 3 Completion verified. Milestone 4: Knowledge sharing session. File 1, “RFC2016016342 Certification Package”, Milestone 4 Evidence, Pages 2 through 8, show a slide deck used at a meeting of SMEs held on October 24, 2016. Page 9, shows the attendee sign-in for that meeting. Proposed Completion Date: October 24, 2016 Actual Completion Date: October 24, 2016 Milestone # 4 Completion verified. Milestone 5: Effectiveness review. File 1, “RFC2016016342 Certification Package”, Milestone 5 Evidence, Page 1, documents an effectiveness review held with and RF staff at the offices on December 7, 2016. Follow-up meetings were held with staff as documented on Pages 3 through 8. A final follow-up with and RF staff was held on a conference call on February 24, 2017. Proposed Completion Date: December 7, 2016 Actual Completion Date: December 7, 2016 Milestone # 5 Completion verified. Milestone 6: Extent of Condition review of patch sources. NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION File 1, “RFC2016016342 Certification Package”, Milestone 6 Evidence. Page 2, contains attestations of completion of an extent-of-condition review by all affected asset types, with the latest completion date of February 28, 2017. While use of attestations as evidence of work completed is considered weak, this evidence us supported by more detailed evidence in Milestone 7. Proposed Completion Date: March 7, 2017 Actual Completion Date: September 29, 2016 Milestone # 6 Completion verified. Milestone 7: Software disposition. File 1, “RFC2016016342 Certification Package”, Milestone 7 Evidence, Pages 2 and 3, show the list of software known by not to be included in patch report. Pages 4 through 40, contain the list of software for each class of asset that is unsure of. states that this list was submitted to for review. response is evidenced in Milestone 8. No date is included in these reports. However, File 1, “RFC2016016342 Certification Package”, Milestone 8 Evidence Page 2, shows opening of a case with support on March 7, 2017 with this information. Proposed Completion Date: March 31, 2017 Actual Completion Date: March 7, 2017 Milestone # 7 Completion verified. Milestone 8: Software verification. File 1, “RFC2016016342 Certification Package”, Milestone 8 Evidence, Pages 2 through 5, show a ticket opened with support to identify information about the packages that were listed as “unknown” in Milestone 7. Pages 43 through 125, show evaluation of response to the questions about whether the software was needed and if so, was it included in NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION patch summary. No dates are included in the evidence. However, File 1, “RFC2016016342 Certification Package”, Page 4, of Milestone 9 Evidence shows initiation of a work order to remove software identified by this process as not needed on July 5, 2017. Proposed Completion Date: June 30, 2017 Actual Completion Date: July 5, 2017 Milestone # 8 Completion verified. Milestone 9: Remove software that is not needed. File 1, “RFC2016016342 Certification Package”, Milestone 9 Evidence, Pages 3 and 4, show a ticket created to remove software identified as not needed. That software is listed on Page 3. PDF Pages 6 through 1157, show a baseline taken after package removal. Examination of a small sample of software shows that the software in the list was removed. Page 6, shows the baseline date of July 24, 2017. Proposed Completion Date: July 14, 2017 Actual Completion Date: July 24, 2017 Milestone # 9 Completion verified. Milestone 10: Determine patch source and update as necessary. File 1, “RFC2016016342 Certification Package”, Milestone 10 Evidence, Pages 2 through 554, shows a detailed patch source listing for each applicable system. This patch list shows changes identified by the process in Milestones 7-9. For example, a patch source list from October 2016 shows software as having as a patch source whereas the list provided for this Milestone shows as the patch source. No date is provided in the PDF. However, the PDF itself shows a creation date of July 31, 2017. Proposed Completion Date: July 31, 2017 NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION Actual Completion Date: July 31, 2017 Milestone # 10 Completion verified. The Mitigation Plan is hereby verified complete. Tony Purgar Manager, Risk Analysis & Mitigation ReliabilityFirst Corporation Date: NON-PUBLIC AND CONFIDENTIAL INFORMATION HAS BEEN REMOVED FROM THIS PUBLIC VERSION |
187564 | https://www.sunyopt.edu/labs/Zaidi/pubs/Alldocuments/Sawada&Zaidi.2018.pdf | Rotational-symmetry in a 3D scene and its 2D image Tadamasa Sawada1 and Qasim Zaidi2 1School of Psychology, National Research University Higher School of Economics 2SUNY Optometry - Graduate Center for Vision Research Abstract A 3D shape of an object is N-fold rotational-symmetric if the shape is invariant for 360/N degree rotations about an axis. Human observers are sensitive to the 2D rotational-symmetry of a retinal image, but they are less sensitive than they are to 2D mirror-symmetry, which involves invariance to reflection across an axis. Note that perception of the mirror-symmetry of a 2D image and a 3D shape has been well studied, where it has been shown that observers are sensitive to the mirror-symmetry of a 3D shape, and that 3D mirror-symmetry plays a critical role in the veridical perception of a 3D shape from its 2D image. On the other hand, the perception of rotational-symmetry, especially 3D rotational-symmetry, has received very little study. In this paper, we derive the geometrical properties of 2D and 3D rotational-symmetry and compare them to the geometrical properties of mirror-symmetry. Then, we discuss perceptual differences between mirror- and rotational symmetry based on this comparison. We found that rotational-symmetry has many geometrical properties that are similar to the geometrical properties of mirror-symmetry, but note that the 2D projection of a 3D rotational-symmetrical shape is more complex computationally than the 2D projection of a 3D mirror-symmetrical shape. This computational difficulty could make the human visual system less sensitive to the rotational-symmetry of a 3D shape than its mirror-symmetry.
1. Introduction The human visual system is sensitive to the following three types of symmetry (Mach, 1906/1959): mirror (or bilateral or reflectional), rotational (or cyclic or radial)1, and translational (or repetition). Each type of symmetry is formally defined as an invariant against a particular transformation (Liu, Hel-Or, Kaplan, Van Gool, 2009; Weyl, 1952; Stewart & Golubitsky, 1992). For example, consider objects with rotational-symmetry. A 3D shape of a rotational-symmetrical object coincides with itself after rotating the object about its axis of symmetry for a particular angle (see Figure 1 for examples). Rotational-symmetry Publisher's Disclaimer: This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final citable form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
6It is possible to assert that this camera rotation transforms the vanishing point vaxis of the symmetry axis to the principal point so that the symmetry axis becomes normal to the image plane ΠI after the camera rotation.
1It is possible to assert that rotational-symmetry in Cartesian-coordinates is translational-symmetry in polar-coordinates (van der Helm & Leeuwenberg, 1996).
HHS Public Access Author manuscript J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Published in final edited form as: J Math Psychol. 2018 December ; 87: 108–125. doi:10.1016/j.jmp.2018.10.001.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript appears in many man-made objects, many flowers (Neal, Dafni, & Giurfa, 1998; Culbert & Forrest, 2016), some animal species (e.g. echinoderm and cnidarian), and local parts of plants and animals (Savriama & Klingenberg, 2011). It is also common in 2D image designs: e.g. texture patterns (Liu, Collins, Tsin, 2004; Clarke, Green, Halley, & Chantler, 2011; Westphal & Fitch, 2012) and logos (Hargittal & Hargittal, 1997).
Rotational-symmetry of a 2D image plays a role in visual perception and cognition, but its effects tend to be weaker than the effects of mirror-symmetry (see van der Helm & Leeuwenberg, 1996; Wagemans, 1995; Swaddle, 1999 for reviews). Rotational-symmetry can be reliably detected (dʹ > 1) in a low-density random-dot pattern even with a brief viewing duration (100 ms) but it is not as easy to detect as mirror-symmetry (Figure 2a, b, Kahn & Foster, 1986; Wagemans, Van Gool, Swinnen, & Van Horebeek, 1993; see also Szlyk, Seiple, & Xie, 1995 for a relevant study). The detection of rotational-symmetry takes longer than the detection of mirror-symmetry (Royer, 1981; Palmer & Hemenway, 1978). Julesz (1971) showed that rotational-symmetry is hard to detect in a high-density random-dot pattern while mirror-symmetry is easily detected (Figure 2c, d). Past studies have shown that figures with rotational-symmetry are rated as “good” (Palmer, 1991; Garner & Clement, 1963), rated as “organized” (Hershenson & Ryder, 1982, see also Wagemans, 1997), and associated with positive words (Makin, Pecchinenda, & Bertamini, 2012) more often than asymmetrical figures are, but less often than mirror-symmetrical figures are (see also Hamada & Ishihara, 1988; Hamada et al., 2016 for inconsistent results).
Results of brain imaging studies suggest that the ventral stream in the visual system is involved in processing 2D rotational-symmetry. Both rotational- and mirror-symmetry induce sustained posterior negativity of the ERP signal measured from two occipital electrodes (PO7 and PO8 according to the international 10–20 system) 300ms after the onset of the stimuli (Makin, Wilton, Pecchinenda, & Bertamini, 2012; Makin, Rampone, Pecchinenda, & Bertamini, 2013). This induced effect is stronger for mirror-symmetry than for rotational-symmetry. Based on a source localization analysis of the ERP signal, the effect is caused primarily by activity in the lateralized extrastriate visual cortex (Makin, Wilton et al., 2012). Kohler, Clarke, Yakovleva, Liu, and Norcia (2016) showed that rotational-symmetry in a texture pattern is parametrically represented in V3 and in later visual areas in the ventral stream (V4, VO1, and LOC) using fMRI and EEG.
The 3D rotational-symmetry of an object can play some role in the perception of the object’s 3D shape. The perception of the center-of-gravity of an object becomes more accurate if the object is rotational-symmetrical (Bingham & Muchisky, 1993a, b). According to Biederman’s Recognition-by-components theory, a complex 3D shape of an object can be decomposed into simpler parts called “geons” (Biederman, 1987; see also Pentland, 1986; Binford, 1971 for analogous ideas). Some of the geons used in past studies (Biederman & Gerhardstein, 1993) were 3D rotational-symmetrical. Note that the perception of 3D rotational-symmetry has been studied much less often than 3D mirror-symmetry.
Now, consider human’s visual perception of mirror-symmetry. Humans can detect the mirror-symmetry of a retinal image efficiently (e.g. Barlow & Reeves, 1979; Jenkins, 1983; Cohen & Zaidi, 2013), of a non-frontoparallel planar figure (Sawada & Pizlo, 2008; Sawada and Zaidi Page 2 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Wagemans, 1992, 1993; van der Vloed, Csathó, & van der Helm, 2005; Szlyk, Rock, & Fisher, 1995), and of a volumetric object (Sawada, 2010). Moreover, mirror-symmetry also plays a critical role in the perception of the shapes of planar figures (Sawada, 2008; Saunders & Knill, 2001) and of volumetric objects (Pizlo, 2008; Li, Pizlo, & Steinman, 2009; Li, Sawada, Shi, Kwon, & Pizlo, 2011; Pizlo, Sawada, Li, Kropatsch, & Steinman, 2010; Pizlo, Li, Sawada, & Steinman, 2014). Mirror-symmetry of the volumetric object allows us to recover a complete 3D shape of the object including its invisible back part from a single 2D image of the object (Mitsumoto, Tamura, Okazaki, Kajimi, & Fukui, 1992; Pizlo et al., 2010, 2014; Michaux, Kumar, Jayadevan, Delp, & Pizlo, 2017). It is important to recover the complete 3D shape of the object for interacting with the object (Varley, DeChant, Richardson, Ruales, & Allen, 2017).
The high sensitivity of human’s to mirror-symmetry is often explained teleologically. For example, many objects around us are mirror-symmetrical and mirror-symmetry serves as an important factor for sexual selection of many animals (Møller & Thornhill 1998; Møller, Thornhill, & Gangestad, 2005). Furthermore, the mirror-symmetry of an object introduces unique geometrical properties into its 3D shape and into its 2D retinal image (e.g. Vetter & Poggio, 1994; Sawada, 2010; Sawada, Li, & Pizlo, 2014). These geometrical properties can play an important role in the perception of mirror-symmetry (Sawada, Li, & Pizlo, 2015; Pizlo et al., 2014).
In this study, we derive the geometrical properties of 2D and 3D rotational-symmetry that correspond to geometrical properties of mirror-symmetry and then compare these properties between the two types of symmetry. This comparison allows us to analytically discuss human perception and cognition of rotational- and mirror-symmetry.
2. Definition In this study, it is assumed that all 2D and 3D curves are “tame”: (i) they are finitely long and are decomposed into a finite number of segments that are also finitely long, (ii) are continuously twice differentiable, (iii) each segment of the 2D curve does not have any intersection with a tangent line at every non-endpoint of the segment (see Latecki & Rosenfeld, 1998 for a further discussion), (iv) each segment of the 3D curve does not intersect with a rectifying plane at every non-endpoint of the segment (A. Michaux, personal communication, May 8, 2013). The rectifying plane is tangent to the segment and is perpendicular to a plane of curvature at the point (Hilbert & Cohn-Vossen, 1952). The XYZ Cartesian coordinate system of a 3D scene and the xy Cartesian coordinate system of a 2D image in the scene are set as follows: (i) the Z-axis of the 3D coordinate system is perpendicular to the image plane ΠI and ΠI is Z = f where f is a constant, (ii) the Z-axis passes through the origin of the 2D coordinate system, and (iii) the X- and Y-axes of the 3D coordinate system are parallel to the x- and y-axes of the 2D coordinate system, respectively. Under an orthographic projection, a 2D orthographic projection xo2D yo2D T of a point X3D Y3D Z3D T in a 3D scene is: Sawada and Zaidi Page 3 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript xo2D yo2D T = X3D Y3D T Under a perspective projection, the origin of the 3D coordinate system is at a “center of projection” F. Note that the Z-axis passes F, intersects with the image plane ΠI at the origin of the 2D coordinate system on ΠI, and is normal to ΠI. Then, the Z-axis is referred as the principal axis and the intersection is referred as the principal point of the perspective projection. If f is the focal distance of the camera, a 2D perspective projection of X3D Y3D Z3D T is xp2D yp2D T = f X3D/Z3D f Y3D/Z3D T and this relation can be written as: xp2D yp2D wp2D = f 0 0 0 f 0 0 0 1 X3D Y3D Z3D where xp2D yp2D wp2D T is called the homogeneous coordinates of xp2D yp2D T and xp2D yp2D T = xp2D /wp2D yp2D /wp2D T .
A 2D rotation can be written as: R2D σ2D = cosσ2D −sinσ2D sinσ2D cosσ2D where σ2D is an angle of the rotations. Note that the 3D XYZ Cartesian coordinate system used in this study is right-handed. Hence, rotations RX, RY, and RZ around the X-, Y-, and Z-axes can be represented by the following rotation matrices: RX σX = 1 0 0 0 cosσX −sinσX 0 sinσX cosσX , RY σY = cosσY 0 sinσY 0 1 0 −sinσY 0 cosσY , RZ σZ = cosσZ −sinσZ 0 sinσZ cosσZ 0 0 0 1 where σX, σY, and σZ are angles of the rotations.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript 3. Theorems and Proofs 3.1. 2D rotational-symmetry Rotational-symmetry can be characterized by its degree of “fold”2. Consider a 2D n-fold symmetrical shape where n ≥ 2 (Figure 3). This shape is invariant against its rotation around its symmetry point for 360i/n degree where i is an integer. The symmetry point is at the center of gravity (CoG) of the shape. A set of n points of the shape are symmetrically corresponded if a position of their j-th point after a rotation for 360i/n degree coincide with a position of their ((j + i) % n)-th point before the rotation where j is an integer and % represents the modulo operation. For a planar symmetrical figure in a 3D scene, its symmetry axis is defined as a line that is normal to the plane of the figure and that passes the symmetry point. If n = 2, the rotation angle of symmetry is 180 degrees and each pair of points symmetrically corresponded in the shape can be connected by a line-segment whose midpoint is at the symmetry point. If n > 2, n corresponding points form a regular n-sided polygon whose CoG appears at the symmetry point. Now let us call these “regular polygons” and the line-segments connecting the corresponding points “symmetry polygons”. If n = 2, the symmetry polygon is a line-segment that is an “open polygon.” The symmetry polygons of 2-fold symmetry are also called symmetry line-segments in this study. When n → ∞, the symmetry polygon becomes a circle, which is the most regular shape (Pizlo, 2008, see also Metzger, 1936/2009).
3.1.1. Skewed rotational-symmetry—Consider a planar n-fold symmetrical figure slanted relative to the observer. The image produced by slanting a planar figure is called skewed symmetry (Kanade, 1981; Kanade & Kender, 1983). Some properties of the symmetry of the figure are preserved in skewed symmetry under both orthographic and perspective projections as model-based invariants (Sawada, Li, & Pizlo, 2015; Rothwell, 1995). The human visual system detects mirror-symmetry of a planar figure and of a volumetric object based on an invariant of mirror-symmetry under an orthographic projection (Sawada & Pizlo, 2008; Sawada, 2010; Wagemans, 1995). It is possible that some invariant of rotational-symmetry could be also important for the visual system to detect rotational-symmetry rotational-symmetry could be also important for the visual system to detect rotational-symmetry We discuss model-based invariants of rotational-symmetry under the projections.
Now, consider an orthographic projection of a planar symmetrical figure to a 2D image plane. The orthographic projection is a 2D compression along the orientation of a slant σslant of the planar figure by a factor of cos(σslant). The 3D orientation of the figure can be computed from the compression of the projection of the symmetry polygon if the number of the symmetry folds of the figure is more than 2. A symmetry point of the figure is projected to the CoG of the image of the figure. Note that if the number of the symmetry folds of the figure is even, the orthographic projection is also 2-fold symmetrical (see images of 2- and 4-fold symmetrical figures in Figure 4).
2From here on, we will use “symmetry” to mean “rotational-symmetry” unless something else is specified.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript Under a perspective projection, the 3D orientation of the figure can be computed from its single symmetry polygon if the number of folds n of the figure is more than 3 (see also Van Gool, Moons, & Proesmans, 1996). The symmetry polygon is an n-sided regular polygon and mirror symmetrical with n symmetry axes. This mirror-symmetry of the symmetry polygon can be used to compute the 3D orientation of the polygon (Hong, Yang, Huang, & Ma, 2004; Yang, Huang, Rao, Hong, & Ma, 2005). Note that the symmetry point is not projected to the CoG of the image of the figure under the perspective projection. The projection of the symmetry point coincides with a projection of the CoG of a symmetry polygon of the figure. A projection of the CoG can be derived from the fact that every symmetry polygon is a regular polygon. Now consider the following four cases for finding a projection of the CoG of a symmetry polygon that depend on the number of folds n of the figure, namely, i) n > 3 and n is even, ii) n > 3 and n is odd, iii) n = 3, and iv) n = 2. First, consider the case in which n > 3 and n is even. Each symmetry polygon of the figure is a regular n- sided polygon. The CoG of the symmetry polygon can be determined by drawing auxiliary line-segments each of which connects the pair of a vertex of the symmetry polygon with the next vertex but (n − 2)/2 (see Figure 5A for n = 4). Specifically, the i-th vertex of a n-fold symmetry polygon is connected with (i + (n − 2)/2)-th vertex. These line-segments intersect with one another at the CoG of the symmetry polygon. A projection of the intersection of the line-segments is an intersection of the line-segments that are projections of these line-segments. It follows that the projection of the CoG can be derived by finding an intersection of the line-segments that connect the vertices of the projection of the symmetry polygon (Figure 5C).
If n > 3 and n is odd, the CoG of the symmetry polygon and its projection can be determined in an analogous way. The CoG of the symmetry polygon is an intersection of line-segments, each of which connects a vertex of the symmetry polygon with a midpoint of an edge between the next vertices of the symmetry polygon but (n−1)/2 and (n−3)/2. Note that the midpoint of the edge, however, is not projected to a midpoint of a projection of the edge under a perspective projection. On the other hand, an intersection of two lines is projected to an intersection of projections of the lines under both the perspective and orthographic projections. Here, instead of using the midpoints, the line-segments that pass the CoG of the symmetry polygon can be drawn by drawing additional auxiliary line-segments and using their intersections (see a case n = 5 in Figure 5A). Each of these additional auxiliary line-segments connects a vertex of the symmetry polygon with the vertex after the next vertex. Namely, the i-th vertex of a n-fold symmetry polygon is connected with the (i±2)-th vertex. With this in place, an n-pointed star appears within the symmetry polygon. Note that the symmetry polygon is a regular n-sided polygon and is 2D mirror-symmetrical with n-symmetry axes. The star is also 2D mirror-symmetrical about the symmetry axes of the symmetry polygon because the auxiliary line-segments connects vertices of the symmetry polygon in a symmetrical manner (the i-th vertex of the symmetry polygon is connected with the (i±2)-th vertex). Each symmetry axis connects the i-th vertex of the star with the (i+n)-th vertex of the star. The symmetry axes intersect at the CoG of the star, which is at the CoG of the symmetry polygon. It is because a symmetry axis of any 2D mirror-symmetrical polygon passes the CoG of this polygon. Recall that an intersection of two lines is projected to an intersection of projections of the lines under both the perspective and orthographic Sawada and Zaidi Page 6 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript projections. The projection of the CoG can be derived by drawing auxiliary line-segments in the same way in the projection of the symmetry polygon (see cases n = 5 in Figure 5B and C). These methods using the auxiliary line-segments for finding the image of the center of gravity for n > 3 can also be applied to an orthographic projection.
If n = 3, the projection of the CoG of the individual symmetry polygon cannot be determined uniquely. Now, assume that the center of the perspective projection from the symmetrical figure to the 2D image is given (calibrated camera, see Li, Sawada, Latecki, Steinman, & Pizlo, 2012). The shape of the 3D symmetry polygon is also known (a regular triangle), and that its 2D projection is a triangle, except when it is presented in a degenerate view. It is often impossible to determine, uniquely, the 3D orientation of a symmetry polygon from a 2D triangle of its projection (Fischler & Bolles, 1981; Gao, Hou, Thang, & Cheng, 2003; Minkov & Sawada, 2018). There can be up to four possible 3D orientations of a symmetry polygon for its 2D triangle and, when there are, its CoG is projected to different positions within the 2D image, depending on the symmetry polygon’s orientation (see Figure 6). If the planar figure has multiple symmetry polygons, their CoGs must coincide with one another and a unique CoG can be determined. Then, the 3D orientation of the figure can be uniquely determined.
If n = 2, the symmetry polygon is a line-segment (symmetry line-segment) and its symmetry point is at the midpoint of the line-segment. The perspective projection of the midpoint, however, is not a midpoint of the perspective projection of the symmetry line-segment. If a planar figure with 2-fold symmetry has multiple symmetry line-segments, these line-segments intersect with one another at their midpoints. Otherwise, a vanishing point vi3 of the symmetry line-segment is required to derive a perspective projection mi of its midpoint Mi. Note that vi and mi are collinear with the projection of the line-segment and their relationship can be written as (see supplemental material): rmi = 2rφirψi rφi + rψi (1) where rmi, rφi, and rψi are distances from vi to mi and two endpoints of the projection of the line-segment. If the planar figure has multiple symmetry line-segments, they intersect at their common CoG. Then, their vanishing points can be computed from Equation (1) and the 3D orientation of the figure can be uniquely determined.
3.2. 3D rotational-symmetry A 3D symmetrical object has a symmetry axis that is normal to the symmetry polygons of the 3D object and passes their CoGs (Figure 7). When n → ∞, the symmetry polygon becomes a circle and the whole symmetrical object becomes a surface-of-revolution (SoR). 3Consider a perspective projection of a line segment in a 3D scene to an image plane. The vanishing point of the segment can be uniquely determined so that the vanishing point and the center of projection can be connected by a line that is parallel to the segment. Any line parallel to the segment is projected to a pair of collinear half-lines that emanate from the vanishing point unless the line passes the center of projection.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript Note that a single 3D object can have multiple symmetry axes. For example, a cube has three axes for a 4-fold symmetry, four axes for a 3-fold symmetry, and six axes for a 2-fold symmetry.
A 2D image of a 3D n-fold symmetrical object becomes 2D n-fold symmetrical only from an accidental view-point. The image is n-fold symmetrical when the symmetry axis is normal to a plane of the image under an orthographic projection and when the symmetry axis coincides with the principal axis under a perspective projection4. If it does not, n-fold symmetry is not present in the image.
3.2.1. A 3D symmetrical object and its 2D image 3.2.1.1. Properties of a 2D image of a 3D symmetrical object: A 3D symmetrical object has multiple symmetry polygons whose CoGs are collinear along its symmetry axis. This collinearity is a model-based invariant under both orthographic and perspective projections. Projections of the CoGs are also collinear with a projection of the symmetry axis in a 2D image of the symmetrical object. It follows that the projection of the symmetry axis can be derived if the projections of the CoGs of the multiple symmetry polygons can be detected in the image (see 3.1.1.). Detecting the symmetry axis from the image allows the visual system to see symmetry of the whole object rather than symmetry of its individual symmetry polygons. The individual symmetry polygons are formed by local features of the object and the whole object is composed of the local features so that their symmetry polygons share the common symmetry axis. If there is no common symmetry axis among the symmetry polygons, the whole object cannot be symmetrical but can have some symmetrical parts.
Consider a perspective projection of the 3D symmetrical object to an image plane. The vanishing point of the symmetry axis in the image plane can be determined so that a line connecting the vanishing point and the center of projection is parallel to the symmetry axis. The symmetry polygons of the 3D object are planar and perpendicular to the symmetry axis. It follows that if n > 2, the symmetry polygons are parallel to one another and their horizons coincide with a single line in the 2D image. A plane connecting the horizon and the center of projection is parallel to the symmetry polygons (see 3.1.1.) and this plane is perpendicular to the line connecting the vanishing point and the center of projection (Figure 8).
The vanishing point of the symmetry axis and the horizon of the symmetry polygons can be determined from the 2D perspective image of the 3D symmetrical object. If n > 3, each symmetry polygon and its auxiliary lines always form two or more than two sets of parallel line segments. The parallel line segments in each set are projected to line segments converging at their vanishing point that is on the horizon of the symmetry polygon. Hence, the horizon can be determined as a line passing the vanishing points of the sets of parallel line segments. If n = 3, a symmetry polygon is a regular triangle and its 3D orientation cannot be always determined uniquely from the projection of the symmetry polygon. The projection is consistent with up to four 3D orientations of the symmetry polygon. The 4With a reduced eye, whose center of projection is at a center of its spherical retina, the retinal image of a 3D n-fold symmetrical object is 2D n-fold symmetrical if the symmetry axis passes through the center of projection.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript orientation can be uniquely determined if there is another symmetry polygon. Their orientations are determined so that they are parallel to one another in a 3D scene.
If n = 2, the symmetry polygons are line-segments (symmetry line-segments). Their vanishing points appear on a horizon of a plane that is perpendicular to the symmetry axis. The symmetry line-segments are perpendicular to a normal to a plane connecting the horizon and the center of projection. A perspective projection of a symmetry line-segment alone is not enough to determine the projection of its CoG. Determining this requires having a vanishing point of the symmetry line-segment. Vanishing points of symmetry line-segments of a single 3D symmetrical object can be derived if the number of the symmetry line-segments is three, or more than three (see also 3.1.1.). In the 2D perspective image of the 3D object, vanishing points of the symmetry line segments are collinear on a line (Figure 9). This line is the horizon of a plane to which the symmetry line-segments are parallel and the symmetry axis is normal. The plane is normal to a line connecting the vanishing point of the symmetry axis and the center of projection. The vanishing point of the symmetry axis can be found by using an optimization process. The space for this optimization process is two dimensional, specifically the 2D position of the vanishing point of the symmetry axis in the image. For a given position of the vanishing point, projections of the midpoints of the symmetry line-segments can be derived. Note that the midpoints of the symmetry line-segments are collinear on the symmetry axis and their projections are also collinear. Then, validity of the given position can be evaluated based on the collinearity of the projections of the midpoints. The projections of the midpoints are derived in the following steps. First, the horizon of the symmetry line-segments is determined from the given position of the vanishing point of the symmetry axis. Then, the vanishing points of the symmetry line-segments are found at intersections of the horizon with lines of the projections of the symmetry line-segments. From the vanishing points of the symmetry line-segments, the projections of their midpoints can be derived (Equation 1).
3.2.2. Recovering a 3D rotational-symmetrical shape from a single 2D image —The 3D shape of a 3D mirror-symmetrical object can be recovered from its single 2D image uniquely under a perspective projection (Gordon, 1990; Rothwell, 1995; Hong, Yang, Huang, & Ma, 2004; Yang, Huang, Rao, Hong, & Ma, 2005) and up to a one unknown parameter under an orthographic projection (Vetter & Poggio, 1994). This geometrical property has been used for modeling veridical perception of the 3D shape of the mirror-symmetrical object (Pizlo, 2008; Pizlo et al., 2010, 2014).
The 3D shape of a 3D rotational-symmetrical object can also be recovered from its single 2D image by using two different methods. The first method is based on Multiple-view geometry (Hartley & Zisserman, 2004). It uses a “virtual image” of the 3D symmetrical object (Vetter & Poggio, 1994). Note that the shape of the object is invariant against a rotation Raxis around the symmetry axis of the object for 360i/n degree, where n is the number of symmetry folds of the object, and i is an arbitrary integer less than n. With this in place, the image of the object is unchanged if the viewpoint of the image is rotated around the symmetry axis for −360i/n degree (Raxis −1). This means that the single 2D image of the n-fold 3D symmetrical object is equivalent to n images of the same object seen from different viewpoints around the symmetry axis of the object. Those additional images are Sawada and Zaidi Page 9 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript called “virtual images” (Vetter & Poggio, 1994) and the 3D shape of the object can be recovered from the original and virtual images by using Multiple-view geometry (Vetter & Poggio, 1994; Hong, Yang, Huang, & Ma, 2004; Yang, Huang, Rao, Hong, & Ma, 2005).
The second method uses the properties of the image of a 3D symmetrical object (see 3.1.1. and 3.2.1.1.). Under an orthographic projection, we assume that projections of the symmetry axis, symmetry polygons, and their CoGs are given. Under a perspective projection, the vanishing point of the symmetry axis is also given. This recovery method can connect a process for detecting 3D symmetry of the object based on its image properties (see 3.2.1.; 3.2.4, see also 3.2.3.) with the recovery process of its 3D shape. We will now show how the 3D shape of a symmetrical object can be recovered from its 2D image under both orthographic and perspective projections.
3.2.2.1. 3D recovery under a 2D orthographic projection: Consider the recovery of the 3D shape of a symmetrical object from its 2D orthographic image. Projections of the symmetry axis and the symmetry polygons of the object are assumed to be given in the image (see 3.1.1. and 3.2.1.1.). Set the 2D and 3D Cartesian coordinate systems so that the x- and X-axes coincide with the projection of the symmetry axis. When this is done, the symmetry axis should be on the ZX-plane of the 3D coordinate system: Xcosθaxis −Zsinθaxis + daxis = 0, Y = 0 (2) where θaxis is an angle between the symmetry axis and the Z-axis (a normal to ΠI) and daxis is a constant. The constant daxis can be arbitrary and it determines the depth position of a 3D shape recovered in the following process. The symmetry axis (2) is normal to a plane: Xsinθaxis + Zcosθaxis − Xx sinθaxis − daxis tanθaxis = 0 (3) where Xx is an arbitrary real number and the plane (3) intersects with the symmetry axis (2) at: Xx 0 Xx tanθaxis + daxis sinθaxis T (4) Consider n = 2. The symmetry polygons are line-segments (symmetry line-segments) and their CoGs (midpoints) project to midpoints of the projections of the symmetry line-segments. The projections of the midpoints are collinear along the projection of the symmetry axis. Then, projections of vertices of a symmetry line-segment i can be written as: xmi ± xdi/2 ± yi T (5) Sawada and Zaidi Page 10 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript where double-sign corresponds and the midpoint of the projection of the symmetry line-segment i is [xmi 0]T. Note that the midpoint of every symmetry line-segment is on the symmetry axis. From equation (2), the midpoint of the symmetry line-segment i is: XMi YMi ZMi T = xmi 0 xmi tanθaxis + daxis sinθaxis T (6) Since every symmetry line-segment is perpendicular to the symmetry axis, the symmetry line segment i is on the plane (3) when Xx = xmi. Then, the two vertices of the symmetry line-segment i can be recovered as: xmi ± xdi 2 ± yi xmi tanθaxis + daxis sinθaxis ∓ xdi 2 tanθaxis T (7) where double-signs correspond to one another. Note that θaxis is a free parameter that changes the aspect ratio of the recovered 3D shape. From equations (6) and (7), the height of the recovered shape along the symmetry axis changes as a function of 1/tanθaxis and the width of the shape changes as a function of tanθaxis along a line that is perpendicular to the symmetry axis and is on the XZ- plane.
If n > 2, an equation similar to equation (7) can be applied to the 2D projections of the symmetry polygons for recovering the 3D rotational-symmetrical shape: xgi + xpij ypij xgi tanθaxis + daxis sinθaxis −xpijtanθaxis T (8) where [xgi 0]T is the projection of the center of gravity of the symmetry polygon i and [xgi +xpij ypij]T is the projection of the j-th vertex of i. If n > 2, the CoGs of the symmetry polygons are used instead of the midpoints of the symmetry line-segments. Note that θaxis is no longer a free parameter. Recall that all of the symmetry polygons of the 3D rotational symmetrical shape are regular n-sided polygons. There exists a unique θaxis that makes all of the recovered symmetry polygons of the 3D shape regular for a given 2D image of a 3D rotational-symmetrical shape.
3.2.2.2. 3D recovery under a 2D perspective projection: With a perspective projection, it is not necessary to distinguish the n = 2 and n > 2 conditions. Now consider the recovery of the 3D shape of a symmetrical object from its 2D perspective image. Projections of the symmetry axis and symmetry polygons of the object are assumed to be given in the image (see 3.1.1. and 3.2.1.1.). Also assume that the vanishing point of the symmetry axis and the projections of the CoGs of the symmetry polygons are given. Now, let the projection of the symmetry axis be laxis, and the vanishing point of the symmetry axis be vaxis. Set the orientation of the 2D xy Cartesian coordinate system so that vaxis is on the x-axis, vaxis = Sawada and Zaidi Page 11 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript [xaxis 0]T. Let the j-th vertex of the i-th symmetry polygon be Pij = [Xij Yij Zij]T, a projection of Pij be pij = [xij yij] = [fXij/Zij fYij/Zij]T, its CoG be Gi, and a projection of Gi be gi = [xgi ygi]T. The symmetry axis is parallel to a line connecting the center of projection F = [0 0 0]T and vaxis = [xaxis 0 f]T. Note that Gi is on the symmetry axis and gi is on laxis. Once this is established, the projection of the CoG Gi of the i-th symmetry polygon can be written as gi = [xgi ygi]T = vaxis + rgi[cosαaxis sinαaxis]T, where rgi is the distance from vaxis to gi and αaxis is an angle between laxis and the x-axis.
Now, consider making the symmetry axis normal to ΠI and all of the symmetry polygons frontoparallel by rotating the camera. This rotation is around the center of projection F and the image plane ΠI and its 2D coordinate system are rotated together with the camera. Let the camera’s rotation be Rv. Note that the symmetry axis is parallel to a line connecting F and vaxis, and vaxis is on the x-axis. It follows that the orientation of the symmetry axis relative to the Z-axis is σv = atan(xaxis/f). The symmetry axis becomes normal to ΠI by rotating the camera around the Y-axis for σv (Figure 10): Rv = RY(σv). The 2D image after Rv can be computed directly by transforming the 2D image before Rv (Kanatani, 1988): x′ = f x cos σv −f sin σv x sin σv + f cos σv y′ = f y x sin σv + f cos σv (9) where [x y]T is a projection of a point in the 3D scene to ΠI before Rv and [xʹ yʹ]T is its projection after Rv. Equation (9) can also be represented as follows: x′ y′ w′ = f 0 0 0 f 0 0 0 1 RY T σv x y f (10) where x y f T and x′ y′ w′ T are the homogeneous coordinates of [x y]T and [xʹ yʹ]T. This equation, which represents a rotation of the camera by Rv (= RY(σv)), is equivalent to rotating the 3D scene by Rv T. After this transformation, the vanishing point of the symmetry axis is located at the origin [0 0]T, and the projection gi of the center of gravity Gi is transformed to: gi ′ = xgi ′ ygi ′ = rgif sin σv rgi cos αaxis sin 2σv −xaxis cos αaxis cos σv sin αaxis (11) The projections of the centers of gravity of the symmetry polygons are still collinear after the transformation along a half-line lʹaxis that is the transformation of laxis. The half-line lʹaxis emanates from the origin and the angle of lʹaxis, which is measured relative to the direction of the x-axis, is α′axis = tan−1(tanαaxis/cosσv).
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript The symmetry axis becomes perpendicular to the image plane ΠI after the camera is rotated This occurs because a line connecting the center of projection and the vanishing point ([0 0]T) in the image is perpendicular to ΠI after the camera is rotated and it is parallel to the symmetry axis. Based on this fact, the center of gravity Gʹi of the symmetry polygon i after the rotation, can be recovered as: Daxis cos αaxis ′ Daxis sin αaxis ′ f Daxis ri ′ T (12) where Daxis is a free parameter and r′gi = g′i . The symmetry polygons are parallel to the image plane ΠI after the rotation Rv and their Z-coordinates are equal to those of their individual CoGs. At this point, the j-th vertex of the i-th symmetry polygon P′ij = X′ij Y′ij Z′ij T after the rotation can be recovered from its perspective projection P′ij = X′ij Y′ij T as: Pij ′ = Xij ′ Yij ′ Zij ′ T = Daxis ri ′ xij ′ yij ′ f T (13) The free parameter Daxis determines the size of the recovered 3D shape and the distance between the principal axis (the Z-axis) and the symmetry axis. Recall that the camera’s rotation Rv is equivalent to the rigid rotation Rv T of the 3D scene, so the vertex Pij before the camera’s rotation Rv, can be derived by applying the rotation Rv = RvTT to P′ij:Pij = RvP′ij. It is worth pointing out that the perspective projections of the symmetry polygons to ΠI after the camera’s rotation Rv are regular n- sided polygons because the symmetry polygons are regular and they are frontoparallel after the rotation.
3.2.3. Any pair of 2D curves is consistent with a 3D rotational-symmetrical interpretation—3D symmetry of an object has to be detected first from its 2D image to recover a 3D shape of the object using its symmetry. However, the symmetry detection is, at least, very difficult. Consider 3D mirror-symmetry. It is almost always possible to find a 3D mirror-symmetrical interpretation of any arbitrary image. Specifically, for a given pair of arbitrary curves in a 2D image, there is always a 3D mirror-symmetrical pair of curves that projects to the given curves under quite general assumptions (Sawada, Li, & Pizlo, 2011, 2014; Hong, Ma, & Yu, 2004). We proved that this is also true for 2-fold, 3D rotational-symmetry. Specifically, there exists a 2-fold, 3D rotational-symmetrical interpretation of a pair of arbitrary curves in a 2D image as well under some general assumptions.
The gist of this proof is as follows: when a pair of curves in a 2D image is given, a set of pairs of lines for establishing correspondence between these curves is the first thing that is determined. The corresponding pairs of points are determined uniquely as the intersections of these lines with the curves. Under an orthographic projection, it is always possible to find Sawada and Zaidi Page 13 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript a one-parameter family of its 2-fold 3D symmetrical interpretation around a common symmetry axis for any corresponding pair of 2D points. The family is controlled by the angle between the symmetry axis and a normal to the image plane. Next, under a perspective projection, it is always possible to find its unique 3D symmetrical interpretation around a common symmetry axis for any corresponding pair of 2D points. We will consider a special perspective projection before we generalize it to a general perspective projection.
3.2.3.1. A 3D symmetrical interpretation under a 2D orthographic projection Theorem-A1.: Let φ and ψ be curves in a 2D image. Assume that four lines lφ1, lφ2, lψ1, and lψ2 that satisfy the following properties can be drawn in the image: (i) lφ1||lφ2||lψ1||lψ2, (ii) lφ1 and lφ2, do not intersect with φ but do share points with φ individually (by being tangent to φ or passing the endpoints or non-differentiable points of φ), (iii) lψ1 and lψ2 do not intersect with ψ but do share points with ψ individually, and (iv) a distance between lφ1 and lφ2 is equal to that between lψ1 and lψ2 (see Figure 11). Then, there exists a one parameter family of a pair of curves Φ and Ψ in 3D space such that Φ and Ψ are 2-fold rotationally-symmetrical with a symmetry axis As and that φ is an orthographic projection of Φ, and ψ is an orthographic projection of Ψ.
Proof:5: In order to prove this theorem, we must show how the correspondence between φ and ψ is established, and how a corresponding pair of points on φ and ψ can be back-projected in 3D space, such that these back-projected points are symmetrical with respect to the same symmetry axis As. Put simply, the line-segment connecting the back-projected points is bisected by As and is perpendicular to As.
The 2D xy Cartesian coordinate system on the image plane ΠI is set so that the x-axis is parallel to the lines lφ1, lφ2, lψ1, and lψ2 and it is coincident with their midline. The lines lφ1, lφ2, lψ1, and lψ2 can be written as y = ylφ1, y = ylφ2, y = ylψ1, and y = ylψ2, respectively. Without loss of generality, assume that ylφ1 > ylφ2 and ylψ1 < ylψ2. Then, ylφ1−ylφ2 = ylψ2−ylψ1, 0 = ylφ1+ylψ1, and 0 = ylφ2+ylψ2. The y- coordinate of any point on φ is between ylφ1 and ylφ2 and that of any point on ψ is between ylψ1 (=−ylφ1) and ylψ2 (= −ylψ1).
Consider a point pi = [xφi yφi]T on φ where ylφ2 ≤ yφi ≤ ylφ1. This point should correspond with the point qi = [xψi yψi]T that is an intersection of ψ with a line y = −yφi where ylψ2 ≥ −yφi ≥ ylψ1. When this is done, their midpoint mi is on the x-axis: mi = [xφi/2+xψi/2 0]T. Note that under the orthographic projection, a 3D point [X Y Z]T projects to a point [X Y]T in the 2D image. It follows that Pi = [XΦi YΦi ZΦi]T = [xφi yφi ZΦi]T, and Qi = [XΨi YΨi ZΨi]T = [xψi −yφi ZΨi]T project to pi and qi, individually.
Recall that Pi and Qi are 3D symmetrical with respect to As, if and only if, they satisfy the following two requirements: i) the line segment connecting Pi and Qi intersects with As at a midpoint Mi of the segment, and ii) is perpendicular to As. A midpoint Mi between Pi and Qi is an invariant of the orthographic projection and projects to the midpoint mi between pi and qi. The Y-coordinate of Mi is 0 because that of mi is also 0. Hence, As is on the ZX-plane of the 3D coordinate system and can be written as: 5A special case of Theorem-A1 that the 2D curves φ and ψ form a closed 2D curve was proved in Sugihara (2016).
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript X cos θaxis −Z sin θaxis + daxis = 0, Y = 0 (14) where θaxis is an angle between the symmetry axis As and the Z-axis and daxis is a constant. The line segment connecting Pi and Qi satisfies the requirements i) and ii) of 3D symmetry if: sin θaxis 0 cos θaxis Pi −Qi = 0 (15) cos θaxis 0 −sin θaxis Pi + Qi 2 = −daxis (16) From (15) and (16), we have: Pi = xϕi yϕi xϕi + xψi cos 2θaxis sin 2θaxis + daxis sin θaxis T (17) Qi = xψi −yϕi xψi + xϕi cos 2θaxis sin 2θaxis + daxis sin θaxis T (18) Note that θaxis is a free parameter; it can be arbitrary, except for sin2θaxis = 0. So, the 3D interpretations of the 2D curves φ and ψ form a one-parameter family characterized by θaxis. The constant daxis determines the depth positions of the 3D interpretations but does not affect their shapes. Equations (17) and (18) imply that the one-parameter family of the 3D symmetrical interpretations Φ and Ψ of the 2D curves φ and ψ always exist.
QED In the proof of Theorem-A1 above, it was assumed that correspondences between the points of the 2D curves φ and ψ are unique. The case with non-unique correspondences was not considered. Even if the correspondences are not unique, the 3D symmetrical interpretations Φ and Ψ of the 2D curves φ and ψ always exist, and they are a pair of continuous curves (Figure 12). Each corresponding pair of points on φ and ψ under the orthographic projection is always established by a pair of lines that are parallel to the x-axis, and equally distant from the x-axis. Consider reflecting ψ about the x-axis. Then, the lines lφ1 and lφ2 that are parallel to the x-axis share points with both φ and the 180° rotation of ψ (ψ−1) but they do not intersect with them (see condition (ii) of Theorem-A1). Then, the corresponding points of φ and ψ−1 can be connected by a single line parallel to the x-axis (and lφ1 and lφ2, Figure 13). The correspondence is unique only if the line intersects only once with φ and ψ−1 Sawada and Zaidi Page 15 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript individually (Figure 13A). This is equivalent to using a way to establish the correspondence between a pair of 2D curves for their 3D mirror-symmetrical interpretations (Sawada, Li, & Pizlo, 2011; see also Rothwell, 1995; Hong, Ma, & Yu, 2004 for a perspective projection). Even if the correspondence between the 2D curves is not always unique, it has been formally proved that the correspondence can be established so that the 3D mirror-symmetrical interpretations are a pair of continuous curves (Theorem 3 in Sawada et al., 2011). The same method can be applied here to establish the correspondence between φ and ψ−1 under the orthographic projection so that Φ and Ψ are a pair of continuous curves.
Assuming that the symmetry axis of a 3D interpretation is perpendicular to the image plane under a perspective projection, the correspondence between a pair of 2D curves is established by a pair of lines that are parallel to the x-axis as well as under the orthographic projection. This special case in a perspective projection, will be discussed in the next section, and the general perspective projection will be considered on the basis of the special case in the following section.
3.2.3.2. A 3D symmetrical interpretation under a 2D special perspective projection Lemma-for-Theorem-A2.: Let φ and ψ be curves in a 2D image. Assume that four lines lφ1, lφ2, lψ1, and lψ2 that satisfy the following properties can be drawn in the image: (i) lφ1||lφ2|| lψ1||lψ2, (ii) lφ1 and lφ2, do not intersect with φ, but do share points with φ individually (by being tangent to φ or passing the endpoints or non-differentiable points of φ), (iii) lψ1 and lψ2 do not intersect with ψ, but do share points with ψ individually, (iv) the midline between lφ1 and lψ1 coincide with the midline between lφ2 and lψ2, and (v) the principal point is on the midline (see Figure 14). Then, for a given center of projection F there exists a pair of curves Φ and Ψ in 3D space such that Φ and Ψ are 2-fold rotationally-symmetrical with respect to a symmetry axis As which is normal to the image plane, and that φ is a perspective projection of Φ and ψ is a perspective projection of Ψ.
Proof: Set the x-axis of the 2D coordinate system of the image plane ΠI to be parallel to lφ1, lφ2, lψ1, and lψ2 and the X-axis of the 3D coordinate system to be parallel to the x-axis. When this is done, the x-axis coincides with the midline between lφ1 and lψ1, and between lφ2, and lψ2 because of condition (v) in the Lemma-for-Theorem-A2. The lines lφ1, lφ2, lψ1, and lψ2 can be written as y = ylφ1, y = ylφ2, y = ylψ1, and y = ylψ2, respectively. Note that ylφ1−ylφ2 = ylψ2−ylψ1, 0 = ylφ1+ylψ1, and 0 = ylφ2+ylψ2. Then, the y-coordinate of any point on φ is between ylφ1 and ylφ2, and the y-coordinate of any point on ψ is between ylψ1 and ylψ2.
Now consider a point pi = [xφi yφi]T on φ where ylφ2 ≤ yφi ≤ ylφ1. This point should correspond with a point qi = [xψi yψi]T that is an intersection of ψ with a line y = −yφi where ylψ2 ≥ −yφi ≥ ylψ1. Then, the midpoint mi of pi and qi is on the x-axis: mi = [(xφi+xψi)/2 0]T. Under a perspective projection, a 3D point [X Y Z]T projects to an image point [fX/Z fY/ Z]T, so Pi = [XΦi YΦi ZΦi]T = [xφiZΦi/f yφiZΦi/f ZΦi]T and Qi = [XΨi YΨi ZΨi]T = [xψiZΨi/f −yφiZΨi/f ZΨi]T project to pi and qi, individually.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript Recall that Pi and Qi are 3D symmetrical with respect to the symmetry axis As, if and only if, they satisfy the following two requirements: (i) the line-segment connecting Pi and Qi intersects with As at a midpoint Mi of the segment, and (ii) it is perpendicular to As. The line-segment connecting Pi and Qi is also perpendicular to the normal of the image plane ΠI because As is parallel to the normal of ΠI. Under this condition, a midpoint Mi between Pi and Qi projects to the midpoint mi between pi and qi. The Y-coordinate of Mi is 0 because the y-coordinate of mi is also 0. So, As can be written as: X = Xaxis′ Y = 0 (19) where Xaxis represents the X-coordinate of an intersection of As with the x-axis. Since Mi is on As: Mi = Xaxis 0 2f Xaxis xφi + xψi T (20) The Z-coordinate of Mi is the same as those of Pi and Qi because the line-segment connecting Pi and Qi is perpendicular to a normal of the image plane ΠI. From (20), we have: Pi = 2xφiXaxis xφi + xψi 2yφiXaxis xφi + xψi 2f Xaxis xφi + xψi T (21) Qi = 2xφiXaxis xφi + xψi −2yφiXaxis xφi + xψi 2f Xaxis xφi + xψi T (22) Equations (21) and (22) imply that the 3D rotationally-symmetrical interpretation Φ and Ψ of the 2D curves φ and ψ always exists. They diverge to infinity if xφi+xψi = 0 and are not tame (see 2. Definition). Note that Xaxis changes the size of Φ and Ψ but does not affect their shapes.
QED Based on the Lemma-for-Theorem-A2, we will consider establishing symmetry correspondence in a 2D image under a general perspective projection. Recall that the correspondence between a pair of 2D curves was established for the lemma by a pair of parallel lines whose midline passes the principal point. Under the general perspective projection, correspondence is established by a pair of half-lines that emanate from a point in an image.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript 3.2.3.3. A 3D symmetrical interpretation under a 2D general perspective projection Theorem-A2.: Let a center of projection F be [0 0 0]T , a 2D image plane ΠI be Z = f, and φ and ψ be curves in ΠI. Assume that four half-lines lφ1, lφ2, lψ1, and lψ2 , which satisfy the following three properties, can be drawn in the image: (i) endpoints of all of the half-lines are at a point vc, that is, not on φ and ψ, (ii) lφ1 and lφ2 do not intersect with φ, but do share points with φ individually (by being tangent to φ or passing the endpoints or non-differentiable points of φ), (iii) lψ1 , and lψ2 do not intersect with ψ but do share points with ψ , individually (see Figure 15). Now let the 2D xy Cartesian coordinate system be set so that the origin is at the principal point in the image and vc is on the x-axis: vc = [xc 0]T. Let a line hc be x = −f2/xc and its intersection with lφ1, lφ2, lψ1, and lψ2 be uφ1, uφ2, uψ1, and uψ2 respectively. Then, additionally assume (iv) xφψ/xc < 1 where xφψ is the x coordinate of any point on φ and ψ and (v) an angle bisector of ∠Uφ1FUψ1 coincides with that of ∠Uφ2FUψ2 where Uφ1= [uφ1 f]T, Uψ1 = [uψ1 f]T, Uφ2 = [uφ2 f]T, and Uψ2 = [uψ2 f]T. With this done, there exists a pair of curves Φ and Ψ in a 3D space such that Φ and Ψ are 2-fold rotationally-symmetrical with respect to a symmetry axis As and φ is a perspective projection of Φ, and ψ is a perspective projection of Ψ.
Proof: In order to prove this theorem, φ and ψ are transformed to ‶φ and x‶ψ by simulating a camera rotation Rc (Kanatani, 1988) around the center of projection F so that ‶φ and ‶ψ satisfy the conditions of the Lemma-for-Theorem-A2. Condition (iv) of Theorem-A2 should be satisfied so that ‶ and of ‶ψ are a pair of tame curves (see 2. Definition). Conditions (iv) and (v) of the Lemma-for- Theorem-A2 is satisfied for ‶φ and of ‶ψ is satisfied if Condition (v) of Theorem-A2 is satisfied. Based on the Lemma-for-Theorem-A2, the 3D symmetrical interpretation ‶Φ and ‶Ψ of ‶φ and of ‶ψ can be constructed. Then, the 3D symmetrical interpretation Φ and Ψ of φ and ψ is generated by rotating Φ‶ and ‶Ψ by Rc.
Set the x-axis of the 2D coordinate system of the image plane ΠI to pass the point vc and the X-axis of the 3D coordinate system to be parallel to the x-axis (Figure 15). Then, any point in ΠI can be represented in a polar coordinate system and can be written as [x y]T = [xc +rcosα rsinα]T, where r is a length of a line-segment between [x y]T and vc and α is an angle of the segment relative to the direction of the x-axis. Any point on a half-line that emanates from vc can be represented with a constant α. For example, α is tan−1(−yuφ1/xc) for lφ1, tan−1(−yuφ2/xc) for lφ2, tan−1(−yuψ1/xc) for lψ1, and tan−1(−yuψ2/xc) for lψ2 where yuφ1, yuφ2, yuψ1, and yuψ2 are y-coordinates of uφ1, u2, uψ1, and uψ2. Then, α is between tan−1(−yuφ1/xc) and tan−1(−yuφ2/xc) inclusive for any point on φ and between tan −1(−yuψ1/xc) and tan–1(−yuψ2/xc) inclusive for that on ψ.
A 3D rotationally-symmetrical interpretation Φ and Ψ of the 2D curves φ and ψ will be constructed so that the vanishing point vaxis of their symmetry axis is on hc and vc is on the horizon haxis of the symmetry axis. The common angle bisector of ∠Uφ1FUψ1 and ∠Uφ2FUψ2 intersects with the image plane ΠI at the position of vaxis.
Consider rotating the camera with the image plane ΠI around the center of projection F to transform φ, ψ, lφ1, lφ2, lψ1, and lψ2 so that their transformations satisfy the conditions of Sawada and Zaidi Page 18 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript the Lemma for-Theorem-A2.6 The image after the rotation can be computed from the image before the rotation as long as the position of the center of projection F is kept constant in the scene (Kanatani, 1988). The rotation Rc is done in two steps: the first rotation RcY around the Y-axis and the second rotation RcX around the X-axis (Figure 16). The first rotation RcY is RY(σcY) where σcY = tan−1(−f/xc). Then, [x y]T = [xc+rcosα rsinα]T before RcY is transformed to (Kanatani, 1988):x y = f x cos σcY −f sin σcY f cos σcY + x sin σcY f y f cos σcY + x sin σcY = −xc − xc 2 + f 2 r cos α −xc tanα cos σcY (23) after RcY. Then, uφ1, uφ2, uψ1, and uψ2 on hc (x = −f2/xc) are transformed to uφ1 = [0 yuφ1cosσcY]T,u2 = [0 yuφ2cosσcY]T, uψ1 = [0 yuψ1cosσcY]T, and uψ2 = [0 yuψ2cosσcY]T on the y-axis. Recall that the angle bisector of ∠Uφ1FUψ1 coincides with that of ∠Uφ2FUψ2 (condition (v) of Theorem-A2) before RcY. Let an intersection of the bisector with ΠI be U0 = [−f2/xc y0, f]T:∠U0FUφ1 =−∠U0FUψ1, and ∠U0FUφ2 = −∠ U0FUψ2. After RcY, u0 = [−f2/xc y0]T is transformed to u0 = [0 y0cosσcY]T. The second rotation RcX around the X-axis is determined so that u0 is transformed tou0 = [0 0]T after RcX: RcX = RX(σcX) where σcX = tan−1(−y0cosσcY/f). Then, a point [x y]T = [xv+rcosα rsinα]T before RcYRcX (= Rc)7 is projected to (Kanatani, 1988): xy = f x f cos σcX −y sin σcX f f sin σcX + y sin σcX f cos σcX −y sin σcX − f 2 y0xctanα −f 2 x cos σcY + xc 2 + f 2 r cos α cos σcX xctanα + y0 cos 2σcY cos σcY (24) after RcYRcX.
Let ‶φ and ‶ψ be transformations of φ, ψ and ‶lφ1, ‶lφ2, ‶lψ1, and ‶lψ2 be transformations of lφ1, lφ2, lψ1, and lψ2 after the camera rotation Rc (= RcYRcX). All the five conditions of Lemma-for-Theorem-A2 are satisfied by curves ‶φ, and ‶ψ and lines ‶lφ1, ‶lφ2, ‶lψ1, and ‶lψ2. From equation (24), ‶y is dependent on α but is independent from r. Then, ‶lφ1, ‶lφ2, ‶lψ1, and ‶l ψ2 are lines parallel to the x-axis (the condition (i) of the lemma) because lφ1, lφ2, lψ1, and lψ2 are represented individually by constant α. Next, from the conditions (ii) and (iii) of Theorem-A2, lφ1 and lφ2 do not intersect with φ, but do share points with φ individually and (iii) lψ1 and lψ2 do not intersect with ψ but do share points with ψ individually. These properties are invariant under the transformation that simulates the camera rotation Rc. Therefore, the conditions (ii) and (iii) of the Lemma are also satisfied.
7A rotation matrix combining the first rotation RcY and the second rotation RcX is written as RcYRcX because the 3D coordinate system rotates when the camera rotates.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript Now, consider a midline between ‶lφ1, and ‶l ψ1 and that between ‶l φ2, and ‶lψ2. These midlines are parallel to the x-axis because ‶lφ1, ‶lφ2, ‶lψ1, and l‶ψ2 are. The y-intercepts of ‶lφ1, ‶lφ2, ‶lψ1, and ‶l ψ2 are ‶uφ1 = [0 ftan∠U0FUφ1]T, ‶u2 = [0 ftan∠U0FUφ2]T, ‶uψ1 = [0 ftan∠U0FUψ1]T, and ‶uψ2 = [0ftan∠U0FUψ2]T, which are transformations of uφ1, uφ2, uψ1, and uψ2 after Rc via uφ1,u2, uψ1, and uψ2.Note that ftan∠U0FUφ1 = − ftan∠U0FUψ1, and ftan∠U0FUφ2 = − ftan∠U0FUψ2 because ∠U0FUφ1 =−∠U0FUψ1, and ∠U0FUφ2 = −∠U0FUψ2 (see Supplemental Materials). With this done, the midline between ‶lφ1, and ‶lψ1 and that between ‶lφ2, and ‶l ψ2 coincide with the x-axis. It follows that the conditions (iv) and (v) of the Lemma are also satisfied.
All the conditions of the Lemma-for-Theorem-A2 are satisfied by curves ‶φ, and ‶ψ with lines ‶l φ1, ‶lφ2, ‶lψ1, and ‶l ψ2. Therefore, from the Lemma-for-Theorem-A2, a 3D rotationally symmetrical interpretation ‶Φ and ‶Ψ of the 2D curves ‶φ and ‶ψ always exists. The camera rotation RcYRcX is equivalent to a rigid rotation of the 3D scene by RcXTRcyT around F so, the 3D interpretation ‶Φ and ‶Ψ project to φ and ψ after being rotated by RcXTRcyT around F.
QED 3.2.4. Model-based invariant of 3D rotational-symmetry with planarity of contours—In the previous section we showed that detecting a 2-fold 3D symmetry from a 2D image is an ill-posed problem. Almost any 2D image can be consistent with some 3D symmetrical interpretation. The best (probably the only) way to transform this ill-posed problem to a well-posed problem is by applying an additional constraint.
Now consider 3D mirror-symmetry. There are model-based invariants of 3D mirror symmetry under both orthographic and perspective projections. Under the orthographic projection, lines connecting pairs of corresponding points in the 2D image are parallel to one another. Under the perspective projection, the lines connecting corresponding points converge at a point in the 2D image. These are the only invariants of 3D mirror-symmetry, but additional invariants can be introduced into the image if 3D mirror-symmetry is used along with another constraint, namely with the planarity of contours (Sawada, Li, & Pizlo, 2014). Consider a 3D mirror-symmetrical pair of planar curves. Under the orthographic projection, the relationship between images of the curves can be represented by a sub-group of the 2D affine transformation. Under the perspective projection, the relationship can also be represented by a transformation that includes the same sub-group of the 2D affine transformation. This planarity constraint plays an important role for the human visual system to detect 3D mirror-symmetry from the 2D image (Sawada, Li, & Pizlo, 2011, 2014).
In this section, we will apply the same approach to 3D rotational-symmetry, specifically we will derive transformations among 2D images of a 3D symmetrical set of planar curves under both orthographic (Figure 17) and perspective projections. These transformations are model-based invariants of 3D symmetry taken together with the planarity of contours under those projections.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript 3.2.4.1. Model-based invariant under a 2D orthographic projection Theorem-B1.: Let Φn1, Φn2, … Φni, … Φnn be an n-fold 3D symmetrical set of planar curves and φon1, φon2, … φoni, … φonn be their orthographic projections in a 2D image plane ΠI where i is a natural number between 1 and n. Assume that an orthographic projection of their symmetry axis is given. Let us set the 2D xy Cartesian coordinate system in ΠI so that the projection of the symmetry axis coincides with the x-axis. Then, the relation between φon1 and φoni can be represented as follows: R2D(π −ζni) xoni yoni 1 = m11 m12 m13 0 1 0 0 0 1 R2D(ζni) xon1 yon1 1 ζni = tan−1 sin 2π(i −1)/n cos 2π(i −1)/n −1 cos θaxis (25) where pon1 = [xon1 yon1]T and poni = [xoni yoni]T are a pair of corresponding points on φon1 and φoni, m11, m12, and m13 are free parameters, and θaxis is an orientation of the symmetry axis relative to a normal of ΠI.
Proof: Consider the n-fold 3D symmetrical set of planar curves Φn1, Φn2, … Φni, … Φnn where i is a natural number between 1 and n. Without loss of any generality, we can assume that its symmetry axis is on the XZ-plane (see 3.1.1. and 3.2.1.1.). A symmetrical pair of points Pn1 and Pni on Φn1 and Φni can be written as: Pn1 = Rz θaxis Xn1 + DX Yn1 Zn1 = Xn1 cos θaxis + Zn1 sin θaxis + DX cos θaxis Yn0 −Xn1 sin θaxis + Zn1 cos θaxis −DX cos θaxis (26) Pni = Rz θaxis Xn1 cos θaxis −Yn1 sin ωni + DX Xn1 sin ωni + Yn1 cos ωni Zni = Xn1 cos ωni −Yn1 sin ωni cos θaxis + Zn1 sin θaxis + DX cos θaxis Xn1 sin ωni + Yn1 cos ωni −Xn1 cos ωni −Yn1 sin ωni sin θaxis + Zn1 cos θaxis −DX sin θaxis (27) where DX and θaxis are constants and ωni is 2π(i−1)/n. An orthographic projection of the symmetry axis is on the x-axis and orthographic projections of Pn1 and Pni are: Sawada and Zaidi Page 21 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript pon1 = xon1 yon1 = Xn1 cos θaxis + Zn1 sin θaxis + DX cos θaxis Yn1 (28) poni = xoni yoni = Xn1 cos ωni −Yn1 sin ωni cos θaxis + Zn1 sin θaxis + DX cos θaxis Xn1 sin ωni + Yn1 cos ωni (29) Then, a pair of points ṗon1 and ṗoni in the image plane ΠI is computed by rotating pon1 for ζni and poni for π−ζni: p ˙oni = R2D(ζni)pon1 = xon1 cos ζni −yon1 sin ζni Xn1 cos θaxis + Zn1 sin θaxis + DX cos θaxis sin ωni + cos ωni −1 cos θaxis sin 2ωni cos ωni −1 2 cos 2θaxis (30) p ˙oni = R2D(π −ζni)poni = −xonicos ζni −yoni sin ζni Xn1 cos θaxis + Zn1 sin θaxis + DX cos θaxis sin ωni + cos ωni −1 cos θaxis sin 2ωni cos ωni −1 2 cos 2θaxis (31) where: ζni = tan−1 sin ωni cos ωni −1 cos θaxis (32) Recall that both Φn1 and Φni are individually planar. Each planar curve in a 3D scene is a 2D curve on a plane. Therefore, the orthographic projections φon1 and φoni of Φn1 and Φni from Sawada and Zaidi Page 22 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript the 3D scene to the 2D image plane ΠI can be represented by 2D orthographic projections from the planes of Φn1 and Φni to ΠI and the 2D orthographic projection is a sub-set of the 2D affine transformation. Then, a relation between the orthographic projections φon1 and φoni can also be represented by a 2D affine transformation. The y-coordinates of ṗon1 and ṗoni are identical to one another in Equations (30) and (31). Therefore, the relation between φon1 and φoni is represented specifically by a subgroup of the 2D affine transformation: R2D π −ζni poni = m11 m12 0 1 R2D(ζni)pon1 + m13 0 (33) Note that m11, m12, and m13 represent 1D scaling, shear, and translation along the x-axis between ṗon1 and ṗoni.
QED Note that ζni is 0 and is independent from θaxis if i = n = 2. The slant θaxis of the symmetry axis is a free parameter for recovering a 3D shape of a 2-fold symmetrical object from its 2D orthographic image (see 3.2.2.1) and does not affect this relation between φon1 and φoni. A relation between φon1 and the 180° rotation of φoni is represented by the 1D scaling, shear, and translation along the x-axis 805 under this condition (see 3.2.3.1, Figure 13).
It is worth pointing out that the subgroup of the 2D affine transformation in Equation (33) also appears in a transformation representing the relationship between perspective projections of a 3D mirror-symmetrical pair of planar curves (Equation (1) in Sawada, Li, & Pizlo, 2014). But, Equation (33) has two rotation matrices R2D(ξni) and R2D(π−ξni) that do not exist in the transformation used for 3D mirror-symmetry. Hence, Equation (33) is more complicated than the transformation for 3D mirror-symmetry.
3.2.4.2. Model-based invariant under a 2D special perspective projection Lemma-for-Theorem-B2.: Let Φn1, Φn2, … Φni, … Φnn be an n-fold 3D symmetrical set of planar curves and φon1, φon2, … φoni, … φonn be their perspective projections in the 2D image plane ΠI where i is a natural number between 1 and n. Assume that their symmetry axis is perpendicular to ΠI and its perspective projection in ΠI is given. Note that a vanishing point of the symmetry axis appears at the principal point of the perspective projection in ΠI. Let us set the 2D xy Cartesian coordinate system in ΠI so that the origin is at the principal point and the projection of the symmetry axis coincides with the x-axis. Then, a relation between φpn1 and φpni can be represented as follows: Rz(ξni −ωni) xpni ypni f = m11 m12 m13 0 1 0 0 0 1 Rz(ξni) xpn1 ypn1 f (34) Sawada and Zaidi Page 23 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript ξni = tan−1 sin ωni cos ωni −1 where [xpn1 ypn1]T and [xpni ypni]T are a pair of corresponding points on φpn1 and φpni and m11, m12, and m13 are free parameters.
Proof: Consider the n-fold 3D symmetrical set of planar curves Φn1, Φn2, … Φni, … Φnn with a symmetry axis that is parallel to the Z-axis and intersects with the X-axis. A symmetric pair of points Pn1 and Pni on Φn1 and Φni can be written as: Pni = Xn1 + DX Yn0 Zn0 T (35) Pni = Xni Yni Zni T = Xn1 cos ωni −Yn1 sin ωni −DX Xn1 sin ωni + Yn0 cos ωni Zn1 T (36) where [DX 0 0]T is the X-intercept of the symmetry axis. Let φpn1, φpni, ppn1, and ppni be perspective projections of Φn1, Φni, Ppn1, and Ppni: ppn1 = f Xn1 + DX Zn1 f Y1 Zn1 T (37) ppni = f Xn1 cos ωni −Yn1 sin ωni + DX Zn1 f Xn1 sin ωni + Yn1 cos ωni Zn1 T (38) Next, consider rotating Φn1 for ξni and Φni for ξni−ωni around the Z-axis where: ξni = tan−1 sin ωni cos ωni −1 (39) Then, Φn1, Φni, Pn1, and Pni are transformed to Φ ˙n1, Φ ˙ni, P ˙n1, and Ṗni: P ˙ n1 = X ˙ n1 Y ˙ n1 Z ˙ n1 = Rz ξni Xn1 + DX Yn1 Zn1 = Xn1 + DX cos ξni −Yn1 sin ξni Xn1 + DX sin ξni + Yn1 cos ξni Zn1 (40) Sawada and Zaidi Page 24 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript P ˙ ni = X ˙ ni Y ˙ ni Z ˙ ni = RZ ξni −ωni Xn1 cos ωni −Yn1 sin ωni + DX Xn1 sin ωni + Yn1 cos ωni Zn1 = Xn1 −DX cos ξni −Yn1 sin ξni Xn1 −DX sin ξni −Yn1 cos ξni Zn1 (41) where tan(ξni−ωni) = sinωni /(1−cosωni). From Equations (40) and (41), the Y- and Z-coordinates of Ṗn1 and Ṗni are identical to one another. The perspective projections ṗpn1 and ṗpni of Ṗpn1 and Ṗpni can be computed also by rotating ppn1 for ξni and ppni for ξni−ωni: P ˙ pn1 = x ˙pn1 y ˙pn1 = R2D ξni ppn1 = f cos ξni Zn1 Xn1 −Yn1tanξni + DX Xn1tanξni + Yn1 + DXtanξni (42) P ˙ pni = x ˙pni y ˙pni = R2D ξni −ωni ppni = f cos ξni Zn1 Xn1 −Yn1tanξni + DX Xn1tanξni + Yn1 + DXtanξni (43) The y-coordinates of ṗpn1 and ṗoni are also identical with one another and the distance between their x-coordinates depends on Zn1. If Φn1 and Φni are individually planar, Zn1 becomes a function of Xn1 and Yn1. Then, both the x- and y-coordinates of ṗpn1 and ṗoni become functions of Xn1 and Yn1. This introduces a systematic relation between perspective projections φ ˙n1 and φ ˙ni of Φ ˙n1 and Φni.
Assume Φ ˙n1 and Φ ˙ni are individually planar. Then, the following equation is satisfied by P ˙n1 = X ˙n1 Y ˙n1 Z ˙n1 T: an1X ˙ n1 + bn1Y ˙ n1 + cn1Z ˙ n1 + dn1 = 0 (44) where an1, bn1, cn1, and dn1 are constants. From Equations (40), (41), and (44) (see Supplemental Materials): x ˙n1 2an1DX cos ξni + dn1 + 2y ˙n1bn1Dx cos ξni + 2f cn1DX cos ξni = dn1x ˙ni (45) Namely, xni can be represented as a weighted sum of xn1, yn1, and a constant. From equations (42), (43), and (45), a relation between ṗpn1 and ṗoni can be written as: Sawada and Zaidi Page 25 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript p ˙pni = 2an1DX cos ξni/dn1 + 1 0 2bn1DX cos ξni/dn1 1 p ˙pn1 + f 2cn1DX cos ξni/dn1 0 (46) Since ṗpn1 and ṗoni are 2D rotations of ppn1 and ppni, a relation between φpn1 and φpni is: R2D ξni −ωni ppni = m11 m12 0 1 R2D ξni Ppn1 + f m13 0 (47) where: m11 = 2an1DX cos ξni/dn1 + 1 m12 = 2bn1DX cos ξni/dn1 m13 = 2cn1DX cos ξni/dn1 Equation (46) shows that, after rotating φpn1 for ξni and φpni for ξni−ωni, a relation between these two curves in ΠI can be represented by a subgroup of the 2D affine transformation. This is a combination of scaling, shear, and translation along the x-axis. It is worth pointing out that the same subgroup of the 2D affine transformation also appeared in Equations (25) and (34), which represent the relationship among orthographic projections of a 3D symmetrical set of planar curves.
QED Note that ζni = 0 and ξni−ωni = −π if i = n = 2. Then, the subgroup of the 2D affine transformation represents a relation between φpn1 and the 180° rotation of φpni (see 3.2.3.2).
3.2.4.3. Model-based invariant under a 2D general perspective projection Theorem-B2.: Let Φn1, Φn2,…Φni,…Φnn be an n-fold 3D symmetrical set of planar curves and φon1, φon2,…φoni,…φonn be their perspective projections in a 2D image plane ΠI where i is a natural number between 1 and n. Assume that a perspective projection of their symmetry axis and a vanishing point vaxis of the axis in ΠI are given. Let us set the 2D xy Cartesian coordinate system on ΠI so that the origin is at the principal point and vaxis is on the x-axis: vaxis = [xaxis 0]T. The projection of the symmetry axis can be written as vaxis + t[xl yl]T where t is a free parameter. Then, a relation between φpn1 and φpni can be represented as follows: RZ ξni −ωni −σa RY T σv xpni sin σv + f cos σv xpni ypni f = m11 m12 m13 0 1 0 0 0 1 RZ ξni −σa RY T σv xpn1 sin σv + f cos σv xpn1 ypn1 f (48) Sawada and Zaidi Page 26 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript ζni = tan−1 sin ωni cos ωni −1 where ppn1 = [xpn1 ypn1]T and ppni = [xpni ypni]T are a pair of corresponding points on φpn1 and φpni, f is a focal distance, σv = atan(xaxis/f), σa = atan(yl/(xlcos σv)), ωni = 2π(i−1)/n, and m11, m12, and m13 are free parameters.
Proof: In the prior sub-section, it was assumed that an axis of an n-fold 3D symmetrical set of planar curves is parallel to the Z-axis. We now consider the more general condition in which this assumption is not necessary. A camera of a perspective projection can be virtually rotated with the image plane ΠI and the 3D coordinate system around the center of projection arranged such that the assumption of the prior sub-section is satisfied. Note that the same procedure was used for the recovery of the 3D shape that was described in an earlier section of this study (see 3.2.2.2.).
Now, consider an n-fold 3D symmetrical set of planar curves (Φn1, Φn2, … Φni, … Φnn) and their perspective projections in the image plane ΠI. A projection laxis of their symmetry axis and a vanishing point vaxis of the axis are given (see 3.1.1. and 3.2.1.1.). Set the orientation of the 2D xy Cartesian coordinate system so that vaxis is on the x-axis; vaxis = [xaxis 0]T. The symmetry axis is parallel to a line connecting the center of projection F (= [0 0 0]T) and [xaxis 0 f]T. Any point on the projection of the symmetry axis can be written as vaxis + t[xl yl]T where t is a free parameter.
The image is transformed by emulating a camera rotation Rv around the Y-axis so that the vanishing point is transformed to the principal point in the image plane ΠI: Rv = RY(σv) where σv = atan(xaxis/f). Then, a point [x y]T in the original image is transformed to [xʹ yʹ]T after Rv (see Supplemental Materials): x′ y′ f = f RY T σv x sin σv + f cos σv x y f (49) A line connecting the center of projection F and the vanishing point ([0 0]T) is parallel to the symmetry axis and is perpendicular to ΠI after Rv. The projection of the symmetry axis after Rv can be written as follows: f t 1 + txaxis sin σv xl cos σv yl (50) The projection of the symmetry axis becomes coincident with the x-axis after another camera rotation Ra (= RZ(σa)) around the Z-axis for σa = atan(yl/(xlcos σv)): Sawada and Zaidi Page 27 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript RZ T σa x′ y′ f = f RZ T σa RY T σv x sin σv + f cos σv x y f (51) A transformation of the image caused by Ra is equivalent to a 2D image rotation: R2DT(σa). After the camera rotation RvRa (= RY(σv)RZ(σa)), the curves Φn1, Φn2, … Φni, … Φnn satisfy all the conditions of the Lemma-for-Theorem-B2. Hence, from Equation (34) of Lemma-for-Theorem-B2, the relation between φpn1 and φpni can be represented as follows: RZ ξni −ωni −σa RY T σv xpni sin σv + f cos σv xpni ypni f = m11 m12 m13 0 1 0 0 0 1 RZ ξni −σa RY T σv xpn1 sin σv + f cos σv xpn1 ypn1 f (52) where [xpn1 ypn1]T and [xpni ypni]T are a pair of corresponding points on φpn1 and φpni and m11, m12, and m13 are free parameters.
QED It is worth pointing out that the subgroup of the 2D affine transformation in Equation (52) also appears in a transformation representing the relationship between perspective projections of a 3D mirror-symmetrical pair of planar curves (Equation (20) in Sawada, Li, & Pizlo, 2014). But, Equation (52) has two rotation matrices RZ(ξni−σa) and RZ(ξni−ωni −σa) that do not exist in the transformation used for 3D mirror-symmetry. Hence, Equation (52) is more complicated than the transformation for 3D mirror-symmetry.
4. General Discussion The study has shown that 3D rotational-symmetry has the following properties: (i) a 3D rotational-symmetrical shape can be recovered from one of its 2D images, (ii) any pair of 2D curves is consistent with a 3D, 2-fold rotational-symmetrical interpretation, and (iii) additional model-based invariants of 3D rotational-symmetry can be introduced under both orthographic and perspective projections if the 3D rotational-symmetrical set of curves are individually planar. Another important property of 3D rotational-symmetry is called a “virtual image” (Vetter & Poggio, 1994), namely, the single 2D image of the n-fold 3D symmetrical object is equivalent to n images of the same object seen from different viewpoints.
These properties are also present in 3D mirror-symmetry. Note, however, that it is computationally much harder to use 3D rotational-symmetry to recover a 3D shape than it is to use 3D mirror-symmetry to perform this kind of recovery. For example, the symmetry axis in 3D rotational-symmetry is specified by 4 parameters, 2 for orientation and 2 for position, but the symmetry plane in 3D mirror-symmetry can be specified by only 3 parameters, 2 for orientation and 1 for position (see 3.2.1.1.).8 Note also that a 2D image of Sawada and Zaidi Page 28 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript a 3D rotationally-symmetrical pair of planar curves satisfy Equations (25), (34), or (48), depending on the type of projection. These equations represent the model-based invariants of the 3D rotational-symmetry, and the planarity of the curves. The model-based invariants of 3D mirror-symmetry and the planarity of the curves are also represented by analogous equations (Equations 1, 8, and 20 in Sawada, Li, & Pizlo, 2014) but, here again, the equations for 3D mirror-symmetry are simpler than those for 3D rotational symmetry. Also note that it is more difficult to find a corresponding pair of points in a 2D image of a 3D rotationally-symmetrical shape than in a 2D image of a 3D mirror-symmetrical shape. Consider two pairs of 2D curves in the image plane: one is an image of a 3D rotational-symmetrical pair of curves and the other is an image of a 3D mirror-symmetrical pair of curves. Each corresponding pair of points on the curves is established by finding intersections of the 2D curves with a pair of lines for 3D rotational-symmetry (see 3.2.3.) and with only a single line for 3D mirror-symmetry (Sawada, Li, & Pizlo, 2011; Rothwell, 1995; Hong, Ma, & Yu, 2004).
These complexities, which are inherent in using 3D rotational-symmetry compared with 3D mirror-symmetry, could actually be critical with respect to their utility within the human visual system. With 3D mirror-symmetry, a pair of curves in a 3D scene is easier to detect from its 2D image if the curves are individually planar (Sawada, Li, & Pizlo, 2011, 2014). This is not the case with 3D rotational-symmetry. We made some subjective observations that suggest that we can detect 3D rotational-symmetry only if the number of symmetry folds is sufficiently large (Figure 18). It also seems worthwhile to point out here that the boundary contour of a 2D image of a 3D rotationally symmetrical shape becomes closer to 2D mirror-symmetrical as the number of the folds increases. As the number of the folds increases, two regular features emerge in a 3D rotationally-symmetrical shape. First, a symmetry polygon of the 3D rotationally-symmetrical shape becomes closer to a circle, which is the most regular shape (Pizlo, 2008). A 2D image of the circle in the 3D scene is always an ellipse under both orthographic and perspective projections (Pizlo & Salach-Golyska, 1994). It is possible that the visual system is sensitive to the circle in the 3D scene (see Zanker & Quenzer, 1999) or the ellipse in the 2D image. The other emerging feature is 2D mirror-symmetry. A 3D rotationally-symmetrical shape becomes closer to a surface-of-revolution as the number of the folds increases, and the boundary contour of a 2D image of a surface-of-revolution is always 2D mirror-symmetrical under an orthographic projection (Figure 19) as well as under a perspective projection at least with the spherical retina of the "reduced" eye9 (Horaud & Brady, 1988). The visual system is very sensitive to 2D mirror-symmetry in a retinal image (e.g. Barlow & Reeves, 1979; Jenkins, 1983; Cohen & Zaidi, 2013). All of these observations suggest that the human visual system is relatively insensitive to 3D rotational-symmetry, at least when the number of the folds in the shape is small.
8There is no difference between 2D rotational- and mirror-symmetry in this aspect. Both the symmetry point of 2D rotational-symmetry and the symmetry axis of 2D mirror-symmetry can be specified by 2 parameters.
9If the image is planar under a perspective projection, the boundary contour can be represented by a Kanatani transformation (Kanatani, 1988) of a 2D mirror-symmetrical shape whose axis passes the principal point (Wong, Mendoça, & Cipolla, 2004).
Sawada and Zaidi Page 29 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript On the other hand, the virtual image of a 3D rotational-symmetrical object is computationally easier to generate than the virtual image of a 3D mirror-symmetrical object with a single symmetry plane. Virtual images of the same objects from different viewpoints can be generated from images of these symmetrical objects (Vetter & Poggio, 1994). The virtual image of a rotational-symmetrical object is identical to the original image, but the virtual image of a mirror symmetrical object is identical to the “reflection” of the original image. The virtual image of a mirror symmetrical object is computationally more complex because of this reflection. Consider the human’s recognition of the 3D shape of an object. Human performance in 3D shape recognition tasks is reliable with a 3D mirror-symmetrical object that has a single symmetry plane, but not with a 3D asymmetrical object (Li & Pizlo, 2011; Chan, Stevenson, Li, & Pizlo, 2006; Liu, Knill, & Kersten, 1995; Liu & Kersten, 2003; Pizlo & Stevenson, 1999; van Lier & Wagemans, 1999; Vetter, Poggio, & Bülthoff, 1994). This superior performance with the mirror-symmetrical object could be explained by a mechanism based only on the 2D template matching of memorized images of the object, or on a 2D image interpolation between the memorized images (e.g. Bülthoff, Edelman, & Tarr, 1995). The virtual image of the mirror-symmetrical object could serve as an additional memorized image for an image -based mechanism. If this applied here, human performance when recognizing a rotational994 symmetrical object should be better than performance when recognizing a mirror-symmetrical object. Note that the virtual image of the rotational-symmetrical object is more easily generated than the virtual image of the mirror-symmetrical object. We know of no psychophysical study that tested recognition with a 3D rotational-symmetrical object, but based on our subjective observations, it is easy to see that it is difficult to recognize 3D rotational-symmetrical objects from different views when they only have a small number of symmetry folds (see Figure 18).
The human visual system can detect the 3D rotational-symmetry of an object reliably from its 2D image only under limited conditions, for example, when there are many symmetry folds (Figure 18), or when the object is close to being flat (Figure 1D), and when the object is viewed from a degenerate viewpoint that makes its 2D retinal image rotational-symmetrical (e.g. a right-bottom flower in Figure 1A). Another possible condition is when the shape of the object (or parts composing the object, see Figure 17) satisfies some other important constraints for recovering the veridical 3D shape of the object from a 2D image. Under this condition, the human visual system can detect the 3D rotational-symmetry of the object, but not from the image itself. It can detect it from the 3D shape perceived from the 2D image. For example, the 3D shape of an object is perceived reliably if the object is 3D mirror-symmetrical (Li et al., 2009, 2011; Pizlo et al., 2010, 2014). This detection is rather common. Some 3D rotational-symmetrical objects in real life are also 3D mirror-symmetrical (see Figure 1DF for examples). Also, any 3D mirror-symmetrical object with multiple symmetry planes is always 3D rotational-symmetrical (Stewart & Golubitsky, 1992; van der Helm & Leeuwenberg, 1996, see Figure 20 for illustration of this relation between rotational- and mirror symmetry using 2D figures). On the other hand, a 3D rotational-symmetrical object with multiple folds is not necessarily 3D mirror-symmetrical (see Figure 1ABC for examples).
In this study, we studied the geometrical properties of 3D rotational-symmetry, specifically, the 3D shape of a 3D rotational-symmetrical object and its 2D projection. Interestingly, 3D Sawada and Zaidi Page 30 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript rotational-symmetry shares many geometrical properties with 3D mirror-symmetry, but based on our subjective observations, these two types of symmetry seem to be perceived in very different ways. We plan to study these perceptual differences in psychophysical experiments in our future work.
Supplementary Material Refer to Web version on PubMed Central for supplementary material.
Acknowledgement This project was supported by National Eye Institute Grants EY013312 to Q. Zaidi.
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Pizlo Z & Salach-Golyska M (1994). Is vision metric? Comment on Lappin and Love (1992). Perception & Psychophysics, 55, 230–234 [PubMed: 8036104] Pizlo Z, Sawada T, Li Y, Kropatsch W, & Steinman RM (2010). New Approach to the Perception of 3D Shape Based on Veridicality, Complexity, Symmetry and Volume. Vision Research, 50, 1–11. [PubMed: 19800910] Pizlo Z, & Stevenson AK (1999). Shape constancy from novel views. Perception & Psychophysics, 61, 1299–1307 [PubMed: 10572459] Rothwell CA (1995). Object recognition through invariant indexing Oxford, England: Oxford University Press.
Royer FL (1981). Detection of symmetry. . Journal of Experimental Psychology: Human Perception and Performance, 7, 1186–1210. [PubMed: 6458647] Saunders JA, & Knill DC (2001). Perception of 3D surface orientation from skew symmetry. Vision Research, 41, 3163–3183. [PubMed: 11711141] Savriama Y & Klingenberg CP (2011). Beyond bilateral symmetry: Geometric morphometric methods for any type of symmetry. BMC Evolutionary Biology, 11:280. [PubMed: 21958045] Sawada T, & Pizlo Z (2008). Detection of skewed symmetry. Journal of Vision, 8(5):14, 1–18 Sawada T (2010). Visual detection of symmetry of 3D shapes. Journal of Vision, 10(6):4, 1–22.
Sawada T, Li Y, & Pizlo Z (2011). Any pair of 2D curves is consistent with a 3D symmetric interpretation. Symmetry, 3, 365–388.
Sawada and Zaidi Page 33 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Sawada T, Li Y, & Pizlo Z (2014). Detecting 3-D Mirror Symmetry in a 2-D Camera Image for 3-D Shape Recovery. Proceedings of the IEEE, 102, 1588–1606.
Sawada T, Li Y, & Pizlo Z (2015). Shape Perception. In Busemeyer J, Townsend J, Wang ZJ, & Eidels A (Eds.), Oxford Handbook of Computational and Mathematical Psychology (pp. 255–276). New York, NY: Oxford University Press Stewart I & Golubitsky M (1992/1993). Fearful Symmetry: Is God a Geometer? London, UK: Penguin books.
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Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 1. Rotational-symmetrical objects in real life.
Sawada and Zaidi Page 35 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 2. Random-dot patterns with (A, C) rotational- and (B, D) mirror-symmetry with two different densities.
Sawada and Zaidi Page 36 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 3. 2D symmetrical figures with 2-, 3-, 4-, and 5-folds. Their symmetry points are indicated by open circles and their symmetry polygons are drawn with dashed lines.
Sawada and Zaidi Page 37 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 4. Orthographic projections of planar symmetrical figures in Figure 3 from viewing directions slanted 60° from their symmetry axes. Projections of their symmetry points are indicated by open circles. Note that the orthographic projections of the 2- and 4-fold symmetrical figures are also 2-fold symmetrical.
Sawada and Zaidi Page 38 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 5. (A) Symmetry polygons with 2-, 3-, 4-, 5-folds and their (B) orthographic and (C) perspective projections. The perspective projections of the symmetry points (open circles) can be derived from the perspective projections of the symmetry polygons only if the number of the folds is more than three. Auxiliary lines for finding the symmetry points are rendered in dotted and dashed lines. The projections of the symmetry points cannot be derived from the 2- or 3-fold symmetry polygons alone.
Sawada and Zaidi Page 39 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 6. Perspective projections of a 3-fold symmetry polygon (equilateral triangle) with its symmetry axis with four different orientations. The four images of the symmetry polygon are identical to one another. The Principal points of the perspective projection are indicated by ‘x’.
Sawada and Zaidi Page 40 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 7. Orthographic projections of 3D symmetrical objects with 2-, 3-, 4-, and 5-folds. Their symmetry axes are indicated by thick line segments and symmetry polygons are drawn in gray.
Sawada and Zaidi Page 41 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 8. A perspective projection of a 4-fold symmetrical object to the image plane ΠI. The symmetry axis is parallel to a line connecting the vanishing point vaxis of the symmetry axis and the center of projection F. A plane including F and the horizon haxis of the symmetry axis is normal to the symmetry axis and to the segment Fvaxis and are parallel to the symmetry polygons.
Sawada and Zaidi Page 42 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 9. A perspective projection of a 2-fold symmetrical object. The vanishing points v1, v2, and v3 of the symmetry polygons are collinear on the horizon haxis of the symmetry axis.
Sawada and Zaidi Page 43 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 10. (A) A perspective projection and another projection after rotating the camera (the principal axis and the image plane ΠI) for σv about the center of projection F so that the symmetry axis becomes normal to ΠIʹ. (C) The original perspective image (solid) and the image after the rotation (dotted). The image after the rotation can be computed directly by transforming the original 2D image. (B) The transformation of the image by rotating the camera is the same as the image transformation by rotating the 3D scene about F in the opposite direction.
Sawada and Zaidi Page 44 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 11. (A) A pair of 2D curves φ and ψ satisfying conditions of Theorem-A1 and (B, C) two views of their 3D symmetrical interpretation. The symmetrical interpretation was constructed by assuming that the slant of its symmetry axis is 45° under an orthographic projection. (B, C) Two orthographic images of the interpretation with its symmetry axis normal to the image plane (B) and with the symmetry axis parallel to the image plane (C). Note that the image in (B) is 2D rotational symmetrical and that in (C) is 2D mirror-symmetrical. These are properties 3D rotational-symmetry under the 2D orthographic projection. See Demo 1 in supplemental material for an interactive illustration of the 3D symmetric curves (the demo is also available at:
Sawada and Zaidi Page 45 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 12. (A) A pair of 2D curves φ and ψ satisfying conditions of Theorem-A1 and (B, C) two views of their 3D symmetrical interpretation. Some point on one curve in (A) corresponds with multiple points on the other curve and vice versa for the 3D rotational-symmetrical interpretation. The symmetrical interpretation was constructed by assuming that the slant of its symmetry axis is 30° under an orthographic projection. (B, C) Two orthographic images of the interpretation with its symmetry axis normal to the image plane (B) and with the symmetry axis parallel to the image plane (C). Note that the 3D curves of the interpretation of (A) are much more complex than the 2D curves in (A). It is complex because multiple segments of the 3D curves in (B, C) are projected to single segments of the 2D curves in (A). See Demo 2 in the supplemental material for an interactive illustration of the 3D symmetric curves (the demo is also available at: rotsym/).
Sawada and Zaidi Page 46 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 13. Visual method of establishing the correspondence between a pair of 2D curves for its 3D symmetrical interpretation. The pair of the 2D curves φ (black, solid) and ψ (grey solid) (A) in Figure 11A and (B) in Figure 12A and the 180° rotation of ψ (ψ−1, black dashed). The curve ψ−1 is translated along lφ1 and lφ2 for the clarity of the images. The correspondence between φ and ψ−1 can be established between intersections (black open circles) of φ and ψ −1 with a line (dotted) parallel to lφ1 and lφ2. In (A), the parallel line that intersects with φ has a unique intersection with ψ−1 and vice versa. In (B), the parallel line that intersects with φ has one or a finite number of intersections with ψ−1 and vice versa. The corresponding points on ψ are also indicated by grey open circles.
Sawada and Zaidi Page 47 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 14. (A) A pair of 2D curves φ and ψ satisfying conditions of Lemma-for-Theorem-A2 and (B, C)two views of their 3D symmetrical interpretation. The symmetrical interpretation was constructed under a perspective projection and its symmetry axis is normal to the image plane. Note that the contours in (A) are identical with those in Figure 11A to allow a comparison between the 3D symmetrical interpretations under the perspective (B, C) and the orthographic (Figure 11B, C) projections. The Principal points of the perspective projection are indicated by ‘x’. (B, C) Two orthographic images of the interpretation with its symmetry axis normal to the image plane (B) and with the symmetry axis parallel to the image plane (C). The orthographic projection is used in (B, C) to show the properties of 3D rotational-symmetry under a 2D orthographic projection (Figure 11): the image in (B) is 2D rotational-symmetrical and the image in (C) is 2D mirror-symmetrical. See Demos 3 and 4 in the supplemental material for an interactive illustration of the 3D symmetric curves (the demos are also available at:
Sawada and Zaidi Page 48 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 15. A pair of 2D curves φ and ψ satisfying conditions of Theorem-A2. The symmetrical interpretation was constructed under a perspective projection (see Demo 5 in the supplemental material for an interactive illustration of the 3D symmetric curves, the demo is also available at: The Principal points of the perspective projection are indicated by ‘x’. The symmetry axis of the 3D interpretation is oriented so that its vanishing point appears at vaxis. The visual angles from vaxis to uφ1 and to uψ2 are equal to one another and those from vaxis to uψ1 and to u φ2 are also equal to one another.
Sawada and Zaidi Page 49 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 16. The transformations of the image in Figure 15 after the camera has rotated (A) RcY and (B) RcYRcX. (A) The transformed image after RcY (solid-black) is superimposed to the original image (dotted-grey). (B) The transformed image after RcYRcX (solid-black) is superimposed to the transformed image after RcY (dotted-grey). The Principal points of the perspective projection are indicated by ‘x’.
Sawada and Zaidi Page 50 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 17. An orthographic projection of a rotational-symmetrical object composed of a pair of wedges. Dotted and dashed contours are projections of a symmetrical pair of planar contours of the object. The relationship between their orthographic projections can be represented as a sub-group of the 2D affine transformation (Theorem-B1).
Sawada and Zaidi Page 51 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 18. Objects composed of planar contours with 2-, 3-, 4-, 10-, and 20-fold symmetry. Three orthographic views of the individual objects are shown in rows.
Sawada and Zaidi Page 52 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 19. Orthographic views of a surface of revolution from three different viewpoints. The image of the surface of revolution is always mirror-symmetrical under the orthographic projection.
Sawada and Zaidi Page 53 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript Figure 20. Figures with 1-, 2-, 3-, 4-, and 5-axes of 2D mirror-symmetry. The mirror-symmetrical figures are also 2D rotational-symmetrical if the number of the symmetry axes are more than one.
Sawada and Zaidi Page 54 J Math Psychol. Author manuscript; available in PMC 2019 December 01.
Author Manuscript Author Manuscript Author Manuscript Author Manuscript |
187565 | https://www-backup.salemstate.edu/skew-lines-geometry | Salem State Vault
Home / Plus / Skew Lines Geometry
Skew Lines Geometry
Ashley
November 4, 2024 8 min read
Skew lines are a fundamental concept in geometry, referring to lines that do not intersect and are not parallel. This concept is crucial in understanding the three-dimensional space and has numerous applications in various fields, including architecture, engineering, and computer graphics. In this article, we will delve into the world of skew lines geometry, exploring their definition, properties, and significance.
To begin with, let’s establish the contextual foundation of skew lines. In geometry, two lines are said to be skew if they do not intersect and are not parallel. This means that skew lines do not lie in the same plane and do not have any common points. The concept of skew lines is essential in understanding the geometry of three-dimensional space, as it helps us describe the relationship between lines that do not intersect or lie in the same plane.
Definition and Properties of Skew Lines
Skew lines can be defined as lines that do not intersect and are not parallel. In other words, two lines are skew if they do not have any common points and do not lie in the same plane. The properties of skew lines are as follows:
Skew lines do not intersect, meaning they do not have any common points.
Skew lines are not parallel, meaning they do not lie in the same plane.
Skew lines are not coplanar, meaning they do not lie in the same plane.
These properties are essential in understanding the behavior of skew lines and their significance in geometry.
Types of Skew Lines
There are two types of skew lines: right skew lines and left skew lines. Right skew lines are lines that do not intersect and are not parallel, and the angle between them is less than 90 degrees. Left skew lines, on the other hand, are lines that do not intersect and are not parallel, and the angle between them is greater than 90 degrees.
| Type of Skew Line | Angle Between Lines |
--- |
| Right Skew Lines | Less than 90 degrees |
| Left Skew Lines | Greater than 90 degrees |
💡 The concept of right and left skew lines is essential in understanding the geometry of three-dimensional space, as it helps us describe the relationship between lines that do not intersect or lie in the same plane.
Significance of Skew Lines in Geometry
Skew lines have numerous applications in geometry, including:
Describing the relationship between lines: Skew lines help us describe the relationship between lines that do not intersect or lie in the same plane.
Understanding three-dimensional space: Skew lines are essential in understanding the geometry of three-dimensional space, as they help us describe the relationship between lines that do not intersect or lie in the same plane.
Applications in computer graphics: Skew lines have numerous applications in computer graphics, including the creation of 3D models and animations.
Key Points
Skew lines are lines that do not intersect and are not parallel.
Skew lines have numerous applications in geometry, including describing the relationship between lines and understanding three-dimensional space.
There are two types of skew lines: right skew lines and left skew lines.
Skew lines have numerous applications in computer graphics, including the creation of 3D models and animations.
Understanding skew lines is essential in understanding the geometry of three-dimensional space.
Real-World Applications of Skew Lines
Skew lines have numerous real-world applications, including:
Architecture: Skew lines are used in architecture to create complex designs and structures.
Engineering: Skew lines are used in engineering to create complex systems and models.
Computer graphics: Skew lines are used in computer graphics to create 3D models and animations.
In conclusion, skew lines are a fundamental concept in geometry, referring to lines that do not intersect and are not parallel. Understanding skew lines is essential in understanding the geometry of three-dimensional space, and they have numerous applications in various fields, including architecture, engineering, and computer graphics.
What are skew lines in geometry?
+
Skew lines are lines that do not intersect and are not parallel. They do not lie in the same plane and do not have any common points.
What are the types of skew lines?
+
There are two types of skew lines: right skew lines and left skew lines. Right skew lines are lines that do not intersect and are not parallel, and the angle between them is less than 90 degrees. Left skew lines, on the other hand, are lines that do not intersect and are not parallel, and the angle between them is greater than 90 degrees.
What are the applications of skew lines in geometry?
+
Skew lines have numerous applications in geometry, including describing the relationship between lines, understanding three-dimensional space, and applications in computer graphics.
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187566 | https://www.reddit.com/r/HomeworkHelp/comments/4iaijp/precalculus_parabola_tangent_line/ | [Pre-Calculus] Parabola Tangent Line : r/HomeworkHelp
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Percy327
[Pre-Calculus] Parabola Tangent Line
Let (x0, y0) be a point on the parabola x2 = 4py. Show that the equation of the line tangent to the parabola at (x0, y0) is y= (x_0/2p)x- y_0
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Writing a compelling essay introduction
Writing a compelling essay introduction is crucial to grab your reader's attention and set the stage for your argument. Here are some tips and strategies based on advice from Redditors:
General Tips
Keep it Concise: Aim for 5-10 sentences or about 10% of the total word count. "A typical introduction is about 5-10 sentences."
Start with a Hook: Use a question, anecdote, quote, or statistic to make your introduction engaging. "Crafting a compelling hook demands a toolbox of strategies."
Provide Context: Briefly explain the background or significance of your topic. "Provide a context for the topic--a short story or anecdote (very short) that tells the reader why this matters."
State Your Thesis: Clearly present your main argument. "Your overarching thesis statement--what you hope to assert in the essay."
Outline Key Points: List the main points you will discuss to support your thesis. "A list if the points you plan to make to support this thesis."
Specific Strategies
Using a Hook
Question: "Have you ever wondered about the untold stories hidden in the forgotten corners of our cities, waiting to be discovered?"
Anecdote: "In the quiet alleys of my childhood hometown, I stumbled upon an old, dusty journal that held the secrets of generations past."
Quote: "As the great Maya Angelou once said, ‘We all should know that diversity makes for a rich tapestry, and we must understand that all the threads of the tapestry are equal.’"
Statistic: "Numbers have a magnetic quality, especially when they highlight a significant issue."
Structuring Your Introduction
Broad to Narrow Approach: Start with a general statement and then narrow down to your specific topic. "Start with a broad statement, then get more narrow, and end your introductory paragraph with your thesis statement."
Context and Background: Provide necessary context to help the reader understand the topic. "Include background information so the reader knows what context your essay is in."
Thesis Statement: Clearly state your main argument. "Your overarching thesis statement--what you hope to assert in the essay."
Avoiding Common Mistakes
Overly Broad Statements: Avoid generic openings like "Throughout history, many people have wondered about..."
Too Much Detail: Keep the introduction concise and focused. "An introduction should not be more than 10% - 15% of the total word count of the overall assignment."
Additional Resources
Subreddits for Further Help:
r/CollegeHomeworkTips
r/writing
r/AskAcademia
r/GetStudying
By following these tips, you can write an introduction that not only grabs your reader's attention but also effectively sets the stage for your essay. Good luck!
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187567 | https://www.instructortoolkit.co.uk/knowledge-base/chart-work-navigation/taking-bearings-on-a-chart/ | Taking Bearings on a Nautical Chart - Instructor Resources
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Taking Bearings on a Nautical Chart
The Electronic Way:
Identify your target:Locate the object or waypoint on the chart that you want to determine the bearing to.
Access the "Measure" or "Bearing and Range" tool:Most chartplotters have a specific function designed for measuring bearings and distances between two points on the chart.
Select your current position:Utilise the chartplotter's GPS or other positioning system to indicate your current location as the starting point for the bearing measurement.
Select the target location:Use the chartplotter's interface to pinpoint the desired location or object (waypoint) on the chart that you want the bearing to.
Read the bearing:The chartplotter will display the calculated bearing from your current position to the selected target, often in degrees relative to true north.
The Traditional Way:
Overview:
A = Make sure the arrow is pointing in the direction you wish to travel
B = Make sure the arrows are pointing to North on the Chart
C = Read off the bearing
Calculating effective bearings involves:
T - True Course in Degrees,
V - Apply Variation,
M - to give us a Magnetic course,
D - Apply Variation,
C - to give us a Compass Course.
Example:
1) PICK TWO POINTS YOU WISH TO TRAVEL BETWEEN ON THE CHART (Ensure there aren't any hazards between!)
2) LINE UP THE PLOTTER BETWEEN THE FIRST AND SECOND POINT, MAKING SURE THE DIRECTION OF BEARING ARROW ON YOUR PLOTTER IS POINTING IN THE DIRECTION YOU WISH TO TRAVEL.
3) LINE UP THE COMPASS ON THE PLOTTER WITH THE GRID LINES ON THE CHART
4) TURN THE PLOTTERS COMPASS UNTIL THE 'N' IS POINTING TO NORTH ON THE CHART
5) THE NUMBER ON THE PLOTTERS COMPASS BELOW THE '0' IS YOUR ‘TRUE’ BEARING. IN THIS EXAMPLE IT IS 291 DEGREES (TRUE)
Now we need to consider Variation:
WHAT IS VARIATION?
Variation is the difference between True and Magnetic North.
The Longitude Lines on our Chart point to the 'True' North Pole
Our compass points to the 'Magnetic' North Pole.
Variation varies dependant on where you are in the world so always check Variation on the Compass Rose on an up to date local chart.
There are plenty of mnemonic's to remember how to calculate Variation. We think the simplest is 'Easy is Least, West is Best'
I.E. when the Variation is East you SUBTRACT it from your True Bearing in order to gain your Magnetic Bearing. ; if the Variation is West you ADD it to your True Bearing in order to gain your Magnetic Bearing.
Image ©theairlinepilots.com
6) CALCULATE VARIATION: LOCATE THE COMPASS ROSE ON YOUR CHART NEAREST TO YOUR POINT OF TRAVEL (BEARING)
7) IN THIS EXAMPLE: THE VARIATION MARKED ON THE CHART IS 2 DEGREES 35’ WEST.
8) TO CALCULATE VARIATION USE THE FOLLOWING:
IN THIS EXAMPLE:
TRUE BEARING = 291 DEGREES
VARIATION = 2 DEGREES WEST(ACCURACY IS ONLY NEED TO BE TO THE NEAREST DEGREE FOR SHORTER DISTANCES)
THEREFORE = 291 (T) + 2 (V) = 289 DEGREES (M) WHICH IS THE MAGNETIC BEARING
AS ITS NEARLY IMPOSSIBLE TO STEER TO 'A DEGREE' WHEN USING THIS AFLOAT (ASSUMING YOU'RE TRAVELLING RELATIVELY SHORT DISTANCES ROUND UP OR DOWN TO THE NEAREST '0' OR '5' TO CREATE A 'USABLE' BEARING, SO IN THIS CASE 290 DEGREES (M)
And now another consideration, Deviation:
WHAT IS DEVIATION?
Compass Deviation is caused by the magnetic fields generated onboard your boat i.e. engines, electronics, metal work etc. Each vessel will have its own unique deviation which is calculated and detailed on a 'deviation card'.
Once Deviation is known we revert back to the mnemonic at the beginning of this page:
T- 'True Bearing'
V- Apply Variation
M- Gives us a 'Magnetic Bearing'
D- Apply Deviation
C- to give us a 'Compass Bearing'
This is also where the mnemonic CADET effectively comes into play.
Example of a Deviation Card:
F.A.Q.s
What is a True North?
True North is where the Longitude Lines meet at the North Pole.
What is Variation?
In a nutshell it is the difference between True North and Magnetic North.
True North is where the Longitude Lines meet at the North Pole. Magnetic North is where our compass points which is towards the Magnetic North Pole; how much variation and whether it lies to the East or West depends on where we are in the world. Variation is shown on the Compass Rose on your Chart.
What is Compass Deviation?
Deviation is created by items such as electronics, metal work, engines etc on your boat all of which create their own magnetic field which in-turn effects your compass. Every vessel will therefore have its own unique variation which can be calculated and logged on a Deviation Table.
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187568 | https://books.core-econ.org/the-economy/macroeconomics/03-aggregate-demand-08-government-net-exports.html | The Economy 2.0
Macroeconomics
Unit 3 Aggregate demand and the multiplier model
3.8 The multiplier model: Including the government and net exports
In this section, we add the government and the rest of the world to the model, and provide a more detailed model of investment. As before, we assume that firms are willing to supply any amount of goods demanded, so that in equilibrium:
output=aggregate demandY=AD
When we include the government and interactions with the rest of the world through exports and imports, aggregate demand can be split into these components:
aggregate demand=consumption+planned investment+government spending+net exports
To understand the aggregate demand function, it is useful to go through each component in turn:
Consumption
To simplify the multiplier model, we ignore the role of transfers such as unemployment benefits. To include them, you can interpret the tax rate as net of income-related transfers, and add an extra term for non income-related transfers to disposable income.
Our model of household consumption is essentially as set out previously in Section 3.6, but now depends on post-tax income. The government charges a tax, t, which we assume is proportional to income. The income left after the payment of tax, (1−t)Y, is called disposable income. The marginal propensity to consume, c1, is now the fraction of disposable income (not pre-tax income) consumed. This means that in the aggregate consumption function:
Spending on consumption is written as: C = c0+c1(1−t)Y.
All of the influences on consumption, apart from current disposable income, are again included in autonomous consumption, c0, and will therefore shift the consumption function in the multiplier diagram. We explore these other determinants in more detail in Section 3.9.
Investment
interest rate, rate of interest
: The price of bringing buying power forward in time, by borrowing; it is the additional amount that the borrower promises to repay; the rate of interest is the amount of interest to be repaid per period, as a proportion of the loan. See also: nominal interest rate, real interest rate.
The determinants of investment are explored further in Sections 3.11 and 3.12, and, in more depth, in Unit 5. At this stage, we introduce just one additional factor: namely a role for the interest rateinterest rate, rate of interest The price of bringing buying power forward in time, by borrowing; it is the additional amount that the borrower promises to repay; the rate of interest is the amount of interest to be repaid per period, as a proportion of the loan. See also: nominal interest rate, real interest rate.. We therefore assume that aggregate investment is given by
I=a0−a1r
Since the price level in the economy changes over time, we need to distinguish between real and nominal GDP. Similarly, inflation affects the cost of borrowing and in Section 5.2 we explain the distinction between real and nominal interest rates. When we refer to the interest rate, r, in this unit, we mean the real interest rate.
autonomous investment
: In a model of investment demand, autonomous investment is planned investment expenditure that does not depend on other variables in the model (such as income, or the interest rate).
where a0 is autonomous investmentautonomous investment In a model of investment demand, autonomous investment is planned investment expenditure that does not depend on other variables in the model (such as income, or the interest rate). and a1 reflects how sensitive investment is to the interest rate.
Ceteris paribus, we assume that a higher interest rate reduces investment spending. At this stage, we simply focus on the basic intuition that a higher cost of borrowing for firms will make them less willing to borrow the funds they need to undertake investment. We examine this process in more detail in Unit 5.
All of the influences on investment apart from the interest rate are included in the autonomous investment, a0. For example, higher expected profits will increase investment spending, which we capture by an increase in a0. Government investment is included in a0.
Government spending
Much of government spending (excluding transfers) is on general public services, health, and education. Government spending does not change in a systematic way with changes in income. It is referred to as exogenous to the model.
An increase in government spending shifts the aggregate demand curve up in the multiplier diagram.
Net exports
marginal propensity to import
: The change in total imports when aggregate income changes by one unit.
The home economy sells goods and services abroad, which are its exports. The amount of foreign goods the home economy demands (its imports) will depend on domestic incomes. In the multiplier model, it is assumed that imports only depend on the level of income. The fraction of each additional unit of income that is spent on imports is termed the marginal propensity to importmarginal propensity to import The change in total imports when aggregate income changes by one unit. (m), which is between 0 and 1. So we have:
net exports=X−M=X−mY
We also ignore the fact that imports can be driven by exports, which is the case for export-oriented countries that buy their components from suppliers abroad. An increase in exports would then be accompanied by an increase in imports.
If a country’s costs of production fall so that it can sell its goods at a lower price on world markets compared to the prices of other countries, the demand for its exports will increase, and domestic demand for imports will fall. The exchange rate affects the prices of a country’s goods on world markets. Growth in world markets also increases exports. However, for now, we will ignore these effects and assume that imports depend only on income and that exports, like government spending, are exogenous (not explained by the model).
The aggregate demand curve
Putting together each of the components of aggregate demand, we have:
\text{AD} = \underbrace{c_0 + c_1(1−t)Y}_{\text{consumption}} + \underbrace{a_0 − a_1r}_{\text{investment}} + G \ + \underbrace{X − mY}_{\text{net exports}}
As in the previous section, aggregate demand depends on income, Y, and we can draw it in a diagram with Y on the horizontal axis. We call this graph ‘the aggregate demand function’ or the ‘aggregate demand curve’, although in our model it is a straight line.
Changes in autonomous consumption, c_0, or government spending, G, lead to a parallel shift in the aggregate demand curve as we analysed before. Likewise, an increase in exports, X, or autonomous investment, a_0, would shift the AD curve upward.
Now, the aggregate demand also depends on the interest rate, r. When we draw the diagram of the AD curve, the level of the interest rate determines how high or low AD is at each level of income. We assume that an increase in the interest rate, r, will reduce investment, and hence would cause a parallel downward shift of the AD curve; a decrease in r would shift it upward.
The multiplier
Previously, the slope of the aggregate demand curve was c_1, the marginal propensity to consume. Now the slope also depends on the tax rate, t, and the marginal propensity to import, m—so they change the multiplier. Exports and government spending, in contrast, are now additional components of autonomous demand.
Both taxes and imports reduce the size of the multiplier. Recall that the multiplier tells us the amount by which an increase in spending (such as a rise in autonomous consumption, investment, government spending, or exports) raises GDP in the economy. When we include taxation and imports in the model, the indirect multiplier effect of a given rise in spending on GDP is smaller. This is because some household income goes straight to the government as taxation, and some is used to buy goods and services produced abroad. But in the model, we assume that the government does not increase its spending when taxes go up, and foreign buyers do not import more of our goods when we buy more of theirs. So some of the initial increase in income resulting from a demand shock does not lead to further indirect income increases in the domestic economy. The result is to reduce the indirect effects on aggregate demand, output, and employment.
Follow the steps in Figure 3.16 to understand how different components of the aggregate demand equation affect the AD curve.
Fullscreen
Goods market equilibrium
The economy is in goods market equilibrium (point D).
Fullscreen
A rise in G
An increase in government spending from G to G′ shifts the aggregate demand curve upwards. The new goods market equilibrium (D_\text{new}G) is at a higher level of output.
Fullscreen
A rise in r
A higher interest rate (r′ instead of r) decreases investment spending, shifting the aggregate demand curve downwards. The amount that the curve shifts depends on a_1, which reflects how sensitive investment is to a change in the interest rate. The new goods market equilibrium (D_\text{new}r) is at a lower level of output.
Fullscreen
A fall in t
A fall in the tax rate (from t to t′) increases the size of the multiplier, because more household income is available to spend on goods and services. Ceteris paribus, the aggregate demand curve is steeper, meaning that the indirect effects of any changes in the intercept (c_0, a_0, G, X, r) on output and employment are now larger. The new goods market equilibrium (D_\text{new}t) is at a higher level of output.
Fullscreen
A rise in m
A rise in the marginal propensity to import decreases the size of the multiplier, because less household income is spent on domestic goods and services. Ceteris paribus, the aggregate demand curve is now flatter, meaning that the indirect effects of any changes in the intercept (c_0, a_0, G, X, r) on output and employment are now smaller. The new goods market equilibrium (D_\text{new}m) is at a lower level of output.
Figure 3.16
Changes in the AD curve.
To summarize:
A higher marginal propensity to import reduces the size of the multiplier: This makes the aggregate demand curve flatter.
An increase in the tax rate reduces the size of the multiplier: This also makes the aggregate demand curve flatter.
An increase in exports or government spending shifts the aggregate demand curve up in the multiplier diagram.
Calculating the multiplier in an economy with a government and foreign trade
To calculate the multiplier in the full model, we can again use the fact that there is equilibrium in the goods market when output is equal to aggregate demand. (Equilibrium is where the aggregate demand curve crosses the 45-degree line in the multiplier diagram.) The aggregate demand equation can be rearranged to solve for output, and consequently, the multiplier:
\begin{align}
\text{output} &= \text{consumption} \
&+ \text{planned investment} \
&+ \text{government spending} \
&+ \text{net exports}
\end{align}
Therefore:
\begin{align}
\text{AD} &= C + I(r) + G + X − M \
&= c_0 + c_1(1-t)Y + a_0 − a_1 r + G + X − mY
\end{align}
Here we have written I(r) as a reminder that investment is a function of r. Rearranging this equation to find Y, we get:
\begin{align}
Y &= c_0 + c_1(1−t)Y + a_0 - a_1 r + G + X − mY \
Y(1−c_1(1−t) + m) &= c_0 + a_0 − a_1 r + G + X \
Y &= \underbrace{\frac{1}{(1 − c_1(1-t) + m)}}_\text{multiplier} \times \underbrace{(c_0 + a_0 − a_1 r + G + X)}_\text{demand that doesn't depend on income} \
&= k \times (c_0 + a_0 − a_1 r + G + X)
\end{align}
So, in the full model, the multiplier is equal to:
k=\frac{1}{1−c_1(1−t)+m}
The multiplier is smaller when we introduce the government and foreign trade:
\begin{align}
\frac{1}{(1−c_1(1−t) + m)} < \frac{1}{(1−c_1)}
\end{align}
The reason is that the denominator on the left-hand side is larger than that on the right:
\begin{align}
1−c_1(1-t) + m > (1−c_1)
\end{align}
To illustrate, suppose as before that c_1=0.6, but now the tax rate is 20% (0.2) and the marginal propensity to import is 0.1. Substituting these numbers in the formula for the multiplier, we find that k = 1.6, compared to 2.5 without including taxation and imports.
In Unit 5, we investigate how economists have estimated the size of the multiplier from data, why their estimates differ, and why it matters.
Exercise 3.5 The multiplier model
Consider the multiplier model discussed in this section.
Compare two economies, which differ only in their share of credit constrained households but are identical otherwise. In which economy is the multiplier larger? Illustrate your answer using a diagram.
On the basis of your comparison of the two economies, would you expect the multiplier in an economy to vary over its business cycle?
3.7 The multiplier model: Aggregate demand shocks cause business cycle fluctuations
3.9 Why is consumption relatively smooth?
The Economy 2.0
Macroeconomics |
187569 | https://www.rd.com/article/lateral-thinking-puzzles/ | Published Time: 2021-07-15T15:29:59Z
50 Lateral-Thinking Puzzles (with Answers) to Challenge Yourself
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50 Lateral-Thinking Puzzles That Are Harder Than They Seem
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ByHedy Phillips
Updated on Jul. 25, 2025
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These lateral-thinking puzzles require some serious out-of-the-box thinking to solve. Are you up to the challenge?
There are riddles, and then there are riddles. Lateral-thinking puzzles fall into the latter category. Among the trickiest riddles out there, these brain games will give your noggin a serious workout. That’s because they’re not as clear-cut as your average riddle—in fact, they require a significant amount of inventiveness to get you to your answer. They might even have multiple answers!
We’re not going to lie: These riddles are challengingand require a bit more creativity than your average tricky brainteasers. But they’re also a whole lot of fun. Ahead, learn exactly what lateral-thinking puzzles are and how to solve them. Then, of course, put yourself to the test and see how many of these puzzles you can figure out. Ready, set … go!
Get Reader’s Digest’sRead Up newsletterfor more brainteasers, humor, cleaning, travel, tech and fun facts all week long.
What is lateral thinking?
Lateral thinking is a creative method of problem-solving in which you look for an out-of-the-box solution. Instead of evaluating the problem (or puzzle, in this case) in a linear way, you come at it with unique angles to make the less-obvious connection. This doesn’t mean your thinking is completely random—you’re making creative connections based on the facts.
The father of lateral thinking, Maltese physician, psychologist and inventor Edward de Bono, believed this method would help people explore more possibilities for outcomes rather than simply accepting things the way they expected them to be. This would ultimately make them more critical thinkers and better problem solvers.
What’s the difference between lateral thinking and vertical thinking?
When you engage in vertical thinking, you’re using conventional logic to go straight for the “right” answer. With lateral thinking, you approach the problem from alternative angles, coming up with a solution that is more complex and less straightforward. There are a few ways you could come up with a “correct” answer—and, as noted earlier, there might even be more than one plausible answer. These solutions are more about creativity than correctness.
How to work on a lateral-thinking puzzle
When you’re working on lateral-thinking puzzles, look at the scenario presented and try to find context clues. Sometimes, the most correct answer is actually the most obvious, which is why you don’t land there immediately. At other times, the correct answer is really hard to find because you have to create the rest of the scenario yourself and fill in the blanks.
We pulled together some lateral-thinking puzzles—a mix of classic setups and brand-new challenges—that will demonstrate how these riddles work. See how many you can get right. No peeking!
Classic lateral-thinking puzzles
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1. A man who lives in a 30-story building decides to jump out of his window. He survives the fall with no injuries. How did that happen?Answer: He may live in a 30-story building, but he jumped out a first-floor window.
2. Laura is restrained all night long, with her hands pinned to her sides, and cries out occasionally, while someone watches her on a video camera. No one is alarmed, and Laura is happy in the morning. Why?Answer: Laura is a baby who is swaddled, and her mother is watching her via a baby monitor. Though she cries out a few times during the night, she goes back to sleep and is well rested in the morning.
3. A man is condemned to death. He has to choose from three rooms to accept his punishment. The first room has a firing squad with guns loaded. The second room has a blazing fire. The third room is full of tigers that haven’t eaten for six months. Which room should he choose?Answer: The room full of tigers. If the tigers haven’t eaten for six months, they’ll be dead.
4. Three-year-old Lily goes missing, and her parents are devastated. They desperately try to find her, searching everywhere and putting up flyers asking for her safe return, but the FBI and even local law enforcement refuse to help. Why?Answer: Lily is a dog.
5. A woman lives on the 30th floor and hates taking the stairs. Every day, she takes the elevator down to the lobby floor to go to work. When she comes home from work, she takes the elevator to the 20th floor and walks the rest of the way up, except on days when it rains. Those days, she takes the elevator all the way up. Why does she walk the last 10 flights of stairs if she hates it so much?Answer: The woman is too petite to reach the button for the 30th floor. She can only reach the 20th-floor button. On days when it rains, she uses her umbrella to hit the button for the 30th floor.
6. A man pushes his car until he reaches a hotel. When he arrives, he realizes he’s bankrupt. What happened?Ans wer: He’s playing Monopoly, and his piece is the car. He lands on a space with a hotel and doesn’t have the money to pay the fee.
7. A woman opened a door, screamed and was found dead a few minutes later. No gunshots were heard in the area. What happened to her?Answer: She was on an airplane, opened the door while the plane was in flight and fell to her death.
8. Sam wouldn’t stop playing video games, despite his father’s pleas. To keep Sam from playing video games all the time, his dad grabbed a hammer and solved the problem. Now he can play video games, but Sam cannot. What did the dad do?Answer: The dad built a shelf out of Sam’s reach and put the video-game console up there. The dad can still reach it to play, but Sam cannot.
9. After a night of partying with her friends, a woman arrives home and finds that she cannot enter the house. She’s sure that she’s at the right home, but she can’t get inside. Why not?Answer: Her friends took away her keys because she’d had too much to drink. They called her a cab to get home and kept her keys, which meant she didn’t have her house key.
10. Every two weeks, a woman sits down and writes two words on 60 sheets of paper. Why does she do this?Answer: The woman owns her own business with 60 employees. Every week, she signs her name on their paychecks.
11. A man is stranded on a deserted island. There are no means of transportation available to him anywhere, yet he somehow manages to escape without swimming away or anyone coming to the island to rescue him. How did he do it?Answer: When the tide went out, it exposed a massive sandbar that led to a populated island, which he was then able to walk to.
12. A woman walked up to a man behind a counter and handed him a book. He looked at it and said, “That will be $4.” She paid him and walked out without the book. He saw her leave but did not call her back. Why?Answer: She was returning an overdue library book.
13. A man and his wife are on the side of the road in their parked car when a stranger suddenly enters the vehicle. However, the couple is not alarmed. Why?Answer: The woman is pregnant and gives birth in the car. The stranger is the baby, and the parents are happy to meet their new bundle of joy.
14. A man walks into a bar and asks the bartender for a glass of water. The bartender pulls out a gun and points it at the man. The man says, “Thank you,” and walks out.Answer: The man had hiccups and the gun scared them out of him, to which he said, “Thank you.”
15. John goes to the same woman every day for advice. He trusts her implicitly and always follows her directions, but he has no intention of ever meeting her, and he never asks her about herself. What’s stopping him from developing a more meaningful relationship with her?Answer: The woman is not a real person. She’s the voice on his car’s GPS.
Hard lateral-thinking puzzles
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16. A woman married 10 different men over the course of her lifetime, but she never got divorced, she wasn’t widowed and she wasn’t a polygamist. How is this possible?Answer: She was an actress, and she “married” 10 different men while playing different roles.
17. A man is racing through the streets at 200 mph. He passes men, women and even children. Cops see him go by, and yet he never gets a ticket. How?Answer: He’s a driver in the Monaco Grand Prix.
18. A woman books a flight to London in the morning, makes reservations in New York for lunch and buys Cirque de Soleil tickets in Vegas for an 8 p.m. show—all for the same day. How will she make it to each of these places in time?Answer: She won’t. She’s a travel agent booking reservations for three different clients.
19. A man stepped on top of a diamond, which caused his leg to break. What happened to him?Answer: He was a skier on a black diamond slope who broke his leg in a crash.
20. A man isn’t alarmed to see sharks swarming around him, despite there being blood in the water. How did he survive this encounter?Answer: He was in a shark-diving cage, and the company that organized the experience put blood in the water to attract the sharks.
21. Tom was walking through the woods when all of a sudden, he was shot repeatedly. He survived the attack without serious injury. How is that possible?Answer: He was playing paintball.
22. Loud voices were heard coming from a room behind a door. However, when the room was checked, there was only one person sitting inside quietly. This person was completely sane, so what’s going on?Answer: The person was watching TV.
23. If you were alone in a dark room, with only one match and an oil lamp, a fireplace and a candle to choose from, which would you light first?Answer: The match. You can’t light any of the rest of the objects without first lighting the match.
24. Joe was born in 1955 and died in 1980 at the age of 75. How can this be?Answer: He was born in hospital room 1955 and died in hospital room 1980.
25. A woman enters a house that isn’t hers. She doesn’t live there and isn’t related to anyone who does. Though the homeowners are home, they say nothing when she leaves with some of their stuff. Who is she, and what did she take?Answer: She’s the housecleaner, and she removed the trash.
26. Two men enter a bar. They are served identical drinks. After taking a few sips, one lives and the other dies. What happened?Answer: One of them is deathly allergic to an ingredient in the drink, and the other is not.
27. A man is driving down the highway at 65 miles per hour in a 65 mph zone. He passes three cars going 60 miles per hour, then gets pulled over by a police officer and is given a ticket. Why?Answer: He’s driving the wrong way down the highway.
28. You are standing in a room in front of three light switches in the off position. They are connected to three lightbulbs behind a door in another room. You can do anything you want with the switches, but only one can remain on at a time. When you walk into the next room, you need to know which switch is connected to which bulb. How can you do that?Answer: Turn on the first switch and leave it on for five minutes. Then turn it off and turn on the next switch, and go into the other room. You now know that the second switch turned on the light that is currently on. The first switch you flipped turns on the bulb that is still warm. The switch you never flipped is connected to the bulb that is still cold.
29. A family has a chicken coop for their 12 chickens. One night, a tornado ripped through the neighborhood and sadly killed all but eight chickens. How many chickens did the family have the next morning?Answer: Still 12—four dead and eight alive.
30. Everyone hates when Susan snores, because it’s loud and annoying, but today in particular, her snoring is causing you to panic. Why?Answer: Susan is a taxi driver, and you’re her passenger.
Dark lateral-thinking puzzles
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31. A man talks in great detail about a gruesome murder he planned and carried out, but he isn’t arrested. Why not?Answer: He’s on a talk show, discussing the plot of his new book.
32.A man aimed his gun at his victim and pulled the trigger. He very quickly realized his mistake, and minutes later, the killer was dead too. What happened?Answer: The killer and the victim were on a snowy mountain, and he fired his gun near a cliff with snow. The shot triggered an avalanche that he couldn’t outrun.
33.Lucy commits a murder every single day, yet no one reports that the victims are missing. She even keeps trophies of the killings and displays them for the rest of her family to see, yet no one calls the police. Why?Answer: Lucy is a cat, and she’s killing mice and birds, then leaving them as “gifts” for her owners.
34. There are explosives in a building, and everyone is told to evacuate. The explosives detonate, and the building is destroyed, but there is no investigation or subsequent arrest. Why?Answer: It was a planned building demolition.
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187570 | https://math.stackexchange.com/questions/62908/how-can-an-ordered-pair-be-expressed-as-a-set | Skip to main content
How can an ordered pair be expressed as a set?
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My book says
(a,b)={{a},{a,b}}
I have been staring at this for a bit and it doesn't make sense to me. I have read several others posts on this, but none made any sense to me.
For example, Definition of an Ordered Pair
Based on how my ignorant brain is viewing this, I don't see why the definition could not be.
(a,b)={{a},{b},{a,b}}
aka the power set. What is the significance of the {a} in that definition? Please keep things simple if possible. Normally definitions have a valid and clear reason for being defined that way.
Clarification
First, I understand what an ordered pair is. I just don't see how the set notation says that.
Second,
(a,b)={{a},{a,b}}={{a,b},{a}}
Sets don't preserve order, but ordered pairs do. How does the third part of the equality apply to the definition?
Third, another issue with the notation that I have starts with the Product Property of Sets
Let X and Y be sets: X={a,b,c} and Y={a,d,e}.
Then X×Y={(a,a),(a,d),(a,e),(b,a),(b,d),(b,e),…,(c,e)}
If we look at the first ordered pair and our given definition we have
(a,a)={{a},{a,a}}
How can this be so, you can't have duplicates in sets? I guess what I am looking for in an answer, is not a proof or a definition of ordered pairs, but rather something like, "This notation says what it says because...". Except for the second to last point I get the terminology, I just don't get the connection between the two different uses of notation.
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edited Jun 3, 2018 at 19:54
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asked Sep 8, 2011 at 18:36
Matthew HogganMatthew Hoggan
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The important thing here is that (a,b)≠(b,a) if a≠b. In your proposed definition (a,b)={{a},{b},{a,b}}=(b,a) for all a and b.
– MartianInvader
Commented
Sep 8, 2011 at 18:48
Hmmm, ignorance is frustrating as hell. So lets forget my definition. If I could only ask my question correctly, I am sure I would get the answer that helped me the most. Let me ponder on what has been said, I will try and rephrase my question in a bit.
– Matthew Hoggan
Commented
Sep 8, 2011 at 18:53
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It has nothing to do with the answer, but your (a,b) is not the power set. Poor ∅ has been neglected, an easy thing to do, since it is so small.
– André Nicolas
Commented
Sep 8, 2011 at 18:54
@Matthew, since I already gave an answer (and won't be able to edit it soon for quite some time), please do not completely edit away the question. If you are planning on a completely different question, I suggest you may ask a new one.
– Asaf Karagila
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Sep 8, 2011 at 19:06
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– Filippo
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Whatever it is we define "(a,b)" to be as a set, what we really want is the following "defining property":
(a,b)=(c,d) if and only if a=c and b=d.
There are many ways to achieve this, but this is what we really want to achieve; once we achieve this via some definition, we want to avoid using the actual "guts" of the definition and stick exclusively to that defining property. (Similar to the points made in this answer and comment about how to represent the real numbers as sets).
One way to achieve this "defining property" is via the Kuratowski definition, by defining
(a,b)={{a},{a,b}}.
We can prove that this is a set, and that this set has the property we want.
There are other ways of achieving the same result; for example, Wiener proposed
(a,b)={{{a},∅},{{b}}},
which also has the "defining property".
The problem with your proposal is that it does not have the defining property we want for ordered pairs: for example, ∅≠{∅}, so we want (∅,{∅})≠({∅},∅). But in your proposal, we have:
(∅,{∅})({∅},∅)={{∅},{{∅}},{∅,{∅}}},={{{∅}},{∅},{{∅},∅}};
so that (∅,{∅})=({∅},∅).
So the proposal, while a perfectly fine definition of a set, does not achieve the ultimate purpose of defining the ordered pair, and so it should not be the definition of "ordered pair".
Proof that the Kuratowski definition has the "defining property".
If a=c and b=d, then
(a,b)={{a},{a,b}}={{c},{c,d}}=(c,d).
Assume conversely that (a,b)=(c,d). Then
⋂(a,b)=⋂(c,d). Since
⋂(a,b)=⋂{{a},{a,b}}={a}∩{a,b}={a}
and
⋂(c,d)=⋂{{c},{c,d}}={c}∩{c,d}={c},
we conclude that a=c.
If b=a, then (a,b)={{a}}=(c,d)={{c},{c,d}}. Therefore, {c,d}∈{{a}}, so d∈{a}, hence d=a=b and we conclude d=b, as desired. Symmetrically, if c=d, then {a,b}∈(a,b)=(c,d)={{c}}, so {a,b}={c}, hence b=c=d and we again conclude d=b as desired.
If b≠a and c≠d, then ⋃(a,b)−⋂(a,b)=⋃(c,d)−⋂(c,d). Since
⋃(a,b)−⋂(a,b)=⋃{{a},{a,b}}−{a}=({a}∪{a,b})−{a}={a,b}−{a}={b}
and
⋃(c,d)−⋂(c,d)=⋃{{c},{c,d}}−{c}=({c}∪{c,d})−{c}={c,d}−{c}={d}
(where we've used that a≠b to conclude that {a,b}−{a}={b} and we've used c≠d to conclude {c,d}−{c}={d}), then we have {b}={d}, hence b=d, again as desired.
Thus, if (a,b)=(c,d), then a=c and b=d.
Addressing the comments added to the question.
This definition is part of a way to try to define a lot of the things that we use in mathematics on the basis of an axiomatic theory; in this case, we start with Axiomatic Set Theory, where the only notions we have (if we are working in Zermelo-Fraenkel Set Theory) are "set" and "is and element of", together with the axioms that tells us properties of sets and things we can do with sets.
We want to have something that works like what we know as "the ordered pair"; but all we have to work with are sets. So we need to find a way of constructing a set that has the properties we want for the ordered pair.
For a metaphor: the ordered pair is like a car; we know how to drive. But in order to actually have a car, there needs to be an engine and gasoline, and the engine has to work. We are trying to construct that engine so that we can later drive it.
So this is not notation, this is a definition of what the ordered pair is in set theory. We are defining an object, which we call "(a,b)", to be the given set. It's not merely how we are writing the ordered pair, is what the ordered pair is if you are interested in actually seeing the engine of the car working. We know what we want "ordered pair" to behave like, but we have to actually construct an object that behaves that way. This is a way of defining an object that does behave that way.
There aren't "two notations" here. We define "the ordered pair with first component a and second component b" to be the set
{{a},{a,b}},
(which one can prove is indeed a set using the Axioms of Set Theory, if a and b are already in the theory).
Then we prove that "the ordered pair with first component a and second component b" is equal to "the ordered pair with first component c and second component d" if and only if a=c and b=d.
Then we abbreviate "the ordered pair with first component a and second component b" by writing "(a,b)" (or sometimes "⟨a,b⟩").
"(a,b)" is notation. The other side is the definition of this set.
The definition is the way it is because it works; that's really all we care about. In fact, we forget about the definition pretty much as soon as we can, and simply use the (a,b) and the "defining property." We can do that, because we know that "under the hood" there actually is an engine that does what we need it to do, even if we don't see it working while we are driving the car.
So, there is only one bit of notation, and it's "(a,b)". The other side is the definition of what that notation actually is.
The "set notation" doesn't "say" the ordered pair is what you think it is. What we are doing is defining what an ordered pair is, in a theory where the only thing we have are sets and the axioms of set theory. Because sets don't respect order, we cannot rely on simply how we write something; in order to be able to define an ordered pair we need to give a purely set-theoretic definition that actually achieves the purpose we want. Kuratowski's definition of an ordered pair (a,b) to be the set given by {{a},{a,b}} achieves this objective, in that the defined object has precisely the property we want an "ordered-pair-whatever-it-may-actually-be" to have. Since this set has that property, we define that set to be what the ordered pair "really is". But we don't actually care about what an ordered pair "really is", we just care about its desired "defining property".
In order for your car to work, there has to be an engine somewhere; but once there is an engine and your car works, you don't need to see the engine working in order to drive the car. The same with the ordered pair: for us to have an "ordered pair" in set theory, we need to be able to construct it somehow using sets. Once we have managed to do that, we don't need to see the actual set, we can just use the fact that there is a set that achieves our desired goal.
2. Yes, two sets are equal if and only if they have the same elements. So {{a},{a,b}}={{a,b},{a}}. This does not matter. What matters is that {{a},{a,b}}={{c},{c,d}} if and only if a=c and b=d, because that's what we are going for. The definition of ordered pair by Kuratowski is specifically designed so that the end result "encodes" an order and distinguishes between the "first component" and the "second component" of (a,b). The definition actually achieves this, as I proved above.
3. There is no problem with "duplicate elements". It's just that the set {{a},{a}} is equal to the set {{a}} by the Axiom of Extension, which says that two sets A and B are equal if and only if for every x, x∈A↔x∈B. The ordered pair (a,a), as a set, is a set which can be written as
{{a},{a,a}} or as {{a},{a}} or as {{a}}.
There is no problem with this, because that set has the property that (c,d) is equal to {{a}} if and only if c=d=a, which is exactly what we want.
Again: the whole point of this definition is only that it satisfies the property
{{a},{a,b}}={{c},{c,d}} if and only if a=c and b=d.
Once we have this property, we abbreviate the set {{a},{a,b}} as (a,b), and simply use the property listed above.
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Arturo MagidinArturo Magidin
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This,to me,is the argument I keep having with people over foundations of mathematics: Axiomatic set theory allows us to be very precise about the ontology of mathematics in a way that category theory without set theory cannot give us. Kuratowski's definition of an ordered pair-and the resulting definition of a function-which was proposed by Giuseppe Peano in a little known paper in 1911-are perfect examples.
– Mathemagician1234
Commented
Sep 9, 2011 at 3:27
But then if we were going the other way, i.e., starting with this Kuratowski definition of an ordered pair, then asked to deduce that it implies ordered pairs - yeah, good luck. I suppose it's possible. Some "cheat" by imposing a "tagged union" convention of just saying we'll have two subsets, the singleton will identify the first element. But that's a deus ex machina kinda thing, IMHO.
– 147pm
Commented
Jan 27, 2022 at 18:39
@147pm : I do not understand what "going the other way" means here. In any case, we knew what ordered pairs needed to be. The Kuratowski definition is a way to reify that notion. There is no deus ex machina because there was a purpose in the definition. It's not like anyone gave a definition and then wondered what it could possibly be good for. There are multiple ways of defining an ordered pair as a set, and all of them start from the viewpoint of "we need a set that will have the following properties." They are all attempts at a solution, not random constructions.
– Arturo Magidin
Commented
Jan 27, 2022 at 18:44
@147pm: In any case, one can certainly start with the definition and prove that {{a},{a,b}}={{c},{c,d}} if and only if a=c and b=d. Heck, that's how you verify the definition does what you want it to do. And you prove it like I did above: c is the "first element" if and only if x∈∩{{a},{a,b}}, and y is the second element if and only if y∈∪{{a},{a,b}}−∩{{a},{a,b}}; no cheats or conventions, but standard set operations.
– Arturo Magidin
Commented
Jan 27, 2022 at 18:47
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You want to be able to "decode" the set {{a},{a,b}} into ⟨a,b⟩.
We do this by saying "The right coordinate is the one which appears only in one of the elements in {{a},{a,b}}, and the left coordinate is the other one, if exists; otherwise it is ⟨a,a⟩ and a=b".
Edit: Since an ordered pair is a pair of elements in which the position within the pair matters, we refer to a in ⟨a,b⟩ as "The left coordinate" and we refer to b as "The right coordinate". This is a matter of terminology only, to allows us to distinguish which element is in each of the entries of the ordered pair.
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Is this definition only to be interpreted based on the (en.wikipedia.org/wiki/Cartesian_coordinate_system)?
– Matthew Hoggan
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Sep 8, 2011 at 19:06
@Matthew: I have added to my answer to explain on this question (which I think is very good).
– Asaf Karagila
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You can also think a pair is a set of two elements where the order is specified with identifying the first element.
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answered Sep 8, 2011 at 20:45
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What are the last two digits of (2^2012)?
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Sridhar Ramesh
Mathematician/Logician/All-Around Great Guy · Upvoted by
Vladimir Novakovski
, silver medals, IOI 2001 and IPhO 2001 · Author has 954 answers and 6.7M answer views
·12y
Anon User is correct. However, we can simplify the cycle-finding.
A number's last two digits are determined by its value modulo 25 and its value modulo 4.
2^10 = 1024, which is -1 modulo 25. Thus, 2^2012 = (2^10)^201 2^2 = (-1)^201 4 = -4 modulo 25.
2^2 = 4, and thus 2^2012 is a multiple of 4.
So 2^2012 is a multiple of 4 which is also 4 below a multiple of 25. This means it's 4 below a multiple of 100. Which makes its last two digits 96.
(For what it's worth, the Euler-Fermat/Carmichael theorems can also help you with this sort of thing, if you're interested in this sort of thing...)
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Ravi Handa
Teaching Number Theory for 10 years on www.handakafunda.com · Author has 758 answers and 11.1M answer views
·Updated 10y
96
For finding out the last two digits of an even number raised to a power, we should first try and reduce the base to a number ending in 24.
After that, we can use the property
Last two digits of 24^Odd = 24
Last two digits of 24^Even = 76
Last two digits of 2^2012
= Last two digits of 2^2 (2^2010)
= Last two digits of 4 (1024^201)
= Last two digits of 424
= 96
I have answered a bunch of very similar questions on remainders. You can get the complete list here: Remainder Theorem and related concepts for CAT Preparation
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Eugene Miya
By degree, and a journal editor. · Author has 17.8K answers and 16.1M answer views
·10y
Unix systems including OS X and most others include a couple arbitrary precision calculators (you can change input and output bases, but the default is 10). Sticking 2^2012 into bc(1) (you could use dc(1)) yields the last 2 digits as
96
in less time that most posters (including me) can write about this. But the really nice thing is that you get all the other digits, too.
You can also use symbolic math packages like Mathematica and Macsyma or Maple.
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Compute 2^2012 mod 100. Use CRT with mod 4 and mod 25.
1) Mod 4: 2^2 = 0 (mod 4) for exponents ≥2, so 2^2012 ≡ 0 (mod 4).
2) Mod 25: φ(25)=20, so 2^20 ≡ 1 (mod 25). Reduce exponent: 2012 ≡ 2012 mod 20 = 12. Thus 2^2012 ≡ 2^12 (mod 25).
2^5 = 32 ≡ 7; 2^10 ≡ 7^2 = 49 ≡ -1 ≡ 24 (mod 25). Then 2^12 = 2^10·2^2 ≡ 24·4 = 96 ≡ 21 (mod 25).
So we need x ≡ 0 (mod 4) and x ≡ 21 (mod 25).
Solve x = 25k + 21 ≡ 0 (mod 4). Reduce 25k+21 ≡ (1)k + 1 (mod 4) because 25≡1, 21≡1. So k+1 ≡ 0 ⇒ k ≡ 3 (mod 4). Take k=3 → x = 25·3+21 = 96. Thus 2^2012 ≡ 96 (mod 100).
Answer: 96.
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Utsav Dey
Information technology in Bachelor of Technology Degrees, Jadavpur University (Expected 2026) · Author has 726 answers and 1.8M answer views
·Updated 7y
Last two digits are the residues of modulus 100.
So 2 2012=4×2 2010≡0 mod 4.2 2012=4×2 2010≡0 mod 4.
Then residue of modulus 25.
2 2010=(2 10)201≡(−1)201=−1≡24 mod 25.2 2010=(2 10)201≡(−1)201=−1≡24 mod 25.
We use an idea of divisibilty
N=100 p+R.N=100 p+R.
N=25×4×p+R.N=25×4×p+R.
Let N=m×n N=m×n.
m=4 p 1+r 1.m=4 p 1+r 1.
n=25 p 2+r 2.n=25 p 2+r 2.
So 100 p+R=(4 p 1+r 1)(25 p 2+r 2)=100 p 1 p 2+4 p 1 r 2+25 p 2 r 1+r 1 r 2.100 p+R=(4 p 1+r 1)(25 p 2+r 2)=100 p 1 p 2+4 p 1 r 2+25 p 2 r 1+r 1 r 2.
Here p 1=1;r 1=0;r 2=24.p 1=1;r 1=0;r 2=24.
So on putting the values we get 100 p 1 p 2+4×1×24≡96 mod 100.■100 p 1 p 2+4×1×24≡96 mod 100.■
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Awnon Bhowmik
I know a little about elementary Number Theory · Author has 3.7K answers and 11.2M answer views
·8y
2 2012 mod 100 2 2012 mod 100
2 10=1024 2 10=1024
2 11=2048 2 11=2048
2 12=4096 2 12=4096
So, 2 12≡96 mod 100.......[i i]2 12≡96 mod 100.......[i i]
Multiplying the two equations
2 2012≡96 mod 100()()2 2012≡96 mod 100
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Harshit Anand
Loves to solve Maths · Author has 425 answers and 375.7K answer views
·Updated 4y
⇒R(2 2012 100)⇒R(2 2012 100)
⇒R(2×2×2 2010 100)⇒R(2×2×2 2010 100)
⇒R(2×2×2 2010 2×2×25)⇒R(2×2×2 2010 2×2×25)
⇒2×2×R(2 2010 25)⇒2×2×R(2 2010 25)
⇒4×R(2 2010 25)⇒4×R(2 2010 25)
By using Euler’s Theorem,
a φ(n)≡1(mod n)a φ(n)≡1(mod n)
⇒2 φ(25)≡1(mod 25)⇒2 φ(25)≡1(mod 25)
⇒2 20≡1(mod 25)⇒2 20≡1(mod 25)
⇒2 20×100+10≡2 10(mod 25)⇒2 20×100+10≡2 10(mod 25)
⇒2 2010≡2 10(mod 25)⇒2 2010≡2 10(mod 25)
\Rightarrow 4\times R\left(\dfr\Rightarrow 4\times R\left(\dfr
Continue Reading
⇒R(2 2012 100)⇒R(2 2012 100)
⇒R(2×2×2 2010 100)⇒R(2×2×2 2010 100)
⇒R(2×2×2 2010 2×2×25)⇒R(2×2×2 2010 2×2×25)
⇒2×2×R(2 2010 25)⇒2×2×R(2 2010 25)
⇒4×R(2 2010 25)⇒4×R(2 2010 25)
By using Euler’s Theorem,
a φ(n)≡1(mod n)a φ(n)≡1(mod n)
⇒2 φ(25)≡1(mod 25)⇒2 φ(25)≡1(mod 25)
⇒2 20≡1(mod 25)⇒2 20≡1(mod 25)
⇒2 20×100+10≡2 10(mod 25)⇒2 20×100+10≡2 10(mod 25)
⇒2 2010≡2 10(mod 25)⇒2 2010≡2 10(mod 25)
⇒4×R(2 10 25)⇒4×R(2 10 25)
⇒4×R(1024 25)⇒4×R(1024 25)
⇒4×R(24 25)⇒4×R(24 25)
Hence, the last two digits of 2 2012 2 2012 is 96 96. ■◼
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Anonymous
12y
96
2^2012 = 4702743327843346531257684792023785406555413307755295541156424650038338606663148805556877255952409681585954671161292647520039399263695074637520614834858611447362764355391990988828212393911912237933728849513000396586254969390956067387282105386617501858648268659022331855214372028646330845165001286531904826627857158512003420389473463697732159852582288445457571951276303394018181453096398081710541532500672782294851437286482812103000229566867586385983114831212629685065931070815832966723996956967593188669660382231909417596041166291739936952685169696107832327185839259681703639892703864986099
Continue Reading
96
2^2012 = 470274332784334653125768479202378540655541330775529554115642465003833860666314880555687725595240968158595467116129264752003939926369507463752061483485861144736276435539199098882821239391191223793372884951300039658625496939095606738728210538661750185864826865902233185521437202864633084516500128653190482662785715851200342038947346369773215985258228844545757195127630339401818145309639808171054153250067278229485143728648281210300022956686758638598311483121262968506593107081583296672399695696759318866966038223190941759604116629173993695268516969610783232718583925968170363989270386498609909150306424324096
That being said, if you wanted to do this without a computer, note that the last digit goes in a cycle 2->4->8->6->2->4, etc. This lets us find that the last digit is 6.
The last two digits go in the cycle 02->04->08->16->32->64->28->56->12->24->48->96->92->84->68->36->72->44->88->76->52->04->08 etc. If you go around 2012 times you wind up on 96, as division will show you. (There are 20 terms in the sequence.)
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9 2
Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW · Author has 8.1K answers and 2.7M answer views
·4y
4321#4321
////
2,4,8,16,32,64,28,56,12,24,48,96,92,2,4,8,16,32,64,28,56,12,24,48,96,92,
84,68,36,72,44,88,76,52,04 84,68,36,72,44,88,76,52,04
2 2≡2 22≡2 2002(mod 100),2 2≡2 22≡2 2002(mod 100),
2 2012≡2 12≡96.2 2012≡2 12≡96.
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Lai Johnny
M. Phil in Mathematics Major, The Chinese University of Hong Kong (Graduated 1985) · Author has 5.8K answers and 11.7M answer views
·4y
By Euler’s Theorem, (2, 25)=1 and ϕ(25)=20,ϕ(25)=20,
2 20≡1(mod 25)2 20≡1(mod 25)
2 2010 2 2010
=(2 20)100⋅2 10=(2 20)100⋅2 10
≡2 10(mod 25)≡2 10(mod 25)
≡2 7⋅2 3(mod 25)≡2 7⋅2 3(mod 25)
≡3⋅8(mod 25)≡3⋅8(mod 25)
≡24(mod 25)≡24(mod 25)
∴2 2010=23 k+24∴2 2010=23 k+24 for some integer k k.
Hence 2 2012=4(25 k+24)=100 k+96 2 2012=4(25 k+24)=100 k+96.
Thus the last two digits of 2 2012 2 2012 is 96.96.
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Goh Kim Tee
Former Tutor at Private (non-agency) (2008–2017) · Author has 4.2K answers and 2.4M answer views
·7y
Originally Answered: What are the last two digits of 2^2012? ·
last two digits are 96
2^10=1024=(1025–1)≡ -1 M25
2^20 ≡ 1 M25
2^2000 ≡ 1 M25
2^2010 =2^2000×2^10≡-1 M25
2^2012=2²×2^2010M25≡-4M100≡96M100
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Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
·4y
Last two digits is the remainder when 2 2012 is divided by 100 Last two digits is the remainder when 2 2012 is divided by 100
100=4×25 100=4×25
2 2012≡0(mod 4)⟹2 2012≡−4(mod 4)(1)(1)2 2012≡0(mod 4)⟹2 2012≡−4(mod 4)
(2, 25) are relatively prime. We can use Euler’s theorem(2, 25) are relatively prime. We can use Euler’s theorem
φ(25)=25(1−1 5)=25(4 5)=20 φ(25)=25(1−1 5)=25(4 5)=20
2 20≡1(mod 25)2 20≡1(mod 25)
⟹2 2000≡1(mod 25)⟹2 2000≡1(mod 25)
⟹2 2012≡2 12(mod 25)⟹2 2012≡2 12(mod 25)
⟹2 2012≡21(mod 25)⟹2 2012≡21(mod 25)
⟹2 2012≡−4(mod 25)(2)(2)⟹2 2012≡−4(mod 25)
(4, 25) are relatively prime. Hence by (1) and (2)(4, 25) are relatively prime. Hence by (1) and (2)
2 2012≡−4(mod 4×25)2 2012≡−4(mod 4×25)
⟹2 2012≡−4±⟹2 2012≡−4±
Continue Reading
Last two digits is the remainder when 2 2012 is divided by 100 Last two digits is the remainder when 2 2012 is divided by 100
100=4×25 100=4×25
2 2012≡0(mod 4)⟹2 2012≡−4(mod 4)(1)(1)2 2012≡0(mod 4)⟹2 2012≡−4(mod 4)
(2, 25) are relatively prime. We can use Euler’s theorem(2, 25) are relatively prime. We can use Euler’s theorem
φ(25)=25(1−1 5)=25(4 5)=20 φ(25)=25(1−1 5)=25(4 5)=20
2 20≡1(mod 25)2 20≡1(mod 25)
⟹2 2000≡1(mod 25)⟹2 2000≡1(mod 25)
⟹2 2012≡2 12(mod 25)⟹2 2012≡2 12(mod 25)
⟹2 2012≡21(mod 25)⟹2 2012≡21(mod 25)
⟹2 2012≡−4(mod 25)(2)(2)⟹2 2012≡−4(mod 25)
(4, 25) are relatively prime. Hence by (1) and (2)(4, 25) are relatively prime. Hence by (1) and (2)
2 2012≡−4(mod 4×25)2 2012≡−4(mod 4×25)
⟹2 2012≡−4(mod 100)⟹2 2012≡−4(mod 100)
⟹2 2012≡96(mod 100)⟹2 2012≡96(mod 100)
The number formed by the last two digits is 96 The number formed by the last two digits is 96
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9 1
Sarthak Dash
Founder at APTITUDE CLUB (2018–present) · Author has 491 answers and 2.8M answer views
·7y
Last two digits have a cyclicity of 40.
A 40 k+r=A r A 40 k+r=A r
So, Last two digits of 2 2012 2 2012
= Last two digits of 2 12 2 12 (As, 2012=40×50+12 2012=40×50+12)
= 40 96
=96 (Answer)
To know more on last two digits, you can follow this post.
Last 2 Digits
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This article is part of the Research TopicRemote Sensing of the CryosphereView all 10 articles
Spatial variability of near-surface ground temperatures in a discontinuous permafrost area in Mongolia
Khurelbaatar Temuujin1,2Avirmed Dashtseren1,3Bernd Etzelmüller4Tsogtbaatar Undrakhtsetseg1Kristoffer Aalstad4Sebastian Westermann4,5
1Institute of Geography and Geoecology, Mongolian Academy of Sciences, Ulaanbaatar, Mongolia
2Department of Environment and Forest engineering, School of Engineering and Technology, National University of Mongolia, Ulaanbaatar, Mongolia
3UNESCO Chair of Environmental Sciences in Eastern Central Asia, Mongolian Academy of Sciences, Ulaanbaatar, Mongolia
4Department of Geosciences, University of Oslo, Oslo, Norway
5Centre for Biogeochemistry in the Anthropocene, Oslo, Norway
In Central Asia, the ground thermal regime is strongly affected by the interplay between topographic factors and ecosystem properties. In this study, we investigate the governing factors of the ground thermal regime in an area in Central Mongolia, which features discontinuous permafrost and is characterized by grassland and forest ecosystems. Miniature temperature dataloggers were used to measure near-surface temperatures at c. 100 locations throughout the 6 km2 large study area, with the goal to obtain a sample of sites that can represent the variability of different topographic and ecosystem properties. Mean annual near-surface ground temperatures showed a strong variability, with differences of up to 8 K. The coldest sites were all located in forests on north-facing slopes, while the warmest sites are located on steep south-facing slopes with sparse steppe vegetation. Sites in forests show generally colder near-surface temperatures in spring, summer and fall compared to grassland sites, but they are warmer during the winter season. The altitude of the measurement sites did not play a significant role in determining the near-surface temperatures, while especially solar radiation was highly correlated. In addition, we investigated the suitability of different hyperspectral indices calculated from Sentinel-2 as predictors for annual average near-surface ground temperatures. We found that especially indices sensitive to vegetation properties, such as the Normalized Difference Vegetation Index (NDVI), show a strong correlation. The presented observations provide baseline data on the spatiotemporal patterns of the ground thermal regime which can be used to train or validate modelling and remote sensing approaches targeting the impacts of climate change.
1 Introduction
Ecosystems in Mongolia are intricately shaped by the interplay between topography and the associated radiation regime, which especially in Central Mongolia determines the small-scale pattern of forests and steppes (Dulamsuren et al., 2011). In particular, the radiation regime determines the evapotranspiration during the summer months, which has a significant impact on soil moisture and thus ecosystem dynamics (Batima et al., 2005). Another crucial factor that exerts a strong control on these ecosystems is the insulating winter snow cover, which plays a vital role in governing ground temperatures (e.g., Zhang, 2005; Rödder and Kneissel, 2012; Gisnås et al., 2014). Furthermore, in Mongolia about 30% of the area is underlain by permafrost (Jambaljav et al., 2022). While the larger-scale permafrost patterns in Mongolia are climate- and elevation-controlled, the radiation regime, snow cover, and properties of the subsurface control the smaller-scale patterns of permafrost occurrence. Permafrost strongly interacts with the structure and function of Mongolia’s ecosystems, influencing vegetation, hydrology, and soil processes (Dulamsuren et al., 2011; Zweigel et al., 2024). The transition from the Siberian boreal forest region in the north to the steppes in the central parts and finally to the Gobi Desert in the south corresponds to a change from discontinuous to sporadic permafrost and ultimately to seasonally frozen ground (Dashtseren et al., 2014). In a warming climate, the response of permafrost is often controlled by ecosystem factors which in some cases can strongly delay thawing (e.g., Shur and Jorgenson, 2007). On the other hand, climate-induced ecosystem changes, e.g., due to fires, also have the potential to initiate or accelerate permafrost thaw (Shur and Jorgenson, 2007) which highlights the importance of local-scale ecosystem interactions for permafrost dynamics in Mongolia and elsewhere.
Mongolia has experienced one of the fastest warming rates globally, with air temperatures increasing by 2.4°C over the past 60 years, approximately three times higher than the global average (IPCC, 2013; MARCC, 2014; Dashtseren et al., 2021). As a consequence, permafrost in the region is rapidly warming and thawing, and it has already disappeared from many marginal areas (Sharkhuu, 2003; Ishikawa et al., 2018; Dashtseren, 2021). However, the local-scale response of permafrost and associated ecosystem types to climate change remains highly uncertain. Therefore, it is of utmost importance to better understand the factors governing the ground thermal regime at spatial scales of tens of meters to kilometres, in particular the interplay between climatic, topographic, and ecosystem-related controls.
In-situ monitoring of ground temperatures in permafrost areas generally relies on boreholes which capture the ground thermal regime on the plot scale (Biskaborn et al., 2019). As borehole installations are logistically challenging and expensive, there are usually only a single or few boreholes measurement available within a certain area, so that the small-scale spatial variability caused by topographic and ecosystem factors remains uncaptured. Temperature measurements near the ground surface can at least partly fill this gap: as they are efficient to conduct using miniature temperature loggers, it is possible to install arrays of dozens or even hundreds of temperature loggers that can indeed capture variations of near-surface ground temperature on spatial scales of tens of meters to kilometres. Studies conducted outside Mongolia have shown that annual averages of near-surface ground temperature can vary by several Kelvin over short spatial distances, driven by environmental factors like snow depth (Gisnås et al., 2014; Zweigel et al., 2021), topographic factors like altitude, slope and aspect (Serban et al., 2023), as well as surface material and vegetation (Rödder and Kneissel, 2012; Luo et al., 2019).
In this study, we present spatially distributed measurements of near-surface ground temperatures from a 6 km2 area in Central Mongolia, which is located in the discontinuous permafrost zone. The dataset allows us to statistically evaluate the importance of different factors for both annual and seasonal temperatures, in particular topography- and ecosystem-related controls. Furthermore, we evaluate the suitability of satellite-derived indices as predictors for mean annual near-surface ground temperature. Finally, we discuss the potential to employ the dataset presented in this study as training and/or validation for both remote sensing and numerical modelling targeting the state of permafrost.
2 Study area
The Terelj climate and permafrost observatory (47°58′, 107°26′; Figure 1) is located in the southern Khentii Mountain in the discontinuous permafrost zone, close to the southern limit of permafrost occurrence in Mongolia. In the study area, the elevations vary between 2,200 m a.s.l. and 1,550 m a.s.l. The area is located in the forest-steppe zone, with forests dominated mainly by larch, birch and occasional pine trees (Batchuluun et al., 2021). At the site, the south-facing slopes are permafrost-free and occupied by steppe vegetation, while the majority of the forests occurs on north-facing slopes underlain by permafrost. Soil properties change rapidly in space; in the forest, the organic soil layer has a thickness of 0.2–0.4 m, whereas it is much thinner or even non-existent in the steppe (Ishikawa et al., 2005; Dashtseren et al., 2014).
Figure 1
Figure 1. Map of the study area (left) with the locations of the ground surface temperature logger, as well as the location of the study area within Mongolia (right).
Three automatic meteorological stations are located in the study area: one in bottom of the central valley (VS), one in the north-facing slope inside the forest (FS), and one on the south-facing slope in the mountain steppe (GS). In this study, we use observations of air temperature data (Table 1).
Table 1
Table 1. Position and site description of automatic meteorological stations.
Figure 2 shows the mean annual air temperatures (MAAT) measured at the Terelj meteorological station from 1986 to 2017, ranging from about −5°C to −2°C. Over the three shown decades, temperatures have increased by about 0.75°C which largely corresponds to the general temperature trend in Mongolia. In the study period, MAAT were relatively warm with around −3°C.
Figure 2
Figure 2. Mean annual air temperature (MAAT) measured at Terelj meteorological station. Dashed line: linear trend from 1986 to 2017.
3 Methods
3.1 Near-surface ground temperature and snow measurements
We installed miniature temperature data loggers, specifically Micro-Temperature Data Loggers (MTDs), 2–3 cm below the ground surface at 100 locations within an area of approximately 6 km2 (Figure 1). These devices, covering two valley sides, were deployed across a variety of elevations, slopes, and aspects, and included different land cover types. The Maxim iButton DS 1922L model was selected for its accuracy of ±0.5°C and a resolution of 0.0625°C, ensuring precise temperature measurements critical for our study (Thermochron iButton, 2014). The measurements were automatically conducted at 4-h temporal resolution. The loggers were installed on 1 Aug 2015, and continued until the logger malfunctioned, which generally occurred by summer 2018. For the main analysis, we concentrated on 59 points for which the two complete years 2016 and 2017 are available. Twenty-four of these points are located in rather dry steppe areas, ten are in the relatively moist grasslands in the central valley, and twenty-five points are located in forest areas (Figure 1). For the analysis, the sites are divided into two categories according to their surface type, i.e., forest and grassland.
While there can be differences between the ground surface temperature (i.e., skin temperature when the ground is not snow-covered) and the measured temperatures 2–3 cm below the ground surface for individual points in time, these largely cancel out for longer-term averages (e.g., Gisnås et al., 2014; Way and Lewkovicz, 2018), making it possible to derive long-term averages of ground surface temperature from our measurements. In line with previous studies using similar measurement setups (e.g., Luo et al., 2019; Zweigel et al., 2021), we therefore use the well-established term “MAGST” (Mean Annual Ground Surface Temperature), when referring to annual averages of near-surface ground temperature derived from our measurements. In all other cases, we use the term “near-surface ground temperature” throughout the manuscript.
A survey with local residents was conducted to establish the start and end dates of a stable winter snow cover, and the results were cross-checked by comparing them to the records of the near-surface ground temperature (Klein et al., 2016). Additionally, in situ measurements of the snow depth and density were conducted on 20 February 2017, at various locations with MTDs. The date of the snow survey was selected to approximately coincide with the annual maximum of snow depths in the study area (Sturm et al., 2010, Sturm and Liston, 2021). In addition, air temperature data from meteorological stations located within and close to the study area (Table 1) are used for interpretation of near-surface ground temperatures.
3.2 Satellite remote sensing data
To further characterize the study area, we use spectrally resolved late summer surface reflectance from the Sentinel-2A satellite and the global digital elevation model (GDEM) from the Advanced Spaceborne Thermal Emission and Reflection Radiometer (ASTER) onboard the Terra satellite (Table 2).
Table 2
Table 2. Satellite data used to calculate indices.
From Sentinel-2A, data, the Normalized Difference Vegetation Index (NDVI), Normalized Difference Water Index (NDWI), Normalized Difference Moisture Index (NDMI), Soil Adjustment Vegetation Index (SAVI), and Modified Soil Adjustment Vegetation Index (MSAVI), Modified Normalized Difference Water Index (MNDWI) have been calculated (Table 3) for the pixel closest to each logger site (see Table 3).
Table 3
Table 3. Indices used to analyse with MAGST (mean annual ground surface temperature).
Furthermore, we determine the elevation, slope, aspect and the potential incoming solar radiation for all logger points from the ASTER GDEM (Abrams, M., 2015) using ArcGIS (©).
3.3 Frost number
To analyse the surface offset between air and near-surface ground temperatures, we calculate thawing degree days (TDD) and frezing degree days (FDD) from air and near-surface ground temperature:
TDDS=∑Dt0¯¯¯TA
FDDS=∑Df0¯¯¯TS;
where, FDDS and TDDS are freezing and thawing degree days near the ground surface, ¯¯¯TA and ¯¯¯Ts are daily mean air and surface temperatures, D is number of days in each month, Df is freezing period and Dt is thawing period.
Furtermore, we determine the frost number Fn (also called “permafrost index”), which is a measure for the likelihood of permafrost occurrence (Smith and Riseborough, 1996):
Fn=√FDDS√FDDS+√TDDS
Frost numbers smaller than 0.5 indicate that permafrost-free conditions are likely, while frost numbers higher than 0.5 are indicative for permafrost.
4 Results
4.1 Spatial variability of near-surface ground temperatures
Of the 59 analysed sites, 25 points are located in grasslands the flat valley and south facing slopes, while 34 points are located in forested areas mainly on north-facing slopes. Figure 3 shows histograms for seasonal average near-surface ground temperature for these two main landcover types for both 2016 and 2017. In winter, near-surface ground temperatures of the “forest” points are on average warmer than the “grassland” points, while summer and spring temperatures are significantly warmer. In autumn, near-surface ground temperatures are more similar, but still on average warmer for the grassland sites. This pattern is largely consistent in both analysed years (Figure 3), as well as for the entire time series containing three winter and summer seasons (Figure 4).
Figure 3
Figure 3. Histogram of mean seasonal near-surface ground temperature of the loggers at the study area in 2016 (N=59) and 2017 (N=59). Mean winter temperature (December, January, February); mean spring temperature (March, April, May); mean summer temperature (June, July, August); mean autumn temperature (September, October, November).
Figure 4
Figure 4. Daily average near-surface ground temperature of forest (mean of 25 sites) and grassland sites (mean of 34 sites).
4.2 Controlling factors of near-surface ground temperature
Figure 5 shows the dependence of mean annual near-surface ground temperature (denoted MAGST, see Section 3) on topographic factors and snow depth. Aspect is a major control for MAGST, with south-facing aspects on average more than 4°C warmer than north-facing expositions (Figure 5), for which MAGST between −1°C and 1°C are observed. Furthermore, north-facing expositions are largely associated with forests, while grasslands dominate the south-facing expositions. In contrast, the elevation is uncorrelated with MAGST (Figure 5), which suggests that elevation-dependent differences in climate are overridden by other factors. Similarly, there is no clear dependence on the slope, especially for the forest sites. However, there is a strong positive correlation with potential incoming solar radiation (Figure 5) which is controlled by both slope and aspect, with R2 values between 0.57 and 0.64.
Figure 5
Figure 5. Correlation between topographic factors and snow depth and mean annual ground surface temperature (MAGST, see Sect. 3) of 59 points in the study area in 2016 and 2017.
For the year 2017, the data illustrate only a very moderate correlation between February snow depth and MAGST (R2 = 0.12), indicating that the impact of the snow cover on annual averages of near-surface ground temperatures is less pronounced compared to solar radiation. The measured snow height varied between 6 and 20 cm, with low snow depths generally associated with grasslands and thus higher values of MAGST (Figure 5). On the contrary, average winter (December to February) near-surface ground temperatures were strongly correlated with snow depth (R2 = 0.72, Figure 6), with low snow depths associated with low average winter near-surface ground temperatures, likely due to the thermally insulating effect of the seasonal snow cover. This shows that while the snow cover is a major control on the ground thermal regime during the winter season, its effect is largely overridden by other factors for annual averages. Table 4 presents the results of the snow cover duration analysis, showing that a stable snow cover forms more than 10 days earlier for the forested compared to the grassland sites. Furthermore, the snow disappears considerably earlier at the grassland sites, with a difference of approximately 1 month, leading an about 40 days longer snow-covered period for the sites located in forest (Table 4) which likely contributes to the overall lower MAGST values in the forested areas.
Figure 6
Figure 6. Correlation between snow depth and mean winter (December, January and February) near-surface ground temperature in 2017.
Table 4
Table 4. Stable snow cover start, melt date, and duration during the study period, based on interview data from local people.
4.3 Remote sensing indices as predictors for MAGST
Figure 7 shows the relationship between the six indices and the measured MAGST. All indices are to some extent correlated with MAGST, with R2 values between 0.45 and 0.75. The main reason for this strong correlation is their ability to distinguish between the grassland and forest which was found to be a major control for near-surface ground temperatures (Section 4.1). In general, the vegetation-related indices perform better than the soil moisture related indices, especially NDVI and SAVI for which R2 values exceeded 0.7 in both years (Figure 7). The weakest correlation was found with MNDWI (R2 = 0.45). It is worth pointing out that the best-performing satellite-derived indices outperform topography-based predictors for MAGST in our study area, although the potential incoming solar radiation shows a similar significant correlation.
Figure 7
Figure 7. MAGST (mean annual ground surface temperature, see Section 3) vs. NDWI (Normalized Difference Water Index), NDVI (Normalized Difference Vegetation Index), MSAVI (Modified Soil Adjusted Vegetation Index), MNDWI (Modified Normalized Difference Water Index), SAVI (Soil Adjusted Vegetation Index), and NDMI (Normalized Difference Moisture Index), for the years 2016 and 2017.
4.4 Implications for permafrost occurrence
Figure 8 shows the distribution of frost number (FN) values across different temperature data loggers placed in grassland and forest ecosystems. For the grassland areas, the frost numbers are exclusively below 0.5, indicating permafrost-free conditions. In contrast, the frost number values for forested areas are predominantly higher than 0.5, thus giving a high probability for permafrost presence. However, the frost numbers are generally close to 0.5 which indicates warm permafrost which may be close to thawing.
Figure 8
Figure 8. Histogram of frost number values of the loggers at the study area in 2016 and 2017 (N=59).
5 Discussion
5.1 Uncertainties and limitations
This study aims to capture the spatial variability of near-surface ground temperatures within the study area, but it is unclear to what extent the 59 analysed temperature loggers can indeed represent the true spatial distribution of near-surface ground temperature. While the measurements show a large variability of 8 K for the annual averages, it is possible that even colder and warmer sites exist, but likely only a few localized spots. With the available measurements, we have performed a correlation analysis with different topographic and environmental factors (Section 4.2), as well as satellite-derived hyperspectral indices (Section 4.3). Hereby, the coarse spatial resolution (30 m pixels) of both the employed DEM and the satellite products can be a limiting factor, as the logger represents only a single point within the footprint and factors like exposition and surface reflectance could vary at smaller spatial scales. While this can be problematic for single sites, the large number of analysed sites moderate this issue, so that we are confident that our analysis can realistically capture the correlations between key variables and the ground thermal regime. Furthermore, it must be emphasized that the different factors are not independent of each other, which makes it challenging to compare the correlation coefficients of different factors and indices. For the same reason, this study does not allow to entangle causal relationships between the ground thermal regime and ecosystem-related factors. Forests, for example, occur almost exclusively on north-facing slopes, but they also strongly affect ground temperatures (Dulamsuren et al., 2011; Dashtseren et al., 2014), so that it is not possible to separate the relative impacts of the ecosystem and the topographic controls. Here, physically-based numerical modelling studies could improve our understanding of the controlling mechanisms. For our study area, Zweigel et al. (2024) demonstrated with a land surface model that the topographic factors are likely the main control on annual average ground temperatures, while the “ecosystem control” by the forest canopy more affected the seasonal temperature amplitude of ground temperatures. The dataset presented in this study can thus be further exploited together with model approaches which rely on established process understanding to aid the interpretation of the field observations.
5.2 Relation between satellite-derived indices and near-surface ground temperature
The correlation analysis between MAGST and satellite-based indices suggests that especially indices like NDVI related to vegetation properties can be excellent predictors for the ground thermal regime. However, this finding is almost certainly restricted to the relatively small study site, or possibly similar areas in the vicinity, but it is unlikely that it can capture climate-related MAGST patterns over large spatial scales. The good correlation is likely related to the fact that basically all tested indices can very well distinguish between forest and grassland ecosystems which are strongly related to the observed near-surface ground temperatures. Nevertheless, there are clear differences in the performance of the different spectral indices, for example, the vegetation-focused indices NDVI, SAVI, and MSAVI perform better than soil moisture and evaporation-focused indices like MSAVI, NDMI, and NDWI. The widely used NDVI index has the overall best performance, on par with the SAVI index which was found to achieve a better separation in regions with low vegetation cover (Bader Almutairi and Vargas, 2013). The MSAVI index (Qi et al., 1994) performs worse than NDVI and SAVI, especially in 2016, but still better than the water-related indices, although the NDMI is developed as a measure for the amount of water contained in plants (Ajay Kumar Taloor, 2021). Surprisingly, the NDWI index has a relatively good performance although it is more suitable for mapping water bodies (Ajay Kumar Taloor, 2021). Overall, the strong performance of especially NDVI and SAVI suggests that they could be employed for upscaling purposes at least on the regional scale, possibly supported by numerical modelling or machine learning (Zweigel et al., 2024). The near-surface ground temperature observations could be further exploited to calibrate and validate remote sensing-based approaches to characterize permafrost, such as statistical models targeting ground temperatures on a continental scale (e.g., Aalto et al., 2020). In addition to multispectral satellite data, the observations presented in this study could be employed to validate and enhance satellite-based observations of land surface temperature (LST), e.g., coarse 1 km scale remote observations from MODIS (which are also employed as input for permafrost modelling, Westermann et al., 2017), or more fine-scale measurements, e.g., from Landsat (Gao, 1996).
5.3 Ground thermal regime and implications for permafrost occurrence
The warmest temperatures are recorded on south-facing slopes within the grassland/steppe ecosystem, with MAGST close to +7°C. On the other hand, the coldest MAGST of close to −1°C occurred in the forest ecosystems on north-facing slopes, thus yielding significant spatial variations of up to 8 K. Our results confirm that vegetation and solar radiation are significant in controlling near surface ground temperature (Cheng, 2004; Etzelmüller et al., 2006; Heggem et al., 2006; Kade and Walker, 2008; Dashtseren et al., 2014), while the snow depth is not a major control for MAGST in our study area, contrary to many Arctic regions (e.g., Gisnås et al., 2014). Our study shows that sites with higher snow depths are characterized by warmer winter temperatures, as expected from the insulating properties of snow (e.g., Sturm et al., 2021; Cheng, 2004). However, as high-snow sites more often occur in forest in north-facing expositions (Figure 5), the warmer winter temperatures are offset by colder spring, summer and autumn temperatures (Figure 3), thus strongly reducing the influence of snow depth on the annual mean. Furthermore, the snow cover persists around 1 month longer in the forested sites which confines near-surface ground temperatures to 0°C or below, while the snow-free grassland sites already experience positive temperatures during this period (Figure 3; Robin et al., 2024), thus contributing to the overall lower MAGST values at the forest sites.
The MAGST variability within the relatively small study area is large compared to the observed temperature increase of +2.4 K in the last 60 years, which could suggest that ecosystems can to a certain degree adapt by spreading to more suitable locations in the immediate vicinity, when conditions at their previous locations become too warm. However, this picture is clearly oversimplified, as atmospheric warming can also affect other key processes, especially precipitation and evaporation which are main controls for soil moisture conditions and ecosystems (Batima et al., 2005). This is especially true for permafrost which is largely restricted to the north-facing slopes in the study area (Jambaljav et al., 2022). The measured MAGSTs between −1 and +1°C indicate the presence of warm permafrost, considering the likely thermal offset between the ground surface and the top of the permafrost (Smith and Riseborough, 1996). This interpretation is supported by the calculated frost numbers higher than 0.5 and further corroborated by borehole measurements at the north-facing slope in the study area (Sharkhuu, 2003). Continued warming could therefore quickly lead to the loss of permafrost which may in turn affect the forest ecosystem and thus lead to a non-linear response to the warming. On the other hand, the presence of a forest canopy can change the response of the ground thermal regime to atmospheric warming (Zweigel et al., 2024). To further disentangle the interplay between climatic, topographic, and ecosystem-related factors, a close integration of field observations (as presented in this study), manipulation experiments (Batima et al., 2005), and land surface modelling (e.g., Zweigel et al., 2024) could be a promising way forward.
6 Conclusion
In this study, we present 2 years of measurements of spatially distributed near-surface ground temperatures from a 6 km2 large study area in Central Mongolia which features a range of different elevations, expositions, and ecosystem types. We analyze observations from more than 50 sites, which constitute a near-representative sample of conditions within the area, so that it is possible to analyze the governing factors of the ground thermal regime. From the study, the following conclusions can be drawn:
• The spatial variability of annual average near-surface ground temperatures is on the order of 8 K, highlighting the large range of ground thermal conditions which shape the local ecosystems.
• Altitude and winter snow depth is largely uncorrelated with annual average near-surface ground temperatures, while slope and aspect, and the related potential incoming short-wave radiation are strong controls.
• Forest ecosystems, which are almost entirely located on local north-facing slopes, feature significantly colder annual average near-surface ground temperatures than grassland ecosystems, indicating the presence of permafrost. Permafrost conditions are less likely for grassland sites, and the warmest sites are located on steep south-facing slopes in grasslands,
• In grasslands, near-surface ground temperatures are warmer in spring, summer, and fall compared to forests, while they are colder in the winter season.
• Spectral indices sensitive to vegetation, such as the Normalized Difference Vegetation Index (NDVI), are good predictors for mean annual near-surface ground temperatures, with square correlation coefficients R2 exceeding 0.7.
The study presents a new observational dataset on the spatial variability of the ground thermal regime in permafrost regions. Especially for Central Asian permafrost, which is strongly dominated by the radiation regime, such datasets are extremely rare. However, they can provide valuable information to better understand the governing factors of the ground thermal regime and assess the impacts of climate change on the spatial patterns of ecosystems and permafrost. Our findings emphasize the importance of vegetation cover and topographic factors in controlling near-surface ground temperatures and permafrost stability. The data can be further utilized in conjunction with modeling studies to improve the predictive understanding of permafrost dynamics under changing climatic conditions. Additionally, the correlation between spectral indices and ground temperatures suggests potential for upscaling in situ observational data using remote sensing techniques, thereby enhancing the spatial coverage and resolution of permafrost monitoring efforts. This can aid in developing adaptive strategies to mitigate the impacts of climate change on permafrost and associated ecosystems.
Author contributions
KT: Conceptualization, Data curation, Formal Analysis, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing–original draft, Writing–review and editing. AD: Data curation, Funding acquisition, Project administration, Resources, Writing–review and editing. BE: Writing–review and editing. TU: Writing–review and editing. KA: Writing–review and editing. SW: Conceptualization, Data curation, Formal Analysis, Funding acquisition, Methodology, Project administration, Resources, Supervision, Writing–original draft, Writing–review and editing.
Funding
The author(s) declare that financial support was received for the research, authorship, and/or publication of this article. We acknowledge funding from Permafrost4Life (Research Council of Norway, grant no. 301639) and the European Space Agency CCI+ Permafrost (grant no. 4000123681/18/I-NB).
Acknowledgments
We would like to express our sincere gratitude to all individuals and organizations that contributed to the success of this research. Firstly, we thank the local residents of Terelj for their invaluable assistance and cooperation during the field surveys and interviews. Your insights and knowledge were crucial for understanding the local climatic and environmental conditions. We are deeply grateful to the Institute of Geography and Geoecology at the Mongolian Academy of Sciences for their support and collaboration.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
Supplementary material
The Supplementary Material for this article can be found online at:
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Zweigel, R. B., Dashtseren, A., Temuujin, K., Aalstad, K., Webster, C., Stuenzi, S. M., et al. (2024). Simulating the thermal regime and surface energy balance of a permafrost-underlain forest in Mongolia. J. Geophys. Res. Earth Surf. 129, e2023JF007609. doi:10.1029/2023jf007609
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Zweigel, R. B., Westermann, S., Nitzbon, J., Langer, M., Boike, J., Etzelmüller, B., et al. (2021). Simulating snow redistribution and its effect on ground surface temperature at a High-Arctic site on Svalbard. J. Geophys. Res. Earth Surf. 126 (3), e2020JF005673. doi:10.1029/2020jf005673
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Keywords: permafrost, ground surface temperature, snow cover, topography, vegetation, remote sensing, climate change
Citation: Temuujin K, Dashtseren A, Etzelmüller B, Undrakhtsetseg T, Aalstad K and Westermann S (2024) Spatial variability of near-surface ground temperatures in a discontinuous permafrost area in Mongolia. Front. Earth Sci. 12:1456012. doi: 10.3389/feart.2024.1456012
Received: 27 June 2024; Accepted: 19 September 2024;
Published: 17 October 2024.
Edited by:
Zhongqiu Sun, Northeast Normal University, China
Reviewed by:
Xiaoying Jin, Northeast Forestry University, China
Rensheng Chen, Chinese Academy of Sciences (CAS), China
Copyright © 2024 Temuujin, Dashtseren, Etzelmüller, Undrakhtsetseg, Aalstad and Westermann. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Khurelbaatar Temuujin, temuujinkh@mas.ac.mn; Sebastian Westermann, sebastian.westermann@geo.uio.no
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. |
187573 | https://www.youtube.com/watch?v=7gMlFiJzyxU | [Math 3003] Using Fleury's Algorithm to Find the Euler Path or Circuit
Kyle Havens
230 subscribers
17 likes
Description
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Posted: 26 May 2020
Transcript:
the penultimate algorithm for this chapters what we refer to as flurries algorithm which is basically a step by step procedure for finding an Euler circuit and Euler path now I personally don't want you to necessarily focus on this algorithm because I think it's it's intuitive enough without really needing it but let's just walk through it and then again as an example we want to actually find an Euler path or Euler circuit in a graph should it exist so what flurries algorithm says to do first is to count the number of odd vertices in the graph and as we practiced right we see the relevance of that because you know the number of even and odd vertices will help us know whether it's possible or not so in the case one when you have no Odysseys it no other circuit exists by euler's circuit theorem in the case two when two odd vertices exist an Euler path exists by Euler path theorem and in case three if there's more neither exists due to both theorems combined and again one and three those are not possible cases so the this is the only thing you there's only three things that could happen and we're either gonna have an Euler circuit or a path or neither then once we've determined whether or not it has them then we can actually find it that's kind of the point of this algorithm is to actually determine the path and theorem and so basically where am i starting is the first question where in the case one if it has an Euler circuit you could start anywhere in the graph but in case two if it's an Euler path it has to start at an odd vertex which is something that's important to remember it will not work if it does not then the Flurry's algorithm says to do and this is why I think it's not particularly helpful it says to travel along the edges of your choosing without dead heading so of course we're not gonna deadhead because that's gonna defeat the whole point right the whole point is to find an answer that does not that had it and basically it's just as do what you want travel the edges when you're choosing it says don't cross a bridge unless you must and that's because a bridge technically you can't have an Euler circuit if there's a bridge but you could have an Euler path with a bridge a bridge is like you're not it's the only edge that connects the two once you cross it you can't go back and basically it it tells us to avoid a bridge but there's other there's other tips which I think are important which are not described by Flurry's Akram which I wanted to show you and I want to actually do a few examples here as well and that's what it says to do as the next example is use Flurry's algorithm to find Euler paths and Euler circuits in the prior examples so let's move back to the packet and do so okay and so in these graphs we've already determined whether they have Euler paths or circuits so that's step one we've already done that so in this first graph we know that this had two odd therefore it had an Euler path and in that case we must start at one of the odd vertices for our starting point so since and if we remember those were G and F were the two odd vertices in that graph we got to start at one of those let's just say we're gonna start at G because we don't have any other information but usually you would know where am i starting if this was a real problem and basically it says to travel along the edges of your choosing so let's just say we go left first and basically our goal is to just formulate a Oyler path basically I want to start a G and I want to end at F because I have to and it will not work otherwise and like a lot of the time we don't really have a choice of where we're going like if I'm gonna go to H first I kind of have to just travel a lot because I have to use adjacent edges there's not really much choice there's no options for where to go but in a case where I have some choice you typically can go whichever direction you want there are it is possible to box yourself into a corner though if you're not careful so my recommendation is to try to completely cover every edge that you see before you move on but usually it's not that big of a deal but anyways if you're wondering where to go next well technically you can just go wherever you want so let's just say I'm gonna move this way in this way next also usually what I recommend doing is labeling your steps that way if someone was to try to follow along with your work they can kind of tell exactly what you're doing by following along with the numbers so based on I'm gonna number each edge that I'm using and use an arrow to follow the direction that I'm traveling in this Euler path and basically I would just keep going I would be step six F would be step seven again here notice that I'm actually at F which was the end goal but I'm not done because again I have to every edge that's what an Euler path is right the definition is a path that hits every edge exactly once starting and ending at a different point but again where should I go what I would recommend trying to be thorough so maybe I'll go this way next and then I could go up to back to be on step 9 step 10 and then I'll kind of keep moving 11 12 13 and really assuming that you check to see whether I had another path or circuit it should just work and notice that I'm not done until I hit every edge so while I again I'm back to F once again I'm not going to be finished until I had every edge so I need to go up to C and back to F for the final step and that would thus finish my other path and again assuming that you check for how many odds it has first it's gonna be it should be pretty easy to determine an actual Euler path or circuit whereas it just sort of works it just sort of flows and that's the beauty of evenness if it's even it flows you don't get stuck um and when we have two odds we can start and end at different points because we're starting one and ending at the other for starting and ending odd is good if we don't want to come back and that's why two odds are necessary in an Euler path but anyways we can see that it worked because everything has been used and it makes sense and this would also be exactly the right way to show your work on an exam because I could go then and I could go and follow along your steps 1 2 3 4 5 6 7 and see does this circuit actually make sense or in this case path so looking at another example in this graph it had zero odds which we concluded would have an Euler circuit instead sorry I wrote that really poorly zero odds and by Flurry's algorithm in that case when I have no odds there's gonna be an order circuit and I can start at any vertex so I could start wherever let's just say down here I did not label them which also just before I move on notice that I could have described this circuit or this path sorry separately by just listing the excuse me listening the vertices so another way graphically I can see it I think this is the easiest way to do it you can also describe a path by listing while G a B G IJ f g b c d e f c/f and that's another way of writing it but you know it gets really long and horribly messy and it to me that's a lot harder to follow than doing it this way so this is kind of my preferred way of writing out a circuit but sometimes if you're thinking about your online homework it's gonna be hard to do this sort of thing on the on a homework and they could ask you to represent it like ABCD efg or whatever instead but it's really the same idea as a path or a circuit that we that I described in the earlier videos anyways in this case they're not labeled which as I said this is bad so if we're trying to write it that way we can't because we have not labeled the vertices which is why you want to label the vertices using letters but do as I say not as I do right anyways you could see that an Euler circuit will work just naturally here because again they're all even notice that technically there is a way to do it wrong and that's what I wanted to mention is when you're moving along randomly as Flurry's algorithm says notice that I could go back to the start here but this would be a problem because in this case I'm returning the start without really finishing the Euler circuit I have left some of it off so what this is what I recommend is to try to be complete and usually you want to avoid going back to the start until the end because once you head back to the start there is it possible you could possibly box yourself in and that's another general principle that you'll see I use when I'm solving harder problems in later guides but this would be our example of an Euler circuit and again you can label that as one two three four five six seven in this case I don't leave this one as an exercise for you to practice yourself but that one has an Euler path and it should have started and ended at the two odds |
187574 | https://www.dictionary.com/browse/cuff | CUFF Definition & Meaning | Dictionary.com
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cuff 1
noun (10)verb (used with object) (3)
cuff 2
verb (used with object) (1)noun (1)
Word History and OriginsIdioms and PhrasesExample SentencesRelated WordsMore AboutWord of the DayQuiz
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View synonyms for cuff
American British
cuff
1
[kuhf]
Phonetic (Standard) IPA
noun
a fold or band serving as a trimming or finish for the bottom of a sleeve.
a turned-up fold, as at the bottom of a trouser leg.
the part of a gauntlet or long glove that extends over the wrist.
a separate or detachable band or piece of fabric worn about the wrist, inside or outside of the sleeve.
an elasticized, ribbed, or reinforced band at the top of a sock or stocking.
a band of leather or other material, wider than a collar, sewed around the outside of the top of a shoe or boot to serve as a trimming or finish.
a handcuff.
I accessorized my costume with cuffs, a badge, and a toy gun.
8. Anatomy.rotator cuff.
Furniture. a horizontal strip of veneer used as an ornament on a leg.
Medicine/Medical. an inflatable wrap placed around the upper arm and used in conjunction with a device for recording blood pressure.
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verb (used with object)
to make a cuff or cuffs on.
to cuff a pair of trousers.
2. to put handcuffs on.
The officer was quick to cuff the suspect and read him his rights.
3. Slang. to start an exclusive relationship with.
You’ve gotta cuff her if you want to keep her.
cuff
2
[kuhf]
Phonetic (Standard) IPA
verb (used with object)
to strike with the open hand; beat; buffet.
noun
a blow with the fist or the open hand; buffet.
cuff
1
/ kʌf /
noun
the part of a sleeve nearest the hand, sometimes turned back and decorative
the part of a gauntlet or glove that extends past the wrist
Also called (in eg Britain):turn-up.the turned-up fold at the bottom of some trouser legs
informal improvised; extemporary
“Collins English Dictionary — Complete & Unabridged” 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
cuff
2
/ kʌf /
verb
(tr) to strike with an open hand
“Collins English Dictionary — Complete & Unabridged” 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
noun
a blow of this kind
“Collins English Dictionary — Complete & Unabridged” 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
Discover More
Word History and Origins
Origin ofcuff 1
First recorded in 1350–1400; Middle English cuffe “mitten”; perhaps akin to Old English cuffie “cap,” from Medieval Latin cuphia;coif 2
Origin ofcuff 2
First recorded in 1520–30; origin uncertain; perhaps from a Scandinavian language; compare Norwegian, Swedish dialect kuffa “to push, shove”; also German cant kuffen “to thrash”
Discover More
Word History and Origins
Origin ofcuff 1
C14 cuffe glove, of obscure origin
Origin ofcuff 2
C16: of obscure origin
Discover More
Idioms and Phrases
Idioms
1. on the cuff,
1. with the promise of future payment; on credit.
2. without charge; with no payment expected.
He enjoyed his meal the more because it was on the cuff.
off the cuff,
extemporaneously; on the spur of the moment.
She made those comments off the cuff, and they came back to haunt her later.
2. unofficially or informally.
I'm telling you this strictly off the cuff.
see off the cuff; on the cuff.
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Example Sentences
Examples have not been reviewed.
With Isaac still bound to the bed, Iris proposes that he stay cuffed for the remainder of the day, giving her 12 hours to persuade him of their mutual destiny.
FromSalon
Then they cuffed his hands and shoved him into the SUV.
FromLos Angeles Times
And: “A few minutes later, an ICE wagon pulls up next to them, agents cuff and stuff them into the back and then summarily send them back to Ireland.”
FromLos Angeles Times
He also drew from the unpleasant times, particularly the three months he was sidelined with a strained rotator cuff.
FromLos Angeles Times
His hands were cuffed in front of his wrinkled yellow jail T-shirt and his ear lobes were stretched with white paper plugs over his tattooed neck.
FromLos Angeles Times
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Related Words
belt
sock
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When To Use
What else does cuff mean?
Cuff can refer to the ends of sleeves or rolled pants, handcuffs, or, in the world of modern love, the act of going steady with someone over the winter months.
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Definitions and idiom definitions from Dictionary.com Unabridged, based on the Random House Unabridged Dictionary, © Random House, Inc. 2023
Idioms from The American Heritage® Idioms Dictionary copyright © 2002, 2001, 1995 by Houghton Mifflin Harcourt Publishing Company. Published by Houghton Mifflin Harcourt Publishing Company.
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187575 | https://www.youtube.com/playlist?list=PLbMVogVj5nJTW50jj9_gvJmdwFWHaqR5J | Engineering Design - Vehicle Dynamics - YouTube
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Engineering Design - Vehicle Dynamics
32 videos 615,598 views Last updated on Aug 11, 2014
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Vehicle Dynamics by Dr.R.Krishnakumar,Department of Engineering Design,IIT Madras.For more details on NPTEL visit
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Engineering Design - Vehicle Dynamics
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Vehicle Dynamics by Dr.R.Krishnakumar,Department of Engineering Design,IIT Madras.For more details o…...More
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excel公式来计算平均增长率的方法？
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回答：
Excel公式是一种非常强大的工具，可以帮助人们在各种情况下进行复杂的计算。其中一项功能是计算平均增长率，这在许多领域中都是非常有用的。本文将从多个角度分析Excel公式来计算平均增长率的方法。一、什么是平均增长率？
平均增长率是指某一时间段内，某个变量的增长率的平均值。例如，在过去5年中，某个公司的年销售额分别为100、110、120、130和140万元，那么这家公司的平均增长率为10%。
二、如何计算平均增长率？
在Excel中，可以使用以下公式来计算平均增长率：
(最终值/初始值)^(1/年数)-1
其中，最终值是某个变量在一段时间内的最终值，初始值是该变量在同一段时间内的初始值，年数是该时间段的年数。例如，在上面的例子中，最终值为140万元，初始值为100万元，年数为5年，那么平均增长率为：
(140/100)^(1/5)-1=0.0803=8.03%
三、如何使用Excel公式计算平均增长率？
在Excel中，可以使用以下步骤来计算平均增长率：
打开Excel，并在一个单元格中输入初始值。
在另一个单元格中输入最终值。
在第三个单元格中输入年数。
在第四个单元格中输入平均增长率公式：=(B2/B1)^(1/B3)-1。
按下Enter键，即可得到平均增长率的值。
四、平均增长率的应用
平均增长率在许多领域中都有广泛的应用。以下是一些例子：
经济学：平均增长率是衡量一个国家或地区经济增长的重要指标。
金融学：平均增长率可以用来计算股票或基金的复合年增长率。
商业：平均增长率可以帮助企业了解其销售、利润等指标的增长情况。
科学研究：平均增长率可以用来研究某种现象的增长趋势，例如人口增长、气温变化等。
五、结论
Excel公式是计算平均增长率的一种有效工具。使用Excel公式可以快速、准确地计算平均增长率，这在许多领域都非常有用。通过计算平均增长率，人们可以更好地了解某个变量的增长趋势，从而做出更好的决策。
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187577 | https://mathoverflow.net/questions/315485/integer-partitions-under-divisibility-constraint | Stack Exchange Network
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Integer partitions under divisibility constraint
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$\begingroup$
Consider integer partitions of $x \in \mathbb{N}$ of size $k$ under the constraint that the partition elements are distinct and the ratio of any element to each smaller element is a natural number.
Example $f(x,k)$:
$f(17,3) = { 1, 4, 12}$
$f(14,3) = { 2, 4, 8 }$
$f(101,4) = { 1, 2, 14, 84 }$
$f(4,3) = \emptyset$
As pointed out by @Henrik, one can express one constraint as: $x = a_1 + a_1 a_2 + \ldots + a_1\cdots a_k$ and then factor $x$, $x-a_1$, $x - a_1 - a_1 a_2$, etc. to find candidates for successive $a_i$. (In the special case of $x$ being prime and $k>1$, then $a_1 = 1$.) However, because there are typically several choices for each successive $a_i$, perhaps one must use some sophisticated search (with backtracking) or variant on linear programming.
Have these partitions (or series) been studied? Is there an efficient algorithm or method for finding them, given $x$ and $k$? When might $f(x,k)$ be empty or not unique?
Influenced by the comment from @Gerhard "always has a clever middle name" Paseman, I thought I'd plot candidates for the case $f(28,2)$. The abscissa is $a_1$ (whose value must be a factor of $28$, i.e., $1, 2, 4, 7, 14, 28$), and the ordinate $a_1 a_2$. Because $a_1 + a_1 a_2$ must ultimately equal $n = 28$, we need consider the smallest partition in the range $0 < a_1 \leq \lfloor {n \over k+1} \rfloor = 9$ (marked by the non-gray region). Because the $a_2 \geq 2$, the region of candidates for $a_1 a_2$ must be $\geq 2 a_1$, as shown in yellow.
Thus we seek a point $(a_1, a_1 a_2)$ on the line $a_1 + a_1 a_2 = n$, as shown by the black line.
For this case, the only solution is $(1, 27)$.
I'm not quite sure how this helps in finding an efficient algorithm, but my problem-solving style is to visualize or graph as much as possible, so perhaps this will shed light for someone else.
Here's a table of candidates for $f(25,3)$, showing there are four solutions: ${ 1, 6, 18}$, ${ 1, 2, 22}$, ${ 1, 4, 20}$, ${ 1, 8, 16 }$. (Actually, we can eliminate the case $a_1 = 5$ a priori, but I include it for completeness.)
$$ \begin{array}{|r|c|l|} \hline a_1 & a_1 a_2 & a_1 a_2 a_3 \ \hline 1 & 2 & 4,8, 12, 16, 20, {\bf 22} \ & 4 & 8, 12, 16, {\bf 20} \ & 6 & 12, {\bf 18} \ & 8 & {\bf 16} \ \hline 5 & 10 & 20 \ \hline \end{array} $$
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edited Nov 17, 2018 at 1:23
David G. StorkDavid G. Stork
asked Nov 16, 2018 at 21:02
David G. StorkDavid G. Stork
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$\endgroup$
6
$\begingroup$ Perhaps analyze the case k=2 and describe all of those, and then use that to understand the case k=3. There may be an alternate characterization which allows you to avoid much backtracking. Gerhard "Don't Go Far On Branch" Paseman, 2018.11.16. $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2018-11-16 22:02:56 +00:00
Commented Nov 16, 2018 at 22:02
2
$\begingroup$ $k=2$ is just the pure factoring problem. So I'm not sure how efficient it can be... $\endgroup$
fedja
– fedja
2018-11-16 22:36:21 +00:00
Commented Nov 16, 2018 at 22:36
$\begingroup$ @fedja Maybe not possible to efficiently get the exact value, but getting estimates from $\sum_{x} f(x,2) y^x = \sum \sigma_0 (n) y^n$ or analogous for $k \neq 2$. $\endgroup$
AHusain
– AHusain
2018-11-16 22:46:04 +00:00
Commented Nov 16, 2018 at 22:46
2
$\begingroup$ One of us is not understanding the situation. For f(25,3) I get 4,20 and 2,22 as continuations. For f(28,2), any divisor of 28 less than 28/3 can be a first candidate. The idea was to use k=2 to simplify analysis, as f(n,3) is a , af(n/a,2) over all appropriate divisors a of n. Gerhard "Need To Avoid Prime Quotients" Paseman, 2018.11.16. $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2018-11-17 01:18:08 +00:00
Commented Nov 17, 2018 at 1:18
$\begingroup$ @GerhardPaseman: Oops... I didn't continue my list far enough to find the cases you gave. (Thanks.) . There might indeed be some recursive method as you suggest. I'll go back to thinking about it. $\endgroup$
David G. Stork
– David G. Stork
2018-11-17 01:21:44 +00:00
Commented Nov 17, 2018 at 1:21
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187578 | https://math.stackexchange.com/questions/4069644/need-a-function-that-measure-the-proximity-of-a-given-value-to-a-target-value | Need a function that measure the proximity of a given value to a target value - Mathematics Stack Exchange
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Need a function that measure the proximity of a given value to a target value
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Like the title says, I'm looking for a function that take a given value, and returns a value between 0 and 1 which measures how much the given value is near to a target value. Ideally, the returned value should increase faster when the target value is almost reached. For example, if the target is 10:
Input = 5; output = 0.5
Input = 8; output = 0.85 (more than 0.8 because we are near to the target)
Input = 10; output = 1 (of course)
Input = 15; output = 0.5 (because we have exceed the target)
Any help?
EDIT 1
The function only takes non negative numbers.
The input number can be at most twice the target number
EDIT 2
This is what have I done so far:
if (amount <= target)
return amount / target;
else if (amount >= 2 target)
return 0;
else
return 1 - ( amount % target ) / target;
It's very rudimental, and it doesn't implement the concept of "speed".
EDIT 3
The goal is to use the function for an automatic optimizer. I have a set of features, and the idea behind is to assign a score to the value assigned to each feature. The more the assigned value is near to the target value, the merrier. I'm using a maximizer to maximize the sum of the scores.
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edited Mar 29, 2021 at 8:50
AndreaAndrea
asked Mar 20, 2021 at 16:55
AndreaAndrea
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There are infinitely many functions that satisfy your requirement. If the function is differentiable at the target, though, the derivative will be 0 0, so your requirement that the speed is increasing won't be satisfied. Therefore, it sounds to me like you should be looking for a function with a cusp at the target, and a vertical tangent there.saulspatz –saulspatz 2021-03-20 17:27:59 +00:00 Commented Mar 20, 2021 at 17:27
Can you provide an example? I have very little mathematical competence, I just need to use this function in an app I'm creating Andrea –Andrea 2021-03-20 18:03:08 +00:00 Commented Mar 20, 2021 at 18:03
mathworld.wolfram.com/CuspMap.htmlsaulspatz –saulspatz 2021-03-20 18:05:55 +00:00 Commented Mar 20, 2021 at 18:05
I still don't get how to implement the target value in the cusp function Andrea –Andrea 2021-03-20 18:20:28 +00:00 Commented Mar 20, 2021 at 18:20
1 You could try something like 1−∣∣a t−1∣∣c 1−|a t−1|c with c c near 1.1 1.1.Raymond Manzoni –Raymond Manzoni 2021-03-29 08:27:54 +00:00 Commented Mar 29, 2021 at 8:27
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I think the reason that there are not answers is because there are many options and it really does depend on how you are going to use this. For example, if it's for an automatic optimizer, usually you want something that goes to zero and you use a minimizer.
You don't care that it's non-differentiable at the cusp?
Based on the description so far, how about
f(x,x 0,a)=exp(−a|x−x 0|)f(x,x 0,a)=exp(−a|x−x 0|)
First plot is for x 0=10 x 0=10 and a=0.4 a=0.4
Or you can try
1 2(f(x,x 0,0.2)+f(x,x 0,2))1 2(f(x,x 0,0.2)+f(x,x 0,2))
to get some accelerated "hot zone".
click for full size
```
import numpy as np
import matplotlib.pyplot as plt
def f(x, x0, a):
return np.exp(-a np.abs(x-x0))
x = np.linspace(0, 20, 1001)
if True:
plt.figure()
plt.plot(x, f(x, 10, 0.4))
plt.plot(x, 0.5(f(x, 10, 0.2) + f(x, 10, 2)))
plt.show()
```
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edited Mar 29, 2021 at 9:02
answered Mar 29, 2021 at 8:35
uhohuhoh
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1 The goal is to use the function for an automatic optimizer. I have a set of features, and the idea behind is to assign a score to the value assigned to each feature. The more the assigned value is near to the target value, the merrier. I'm using a maximizer to maximize the sum of the scores.Andrea –Andrea 2021-03-29 08:49:49 +00:00 Commented Mar 29, 2021 at 8:49
@Andrea the more information in the question to begin with, the merrier also! Do you like the pointy cusp on top? Are you sure you don't want a rounded top? If you add a bunch of cusps the sum looks like a forest of cusps and the maximizer just might get stuck sitting on one cusp and ignore the others totally. You are looking for a merit function, see also What is the merit function?uhoh –uhoh 2021-03-29 08:54:25 +00:00 Commented Mar 29, 2021 at 8:54
1 I've implemented the function you provided, and as you guessed the pointy cusp is problematic. A rounded top may function better. However, I've read the description of merit function, and it looks like it is exactly what I'm trying to achieve! Do you have any example I could test?Andrea –Andrea 2021-03-29 09:50:40 +00:00 Commented Mar 29, 2021 at 9:50
@Andrea its a general term for what you want, that's all. We all still have to write our own. Least squares minimization works so well in so many cases, it's hard for me to think about using maximization. You square the differences add them, e.g. (x−x 0)2+(y−y 0)2+(x−x 0)2+(y−y 0)2+... and then minimize that. You can weight then differently by putting a different constant in front of each one if one is more sensitive than another. Are you sure that you absolutely must use a maximizer? Least squares is the gold standard; it's simple, universal and robust.uhoh –uhoh 2021-03-29 10:00:06 +00:00 Commented Mar 29, 2021 at 10:00
1 I've started with a maximizer because I had in my mind the idea of score, but I can switch to a minimizer in no time!Andrea –Andrea 2021-03-29 10:39:31 +00:00 Commented Mar 29, 2021 at 10:39
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Algebra 2 (FL B.E.S.T.)11 units · 156 skillsUnit 1 Properties of functionsUnit 2 Linear equations, inequalities, and systemsUnit 3 Quadratic functions & equations introductionUnit 4 More on quadratics & complex numbersUnit 5 Polynomial equations & functions introductionUnit 6 More on polynomial equations & functionsUnit 7 Inverse functionsUnit 8 Radical functions & equationsUnit 9 Exponential functionsUnit 10 Logarithmic functionsUnit 11 Rational functions
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187580 | https://towardsdatascience.com/the-one-stop-guide-for-transformation-matrices-cea8f609bdb1/ | Skip to content
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The one-stop guide for transformation matrices
In this article, I'll explain how to create transformation matrices and use them for converting from one reference frame to another.
Manpreet Singh Minhas
9 min read
I’ve been recently learning about 3D to 2D projection and back projection. One of the most confusing things that I found was constructing transformation matrices from camera to the world using Euler angles. Although rotation as a concept is easy to understand, building transformation matrices and using them can be extremely confusing. In this article, I’ll explain how to create transformation matrices and use them for converting from one reference frame to another. We’ll also visualize the transformations and few sample points by plotting them.
Introduction
In computer vision, robotics, aerospace, etc. we require the usage of transformation matrices (rotation and translation) to go from one frame of reference to the other. This is a very important concept if you want to work with geometric computer vision and stereo vision (epipolar geometry). However, as was my recent experience, it can be challenging to understand how to use these transformations and go from one frame of reference to another. First I want to make a distinction between frames of reference and convention.
Frames of reference:
It is a coordinate system for measuring points in a 3D (or N-D) space. It can be oriented in any way and translated by any amount w.r.t. other frames of reference. To take an example, for an autonomous car we can have the car frame of reference, camera frame of reference, IMU frame of reference etc. But it is important to note that the point remains at the same location physically no matter how the numbers change in the different frames. The motivation behind moving between different frames of reference is that in a local frame of reference you can measure points w.r.t. the origin and orientation of that frame of reference, which is easier to manage. Say it is easier to say 1m forward, 2m right, 3m above w.r.t. a robot than defining the same space in a global frame of reference with those coordinates shifted and rotated.
Convention:
This is only the order of x, y and z values and how they relate to the actual physical directions. It is not a frame of reference and shouldn’t be confused with it. Although we need to ensure that the convention of the library/application/equations that you are using is followed, it doesn’t have anything to do with the frame of references themselves. An example of a convention can be NED (Northing, Easting, Down) which implies x is forward, y is right and z is down. Another one is ENU (Easting, Northing, Up) which implies that x is right, y is forward and z is up. OpenCV uses EDN (Easting, Down, Northing) convention which implies that x is right, y is down and z is forward. So someone in the ENU system saying (1,2,3) would be implying 1m right, 2m forward and 3m up. But if these numbers would be plugged into say the EDN convention without proper rotations, we would end up with 1m right, 2m down and 3m forward which wouldn’t be the intention of the of these numbers. Please use only one convention in your code if possible to make things clearer. Otherwise, have clear documentation stating which different conventions are being followed and which specific rotations are being applied to convert between them.
Right-handed system v/s Left-handed system
This is also an important concept that needs to be understood. The handedness of a system is nothing but which hand do we use to get the direction of the z-axis from the cross product of x and y axes.
Right-handed system:
x x y = z, where x,y,z are orthogonal unit vectors for the coordinate axis. The direction of z can be obtained by the direction obtained by curling the fingers of your right hand from x to y and the thumb would point in the z-direction. The direction of positive rotation would be obtained by placing our curled right hand along the z-direction. Positive rotation is counterclockwise if you view from the top of the +ve direction of the axis of rotation in this system.
Left-handed system:
x x y = -z (in right-handed convention notation), where x,y,z are orthogonal unit vectors for the coordinate axis. The direction of z can be obtained by the direction obtained by curling the fingers of your left hand from the direction of x to y and the thumb would point in the +ve z-direction. The direction of positive rotation would be obtained by placing our curled left hand along the z-direction. Positive rotation is clockwise if you view from the top of the +ve direction of the axis of rotation in this system.
Transformation Matrix
I’ll be sticking to the homogeneous coordinates for constructing the transformation matrices. Explaining these coordinates is beyond the scope of this article. But the main point is that these coordinates allow projective transformations to be represented as a 4×4 matrix. Also, points and vectors (directions) are nicely differentiated in this system. You can read more about it here: Homogeneous coordinates
We won’t be using these to perform 3D to 2D projections in this article, but rather only convert between different frames of reference.
Before starting with constructing the matrix I’ll briefly talk about row and column vector notations and their effect on how to use the transformation matrix. An nx1 matrix is called a column vector and a 1xn matrix is called a row vector. Depending on how you define your x,y,z points it can be either a column vector or a row vector.
For a column vector, we pre-multiply the rotation/transformation matrix which is in a column-major format. The result is a linear combination of the columns of the rotation matrix.
For a row vector, we post-multiply the rotation/transformation matrix which is in a row-major format. The result is a linear combination of the rows of the rotation matrix.
So if you construct your matrix in a column-major format you need to take a transpose of the rotation matrix before post multiplying with a row vector.
For this article, I’ll be sticking to column vectors. Next, we look at how to construct the transformation matrix. I’ll be using the scipy library for making the rotation matrices from Euler angles. You can read about how these angles work here: Better rotation representations for accurate pose estimation by Dmitry Kostyaev and watch this video: Euler Angles and the Euler Rotation Sequence [Awesome explanation!]. So assume that you know the three Euler angle values of your frame B w.r.t. frame A and the translation (tx, ty, tz) of the origin of frame B from frame A in the frame A coordinates. The transformation matrix for converting from the frame of reference B to A is given as:
Note the order AB in the name of the transformation matrix. This is because these matrices are multiplied from right to left. So if we were to cascade a few of them we would know from this convention whether we are following the correct conversion order. In the following example, we cascade three transformation matrices to take a point p0 from frame of reference D to frame of reference A. Note that a point in homogenous coordinates is represented as [x, y, z, 1] (All scaled versions of these coordinates are equivalent i.e. [wx, wy, wz, w]).
Now let us look at the code required to do this in python. I’ll be using scipy for building the rotation matrix and pytransform3d for visualizing the various transformations. First let us see which all libraries are required to be imported.
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| | |
--- |
| | import numpy as np |
| | import matplotlib.pyplot as plt |
| | from pytransform3d.plot_utils import make_3d_axis |
| | from pytransform3d.transform_manager import TransformManager |
| | from scipy.spatial.transform import Rotation |
view raw
transformation_matrices1.py
hosted with ❤ by GitHub
Next, we define transformation matrices for a few different frames. Note that the names in the code don’t follow the on paper convention. I use A2B to mean that I go from the frame of reference A to B. The key function that we use is Rotation.from_euler, it takes a string of valid Euler angle rotations and a list of rotation angles. You can specify the rotation in degrees or radians. If you want to use radians, set degrees=False. I use the "XYZ" rotation sequence in this code. I define four frames of reference A, B, C and D w.r.t each other. Finally, I also define few points to plot in frames of reference A and B.
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| | |
--- |
| | TB2A = np.eye(4) |
| | R = Rotation.from_euler("XYZ",[75,0,-45], degrees=True).as_matrix() |
| | TB2A[:3,:3] = R |
| | TB2A[:3,3] = np.array([1,5,3]) |
| | TA2B = np.linalg.inv(TB2A) |
| | |
| | TD2A = np.eye(4) |
| | R = Rotation.from_euler("XYZ",[-2,-90,-45], degrees=True).as_matrix() |
| | TD2A[:3,:3] = R |
| | TD2A[:3,3] = np.array([2,-5,2]) |
| | TA2D = np.linalg.inv(TD2A) |
| | |
| | TC2B = np.eye(4) |
| | R = Rotation.from_euler("XYZ",[45,45,0], degrees=True).as_matrix() |
| | TC2B[:3,:3] = R |
| | TC2B[:3,3] = np.array([1,2,3]) |
| | TB2C = np.linalg.inv(TC2B) |
| | |
| | # Define few points from different frames to frame A |
| | origin_c_in_a = TB2A@TC2B@np.array([0,0,0,1]).reshape(4,1) |
| | point_P_c_in_a = TB2A @ TC2B @ np.array([2,0,0,1]).reshape(4,1) |
| | origin_d_in_a = TD2A @ np.array([0,0,0,1]).reshape(4,1) |
| | |
| | # Define few points from different frames to frame B |
| | origin_a_in_b = TA2B@np.array([0,0,0,1]).reshape(4,1) |
| | origin_c_in_b = TC2B@np.array([0,0,0,1]).reshape(4,1) |
| | origin_d_in_b = TA2B@TD2A@np.array([0,0,0,1]).reshape(4,1) |
| | point_P_c_in_b = TC2B @ np.array([2,0,0,1]).reshape(4,1) |
view raw
transformation_matrices2.py
hosted with ❤ by GitHub
We construct a transformation manager from these transformation matrices and then plot the points and frames w.r.t frame of reference B and A. Pytransform3d is a handy package to work with 3d transformations. The colour coding for the axes is RGB correspond to XYZ respectively.
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| | |
--- |
| | tm = TransformManager() |
| | tm.add_transform("A", "B", TA2B) |
| | tm.add_transform("B","C",TB2C) |
| | tm.add_transform("A", "D", TA2D) |
| | |
| | plt.figure(figsize=(8, 12)) |
| | |
| | ax = make_3d_axis(6, 211) |
| | ax = tm.plot_frames_in("B", ax=ax, s=3) |
| | ax.plot(origin_a_in_b[:3],"rx") |
| | ax.plot(origin_c_in_b[:3],"bx") |
| | ax.plot(origin_d_in_b[:3],"yx") |
| | ax.plot(point_P_c_in_b[:3],"gx") |
| | ax.view_init(30, 20) |
| | |
| | ax = make_3d_axis(6, 212) |
| | ax = tm.plot_frames_in("A", ax=ax,s=3) |
| | ax.plot((1,5,3), "yx") |
| | ax.plot(origin_c_in_a[:3], "rx") |
| | ax.plot(point_P_c_in_a[:3], "gx") |
| | ax.plot(origin_d_in_a[:3],"bx") |
| | ax.view_init(30, 20) |
| | plt.show() |
view raw
transformation_matrices3.py
hosted with ❤ by GitHub
Resultant plots are shown below. We see that the frames are correctly oriented w.r.t each other. And we successfully transformed points from one frame of reference to another by either using a direct transformation matrix or by cascading a couple of them.
If you want to see a cool application of transformation matrices in action do check out: A Hands-On Application of Homography by Daryl Tan.
Conclusion
Working with transformation matrices is the basic step for various fields like robotics, aerospace, autonomous driving, epipolar geometry etc. In this article we saw several concepts namely frame of reference, convention, right v/s left-handed systems. We saw how to handle column and row vectors in transformation matrices. And finally how to create these matrices in Python and visualize them in graphs. I hope you enjoyed reading the article as much as I enjoyed writing it. If you spot any errors or have any other questions or suggestions please feel free to comment here or message me on LinkedIn.
References
Right-hand rule
scipy.spatial.transform.Rotation – SciPy v1.7.0 Manual
Euler angles
Rotation matrix
pytransform3d – pytransform3d 1.9 documentation
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Manpreet Singh Minhas
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187581 | https://physics.stackexchange.com/questions/79843/rate-of-effusion-in-kinetic-molecular-theory | Skip to main content
Rate of effusion in kinetic molecular theory?
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According to the kinetic molecular theory obeying Maxwell-Boltzmann distribution of speeds, the rate of effusion through a pinhole of area A is
R=PA2πMRT−−−−−−−√
where M is the molecular weight, R is the gas constant and T the absolute temperature.
To derive this, I consider the collision frequency on any small area (A):-
using
vavg=8RTπM−−−−−√
I get the result, that the atoms in volume vavgA can hit the area. The number of particles in this volume is nvavgA (n=number density). But the derivation includes a factor of 14 before this term to find the actual number of atoms in this volume hitting the wall. I want to know how that factor of 14 came into picture to make the collision frequency per unit area as 14nvavg=P2πMRT√.
I know the origins of a factor of 12 before the pressure term while calculating it, considering change in momentum of a molecule on collision with the wall, which is to account for the fact that includes both the vx terms, going towards and going away from the wall(positive and negative directions) but the ones colliding are just the half of these 12(going in any one direction).
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edited Oct 7, 2013 at 11:35
stochastic13
asked Oct 6, 2013 at 17:44
stochastic13stochastic13
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In considering the flow passing by the area A, you have a 12 factor coming by selecting one way, and I think that the other 12 comes because you have to consider the projection of the speed on the normal to the surface A , see also this answer. The problem is different, but just look at the difference between integrals in θ in equations (2) and (4), and you get your 14 factor.This is the difference between ∫π0sinθ dθ and ∫π/20dθsinθ cosθ
– Trimok
Commented
Oct 7, 2013 at 18:18
@Trimok For my problem, i calculated the integral to be:- n∫π/2012vavgAtcosθdθ which does not yield the given answer and hence must be wrong. Can you tell me what the integral for my problem should be?
– stochastic13
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If your Maxwell-Boltzmann distribution is μ(v⃗ )=μ(v)=(m2πkT)3/2e−mv22kT, then, if I am not mistaken, you should have to perform an integral with θ limited between 0 and π/2 of kind I=nA∫0≤θ≤π/2d3v⃗ μ(v⃗ )(v⃗ .n⃗ ), where n⃗ is the unit normal to A, and we choose θ such as v⃗ .n⃗ =vcosθ,so you would have : I=nA∫2π0dϕ∫π/20dθsinθcosθ∫+∞0dv v3 μ(v). Note that vavg=∫2π0dϕ∫π0dθsinθ∫+∞0dv v3 μ(v). Your result should be I=nAvavg4
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answered Oct 8, 2013 at 15:52
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187582 | https://www.youtube.com/watch?v=Z65mz__8DQ0 | Slope and y intercept from equation
Khan Academy
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Keep going! Check out the next lesson and practice what you’re learning:
Slope and y intercept from equation
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Transcript:
what i'd like to do in this video is a few more examples recognizing the slope and y-intercept given an equation so let's start with something that we might already recognize let's say we have something of the form y is equal to 5 x plus 3. what is the slope and the y-intercept in this example here well we've already talked about that we can have something in slope-intercept form where it has the form y is equal to the slope which people use the letter m for the slope times x plus the y-intercept which people use the letter b for so if we just look at this m is going to be the coefficient on x right over there so m is equal to 5 that is the slope and b is just going to be this constant term plus 3 so b is equal to 3. so this is your y intercept so that's pretty straightforward but let's see a few slightly more involved examples let's say if we had form y is equal to 5 plus 3x what is the slope and the y-intercept in this situation well it might have taken you a second or two to realize how this earlier equation is different than the one i just wrote here it's not 5x it's just 5 and this isn't 3 it's 3x so if you want to write it in the same form as we have up there you can just swap the 5 and the 3x it doesn't matter what or which one comes first you're just adding the two so you could rewrite it as y is equal to 3x plus 5 and then it becomes a little bit clearer that our slope is 3 the coefficient on the x term and our y-intercept is 5 y-intercept let's do another example let's say that we have the equation y is equal to 12 minus x pause this video and see if you can determine the slope and the y-intercept all right so something similar is going on here that we had over here the standard form slope intercept form we're used to seeing the x term before the constant term so we might want to do that over here so we could rewrite this as y is equal to negative x plus 12 negative x plus 12. and so from this you might immediately recognize okay my constant term when it's in this form that's my b that is my y intercept so that's my y intercept right over there but what's my slope well the slope is the coefficient on the x term but all you see is a negative here what's the coefficient well you could view negative x as the same thing as negative one x so your slope here is going to be negative one let's do another example let's say that we had the equation y is equal to 5 x what's the slope and y intercept there at first you might say hey this looks nothing like what we have up here this is only i only have one term on the right hand side of the equality sign here i have two but you could just view this as 5x plus zero and then it might jump out at you that our y intercept is zero and our slope is the coefficient on the x term it is equal to five let's do one more example let's say we had y is equal to negative seven what's the slope and y-intercept there well once again you might say hey this doesn't look like what we had up here how do we figure out the slope or the y-intercept well we could do a similar idea we could say hey this is the same thing as y is equal to zero times x minus seven and so now it looks just like what we have over here and you might recognize that our y intercept is negative seven y intercept is equal to negative seven and our slope is the coefficient on the x term it is equal to zero and that makes sense for a given change in x you would expect zero change in y because y is always negative seven in this situation |
187583 | https://engineeringsurvey899633060.wordpress.com/2019/01/06/taping-corrections/ | Skip to content
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Let Us Resolve Your Surveying Dilemmas
Taping Corrections
Posted by Syed Fahadullah Hussaini on
In surveying, tape correction refer(s) to correcting measurements for the effect of slope angle, expansion or contraction due to temperature, and the tape’s sag, which varies with the applied tension. Not correcting for these effects gives rise to systematic errors, i.e. effects which act in a predictable manner and therefore can be corrected by mathematical methods.
Correction Due To Slope
When you take a measurement with a tape along an inclined plane (along the natural slope of the ground), obviously, the taped distance is greater than the horizontal distance. The difference between the slope distance and the horizontal distance (s – d) is called the slope correction. This correction is always subtracted from the slope distance. To compute for the slope correction, you should know either the vertical.
Measured Distance with tape= L Elevation= d Corrected Tape distance= H
Corrected Horizontal Distance can be calculated using Pythagoras Theorem, where L is the hypotenuse, d is the perpendicular and H is the adjacent. Hence H that is The corrected slope distance can be calculated via the formula above
Measured Tape Distance (L) = 350 m Elevation (d) = 2 m H= (350^2 – 2^2)^1/2 H= 349.994 m
Temperature Correction
When measuring or laying out distances, the standard temperature of the tape and the temperature of the tape at time of measurement are usually different. A difference in temperature will cause the tape to lengthen or shorten, so the measurement taken will not be exactly correct. A correction can be applied to the measured length to obtain the correct length.
The correction of the tape length due to change in temperature is given by
Ct = A (Tm – Ts)
Where,
A = coefficient of thermal expansion per one degree Fahrenheit or Celsius
Tm = temperature during the measurement in Fahrenheit or Celsius
Ts = temperature at which the tape standardized in F/C
For common tape measurements, the tape used is a steel tape with coefficient of thermal expansion C equal to 0.000,011,6 units per unit length per degree Celsius change. This means that the tape changes length by 1.16 mm per 10 m tape per 10 °C change from the standard temperature of the tape. For a 30 meter long tape with standard temperature of 20 °C used at 40 °C, the change in length is 7 mm over the length of the tape.
Steel tapes are standardized at 68° F (20° C).A temperature higher or lower will change the length of the tape.Coefficient of thermal expansion for steel
.0000116 per unit length per degree Celsius
.0000065 per unit length per degree Fahrenheit
Temperature of the tape can be quite different from the air temperature measured. This error can be accounted for by correcting the value of measured lengths.
Worked Example
Correction Due To Incorrect Tape Length
Manufacturers of measuring tapes do not usually guarantee the exact length of tapes, and standardization is a process where a standard temperature and tension are determined at which the tape is the exact length. The nominal length of tapes can be affected by physical imperfections, stretching or wear. Constant use of tapes cause wear, tapes can become kinked and may be improperly repaired when breaks occur.
The correction due to tape length is given by:
Correction= Measured Length Correction per measurement number of measurements Corrected Length: CL Number of Measurements (n): Measured Distance/ Maximum Tape Length Correction per unit Length: Corr Measured Distance: ML
Numerical: A measurement of 171.278 m was recorded with a tape of 30 m that was actually of length 29.996 m. Corr= -.004 n= 171.278/30 Total correction= -.004 171.278/30 Total correction= -.023 CL= 171.278- .023 CL= 171.255 m
Worked Example
Correction Due To Tension
Some tapes are already calibrated to account for the sag at a standard tension. ]In this case, errors arise when the tape is pulled at a tension which differs from the standard tension used at standardization. The tape will stretch less than its standard length when a tension less than the standard tension is applied, making the tape too short. A tape stretches in an elastic manner until it reaches its elastic limit, when it will deform permanently and ruin the tape.
Worked Example
Correction Due To Sag
A tape not supported along its length will sag and form a centenary between end supports. The correction due to sag must be calculated separately for each unsupported stretch separately and is given by.
A tape held in catenary will record a value larger than the correct measurement. Thus, the correction Cs is subtracted from L to obtain the corrected distance:
d=L-Cs
Note that the weight of the tape per unit length is equal to the weight of the tape divided by the length of the tape:
Combined Corrections
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187584 | https://www.youtube.com/watch?v=JnFykw00HvE | Deriving Kinematics Equations Using Calculus
lasseviren1
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Posted: 20 Sep 2009
This video uses some basic calculus to derive the kinematics equations for uniform motion. It is at the AP physics level.
For a complete index of these videos visit
157 comments
Transcript:
hi um we're going to use some calculus to derive these three kinematics equations now so um let's start off with just your basic the acceleration is um equal to dvdt now those equations only work when acceleration is constant so a is a constant so what I'm going to do is I'm going to bring the DT over to the other side so a DT is equal to DV now I want to tell you how you interpret this this is a very very very tiny change in velocity when will you get a very very tiny change in velocity when you have an acceleration for a very very very tiny time so if you only let something accelerate for a very very very tiny time infinite teson small you'll get a very very tiny change in velocity okay now what we do when we when we bring those over like that we separate these um I'm going to take the integral of both sides so if they're equal then the the integrals of them should be equal now I'm going to integrate this from V initial to V final my boundary conditions always have to be in terms of the differential so this differential is T so this would be T initial zero to T final I'll just call it t okay now remember this is a constant so I'm going to um that's just like the number five or something and I'm only taking the anti-derivative of the of this integrant so this is just with respect to time so that would be a t from Zer to T remember with these boundaries you just put them like that we're going to subtract we're going to put that in and then subtract what ever we get when we put that one in so that's equal to now there's a one in front of the DV so you're really taking the anti-derivative of one with respect to V so that that turns out to be just V and I'll put in from V initial to V final okay if I throw in these these um times then I'm going to have a T minus now I put in that one a 0 is equal to now throw that in V final minus throw that in minus V initial so if you come all the way up here with me it looks like I have if I bring the V initial on the other side I have V final equals V initial plus a so that's the first one I'll box that that was one of the equations that we were after okay let's take this equation that V final and um since V fi V final the V at any time this is the V at any time T and so um V at any time is dxdt so instead I'll write V is equal to V initial plus a t but that is DX DT that's what that V is V is the derivative of x with respect to time let me bring the DT on the other side so we have DX is equal to V initial plus a t DT I brought the DT on the other side and now I'll integrate both sides this has to be with respect to time so I'll say tal 0 to T and this has to be with respect to X so I'll say x initial and X final okay there's a one here so when I that's that's what I'm taking the anti-derivative of so it turns out that um when I take the anti-derivative of one the the value of that is just going to be X X and I'm going to subtitute these in in a second and then if I take the anti-derivative of this with respect to time this is going to be um v i t plus now a is just a constant so it's going to be um a over 2 t^2 take that derivative and see if you don't get the what's what's here what's the you see if you don't get the integrant okay and I remember I'm going to substitute in for that I'm going to substitute in um X initial and X final okay well when I do that when I do that I'm going to get the following I'm going to get um X final minus X initial is equal to oh I'm sorry I messed up here this is I have to substitute in with respect to time here so this is T this is zero for T initial and this is some other time T yeah I was taking that with respect to time okay so that's going to be X final minus X initial when I put in time for T first I'm going to get this I'm going to make that 12 a t^2 when I now I'm going to subtract from that what I get when I put in zero now when I put in zero to each term here they both give me zero so I'll be darned if that doesn't give me Delta X is equal to VI t + 12 a t^ 2 that's our third equation okay finally the second equation is a little trickier there's a trick you have to do to get that one and that is um you start off with a equal dxdt I'm sorry dvdt a equals dvdt and um I'm going to say that then a is equal to dvdx DX DT do you see how when we if we cross multiplied that would give us dvdt I know you wouldn't think to do this but this is how you derive that equation this convince yourself that that is that now this right here that's V so I'm going to say a is equal to V DV DX so for that I just put in V let me bring the DX on the other side a DX is equal to V DV we'll integrate both sides I'm going to start at V initial and go to V final here and I'm going to start at um X initial and go to x final okay the derivative of that with respect to X is just going to be Ax from X initial to X final let me put that up so you can see that that's just ax because a is a constant and the derivative of that with respect to V is going to be um I'm thinking that that's going to be v^ 2times 12 and we'll go from V not to V final okay putting in the terms here that would be a x final minus a x initial and for these I put in 12 V final squared minus 12 V initial squared well this side is a Delta X and this side if I um if I bring the half out in fact let me multiply both sides by two bring a two over here that gets rid of that half then I'm left with v final squar minus V initial squar lo and behold I get if I solve for V final squared it looks like it's V initial S + 2 a Delta X so there you have it |
187585 | https://www.uwyo.edu/mechanical/_files/docs/meref/propagation-of-errors.pdf | 1 PROPAGATION OF ERRORS 1. PREFACE It is seldom possible to directly measure the quantity R that an experiment is designed to determine. Instead, R will normally be computed from related physical quantities (x,y...) that can be experimentally measured, R(x,y,...). For example, the volume of a cylinder (V) can be computed by determining its diameter (D) and length (L); that is V(D, L). These two length measurements have errors associated with them and these errors will propagate into the computed volume. A technique for estimating how errors propagate from measured quantities into the computed results of an experiment is discussed in this treatise. 2. USING DIFFERENTIAL CALCULUS TO DETERMINE THE PROPAGATION OF ERRORS Suppose that an experiment is performed in which only two physical quantities (x, y) are measured so that the desired result quantity can be computed, R(x, y). As described in Experimentation and Uncertainty Analysis for Engineers1, each of these measured quantities, x and y, has a best estimate and probable error associated with it. The computed experimental result, R, should be reported in the same manner as the measured quantities, i.e. R U R R ± = where R U is the expanded uncertainty of R at a given percent confidence, normally 95%. The goal is to develop a relationship that relates the statistics of the measured quantities, x and y, to the statistics of the result, R. The best estimate of R is obtained by using the mean values of the independent variables, R R(x y) ≡ , . This is not equivalent to finding the mean of the calculated values, ∑R( x , y ) / n i i . For example, if R(x, y).= x + y2 and the following values are used: x y 1 2 1 7 4 1 13.11 , = ) y x R( , while 20.0 = ∑ )/n y , x R( i i . The recommendations of the ISO Guide2 is that R U be estimated from the total variance, i.e. the sum of the variances from systematic error and random error, times a coverage factor. The ISO Guide2 recommends using the values from the t distribution for this coverage factor assuming that the distribution for the total errors is Gaussian. Therefore, the expanded uncertainty is: where 2 i Sβ is the variance from systematic errors and 2 i S is the variance from random error. Variance is a statistical value defined as: 2 2 % % i i R S S t U + = β (1) ( ) 1 1 2 2 − − = ∑ = n v v S n i i (2) 2 where i v is an individual observation, v is the sample mean and n is the number of observations. The value of an individual point, Ri , can be found by expanding ) y x R( , about x and y using a Taylor series. The remainder contains second and higher order derivatives and error terms. Assuming the derivatives are of reasonable values and the errors are small, the remainder will approach zero much faster than the first order terms, and can be neglected. This approximation assumes R(x, y) is a continuous and differentiable function. Squaring both sides of the equation, summing over i and dividing both sides of the equation by (n-1) gives: The last expression in the above equation is called the covariance term. This term should approach zero if the independent variables x and y are statistically independent since there are approximately the same number of positive and negative deviations from the mean. From Equation (2) the LHS of the equation can be recognized as the variance of R and the variance of x and y can be identified in the RHS of the equation. The variance of x and y are assumed to be the total variance. Substituting these values yields: Multiplying both sides of the equation by the square of the table t value for the given percent confidence gives: ( ) Remainder , + ∂ ∂ ∂ ∂ + = ) y -y ( y R + ) x -x ( x R y x R R i i i (3) ) y -y )( x -x ( y R x R 1 -n 2 + y -y ( 1 -n 1 y R + ) x -x ( 1 -n 1 x R ) R -R ( 1 -n 1 i i n 1 = i y , x 2 i n 1 = i y , x 2 2 i n 1 = i y , x 2 2 i n 1 = i ∑ ∑ ∑ ∑ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ≈ ) (4) 2 2 2 y y , x 2 x y , x 2 R S y R + S x R S ∂ ∂ ∂ ∂ ≈ (5) 2 2 % 2 2 % 2 2 % y y , x 2 x y , x 2 R S t y R + S t x R S t ∂ ∂ ∂ ∂ ≈ (6) 3 Finally, substituting Equation (1) and taking the square root of both sides of the equation yields an expression for the probable error of the computed R . The extension of Equation (7) to more than two independent variables should be obvious. 3. ANALYTIC EXAMPLE Using the example of the volume of a cylinder, V = D L / 4 2 π cited above, taking the partial derivatives of V with respect to D and L ∂ ∂ V D = DL / 2 D L , π where L is held constant in this partial differential and ∂ ∂ V L = D / 4 D L 2 , π where D is constant in this differential. The uncertainty V U can now be written in terms of the mean values of D and L and their respective uncertainties. Thus the determination of V U V = V ± is complete. 4. SYMBOLIC LOGIC PROGRAMS Evaluation of the partial derivatives in the above example is simple and straight forward, but this is not usually the case. Most propagation of error analyses are quite complicated with many independent measurements involved and complex mathematical relationships between the dependent and independent variables. Symbolic logic programs like DERIVE and Maple are capable of evaluating the partial derivatives that are required in these analyses and therefore help eliminate the drudgery associated with these calculations. Finding these partial derivatives is quite useful in experimental design, since the sensitivity of the probable error of the calculated results to the accuracy of the various measurements that will be made can be investigated ahead of time. 5. NUMERICAL APPROXIMATION Since any particular measurement, xi, is the sum of the mean value plus the total associated error, x i U x x + = , an approximation of the contribution of x U to the overall R U can be found by adding x U 2 2 y y , x 2 x y , x 2 R U y R + U x R U ∂ ∂ ∂ ∂ ≈ (7) 2 2 2 L 2 2 2 V U ) /4 D ( U ) /2 L D ( U D π π + ≈ (8) 4 to x , computing the resulting R using ) ( x U x + , subtracting R and squaring the result. The summation of the contributions from all independent variables computed in the same manner gives a close approximation of R U . In this case we replace the first order Taylor series analytic approximation, Equation 7, with the following finite difference approximation: 2 2 )] , ( ) , ( [ )] , ( ) , ( [ y x R U y x R y x R y U x R U y x R − + + − + ≈ (9) This finite difference approximation may be much easier to evaluate. For example, the probable error , V U of a cylinder is approximated as: An example of this method is also presented in the following Spreadsheet. 6. SPREADSHEET EXAMPLE The file Prop Errors Cylinder.xls presents an experimental design example estimating the maximum probable error both analytically and numerically. The example also illustrates the use of a Visual Basic® function module. To see the function module, click the menu sequence Tools, Macro and Visual Basic Editor. Visual Basic® function modules are a useful tool for these calculations. The Prop Errors Cylinder.xls example can serve as a detailed guide on how to utilize spreadsheets for propagation of error analysis. The file Prop Errors Cantilevered Beam.xls presents an experimental design example with four independent variables. 7. RELATIVE UNCERTAINTIES Sections 2, 3, 4, 5 and 6 above deal with absolute uncertainties. Equation 7 can be made non-dimensional by dividing both sides of the equation by R, squaring both sides and multiplying each term on the RHS by the appropriate factor etc. , 1 or 1 2 2 = = y y x x The resulting equation is: The factors y U x U y x and are the relative uncertainties for the two independent variables, and these will generally be numbers much less than one. The factors that multiply the relative uncertainties, 2 2 2 2 2 2 ] 4 / 4 / ) ( [ ] 4 / 4 / ) ( [ L D U L D L D L U D U L D V π π π π − + + − + ≈ (10) 2 2 2 2 ∂ ∂ + ∂ ∂ ≈ y U y R R y x U x R R x R U y y , x 2 x y , x 2 R (11) 5 ∂ ∂ ∂ ∂ y R R y x R R x y , x y , x and , are called the uncertainty magnification factors (UMFs), and these factors indicate the influence of the uncertainty of a particular variable on the uncertainty of the result. A UMF value greater than 1 indicates the influence of the variable is magnified as it propagates through the result calculation equation. A UMF value of less than 1 indicates the influence diminishes as it propagates through the data equation into the result. The UMFs are particularly useful for identifying those factors that are most important in reducing the overall uncertainty. The uncertainties of the individual variables do not have to be known to analyze the UMFs. Since the UMFs do not depend on the uncertainties of the variables, a second normalized form of Equation 7 is useful for finding the uncertainty percentage contributions (UPCs) from the variables to the uncertainty of the result squared. To obtain the UPCs divide both sides of Equation 7 by R U and square both sides to give: The UPC for the variable x is then defined as: The UPCs for the remaining variables have similar form. These two non-dimension forms of the uncertainty equation are particularly useful for designing experiments to identify the major sources of error and to devise strategies to minimize the impact of the errors on the final result. 8. REFERENCES 1 Coleman, Hugh W., Steele, W. Glenn, Experimentation and Uncertainty Analysis for Engineers, Second Edition, John Wiley & Sons, Inc., New York, 1999. 2 International Organization for Standardization, Guide to the Expression of Uncertainty in Measurement, ISO, Geneva, 1993. 3Physics 1210/1220 Lab Manual, Department of Physics and Astronomy, University of Wyoming. 2 2 2 2 1 R y y , x 2 R x y , x 2 U U y R U U x R ∂ ∂ + ∂ ∂ ≈ (12) ( ) 2 2 2 2 2 2 2 2 2 x UMF 100 x 100 x UPC = ∂ ∂ = ∂ ∂ = R U x U R U x U x R R x U U x R R x R x R x (12) 6 9. EXAMPLE PROPAGATION OF ERROR ANALYSIS File: Prop_of_Errors.xls The theoretical formula for the tip deflection of an end-loaded, circular cantilevered beam is: where: y = tip deflection (m) F = tip loading (N) L = length of rod (m) E = Young's Modulus (Pa) D = diameter of the rod (m) Solving for Young's Modulus gives: Calculate the expected minimum deflection using Equation 1: E = 1.97E+11 Pa F = 15 N L = 0.250 m D = 0.005 m y = 0.01293 m y=64FL3/(3PI()ED4) Calculate the dimensionless UMFs for Equation 2: UMFF= 1.00 UMFF= (F/E)(∂E/∂F) = F/E64L^3/(3PI()yD^4) UMFL= 3.00 UMFL= (L/E)(∂E/∂L) = L/E64FL^2/(PI()yD^4) UMFD= -4.00 UMFD= (D/E)(∂E/∂D) = D/E-256FL^3/(3PI()yD^5) UMFy= -1.00 UMFy= (y/E)(∂E/∂y) = y/E-64FL^3/(3PI()y^2D^4) The smallest measuring units for the instruments to be used in the experiment are: smu_F = 0.5 N smu_L = 0.01 m smu_D = 0.001 m smu_y = 0.002 m Assuming a uniform distribution between graduations, the corresponding uncertainties are: t95% = 1.96 UF = 0.28 N UL = 0.006 m UD = 0.0006 m Uy = 0.0011 m Disregarding signs, the UMFs for the length and the diameter measurements indicate the uncertainties for these parameters will be magnified through the calculation of Young's Modulus.
PROPAGATION OF ERROR ANALYSIS EXAMPLE Assuming: – the rod is 1018 steel, – Young's Modulus is about 197 Gpa, – the minimum force is 15 N, – the minimum rod length is 0.250 m and – the minimum rod diameter is 0.005 m, find the expected uncertainty for determining Young's Modulus by measuring the rod and the tip deflection.
(1) 3 64 4 3 ED FL y π = (2) 3 64 4 3 yD FL E π = 4 2 3 5 3 4 2 4 3 3 64 3 256 64 3 64 D y FL D E yD FL D E yD FL L E yD L F E π π π π − = ∂ ∂ − = ∂ ∂ = ∂ ∂ = ∂ ∂ 7 Example propagation of error analysis (continued) Calculate the expected uncertainty for the calculated Young's Modulus from Equation 2: UE = 9.19E+10 Pa UE/E x 100 = 46.6% Calculate the dimensionless UPCs for Equation 2 to find which variable(s) contribute the largest percentage of the errors: UPCF= 0.2 % UPCL= 2.1 % UPCD= 94.2 % UPCy= 3.5 % 100 % Since the uncertainty of the result is about 50% of the magnitude of the estimated value, this experiment would have to be refined to achieve reasonable results.
For this experiment the diameter contributes the majority of the uncertainty. |
187586 | https://www.youtube.com/watch?v=ImhUKiKY1rU | What are nematocysts?
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What are nematocysts?
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बच्चों लेट्स री द क्वेश्चन फर्स्ट व्हाट आर नेमेटोमोरफा के अंदर जितने भी ऑर्गेनिस्ट म होते हैं उनके पास नमेट सिस्ट प्रेजेंट होते हैं नमो सिस्ट नमेट सिस प्रेजेंट होते हैं स्पेशल सेल्स जिनको नीडो ब्लास्ट बोला जाता है नीडो ब्लास्ट के अंदर होते हैं नीडो ब्लास्ट को हम स्टिंग सेल्स भी बोलते हैं स्टिंग सेल्स इन सेल्स के अंदर यह होते हैं कैप्सूल शेप के कैप्सूल शेप के नेमेटोमोरफा के इंटरेक्शन में आते हैं ये स्टिमुलेटिंग इनका एनिमी भी हो सकता है अब इन्हीं थ्रेड्स के अंदर टॉक्सिंस प्रेजेंट होते हैं जो न्यूरो टॉक्सिन होते हैं न्यूरो टॉक्सिन न्यूरो टॉक्सिन प्रेजेंट होते हैं जो डायरेक्टली इनके प्रे इनके एनेमी पे अटैक करके उसको पैरालाइज कर देते हैं जिससे कि यह डिफेंस मैकेनिज्म के लिए डिफेंस या अटैक मैकेनिज्म दोनों के लिए यूज करते हैं डिफेंस और अटैक मैकेनिज्म के लिए इनको एक बार स्ट्रक्चर के थ्रू भी देख लेते हैं सो फॉर एग्जांपल अगर ये है नीडो ब्लास्ट नीडो ब्लास्ट तो इसके अंदर ऐसे कैप्सूल शेप की सेल्स होती है कैप्सूल शेप्स प्रेजेंट होती है इसके अंदर यहां पर ऐसे थ्रेड लाइक कॉइल्ड स्ट्रक्चर प्रेजेंट होते हैं एंड इन्हीं को बोला जाता है नेमेटोमोरफा |
187587 | https://calcworkshop.com/sequences-series/telescoping-series/ | Telescoping Series
// Last Updated: - Watch Video //
Now it’s time to look at a genuinely unique infinite series.
The Telescoping Series!
This type of infinite series utilizes the technique of Partial Fractions which is a way for us to express a rational function (algebraic fraction) as a sum of simpler fractions.
In this case, we are going to change our function into the sum of two
“smaller, easier” fractions, where one is positive, and the other is negative.
Having opposite signs is a significant detail for this series because as we will soon discover, we are hoping that terms will cancel in pairs.
In fact, this is how the series gets its name. The partial sums collapse like an old-fashioned collapsable telescope leaving just two or three terms for us to find the sum of the series, as Paul’s Online Notes nicely explains.
Example of a Telescoping Sum
But another way to think about it is that we can’t see the end of an infinite series, but by using our “telescope” we can see that all the terms are canceling, leaving just the first and last terms of the series.
It’s like being able to see the future!
So together we will review the necessary skills for creating our Partial Fractions, and practice generating a few partial sums to determine that terms are indeed canceling.
And then apply a limit approaching infinity to our remaining terms and find what the series converges to.
Now, it is important to note that if we are just trying to determine if series converges or diverges, then applying the Telescoping Series Test will probably not be our first choice. But, if we are asked to find the sum of the series, and it’s not a geometric series then this is a good test to use!
Telescoping Series Video
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187588 | https://www.media4math.com/library/lesson-plan-systems-equations-lesson-3-solving-elimination | Lesson Plan--Systems of Equations--Lesson 3--Solving by Elimination | Media4Math
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Lesson Plan--Systems of Equations--Lesson 3--Solving by Elimination
Lesson Plan: Solving Systems of Equations by Elimination
In this third lesson of the Systems of Equations Unit, students will learn how to solve systems of equations using the elimination method. This algebraic technique allows students to eliminate one variable by adding or subtracting equations, making it easier to find the solution.
Students will:
Understand when elimination is the most effective method for solving a system
Learn step-by-step how to add or subtract equations to eliminate a variable
Use multiplication to create equivalent equations when necessary
Practice solving systems of equations using elimination in various scenarios
Apply elimination to solve real-world word problems
Designed for Algebra 1 and Algebra 2 students, this lesson aligns with Common Core standards and reinforces students’ ability to manipulate equations efficiently. Mastering elimination helps students develop strong algebraic reasoning and prepares them for more advanced mathematical concepts.
This is the third lesson in an eight-part unit on systems of equations. Future lessons will focus on comparing solution methods and applying systems of equations to real-world contexts.
Subscribers to Media4Math can download these lesson plans in PDF format for easy classroom use.
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| Common Core Standards | CCSS.MATH.CONTENT.HSA.REI.C.6 |
| Grade Range | 9 - 12 |
| Curriculum Nodes | Algebra •Linear Systems •Solving Systems of Equations |
| Copyright Year | 2025 |
| Keywords | linear systems, elimination method, solving a system through elimination |
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187589 | https://www.youtube.com/watch?v=A7zXamWQY-4 | Do You Simplify Probability Fractions? - The Friendly Statistician
The Friendly Statistician
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Posted: 27 May 2025
Do You Simplify Probability Fractions? Have you ever wondered how to make sense of probability in everyday situations? In this informative video, we break down the concept of probability and how simplifying fractions can help clarify your understanding. We’ll start by explaining what probability is and how it is often represented as a fraction. You’ll learn the steps to simplify these fractions, making it easier to grasp the chances of various events occurring.
We will also discuss the importance of comparing different probabilities and how simplification plays a key role in decision-making. By simplifying fractions, you can quickly identify which event is more likely to happen, allowing for better choices in games, activities, or any scenario involving chance.
Whether you're a student trying to grasp the fundamentals of probability or someone looking to improve your decision-making skills, this video is designed for you. Join us as we simplify the world of probability and make it more accessible. Don’t forget to subscribe to our channel for more engaging content on measurement and data.
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Transcript:
Do you simplify probability fractions? Imagine you are at a carnival and you see a game where you can win a prize by tossing a ring onto a bottle. The sign says you have a one-in for chance of winning. But what does that really mean? Understanding probability can sometimes feel like navigating a maze, but simplifying those numbers can make it much clearer. When we talk about probability, we often express it as a fraction. For example, if there are three red balls and one blue ball in a bag, the probability of drawing a red ball is three out of four. This fraction can be simplified if possible, just like any other fraction in mathematics. To simplify a probability fraction, you look for the greatest common divisor of the numerator and the denominator. In our example, three and for do not have any common factors other than one. So the fraction remains 3 over 4. However, if you had a scenario where the probability was 6 out of 8, you could simplify that to three out of for by dividing both the numerator and the denominator by two. Simplifying probability fractions can make it easier to understand the likelihood of an event occurring. It helps in comparing different probabilities and making decisions based on those comparisons. For instance, if you have a probability of one out of to four one event and three out of for four another, it is easier to see that the second event is more likely to happen when the fractions are in their simplest forms. In conclusion, simplifying probability fractions is not only possible but also beneficial. It clarifies the chances of events and aids in better decision-m. So the next time you encounter a probability fraction, consider simplifying it for a clearer perspective. |
187590 | https://mowrator.com/blogs/lawn-mower/standard-decibel-level | What is the Standard Lawn Mower Decibel Level? – MOWRATOR
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Contents
What is the Standard Lawn Mower Decibel Level?
September 26, 2024
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Every spring and summer, the sound level of neighborhoods across America revs up as homeowners start their annual lawn care routine. While the sounds of a lawn mower are irritating and can disrupt important meetings or your relaxation, you may not have known they can also lead to hearing loss over time. The average gas-powered lawn mower decibel level is 90 dB, which is above the 85 dB threshold for hearing damage. For a quieter and more efficient alternative, a remote controlled lawn mower allows you to maintain your yard with minimal noise and maximum convenience.
If you don't have an electric lawn mower like Mowrator'sS1 Remote Control Lawn Mower 4WD, you’ll need to protect your hearing. We've put together this article to help you understand the importance of lawn mower decibels, how they affect your hearing, and what you can do to reduce your noise levels.
Key Takeaways
Most lawn mowers emit a noise level of 85 to 100 decibels (dB), except electric lawn mowers, which only emit around 56 to 80 dB.
The sound your lawn mower makes is impacted by the RPM, blade design, sound insulation, and deck size.
Listening to 88 dB for at least 4 hours within a week can cause hearing damage.
Installing sound insulation, wearing ear protection, and switching to an electric mower can reduce your chances of hearing loss in the long term.
What is the Decibel Level of Different Lawn Mowers?
| Lawn Mower | Noise Level | Ear Protection Required? |
---
| Gas-Powered Lawn Mowers | 85 to 100 dB | Yes |
| Push Mowers | 86 to 92 dB | Yes |
| Riding Lawn Mowers | 88 to 96 dB | Yes |
| Electric Lawn Mowers | 56 to 80 dB | No |
What Affects Lawn Mower Decibel Levels?
The sound intensity of your lawn mower can be affected by several factors, including:
Revolutions Per Minute (RPM): The higher the RPM, the louder the noise your lawn mower produces. This is caused by your engine and your mower blades operating at higher speeds, which increases the noise pollution you hear. Your lawn mower will likely be optimized at around 3,600 RPM which balances performance with decibel level, but any minor deviations from this RPM can lead to noise-induced hearing loss if heard over an extensive period of time.
Blade Design: The sound of your lawn mower's blades cutting through air and grass is one of the primary sources of noise pollution while you mow. While the RPM drives the blades and influences how loud they can be, the design of your lawn mower blades can also impact how many decibels you are able to hear. Modern blade designs are trending towards flatter and thinner blades to reduce their dB sound levels.
Sound Insulation: You can purchase sound-insulating materials that can make your lawn mower quieter and easier on the ears. These are applied to the engine housing or mower deck and can reduce noise pollution by 6 to 8 dB. While using sound insulation is an effective way to reduce noise, your lawn mower's functionality may be affected as well.
Deck Size: Scientific literature has shown that a larger deck size will amplify the mechanical vibrations that your lawn mower makes, often ranging between86.4 and 96.4 dBA. This means that choosing a lawn mower that has a smaller deck size can spare your ears at the cost of reducing the width at which you can cut the grass.
How does the Noise Level of a Lawn Mower Affect Your Hearing?
Prolonged exposure to at least 85 dB or higher will cause noise-induced hearing loss (NI HL). Decibels work on an exponential scale, so while listening to sounds at 85 dB for 8 hours will cause hearing loss, you only need to be exposed to 4 hours of sounds at 88dB to damage your hearing. Any noises above 120 dB will cause immediate damage to your inner ear, potentially leading to tinnitus and permanent hearing loss.
While mowing your lawn will likely take less than 4 hours, certain lawn mowers like riding mowers and gas-powered lawn mowers can reach the 95 dB range. At this decibel range, you may experience hearing damage after only an hour if you are not wearing ear equipment. However, high decibels aren't the only issue you have to worry about if you want to protect your hearing — low-range frequencies may also be disrupting your day.
How do Low Frequencies Affect You?
Lawn mowers also produce a large amount of low-frequencies while you use them, which can impact your cognitive performance and health issues. Low-frequency noises travel further than high frequencies, which means they can be heard further away and even felt through buildings and other structures. These low frequencies can disrupt your entire community, increase stress, and disturb sleep.
If your entire neighborhood is using gas-powered lawn mowers, the ambient noise level will not only make it more likely that you will have exposure to loud decibels, but you will also feel it. These low-frequency noises are present up to around 500 Hz, which falls into the range that most lawn mowers and blades emit (500 to 2,000 Hz).
What can you do to Lower Your Lawn Mower Decibel Levels?
There are three easy ways to lower your exposure to loud lawn mower decibel levels: alter your lawn mower to be quieter, wear ear protection, or switch to an electric lawn mower. No matter what choice you make, it's still recommended you use a sound level meter to monitor the decibels you are experiencing during lawn care. You can opt to invest in a device to tell you sound pressure levels, or just use an app on your phone.
Alter Your Lawn Equipment to Reduce Noise
As we've mentioned, one solution to lowering your exposure to high noise pollution is to use a quieter lawn mower. Here are some ways you can limit the noise your lawn mower makes:
Regular maintenance
Using a high-quality muffler
Soundproofing your mower deck
Silencing the exhaust
Using a quieter blade design
Altering your lawn mower can often be a costly procedure, but with multiple modifications, there is a chance you'll be able to get your lawn mower to run at around 75 decibels.
Wear Ear Protection
The easiest way to lower your exposure to lawn mower noise pollution is to wear hearing protection. Wearing earplugs will decrease your exposure by around 10 to 18 dB. This is measured using the noise reduction rating, although improper fitting and usage will make this reduction far less than normal. Nevertheless, a high performance model earplug will significantly lower noise exposure and can help reduce the likelihood of hearing damage (especially if you work with lawn mowers commercially).
Switch to an Electric Lawn Mower
Switching to an electric mower doesn't only offer you a less noisy mowing experience, but also several other benefits including lower maintenance costs, lower environmental impact, and improved quality of life features. If you want to reduce your exposure to decibels even further, you can invest in innovative remote controlled lawn mowers like theS1 Remote Control Lawn Mower 4WD. Mowrator's premier remote control model means you won't have to worry about gas, oil, or emissions. You’ll be able to precisely control your lawn care from your back porch.
FAQs What is the Decibel Level of a Lawn Mower?
Does weather affect your lawn mower's decibels?
Yes, the weather does affect how loud or quiet your lawn mower is. At a fixed temperature of 60℉ with relative humidity between 20 and 80%, your lawn mower will sound 3 decibels quieter if you are 900 yards away. The colder the weather, the louder garden equipment like a lawn mower or leaf blower will sound.
What is noise-induced hearing loss?
Noise-induced hearing loss occurs when the hairs in your inner ear become too damaged to transmit sound to your brain. If you are exposed to lawn and garden equipment that exceeds 85 decibels for a long period of time without ear protection, you will experience some degree of noise-induced hearing loss.
Back to Top: What is a Standard Lawn Mower Decibel Level?
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187591 | https://stackoverflow.com/questions/34424719/number-of-broadcasts-to-allow-95-of-node-pairs-to-exchange-in-directed-graph | Number of broadcasts to allow 95% of node pairs to exchange in directed graph - Stack Overflow
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Number of broadcasts to allow 95% of node pairs to exchange in directed graph
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Asked 9 years, 9 months ago
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I have a directed unweighted graph with N nodes and E edges. Nodes are of an average degree 2E/N.
In the first round, nodes each broadcast their information to all their neighbours. In subsequent rounds, nodes broadcast the information received from their neighbours during the previous round to all other neighbours, and so forth.
The graph is not guaranteed to be acyclic.
My question is: how many consecutive rounds of broadcast are required, on average, for 95% of node pairs to have reached one another? Is it possible to calculate an approximate figure based on the average degree of the graph?
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edited Dec 25, 2015 at 19:53
user2398029user2398029
asked Dec 22, 2015 at 21:29
user2398029user2398029
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1 Since your graph is directed, there are cases where information won't propagate. Therefore the average is infinite.Lior Kogan –Lior Kogan 2015-12-30 09:22:55 +00:00 Commented Dec 30, 2015 at 9:22
average is a very broad concept -- say you have a worst case a->b->c->d and all other topologies are "nice", what should the weight / frequency of worst case be? 0.53? 50%? 1%?Dima Tisnek –Dima Tisnek 2015-12-31 10:31:37 +00:00 Commented Dec 31, 2015 at 10:31
@LiorKogan good observation, e.g. in a case like a->x; b->x; c->x node x will learn entire graph, by a,b,c will never learn their siblings. That in itself doesn't guarantee infinite 95-percentile though, does it? For example OP could constrain graph to have less than 5% leaf (and thus root) nodes. Or to typically have such guarantee Dima Tisnek –Dima Tisnek 2015-12-31 10:41:09 +00:00 Commented Dec 31, 2015 at 10:41
I'm looking for the typical case.user2398029 –user2398029 2015-12-31 15:05:25 +00:00 Commented Dec 31, 2015 at 15:05
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By average, I assume you mean average over all possible (N,E) directed graphs with no multiple edges.
Theorem 1
If E <= (N-1)^2, there will be at least one graph where information won't propagate.
Proof
A directed graph with N nodes has up to N(N-1) edges. Consider a complete graph, select a node, and remove all its outgoing edges (Alternatively, we can remove all its incoming edges). Information from this node cannot propagate, and we are left with N(N-1)-(N-1) = (N-1)^2 edges.
Corollary 1
When E <= (N-1)^2, there is at least one graph where information cannot propagate, therefore the average number of rounds is infinite.
Theorem 2a
If E > (N-1)^2 the maximal number of rounds is 2.
Proof
A directed graph with N nodes and E > (N-1)^2 edges is a complete graph where up to (N-2) edges removed.
If we want to remove edges from a complete graph such the the number of rounds will be 3 (e.g. from node A to node B), we'll need to make sure there is no node B and edges A->B and B->C. This means that we need to remove at least one edge (either A->B or B->C) for each of the (N-2) possible 'B' nodes. We also need to remove the direct A->C edge. In total we need to remove (N-3) edges.
Theorem 2b
If E > (N-1)^2 the minimal number of rounds is 2.
Proof
Trivial. The graph is incomplete, therefore there is at least one path of length 2.
Corollary 2
if (N-1)^2 < E < N(N-1), the number of rounds is 2.
Theorem 3
If E = N(N-1), the number of rounds is 1
Proof
Trivial. Complete graph.
Now, you are asking about more than 95% of the node pairs.
Of course we can build some (N-1)^2 < E < N(N-1) graphs, where >= 95% of ordered-node-pairs can communicate in 1 round, but the other ordered-node-pairs communicate in 2 rounds.
This is trivial if you consider a complete directed graph of 6 nodes where only one edge is removed. (65-1) / (65) = 96.66% of the ordered-node-pairs can communicate in one round.
Why do you ask specifically about 95%? Is it important to derive calculations for exactly this number? Let us know. I don't think that you can derive a simple accurate generic formula, especially when N and E are small. Maybe we can formulate something asymptotically (for very large N).
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edited Dec 31, 2015 at 12:30
answered Dec 31, 2015 at 12:21
Lior KoganLior Kogan
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user2398029Over a year ago
I'm modelling a graph algorithm using message passing. Accuracy of the solution will depend on how complete the message exchange process is. The number 95% is not important, I'd just like to be able to calculate a typical case bound on the number of nodes that might be left out from the calculation.
2015-12-31T15:04:43.167Z+00:00
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Lior KoganOver a year ago
If so, i believe i answered your question ;-)
2015-12-31T15:08:30.363Z+00:00
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187592 | https://brilliant.org/wiki/interest-rate/ | Interest Rate
Ashish Menon,
Sandeep Bhardwaj,
Christopher Williams,
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Contents
Simple Interest
Compound Interest
Continuously Compounded Interest
Rule of 72
Annual Percentage Rate (APR)
Annual Percentage Yield (APY)
Simple Interest
Simple interest (SI) can occur either when a person borrows money or invests it. When a borrower receives a certain sum of money over a period of time, they agree to pay it back, along with a fee, known as the interest owed. Interest in an investment is the money one can earn by initially investing some money (called principal) and receiving a return on that investment. The return is a percentage of the principal (interest) and is added to the principal, making the initial investment grow.
Simple Interest is a type of interest that is applied to the amount borrowed or invested for the entire duration of the loan, without taking any other factors into account, such as past interest (paid or charged) or any other financial considerations. Simple interest is paid only on the original principal, it does not compound. Simple interest is generally applied to short-term loans, usually one year or less, that are administered by financial companies, or money invested for a similarly short period of time.
Simple interest (SI) is calculated by using the formula
SI=100P×R×T.
Here P is principal amount, R is rate of interest, and T is time period of interest. The final amount to be paid is the initial principle plus the simple interest, P+SI.
Here are some illustrations of the concept of simple interest through the following examples:
An invested sum fetched a total simple interest of 4016.25 rupees in 1 year at the rate of 9% per annum. What was the initial principal?
Let the principal amount be P, rate of interest R% per annum, SI the simple interest, and T the time period, then
[\begin{align}
SI & = \dfrac{PRT}{100}\ \
4016.25 & = \dfrac{P× 9 × 1}{100}\ \
\Rightarrow P & = \dfrac{4016.25 × 100}{9} =44625.\ _\square
\end{align}]
In how many years will an amount of Rs.1000 generate an equal return, if interest is 10% per annum?
(How many years will it take to double?)
Let the principal amount be P, rate of interest be R% per annum, SI be the simple interest, and T be the time period. Then
[\begin{align}
\text{SI} & = \dfrac{\text{PRT}}{100}\ \
\require{cancel}{\cancel{\text{P}}} & = \require{cancel}{\dfrac{\cancel{\text{P}}\text{RT}}{100}} \qquad \left(\text{because P = SI}\right)\ \
1 & = \dfrac{10 × \text{T}}{100}\ \
\Rightarrow \text{T} & = \dfrac{100}{10}=10.\ _\square
\end{align}]
A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. What is the rate of interest (in percent) per annum?
Let the principal amount be P, rate of interest R% per annum, SI the simple interest, and T the time period. Then
SI for 3 yearsSI for 5 years=Rs.(12005−9800)=Rs.2205=Rs.32205×5=Rs.3675.
Therefore, the principal is (9800−3675)=6125 rupee, which implies
T=6125×53675×100=12. □
Abhay lent me some amount of money 3 years ago. If what I owe him now amounts to Rs.127 taking into consideration the simple interest of 9% per annum, find the amount of money that he lent me.
The correct answer is: 100
18 years
15 years
10 years
20 years
When calculating with simple interest, a sum of money doubles itself in 10 years.
In how many years would it triple itself (with simple interest)?
The correct answer is: 20 years
Fill in the blank:
Consider a bank that gives a simple interest rate of i%.
If I deposit $6500 into this bank, the money will be accumulated to $8840 after 4 years.
If I deposit $_____ into this bank, the money accumulated will be $10200 after 4 years.
The correct answer is: 7500
Compound Interest
Typically interest rates on investments and loans are compound interest rates; the interest is not calculated off of just the initial principle but on the amount invested or owed at the time of the calculation. In this way, interest is said to compound. This works to an investor's advantage, allowing their returns to generate more returns, not just their initial investment.
Interest can be compounded discretely at many different time intervals. The number of and the distance between compounding periods are explicitly defined with discrete compounding. For example, interest that compounds on the first day of every month is discrete. Interest that is calculated at the end of the year is said to "compound annually," interest that is calculated at the end of the month is said to "compound monthly," and so on. Interest can also compound continuously.
Suppose that you open an account at the beginning of Year 1 with a principal of $1000, and that the bank provides an annually compounding interest rate of 5% (this is much higher than most savings rates today). At the end of the year, the bank will add $50 to your account, and your new balance will be $1050.
At the end of Year 2, instead of using the original balance of $1000, the bank will use the new balance of $1050 to determine the interest. They multiply $1050×5%=$52.50. This is $2.50 more interest than you earned last term.
At the end of Year 3 the new balance will be $1102.50. So to figure that term's interest, the bank uses that new, higher balance instead of the original $1000 balance, now earning $55.12 (if they truncate). See how this works in favor of the person making the interest? The interest builds up much more quickly because the bank is using a higher balance every year to calculate the new interest.
The formula for calculating the amount returned when interest is compounded annually is
A=P(1+100R)t,
where A is the amount obtained, P is the principal amount, R is the rate of interest, and t is the number of times the interest compounds in this time period (for annual compounding this is the number of years). Then, the total compound interest over the term calculated is
Compound Interest=A−P.
Interest at the end of the first year is
100P1RT=100PR×1=100PR.
So, the amount at the end of the first year is
P1+I1=P+100PR=P(1+100R).
Interest at the end of the second year is
100P2RT=100P(1+100R)R×1=P(1+100R)100R.
So, the amount at the end of two years is
P2+I2=P(1+100R)+P(1+100R)100R=P(1+100R)(1+100R)=P(1+100R)2.
Interest at the end of the third year is
100P3RT=100P(1+100R)2R×1=P(1+100R)2100R.
So, the amount at the end of three years is
P3+I3=P(1+100R)2+P(1+100R)2100R=P(1+100R)(1+100R)(1+100R)=P(1+100R)3.
So, we see that the amount in each case follows a geometric progression with first term P and common ratio (1+100R).
Therefore, the amount at the end of the nth year is the(n+1)th term of the geometric progression, which is
P(1+100R)n+1−1=P(1+100R)n. □
Find the interest on Rs.500 for 2 years if the interest is compounded annually at a rate of 5% per annum.
Let A be the amount obtained, P the principal amount, R the rate of interest, t the time period, and CI the compound of interest. Then
[\begin{align}
\text{A} & = \text{P}{\left(1 + \dfrac{\text{R}}{100}\right)}^{\text{t}}\ \
& = 500 × {\left(1 + \dfrac{5}{100}\right)}^{2}\ \
& = 500 × {\left(\dfrac{21}{20}\right)}^{2}\ \
& = 500 × \dfrac{441}{400}\ \
& = 551.25\ \
\text{CI} & = \text{A} - \text{P}\
& = 551.25 - 500\
& = 51.25.\ _\square \end{align}]
What would Rs.100 amount to in a year if the interest is compounded at a rate of 10% half-yearly (every six months)?
Let A be the amount obtained, P the principal amount, R the rate of interest, and t the number of times that interest compounds. Then
[\begin{align}
\text{A} & = \text{P}{\left(1 + \dfrac{\text{R}}{200}\right)}^{\text{t}}\ \
& = 100 × {\left(1 + \dfrac{10}{200}\right)}^{2}\ \
& = 100× {\left(\dfrac{21}{20}\right)}^{2}\ \
& = 100 × \dfrac{441}{400}\ \
& = \dfrac{441}{4}\ \
& = 110.25.\ _\square \end{align}]
Try the following problems:
Once Abhay lent Rs.360 to me for 2 years at an interest rate of 9% per annum. Being my friend he lent the money on simple interest. How much more would I have to pay if the interest was compounded annually?
If the answer is Rs.A, submit the value of 1000A.
The correct answer is: 2916
Skanda borrowed a total amount of $20,000 from Sravanth.
Sravanth asked him to return back his money within 2years at the rate of 5%.
Calculate the interest which should be paid by Skanda to Sravanth, if Sravanth likes to receive interest compounded annually.
The correct answer is: 2050
Continuously Compounded Interest
Discrete and continuous compoundings are closely related terms. An interest rate is discretely compounded whenever it is calculated and added to the principal at specific intervals (such as annually, monthly or weekly). Continuous compounding means that an amount is always said to be "accruing" interest. Continuously compounding interest uses a natural log-based formula (natural logs being logarithms with a base of e) to calculate and add back accrued interest at the smallest possible intervals.
Continuously compounded interest means that principal is constantly earning interest with the interest immediately calculating and compounding. Graphically, it can be demonstrated as below.
Mathematically, continuously compounding interest can be expressed as follows:
A=Pert,
where A represents the total amount at a given time, P represents the initial principal, r is interest rate (expressed as a decimal), and t is the time.
This is similar to taking the discrete compound interest formula
A=P(1+xr)xt,
where A represents the amount earned, P represents the initial principal, r is interest rate, x is the number of times the interest is credited in a year, and t is the time. Then transforming it from discrete to continuous by taking x→∞, that is "constantly earning interest with the interest immediately calculating and compounding":
A=Px→∞lim(1+xr)xt=Pert.
If you invest $2400 at an annual interest rate of 10% compounded continuously, calculate the final amount you will have in the account after 5 years.
Let P be the principal amount, A the amount, r the interest rate, and t the time period. Then
[\begin{align}
\text{A} & = \text{P}e^{\text{rt}}\
& = 2400 × e^{\tfrac{10×5}{100}}\
& = 2400 × e^{0.5}\
& = 2400 × 1.6487\
& = 3956.93.\ _\square
\end{align}]
Note: 10% annual interest compounded discretely for five years on a principal of $2400 would result in a total of $3865.22, a difference of $91.70. □
If you invest $500 at a rate of 9% continuously compound interest, how much would you have to pay back after 10 years?
Give your answer (in $) to two decimal places.
The correct answer is: 1229.80
Rule of 72
The rule of 72 states that at r% annual interest rate, an investment would take approximately r72 years to double.
Solve for t from the exponential equation:
[\begin{aligned} \text{A} &= \text{P}{\left(1 + r\% \right)}^{\text{t}} \
2\text{P} &= \text{P}{\left(1 + r\% \right)}^{\text{t}} \
\ln 2 &= \text{t}\ln (1+r\%) \
\text{t} &= \frac{\ln 2}{\ln (1+r\%)}.
\end{aligned}]
Applying the Maclaurin series to ln(1+r%)=r%+… and using ln2≈0.69, we obtain
ln(1+r%)ln2≈r%0.69=r69.
We then change the numerator slightly to 72, because 72 is easier to divide for small values of r. □
What is the error of applying rule of 72 with the actual calculation for an annual interest rate of 9%?
From rule of 72,
t=972=8.
Actual result is
t=ln(1+0.09)ln2=8.04.
The error is calculated with the formula 8.048.04−8×100%=0.50%. □
Below is the analysis table for the accuracy of rule of 72.
| | | | | |
--- ---
| Rate of Interest (r) | Rule of 72 | Actual Number | Difference | Error |
| 1% | 72 | 69 | 3 | +4.35% |
| 2% | 36 | 35 | 1 | +2.86% |
| 3% | 24 | 23.4 | 0.6 | +2.56% |
| 5% | 14.4 | 14.2 | 0.2 | +1.41% |
| 8% | 9 | 9.01 | 0.01 | −0.11% |
| 12% | 6 | 6.12 | 0.12 | −1.96% |
| 16% | 4.5 | 4.67 | 0.17 | −3.64% |
| 25% | 2.88 | 3.11 | 0.23 | −7.40% |
| 50% | 1.44 | 1.71 | 0.27 | −15.79% |
| 72% | 1 | 1.27 | 0.27 | −21.26% |
| 100% | 0.72 | 1.00 | 0.28 | −28.00% |
For an error margin between −3% and +3%, the rule of 72 is fairly accurate for rate of interest in range 2%~16%, which turns out to be most of the cases in real world.
24
2
72
3
18
36
Based on the rule of 72, approximately how many years will it take to double the principal in an account which earns 3% of interest per year?
The most powerful force in the universe is compound interest. - Albert Einstein
As r increases, the approximation from Maclaurin series starts to diverge. Hence, this provides a way of estimating the extent of error. For example, since ln(1+r)=r−2r2+⋯, the approximation by r% introduces an error of 2r2%.
For more details, see this discussion.
Annual Percentage Rate (APR)
The annual percentage rate (APR) of a loan is the cost of the credit per year, expressed as a percentage of the loan amount. It is the annual rate that is charged for borrowing (or made by investing), expressed as a single percentage number that represents the actual yearly cost of funds over the term of a loan. It refers to the rate paid to a financial institution by a borrower.
For instance, a credit card might express their APR as 20%. This means that a card holder who uses their credit card to make a $100 purchase, and does not pay it back for 365 days can expect to owe $20 in interest at the end of that time period.
APR does not specify the term that interest is compounded at, merely the annual rate of whatever the interest is (called the nominal interest rate) at whatever term length it's compounded plus fees and additional costs. Yes, this means that the APR is typically higher than just the nominal rate, as this includes any fees or additional costs associated with the transaction.
Annual Percentage Yield (APY)
The APY of a loan is the amount that is earned on an interest-bearing investment per year, expressed as a percentage of the total due. The APY is typically higher than the interest rate, as the interest is being compounded. However, it does not account for the possibility of account fees affecting the net gain. APY generally refers to the rate paid to a depositor by a financial institution.
The formula for calculating the APY is given as
APY=100(1+PrincipalInterest)Days in terms365−1.
Cite as:
Interest Rate.
Brilliant.org.
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02:43, September 28, 2025,
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Joanne Bowers,
Joanne Bowers
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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Naayema Hussaini,
Naayema Hussaini
School of Medicine, Trinity Biomedical Sciences Institute, Trinity College Dublin, Dublin, Ireland
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Maria Charalambides,
Maria Charalambides
Birmingham Medical School, College of Medical and Dental Sciences, University of Birmingham, Birmingham, England
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Jack Coumbe,
Jack Coumbe
Henriette Raphael House, King's College London, Guy's Campus, London, England
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Carsten Flohr,
Carsten Flohr
St John's Institute of Dermatology, Guy's and St Thomas' NHS Foundation Trust, London, England
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Alison M. Layton,
Alison M. Layton
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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Aoife U. Daly,
Corresponding Author
Aoife U. Daly
adaly27@qub.ac.uk
School of Medicine, Dentistry and Biomedical Sciences, Queens University Belfast, Belfast, Northern, Ireland
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
Correspondence Aoife U. Daly, School of Medicine, Dentistry and Biomedical Sciences, Queens University Belfast, University Rd, Belfast BT7 1NN, Northern Ireland.
Email: adaly27@qub.ac.uk
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Rui Baptista Gonçalves,
Rui Baptista Gonçalves
orcid.org/0000-0002-4373-5379
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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Eva Lau,
Eva Lau
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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Joanne Bowers,
Joanne Bowers
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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Naayema Hussaini,
Naayema Hussaini
School of Medicine, Trinity Biomedical Sciences Institute, Trinity College Dublin, Dublin, Ireland
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Maria Charalambides,
Maria Charalambides
Birmingham Medical School, College of Medical and Dental Sciences, University of Birmingham, Birmingham, England
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Jack Coumbe,
Jack Coumbe
Henriette Raphael House, King's College London, Guy's Campus, London, England
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Carsten Flohr,
Carsten Flohr
St John's Institute of Dermatology, Guy's and St Thomas' NHS Foundation Trust, London, England
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Alison M. Layton,
Alison M. Layton
Dermatology Department, Harrogate and District NHS Foundation Trust, Harrogate, England
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First published: 26 April 2023
Citations: 9
Abstract
How does low-dose oral isotretinoin compare with recommended and high-dose oral isotretinoin in terms of efficacy, adverse effect profile, duration of remission, economic profile and rate and severity of relapse? A systematic literature search of EMBASE, MEDLINE, PubMed and The Cochrane Library from 1983 to July 2022 was conducted to identify randomised control trials (RCTs), cohort studies and cross-sectional studies investigating the treatment of acne vulgaris with oral isotretinoin. This systematic review of the literature sought to explore and compare the use of different isotretinoin dosing regimens. Primary outcomes were efficacy, relapse, and safety profiles of oral isotretinoin of varying doses. Secondary outcomes included adverse events and economic considerations. The quality of studies, including risk of bias was assessed using GRADE (Grading of Recommendations, Assessment, Development and Evaluations). A total of 32 studies were included. Severe acne responds better to conventional or high fixed daily doses of isotretinoin. In cases of mild to moderate acne, where oral isotretinoin was prescribed, clearance rates were comparable with low, conventional or high dosage regimens. Despite this, relapse was more frequent in those treated with a lower dose. The severity of mucocutaneous adverse effects worsens as the dose of isotretinoin is increased. Comparison between studies was challenging due to differing methods of assessment, outcome measures and duration of follow-up. This review highlights the need for an adequately powered RCT comparing low, conventional and high doses of oral isotretinoin to establish optimal dosing for treatment and prevention of relapse in acne vulgaris.
INTRODUCTION
Acne vulgaris (syn acne) is a very common chronic inflammatory skin disease affecting an estimated 9.4% of the world's population.1 Acne is a disease of the pilosebaceous follicle and predominantly affects the face, chest and back. The main aetiological factors relate to an androgen-mediated increase in sebum production, altered keratinisation within the intrafollicular duct, colonisation with Cutibacterium acnes (C.acnes) and inflammatory changes. The disease manifests as seborrhoea, closed and open comedones, papules, pustules and nodules. Acne, and the scarring that can result from it, can have a profound impact on psychosocial aspects as well as health-related quality of life for patients.2, 3 Multiple treatment options exist, ranging from topical treatments such as benzoyl peroxide, retinoids or antibiotics used as monotherapy or fixed combination products for mild, moderate and, in some cases, severe disease, as well as oral antibiotics used in combination with topical agents for more moderate–severe disease.4 Hormonal treatments can be considered in adult females including the oral contraceptive pill5 and oral spironolactone.6 Oral isotretinoin is recommended for severe forms of acne including acne with evidence of and/or at risk of permanent scarring that has failed adequate courses of standard therapy with systemic antibiotics and topical therapy.4, 7
Oral isotretinoin (13-cis-retinoic acid) is a synthetic Vitamin A analogue, which is involved in sebaceous gland apoptosis and shrinkage via p21 protein expression and decreased cyclin D.8, 9 13-cis-retinoic acid has been found to have antiandrogenic effects via oxidation processes10 and reduces insulin-like growth factor 1, insulin-like growth factor-binding protein 3, luteinising hormone, prolactin, adrenocorticotropic hormone and free T3.11 Oral isotretinoin is highly lipophilic and taking the medication with a fatty meal can enhance absorption by 60%.12 It is associated with adverse effects including increased serum triglycerides, deranged liver function tests (LFTs), xerosis of the skin and mucous membranes and teratenogenicity.13-15 Although it remains controversial and no causal relationship has been firmly established, a range of neuropsychiatric effects have been described, and isotretinoin may induce changes in brain functioning.16
Current UK recommendations and evidence-based guidelines suggest oral isotretinoin should be prescribed at a dose of between 0.5 and 1.0 mg/kg/day with a desired cumulative dose of between 120 and 150 mg/kg.4 This is corroborated in the indication by the European Medicines Agency and the European Evidence-Based Guidance that isotretinoin should be started at a dose of between 0.3 and 0.5 mg/kg.17 Although these dosage regimens have been established as an effective treatment for severe acne and acne unresponsive to more conventional therapies, there is debate about optimum dosing regimens. There have been reports in recent years suggesting that lower dosing regimens are effective, better tolerated13-15, 18 and more economical.15, 19, 20 However, the comparable efficacy, prevalence of adverse events and long-term remission of lower doses compared with higher dosing regimens is still unclear.13, 14
This systematic review seeks to evaluate, through the available literature, treatment of acne by comparing the use of different dosing regimens of oral isotretinoin. Primary outcomes were efficacy, relapse, and safety profiles of oral isotretinoin of varying doses. Secondary outcomes sought to examine side effects and adverse events of low-dose isotretinoin compared with standard dosing regimens.
METHODS
Systematic searches were performed on EMBASE, MEDLINE, PubMed and The Cochrane Library, from 1983 (year in which isotretinoin was licensed for medical use in the European Union) to July 2022 to identify English-language studies that presented data on differing doses of isotretinoin and its efficacy and safety for treating acne vulgaris. Our strategy searched for the words “isotretinoin” OR “roaccutane” within the title or abstract OR “ISOTRETINOIN” in the subject heading. The previous terms were combined with the following “acne” terms: “acne” within the title or abstract OR exp ACNE. And combined with the following “dosage” terms: “DRUG DOSAGE COMPARISON” in the subject heading OR (high ADJ3 dos OR low ADJ3 dos OR conventional ADJ3 dos) in the title or abstract.
This study was registered on PROSPERO (CRD42021215665). Figure 1 shows the results of the reviewing process as a PRISMA21 flowchart. Further details of the search strategy, study selection and risk of bias assessment can be found in Supporting Information: Appendix 1.
The following data were extracted from each included study: first author, year, study design, number of participants and their characteristics (including age, ethnicity, sex), dose per day (mg), duration of treatment, number of isotretinoin courses, cumulative dose (mg), definitions of efficacy, efficacy rate, relapse and relapse rate, site of acne, acne grading system used, acne severity, dietary factors, adverse side effects, including biochemical, cheilitis, acne flare and mood disturbance, follow-up period, economic considerations. A summary of key data is displayed in Table 1. The quality of studies was assessed using GRADE (Grading of Recommendations, Assessment, Development and Evaluations)47 and an overall quality score given (summarised in Table 2).
Table 1.
Summary of all studies included: Design, isotretinoin dosing, number of participants, groups included, key results including efficacy and relapse rates, as well as definition of relapse and follow-up time.
| References | Study type | Dosing | N | Dose groups of isotretinoin | Efficacy rate | Relapse rate | Relapse definition | Follow up duration posttreatment |
--- --- --- ---
| Agarwal et al.22 | RCT | Conventional versus intermittent and/or low doses | 120 | A. 1 mg/kg/day B. 1 mg/kg/day alternate day C. 1 mg/kg/day 1 week/4 weeks D. 20 mg alternate days 16 weeks (treatment duration 24 weeks) | A: 96.03% B: 90.33% C: 76.85% D: 93.89% (efficacy defined as reduction in total acne load) | 0% | Emergence of pretreatment acne score | 2 months |
| Ahmad23 | RCT | Conventional OD versus divided dosing | 58 | A. 0.5–1 mg/kg/day once daily dose B. 0.5–1 mg/kg/day dived doses (mean treatment duration 22 weeks) | A: median GAGS decrease from 34 to 0 B: median GAGS decrease from 31 to 0 | Not assessed | N/A | None |
| Akman et al.24 | RCT | Fixed versus intermittent conventional dose | 66 | A. 0.5 mg/kg/day of isotretinoin first 10 days of each month for 6 months B. 0.5 mg/kg of isotretinoin each day for 1st month then the first 10 days of each month for 5 months C. 0.5 mg/kg/day isotretinoin daily for 6 months | Significantly decreased global grade in all groups p < 0.001 | 14% (all in Group A) | Emergence of pretreatment acne score | 1 year |
| Amichai et al.19 | Prospective noncomparative | Low dose | 638 | All received 20 mg/day daily for 6 months Group 1. Ages 12–20 years Group 2. Ages 21–35 years | Group 1. 94.8% achieved significant improvement or complete remission Group 2. 92.6% achieved significant improvement or complete remission (No statistically significant difference p = 0.36) | Group 1. 3.9% Group 2. 5.9% (No statistically significant difference p = 0.35) | Not defined | 4 years |
| Bellosta et al.25 | Prospective noncomparative | Conventional dose divided | 60 | 0.5 mg/kg/day in divided doses for 12–20 weeks | Statistically significant reduction in pustules at 2 weeks and nodules at 4 weeks (p = 0.01) | 5% | Not defined | 1 year |
| Blasiak et al.26 | Prospective observational | High cumulative dose | 180 | A: cumulative dose <220 mg/kg B: cumulative dose >220 mg/kg | 97.4% reported improvement | Relapse - Group A 47.4%, Group B 26.9% (Statistically significant p < 0.3) Retrial - Group A 0%, Group B 1.72% | Relapse- treatment with any prescription medication other than isotretinoin retrial- retreatment with isotretinoin | 1 year |
| Borghi et al.48 | Prospective noncomparative | Low cumulative dose | 150 | 0.2 mg/kg/day increased by 5 mg fortnightly until maximum dose tolerated mean cumulative dose of 80.92 mg/kg | Aim of the study was to determine relapse rate | 9.35%. When divided into groups of cumulative dose <40 mg/kg, 40–79 mg/kg, 80–119 mg/kg and >120 mg/kg there was no statistically significant difference | Emergence of acne >0.5 Leeds Acne Grade or requiring systemic treatment | 2 years |
| Boyraz and Mustak53 | Prospective observational | Fixed versus intermittent low dose | 60 | A: 20 mg daily B: 0.5–0.8 mg/kg/day for 1 week every month | 100% improvement in both groups at 8 months | Group A 0% Group B % 10% | Not defined | 6 months |
| Cyrulnik et al.27 | Retrospective noncomparative | High dose | 80 | Mean daily dose 1.6 mg/kg/day | 80% of patients had no active disease at 150 days of treatment | 12.50% | Further course of isotretinoin required | 3 years |
| Del Rosso et al.50 | Prospective noncomparative | Conventional dose lidose-isotretinoin | 201 | 0.5 mg/kg/day for 4 weeks then 1 mg/kg/day for 16 weeks of lidose-isotretinoin | aim of the study was to determine relapse rates | Retreatment with isotretinoin 4.2% retreatment with any acne therapy 15.1% (topical, oral, OTC) | The point at which patient required any form of acne treatment, topical or oral, OTC | 2 years |
| Dhaked et al.28 | RCT | Low dose daily versus alternate days | 240 | A: 20 mg daily B: 20 mg alternate days | Decrease in acne load from baseline Group A: 98.99%, Group B: 97.69% (p < 0.01) | 0% | Emergence of near pretreatment severity of acne in treated patients within 12 weeks of follow up | 3 months |
| El-Sherif et al.29 | RCT | Low dose versus low intermittent dose | 55 | Group 1 20 mg OD Group 2 20 mg alternate days | >50% reduction of lesions in both groups | 21.9% in Group 1, 39.1% in Group 2. (Not statistically significant) | Not defined | 6 months |
| Faghihi et al.30 | RCT | Low versus conventional dose | 60 | A: 0.25 mg/kg/day B: 0.5 mg/kg/day | Both groups showed significant difference in improvement of severity of acne at baseline,6 and 12 months but no significance when comparing both gorups (p > 0.005) | Not assessed | Not defined | 6 months |
| Ghaffarpour et al.31 | Retrospective noncomparative | Conventional dosing | 132 | Mean daily dose 0.56 mg/kg/day mean duration 6.6 months | Mean improvement of 96.7% | 18.35% (9.17% required further course isotretinoin) | Not defined | Mean 4.4 years |
| Goulden et al.32 | Prospective noncomparative | Intermittent conventional dose | 80 | 0.5 mg/kg/day for 1 week every 4 weeks for 6 months | Both total acne grades and inflamed lesion count significantly reduced (p < 0.0001), acne resolved in 88% | 39% | Acne grade >0.25 at any single affected area and lesion count >5 | 1 year |
| Hermes et al.33 | Retrospective noncomparative | Low dose isotretinoin Incremental increase | 94 | mean daily dose 31.4 mg or 0.43 mg/kg started at 10 mg/day then increased by 10 mg weekly until dose of 50 mg/day if tolerated (treated until 1 month after clearance) | Clinical improvement 4-point scale - very good, good, satisfactory, not satisfactory 62.8% very good, 31.9% good, 5.3% satisfactory | Relapse rate overall 33% of which 21.3% required retreatment with isotretinoin | Not defined | Range of 1–6 years |
| Hafeez et al.34 | RCT | Low versus conventional dose | 140 | Group A 20 mg/day Group B 80 mg/day (mean treatment duration 8.3 months) | A 50%, B 71.4% (p = 0.009) | Not assessed | N/A | None |
| Kapadia et al.35 | RCT | Low versus conventional | 60 | A: 20 mg/day B: 40 mg/day (treatment duration 6 months) | 100% improvement in both groups at 6 months | Not assessed | N/A | None |
| Kaymak et al.51 | Prospective, noncomparative | Intermittent dosing | 60 | A: mild acne (Leeds 0.75–1):0.5 mg/kg/day B: moderate acne (Leeds 1.25–1.5): 0.6–0.75 mg/kg/day for 1 week out of every 4 for 24 weeks | 82.9% complete resolution | Not assessed | N/A | None |
| Lee et al.36 | RCT | Conventional fixed versus low dose fixed versus conventional intermittent | 60 | A: 0.5–0.7 mg/kg daily conventional B: 0.25–0.4 mg/kg daily low dose C: 0.5–0.7 mg/kg daily for 1 week every 4 weeks intermittent Treated for 6 months | differences in GAGS score, significant between A and C (p < 0.001) B and C (<0.044), not significant between A and B | Group A 13%, Group B 18%, Group C 56% | Moderate or severe acne according to GAGS score | 1 year |
| Mandekou- Lefaki et al.37 | Prospective interventional | Low versus conventional | 64 | A: 0.15–0.40 mg/kg/day low, mean daily dose 0.24 mg/kg/day B: 0.5–1.0 mg/mg/day high, 0.67 mg/kg/day Mean duration treatment Group A 8 months Group B 5.7 months | Group A 69%, Group B 91% No significant difference between groups (χ2 = 1.002) | Group A 9.37%, Group B 3.1% | When acne severe enough to require any form of oral treatment | 7 years |
| Niazi and Shehzad52 | RCT | Low dose daily versus alternate days | 60 | A: 20 mg/day B: 20 mg alternate days | Group A - efficacy 90%, mean percentage decrease in GAGS 73.95 ± 14.04 Group B - efficacy 86.7%, mean percentage decrease in GAGS 66.57 ± 14.97 no significance- p > 0.05 | Not assessed | N/A | 1 year |
| Rademaker et al.49 | RCT | Low dose | 60 | A: 5 mg daily for 32 weeks B: placebo for 16 weeks, then 5 mg open-label isotretinoin daily for 16 weeks | Significant differences in lesion count 3.6 versus 7.2 after 16 weeks on iso compared with placebo p < 0.0001 DLQI showed significant difference between groups at Week 16 (p < 0.001) but no difference by Week 32. | Not assessed | N/A | 2.5 months |
| Rademaker38 | Retrospective data analysis | Effects of daily and cumulative dose on relapse | 1453 | Total: mean dose 44.2 m ± 19.8 = 0.66 mg/kg ± 0.32 control: 42.7 ± 20.1 = 0.58 mg/kg ± 0.32 study: 49.2 ± 17.9 = 0.71 mg/kg ± 0.30 | Study group: 95.5% clear after 1 course | 22.40% | Defined as further course of isotretinoin. Restarted at patient request | 1–5 years |
| Rao et al.39 | Prospective noncomparative | Low dose | 50 | 20 mg/day (0.3–0.4 mg/kg/day) for 3 months | At end of 12 weeks, 6% of patients clear, 90% had >75% clearance and 4% had 50%–75% clearance Overall efficacy 90% | 4% | Recurrence of GAGS grade 1 or more | 6 months |
| Rasi et al.40 | Retrospective noncomparative | Low dose | 146 | 20 mg/day until a cumulative dose of 120 mg/kg given | 96.4% complete clearance | 7.90% | Not defined | 5 years |
| Sardana et al.41 | Prospective noncomparative | Low-dose alternate days | 320 | 20 mg alternate days with topical clindamycin 1% gel BD for 6 months | Very good response (greater than 80% improvement) in 68.2% | 16.39% | Emergence of pretreatment severity of acne | 6 months |
| Shetti et al.42 | RCT | Low-dose fixed versus low-dose intermittent | 100 | A: 20 mg OD for 4 months B: 20 mg OD 1 week every 4 weeks for 4 months | Group A = 86%, Group B = 70% (difference in GAGS significant p < 0.005) | Relapse rate higher in Group B not significant | Not defined | 8 months |
| Strauss et al.43 | RCT | Low versus conventional higher and lower-end range | 150 | A: 0.1 mg/kg/day B: 0.5 mg/kg/day C: 1 mg/kg/day for up to 20 weeks or when 70%–80% of lesions had resolved | Improvement of lesion count with all groups p < 0.01 at all treatment and posttreatment periods for face and trunk | Group A 30.0%, Group B 31.4%, Group C 6.7% 42% of patients in Group A require retreatment compared with 10% in Group C | Not defined | 12–18 month |
| Ullah et al.44 | Prospective observational | Low dose isotretinoin + doxycycline alternate days | 30 | Isotretinoin 20 mg/doxycycline 100 mg BD alternate days up to 6 months | 85.71% complete clearance at 3 months- significant (p < 0.01), a further 10.71% clear at 5 months | Not assessed | N/A | None |
| Van der Meeren et al.45 | RCT | Conventional doses versus each other | 58 | A: 0.5 mg/kg/day - 31 patients B: 1.0 mg/kg/day - 27 patients for 6 months | No significant correlation between two doses and clinical efficacy after 12 weeks (p > 0.05) | not recorded | N/A | Up to 15 months |
| Yap46 | Prospective noncomparative | Fixed low dose | 150 | 10 mg/day fixed-dose until cumulative dose of 90–110 mg/kg | All patients achieve complete clearance of acne lesions (mean duration 24 weeks) | 4% | Requiring topical or oral treatment | 1 year |
Abbreviations: BD, twice daily; DLQI, Dermatology Life Quality Index; GAGS, Global Acne Grading System; OD, once daily; OTC, over the counter; N/A, not applicable; RCT, randomised controlled trial.
Table 2.
GRADE evidence profile for the 32 studies included in the systematic review.
| References | Study design | Risk of bias | Inconsistency | Indirectness | Imprecision | Other considerations | | | | N | Score |
--- --- --- --- --- --- |
| Publication bias | Large effect | Plausible confounding | Dose-response gradient |
| Dhaked et al.28 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 240 | Low |
| Shetti et al.42 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 100 | Low |
| Ahmad23 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 58 | Low |
| Ullah et al.44 | Observational study | Serious | Not serious | Serious | Serious | None | None | None | None | 30 | Low |
| Faghihi et al.30 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 40 | Low |
| Niazi and Shehzad52 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Boyraz and Mustak53 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Lee et al.36 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 40 | Low |
| Rasi et al.40 | Retrospective noncomparative | Serious | Not serious | Serious | Serious | None | None | None | None | 146 | Low |
| Sardana et al.41 | Prospective noncomparative | Serious | Not serious | Serious | Not serious | None | None | None | None | 320 | Low |
| Blasiak et al.26 | Prospective interventional | Serious | Not serious | Not serious | Not serious | None | None | None | None | 180 | low |
| Cyrulnik et al.27 | Retrospective noncomparative | Serious | Not serious | Not serious | Not serious | None | None | None | None | 80 | Low |
| Bellosta et al.25 | Prospective noncomparative | Serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Hermes et al.33 | Retrospective noncomparative | Serious | Not serious | Serious | Serious | None | None | None | None | 94 | Low |
| Ghaffarpour et al.31 | Retrospective noncomparative study | Serious | Not serious | Not serious | Not serious | None | None | None | None | 134 | Low |
| Rademaker 38 | Retrospective data analysis | Serious | Not serious | Serious | Not serious | None | None | None | None | 1219 | Low |
| Mandekou Lefaki et al.37 | Prospective interventional | Serious | Not serious | Not serious | Not serious | None | None | None | None | 64 | Low |
| Borghi et at.48 | Prospective noncomparative | Serious | Not serious | Not serious | Not Serious | None | None | None | None | 150 | Low |
| Amichai et al.19 | Prospective noncomparative | Serious | Not serious | Not serious | Not serious | None | None | None | None | 638 | Low |
| Agarwal et al.22 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 120 | Low |
| Yap46 | Prospective noncomparative | Serious | Not serious | Not serious | Not serious | None | None | None | None | 150 | Low |
| Rao et al.39 | Prospective noncomparative | Serious | Not serious | Not serious | Not serious | None | None | None | None | 50 | Low |
| Rademaker et al.49 | RCT | serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Strauss et al.43 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 150 | Low |
| Kapadia et al.35 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Goulden et al.32 | Prospective noncomparative cohort | Serious | Not serious | Not serious | Not serious | None | None | None | None | 80 | Low |
| Del Rosso et al.50 | Prospective noncomparative cohort | Serious | Not serious | Not serious | Not serious | None | None | None | None | 201 | Low |
| Kaymak et al.51 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 60 | Low |
| Akman et al.24 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 66 | Low |
| El-Sherif et al.29 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 75 | Low |
| Hafeez et al.34 | RCT | Serious | Not serious | Not serious | Not serious | None | None | None | None | 140 | Low |
Abbreviation: RCT, randomised controlled trial.
RESULTS
In this review, 32 studies were included, with a total sample size of 5245 patients with mild to very severe acne vulgaris (Table 1). Of these, 15 were randomised controlled trials (RCT), 10 were prospective noncomparative studies, 2 were prospective comparative studies, and 5 were retrospective, noncomparative series. When assessing clinical severity of acne 10 studies focused on mild–moderate acne, 12 on moderate to severe and 5 on severe only. Three studies included a wider range, from mild to severe, whereas one focused solely on resistant acne.
The grading systems adopted were inconsistent across the studies: 9 studies used GAGS (Global Acne Grading System) as an assessment scale, 8 used the Leeds Revised photometric Grading System, 2 the Pochi scale, 1 IGA (Investigator Global Assessment) and the remaining 12 used various acne scale based on lesion counting.
In the large majority of the studies, low dose was defined as 0.1–0.5 mg/kg/day or a cumulative dose <120 mg/kg. Conventional dose was defined as either 0.5–1 mg/kg/day or cumulative dose 120–150 mg/kg and high dose >1 mg/kg/day or cumulative dose >150 mg/kg.
This constitutes a pharmacological conundrum as low daily dosage may still result in high cumulative dose if the duration of treatment spans across multiple months. As such, studies that focused on cumulative dosing will be analysed separately.
Intermittent dosing regimens varied but included alternate day regimens, as well as 0.5–0.8 mg/kg/day 1 week or 10 days per month, for 4 or 6 months.
Duration of treatment ranged from 16 weeks22 to 24 months.46 Follow-up as a means of assessing relapse ranged from 10 weeks49 to 7 years.37
The primary outcome describing success of treatment was defined as the percentage decrease in acne severity (n = 15) or improvement of the acne grading score (n = 7) or described as ‘clear or almost clear’ in eight studies. Two studies did not define any outcome measures.
Relapse was also captured in 23 studies. The definition of relapse varied but could be summarised into two large groups: relapse as emergence of pretreatment acne scores (with variable indicative thresholds) or relapse as the need for treatment with oral or systemic medication.
ISOTRETINOIN DOSING REGIMENS
Low dose
A total of eight studies focused solely on assessing the efficacy of low-dosing regimens alone. Amichai et al.19 found that a dose of 20 mg/day for 6 months regimen had no statistical difference between two age groups with respect to improvement (p = 0.36) or relapse rate (p = 0.35). Rao et al.39 also used 20 mg/day for 3-month period and reported that 90% had >75% clearance. Similarly, another40 retrospective study confirmed 96.4% clearance rate, which was achieved at a mean time of 4.5 months, with a 7.9% relapse rate in their 5-year follow-up period. Ullah et al.44 found that 85.71% had achieved complete clearance at 3 months (p < 0.01) with a 20 mg dose on alternate days combined with doxycycline 100 mg twice daily. Another study46 used a lower dose of 10 mg/day alongside topical clindamycin. All patients achieved complete clearance of acne lesions after a mean period of 24 weeks. However, those with more severe acne required a higher cumulative dose (p < 0.001) and longer duration of treatment to achieve clearance (p < 0.001).
Golden et al.32 used low-dose intermittent isotretinoin (0.5 mg/kg for 1 week in 4 for 6 months) and achieved clearance in 88% of subjects despite reporting higher relapse rates at 12 months in those with predominantly truncal acne (p < 0.01), those with more severe acne initially (p < 0.0001) and those with higher sebum excretion rates (p < 0.001).
Low dose versus conventional regimens
One study30 compared low (0.25 mg/kg/day) versus conventional (0.5 mg/kg/day) dose (alongside a 2-week regimen of azithromycin and 1 week of prednisolone), demonstrating nonsignificant improvement of acne severity in all groups.
Similarly, there was no significant difference in GAGS score and lesion count between conventional daily and low daily dosing at 24 weeks in a different study.36 It also found a similar efficacy between low and conventional doses but significantly reduced efficacy in the intermittent group. Another trial34 also showed a statistically significant increased efficacy in those taking a dose of 80 mg/day compared with 20 mg/day.
The study by Strauss et al.43 compared three groups: low, conventional 0.5 mg/kg and conventional 1 mg/kg. All demonstrated a reduction in number of flares after 20 weeks, with no differences between dosages (p < 0.01).
Kapadia et al.35 reported that patients treated with higher doses achieved clearance faster than those treated with lower doses, but all achieved clearance at 24 weeks (p > 0.5). However, in a different study,43 although clearance was similar in all groups, the number of relapses was higher (42% vs. 10%) in the group receiving 0.1 mg/kg/day.
Three studies used a different approach: two studies started at a lower dose and then offered weekly increments until the highest dose tolerated was reached,33, 48 and the other study reached a dose of 1 mg/kg/day.50 Borghi et al.48 treated patients for 1 month after complete clearance and noted a relapse rate of 9.35%. In contrast, another study50 using a target cumulative dose between 120 and 150 mg/kg for 20 weeks regardless of whether clearance had occurred, had a relapse rate of 7.8%.
Four studies30, 34, 36, 37, 43 demonstrated no statistical difference in clinical efficacy between the low dose versus conventional dose at the end of the treatment period (24–30 weeks), with a similar number and timing of relapses in each group.
Low dose versus intermittent doses
Low dose daily versus intermittent (alternate days)
One study41 used a dose of 20 mg on alternate days achieving very good results (more than 80% resolution of baseline acne) in 68.2% of patients, despite a failure of treatment in 12.46% and a relapse rate of 16.4%.
When comparing this intermittent dosing regimen to a standard 20 mg fixed daily dose, two studies have offered further insight: a daily fixed dose provided more rapid response22, 52 but there was no difference in overall efficacy at the end of treatment. One study52 reported that those with severe acne responded better to a daily treatment at 36 weeks (p < 0.001).
Low dose daily versus intermittent (1 week per month)
Three studies29, 42, 53 compared a daily 20 mg dose to a 1-week per month regimen. No statistically significant difference was noted, although faster clinical response and lower relapse was found in the groups using a daily regimen.
Daily conventional versus intermittent conventional doses
Five studies compared conventional daily dose with intermittent conventional dose: four studies included moderate acne (one examined mild, moderate and severe acne, one examined moderate to severe, one included mild to moderate and one just focused on moderate disease) and two studies included severe acne.
One prospective study examined an intermittent regimen. The authors found that a regimen of 0.5 mg/kg/day for 1 week every 4 weeks for 6 months achieved a satisfactory response in 88% (p < 0.0001). Despite this, there was a reported 9% treatment failure rate, and 39% of the lesions relapsed after 12 months.32
The remaining studies offered a comparative analysis between intermittent and conventional daily regimens, which was found to be around 20% more efficacious and with a quicker response than the intermittent option.
Another study24 showed that, whilst acne was significantly reduced in all groups (p < 0.001), only those with severe acne were found to have responded to the different regimens (p = 0.0013). This suggests that a fixed daily dose may be better at treating severe acne, whereas intermittent doses can be used to treat moderate acne. A fixed daily dose may also be preferable, as relapse was only found in those receiving an intermittent dose.24, 36
Daily conventional doses compared with each other
Two studies compared daily dosing regimens where patients received doses at either end of the conventional range of 0.5 and 1 mg/kg. One study45 compared doses of 0.5–1 mg/kg/day and found no difference in clearance between groups, despite finding a statistically significant increased frequency of side effects in the 1 mg/kg/day group (p = 0.002).
Another study35 compared three groups, two of which were also on 0.5 and 1 mg/kg/day and found an increased rate of side effects in the 1 mg/kg/day group (p < 0.05). Those in the 0.5 mg/kg group had double the recurrence rates (20%) of those in the 1 mg/kg group (10%). Both studies only included patients with either acne conglobata45 or severe nodulocystic acne43 which may affect the comparability of these studies.
Conventional doses given in divided doses
Three studies looked at conventional doses divided in two and given twice instead of once daily.23, 25, 50 One compared once daily to twice daily dosing of conventional doses and did not find any difference between reduced global acne scores (p = 0.8) but highlighted that side effects were more common in those receiving once daily dosing (p = 0.001).23 Another found twice daily dosing of 0.5 mg/kg/day resulted in significant improvement of nodular, pustular and cystic lesions.25 The third used lidose-isotretinoin twice daily starting at 0.5 mg/kg for the first 4 weeks and then increasing to a dose of 1 mg/kg for the remaining 16 weeks of treatment. It demonstrated statistically significant improvement in mean acne lesion count but did not compare this to standard isotretinoin or once daily dosing.50
High dose regimens
Only one study27 looked at high daily doses. Participants were given a mean dose of 1.6 mg/kg/day of isotretinoin with a cumulative dose of 290 mg/kg. All patients cleared with this regimen, with a relapse rate of 12.5%. This included patients who had required previous courses of isotretinoin, a group that was mostly excluded from other studies. When these patients are removed from this sample, the relapse rate falls to 8.7%.
Cumulative dosing considerations
Borghi et al.48 pointed out that those with acne in more than three bodily areas required higher cumulative doses. Hermes et al.33 found no statistical significance between patient demographics, type of acne, duration of treatment or cumulative dose and relapse rate. This study had the highest relapse rate of the three, with 33% mean relapse rate—although relapse was determined by new emergence of acne of greater than 0.5 grade, which was a lower threshold than most of the other studies.
Another study26 used standard daily doses of isotretinoin until 1 month after no new lesions occurred, then divided them into groups of those that received more or less than 220 mg/kg as a cumulative dose. They found that relapse was higher (47.4%) in those taking the lower cumulative doses (<220 mg/kg) compared with those who received higher cumulative doses (26.9%) (>220 mg/kg) (p = 0.03). However, this study defined relapse as requirement of any further prescribed treatment including topical treatments. When patients requiring a further course of isotretinoin were identified, these were only two, and belonged to the group with higher cumulative dose.
With respect to adverse effects, one comparative study found no statistically significant difference in deranged LFTs and different cumulative doses.26 Similarly, no significant difference was found in triglycerides between those that received more or less than 220 mg cumulative dose of isotretinoin.26 One study demonstrated a dose-dependent increase in lipids and triglycerides43 and one study reported a significantly higher rate of retinoid dermatitis in their group that received more than 220 mg of isotretinoin as a cumulative dose compared with less than 220 mg.26
ADVERSE EFFECTS
Biochemical
Most studies reported some derangement in LFTs or lipids on blood monitoring. The frequency of LFTs above the upper limit of normal ranged from 0%32, 35 to 22%.27 AST was most frequently found to be abnormal but a rise in transaminases was transient and resolved on cessation of isotretinoin. No statistically significant difference in ALT rises was found in a study of two groups receiving 0.1 and 0.5 mg/kg/day, despite there being a significant difference in the group receiving 1 mg/kg/day.43 Similarly, one study only reported deranged LFTs in the 1 mg/kg/day treatment group22 and the study using the highest dose of isotretinoin at 1.6 mg/kg/day also reported the highest rate of ALT derangement at 22%.27
Lipids
Studies measured various lipid profiles including serum lipids, cholesterol and triglycerides. Triglycerides were most frequently elevated. Three studies reported normal lipid profiles despite differing doses of isotretinoin.32, 35, 38 18.8% of patients receiving 1.6 mg/kg/day of isotretinoin in the highest dosing study developed abnormal triglycerides.27
Mucocutaneous side effects
Cheilitis and xerosis were the most common side effects and were reported in all studies. Several studies demonstrated a dose-dependent relationship between cheilitis and isotretinoin.24, 29, 36, 43, 45 Acne flare was identified as a side effect in four studies.33, 37, 46, 51 One study found 15.6% of participants experienced acne flare in their lower dose group (0.15–4.0 mg/kg day of isotretinoin) compared with 25% in their conventional dosing group.37 Hermes et al.33 found 27% of patients had an acne flare but this was not related to dose. Another study46 reported 5.3% of patients experienced an acne flare on a low-dose regimen of 10 mg of isotretinoin per day. Long-term follow-up to assess if side effects other than flares were persistent after cessation of isotretinoin was only performed by one study26 that found 20.7% of patients continued to have cheilitis and 17.2% had xerosis a year after treatment.
Mood disturbance
In seven studies, patients reported psychiatric or psychological symptoms whilst taking isotretinoin.26, 31, 35, 38, 40, 46, 51 Reported symptoms included irritability, nervousness or anxiousness and depressive symptoms, with a frequency ranging from 0% to 7.34% across all studies. Only 24 participants out of the total of over 5050 across all studies developed mood disturbance. Most studies did not enquire specifically about mood changes, but the three that did, reported no form of mood disturbance in their sample,19, 27, 39 including the study that used the highest recorded dose.27 One study26 divided their patients into groups who had received a cumulative dose of more or less than 220 mg/kg and reported mood disturbance in 2.6% of their participants, despite there being no statistically significant difference between the two groups. In addition, even very low doses of isotretinoin of 5 mg a day, resulted in improvements in Dermatology Quality of Life Index (DLQI) scores.38
ECONOMIC CONSIDERATIONS
Most studies overlooked the economic impact of the dosing regimens used. A comparative study in which patients with moderate acne were given 20 mg daily of isotretinoin found their results were comparable with standard doses of isotretinoin in other studies, at a lower cost.19
Another comparative study demonstrated a mean cost of isotretinoin treatment of €376.69 versus €684.45 in their two treatment groups. However, although the cost of treatment was lower in the group receiving the lower dose (mean dose of 0.24 mg/kg) compared with the recommended dose (mean dose 0.67 mg/kg), it did not result in greater resolution of acne (68.8% compared with 87.5%, respectively).37 Another study reported a 20% saving using an intermittent regimen of 0.5 mg/kg 1 week per month, for 6 months, compared with daily dosing.32
DISCUSSION
This systematic review has examined studies conducted over 50 years to clarify whether there is an optimal isotretinoin dosing regimen to achieve efficacy and sustained improvement in treating acne vulgaris. It also sought to establish if low-dose regimens are as clinically effective, lead to fewer adverse effects and result in sustained improvement when compared with high-dose regimens.
Thirty-two studies were identified and critically appraised regarding study design, number of participants and their characteristics, dose, duration of treatment, number of isotretinoin courses, cumulative dose, definitions of efficacy, efficacy and relapse, site of acne, acne grading system used, acne severity, dietary factors, adverse side effects, including biochemical, cheilitis and mood disturbance, follow-up period, economic considerations and study quality score.
The review of these published studies has demonstrated several issues or omissions which hinders comparison between studies and interpretation of results. GRADE assessment revealed all studies had a serious risk of bias and a low overall assessment score. With respect to study design, 10 different grading systems have been used to define treatment outcomes which has resulted in inconsistency when comparing the clinical efficacy of low or intermittent regimens to high-dose isotretinoin treatment.
Dosage regimens were variable across studies. Although isotretinoin is licenced for severe acne, the grade of acne across studies included mild, moderate and severe disease and the definition of relapse was either not clearly defined or inconsistent across studies. Relapse rates may be influenced by multiple factors including the severity of acne, younger age of onset, hyperandrogenism, smoking, cessation of treatment before acne clearance and excessive seborrhoea. It was not clear from these studies whether these prognostic risk factors were taken into consideration. Previous large studies have specifically analysed relapse rates and found conflicting results: one study reported relapse rates greater in those taking isotretinoin, but those in the study appear to have been treated with a subtherapeutic dose.54 Another large cohort study showed that relapse was linked to young age and facial acne grade greater than 3, while mean daily dose and total cumulative dose had no prognostic value for relapse after isotretinoin-induced clearing.55
The studies were carried out across 16 different countries, and most did not break down results by ethnic group. Many studies were not controlled and/or did not use a comparator and many were underpowered. Only three studies32, 37, 56 considered the economic impact of different dosing regimens, which have important implications for national health services, insurance companies and self-funded patients. Patient reported outcome measures were not employed, which would have been useful to elicit the patients' perspective on their treatment efficacy. Patient acceptability and adherence to longer courses of isotretinoin has not been adequately explored either.
Oral isotretinoin is known to be highly lipophilic, however, none of the studies controlled for dietary factors and many of the studies were not adequately powered to allow for differences between the different generic formulations that may have been dispensed.
With respect to adverse effects, data were not collected in a consistent manner. When this was recorded, it only identified the more common reported adverse effects. The duration of long-term follow-up was variable between none23, 34, 35, 44, 51 and 7 years,37 and long-term adverse effects were only recorded in one study.26 Fifteen studies provided follow-up of a year or less and this was often self-reported via survey.
Notwithstanding this, there were several consistent results throughout. First, daily dosing regimens resulted in better efficacy than intermittent dosing regimens in severe acne. Second, patients with more severe acne responded better to conventional or high fixed-dose regimens of isotretinoin. Third, the severity of some common, particularly mucocutaneous, side effects were dose-dependent and adopting a low-dose regimen avoided some of the mucocutaneous side effects which are associated with standard or high-dose isotretinoin. Fourth, relapse rates were higher in patients assigned to low-dose and intermittent-dosing regimens when compared with those on high-dose and conventional regimens. Fifth, low dose extended courses of isotretinoin, providing they are continued until clearance or near clearance, appear to have a similar efficacy to conventional doses of isotretinoin. Finally, higher cumulative doses were not associated with increased efficacy in studies of mild to moderate acne in which isotretinoin was being used off-license.
CONCLUSION
This systematic review has identified some trends across studies. Daily dosing was better than intermittent dosing and in severe disease. Low doses resulted in fewer mucocutaneous adverse effects and less initial acne flare. Relapse rates were lower in patients taking higher dosages. However, low-dose extended courses of isotretinoin, providing they are continued until clearance or near clearance, appear to have a similar efficacy to conventional doses of isotretinoin. Higher cumulative doses of isotretinoin were not associated with increased efficacy in studies of mild to moderate acne in which isotretinoin was being used off-license, however, it was not clear if this was the case in severe disease.
Despite these trends, the overarching conclusion demonstrates the need for a well-designed randomised controlled study which is adequately powered to answer the question: can a low-dose oral isotretinoin regimen for severe acne result in similar efficacy, reduced adverse effects and long-term resolution when compared with a conventional dosing regimen?
AUTHOR CONTRIBUTIONS
Study conception and design: Alison M. Layton, Carsten Flohr. Preliminary searches: Naayema, Jack Coumbe, Maria Charalambides. Final search strategy: Aoife U. Daly, Rui Baptista Gonçalves, Eva Lau, Joanne Bowers, Alison M. Layton. Data collection: Aoife U. Daly, Rui Baptista Gonçalves, Eva Lau, Joanne Bowers. Analysis and interpretation of results: Aoife U. Daly, Rui Baptista Gonçalves, Eva Lau, Joanne Bowers, Alison M. Layton. Draft manuscript preparation: Aoife U. Daly, Rui Baptista Gonçalves, Alison M. Layton. All authors reviewed the results and approved the final version of the manuscript.
ACKNOWLEDGEMENTS
The authors would like to thank Helen Weir and Daniel Park, Librarians at Harrogate District General Hospital, for her greatly appreciated support throughout this project with the search strategy and data extraction. The authors declare no funding for this article.
CONFLICT OF INTEREST STATEMENT
All authors declare no conflict of interest.
ETHICS STATEMENT
Not applicable.
Open Research
DATA AVAILABILITY STATEMENT
The data that support the findings of this study are available in the supplementary material of this article.
Supporting Information
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| jvc2154-sup-0001-Appendix_1_ISO.docx14 KB | Supporting information. |
| jvc2154-sup-0002-PRISMA_Checklist_Isotretinoin_Dosing.docx23.7 KB | Supporting information. |
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187595 | https://www.kristakingmath.com/blog/chain-rule-with-trig-functions | Chain rule with trig functions — Krista King Math | Online math help
Chain rule with trig functions
All derivative rules apply when we differentiate trig functions
Let’s look at how chain rule works in combination with trigonometric functions.
Keep in mind that everything we’ve learned about power rule, product rule, and quotient rule still applies.
Applying chain rule to the derivatives of trigonometric functions
Take the course
Want to learn more about Calculus 1? I have a step-by-step course for that. :)
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Using chain rule to differentiate a sine function
Example
Use chain rule to find the derivative.
y=sin8x2
Using substitution, we see that u=8x2 and u′=16x.
Our original equation becomes
y=sinu
and the derivative is
y′=cosu(u′)
Back-substituting for u and u′, we get
y′=cos8x2(16x)
y′=16xcos8x2
Let’s look at another example.
Example
Use chain rule to find the derivative.
y=(tan3x4)3
In this case we want to use a double substitution, where
u=3x4
u′=12x3
and
v=tanu
v′=sec2u(u′)
Our original equation is
y=[v]3
and its derivative is
y′=(3v2)v′
Back-substituting for v and v′, we get
y′=[3(tanu)2][sec2u(u′)]
Back-substituting again, but this time for u and u′, we get
y′=[3(tan3x4)2][sec23x4(12x3)]
y′=36x3tan23x4sec23x4
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Education Standards
Wyoming Standards for Mathematics
Learning Domain: Ratios and Proportional Relationships
Standard: Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
Maryland College and Career Ready Math Standards
Learning Domain: Ratios and Proportional Relationships
Standard: Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak."ť "For every vote candidate A received, candidate C received nearly three votes."ť
Common Core State Standards Math
Cluster: Understand ratio concepts and use ratio reasoning to solve problems
Standard: Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
Egginess Part 2
Comparing Numbers with Ratios
Overview
This lesson formally introduces and defines a ratio as a way of comparing numbers to one another.
Key Concepts
A ratio is defined by the following characteristics:
Other important features of ratios include the following:
Goals and Learning Objectives
Introduction to Ratios
Lesson Guide
In this lesson, students learn the formal definition of a ratio and then use it to solve problems. The start of the lesson introduces the concept of a ratio as a way of comparing numbers of objects using division, which is an alternative to comparing numbers using subtraction. Have students look at the picture of stars and triangles and then read the text.
Ask:
ELL: Keep in mind that some students may not feel comfortable reading aloud. Be prepared to support them in this task.
Opening
Introduction to Ratios
Look at the picture of stars and triangles and read the following information.
Ratio of Egginess
Lesson Guide
Have a volunteer read the definition of a ratio aloud.
Review the definition of ratio: A ratio is a comparison of two numbers by division.
The value of a ratio is the quotient that results from dividing the two numbers. For example, the value of the ratio 35:7 is 5, which you find by computing 35 ÷ 7 = 5.
Have students review the video of the egginess problem from the previous lesson. Discuss the ratio in the egginess problem. Demonstrate how to write the ratio. Talk about the fact that the ratio could be flour to eggs or eggs to flour. The ratio of flour to eggs is 3:2; the ratio of eggs to flour is 2:3.
ELL: When showing the video, be sure that ELLs are following the explanations. Pause the video at key times to allow ELLs time to process the information. Ask students if they need to watch it a second time. Remind students that they are finding the ratio in the egginess problem.
Opening
Ratio of Egginess
A ratio is a comparison of two numbers by division.
The value of a ratio is the quotient that results from dividing the two numbers. For example, the value of the ratio 35:7 is 5, which you find by computing 35 ÷ 7 = 5.
In the previous lesson, you looked at how to fix the egginess in a mixture. Watch the Egginess
Part 2 video.
VIDEO: Egginess Part 2
Egginess Part 2
Math Mission
Lesson Guide
Discuss the Math Mission. Students will explain how ratios are used to compare quantities.
ELL:
Discuss the concept of ratio. Sketch diagrams of the examples so students can make associations with comparison of quantities. Use manipulatives to demonstrate the comparison of two quantities using subtraction and division. Use illustrations or manipulatives to model effective learning to explain ratios. It is important to emphasize that you can describe ratios by saying “for every” or “per” since these terms will be used interchangeably throughout the unit. Underline the two quantities so that students know exactly which quantities you are comparing.
Opening
Explain how ratios are used to compare quantities.
Ms. Lee’s Class
Lesson Guide
Have students work in pairs on the problems and the presentation.
SWD: Help students with disabilities build their mathematical vocabulary by continually modeling the use of new terms in the context of classroom work and activities.
Mathematical Practices
Mathematical Practice 2: Reason abstractly and quantitatively.
Listen for students who use the problem situations to help them make sense of the values they are working with.
Mathematical Practice 6: Attend to precision.
Listen also for students who use the term ratio correctly or who discuss the correct usage of the term as they work together to solve the problems.
Interventions
Student thinks that you write a ratio as a subtraction.
Student reverses the boys and girls in the ratio.
Answers
Work Time
Ms. Lee's Class
There are 15 boys and 17 girls in Ms. Lee's math class.
Ask yourself:
A Tennis Game
Mathematical Practices
Mathematical Practice 2: Reason abstractly and quantitatively.
Listen for students who use the problem situations to help them make sense of the values they are working with.
Mathematical Practice 6: Attend to precision.
Listen also for students who use the term ratio correctly or who discuss the correct usage of the term as they work together to solve the problems.
Interventions
Student thinks there are exactly 3 females and 2 males watching the tennis game.
Student thinks there is 1 more female than male watching the tennis game.
Student believes the number of females, males, or people watching the tennis game can be calculated with only the ratio.
Possible Answers
Work Time
A Tennis Game
The ratio of the number of females watching a tennis game to the number of males watching the tennis game is 3 to 2. You can write that as 32, or 3:2.
Ask yourself:
Prepare a Presentation
Preparing for Ways of Thinking
Listen and look for the following student thinking to highlight during the Ways of Thinking discussion:
Challenge Problem
Possible Answer
Work Time
Prepare a Presentation
Explain what types of conclusions you can and cannot make based on the tennis game ratio.
In your own words, explain what a ratio is.
Challenge Problem
As two quantities get closer to each other, the value of the ratio of the quantities approaches 1.
Make Connections
Mathematics
Have students share their presentations. If there are any misunderstandings, facilitate a class discussion that focuses on the meaning of a ratio and what it does and does not tell us.
Ask:
Compare these examples to the situation about the number of boys and girls in Ms. Lee’s class, in which the actual numbers of students are given rather than a ratio between numbers. Highlight correct use of the term ratio, or provide opportunities for students to revise their use of the term.
Conclude the discussion with a focus on whether the ratio tells you whether more males are watching the game or more females are watching the game. If there is a student who divided 3 by 2 to get 1.5, ask them to share their strategy. Ask the class what 1.5 means in this situation:
Check that all students understand that the order of values in a ratio matters. The ratio of males to females (2:3) is different from the ratio of females to males (3:2).
Mathematical Practices
Mathematical Practice 2: Reason abstractly and quantitatively.
Ask students to talk about how they can use the context of a problem situation to help them make sense of the values they are working with.
Mathematical Practice 6: Attend to precision.
Call attention to correct uses of the term ratio, or ask for clarifications about its use and meaning as needed as students present their work and ask questions of presenters.
Performance Task
Ways of Thinking: Make Connections
Take notes about your classmates' explanations of the conclusions that can and cannot be made based on the tennis game ratio, and their explanations of what a ratio is.
As your classmates present, ask questions such as:
What I Know about Ratios
A Possible Summary
Ratios allow you to compare quantities, but by themselves, they do not tell you the actual values of the quantities. Using a ratio to compare quantities is different from using subtraction to find the difference between quantities because a ratio tells you the value of one quantity for a given value of the other quantity. For example, for a ratio of 3:2, you know that if the first quantity has a value of 6, the second quantity has a value of 4.
SWD: Create a resource for some students that includes this explanation in simpler language (perhaps with visual illustrations). Annotate illustrations that show ratios and what they represent.
Additional Discussion Points
If there is time, discuss the following:
Formative Assessment
Summary of the Math: What I Know about Ratios
Write a summary of what you learned about ratios.
Check your summary.
Reflect On Your Work
Lesson Guide
Have each student write a brief reflection before the end of class. Review the reflections to find out what students wonder about ratios.
Work Time
Reflection
Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.
Something I wonder about ratios is …
or
or |
187597 | http://www.math.tau.ac.il/~michas/fp013-sharir.pdf | Random Triangulations of Planar Point Sets Micha Sharir ∗ School of Computer Science Tel Aviv University, Tel Aviv 69978, Israel Courant Institute of Mathematical Sciences New York University, New York, NY 10012, USA michas@tau.ac.il Emo Welzl Institut f¨ ur Theoretische Informatik ETH Z¨ urich CH-8092 Z¨ urich, Switzerland emo@inf.ethz.ch ABSTRACT Let S be a finite set of n + 3 points in general position in the plane, with 3 extreme points and n interior points. We consider triangulations drawn uniformly at random from all triangulations of S, and investigate the expected number, ˆ vi, of interior points of degree i in such a triangulation. We provide bounds that are linear in n on these numbers. In particular, n/43 ≤ˆ v3 ≤(2n + 3)/5.
Moreover, we relate these results to the question about the maximum and minimum possible number of triangulations in such a set S, and show that the number of triangulations of any set of n points in the plane is at most 43n, thereby improving on a previous bound by Santos and Seidel.
Categories and Subject Descriptors: G.2 [Discrete Math-ematics]: Combinatorics—Counting problems General Terms: Theory Keywords: Random triangulations, counting, degree se-quences 1.
INTRODUCTION Given a set S of n points in the plane, a triangulation is a maximal crossing-free geometric graph on S (in a geometric graph the edges are realized by straight line segments). Here we consider random triangulations, where “random” refers to uniformly at random from the set of all triangulations of S. We are primarily interested in the degree sequences of such random triangulations.
To be precise, we assume that S is a set of n + 3 points in general position in the plane so that the convex hull of S is a triangle. For such a set and i ∈N, we let ˆ vi denote the expected number of interior points of degree i in a random triangulation. While—for n large enough—the number of ∗Supported by NSF Grants CCR-00-98246 and CCF-05-14079, by a grant from the U.S.-Israel Binational Science Foundation, by Grant 155/05 from the Israel Science Fund, and by the Hermann Minkowski–MINERVA Center for Ge-ometry at Tel Aviv University.
Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee.
SCG’06, June 5–7, 2006, Sedona, Arizona, USA.
Copyright 2006 ACM 1-59593-340-9/06/0006 ...$5.00.
vertices of degree 3 in a triangulation may be any integer between 0 and roughly 2n 3 , we show that n 43 ≤ˆ v3 ≤2n + 3 5 .
Note that general position is Figure 1: Point set with unique triangula-tion.
essential for the lower bound.
Consider the case where the n interior points lie on a common line containing one of the ex-treme points in S, see Fig. 1.
Then there is a unique trian-gulation and this triangulation has one interior point of degree 3; hence, ˆ v3 = 1.
We relate these results to the question about the maximum and minimum possible number of triangulations in a set of n points in the plane. We show that the number of trian-gulations of any such set is at most 43n, thereby improving on a previous bound of 59n by Santos and Seidel . We can also use the upper bound on ˆ v3 to infer a lower bound of roughly 2.5n on the number of triangulations every set of n + 3 points in general position with triangular convex hull has. However, this is inferior to the recent 0.093·2.63n-bound by McCabe and Seidel .
Our results use charging schemes among vertices in tri-angulations that heavily build on the structure imposed by edge flips on the set of all triangulations (see also the dis-cussion of (dis-)charging below). Our approach should be regarded as a continuation of the proof by Santos and Seidel for the 59n upper bound for the number of triangula-tions. This connection may not be obvious in our presenta-tion, since we deal with a different scenario, but it should become more apparent when we get as an intermediate re-sult a lower bound of n/59 for ˆ v3.
The two 59’s are the “same”! Still, we believe that it was the new setting that allowed us to proceed further and derive a better bound for the number of triangulations.
Little seems to be known about random triangulations of (fixed) point sets, although the generation of random tri-angulations has raised some interest (see, e.g., [1, Section 4.3]). Moreover, it is a folklore open problem to determine the mixing rate of the Markov process that starts at some triangulation and keeps flipping a random flippable edge; see [13, 12] where this is treated for points in convex po-sition.We are currently investigating whether our methods have anything to say about this problem. Finally, for ab-stract graphs (without enforced straight line embedding on a given point set), there are results about random planar graphs1, see, e.g., [9, 11, 7]; it is not clear how those com-pare to our setting (see also the discussion of a result by Tutte below).
Number of Triangulations—History.
David Avis was perhaps one of the first to ask whether the maximum number of triangulations of n points in the plane is bounded by cn for some c > 0, see [3, page 9]. This fact was established in 1982 by Ajtai, Chv´ atal, Newborn, and Szemer´ edi , who show that there are at most 1013n crossing-free graphs on n points—in particular, this bound holds for triangulations.
Further developments have yielded progressively better upper bounds for the number of triangulations2 [20, 5, 18], so far culminating in the previously mentioned 59n bound in 2003. This compares to Ω(8.48n), the largest known number of triangulations for a set of n points, recently de-rived by Aichholzer et al. ; this improves an earlier lower bound of about 8n (up to a polynomial factor) given by Garc´ ıa et al. .
For n points in convex position, the number of triangu-lations is known to be Cn−2, where Cm := 1 m+1 2m m ´ = Θ(m−3/24m), m ∈N0, is the mth Catalan number (the Euler-Segner problem, cf. [21, page 212] for a discussion).
Other Crossing-free Graphs.
Besides the intrinsic in-terest in obtaining bounds on the number of triangulations, they are useful for bounding the number of other kinds of crossing-free geometric graphs on a given point set, exploit-ing the fact that any such graph is a subgraph of some tri-angulation. For example, the best known upper bound on the number of crossing-free straight-edge spanning trees on a set of n points in the plane is O((5.˙ 3 τ)n), if τ n is a bound on the number of triangulations; with τ = 43 this is now O(229.˙ 3n). This follows from a result by Rib´ o and Rote, [14, 16], who show that any planar graph on n vertices con-tains at most 5.˙ 3n spanning trees. Similar results have been observed for crossing-free spanning cycles, where a bound of O(( √ 6τ)n) = O((2.45 τ)n) can be obtained, as communi-cated by Raimund Seidel; the resulting bound of O(105.33n) falls still short of the bound of O(86.81n) for cycles given in , though. The total number of crossing-free planar graphs on n points is at most 23n−6τ n < (8 τ)n. So this is now improved to 344n (from 472n).
Next we mention a result and a notion, both seemingly related to what we are doing; hence, they were popping up repeatedly when presenting our result. While we want to take the opportunity to clarify in this way, a fruitful closer connection may be established in the end.
Tutte’s Number of Rooted Triangulations.
Let us briefly discuss a classical result from 1962 by Tutte in his census-series in the Canadian Journal of Mathematics .
He considers so-called rooted triangulations, i.e., maximal planar graphs, with a fixed face with vertices a, b, and c and n additional vertices. Two such triangulations are consid-ered to be equal if there is an isomorphism between them, 1Here one has to discriminate between the labeled and the unlabeled case.
2Interest was also motivated by the obviously related prac-tical question (from geometric modeling ) of how many bits it takes to encode a triangulation of a point set.
which maps each of the points a, b, and c to itself, though.
The number of such triangulations is easily seen to be 1 for n = 1 and 3 for n = 2.
Based on an ingenious analysis employing generating functions, Tutte shows that for n ≥2 the number of such triangulations is exactly 2 n(n + 1) 4n + 1 n −1 !
= Θ „ 1 n5/2 9.481n « .
a a b b c c Figure 2: Two distinct triangulations of a point set that are equal in Tutte’s setting.
How does this relate to the number of triangulations of given n + 3 points? On the one hand, Tutte’s model counts more triangulations, because there are fewer constraints: “The interior points can be moved arbitrarily.” On the other hand, distinct triangulations in the geometric setting may be equal in Tutte’s setting; see Fig. 2. Thus the results are incomparable, although we cannot rule out that a connec-tion may be established.
(Dis-)Charging.
The notion of “charging” does ring a bell in the context of planar graphs. The proof of the cel-ebrated Four-Color-Theorem employs Heesch’s idea of dis-charging (Entladung, ) in order to prove that certain con-figurations are unavoidable in a maximal planar graph, cf.
or a later proof in . There one initially puts charge 6−i on each vertex of degree i in a maximal planar graph— thus the overall charge is 12. Now vertices of positive charge push their charge to other vertices (they discharge) without changing the overall charge. Given that a certain set of con-figurations L does not occur, one proves that all vertices can discharge with a nonpositive charge in the end—a contra-diction and thus the configurations in L are unavoidable.
Our scheme differs in two respects. First of all we need a quantitative version. We let every vertex have a value of 7 −i, in this way we can make sure that the overall value in a maximal planar graph is at least n, or, equivalently, there is at least 1 for every vertex on the average. Secondly, the “discharging” goes across a family of planar graphs, the set of all triangulations of a given point set. We show that the charge can be redistributed so that no vertex of degree exceeding 3 has positive charge, and degree-3 vertices have charge at most 43. This allows us to conclude that at least 1 43 of all vertices over all triangulations have degree 3. Again, we have to leave it open to which extent the rich knowledge on discharging from the 4-Color-Theorem may be useful for our purposes.
Further Steps.
We know that the “43” in the bounds is not tight for our approach, and we are currently working on a more exhaustive analysis, which seems to suggest that the best constant that the technique yields gets close to 30. We hope to report on this in the full version of this paper. There we also plan to provide an argument that, for all i ≥3, there is a positive constant δi so that ˆ vi ≥δin, provided n is large enough (if n < i −2, there is no vertex of degree i).
2.
DEGREES IN TRIANGULATIONS We fix a triple H of non-collinear points in the plane, and, without further mention, restrict ourselves to finite point sets P that are contained in the convex hull of H. We say that P is in general position, if no three points in P + := P ∪H are collinear (P + is what we used to denote by S in the introduction). Let T +(P) denote the set of all triangulations of P +. Recall that a triangulation of N points whose convex hull is a triangle has exactly 3N −6 edges and 2N −5 inner faces, all triangular.
Degrees in Triangulations of P.
For i ∈N and trian-gulation T ∈T +(P), we let vi = vi(T) denote the number of points in P (not P +) that have degree i in T. Obviously, vi ∈N0, v1 = v2 = 0, and P i vi = n := |P|. Moreover, X i i vi ≤6n −5 if n ≥2.
(1) For the latter inequality, note that if d1, d2, and d3 are the degrees in T of the three points of H, then d1 + d2 + d3 + X i i vi = 2(3(n + 3) −6) = 6n + 6 , and d1 + d2 + d3 ≥11, since in a triangulation of at least 5 points all points have degree at least 3, and no two vertices of degree 3 are adjacent.
The vector (vi)i∈N, however, is constrained beyond (1).
For example, v3 ≤2n+1 3 , which can be seen as follows. Given T ∈T +(P) remove all the v3 points from P in T that have degree 3. Note that no two such points can be adjacent in T. Therefore, the resulting graph is a triangulation T ′ of the remaining points P ′+, and each of its faces contains at most one point in P \P ′. So for k := |P ′| = n−v3, the number of points removed is at most 2(k + 3) −5 = 2k + 1. Therefore, v3 ≤2(n−v3)+1; that is, v3 ≤2n+1 3 as claimed. In order to see that this bound of ⌊2n+1 3 ⌋is tight, set k = ⌈n−1 3 ⌉, choose any triangulation in T +(Q) for any set Q of k points, and place another n −k = ⌊2n+1 3 ⌋points, no two in the same face of the triangulation. This is possible by the choice of k. Connect all added points to the three vertices of their respective faces, and we are done. We summarize 0 ≤v3 ≤ —2n + 1 3 (2) which is tight except for the lower bound when n is small.
Degrees in Random Triangulations and the Number of Triangulations.
For i ∈N let ˆ vi = ˆ vi(P) := E(vi(T)) for T uniformly at random in T +(P).
Due to linearity of expectation, any linear identity or in-equality in the vi’s (such as (1) or (2)) will also be satisfied by the ˆ vi’s. However, as we will show, the ˆ vi’s are signifi-cantly more constrained than the vi’s. In particular, there is a constant δ > 0 such that ˆ v3 ≥δ n if n > 0 and the point set is in general position; recall Fig. 1 to see that gen-eral position is indeed necessary here. Before we establish this bound, let us relate it to the question about the num-ber of triangulations. For that, let tr +(P) := |T +(P)| and tr +(n) := max|P |=n tr +(P).
Lemma 2.1. (i) If δ > 0 is a real constant such that, for all n ∈N, ˆ v3 ≥δ n for any set of n points in general position, then, for all n ∈N0, tr +(n) ≤ 1 δ ´n .
(ii) If δ′ > 0 is a real constant and n0 ∈N such that, for all n, n0 ≤n ∈N, ˆ v3 ≤δ′ n for any set of n points in gen-eral position, then for any set P of n ∈N points in general position, tr +(P) = Ω ` 1 δ′ ´n´ .
Proof.
(i) Let P be a set of n > 0 points with tr +(P) = tr +(n). Without loss of generality, let P be in general po-sition (a small perturbation of a point set cannot decrease the number of triangulations).
Note that we can get triangulations of P + by choosing a triangulation of P + \ {q} for some q ∈P, and then inserting q as a vertex of degree 3 in the unique face it lands in. In fact, a triangulation T ∈T +(P) can be obtained in exactly v3(T) ways in this manner. In particular, if v3(T) = 0, T cannot be obtained at all in this fashion. This is easily seen to imply that X T ∈T +(P ) v3(T) = X q∈P tr +(P \ {q}) .
The left hand side of this identity equals ˆ v3(P) · tr +(P), and its right hand side is upper bounded by n·tr +(n−1). Hence, tr +(P) ≤ n ˆ v3(P) · tr +(n −1) ≤1 δ · tr +(n −1) (since we assume ˆ v3(P) ≥δ n), and thus tr +(n) ≤1 δ ·tr +(n−1) for all n ∈N. Since tr +(0) = 1, the lemma follows.
(ii) Along the same lines—omitted.
tr +(n) is also an upper bound for the number of triangula-tions of an arbitrary point set S of n points, without restrict-ing it to be contained in the convex hull of H, and without adding H to make its convex hull triangular. To see this, take S and apply an affine transformation so that it lies in the convex hull of H. This does not change the number of triangulations, and adding H cannot decrease the number of triangulations.
An Example.
Suppose P lies Figure 3: Points on a convex arc.
on a convex arc in the convex hull of H as depicted in Fig. 3.
Then all edges indicated there have to be present in all trian-gulations of P + and all that re-mains is to fill in a triangula-tion of a convex polygon with n + 2 vertices, n := |P|. The number of such triangulations is Cn, thus tr +(P) = Cn. Now consider some point in P. For it to have degree 3, its adja-cent vertices in the convex polygon have to be connected to each other, which leaves an (n + 1)-gon to be triangulated in Cn−1 ways.
Therefore, the probability that this point has degree 3 is exactly Cn−1 Cn = n+1 2(2n−1) = 1 4 + O 1 n ´ and ˆ v3 = n 4 + O(1). It is easy to show that ˆ v4 = ˆ v3 for these point sets, provided n ≥2.
3.
LOWER BOUND ON ˆ v3 The basic idea of our proof is to have all vertices of tri-angulations charge to vertices of degree 3. If every vertex charges at least 1 and each vertex of degree 3 is charged at most c, then we know that ˆ v3 ≥n c . The actual charging scheme is more involved. First, since there are triangulations that have no degree 3 vertices, the charging has to go across triangulations. Moreover, vertices will charge amounts dif-ferent from 1 (even negative charges will occur). However, on average, each vertex will charge at least 1. The difficulty in the analysis will be to bound the maximum charge c to a vertex of degree 3.
Vints and Flipping.
We consider the set P × T +(P) and call its elements vints (vertex-in-triangulation). The degree of a vint (p, T) is the degree (number of neighbors) of p in T; a vint of degree i is called an i-vint. The overall number of vints is obviously n · tr +(P), and the number of i-vints is ˆ vi · tr +(P).
We define a relation on the set of vints. If u and v are vints, then we say that u →v if v can be obtained by flip-ping one edge incident to u in its triangulation. That is, u and v are associated with the same point but in different tri-angulations, and u has to be an (i + 1)-vint and v an i-vint, for some i ≥3. We denote by →∗the transitive reflexive closure of →, and if u →∗v, we say that u can be flipped down to v. Charges will go from vints to 3-vints they can be flipped down to.
The support of a vint u is the number of 3-vints it can be flipped down to, i.e.
supp(u) := |{v | v is 3-vint with u →∗v}| .
1 2 1 2 1 v v′ u1 u2 Figure 4: A 3-vint v that is charged ch4(v) = 1 2 + 1 by 4-vints u1 and u2 in the provisional charging scheme.
A Provisional Charging Scheme.
Given our original plan, a natural charging scheme would let a vint u charge 1 supp(u) to each 3-vint it can be flipped down to—in this way it will charge a total of 1. Let us call this the provisional charging scheme; see Fig. 4. Since every vint can be flipped down to some 3-vint, the charges are well-defined in this way. For technical reasons, our final charging scheme will be somewhat different.
Let us gain some understanding of the notion of supp(u).
Note that the removal of an interior point p and its incident edges in a triangulation T creates a star-shaped polygon (with respect to p). We call this the hole of the vint (p, T).
Lemma 3.1. For a vint u, supp(u) equals the number of triangulations of the hole of u. Therefore, (i) if u is an i-vint, 1 ≤supp(u) ≤Ci−2, where the upper bound attained iffthe hole is convex (see Section 1 for the definition of the Catalan numbers Cm), and (ii) if u →∗u′ for vints u and u′, then supp(u) ≥supp(u′).
Proof.
(i) follows from the fact that a convex i-gon has Ci−2 triangulations, which is the maximum for all i-gons. (ii) uses the fact that if u →u′ then the hole of u′ is contained in the hole of u, with the vertices of the former a subset of the vertices of the latter; i.e. every triangulation of the hole of u′ can be extended to at least one triangulation of u.
For a 3-vint v and i ∈N, we let chi(v) be the amount charged to v by i-vints in the provisional charging scheme described above.
Lemma 3.2. For every 3-vint v and all i ≥3, we have 0 ≤ chi(v) ≤Ci−1 −Ci−2. In particular, ch3(v) = 1, ch4(v) ∈ {0, 1 2, 1, 3 2, 2, 5 2, 3}, ch5(v) ≤9, ch6(v) ≤28, etc.
Proof.
It follows from an analysis in [17, Lemma 4] that the number of i-vints that can charge a 3-vint is at most Ci−1 −Ci−2, and since a vint can charge at most 1 to a 3-vint, the bound follows.
ch3(v) = 1 is obvious. For the claim on ch4(v) it suffices to observe that there are at most three 4-vints that can charge a given 3-vint v, and that the support of a 4-vint is either 1 or 2. The remaining numbers simply evaluate the expression Ci−1 −Ci−2, and are given for future reference.
The Actual Charging Scheme.
In our provisional charging scheme, a 3-vint is charged P i chi(v).
We note that the bounds in Lemma 3.2 are tight (provided n is large enough compared to i). This will follow from the analysis given below, and is illustrated in Fig. 5 for the case i = 5 (the figure too will be better understood after the following analysis). Therefore, there is no uniform upper bound on the amount charged to individual 3-vints in the provisional scheme. For that reason, we switch to a charging where an i-vint u charges 7−i supp(u) to each 3-vint v with u →∗v.
Note that in this scheme, a 3-vint charges 4 to itself (so that sounds like bad news), but 7-vints do not charge at all and all i-vints with i ≥8 charge a negative amount, so that is good news for the 3-vints.
There is Enough Charge for Everybody.
The overall charge of an i-vint is 7−i, so the overall charge accumulated for all vints associated with a triangulation T is exactly P i(7 −i)vi(T) = P i 7vi(T) −P i i vi(T) ≥7n −6n = n, where we have used (1). So vints charge so that, on average, each gets to charge at least 1.
No 3-Vint Gets Charged too Much.
For a 3-vint v, we set charge(v) := P i(7 −i)chi(v) (3) = 4 ch3(v) + 3 ch4(v) + 2 ch5(v) + ch6(v) −ch8(v) −2 ch9(v) −· · · For an initial upper bound, we can ignore the negative terms and invoke the bounds on the chi(v)’s from Lemma 3.2, to get charge(v) ≤4 · 1 + 3 · 3 + 2 · 9 + 28 = 59, 1 1 v u1 u2 Figure 5: A 3-vint v that gets charged 1 by nine 5-vints (two of which are displayed) in the provisional charging scheme. Hence, ch5(v) = 9.
which implies ˆ v3 ≥ n 59, and by Lemma 2.1, this gives an up-per bound of 59n for the number of triangulations of any set of n points. This is the Santos-Seidel bound which we have derived now with ideas similar to theirs but in a different setting.
We improve on this by observing that if all ch4(v), ch5(v), and ch6(v) are large, then the chi(v), i ≥8, are large as well, and therefore charge(v) is not so large after all. For example, if indeed ch4(v) = 3, ch5(v) = 9, and ch6(v) = 28 (which is possible), then charge(v) is extremely small: at most −142636 (the analysis below will clarify this state-ment).
How do we find those vints that flip down to a given 3-vint v = (pv, Tv)?
Clearly, there is v itself.
If an edge in a triangle incident to pv can be flipped in Tv (such an edge cannot be incident to pv!), then flipping such an edge yields a 4-vint u = (pv, Tu) that can be flipped down to v (by reversing the preceding flip). If in the triangulation Tu there is a flippable edge that is not incident to pv but part of a triangle incident to pv, then we can flip this edge to get a 5-vint that can be flipped down to v, etc.
In order to represent this structure, we associate with a 3-vint v = (pv, Tv) a flip-tree τ(v) as follows. The root of the tree is labeled by the pair (tv, Nv), where tv is the hole of v (a triangle) and Nv is the set of its three vertex points (the neighbors of pv in Tv). All other nodes of the tree are associated with a pair (t, q), where t is a face of Tv and q is a point incident to that face (note that tv from the root is not a face of Tv—it contains pv and its incident faces).
(i) Every edge e of tv gives rise to a child if this edge can be flipped in Tv. If so, this child is labeled by the triangle incident to e that is not incident to pv, and by the point in this triangle which is not incident to e. So the root has at most three children.
(ii) Consider now a non-root node of the tree labeled by (t, q) and an edge e of t incident to q. If e is a boundary edge, no child will be obtained via e. Otherwise, let t′ be the other triangle incident to e. If t′ together with the triangle formed by e and pv is a convex quadrilateral (where e can be flipped), then this gives rise to a child of (t, q) labeled by (t′, q′) where q′ is the point of t′ that is not incident to e.
rigid core Figure 6: The tree of a 3-vint. Bold edges are rigid edges.
So a non-root node has at most two children.
Note that the union of all triangles of the nodes of any subtree of τ(v) (containing the root) form a polygon that is star-shaped with respect to pv; this follows easily by the inductive definition of τ(v). The triangles form a triangu-lation of the polygon, and the subtree is actually the dual tree of this triangulation. If we retriangulate this polygon in Tv by connecting pv to all vertices of the polygon, we get a vint that flips down to v. And we get all vints that flip down to v in this way. That is: Lemma 3.3. The subtrees of τ(v) containing its root are in bijective correspondence with the vints that flip down to v.
The next step is to determine how much these vints charge to v. This depends on the number of triangulations of the holes of these vints—the fewer triangulations, the more v is charged in the provisional scheme.
The analysis given here only discriminates between vints that charge 1 to v in the provisional scheme, and all other vints (which charge at most 1 2 in that scheme).
We first define rigid edges of τ(v): An edge of the tree connects two nodes labeled by two triangles t and t′ with a common edge e. If e cannot be flipped in the union of these two triangles, then we call the “dual” tree edge rigid.
Beware that e may be flippable in Tv while it is not flippable in t∪t′—this may happen if one of the two triangles is tv (and thus not a triangle of Tv). Now the rigid core, τ ∗(v), of τ(v) is defined to be the maximal subtree of τ(v) that includes the root and consists exclusively of rigid edges.
τ ∗(v) is non-empty, since it always contains the root of τ(v).
Lemma 3.4. The subtrees of the rigid core τ ∗(v) contain-ing the root are in bijective correspondence with the vints u that flip down to v and provisionally charge 1, i.e. supp(u) = 1.
Proof.
Consider a vint u that flips down to v. We recall that supp(u) = 1 iffthe hole of u has exactly one triangu-lation. Note that one triangulation of this polygon can be obtained by taking the set of triangles in the subtree cor-responding to u. If all edges in this subtree are rigid, then none of the dual edges in the triangulation can be flipped.
That is, there is only one triangulation of the hole, since the set of triangulations of a polygon is connected via edge-flips.
Also, if any of the edges is not rigid, then its dual edge can be flipped, and so obviously there are at least two triangu-lations.
In order to upper bound charge(v), we first restrict ourselves to vints that correspond to subtrees of τ(v) of depths at most 3. Note that in this way we do not lose any 3-, 4-, 5-, or 6-vints, i.e., no vint that charges a positive amount in the actual scheme is lost. Moreover, we let all i-vints, i = 4, 5, 6, whose subtree is not part of the rigid core charge 7−i 2 ; this is an upper bound on the actual charge. Finally, we include in the charge only the negative charges that come from i-vints, i ≥8, whose subtrees are part of the rigid core, and thus charge 7 −i. These modifications cannot decrease the overall charge made to the 3-vint v.
rigid core Figure 7: The rigid core that gives 43 with the five subtrees corresponding to 5-vints that provisionally charge 1.
How much can be charged with these restrictions?
We further simplify the analysis, by assuming that our tree is complete3 up to level 3. If not, we can extend the tree with non-rigid edges, and thus increase the modified charge (since those edges will not be used for negative charges). Now we simply have to maximize the modified charge over all possi-bilities of rigid cores of complete trees of depth 3. We have 3“Complete” means that the root has three children, and all other non-leaf nodes have two children.
written a small program to determine the maximum charge, which shows that this charge is at most 43. The maximiz-ing rigid core is shown in Fig. 7. The 3-vint is provisionally charged 1 by one 3-vint (itself), three 4-vints, five 5-vints (out of possible 9), six 6-vints (out of 28), and one 8-vint.
Its modified charge is thus 4 · 1 + 3 · 3 + 2 · 5 + 4 2 ´ + 6 + 22 2 ´ −1 = 43 One can also bypass the program, and argue, using a tedious case analysis, that this is indeed the maximum (modified) charge. Thus charge(v) ≤43 for every 3-vint v and Theorem 3.5. ˆ v3 ≥ n 43 for every set of n points.
The modified charge used in the last step of the analysis has a lot of room for improvement. First, we have assumed that each 3-, 4-, 5-, and 6-vint that does not come fully from the rigid core charges 1 2. However, to really charge 1 2, the associated hole must have only two triangulations, and thus only one flippable edge. Any other vint charges at most 1 3 to the 3-vint. One should therefore examine all rigid cores and all possible ways to attach to them non-rigid children, and count separately the number of vints with charge 1, those with charge 1 2, and bound pessimistically the num-ber of remaining positively-charging vints (which charge at most 1 3).
Initial exploration with this approach suggests that the bound drops to 38. A more careful analysis, that includes also vints with negative charges should decrease the bound further. Of course, the ultimate manifestation of the technique would be to test by a program all possible neigh-borhoods (up to level 3) and calculate exactly the maximum charge possible.
4.
MISCELLANEOUS BOUNDS We exhibit here a number of further restrictions on the expected degree sequences (ˆ vi)i∈N of finite planar point sets.
Lemma 4.1. For all integers 3 ≤i ≤j there is a positive integer δi,j such that ˆ vi ≥ ˆ vj δi,j . In particular, ˆ vi ≥ ˆ vi+1 i , ˆ vi ≥ 2ˆ vi+2 i(i+3), ˆ v3 ≥ ˆ vi Ci−1−Ci−2 , ˆ v4 ≥ ˆ vi Ci−1−2Ci−2 .
Proof.
For the inequality ˆ vi ≥ ˆ vi+1 i , we let every (i+1)-vint charge some i-vint it can be flipped down to. Since every vertex of degree at least 4 is incident to a flippable edge, such an i-vint is always available. Note that an i-vint can be reached at most i times in this way.
For the general inequality we observe that we can choose δi,j = ti,j−i+1 where ti,k denotes the number of binary trees with k nodes with an exceptional root of degree i (just like the binary nodes distinguish between a left and a right child, the root discriminates its children via an in-dex in {1, 2, . . . , i}). To see this, consult a generalization of the flip-trees from the previous section. It is known that t2,k = Ck (for the generic binary trees), which yields also t1,k = Ck−1. ti,1 = 1, ti,2 = i, and ti,3 = `i 2 ´ + 2i = i(i+3) 2 can be easily seen.
The number observes the recurrence ti,k = ti−1,k+1 −ti−2,k+1 (proof omitted, generalizes an ar-gument in ).
Now the asserted values for δi,j can be readily obtained: δi,i+1 = ti,2 = i, δi,i+2 = ti,3 = i(i+3) 2 , δ3,j = t3,j−2 = t2,j−1 −t1,j−1 = Cj−1 −Cj−2, and, finally, δ4,j = t4,j−3 = t3,j−2 −t2,j−2 = Cj−1 −2Cj−2.
Theorem 4.2. ˆ v3 ≤2n+3 5 for every set of n points.
Proof.
We apply a scheme where every 3-vint charges 3 units to vints of larger degrees or to boundary edges (there are three). We show that no vint is charged more than 2, and no boundary edge more than 1. This will imply that 3ˆ v3 ≤3 + 2 X j≥4 ˆ vj = 2(n −ˆ v3) + 3, (4) which yields the asserted inequality.
Let v = (p, T) be a 3-vint, and let tv denote its hole, which is a triangle.
For each edge e of tv we do the following, depending on the properties of e; see Fig. 8.
(1) e is a boundary edge.
Then we let v charge 1 to e, called boundary-charge.
(2) There is a triangle t incident to e on its other side.
(2.1) t forms with p a convex quadrilateral. We can flip e to get a 4-vint (p, T ′) to which v charges 1. We call this a flip-charge.
(2.2) t forms with p a non-convex quadrilateral. Let a the endpoint of e which is reflex in this quadrilateral; note that a cannot lie on the boundary, and it has to be of degree at least 4, since interior vertices of degree 3 are never adjacent (the “interior” condition is necessary only in case n = 1). Here v charges 1 to vint (a, T ), called neighbor-charge. Let us label such a charge with the responsible edge e.
Consider now a vint w = (q, T ). We call an edge ρ incident to q in T a separable edge at w if it can be separated from the other edges incident to q by a line that passes through q. An equivalent condition is that the two angles between ρ and its clockwise and counterclockwise next edges (at q) sum up to more than π. In the context of the neighbor-charge as described above, the responsible edge e is separable at (a, T ). We observe the easy following properties (see Fig. 9 for an illustration).
(S0) No edge is separable at both vints induced by its end-points.
(S1) If w has degree 3, every edge incident to its point is separable at w; (recall here that points of vints are interior).
(S2) If w has degree at least 4, at most two incident edges can be separable at w.
(S3) If w is of degree at least 4 and there are two edges separable at w, then they must be consecutive.
We note that the charges resulting from the three edges of a hole tv are all different. This is clear for charges ob-tained by edge flips. For neighbor-charges, it is impossible that (a, T ) is charged twice, by each of its incident edges in tv, because these two edges cannot both be separable (as follows, e.g., from (S3)).
We are now ready to show that no vint u can be charged more than twice.
Consider first the case of a 4-vint u = (pu, Tu). Let hu denote the quadrangular hole of pu. We note that at most two edges incident to pu are flippable: One out of each pair of opposite edges is separable at u and thus unflippable; see Fig. 10(a).
Figure 9: Illustrating the properties of separable edges.
(a) (b) Figure 10: (a) Only two edges incident to a 4-vint can be flippable. (b) No neighbor of a 4-vint with a convex hole can be of degree 3.
(a) If u receives two flip-charges, it cannot be charged as a neighbor, because in this case hu must be convex, and then no vertex of hu can be interior and of degree 3; see Fig. 10(b).
(b) u can be charged at most once as a neighbor. Indeed, if a is a vertex of hu of degree 3, then it must be a reflex vertex of hu, and there can be at most one such vertex.
(a) and (b) establish the Figure 11: A 4-vint with a non-convex hole is charged twice.
claim for 4-vints. (We note the following stronger prop-erty: If the hole of pu is con-vex, then u is charged ex-actly twice (by edge flips).
On the other hand, if the hole of pu is non-convex then it can be charged twice, once by an edge flip and once as a neighbor, if and only if pu and its charging neighbor are enclosed in a triangle as in Fig. 11.) boundary-charge flip-charge neighbor-charge p p p p e e e a T T T T ′ Figure 8: The various types of charges of a 3-vint in the proof of Theorem 4.2.
Consider next the case where pu Figure 12: Neighbor-charges to a vint u with two separable edges.
u = (pu, Tu) is a vint of degree at least 5.
Each flip-charge is to a 4-vint and therefore vints of degree at least 5 can receive neighbor-charges only. We claim that in this case pu can be a neighbor of at most two points of degree 3 that charge it as a neighbor. Recall the ingredients necessary for such a neighbor-charge to u: (i) an egde e that is separable at u and (ii) a neigh-bor a of pu that has degree 3 so that the edges e and pua are consecutive around pu. Clearly, if there is only one edge separable at u then there are at most two such constellations; see Fig. 13. If there are two separa-ble edges at u, then they have to be consecutive around pu (recall (S3)).
This rules out the possibility that any of these two edges is involved in more than one neighbor-charge, since an edge cannot be both, separable at pu and connect to a point of degree 3; see Fig. 12.
The weakness in the proof of Figure 13: Neighbor-charges to a vint u with one separable edge.
Theorem 4.2 is that it “assumes” that every vint of degree at least 4 is charged exactly twice. We can show that this cannot be the case which gives a slight improve-ment on the result—omitted here.
We also note that there is a limit on how small ˆ v3 can be: A construction in gives sets of n points, with n arbitrarily large, with only 3.17n triangu-lations.
Hence, the best upper bound that we can hope to prove is ˆ v3 ≤n/3.17.
Finally, we derive a lower bound on ˆ v4; it follows from the previously obtained linear constraints, without further reference to the underlying problem necessary.
Lemma 4.3. For n ≥2, ˆ v4 > n 540 .
Proof.
Given some value for ˆ v4 we have a supply of upper bounds for all the other ˆ vi’s due to Lemma 4.1 and Theo-rem 4.2, namely ˆ v3 ≤2n+3 5 ˆ v5 ≤(C4 −2C3)ˆ v4 = 4ˆ v4 ˆ v6 ≤(C5 −2C4)ˆ v4 = 14ˆ v4 ˆ v7 ≤(C6 −2C5)ˆ v4 = 48ˆ v4 ˆ v8 ≤(C7 −2C6)ˆ v4 = 165ˆ v4 . . .
Moreover, we have X i ˆ vi = n (5) X i iˆ vi ≤ 6n −5 (6) Now consider the ˆ vi’s as nonnegative real variables that have to obey the constraints listed. Clearly, in order to satisfy (6) we will push as much of the value n to be distributed among the ˆ vi’s to those of smaller index. Along those lines, if we suppose that ˆ v4 = n 540 , then we will choose ˆ v3 = 2n+3 5 , ˆ v5 = 4n 540 , ˆ v6 = 14n 540 , ˆ v7 = 48n 540 , ˆ v8 = 165n 540 , ˆ v9 = 92n 540 −3 5, and ˆ vi = 0 for i ≥10; (in this way (5) is fulfilled).
Now the sum in (6) evaluates to 3 2n+3 5 + n 540 (4+5·4+6·14+7·48+8·165+9·92)−9·3 5 = 6n−18 5 , a contradiction.
5.
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187598 | https://ouhsc.edu/bserdac/dthompso/web/namics/firstcmc.htm | CMC Joint of the Thumb Axes and movements Close-packed positions Arthrokinematics Naming of movements at the first CMC joint Axes and movements(Smith, Weiss, & Lehmkuhl, 1996, Fig. 6-21, p.212)
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| 1. The thumb's MP and CMC joints abduct and adduct in a plane perpendicular to the palm. Some therapists also refer to abduction as "palmar abduction."| | | 2. The thumb's MP and CMC joints flex and extend in a plane parallel to the palm. Some therapists refer to extension as "radial abduction," because the thumb moves toward the hand's radial side. | |
Why do we name CMC movements in this manner?
The CMC joint is biaxial, not triaxial. However, its loose capsule permits rotation, and the metacarpal rotates automatically when it moves in the other two planes. Specifically:
combined CMC flexion and abduction produces CMC opposition
combined CMC extension and adduction produces CMC reposition.
Close-packed positionsYou cannot passively rotate the CMC when you place it in full opposition or full reposition; these are the CMC joint's close packed positions. In these positions, the capsular fibers are maximally elongated and taut, and so prevent CMC rotation.
ArthrokinematicsThe first metacarpal articulates with the trapezium.
The diagram (Norkin & Levangier, 1992, p. 287) depicts the plane for CMC abduction/adduction. Compare it with the figure in your text (Hertling & Kessler, 1996, Fig. 11-7).
| | The saddle-shaped trapezium is: concave in the plane of CMC abduction/adduction During CMC abduction or adduction (in a direction perpendicular to the palm), the metacarpal rolls and glides in opposite directions. convex in the plane of CMC flexion/extension During CMC flexion or extension (in a direction parallel to the palm), the metacarpal rolls and glides in the same direction. |
--- |
Naming of movements at the first CMC jointWe name thumb movements just as we do movements in the other finger joints. Abduction and adduction occur around an antero-posterior axis. Flexion and extension occur around a lateral axis.
However, because the thumb attaches to the hand at a different angle than do the other digits, the thumb's AP and lateral axes are oriented differently than in the other digits.
The thumb or first ray, which comprises the metacarpal, proximal phalanx, and distal phalanx, attaches to the trapezium and the rest of the hand at the first carpo-metacarpal (CMC) joint. This attachment is at nearly a right angle to those of the second, third, fourth, and fifth digits.
| To appreciate this, examine the orientation of the fingernails on your relaxed hand as you point your fingertips toward yourself. The thumbnail is oriented at a right angle to the nails of the other fingers. || The familiar lateral axes, around which the fingers flex and extend, are turned 90 degrees at the thumb's CMC and MP joints. These joints flex and extend in a plane that is parallel to the palm. || The familiar AP axes, around which the MP joints abduct and adduct, are also turned 90 degrees at the thumb's CMC and MP joints. These joints abduct and adduct in a plane that is perpendicular to the palm. | | | |
--- --- --- |
You can always orient yourself to the thumb's joint axes by considering the plane in which the thumbnail rests to be the thumb's frontal plane. Abduction and adduction of the first CMC and MP joints occur in this plane. Flexion and extension of the first CMC, MP, and IP joints occur in a plane that is perpendicular to the thumbnail. Orienting yourself to the thumb's unique planes of movement in this way lets you name the thumb's movement regardless of its position, and regardless of how the first ray might be rotated during opposition or reposition. References: Hertling, D., & Kessler, R. M. (1996). Management of common musculoskeletal disorders: Physical therapy principles and methods. (3rd ed.). Philadelphia: J.B. Lippincott.
Norkin, C.C., & Levangie, P.K. (1992). Joint structure and function (2nd ed.). Philadelphia: F.A. Davis.
Last updated 3-2-01 ©Dave Thompson PT return to Control of Human Movement 1 and 2 lecture schedule |
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第一章 概率与统计
一 随机变量 1.1 离散型随机变量的分布列
1.2 离散型随机变量的期望与方差
二 统计 1.3 抽样方法
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阅读材料 回归直线方程的推导
实习作业 通过抽样调查，研究实际问题
小结与复习
复习参考题一
第二章 极限
一 数学归纳法 2.1 数学归纳法及其应用举例
阅读材料 不完全归纳法与完全归纳法
研究性学习课题：杨辉三角
二 极限 2.2 数列的极限
2.3 函数的极限
2.4 极限的四则运算
阅读材料 无穷等比数列（|q|<1）的和
2.5 函数的连续性
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