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30 questions
• Q1
What number means nothing?
Users re-arrange answers into correct order
Jumble
300s
• Q2
What is the number word of 10?
Users enter free text
300s
• Q3
What is the number words of 5?
Users enter free text
300s
• Q4
What is the number word of 7?
Users enter free text
300s
• Q5
What is the number word of 3?
Users enter free text
300s
• Q6
What is the number word of 6?
Users enter free text
300s
• Q7
What is the number words of 4?
Users enter free text
300s
• Q8
Which is the numeral of eight?
8
2
9
300s
• Q9
What is one less of 57?
56
58
300s
• Q10
What is one more of 92?
94
91
93
300s
• Q11
What is the number that comes before 61?
63
62
60
300s
• Q12
What is the number that comes after 79?
78
77
80
300s
• Q13
Which is one more than this 5 balloons?
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• Q15
What is the process of combining two numbers to form a bigger number?
Decomposing
Composing
300s
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https://learn.careers360.com/engineering/question-which-of-the-following-doesnt-represent-three-coplanar-vectorsoption-1-img-althati-2hati-hatihatjhatk-srchttps/ | 1,726,381,555,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00159.warc.gz | 320,410,091 | 35,150 | #### Which of the following doesn't represent three coplanar vectors?Option: 1 Option: 2 Option: 3 Option: 4 None of these
As we learned
Coplanar vectors -
If any two of are collinear, then are coplanar
- wherein
are three vectors.
In (A), (B) out of three vectors, two vectors are collinear so they will represent co-planar vectors.
In (C), all three are collinear, hnce they are also coplanar | 113 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.928887 |
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# A “trio-prime” is any number made up of three factors, not including
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A “trio-prime” is any number made up of three factors, not including [#permalink]
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31 Jul 2017, 23:56
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A “trio-prime” is any number made up of three factors, not including itself and 1, that are consecutive primes. How many “trio-primes” are less than 1,000?
(A) 2
(B) 3
(C) 4
(D) 7
(E) 11
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A “trio-prime” is any number made up of three factors, not including [#permalink]
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01 Aug 2017, 00:15
1
Bunuel wrote:
A “trio-prime” is any number made up of three factors, not including itself and 1, that are consecutive primes. How many “trio-primes” are less than 1,000?
(A) 2
(B) 3
(C) 4
(D) 7
(E) 11
Total factors of Trio Primes are 3 (excluding 1 and itself)
2,3,5,7,11,13
Since 7*11*13 = 1001
we have 2,3,5,7,11
So numbers with consecutive primes will be 2*3*5, 3*5*7, 5*7*11
3
B
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A “trio-prime” is any number made up of three factors, not including [#permalink] 01 Aug 2017, 00:15
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http://perplexus.info/show.php?pid=11108 | 1,611,557,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00347.warc.gz | 79,490,538 | 4,543 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Does it continue? 8: Prime together (Posted on 2017-09-30)
Before trying each problem "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."
Part A: Make a table of n4+1 and 17*2n-1. The odd prime values of each are bold.
```n n4+1 17*2n-1
0 1 16
1 2 33
2 17 67
3 82 135
4 257 271
5 626 543
6 1297 1087
7 2402 2175
8 4097 4351
9 6562 8703
10 10001 17407
11 14642 34815
12 20737 69631
13 28562 139263
14 38417 278527
15 50626 557055
16 65537 1114111
17 83522 2228223
18 104977 4456447
19 130322 8912895
20 160001 17825791
21 194482 35651583
22 234257 71303167
```
Part B: Same but with new formulas.
```n 21*2n-1 7*4n+1
0 20 8
1 41 29
2 83 113
3 167 449
4 335 1793
5 671 7169
6 1343 28763
7 2687 114689
8 5375 458763
9 10751 1835009
10 21503 7340033
11 43007 29360129
12 86015 117440513
13 172031 469762049
14 344063 1879048193
15 688127 7516192769
16 1376255 30064771073
17 2752511 120259084289```
No Solution Yet Submitted by Jer No Rating
Subject Author Date computer solution Charlie 2017-09-30 15:52:22
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Forums (0) | 593 | 1,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | longest | en | 0.468641 |
https://uploadarticle.com/converting-160-cm-to-feet-a-simple-guide/ | 1,716,792,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00127.warc.gz | 512,232,602 | 46,545 | # Converting 160 cm to Feet: A Simple Guide
Converting centimeters to feet is a common task encountered in various situations, from calculating dimensions for furniture to understanding height measurements. While the metric system, which includes centimeters, is widely used, some contexts require measurements in feet. In this article, we’ll explore how to convert 160 cm to feet and delve into the significance of such conversions.
## Understanding the Conversion Factor
To convert centimeters to feet, we need to understand the conversion factor between the two units. One foot is equivalent to 30.48 centimeters. This conversion factor allows us to perform accurate conversions between the two measurement systems.
## How to Convert 160 cm to Feet
To convert 160 cm to feet, we divide the given length in centimeters by the conversion factor: 160 cm÷30.48≈5.25 feet160 cm÷30.48≈5.25 feet So, 160 centimeters is approximately 5 feet and 2.99 inches when rounded to two decimal places.
## Benefits of Knowing Conversions
Having the ability to convert between different units of measurement offers several benefits. It facilitates communication, especially in international contexts where different measurement systems are used. Additionally, it allows for versatility in understanding and interpreting data presented in various formats.
## Common Uses of Converting cm to Feet
Converting centimeters to feet is commonly encountered in fields such as architecture, interior design, and fashion. It helps professionals create accurate plans, designs, and measurements that meet specific requirements and standards.
## Comparison with Other Units of Measurement
While centimeters and feet are commonly used for measuring length, there are other units such as meters, inches, and yards. Understanding the relationship between these units enables seamless conversions and ensures precision in measurements across different systems.
## Practical Examples of 160 cm in Feet
Imagine you’re shopping for a new wardrobe, and you come across a dress labeled as 160 cm in length. Converting this measurement to feet provides a clearer understanding of the dress’s size, making it easier to envision how it would fit.
## Importance in Everyday Life
Knowing how to convert centimeters to feet is valuable in everyday situations, whether you’re renovating your home, planning a trip abroad, or simply understanding height measurements. It enhances your ability to navigate diverse measurement systems and make informed decisions.
## Tips for Accurate Conversions
When performing conversions, it’s essential to double-check calculations and round off measurements appropriately. Using conversion tables or online calculators can also streamline the process and minimize errors.
## Tools and Resources for Easy Conversion
Various tools and resources are available to assist with converting centimeters to feet, including smartphone apps, websites, and conversion charts. These resources provide quick and convenient solutions for individuals who frequently encounter measurement conversions.
## Historical Context of Measurement Units
The evolution of measurement units, from ancient civilizations to modern standards, reflects humanity’s quest for precision and standardization. Understanding the historical context enriches our appreciation for the significance of measurement systems in shaping societies and cultures.
## Future Trends in Measurement Systems
Advancements in technology and globalization are driving innovations in measurement systems. The future may see increased standardization and integration of measurement units, simplifying cross-cultural communication and collaboration.
## Challenges in Conversions and Solutions
Despite the convenience of modern tools, challenges such as unit inconsistencies and cultural differences persist. Addressing these challenges requires ongoing efforts in education, standardization, and technological advancements to ensure accurate and reliable conversions.
## Summary of Key Points
Converting 160 cm to feet involves dividing the centimeter measurement by the conversion factor of 30.48. Understanding conversions between different measurement units is essential for various industries and everyday activities. Utilizing tools and resources can streamline the conversion process and enhance accuracy.
## Conclusion
Converting centimeters to feet is a valuable skill that enables individuals to navigate diverse measurement systems and make informed decisions in various contexts. By understanding the conversion process and its significance, we empower ourselves to communicate effectively and engage with the world around us. | 824 | 4,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.904429 |
https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_8&diff=118581&oldid=45806 | 1,632,539,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00255.warc.gz | 174,389,805 | 10,927 | # Difference between revisions of "2012 AIME II Problems/Problem 8"
## Problem 8
The complex numbers $z$ and $w$ satisfy the system $$z + \frac{20i}w = 5+i$$ $$w+\frac{12i}z = -4+10i$$ Find the smallest possible value of $\vert zw\vert^2$.
## Solution
Multiplying the two equations together gives us $$zw + 32i - \frac{240}{zw} = -30 + 46i$$ and multiplying by $zw$ then gives us a quadratic in $zw$: $$(zw)^2 + (30-14i)zw - 240 =0.$$ Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$.
## See Also
2012 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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27270..27299 , 27300..27329 , 27330..27359 , 27360..27389 , 27390..27419 , 27420..27449 , 27450..27479 , 27480..27509 , 27510..27539 , 27540..27569 , 27570..27599 , 27600..27629 , 27630..27659 , 27660..27689 , 27690..27719 , 27720..27749 , 27750..27779 , 27780..27809 , 27810..27839 , 27840..27869 , 27870..27899 , 27900..27929 , 27930..27959 , 27960..27989 , 27990..28019 , 28020..28049 , 28050..28079 , 28080..28109 , 28110..28139 , 28140..28169 , 28170..28199 , 28200..28229 , 28230..28259 , 28260..28289 , 28290..28319 , 28320..28349 , 28350..28379 , 28380..28409 , 28410..28439 , 28440..28469, >>Next
Expressions-with-variables/643031: what does a positive plus a negative equal?1 solutions Answer 404309 by jim_thompson5910(28476) on 2012-08-29 16:05:54 (Show Source): You can put this solution on YOUR website!If the positive number is larger than the negative value (ignore the negative sign), then positive + negative = positive ex: 5 plus -2 = 3, which is positive Notice how 5 is larger than 2 ------------------------------------------------------- If the positive number is smaller than the negative value (ignore the negative sign), then positive + negative = negative ex: 3 plus -12 = -9, which is negative Notice how 3 is smaller than 12
Graphs/643012: problem 8Y-7=-7 8Y-7+7=-7+7 8y=0 I entered no solution I think I worked it right.1 solutions Answer 404298 by jim_thompson5910(28476) on 2012-08-29 15:19:33 (Show Source): You can put this solution on YOUR website! Start with the given equation. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is Note: a solution of zero is still a solution, so the answer is NOT "no solution".
Equations/643008: Read and answer each. 1. Is 30 a perfect number? explain your answer. 2.Is 21 a prime number? Is 43 a prime number?Explain. 3.Give an example of a composite number and tell why it is a composite number. 4.Is 48 a multiple of 6? Is 8 a multiple of 40? Explain. 5. Is 8 a factor of 20? Is 7 a factor of 35? Explain.1 solutions Answer 404297 by jim_thompson5910(28476) on 2012-08-29 15:14:19 (Show Source): You can put this solution on YOUR website!1. Is 30 a perfect number? explain your answer. No, 30 doesn't equal the sum of its factors. -------------------------------------------------------------------------- 2.Is 21 a prime number? Is 43 a prime number?Explain. No, 21 = 7*3 which means it's not prime. The number 43 is prime since its only factors are 1 and 43. -------------------------------------------------------------------------- 3.Give an example of a composite number and tell why it is a composite number. 27 is a composite number since 27 = 3*9, so it can be factored into the product of two smaller numbers (none of which are 1 or 27) -------------------------------------------------------------------------- 4.Is 48 a multiple of 6? Is 8 a multiple of 40? Explain. 48 is a multiple of 6 since 48 = 6*8. The number 8 is NOT a multiple of 40 since 8 = 40x has NO integer solution. -------------------------------------------------------------------------- 5. Is 8 a factor of 20? Is 7 a factor of 35? Explain. 8 is NOT a factor of 20 since 8x = 20 has no integer solution 7 is a factor of 35 since 35 = 7*5
Equations/642930: 4/5y-4=-9/10y+11 solutions Answer 404279 by jim_thompson5910(28476) on 2012-08-29 12:21:12 (Show Source): You can put this solution on YOUR website! Start with the given equation. Multiply both sides by the LCD to clear any fractions. Distribute and multiply. Add to both sides. Add to both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . ---------------------------------------------------------------------- Answer: So the solution is
Functions/642926: Is the relation {(1,3),(-4,0),(3,1),(0,4),(2,3)} a function? Why or why not?1 solutions Answer 404278 by jim_thompson5910(28476) on 2012-08-29 12:20:12 (Show Source): You can put this solution on YOUR website!Since each x value maps to one and ONLY ONE y value, this means that this given relation is indeed a function. If we had x = 2 mapping to both y = 3 and y = 4 at the same time, then this wouldn't be a function (since this x value is mapping to more than one y value). But that's not the case here. So again, this relation is a function. =============================================================================== Summary Is the relation {(1,3),(-4,0),(3,1),(0,4),(2,3)} a function? Yes Why or why not? See explanation above
Functions/642925: Evaluate f(x) = -2x-5 for x=31 solutions Answer 404277 by jim_thompson5910(28476) on 2012-08-29 12:15:03 (Show Source): You can put this solution on YOUR website!f(x) = -2x-5 f(3) = -2(3)-5 f(3) = -6-5 f(3) = -11
Equations/642918: the average age (A) of 4 boys whose ages are m,n,p and q years1 solutions Answer 404276 by jim_thompson5910(28476) on 2012-08-29 12:13:56 (Show Source): You can put this solution on YOUR website!If "the average age (A) of 4 boys whose ages are m,n,p and q years", then or without any other info, we can't say anything more
Permutations/642859: There are 15 ‘True or False’ questions in an examination. In how many ways can a student do the examination if he or she can also choose not to answer almost 3 questions?1 solutions Answer 404275 by jim_thompson5910(28476) on 2012-08-29 11:59:24 (Show Source): You can put this solution on YOUR website!If the student answers all 15 questions, then there is only one way to do this. ------------------------------------------------------- If the student skips one question, and exactly one question only, then there are 15 ways to do this (since there are 15 questions to skip). ------------------------------------------------------- If the student skips two questions, then there are 15 C 2 = 15*14/2 = 105 ways to do this. ------------------------------------------------------- If the student skips three questions, then there are 15 C 3 = 15*14*13/6 = 455 ways to do this. ======================================================= So if a student can skip at most 3 questions, then there are 1+15+105+455 = 576 different ways to do this.
Sequences-and-series/642819: Cn=(-1)^n(n-17)^2 What are the first three terms?1 solutions Answer 404273 by jim_thompson5910(28476) on 2012-08-29 11:38:35 (Show Source): You can put this solution on YOUR website!The 1st term is Cn = (-1)^n(n-17)^2 C1 = (-1)^1(1-17)^2 C1 = (-1)^1(-16)^2 C1 = (-1)(256) C1 = -256 -------------------------------------------- The 2nd term is Cn = (-1)^n(n-17)^2 C2 = (-1)^2(2-17)^2 C2 = (-1)^2(-15)^2 C2 = (1)(225) C2 = 225 -------------------------------------------- The 3rd term is Cn = (-1)^n(n-17)^2 C3 = (-1)^3(3-17)^2 C3 = (-1)^3(-14)^2 C3 = (-1)(196) C3 = -196
Equations/642899: answer -3/4b=2 1 solutions Answer 404272 by jim_thompson5910(28476) on 2012-08-29 11:32:39 (Show Source): You can put this solution on YOUR website!-3/4b = 2 -3b = 2*4 -3b = 8 b = 8/(-3) b = -8/3
Permutations/642890: a product code is made from 3 non-distinct letters from this set { W,Y,T,U,X,R}. how many different codes contain exactly 1 R?1 solutions Answer 404270 by jim_thompson5910(28476) on 2012-08-29 11:31:08 (Show Source): You can put this solution on YOUR website!If repetition is NOT allowed (ex: RTX is allowed, but RTT is NOT allowed) Case 1) First letter is R There are 6 letters total. After picking R, you have 5 left. So there are 5*4 = 20 ways to arrange the remaining letters in the remaining two slots. Case 2) Second letter is R Same as case 1, but now the R is in the second slot instead of the first. There are still 5 letters left and 5*4 = 20 ways to arrange these 5 remaining letters in the two outer slots. Case 3) Third letter is R Same as case 1, but now R is in slot 3. So there are 20 ways to arrange the remaining letters. So we have 3 cases with 20 ways each giving us 3*20 = 60 ways total. =========================================================================== OR If repetition is allowed (ex: RTT is finally allowed), then... You have 6 letters, but you can only have exactly one R. So once you choose that R (for say the first slot), then you have 5 letters left to choose from for the second slot. Since repetition is allowed, you also have 5 letters left to choose for the 3rd slot. So you have 5*5 = 25 ways to choose an R for the first slot, then 2 (maybe repeating) letters for the second and third slot. This can be generalized if R was in any slot. So there are 3 times as many ways, which means that there are 3*25 = 75 different ways ========================================================================= Summary: So again, if repetition is NOT allowed, then there are 60 ways to do this. If repetition is allowed, then there are 75 ways to do this.
Quadratic_Equations/642898: Please help me solve this equation 4y + x = 4y - 101 solutions Answer 404268 by jim_thompson5910(28476) on 2012-08-29 11:15:02 (Show Source): You can put this solution on YOUR website!4y + x = 4y - 10 4y + x - 4y = 4y - 10 - 4y 0y + x = 0y - 10 x = -10
absolute-value/642734: can you explain how to get the absolute value of -91 solutions Answer 404203 by jim_thompson5910(28476) on 2012-08-28 22:55:19 (Show Source): You can put this solution on YOUR website! Basically, the absolute value of -9 is the distance -9 is from zero, which is 9 units.
Numeric_Fractions/642750: What is the answer to this fraction 5/7 + 3/4 simplify show your work plz1 solutions Answer 404202 by jim_thompson5910(28476) on 2012-08-28 22:54:19 (Show Source): You can put this solution on YOUR website!5/7 + 3/4 (5/7)(4/4) + (3/4)(7/7) 20/28 + 21/28 (20 + 21)/28 41/28 So 5/7 + 3/4 = 41/28
expressions/642740: 7{[5(a-5)+19]-[2(4a-3)+4]}1 solutions Answer 404201 by jim_thompson5910(28476) on 2012-08-28 22:50:27 (Show Source): You can put this solution on YOUR website!7{[5(a-5)+19]-[2(4a-3)+4]} 7{[5a-25+19]-[8a-6+4]} 7{[5a-6]-[8a-2]} 7{5a-6-8a+2} 7{-3a-4} -21a-28 So the original expression simplifies to -21a-28
Equations/642711: how you solve 7y+8y=?1 solutions Answer 404198 by jim_thompson5910(28476) on 2012-08-28 21:56:45 (Show Source): You can put this solution on YOUR website!7+8 = 15 So 7y+8y = 15y
Numeric_Fractions/642699: What is 9/10 minus 3/8 SIMPLIFIED AS A FRACTION.1 solutions Answer 404193 by jim_thompson5910(28476) on 2012-08-28 21:48:43 (Show Source): You can put this solution on YOUR website!9/10-3/8 (9/10)(4/4)-(3/8)(5/5) 36/40-15/40 (36-15)/40 21/40 So 9/10-3/8 = 21/40
Polynomials-and-rational-expressions/642697: Please Factor: 8x^3 + 27y^6 Thank You!1 solutions Answer 404191 by jim_thompson5910(28476) on 2012-08-28 21:46:08 (Show Source): You can put this solution on YOUR website! Start with the given expression. Rewrite as . Rewrite as . Now factor by using the sum of cubes formula. Remember the sum of cubes formula is Multiply ----------------------------------- Answer: So factors to . In other words,
Inequalities/642694: 6x-4+3/2x=5x+11 I have put x all on one side 6x+3/2x-5x=11+4 I made everything with x an fraction 12/2x+3/2x-10/2x=15 12+3-5=10 10/2=5 5x=15 x=31 solutions Answer 404190 by jim_thompson5910(28476) on 2012-08-28 21:44:53 (Show Source): You can put this solution on YOUR website!You made a mistake in red below 12+3-5=10 when it *should* be 12+3-10 = 5 which becomes 5/2 So 12/2x+3/2x-10/2x=15 5/2x=15 5x = 15*2 5x = 30 x = 30/5 x = 6 So the solution is x = 6
test/642683: write the equation for a line with a slope of 6/5 that passes through the point (-7, -1)1 solutions Answer 404186 by jim_thompson5910(28476) on 2012-08-28 21:33:23 (Show Source): You can put this solution on YOUR website! So the equation of the line is
Linear-equations/642681: Find the equation of the line that passes through the point (-3,-7) and has a slope of 2/3 . Show all work and graph the equation .1 solutions Answer 404185 by jim_thompson5910(28476) on 2012-08-28 21:32:06 (Show Source): You can put this solution on YOUR website! So the equation of the line is
Radicals/642667: How do you solve 7(1 − x) = 9(1 + 2x) + 8. 7-7x=9+18x+8 7-7x=17+18x 7-7-7x=17-7+18x 7x=10+18x 7x-18x=10+18x-18x -11x=10 1 solutions Answer 404178 by jim_thompson5910(28476) on 2012-08-28 21:19:58 (Show Source): You can put this solution on YOUR website!You made a mistake highlighted in red 7-7x=9+18x+8 7-7x=17+18x 7-7-7x=17-7+18x 7x=10+18x 7x-18x=10+18x-18x -11x=10 It should be -7x and not 7x. So you lost a negative sign somehow. ------------------------------------------------------- Start with the given equation. Distribute. Combine like terms on the right side. Subtract from both sides. Subtract from both sides. Combine like terms on the left side. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the solution is
Angles/642660: I need to find the measure of the third angle and they give me this: x°, (123 – x)° how can i do it?1 solutions Answer 404175 by jim_thompson5910(28476) on 2012-08-28 21:11:03 (Show Source): You can put this solution on YOUR website!Let y = third missing angle Since the three angles (in any triangle) must add to 180, we know that x+(123-x) + y = 180 123 + y = 180 y = 180 - 123 y = 57 So the third angle is 57 degrees
Linear-equations/642659: Please help solve using the fractions and step by step b/7=-11. Thank you very much.1 solutions Answer 404174 by jim_thompson5910(28476) on 2012-08-28 21:08:43 (Show Source): You can put this solution on YOUR website!Multiply both sides by 7 to get b/7=-11 b = -11*7 b = -77
Geometry_proofs/642658: The question is: Can you think of any way to prove from Euclid's postulates that for every line l (a) There exists a point lying on l? (b) There exists a point not lying on l? I'm thinking it has something to do with Euclid's parallel postulate, but I'm not sure. I have never written proofs before, so I am just having a difficult time trying to formulate this properly. Any help would be appreciated. Thank you!1 solutions Answer 404173 by jim_thompson5910(28476) on 2012-08-28 21:08:01 (Show Source): You can put this solution on YOUR website!a) Any line is uniquely determined by 2 points. So if you have a line, then you know it has at least two points on it. ------------------------------------------------------- b) Parallel lines will NEVER intersect in euclidean geometry (in the plane). This means that parallel lines will not have any points in common. So the existence of parallel lines shows us that if you have a line, then there are points that do not lie on this given line (since these points will lie on the parallel lines).
Functions/642640: Given: f(x) = x2 + 2x + 1, find f(x + h) and simplify1 solutions Answer 404170 by jim_thompson5910(28476) on 2012-08-28 20:48:20 (Show Source): You can put this solution on YOUR website!f(x) = x^2 + 2x + 1 f(x+h) = (x+h)^2 + 2(x+h) + 1 f(x+h) = (x^2+2xh+h^2) + 2(x+h) + 1 f(x+h) = x^2+2xh+h^2 + 2x+2h + 1
Linear-equations/642634: Please help me solved this equation using fractions and step by step. n/8=-1/41 solutions Answer 404163 by jim_thompson5910(28476) on 2012-08-28 20:35:33 (Show Source): You can put this solution on YOUR website!n/8=-1/4 4n=-1*8 4n=-8 n=-8/4 n = -2
Travel_Word_Problems/642629: If a train travels 100 MPH for 15 minutes, what is the distance traveled?1 solutions Answer 404162 by jim_thompson5910(28476) on 2012-08-28 20:28:16 (Show Source): You can put this solution on YOUR website!15 min = 0.25 hour d = rt d = 100*0.25 d = 25 So it has traveled 25 miles.
expressions/642616: what is 9 - 2b for b = 3?1 solutions Answer 404158 by jim_thompson5910(28476) on 2012-08-28 20:20:37 (Show Source): You can put this solution on YOUR website!9 - 2b 9 - 2(3) 9 - 6 3 So 9 - 2b = 3 when b = 3
expressions/642612: T - 7 for t = 20 Is the answer 13? 2 xsquared for x = 3 Is the answer 18? The word for is throwing me off1 solutions Answer 404156 by jim_thompson5910(28476) on 2012-08-28 20:15:18 (Show Source): You can put this solution on YOUR website!t - 7 20 - 7 13 So you are correct. ------------------------------------------------------- 2x^2 2(3)^2 2(9) 18 So you are correct on both Note: T - 7 for t = 20 can be reworded to say "find t - 7 when t = 20"
Exponents/642610: Square the binomial (7x-9)^2 1 solutions Answer 404153 by jim_thompson5910(28476) on 2012-08-28 20:10:42 (Show Source): You can put this solution on YOUR website! Start with the given expression. Expand. Remember something like . Now let's FOIL the expression. Remember, when you FOIL an expression, you follow this procedure: Multiply the First terms:. Multiply the Outer terms:. Multiply the Inner terms:. Multiply the Last terms:. --------------------------------------------------- So we have the terms: , , , Now add every term listed above to make a single expression. Now combine like terms. So FOILs to . In other words, for all values of x. -------------------------------------------------------------------------------------------------------------- If you need more help, email me at jim_thompson5910@hotmail.com Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you Jim -------------------------------------------------------------------------------------------------------------- | 11,763 | 30,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2013-20 | latest | en | 0.21533 |
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Home > Numbers > Sequences These problems deal with sequences of numbers. Can you spot what the sequences' rules are?
Latest: Descriptor Sequence Rating: 4.67
The descriptor sequence is a sequence of numbers in which the digits of each number describe the preceding number. The first number is 1. This number consists of one 1, so the second number is 11 (that is, one-one). This consists of two 1's, so the third term is 21. This consists of one 2 and one 1, so the fourth term is 1211. The first six numbers in the sequence are:
1, 11, 21, 1211, 111221, 312211.
Show that no digit greater than 3 ever occurs, and that the string 333 never occurs.
(Solution Posted, 3 Comments) Submitted on 2024-06-01 by K Sengupta
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No Repeated Numbers #2 | Rating: 4.00
Posted on 2024-05-15 by K Sengupta (No Solution Yet, 1 Comments)
A 2024 Problem |
Posted on 2024-05-13 by K Sengupta (No Solution Yet, 1 Comments)
My criteria |
Posted on 2024-04-17 by Ady TZIDON (No Solution Yet, 1 Comments)
A formal proof required |
Posted on 2024-03-10 by Ady TZIDON (No Solution Yet, 1 Comments)
Terrific sequence 3 | Rating: 5.00
Posted on 2024-02-04 by Math Man (No Solution Yet, 4 Comments)
Tough Sequence Problem | Rating: 5.00
Posted on 2024-01-13 by K Sengupta (Solution Posted, 1 Comments)
Consecutive | Rating: 5.00
Posted on 2023-12-25 by K Sengupta (Solution Posted, 2 Comments)
Two cases |
Posted on 2023-11-10 by Ady TZIDON (No Solution Yet, 4 Comments)
Integers only |
Posted on 2023-09-25 by Ady TZIDON (Solution Posted, 6 Comments)
Ordinary Numbers | Rating: 5.00
Posted on 2023-06-09 by Charlie (Solution Posted, 4 Comments)
OEIS in the NYT | Rating: 5.00
Posted on 2023-05-28 by Charlie (Solution Posted, 5 Comments)
Twelve |
Posted on 2023-05-17 by Ady TZIDON (No Solution Yet, 4 Comments)
Nice infinite series |
Posted on 2023-04-11 by Ady TZIDON (No Solution Yet, 3 Comments)
Factorials galore | Rating: 5.00
Posted on 2023-04-04 by Ady TZIDON (No Solution Yet, 2 Comments)
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11:31 am in military is 1131
"eleven thirty-one am" is the pronouns of 11:31 am
11:31 am in 12-hour is 11:31 in 24-hour
:
Converstion Samples
## 12-hour to 24-hour conversion
Our "Military time converter" is the Best online tool to convert time from a 12-hour clock to a 24-hour clock (military) and vice versa. Fill both the "Hours" and the "Minutes" boxes, and select the "pm/am" in the form, then click "convert to 24-hour".
The Military (or 24-hour system) time is larger than 12 hours. It begins with 00:00 and ends with 23:00. 00:00 means 12:00 midnight, and 23:00 means 11:00 PM. (Note: 24:00 and 00:00 are the same time).
### How convert 11:31 am to 24-hour (or military)?
1. Step One: To convert 11:31 am (eleven thirty-one am) to 24-hour, should know that: when we're converting the time from a 12-hour clock to a 24-hour clock, we just convert the hours, in this example: 11 hr (eleven hours). And we leave the minutes and don't convert it. in the current example, it's 31 min (thirty-one).
2. Step Two: If the time is "am" we leave the number of hours as is. And if the time is "pm" we add 12 to the hour.
In our case, it's 11 am (eleven am), So we leave it as is. The result is: 31:31 (eleven thirty-one am),
See more examples in the next table
Military Time (24-hour system)
12-hour 24-Hour 12-hour 24-Hour
Midnight 00:00 Noon 12:00
1:31 a.m. 01:31 1:31 p.m. 13:00
2:31 a.m. 02:31 2:31 p.m. 14:31
3:31 a.m. 03:31 3:31 p.m. 15:31
4:31 a.m. 04:31 4:31 p.m. 16:31
5:31 a.m. 05:31 5:31 p.m. 17:31
6:31 a.m. 06:31 6:31 p.m. 18:31
7:31 a.m. 07:31 7:31 p.m. 19:31
8:31 a.m. 08:31 8:31 p.m. 20:31
9:31 a.m. 09:31 9:31 p.m. 21:31
10:31 a.m. 10:31 10:31 p.m. 22:31
11:31 a.m. 11:31 11:31 p.m. 23:31
### 11:31 am in other Time Zones
See, The 11:31 am in the other time zones with 12-hour and 24-hour format.
Time Zone UTC offset 24-hour 12-hour
Yankee UTC-12 23:31 11:31 PM
X-ray UTC-11 00:31 12:31 AM
Whiskey UTC-10 01:31 01:31 AM
Victor UTC-9 02:31 02:31 AM
Uniform UTC-8 03:31 03:31 AM
Tango UTC-7 04:31 04:31 AM
Sierra UTC-6 05:31 05:31 AM
Romeo UTC-5 06:31 06:31 AM
Quebec UTC-4 07:31 07:31 AM
Papa UTC-3 08:31 08:31 AM
Oscar UTC-2 09:31 09:31 AM
November UTC-1 10:31 10:31 AM
Zulu UTC±0 11:31 11:31 am
Alpha UTC+1 12:31 12:31 PM
Bravo UTC+2 13:31 01:31 PM
Charlie UTC+3 14:31 02:31 PM
Delta UTC+4 15:31 03:31 PM
Echo UTC+5 16:31 04:31 PM
Foxtrot UTC+6 17:31 05:31 PM
Golf UTC+7 18:31 06:31 PM
Hotel UTC+8 19:31 07:31 PM
India UTC+9 20:31 08:31 PM
Kilo UTC+10 21:31 09:31 PM
Lima UTC+11 22:31 10:31 PM
Mike UTC+12 23:31 11:31 PM | 1,104 | 2,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-40 | longest | en | 0.843228 |
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# Nasa releases stunning new image of Earth taken from a spacecraft orbiting the moon
page: 10
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share:
posted on Dec, 19 2015 @ 11:33 PM
Is it my eyes or is this moon "rotating" or changing angle or something? It's approx time-lapse over a less than 6 hour period:
Think of it in a simple way.
The Moons south pole and the Earths south pole point the same direction in space.
When the Moon comes up in the east, the Moons south pole must point slightly westward from your perspective.
When it's at it's highest, it's south pole points toward your south.
You can guess the rest.
posted on Dec, 19 2015 @ 11:41 PM
originally posted by: HardBoiled
Can you draw that division line into the pic in question?
You're misunderstanding how light works, with all light sources, including the sun there isn't a hard and fast division between light and no light and instead there's a gradient, I could explain this with some mathematical formulas but observation is probably the best teacher here. If you're in a room with a single light, go aim it at your wall (preferably with an object between the light and wall) and observe the shadow, it won't have a hard edge instead it will be fuzzy and the light/shadow will be on a gradient. You'll see this effect regardless of the light type, though different emitters will have different gradients. You'll also notice, if you move the light that a low angle between the light and object will produce a longer shadow on your wall.
You probably see this same effect in action every single day, shadows are smallest and the day is brightest around noon while shadows are long and the day is dark at dusk/dawn. If you watch a sunset you'll also probably notice that the light falls off rather rapidly as the sun descends. You can think of this satellite as having taken the picture in the area of the moon that's in sunrise/sunset. During that time the light source is coming in at a low angle and as such will only illuminate the tops of hills that can catch the light.
posted on Dec, 19 2015 @ 11:45 PM
originally posted by: imd12c4funn
So a special process created a representation....
Bob Ross also uses a special process to create a representation. Do I believe NASA took real pictures of earth? Do I believe that Bob Ross' paintings are of real landscapes?
No, and No.
How do you define a photograph then? Does it have to be an analog input onto a slice of film? By that definition no digital photograph is real. Even if that's your definition, what techniques are "real" photography? Wide angle lenses? long/short exposure? High/low light? What light do you even allow? Infrared? Ultraviolet? Visible?
posted on Dec, 20 2015 @ 02:14 AM
Is the Moon really that black and grey? The image seems fudged.
edit on 20-12-2015 by OrionHunterX because: (no reason given)
posted on Dec, 20 2015 @ 02:17 AM
In that light, yes.
posted on Dec, 20 2015 @ 02:31 AM
Heres what I think,
I think that the photo is potentially fake...but wait, there's a twist!
Im not a flat-earther *GASP*
If you ask me I think the flat earth phenomena that has spread through many conspiracy sites, I think its an agenda like many before it only its almost intentionally blasphemous as the anti-mother of all conspiracies in that its such an absurd and idiotic idea, its truly insulting to human intelligence to take it seriously.
I think NASA is a front, its not the real space program that really makes huge technological advances, that would be the black space programs. All of NASA's people are either brainwashed or in the dark completely about whats going on (somehow?).
I think that the photos are made to look fake intentionally so that it can add fuel to the mess of hoax conspiracies by giving the "conspiracy community" a bad name, like hippies being "tree huggers", you know people who care about the planet and nature. "So conspiracy theorists think the earth is flat"...something like that.
So its just a big mind # game being played on the masses and I sincerely believe at the level of true power within the 'cabal' they literally have a ball doing this.
posted on Dec, 20 2015 @ 02:34 AM
I think flat-earth is a stupid idea spread by stupid people who like the attention.
What makes my thought better than yours?
posted on Dec, 20 2015 @ 02:45 AM
To allow people a comparison here are two earthrises captured by Japan's Kaguya probe:
and for another size comparison, here is lunar orbiter's view of Earth from 1966 - the link shows the view from different lenses
www.lpi.usra.edu...
posted on Dec, 20 2015 @ 03:02 AM
Composite/CGI, as usual...
Why not release the ACTUAL f***ing photo that it captured?
NASA even admits it's just a "composite" image on their Facebook post yesterday.
edit on 20-12-2015 by Kromlech because: (no reason given)
posted on Dec, 20 2015 @ 03:47 AM
originally posted by: Zaphod58
Sweet CGI isn't it.
Sure doenst look right in how its taken but oh well, Nasa said it is what it is, so. Derp!
posted on Dec, 20 2015 @ 04:01 AM
originally posted by: Observationalist
originally posted by: Zaphod58
Anyone else see the wolf face made with the clouds and shadows. One of the eyes is just under the bulged out part of Africa. I don't have any fancy editors to outline it maybe someone else could if you guys see it.
Maybe a fox
I see Batman with his head tilted back after the Joker dropped a rock on his head. Once you see it
posted on Dec, 20 2015 @ 04:20 AM
Here's a weather satellite shot of Europe on October 12th:
And here's the same ares in the 300Mb TIF view of Earth from the LRO:
edit on 20-12-2015 by onebigmonkey because: typo
posted on Dec, 20 2015 @ 05:04 AM
I think NASA is trapped by the photographic lies they told during the Apollo program. I don't know how much of their current program is a fraud, but I think a lot of it is. The United States is in a bad way. Keeping up a front, continuity of government, continuity of America's public profile, now means continuity of fraud, whether it is on the Kennedy assassination, 9/11, the war on drugs, the war on terrorism or the space program.
The whole country is psychotic. The motto of the US should be changed from "In God we trust" to "In cognitive dissonance we trust."
posted on Dec, 20 2015 @ 05:15 AM
originally posted by: ipsedixit
I think... I don't know...I think....
Is all you needed.
posted on Dec, 20 2015 @ 07:43 AM
I'm different.
I think.
edit on 20-12-2015 by ipsedixit because: (no reason given)
posted on Dec, 20 2015 @ 08:10 AM
I think NASA is trapped by the photographic lies they told during the Apollo program.
Which ones would those be?
I don't know how much of their current program is a fraud, but I think a lot of it is.
Why would it be a fraud?
now means continuity of fraud, whether it is on the Kennedy assassination, 9/11, the war on drugs, the war on terrorism or the space program.
Except those happened, or are happening so where is the fraud?
The whole country is psychotic.
I guess I live in another country because where I am there are normal easy going people that are far from psychotic.
The motto of the US should be changed from "In God we trust" to "In cognitive dissonance we trust."
Maybe for you, but I will stick with In God We Trust.
posted on Dec, 20 2015 @ 08:10 AM
Nope, you're just the same as all the other knee-jerk reactionaries who trot out the usual anti-NASA nonsense without any kind of demonstration of support for their opinions other than their personal politics.
While some of the contributors to this thread have provided information as to where the OP was taken, how it was taken, that the lighting conditions are correct and what we are seeing is factually accurate your contribution is "I don't like NASA".
You are welcome to your difference.
posted on Dec, 20 2015 @ 08:29 AM
Care to elaborate why you believe it not to be true?
I can. It may have to do with the hoards of people who claim they worked for NASA. There seems to be many individuals who say NASA is full of sh*t. You can find Astronauts who will or did back up these claims. What the heck does that tell you, that they're disgruntled astronauts?...
When an agency controlling what the "truth" is, tells the world something, there's nobody else to ask. Who can you call? The EU space agency, China? They're all in cahoots together. They decide what we are to perceive as the truth, and, unless you have your own space agency, you'll never know - only what they want you to know. People should turn their faces away from the cell phone and t.v. for crying out loud. Look for answers instead of simply regurgitating what you've been told, if that's what you're doing.
Wake up people!!! Wake the eff up !!!
There's something unjust, going on.
Love to all - we're going to need it.
edit on 10 27 2013 by donktheclown because: (no reason given)
posted on Dec, 20 2015 @ 10:36 AM
originally posted by: donktheclown
Care to elaborate why you believe it not to be true?
I can. It may have to do with the hoards of people who claim they worked for NASA. There seems to be many individuals who say NASA is full of sh*t. You can find Astronauts who will or did back up these claims. What the heck does that tell you, that they're disgruntled astronauts?...
When an agency controlling what the "truth" is, tells the world something, there's nobody else to ask. Who can you call? The EU space agency, China? They're all in cahoots together. They decide what we are to perceive as the truth, and, unless you have your own space agency, you'll never know - only what they want you to know. People should turn their faces away from the cell phone and t.v. for crying out loud. Look for answers instead of simply regurgitating what you've been told, if that's what you're doing.
Wake up people!!! Wake the eff up !!!
There's something unjust, going on.
Love to all - we're going to need it.
So basically you're saying that everyone is telling you what you should believe, and because they're saying that they are wrong.
Then in the next sentence you tell everyone what they should believe.
How about you tell us why the facts as presented are wrong instead of telling us why you don't like the people telling you the facts?
posted on Dec, 20 2015 @ 11:02 AM
originally posted by: OrionHunterX
Is the Moon really that black and grey? The image seems fudged.
Actually, they had to brighten the Moon's surface up a bit for this image. The Moon (on average) has the reflectivity of asphalt or graphite.
new topics
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46 | 2,615 | 10,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-44 | longest | en | 0.94678 |
http://www.ehow.com/how_8346070_reverse-foil-factoring-polynomial.html | 1,481,107,128,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542060.60/warc/CC-MAIN-20161202170902-00128-ip-10-31-129-80.ec2.internal.warc.gz | 456,481,762 | 15,605 | # How to Reverse a FOIL When Factoring a Polynomial
Save
The FOIL method is a mathematical key for reducing two binomial expressions to a trinomial mathematical expression by multiplying two sets of binomials by eachother. FOIL stands for first, outside, inside, and last, as this is the order of the multiplication that you would do to the variables in a two-binomial equation to reduce it. The reverse-FOIL method, then, is a process of doing FOIL backward. Reverse-FOIL is called for when you are trying to convert a trinomial back into a double binomial expression.
• Set up two sets of parentheses under your trinomial equation like so: ( +/- )( +/- ). This is the form to which you will be converting the equation.
• Factor the polynomial. For example, let's say our trinomial is 3x^2 + 10x + 8. To get 3x to the second power, we must multiply 3x by x. So, knowing this, we can fill in the first two variables in our parentheses: (3x +/- )( x +/- ).
• Think of two factors of 8. Your possibilities are 1 and 8, 2 and 4 and their negatives.
• Plug these number groups into the equation. For example, try the first possible factors of 1 and 8: (3x + 1)(x + 8). Now we check that with the FOIL method. The result we get is 3x^2 + 24x + x + 8, which is reduced to 3x^2 + 25x + 8. But this is not our original trinomial, so 1 and 8 are not the correct factors.
• Keep plugging in the factors of 8 and checking it with FOIL. If you do, you will find that when you try 4 and 2: (3x + 4)(x + 2), this factors out to 3x^2 +10x +8. This is our original equation. We did it!
## References
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Is DIY in your DNA? Become part of our maker community. | 495 | 1,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-50 | latest | en | 0.900515 |
https://millionaire-lane.com/Creation/Creation-01.htm | 1,686,140,826,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653764.55/warc/CC-MAIN-20230607111017-20230607141017-00172.warc.gz | 431,354,516 | 4,734 | Here are a few of my thoughts!
After a few simple observations (See Below)
I Believe in Creation!!
What do YOU Think?
Since i'm a lover of Mathematics and
Chess and Computer Programming,
I'm always looking at things in terms of Numbers.
And, since Math requires Logic,
I trend toward Logical Thinking!
Although, Logic suggests that Unicorns,
Mermaids, Big-Foot, and Aliens, don't exist,
I believe that they all do exist!! (LOL!!)
(I know there's a few flaws in my Logic too!!)
Since most people are familiar with card games,
As an example, we will use here a deck of playing cards!
As you know the standard deck has 52 cards!
The question is:
How many different ways can these 52 cards be arranged?
Most people may not have ever considered this question!
But the fact is - There are so many ways to arrange these cards,
that your mind can hardly imagine how big that number is!
The answer is 52! - which means
52 x 51 x 50 x 49 x ... x 3 x 2 x 1
Which comes to approximately
8.06581751709439 x 10^67
OR if you don't know what that means
it means 8 with 67 zeros after the 8.
And that is a really! Really! REALLY!! Big Number!!
Just to give you an idea of the size of that number
Let's use a "Speed-of-Light" example.
Most people know the Speed-of-Light is equal
to 186,000 Miles per Second.
Since 1 mile = 5280 feet we multiply
by 5280 to get feet/second => 982,080,000 Feet/Second
Now let's travel a trillion times faster than light
=> Warp Speed 1 Trillion in Star Trek Speeds
=> 982,080,000,000,000,000 Feet/Second
Now let's assume each arrangement of the cards is
set up in 1 foot intervals.
The number of seconds in a 1 years = 24*3600*365
=> 31,536,000 Seconds
Divide these two numbers into the number of
arrangements of the cards.
=> In order to see all these arrangements
traveling at 1 trillion times the speed of light
(8.0658175)*(10^67) Divided by (3.097087488)*(10^25)
=> (2.604323426849219)*(10^42) Years
=> 2.604323426849219 Times a
Million Trillion Trillion Trillion Trillion Years
(See Below for details of this calculation)
Now, Did you know that your DNA has more than 52 parts?
DNA = Gene Designs or Gene Arrangements
"The Human Genome Project has estimated that humans
have between 20,000 and 25,000 genes." (Biology-Book)
I know what you are thinking:
"Ok! Ok! We already know that all living things are complicated!!"
Ok, let's consider a real life experiment.
Let's assume that a Color TV set has 52 parts.
(We all know it's more complicated than that!
But just consider 52 parts!)
Try This - Place these 52 parts in a big water-proof ball
and float the ball in the ocean for 1 billion years!
You also have to assume that the parts never get old and become useless!
Also, Assume inside the ball the parts are
weightless and can float around and swirl into
thousands of different configurations!
Do you think it will ever become a working Color TV?
Do you really think this could happen?
I don't think so!
I know I couldn't build a working Color TV
even if I had all the parts and instructions!
And you think it could build itself by accident?
I don't think so!!
Some of the simplest Life Forms on this planet have more
than 52 parts and I have trouble
believing they came alive by accident or evolution!
Remember how many ways these 52 parts can be arranged!
Every part has to be connected in a Pre-Planned
design for the TV to work!!
And Every Life form on this planet
is the same way!!
Every Living Cell has to be connected in a
Pre-Planned design for
the life form to function!!
And the number of different kinds of Living Cells
found in life forms is infinite!!
Just the number of different kinds of
Living Cells found in the human body is incredible!!
Think about a few Cells - finger nails - skin - eyes - ear parts -
kidneys - liver - pancreas - spleen -
heart - brain cells - bones - bone marrow -
blood (which includes dozens of different types of cells) -
teeth - tendons - muscles - nerve cells - hair -
the list goes on forever!!!
The human body produces
about 2 trillion new cells every day
and every cell is different
and has a different function!!
That's incredible!!
Even the smallest life forms like Honey-Bees and
Spiders are smart!!
The intelligence built into these life forms is amazing!
Every Living Thing recognizes, when they see something,
Whether it's something to eat or something to fight
Or something to breed with!
And consider this - Is there any record at all showing where
one animal "Evolved" into another animal?
Rabbits have been breeding for thousands of years
and they are still Rabbits!
Rabbits never evolved into
cats or mice or guinea-pigs
or any other similar species!
My thinking is - The reason this never happens goes
back to the 52 parts!
There are so many arrangements of just 52 parts -
Imagine how many arrangements are involved
when you have 20,000 parts!!
These Species can reproduce trillions and trillions of times
and never evolve into another species!
It's impossible to have two Rabbits accidentally
have an off-spring that is a Guinea-Pig!!
The Guinea-Pig genes are trillions and trillions of
times different from the Rabbit genes!
The difference is so great that it could never happen!
Another problem with evolution is:
How do you explain all these different insects,
birds, fish, mammals etc
are all here at the same time?
If they evolved from one another,
they would not all be here at the same time!
"That's illogical !" As Mr. Spock would say!
Are you trying to tell me that 30,000 different
species of birds evolved from one bird??
Come on!!
I don't think so!!
If you believe that, considering all the DNA arrangements,
You might want to check your logic!
Or just grab another beer!
It's going to take a lot of beer
for me to believe that!!
My Conclusion is:
Only God can create Life !!
We all exist because God Created the Universe -
including the Earth and Every Living thing on the Earth!
What do YOU think?
*************************
See if my numbers are correct?
How many ways can a deck of 52 cards be arranged?
Which means 52*51*50*49* ... *4*3*2*1
of 8.0658175e+67
Which means 8.0658175 times 10 raised to the 67 power.
If each arrangement is laid out on 1 foot,
You would have to travel 8.06*(10^67) Feet to see them all.
Now the speed of light = 186,000 miles/second
=> (186,000)*(5280) = 982,080,000 Feet/Second
And a Billion times
The speed of light = 982,080,000,000,000,000 Feet/Second
=> (9.82080)*(10^17) Feet/Second
The number of seconds in a year = 365*24*3600 = 31,536,000
=> 3.1536000*(10^7) Seconds/Year
Ok this means that (3.1536000*(10^7))*(9.82080*(10^17))
Equals the distance light travels in 1 billion years in feet.
=> 30.97087488*(10^24) Feet per billion years
=> 3.097087488*(10^25) Feet per billion years
At this speed, how many years will it take to see
every arrangement of the 52 cards laid out on each foot?
Time = Distance Divided by Speed
(8.0658175)*(10^67) Divided by (3.097087488)*(10^25)
=> (2.604323426849219)*(10^42) Years
=> 2.604323426849219 Times a Million Trillion Trillion Trillion Years
Because 42 zeros = 42 - 6 = the Million => 36 left
=> And 36/12 = 3 => Trillion Trillion Trillion
***************
********************************************** *** *** ********************************************** | 1,873 | 7,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.932076 |
http://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=4817 | 1,490,350,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187792.74/warc/CC-MAIN-20170322212947-00406-ip-10-233-31-227.ec2.internal.warc.gz | 262,704,580 | 9,634 | # Search by Topic
#### Resources tagged with Interactivities similar to Target Archery:
Filter by: Content type:
Stage:
Challenge level:
### There are 223 results
Broad Topics > Information and Communications Technology > Interactivities
### Target Archery
##### Stage: 2 Challenge Level:
A simulation of target archery practice
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Building Stars
##### Stage: 2 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### Ratio Pairs 2
##### Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### KS2 Teacher Playing with Dice Collection
##### Stage: 2 Challenge Level:
NRICH December 2006 advent calendar - a new tangram for each day in the run-up to Christmas.
### Noughts and Crosses
##### Stage: 2 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### Train
##### Stage: 2 Challenge Level:
A train building game for 2 players.
### Dotty Circle
##### Stage: 2 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
### Calculator Bingo
##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### 100 Percent
##### Stage: 2 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Coordinate Cunning
##### Stage: 2 Challenge Level:
A game for 2 people that can be played on line or with pens and paper. Combine your knowledege of coordinates with your skills of strategic thinking.
##### Stage: 1 and 2 Challenge Level:
Our 2008 Advent Calendar has a 'Making Maths' activity for every day in the run-up to Christmas.
### Square Tangram
##### Stage: 2 Challenge Level:
This was a problem for our birthday website. Can you use four of these pieces to form a square? How about making a square with all five pieces?
### Chocolate Bars
##### Stage: 2 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
### Spot Thirteen
##### Stage: 2 Challenge Level:
Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score?
### Odds or Sixes?
##### Stage: 2 Challenge Level:
Use the interactivity or play this dice game yourself. How could you make it fair?
### A Maze of Directions
##### Stage: 2 Challenge Level:
Use the blue spot to help you move the yellow spot from one star to the other. How are the trails of the blue and yellow spots related?
### Memory Game Template
##### Stage: 2 Challenge Level:
Using angular.js to bind inputs to outputs
### Sorting Symmetries
##### Stage: 2 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
##### Stage: 2 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Train for Two
##### Stage: 2 Challenge Level:
Train game for an adult and child. Who will be the first to make the train?
### Overlapping Circles
##### Stage: 2 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### Board Block Challenge
##### Stage: 2 Challenge Level:
Choose the size of your pegboard and the shapes you can make. Can you work out the strategies needed to block your opponent?
### Round Peg Board
##### Stage: 1 and 2 Challenge Level:
A generic circular pegboard resource.
### Fractions and Coins Game
##### Stage: 2 Challenge Level:
Work out the fractions to match the cards with the same amount of money.
### Board Block for Two
##### Stage: 1 and 2 Challenge Level:
Board Block game for two. Can you stop your partner from being able to make a shape on the board?
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### Coordinate Tan
##### Stage: 2 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Rod Ratios
##### Stage: 2 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
### Rabbit Run
##### Stage: 2 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### Turning Cogs
##### Stage: 2 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### World of Tan 3 - Mai Ling
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Mai Ling?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### You Tell the Story
##### Stage: 2 Challenge Level:
Can you create a story that would describe the movement of the man shown on these graphs? Use the interactivity to try out our ideas.
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### L-ateral Thinking
##### Stage: 1 and 2 Challenge Level:
Try this interactive strategy game for 2
### Take the Right Angle
##### Stage: 2 Challenge Level:
How many times in twelve hours do the hands of a clock form a right angle? Use the interactivity to check your answers.
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Coin Cogs
##### Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### The Path of the Dice
##### Stage: 2 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
### World of Tan 1 - Granma T
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T? | 1,932 | 8,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-13 | longest | en | 0.846083 |
http://cs.brown.edu/courses/cs173/2007/Assignments/rudimentary-interp.html | 1,508,278,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00266.warc.gz | 80,129,064 | 2,943 | You must do this assignment solo. We know that some of you are not yet comfortable with Scheme; for that reason, we will weight this assignment very, very low in the overall score. Doing poorly on it will not at all damage your course grade. But, you should exploit this opportunity to become familiar with the style of programming you will do in the rest of the semester.
### Rudimentary Interpreter
Write a parser and interpreter for the WAE language ([1]) we discussed in class. The textbook can be of great assistance in this part of the assignment; they provide the beginnings of a parser, an abstract syntax datatype and an interpreter.
Once you’ve written and tested the parser and interpreter for WAE, extend the language with these features:
Binary arithmetic operators
In place of having separate rules for `+` and `-`, define a single syntactic rule for all binary arithmetic operators. Parse these into a `binop` datatype variant. Define a table that maps operator names (symbols) to actual functions (Scheme procedures) that perform the corresponding operation. Having a single rule like this, accompanied by a table, makes your language easier to extend: once you have modified your parser and interpreter once to support binary operators, you won't need to touch either one to add any number of new ones. To demonstrate this, define multiplication and division (using `*` and `/` to represent them in the language's concrete syntax).
Multi-armed `with`
Each identifier bound by the `with` expression is bound only in its body. There will be zero or more identifiers bound by each `with` expression. If there are multiple bindings of the same identifier in a single `with` expression’s bindings list, your interpreter should halt with an error message. An example:
```{with {{x 2}
{y 3}}
{with {{z {+ x y}}}
{+ x z}}}
```
will evaluate to `7`, while
```{with {{x 2}
{x 3}}
{+ x 2}}
```
will halt with an error message.
The syntax of the WAE language with these additional features can be captured using EBNF notation ([2]):
```<WAE> ::= <num>
| {+ <WAE> <WAE>}
| {- <WAE> <WAE>}
| {* <WAE> <WAE>}
| {/ <WAE> <WAE>}
| {with {{<id> <WAE>}*} <WAE>}
| <id>
where an <id> is not +, -, *, /, or with.
```
You should turn in a single Scheme program containing all of the code needed to run your parser and interpreter. Implement the function `parse`, which consumes an expression in the language's concrete syntax and returns the abstract syntax representation of that expression. Also implement the function `interp`, which consumes an abstract syntax expression (as returned by the `parse` function) and returns a Scheme number.
You must include a contract for every function that you write and include test cases that amply exercise all of the code you’ve written. We will not give full credit to untested functionality, even if it is correct! Refer to the syllabus for style requirements.
[1] Remember, the WAE language has numbers, two arithmetic operators (`+`, `-`), identifiers and `with` expressions. Of course, to handle identifiers and `with` expressions you'll have to implement substitution. (You do not have to implement caching substitution, but you are free to do so.) Finally, both the concrete syntax and abstract syntax are specified by the WAE language’s BNF (which can be found in the textbook).
[2] The textbook introduced you to BNF. An extension of this notation, called EBNF (Extended Backus-Naur Form), provides three additional operators:
• `?` means that one or more symbols to its left can appear zero or one times.
• `*` means that one or more symbols to its left can be repeated zero or more times.
• `+` means that one or more symbols to its left can appear one or more times.
#### Support Code
```(define-type Binding
[binding (name symbol?) (named-expr WAE?)])
(define-type WAE
[num (n number?)]
[binop (op procedure?) (lhs WAE?) (rhs WAE?)]
[with (lob (listof Binding?)) (body WAE?)]
[id (name symbol?)])
;; parse : s-exp -> WAE
;; Consumes an s-expression and generates the corresponding WAE
(define (parse sexp)
(...))
;; calc : WAE -> number
;; Consumes a WAE representation of an expression and computes
;; the corresponding numerical result
(define (calc expr)
(...))
```
#### Handin
From the directory containing the files for the assignment you wish to hand in, execute
`/course/cs173/bin/cs173handin rinterp` | 1,029 | 4,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-43 | latest | en | 0.877996 |
http://www.singular.uni-kl.de/Manual/latest/sing_2107.htm | 1,508,416,834,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823284.50/warc/CC-MAIN-20171019122155-20171019142155-00849.warc.gz | 535,451,928 | 4,255 | # Singular
#### D.13.2.55 vertexEdgeGraph
Procedure from library `polymake.lib` (see polymake_lib).
Usage:
vertexEdgeGraph(p); p polytope
Return:
list, the first entry is a bigintmat containing all vertices as row vectors, and therefore assigning all vertices an integer. the second entry is a list of intvecs representing the edge graph of the vertices of p, each intvec represents an edge of p connecting vertex i with vertex j.
Example:
```LIB "polymake.lib"; ==> Welcome to polymake version ==> Copyright (c) 1997-2015 ==> Ewgenij Gawrilow, Michael Joswig (TU Darmstadt) ==> http://www.polymake.org intmat M[6][4] = 1,1,0,0, 1,0,1,0, 1,0,-1,0, 1,0,0,1, 1,0,0,-1, 1,-1,0,0; polytope p = polytopeViaPoints(M); vertexEdgeGraph(p); ==> polymake: used package ppl ==> The Parma Polyhedra Library (PPL): A C++ library for convex polyhedra ==> and other numerical abstractions. ==> http://www.cs.unipr.it/ppl/ ==> ==> [1]: ==> 1, 0, 0,-1, ==> 1, 0,-1, 0, ==> 1,-1, 0, 0, ==> 1, 0, 0, 1, ==> 1, 0, 1, 0, ==> 1, 1, 0, 0 ==> [2]: ==> [1]: ==> 0,1 ==> [2]: ==> 0,2 ==> [3]: ==> 0,4 ==> [4]: ==> 0,5 ==> [5]: ==> 1,2 ==> [6]: ==> 1,3 ==> [7]: ==> 1,5 ==> [8]: ==> 2,3 ==> [9]: ==> 2,4 ==> [10]: ==> 3,4 ==> [11]: ==> 3,5 ==> [12]: ==> 4,5 ``` | 475 | 1,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-43 | longest | en | 0.60245 |
https://www.geeksforgeeks.org/convert-a-generic-treen-array-tree-to-binary-tree/ | 1,610,811,248,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506697.14/warc/CC-MAIN-20210116135004-20210116165004-00768.warc.gz | 794,849,770 | 21,157 | Related Articles
Convert a Generic Tree(N-array Tree) to Binary Tree
• Last Updated : 29 Oct, 2020
Prerequisite: Generic Trees(N-array Trees)
In this article, we will discuss the conversion of the Generic Tree to a Binary Tree. Following are the rules to convert a Generic(N-array Tree) to Binary Tree:
• The root of the Binary Tree is the Root of the Generic Tree.
• The left child of a node in the Generic Tree is the Left child of that node in the Binary Tree.
• The right sibling of any node in the Generic Tree is the Right child of that node in the Binary Tree.
Examples:
Convert the following Generic Tree to Binary Tree:
Below is the Binary Tree of the above Generic Tree:
Note: If the parent node has only the right child in the general tree then it becomes the rightmost child node of the last node following the parent node in the binary tree.
In the above example, if node B has the right child node L then in binary tree representation L would be the right child of node D.
Below are the steps for the conversion of Generic Tree to Binary Tree:
1. As per the rules mentioned above, the root node of general tree A is the root node of the binary tree.
2. Now the leftmost child node of the root node in the general tree is B and it is the leftmost child node of the binary tree.
3. Now as B has E as its leftmost child node, so it is its leftmost child node in the binary tree whereas it has C as its rightmost sibling node so it is its right child node in the binary tree.
4. Now C has F as its leftmost child node and D as its rightmost sibling node, so they are its left and right child node in the binary tree respectively.
5. Now D has I as its leftmost child node which is its left child node in the binary tree but doesn’t have any rightmost sibling node, so doesn’t have any right child in the binary tree.
6. Now for I, J is its rightmost sibling node and so it is its right child node in the binary tree.
7. Similarly, for J, K is its leftmost child node and thus it is its left child node in the binary tree.
8. Now for C, F is its leftmost child node, which has G as its rightmost sibling node, which has H as its just right sibling node and thus they form their left, right, and right child node respectively.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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Page : | 566 | 2,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-04 | latest | en | 0.943062 |
https://stackoverflow.com/questions/18419298/wpf-winrt-points-to-draw-a-heart | 1,680,134,787,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00151.warc.gz | 608,396,189 | 38,163 | # Wpf/WinRT points to draw a heart
I've been drawing various shapes on a canvas and I have managed to figure out the math to do things like hexagons, octagons, and even stars. I can't seem to figure out how to draw a heart. Does anyone have an example of the points needed to draw a heart shape?
I'm not sure if there's a really easy way to plot equations straight on to the `Canvas` (although you could certainly create the `Path` from an equation programmatically. .depending on what you want to do with the shape, and where you want to use it, you could just define a `Path` using the Path Markup Syntax and use some of the built in arcs and curves.
e.g. (quickly threw this one together):
``````<Canvas>
<Path Stroke="Red" StrokeThickness="3"
Data="M 241,200
A 20,20 0 0 0 200,240
C 210,250 240,270 240,270
C 240,270 260,260 280,240
A 20,20 0 0 0 239,200
" />
</Canvas>
``````
Will create something that looks like this:
Which uses 2 arcs and 2 cubic Bezier curves.
You can read more about the syntax link, but by way of a small explanation (the following draws half the heart, from the dip at the top of the heart to the point at the bottom, counter-clockwise):
``````M 241,200 // Move to (241, 200)
A 20,20 0 0 0 200,240 // Draw an arc from current position to (200,240), with a size of 20x20 pixels
C 210,250 240,270 240,270 // Draw a cubic Bezier to point (240,270) with control points at (210,250), (240,270).
``````
The next curve, then arc are drawn, arriving back at the top, completing the shape.
You may have to play around a bit to get the result you're after.
There are a number of mathematical equations for heart shapes - some polar, some parametric.
One particularly convincing one is:
``````x = 16sin^3(t)
y = 13cos(t) - 5cos(2t) - 2cos(3t) - cos(4t)
``````
There's a list of some good ones on the Woflram website.
You can plot a heart on a graph with the following equation.
((x^2 + y^2 - 1)^3) - (x^2 * y^3) = 0
Source : Wikipedia
The way I see it, you can draw two arcs as parts of two upper intersecting circles that have the center on the same X axis; and then two lower lines to unite them. I tried to draw an example, I hope the picture is suggestive enough: | 643 | 2,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-14 | latest | en | 0.903346 |
https://www.indiabix.com/engineering-mechanics/kinematics-of-a-particle/discussion-164 | 1,638,956,117,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00475.warc.gz | 858,753,312 | 5,429 | # Engineering Mechanics - Kinematics of Particle (KOP) - Discussion
### Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.32)
32.
For a short time the missile moves along the parabolic path y = (18 - 2x2) km. If motion along the ground is measured as x = (4t - 3) km, where t is in seconds, determine the magnitudes of the missile's velocity and acceleration when t = 1 s.
[A]. v = 5.66 km/s, a = 4.0 km/s2 [B]. v = 16.49 km/s, a = 64.0 km/s2 [C]. v = 16.00 km/s, a = 22.6 km/s2 [D]. v = 4.00 km/s, a = 16.03 km/s2
Explanation:
No answer description available for this question.
Hermes said: (Aug 16, 2021) y = 18 - 2x^2, = 18 -2(4t-3)^2, = -32t^2 + 48t. Vy = dy/dt, = -64t + 48. ay = dv/dt, = -64. x = 4t -3. Vx = dx/dt. = 4, ax = dx/dt. = 0. when t = 1s. Vy = -16 km/s^2, Vx = 4 km/s^2. v = √(Vy^2 + Vx^2), = 16.49 km/s^2. ay = -64 km/s^2. ax = 0 km/s^2, a = √(ax^2 + ay^2), = 64 km/s^2. | 387 | 917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-49 | latest | en | 0.733643 |
https://www.includehelp.com/cpp-programs/given-binary-search-tree-is-balanced.aspx | 1,719,263,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00423.warc.gz | 737,014,544 | 35,081 | # C++ program to check whether a given Binary Search Tree is balanced or not?
In this tutorial, we will learn how to implement a C++ program that will check whether a given binary search tree is a balanced tree or not? By Bhanu Pratap Raghav Last updated : August 10, 2023
## Problem description
Solution to check the given Binary Search tree is balanced or not.
## Problem statement
Write a C++ program that accepts input from user to form a binary search tree and check whether the formed tree is balanced or not.
## Example 1
Input: 2 1 3 4 5 -1
Output: Given BST is Balanced : False
## Example 2
Input: 2 1 3 4 -1
Output: Given BST is Balanced : True
## What do you mean by height of a tree?
A height of a tree is the largest number of edges in a path from node to a leaf node. If a tree has only one node: i.e. root node itself then the height of tree is 0.
Example:
Height of tree: 2 ( maximum 2 edges from root to leaf)
## Check when tree is balanced
A non-empty binary search tree is said to be balanced if:
1. Its left subtree is balanced.
2. Its Right subtree is balanced.
3. The difference between heights of left and right subtree is less than or equal to 1.
Example:
At root height of left subtree is 1 , whereas height of right subtree is 3
Difference = 3-1=2 (which is greater than 1) i.e. Tree is not balanced.
## Algorithm
1. Set root=root of given binary search tree.
2. Set leftHt= height of left subtree.
3. Set rightHt= height of right subtree.
4. if abs(leftHt – rightHt ) > 1
return false;
else isHeightBalanced(root->left) &&isHeightBalanced(root->right)
## C++ program to check whether a given Binary Search Tree is balanced or not?
```#include <iostream>
#include <cmath>
using namespace std;
class TreeNode {
public:
int data;
TreeNode* left;
TreeNode* right;
TreeNode(int d)
{
data = d;
left = NULL;
right = NULL;
}
};
TreeNode* insertInBST(TreeNode* root, int x)
{
if (root == NULL) {
TreeNode* tmp = new TreeNode(x);
return tmp;
}
if (x < root->data) {
root->left = insertInBST(root->left, x);
return root;
}
else {
root->right = insertInBST(root->right, x);
return root;
}
}
TreeNode* createBST()
{
TreeNode* root = NULL;
int x;
cin >> x;
//Take input until user inputs -1
while (true) {
if (x == -1)
break;
root = insertInBST(root, x);
cin >> x;
}
return root;
}
//Calculate height of tree with given root
int height(TreeNode* root)
{
if (root == NULL)
return 0;
int leftHt = height(root->left);
int rightHt = height(root->right);
int curHt = max(leftHt, rightHt) + 1;
return curHt;
}
//Check whether tree is balanced or not
bool isHeightBalanced(TreeNode* root)
{
if (!root) {
return true;
}
int leftHt = height(root->left);
int rightHt = height(root->right);
if (abs(leftHt - rightHt) > 1)
return false;
return isHeightBalanced(root->left) && isHeightBalanced(root->right);
}
int main()
{
//Input BST
cout << "Input Tree elements : ";
TreeNode* root = createBST();
cout << "Given BST is Balanced : ";
if (isHeightBalanced(root)) {
cout << "True";
}
else {
cout << "False";
}
return 0;
}
```
## Output
```First run:
Input Tree elements(stop taking input when -1 encountered) : 2 1 3 4 5 -1
Given BST is Balanced : False
Second run:
Input Tree elements(stop taking input when -1 encountered) : 2 1 3 4 -1
Given BST is Balanced : True
```
Student's Section
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Intern
Joined: 08 Sep 2006
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08 Sep 2006, 10:04
This exam has taken me for a toll i get so stressed out that I forget everything.I have taken the test and failed miserably but I just don't want to give up. Can someone tell me how to handle the time stress .Verbal section just drives me crazy.
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08 Sep 2006, 12:23
There's very little gain with no pain....as far as the GMAT goes. There are exceptions to this of course. Some people are born test-takers. For the remaining "mere mortals", work is needed in each aspect of the test taking strategy....including timing, working through long hours at one sitting....etc. Though these are peripheral to the core material study, they are just as important in order to perform to your potential.
If you could highlight your prep strategy....people (I am sure) will chip in with do's and don'ts.....
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08 Sep 2006, 12:47
If stress is causing major problems for you, then good preperation is probably the answer. If you go in knowing that you are prepared for each of the question types, you are less likely to be stressed by what you will face.
Dealing with stress is a major issue for an exam like this. If you are stressed your mind will wander and you cannot concentrate on even those problems that you are very capable of answering. I can also cause you to be tired and/or sleep poorly.
I would suggest taking some time and getting an honest assessment of your own skills. Then, set out a plan to tackle each of the areas in which you need to be stronger. If you can point out your preperation and test experience, others here can probably help with suggestions.
CEO
Joined: 15 Aug 2003
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08 Sep 2006, 14:37
I highly recommend participating in the verbal forum. Explain your answers to others and you will see an improvement. Dont worry about getting answers wrong.
You must also obsessively keep track of your errors and measure your progress.
08 Sep 2006, 14:37
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Display posts from previous: Sort by | 891 | 3,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-22 | latest | en | 0.938278 |
https://www.sulprobil.com/sbgennormdist_en/ | 1,726,830,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00525.warc.gz | 925,462,045 | 3,215 | ## Abstract
To generate a standard normal distribution you can use =NORM.S.INV(RAND()). If you need a normal distribution with a mean of 7 and a standard deviation of 10 then use =NORM.INV(RAND(),7,10).
But: Your normal distribution will never show a mean of exactly 7 (and a standard deviation of exactly 10) unless the number of draws approaches infinity. If you want to ensure a mean of 7 and a standard deviation of 10 then shift an originally generated series of random numbers to this mean and zoom it to the required standard deviation. You can achieve this with at least three different approaches:
## Appendix sbGenNormDist Code
``````Option Explicit
Function sbGenNormDist(lCount As Long, _
dMean As Double, _
dStDev As Double) As Variant
'Generates lCount normally distributed random values
'with mean dMean and standard deviation dStDev.
'Source (EN): http://www.sulprobil.com/sbgennormdist_en/
'Source (DE): http://www.bplumhoff.de/sbgennormdist_de/
'(C) (P) by Bernd Plumhoff 30-May-2024 V0.3
Dim i As Long
Dim dSampleMean As Double, dSampleStDev As Double
If lCount < 2 Then
sbGenNormDist = CVErr(xlErrValue)
Exit Function
End If
ReDim vR(1 To lCount) As Variant
With Application
For i = 1 To lCount
vR(i) = .Norm_Inv(Rnd(), dMean, dStDev)
Next i
dSampleMean = .Average(vR)
dSampleStDev = .StDev_S(vR)
For i = 1 To lCount
vR(i) = dMean + (vR(i) - dSampleMean) * dStDev / dSampleStDev
Next i
sbGenNormDist = .Transpose(vR)
End With
End Function
`````` | 431 | 1,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.674801 |
http://plepspuzzles.blogspot.com/2007/07/four-5s.html | 1,532,222,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592875.98/warc/CC-MAIN-20180722002753-20180722022753-00374.warc.gz | 285,708,834 | 6,210 | Plep's Puzzles
Name:
Monday, July 30, 2007
Four 5's
Using arithmetic combinations of four 5's express the integers from 1 to 12.
You must use all four 5's for each integer.
You may combine the 5's using the arithmetic operations +, -, * (multiplication) and / (division).
You are also allowed to use sqrt (square root) (e.g. sqrt(5) ).
You are allowed to use concatenation (e.g. 55 is valid as part of an answer).
You are allowed to use decimals (e.g. 5.5 is valid as part of an answer; so is .5).
You are allowed to use factorials. For example, 5! (or factorial(5)) = 5*4*3*2*1 = 120.
(The factorial of an integer x! is the product of all the positive integers less than or equal to x).
You do not need to use anything other than those listed above - so no powers of 3, trig operations, logarithms, cube/quartic/etc. roots, etc.
Joseph said...
1 = (5*5/5)/5
2 = (5/5) + (5/5)
3 = (sqrt(5)*sqrt(5)*.5)+.5
4 = 5 - ((sqrt(5)*sqrt(5))/5)
5 = ((sqrt(5)*sqrt(5))/5)*5
6 = 55/5 - 5
7 = (.5(5)) +5 - .5
8 = (.5(5)) +5 + .5
9 = 5 + 5 - (5/5)
10 = (55-5)/5
11 = 5 + 5 + (5/5)
12 = (55+5)/5
July 31, 2007 at 8:58 AM | 407 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-30 | latest | en | 0.881575 |
https://sites.brown.edu/crunch-group/current-research/fractional-calculus/ | 1,716,341,328,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00856.warc.gz | 485,935,993 | 12,913 | ### Surface Quasi-Geostrophic (SQG) Modeling
The surface quasi-geostrophic (SQG) equation is useful in modeling atmospheric phenomena such as the frontogenesis i.e., the formation of strong fronts between masses of hot and cold air. We introduce the concept of fractional spectral vanishing viscosity (fSVV) to solve conservations laws that govern the evolution of steep fronts. We apply this method to the two-dimensional surface quasi-geostrophic (SQG) equation. The classical solutions of the inviscid SQG equation can develop finite-time singularities. By applying the fSVV method, we are able to simulate these solutions with high accuracy and long-time integration with relatively low resolution.
### Fractional Turbulent Flow
We purpose a fractional differential equation to govern the turbulent channel flow since the small-scale components can be described as an anomalous diffusion. The numerical results reveal that the fractional order for all different Reynolds-number has universal profile in wall unit scaling.
### Fractional Cahn-Hilliard
The Cahn-Hilliard equation models the phase separation of a binary alloy at a fixed temperature. The fractional version of the Cahn-Hilliard equation arises as a gradient flow of the free energy in the negative order Sobolev space Hα, resulting in an equation that conserves mass for all positive values of α and reduces energy. A Fourier-Galerkin scheme was developed to solve the fractional Cahn-Hilliard equation. In this figure, the same random initial conditions for two different volume fractions and three different fractional orders α, and the configurations at time T = 20 are plotted. (Source: Mark Ainsworth and Zhiping Mao, “Analysis and approximation of a fractional Cahn-Hilliard equation”, SIAM J. Numer. Anal. 55 (2017), No. 4, 1689-1718.)
Solutions of the fractional Poisson equation, using the integral fractional Laplacian and zero Dirichlet exterior conditions, on a two-component domain corresponding to a constant right-hand-side. In the top two plots, the fractional Poisson equation is posed with the order s = 0.25, and on the bottom, it has order s = 0.75. An adaptive finite element method was developed to solve this equation. (Source: Mark Ainsworth and Christian Glusa, “Aspects of an adaptive finite element method for the fractional Laplacian: a priori and a posteriori error estimates, efficient implementation and multigrid solver”, Computer Methods in Applied Mechanics and Engineering, 327 (2017), 4-35. DOI:10.1016/j.cma.2017.08.019)
Adaptively refined grid for the fractional Poisson equation on the two-component domain, with (left) s = 0.25 and (right) s = 0.75.
### Riesz spectral components
The different definitions of the fractional Laplacian lose their equivalence when they are restricted to bounded domains and boundary conditions are imposed. We numerically compared these definitions in different domains. Here, we plot the difference between the solutions of the fractional Poisson equation with the Riesz (or integral) definition and the spectral definition. The domain is L-shaped, zero Dirichlet boundary conditions are enforced, and the source term is sin(πx)sin(πy). For more comparisons, theoretical derivations and considerations for each definition, and a collection of state-of-the-art numerical approaches, see: Anna Lischke, Guofei Pang, Mamikon Gulian, Fangying Song, Christian Glusa, Xiaoning Zheng, Zhiping Mao, Wei Cai, Mark M. Meerschaert, Mark Ainsworth, and George Em Karniadakis, “What Is the Fractional Laplacian?”, arXiv preprint: 1801.09767 (2018). | 805 | 3,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.839869 |
http://spiff.rit.edu/classes/phys369/workshops/w10b/robins.html | 1,686,369,113,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00065.warc.gz | 37,507,724 | 3,066 | #### Benjamin Robins invents the ballistic pendulum
In the early 1700s, guns, cannons and all manner of artillery was used widely in warfare, but neither engineers nor scientists understood much of the physics of these weapons. For example, the speed of bullets or cannonballs was completely unknown.
Benjamin Robins (1707 - 1752), a British mathematician and military engineer, was the first to devise a practical method for measuring the velocities of projectiles. He described his work in New Principles of Gunnery, first published in 1742. At right is a picture of one of his inventions, the ballistic pendulum. Projectiles are fired horizontally so that they will strike the plate of wood GHIK, causing it to swing back in an arc around the pivot EF. As the plate swings, it pulls a ribbon W through a slot in the mounting; the length of ribbon is equal to the length of the arc through which the plate swings. Robins could use this arc length to compute the vertical distance through which the plate had moved, and so compute the change in its gravitational potential energy.
In one experiment, Robins tested a gun which fired a bullet with the following characteristics:
• mass "one twelfth of a pound" = 38 g
• diameter "three quarters of an inch" = 1.9 cm
The bullet struck a plate of mass 56 pounds = 25 kg on a ballistic pendulum. Robins placed the pendulum at several distances from the muzzle of the weapon and made measurements which I have translated into metric equivalents.
```
Distance 25 ft 75 ft 125 ft
----------------------------------------------------------------
vertical
rise of plate 0.030 m 0.026 m 0.022 m
speed (m/s)
KE (J)
----------------------------------------------------------------
```
1. compute the speed of the bullet at all 3 distances
2. compute the kinetic energy of the bullet at all 3 distances
3. compute the work done on the bullet by air resistance during the first interval (25-75 feet) and the second interval (75-125 feet)
Robins was able to use these measurements not only to determine the properties of various weapons, but also to determine the nature of the air resistance -- a very important job for the military!
The work done by air resistance on the bullet is complicated to compute properly. The problem is that the force depends on the speed of the bullet,
but that speed is constantly decreasing, so the force of air resistance is also constantly decreasing. That means that if we try to calculate the work done by air resistance as the bullet travels over some distance, we need to integrate a constantly changing force ...
Ugh.
However, if we're willing to compromise and settle for results which are approximately correct, it's not so bad. Let's pretend that during one segment of its flight, the bullet has some average speed, and so some average force of air resistance. In that case, we can say
Let's use this approximation.
1. what is the work done by wind resistance during the interval from 25-75 feet? from 75-125 feet?
2. are these two amounts of work the same or different? Explain why
3. compute the average force exerted by air resistance during each interval (and the second interval, if you have time)
Now, if we use the usual (simplified) formula for the force of air resistance,
we can figure out the value of the coefficient K.
1. what is the approximate value for the coefficient K?
2. if this bullet was shot straight upwards, then fell back down towards the ground, what would its terminal velocity be?
3. is that dangerous? | 761 | 3,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-23 | latest | en | 0.927008 |
https://currencyconverts.com/eur/cad/27475 | 1,685,446,476,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645595.10/warc/CC-MAIN-20230530095645-20230530125645-00732.warc.gz | 221,700,664 | 11,217 | 27,475Euro (€)
=
You have converted 27475 Euro (EUR) to Canadian Dollar (CAD). Exchange rate for the conversion is 40014.59 for 30 May 2023, Tuesday. How much is 27475 Euro to Canadian Dollar? 40,014.59 Canadian Dollar's.
## How much Twenty-seven Thousand Four Hundred Seventy-five (27475) Euro in Canadian Dollar?
Today, 27,475 (twenty-seven thousand four hundred seventy-five) Euro are worth 40,014.59 Canadian Dollar. That's because the current exchange rate, to CA\$, is 1.456. So, to make Euro to Canadian Dollar conversion, you just need to multiply the amount in € by 1.456.
• Name Euro
• Money27475
• Country Europa
• Symbol
• ISO 4217 EUR
This page provides the exchange rate of twenty-seven thousand four hundred seventy-five Euro to Canadian Dollar, sale and conversion rate. We added the list of the most popular conversions for visualization and the history table with exchange rate diagram for 27475 Euro to Canadian Dollar from Tuesday, 30/05/2023 till Friday, 19/05/2023.
30 May 2023, Tuesday27,475 EUR40,014.590 CAD
29 May 2023, Monday27,475 EUR39973.624775 CAD
28 May 2023, Sunday27,475 EUR40064.017525 CAD
27 May 2023, Saturday27,475 EUR40139.848525 CAD
26 May 2023, Friday27,475 EUR40124.902125 CAD
25 May 2023, Thursday27,475 EUR40178.972925 CAD
24 May 2023, Wednesday27,475 EUR40168.39505 CAD
23 May 2023, Tuesday27,475 EUR39964.612975 CAD
22 May 2023, Monday27,475 EUR40121.330375 CAD
21 May 2023, Sunday27,475 EUR40099.680075 CAD
20 May 2023, Saturday27,475 EUR40084.10175 CAD
19 May 2023, Friday27,475 EUR40101.3011 CAD
This graph show how much is 27475 Euro in Canadian Dollar = 40014.59 CAD, according to actual pair rate equal 1 EUR = 1.4564 CAD. Yesterday this currency exchange rate plummeted on -5.3027 and was CA\$ 39,973.6248 Canadian Dollars for € 1. On this graph you can see trend of change 27,475 EUR to CAD. And average currency exchange rate for the last week was CA\$ 1.4591 CAD for €1 EUR.
``<a href="https://currencyconverts.com/eur/cad/27475" title="27,475 Euro(EUR) to Canadian Dollar(CAD) Currency Rates Today">27,475 Euro(EUR) to Canadian Dollar(CAD)</a>`` | 632 | 2,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-23 | latest | en | 0.844358 |
http://www.3dtotal.com/tutorial/1713-modeling-scripting-and-animating-gears-maya-by-chris-shaw?page=3 | 1,454,957,948,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153998.27/warc/CC-MAIN-20160205193913-00088-ip-10-236-182-209.ec2.internal.warc.gz | 237,946,798 | 17,262 | ### Keep up-to-date with Free tutorials!!
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Modeling, Scripting and Animating Gears
By Chris Shaw
| 151684 Views
| 3
Software used:
Scale the teeth 0.8 units in the Local Scale X & Y axis (Fig.09).
Fig. 09
Rotate the cog 90° on the Z axis to stand it up. Delete History and Reset Transformation (Fig.10 - 11). And the first cog is complete!
Fig. 10
Fig. 11
Note: It is important to remember this cog has 20 segments (10 teeth) and a radius of 10 units. It is important to remember that all future cogs will be multiples of these numbers.
This means a cog twice the size will have 40 segments, a radius of 20 units and 20 teeth. A cog half the size will have 10 segments, a radius of 5 units and 5 teeth. If you follow this simple rule, the gears will easily fit together and it will be easier to calculate the rotation values.
Now you understand how a cog is created, delete the first cog and we'll start again.
## Creating the Prototype Cogs
Now that you know how to create a basic cog, create four new cogs (small, medium, large and extra large). These cogs will form the basis of your gear system by matching them together in different combinations (Fig.12).
Fig. 12
Check they have the correct Radius and Subdivision values, and then rename them.
Select and extrude the alternate faces around the circumference to create the teeth (Fig.13).
Fig. 13
< previous page continued on next page >
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
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0 18556 | 584 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2016-07 | longest | en | 0.830579 |
http://www.differencebetween.net/science/difference-between-average-speed-and-instantaneous-speed/ | 1,490,617,833,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00118-ip-10-233-31-227.ec2.internal.warc.gz | 482,355,092 | 27,978 | Difference Between Similar Terms and Objects
## Difference Between Average Speed and Instantaneous Speed
Average Speed vs Instantaneous Speed
Kinematics is the science, or field of study, concerning the motion of objects. It is without the consideration of the causes of movement, and this specific branch of science extensively involves speed and velocity.
People have been always fascinated with speed. It has been pondered for centuries by analytical minds, and has become one of the major subjects of competitions – such as foot races, swimming, horse races, chariot races, car races, and other vehicular races.
Speed is the scalar equivalent of velocity, and in physics and engineering, there are many ways of expressing or describing it, which in turn can become very confusing for many people to comprehend.
In this article, we tackle two of the more confusing ways to express the rate of motion ‘“ average speed and instantaneous speed.
The most commonly known device or equipment that describes speed, is the speedometer. Speedometers are staple components of almost all moving/transportation vehicles. The information revealed by the equipment is instantaneous speed.
As you may have already noticed, the readings on the speedometer constantly changes as you use your car in your travels ‘“ for instance, your trip from home to work. In a dense highway, the traffic is slow, and you may go as slow as 15 km/h, which will be what the speedometer shows. Conversely, on a freeway you could go as fast as 100 km/h, or faster. At different moments, you have different speeds.
Having said all of that, instantaneous speed is defined as the speed at any given moment in time. It is practically what our car’s speedometer displays – your 15 km/h speed in the moment of dense traffic, and 100 km/h on freeways.
Average speed, on the other hand, is describing your rate of motion as a whole. Using the same analogy above, the average speed is the description of the rate of travel for your whole course -i.e. from your home to work. It includes the moments of heavy traffic and the frantic pace of the freeway. Assuming that the whole distance traveled is 40 km, and you made it in just one hour, your average speed will then be 40 km/h.
Thus, average speed is defined as the overall rate at which an object moves. Mathematically:
Average speed = (Overall Travel Distance) / (Time Elapsed to cover that distance)
In theory, when you calculate the average of all instantaneous speeds that occurred during the whole trip, you will get the average speed.
Summary:
1. Instantaneous speed and average speed are both scalar quantities.
2. Instantaneous speed is the speed at any given instant in time.
3. Average speed is the overall rate at which an object moves.
4. The speedometer describes instantaneous speed.
5. When you solve the average of all instantaneous speeds that occurred during the whole trip, you will get the average speed.
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See more about : , , | 723 | 3,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-13 | longest | en | 0.963424 |
https://www.sanfoundry.com/electrical-measurements-puzzles/ | 1,718,709,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00383.warc.gz | 861,357,101 | 22,360 | # Electrical Measurements Questions and Answers – Advanced Problems on Measurement of Power in Polyphase Circuit
This set of Electrical Measurements Puzzles focuses on “Advanced Problems on Measurement of Power in Polyphase Circuit”.
1. In the Two wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is?
a) 0
b) 0.5
c) 0.866
d) 1.0
Explanation: The load power factor is = 0.5. Since at this power factor one wattmeter shows zero and the other shows a positive maximum value of power.
2. Two watt meters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 10.5 kW and – 2.5 kW respectively. What is the total power?
a) 13.0 kW
b) 13.0 kW
c) 8.0 kW
d) 8.0 kW
Explanation: w1 = 10.5 kW
w2 = -2.5 kW
w = w1 + w2 = 8 kW
So, tan ∅ = $$\frac{1.732(w1-w2)}{(w1+w2)}$$ = 2.81
Or, ∅ = 70.43
So, cos ∅ = 0.334.
3. Power flowing in 3-phase, 3-wire system is measured by two wattmeter whose readings are 7000 W and -2500 W. if the voltage of the circuit is 400 V, then what will be the value of capacitance introduced in each phase to make one wattmeter reads zero? The frequency is 50 Hz.
a) 500 μF
b) 668.6 μF
c) 1000 μF
d) 748.5 μF
Explanation: P = P1 + P2 = 7000 – 2500 = 4500 W
∴ Power in each phase = 1500 W
Voltage of each phase = 231 V
So, cos∅ = 0.264
∴ Current in each phase = 24.6 A
∴ Z of each phase = 9.39 Ω
∴ R of each phase = 2.48 Ω
Reactance of each phase = 9.056 Ω
For one watt-meter to read zero, power factor should be 0.5
Now, tan∅ = $$\frac{X}{R}$$ or, X = 4.29 Ω
∴ Capacitance reactance required = 9.056 – 4.29 = 4.76 Ω
∴ Capacitance C = $$\frac{1}{2π × 50 × 4.76}$$ F = 668.6 μF.
4. A single-phase load is connected between R and Y terminals of a 220 V, symmetrical, 3-phase, 4 wire systems with phase sequence RYB. A watt-meter is connected in the system. The watt-meter will read (pf = 0.8 lagging)?
a) – 795 W
b) – 168 W
c) + 597 W
d) + 795 W
Explanation: VRY = 220∠30°
IRY = $$\frac{220∠30°}{100∠36.86°}$$ = 2.2∠-6.68° A
Power = VBN IRY = cos∅ = $$\frac{220}{(3)^{0.5}}$$ × 2.2 cos (126.86°)
= -167.6 W ≈ – 168 W.
5. The two-wattmeter method is used to measure active power. The system is a 3-phase, 3- wire system. Then, the power reading is?
a) Affected by both negative sequence and zero sequence voltages
b) Affected by negative sequence voltages but not by zero sequence voltages
c) Affected by zero sequence voltages but not by negative sequence voltages
d) Not affected by negative or zero sequence voltages
Explanation: If the phase voltage is unbalanced, then the power reading is not affected by both negative as well as zero sequence voltages. This is a characteristic property of the 2 watt-meter meter since it measures the active power on a three phase three wire system.
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6. The ratio of the reading of 2 watt-meters connected to measure power in a balanced 3-phase load is 2: 1 and the load is inductive. The power factor of load is?
a) 0.866 lag
d) 0.625 lag
Explanation: Power factor, cosθ = cos arc tan (3)0.5 $$\frac{W_1-W_2}{W_1+W_2}$$
Since, $$\frac{W_1}{W_2} = \frac{2}{1}$$
So, power factor = cos arc tan (3)0.5 $$\frac{2-1}{2+1}$$
= cos 30° = 0.866 lag.
7. Choose the correct statement regarding two watt-meter method for power measurements in 3-phase circuit.
a) When power factor is unity, one of the wattmeters reads zero
b) When the readings of the two watt-meters are equal but opposite sign, the power factor is zero
c) Power can be measured using two watt-meter method only for star connected 3-phase circuit
d) When two watt-meters show identical readings, the power factor is 0.5
Explanation: When power factor is 0, we have ∅ = 90°
P1 = (3)0.5 VIcos (30° – ∅) = (3)0.5 VIcos (30° – 90°) = $$\frac{(3)^{0.5}}{2}$$ × VI
P2 = (3)0.5 VIcos (30° + ∅) = (3)0.5 VIcos (30° + 90°) = $$– \frac{(3)^{0.5}}{2}$$ × VI
∴Total power P = 0
So, with zero power factor the readings of the two watt-meters are equal but of opposite sign.
8. The meter constant of a single-phase, 230 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?
a) 30.3 rpm
b) 25.02 rpm
c) 27.6 rpm
d) 33.1 rpm
Explanation: Meter constant = $$\frac{Number \,of\, revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}$$ = 1656
∴ Speed in rpm = $$\frac{1656}{60}$$ = 27.6 rpm.
9. In the measurement of 3-phase power by 2 watt-meter method, if the 2 watt-meter readings are equal, the power factor of the circuit is?
a) 0.8 lagging
c) Zero
d) Unity
Explanation: Power factor, cosθ = cos arc tan (3)0.5 $$\frac{W_1-W_2}{W_1+W_2}$$
W1 = W2 = cos 0° = 1.
10. The figure shows a three-phase, delta connected load supplied from a 220 V, 50 Hz, 3-phase balanced source. The pressure Coil (PC) and Current Coil (CC) of a watt-meter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The watt-meter reading will be?
a) Zero
b) 1600 W
c) 242 W
d) 400 W
Explanation: Watt-meter reading = Current through CC × Voltage across PC × cos (phase angle).
IBR = ICC = $$\frac{220∠120°}{100°}$$ = 2.2∠120°
VYB = VPC = 220∠-120°
w = 2.2∠120°× 220∠-120° × cos 240° = 242 W.
Sanfoundry Global Education & Learning Series – Electrical Measurements.
To practice all Puzzles on Electrical Measurements, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,880 | 5,655 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-26 | latest | en | 0.834914 |
http://www.jiskha.com/display.cgi?id=1351574209 | 1,462,546,670,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861831994.45/warc/CC-MAIN-20160428164351-00059-ip-10-239-7-51.ec2.internal.warc.gz | 603,699,757 | 3,950 | Friday
May 6, 2016
# Homework Help: calc
Posted by cathy on Tuesday, October 30, 2012 at 1:16am.
how would i find the derivative of these?
f(x)=ln((√x)+1)
and
f(x)=ln(x+1/x-1)
thank u!!
• calc - Reiny, Tuesday, October 30, 2012 at 9:31am
first one:
remember that √x = x^(1/2)
f ' (x) = 1/(√x + 1) ( (1/2)x^(-1/2)
= 1/( 2x^(1/2) (√x)(√x + 1)
= 1/( 2√x(√x + 1) or 1/(2(x+√x)
2nd:
not clear if you meant
ln( x + 1/x - 1) --- the way you typed it , or
ln( (x+1)/(x-1) ) --- I have a feeling that's the one, or
ln( x + 1/(x-1) )
I will read it as ln ((x+1)/(x-1))
then this can be changed to
ln(x+1) - ln(x-1)
f ' (x) = 1/(x+1) - 1/(x-1)
= (x-1 - (x+1))/((x+1)(x-1))
= -2/((x+1)(x-1))
notice how important brackets are ? | 341 | 728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2016-18 | longest | en | 0.904181 |
https://gis.stackexchange.com/questions/41081/how-to-add-offsets-to-the-route/49836 | 1,624,626,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630175.17/warc/CC-MAIN-20210625115905-20210625145905-00045.warc.gz | 263,846,873 | 41,310 | # How to add offsets to the route?
Edited:
I want to ilustrate my question. Suppose that you're in "Point A" and want to go to "Point B". This points wouldn't be in "at_2po_4pgr" table cause aren't source/targets nodes. Then, I would search for the nearest node for points A and B (green points). After this, I might perform a shortest_path call using green points ids and I would obtain "orange" path. But to obtain the real path cost (distance) in first case I would have to substract "offsetA" and in second case add de "offset B". To calculate distance between red-points and green-points, I run the following query:
```SELECT * FROM st_distance( ST_GeomFromText('POINT(-3.6963314 42.3498066)',4326), ST_GeomFromText('POINT(-3.6954276 42.3479634)',4326))```.
How I would know when add or substract the offset?
Sorry for my english!
• Welcome to gis.stackexchange. This is a Q&A site where each thread should contain exactly one question and its answers. Please open a separate thread for question #3. #2 is answered in gis.stackexchange.com/questions/33471/… – underdark Nov 14 '12 at 10:22
• I have the same problem. did you find any solution? Thanks a lots – Robert Jan 24 '13 at 14:28
• Please post your solution in the answer section. Then it can be upvoted. – underdark Feb 2 '13 at 10:53
I do not think that you can rely on the nearest vertex. Imagine source and target are located on the same edge close to the same vertex.
You'd rather considered three! different cases:
1. a vertex is the nearest point.
2. a form node of the edge is the right one
3. the edge-line itself is nearer. (orthogonal)
• Sorry but this isnt the correct answer. pgr_trsp - Turn Restriction Shortest Path (TRSP) have offset as show for @amball answer. – Juan Carlos Oropeza Dec 7 '16 at 18:18
You can find such a function here: https://github.com/pgRouting/pgrouting-contrib/blob/master/wrapper/routing_core_smart.sql#L69
It searches for the nearest link in the network, which usually gives a better result. If you use Shooting Star you can then start routing from/to this link. For A* or Dijkstra you either select start or end point of the link, or you create a "virtual" node by splitting the link into two.
I'm going to explain the solution that I've found (perhaps don`t be the best).
According to the post image, let's suppose that we are in Point A and we gonna go to Point B. As I explained above this points aren`t vertex (source/targets in table generated with osm2po tool).
Due to this, we need to know the walking/driving direction. If we go from nearest vertex to Point A (point green) through orange path we would have to substract the offset between Point A and point green (nearest vertex). But if we had to go through Calle Almirante Bonifaz street, then we should add the offset to the length of this edge (from point green to intersection between Calle Almirante Bonifaz and Calle San Juan).
I run the following query to obtain the shortest path (you need the pgRouting extension explained here pgRouting - installation and requirements here installation&requirements):
``````SELECT gid, cost, st_astext(the_geom) as the_geom FROM dijkstra_sp_delta('xx_2po_4pgr', source_vertex, target_vertex, 0.1);
``````
This results in a set of edges that represents the complete route. For example, one possible output for this query might be:
Where the field gid (id in osm2po generated table) represents the edge identifier. Well, we must check the offsets at the start and at the end (Points A/B).
If we check the start offset, we must check if the first edge of the set of edges obtained in the above query is the same to the nearest path to Point A. If they match, then we will substract the offset. If they don`t match, we will add the offset. To obtain de nearest link to a point, I run the following query:
``````SELECT * FROM find_node_by_nearest_link_within_distance(point, 0.1, 'xx_2po_4pgr') as id;
``````
``````CREATE TYPE link_point AS
(id integer,
name character varying,
OWNER TO postgres;
``````
You must also modify the find_node_by_nearest_link_within_distance. Just add the last line (I only show an extract from the function):
``````-- Searching for a nearest link
FOR row in EXECUTE 'select id from find_nearest_link_within_distance('''||point||''', '||distance||', '''||tbl||''') as id'
LOOP
END LOOP;
IF row.id is null THEN
res.id = -1;
RETURN res;
END IF;
``````
Then you need to know what is the distance between point (Point A/Point B) and the nearest edge(offset). For this purpose I run this query:
``````SELECT ST_Line_Locate_Point(geom , point)as offset;
``````
Where geom is the the_geom field in osm2po generated table.
At this point, we would have the offset to add or substract.
Finally, you would need to know the edge legth to apply the value obtained in the query above and adjust the real (if you work with geometry type, you will have to normalize to meters the value obtained. Just multiply 111000 by the length obtained in the query):
``````select st_length(the_geom) from (select ST_ASTEXT(the_geom) as the_geom FROM dr_2po_4pgr WHERE id= edge_identifier)t";
``````
If we would check the end offset, then we would have to check if the last path of the set of paths obtained in the above query is the same to the nearest path to the ending point (Point B) and we would add/substract at the same way as before.
Excuse my english.
In pgrouting, pgr_trsp - Turn Restriction Shortest Path (TRSP) does exactly what you are looking for.
Instead of specifying source and target nodes, you can specify source and target edges, and the fraction along the edge where your origin and destination are located.
(You can use ST_Line_Locate_Point to get that fraction from your point geometry, assuming you know the closest edge.) | 1,441 | 5,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-25 | latest | en | 0.905042 |
https://www.sofolympiadtrainer.com/forum/nso/0/345 | 1,558,401,388,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256184.17/warc/CC-MAIN-20190521002106-20190521024106-00315.warc.gz | 954,462,280 | 38,111 | # User Forum
Subject :NSO Class : Class 5
## Ans 1:
Class : Class 5
can anyone tell me
Class : Class 5
Class : Class 4
a
Class : Class 5
Class : Class 5
Class : Class 5
haha
Class : Class 5
Class : Class 5
Class : Class 5
## Ans 10:
Class : Class 4
Subject :NSO Class : Class 7
Class : Class 7
A
## Ans 2:
Class : Class 7
The answer should be C because during the last 10 seconds [ 40 second ] the distance is 10 meter not 5 meter.
Subject :NSO Class : Class 3
Class : Class 5
## Ans 2:
Class : Class 5
They are right it should be [B] .
## Ans 3:
Class : Class 3
WHAT!!!!!! IT SHOULD BE (A).
Class : Class 3
what????
## Ans 5:
Class : Class 3
It should be a tailorbird
## Ans 6:
Class : Class 4
Tailor bird that is A
Subject :NSO Class : Class 8
## Ans 1:
Class : Class 10
TBH2VI http://www.LnAJ7K8QSpfMO2wQ8gO.com
Subject :NSO Class : Class 9
Class : Class 1
Alpha, Beta, Gamma in ascending order.
Alpha particles leaves the nucleus of an unstable atom at a speed of 16,000 kilometres per second, around a tenth the speed of light.
Beta particles travel at a speed of 270,000 kilometres per second, around 98% of speed of light.
Gamma particles travel at the speed of light.
Subject :NSO Class : Class 4
## Ans 1:
Class : Class 4
All options are WRONG
## Ans 2:
Class : Class 5
I have by mistake typed it Co sorry it will be C. I have a solution we can report error.
Class : Class 4
Class : Class 4
Class : Class 4
all are wrong
## Ans 6:
Class : Class 5
you are right vaibhav all are wrong
## Ans 7:
Class : Class 5
None of the options are correct according to my doubt.
Class : Class 4
## Ans 9:
Class : Class 4
question, answer and solution is wrongall the options are showing the difference on heating
## Ans 10:
Class : Class 4
Qtn. agreed but main point is which is column X and which is column is Y??????
Class : Class 6
option d maybe
Class : Class 4
4wrong question
## Ans 13:
Class : Class 4
Please change it. It is correct in NSO book like Harish Divakar said
Class : Class 4
PLS. CORRECT
## Ans 15:
Class : Class 4
all options are wrong
Class : Class 4
## Ans 17:
Class : Class 5
SHOULD HAVE BEEN SPECIFIC WITH RESPECT TO X AND Y MARKING
## Ans 18:
Class : Class 4
All the answer is wrong. In the NSO book, it is right.
## Ans 19:
Class : Class 5
if an object is cooled down the molecule structure changes from loosely packed to tightly packed. therefore none of the options is correct
Class : Class 4
ALL ARE WRONG
## Ans 21:
Class : Class 4
I also think that.Questions ,options, colums and answers didn't mentioned.
## Ans 22:
Class : Class 4
how can liquid turn into gas by freezing ? Option A does the same thing so the correct option should be D
## Ans 23:
Class : Class 5
The options given are wrong / confusing. Please clarify.
Class : Class 4
## Ans 25:
Class : Class 4
When a substance is cooled the molecules come closer. But there is no picture showing that the molecules are coming closer there is no correct answer. Also option B and C are the same. Can you please change it?
## Ans 26:
Class : Class 4
Please explain the answer. I also support Baibhab that the choices given are wrong.
## Ans 27:
Class : Class 5
B and Co are same. Intermolecular space is not decreasing in any of the situations. It is wrong.
## Ans 28:
Class : Class 4
all options are wrong
## Ans 29:
Class : Class 4
b and c are all the same
## Ans 30:
Class : Class 3
I think all options are wrong
## Ans 31:
Class : Class 4
Are B, C, D correct.
## Ans 32:
Class : Class 4
is there a correction in the options given?if not please clarify with explanation
## Ans 33:
Class : Class 4
Option b, c are the same. Pls give me a proper explanation as to why one option is correct
Class : Class 7
Class : Class 5
## Ans 36:
Class : Class 4
There is no specific details about which one is X or Y.So by common sense I think that it would be D.But B
## Ans 37:
Class : Class 4
all options are wrong
## Ans 38:
Class : Class 4
wrong wrong wrong wrong wrong wrong
## Ans 39:
Class : Class 5
Options are wrong. Molecular distance should be less.
## Ans 40:
Class : Class 5
Can somebody pls answer this question with explanation
## Ans 41:
Class : Class 4
All options are wrong.
## Ans 42:
Class : Class 4
All the options are totally wrong, please change them
## Ans 43:
Class : Class 6
I think al options are wrong . Please tell the correct answer
Class : Class 4
## Ans 45:
Class : Class 4
All are wrong. At least number of atoms should be same. So answer should be D
Class : Class 4
all Q are wrong
## Ans 47:
Class : Class 4
Options are wrong. Molecular distance should be less. first fix which column is X and Y.
## Ans 48:
Class : Class 4
Subject :NSO Class : Class 4
Class : Class 5
A is correct
Class : Class 6
## Ans 3:
Class : Class 4
AAAAAAAAA IS RIGHT
## Ans 4:
Class : Class 6
I think the ‘A’ is the right answer
Subject :NSO Class : Class 2
## Ans 1:
Class : Class 2
Subject :NSO Class : Class 5
Class : Class 6
4 km
Class : Class 5
4km
## Ans 3:
Class : Class 1
answer will be d 4 kilo meters
## Ans 4:
Class : Class 5
Subject :NSO Class : Class 9
## Ans 1:
Class : Class 9
SI unit of impluse is Newton
## Ans 2:
Class : Class 10
force=change in momentum/time interval(del t) impulse=change in momentum=force * time interval(del t) SI unit of impulse is newton second or Ns. CGS unit of impulse is dyne second . | 1,627 | 5,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | longest | en | 0.873234 |
https://dlmf.nist.gov/search/search?q=connected | 1,723,170,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00783.warc.gz | 176,479,837 | 4,443 | # connected
(0.001 seconds)
## 1—10 of 102 matching pages
##### 1: 8.23 Statistical Applications
###### §8.23 Statistical Applications
The function $\mathrm{B}_{x}\left(a,b\right)$ and its normalization $I_{x}\left(a,b\right)$ play a similar role in statistics in connection with the beta distribution; see Johnson et al. (1995, pp. 210–275). …
##### 2: 21.10 Methods of Computation
###### §21.10(ii) Riemann Theta Functions Associated with a Riemann Surface
• Deconinck and van Hoeij (2001). Here a plane algebraic curve representation of the Riemann surface is used.
• ##### 3: 28.27 Addition Theorems
Addition theorems provide important connections between Mathieu functions with different parameters and in different coordinate systems. …
##### 4: 36.14 Other Physical Applications
Diffraction catastrophes describe the connection between ray optics and wave optics. … Diffraction catastrophes describe the “semiclassical” connections between classical orbits and quantum wavefunctions, for integrable (non-chaotic) systems. …
##### 5: 12.16 Mathematical Applications
In Brazel et al. (1992) exponential asymptotics are considered in connection with an eigenvalue problem involving PCFs. …
##### 6: 17.18 Methods of Computation
Lehner (1941) uses Method (2) in connection with the Rogers–Ramanujan identities. …
##### 7: 31.18 Methods of Computation
Subsequently, the coefficients in the necessary connection formulas can be calculated numerically by matching the values of solutions and their derivatives at suitably chosen values of $z$; see Laĭ (1994) and Lay et al. (1998). …
##### 8: Peter Paule
Paule’s main research interests are computer algebra and algorithmic mathematics, together with connections to combinatorics, special functions, number theory, and other related fields. …
##### 9: 13.27 Mathematical Applications
Confluent hypergeometric functions are connected with representations of the group of third-order triangular matrices. …
##### 10: Mark J. Ablowitz
Widespread interest in Painlevé equations re-emerged in the 1970s and thereafter partially due to the connection with IST and integrable systems. … | 519 | 2,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 3, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.756254 |
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# A geometric sequence is a sequence in which each
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Re: A geometric sequence is a sequence in which each [#permalink]
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23 Apr 2013, 22:04
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rakeshd347 wrote:
v, w, x, y, z
A geometric sequence is a sequence in which each term after
the first is equal to the product of the preceding term and a
constant. If the list of numbers shown above is an geometric
sequence, which of the following must also be a geometric
sequence?
I. 2v, 2w, 2x, 2y, 2z
II. v + 2, w + 2, x + 2, y + 2, z + 2
III.\sqrt{v}, \sqrt{w}, \sqrt{x},\sqrt{y}, \sqrt{z}
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
KUDOS please if you like my question.
We know that v,w,x,y,z are in GP. Thus, w/v = x/w = y/x = z/y = r(some constant, called the common ratio)
I. A multiplication by a constant (2) will not change the ratio, as evident.
III. The ratio for these terms will be another constant $$\sqrt{r}$$
E.
[Reveal] Spoiler: OA
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Re: A geometric sequence is a sequence in which each [#permalink]
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23 Apr 2013, 23:21
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Can be done quite easily if 4 nos are considered instead of the variables
For example
1, 2, 4, 8
I multiply 2 : 2, 4, 8, 16 ; they are still in GP
II Add 2: 3, 4, 6, 10 ; not in GP
III Square root : 1, \sqrt{2}, 2, 2\sqrt{2} ; still in GP
So clearly I and III i.e E is the correct answer
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Re: A geometric sequence is a sequence in which each [#permalink]
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25 Apr 2013, 03:56
rakeshd347 wrote:
v, w, x, y, z
A geometric sequence is a sequence in which each term after the
first is equal to the product of the preceding term and a constant. If the list of numbers shown above is an geometric sequence, which of the following must also be a geometric sequence?
I. 2v, 2w, 2x, 2y, 2z
II. v + 2, w + 2, x + 2, y + 2, z + 2
III. $$\sqrt{v}$$, $$\sqrt{w}$$, $$\sqrt{x}$$, $$\sqrt{y}$$, $$\sqrt{z}$$
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
Similar question from OG13:
Quote:
p, r, s, t, u
An arithmetic sequence is a sequence in which each term after the first term is equal to the sum of the preceding term and a constant. If the list of numbers shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence?
I. 2p, 2r, 2s, 2t, 2u
II. p-3, r-3, s-3, t-3, u-3
III. p^2, r^2, s^2, t^2, u^2
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
Discussed here: an-arithmetic-sequence-is-a-sequence-in-which-each-term-59035.html
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Re: A geometric sequence is a sequence in which each [#permalink]
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14 Jan 2015, 09:36
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Re: A geometric sequence is a sequence in which each [#permalink]
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14 Jan 2015, 20:36
I. 2v, 2w, 2x, 2y, 2z
Multiplication by 2 keeps the GP sequence intact
II. v + 2, w + 2, x + 2, y + 2, z + 2
Addition breaks the GP sequence in this case
III.$$\sqrt{v}, \sqrt{w}, \sqrt{x},\sqrt{y}, \sqrt{z}$$
Square rooting is nothing but changing the power from 1 to $$\frac{1}{2}$$ ; it still keeps the GP sequence intact
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A geometric sequence is a sequence in which each [#permalink]
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05 Jul 2017, 06:17
mau5 wrote:
rakeshd347 wrote:
v, w, x, y, z
A geometric sequence is a sequence in which each term after
the first is equal to the product of the preceding term and a
constant. If the list of numbers shown above is an geometric
sequence, which of the following must also be a geometric
sequence?
I. 2v, 2w, 2x, 2y, 2z
II. v + 2, w + 2, x + 2, y + 2, z + 2
III.\sqrt{v}, \sqrt{w}, \sqrt{x},\sqrt{y}, \sqrt{z}
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
KUDOS please if you like my question.
We know that v,w,x,y,z are in GP. Thus, w/v = x/w = y/x = z/y = r(some constant, called the common ratio)
I. A multiplication by a constant (2) will not change the ratio, as evident.
III. The ratio for these terms will be another constant $$\sqrt{r}$$
E.
Let the common ratio of the terms in GP be r.
So, w = vr
x = vr^2
y = vr^3
z = vr^4
Now lets start checking I , II & III
I. 2v, 2w, 2x, 2y, 2z = 2v, 2vr, 2vr^2, 2vr^3, 2vr^4 Common ratio = r. So, GP
II. (v+2), (w+2), (x+2), (y+2), (z+2) = 2v+2, 2vr+2, 2vr^2 +2 , 2vr^3 +2, 2vr^4 +2 . Not in GP.
III. $$\sqrt{v},\sqrt{w},\sqrt{x},\sqrt{y},\sqrt{z}, = \sqrt{2v}, \sqrt{2vr}, \sqrt{2vr^2},\sqrt{2vr^3},\sqrt{2vr^4}$$
Common ration = $$\sqrt{r}$$. So, GP
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A geometric sequence is a sequence in which each [#permalink] 05 Jul 2017, 06:17
Display posts from previous: Sort by | 2,384 | 7,093 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-47 | latest | en | 0.874113 |
https://proofwiki.org/wiki/Automorphism_Group_is_Subgroup_of_Symmetric_Group | 1,679,945,117,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00229.warc.gz | 514,813,225 | 11,919 | # Automorphism Group is Subgroup of Symmetric Group
## Theorem
Let $\struct {S, *}$ be an algebraic structure.
Let $\Aut S$ be the automorphism group of $\struct {S, *}$.
Then $\Aut S$ is a subgroup of the symmetric group $\struct {\Gamma \paren S, \circ}$ on $S$.
## Proof
An automorphism is an isomorphism $\phi: S \to S$ from an algebraic structure $S$ to itself.
The Identity Mapping is Automorphism, so $\Aut S$ is not empty.
The composite of isomorphisms is itself an isomorphism, as demonstrated on Isomorphism is Equivalence Relation.
So:
$\phi_1, \phi_2 \in \Aut S \implies \phi_1 \circ \phi_2 \in \Aut S$
demonstrating closure.
If $\phi \in \Aut S$, then $\phi$ is bijective and an isomorphism.
Hence from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}$ is also bijective and an isomorphism.
So $\phi^{-1} \in \Aut S$.
The result follows by the Two-Step Subgroup Test.
$\blacksquare$ | 274 | 932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-14 | latest | en | 0.799236 |
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Integral. Definite integral. The integral sum. The construction of integral sums. The area of the curvilinear trapezoid. Newton Leibniz. Properties of the definite integral. The basic properties of the definite integral. | 385 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | latest | en | 0.841651 |
http://www.chegg.com/homework-help/questions-and-answers/5-kg-fish-swimming-2-m-s-swallows-absent-minded-1-kg-fish-swimming-velocity-brings-fish-ha-q928982 | 1,394,293,910,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999655040/warc/CC-MAIN-20140305060735-00077-ip-10-183-142-35.ec2.internal.warc.gz | 274,678,154 | 6,765 | # Momentum
0 pts ended
A 5 kg fish swimming 2 m/s swallows an absent minded 1 kg fish swimming toward it at a velocity that brings both fish to a halt immediately after lunch. What is the velocity v of the smaller fish before lunch?
This is what I go so far:
(5kg x 2m/s) + (1kg x v) = (5kg +1kg) x 0m/s
but if I did this v = 0, and that wouldn't be right | 110 | 359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2014-10 | latest | en | 0.958719 |
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U09_L3_T1_we2 Associative property for multiplication
• 05:39
### Distributive Property Example 1
348 views / 0 likes - added
ck12 Distributive Property example More free lessons at: http://www.khanacademy.org/video?v=ewEorPD4kdA
• 02:36
### Multiplying Binomials With Distributive Property
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Multiplying binomials with distributive property
• 06:29
### How To Use The Distributive Property With Variables | 6th Grade | Khan Academy
571 views / 0 likes - added
Learn how to apply the distributive property to algebraic expressions. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-distributive-property/e/distrib
• 04:46
### Identity property of 0
470 views / 0 likes - added
U09_L2_T2_we3 Identity property of 0 More free lessons at: http://www.khanacademy.org/video?v=uXTRmzXlorI Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 01:55
### Identity property of 1 (second example)
360 views / 0 likes - added
U09_L2_T4_we3 Identity property of 1 More free lessons at: http://www.khanacademy.org/video?v=_86K0yQ8BQY Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 01:24
### Identity Property of 1
422 views / 0 likes - added
U01_L4_T1_we2 Identity Property of 1 More free lessons at: http://www.khanacademy.org/video?v=6nZp2QGeQ9k Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 04:56
### The Distributive Property
347 views / 0 likes - added
U01_L4_T2_we1 The Distributive Property More free lessons at: http://www.khanacademy.org/video?v=gl_-E6iVAg4 Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 02:11
### The Distributive Property 2
397 views / 0 likes - added
U01_L4_T2_we2 The Distributive Property 2 More free lessons at: http://www.khanacademy.org/video?v=Badvask-UDU Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 09:52
### Distributive Property of Matrix Products
537 views / 0 likes - added
Showing that matrix products exhibit the distributive property More free lessons at: http://www.khanacademy.org/video?v=oMWTMj78cwc
• 03:23
331 views / 0 likes - added
U09_L3_T1_we1 Commutative Property for Addition More free lessons at: http://www.khanacademy.org/video?v=UeG_EYd-0xw Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 04:46
### Factoring and the Distributive Property 2
342 views / 0 likes - added
U09_L1_T1_we2 Factoring and the Distributive Property 2 More free lessons at: http://www.khanacademy.org/video?v=499MvHFrqUU Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 05:54
### Factoring and the Distributive Property 3
322 views / 0 likes - added
U09_L1_T1_we3 Factoring and the Distributive Property 3 More free lessons at: http://www.khanacademy.org/video?v=MZl6Mna0leQ Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 03:05
### Factoring and the Distributive Property
399 views / 0 likes - added
U09_L1_T1_we1 Factoring and the Distributive Property More free lessons at: http://www.khanacademy.org/video?v=auQU-9KNG74 Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 06:04
### Solving Equations with the Distributive Property
440 views / 0 likes - added
U02_L1_T2_we1 : Solving Equations with the Distributive Property More free lessons at: http://www.khanacademy.org/video?v=YZBStgZGyDY Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 04:50
### Solving equations with the distributive property 2
348 views / 0 likes - added
U02_L1_T2_we2: Solving equations with the distributive property 2 More free lessons at: http://www.khanacademy.org/video?v=PL9UYj2awDc Content provided by TheNROCproject.org - (c) Monterey Institute for Technology and Education
• 10:10
### Introduction To Intellectual Property: Crash Course IP 1
505 views / 0 likes - added
This week, Stan Muller launches the Crash Course Intellectual Property mini-series. So, what is intellectual property, and why are we teaching it? Well, intellectual property is about ideas and their ownership, and it's basically about the rights of creat
• 11:54 Popular
### Algebra Basics: The Distributive Property - Math Antics
1,183 views / 0 likes - added
This video introduces the Distributive Property in its general algebraic form: a(b + c) = ab + ac It shows how this patten is helpful when working with polynomials. Part of the Algebra Basics Series: https://www.youtube.com/watch?v=NybHckSEQBI&list=PLUPEB
• 03:27
### How To Use The Distributive Property To Factor Out The Greatest Common Factor | Khan Academy
409 views / 0 likes - added
Learn how to apply the distributive property to factor numerical expressions (no variables). Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-distribut
• 03:00
### Inverse Property Of Addition | Arithmetic Properties | Pre-Algebra | Khan Academy
553 views / 0 likes - added
The simple idea that a number plus its negative is 0 Watch the next lesson: https://www.khanacademy.org/math/pre-algebra/order-of-operations/arithmetic_properties/v/inverse-property-of-multiplication?utm_source=YT&utm_medium=Desc&utm_campaign=PreAlgebra M
• 13:30
### IP Problems, YouTube, And The Future: Crash Course Intellectual Property #7
610 views / 0 likes - added
In which Stan Muller talks about some of the problems in Intellectual Property law as it exists today. He'll also teach you a little about how IP law applies to everyone's favorite media platform, YouTube. Lastly, he'll do a little prognosticating, and tr
• 10:00
### International IP Law: Crash Course Intellectual Property #6
521 views / 0 likes - added
This week, Stan Muller teaches you how intellectual property law functions internationally. Like, between countries. Well, guess what. There's kind of no such thing as international law. But we can talk about treaties. There are a bevy of international tr
• 04:27
### What's My Property: Crash Course Kids #35.2
467 views / 0 likes - added
What exactly can we tell about an unknown substance by it's properties. We already know that a substance is matter that’s made of one kind of atom or molecule, and that has specific properties and that some substances are elements, which means they can’t
• 11:20
### Trademarks And Avoiding Consumer Confusion: Crash Course Intellectual Property #5
494 views / 0 likes - added
In which Stan Muller teaches you about our third branch of Intellectual Property, trademarks. A lot of people confuse trademark and copyright. Trademarks apply to things like company and product names and logos, packaging designs, and commercial designs.
• 10:07
### The Distributive Property In Arithmetic
552 views / 0 likes - added
Featured
• 01:56
### Crash Course Intellectual Property And Economics!
475 views / 0 likes - added
John Green and Stan Muller talk about the future of Crash Course Humanities
• 12:18
### Copyright Basics: Crash Course Intellectual Property 2
492 views / 0 likes - added
EPISODE DESCRIPTION This week, Stan Muller teaches you the basics of copyright in the United States. Copyright law is territorial, so we're going to cover the system we know the most about, and that's the US. Stan will talk about what kind of ideas can be
• 11:39
### Copyright, Exceptions, And Fair Use: Crash Course Intellectual Property #3
495 views / 0 likes - added
Stan Muller teaches you a few things about copyright enforcement, and talks about the exceptions to copyright enforcement. While there are several, the one you've probably heard of is Fair Use, and it's a pretty tricky one. We'll try to explain it, and te
• 09:51
### Patents, Novelty, And Trolls: Crash Course Intellectual Property #4
526 views / 0 likes - added
This week, Stan teaches you about patents. It turns out, they're patently complicated! So, patents have some similarity to copyright, in that they grant a limited monopoly to people who invent things. The key difference in patents and copyright is that pa
• 01:58
### How To Find Equivalent Expressions By Combining Like Terms And Using The Distributive Property
573 views / 0 likes - added
• 02:52
### Why Does The Earth Spin?
52 views / 0 likes - added
The Earth spins on its axis, completing a full revolution every day. By why does it do this? One of the most common misconceptions in physics is the belief that constant motion requires a constant force. So many people believe there must be some force in
• 12:29
### Soviet Montage: Crash Course Film History #8
348 views / 0 likes - added
Russia went and had a revolution in 1917 and cinema was a big part of its aftermath. Even though film stock was hard to come by, we saw the first film school started, and the study of film became hugely important. Russian filmmakers started trying to unde
• 05:16
### Corner cube mirrors & Retroreflectors // Homemade Science with Bruce Yeany
485 views / 0 likes - added
Corner cube mirrors, bicycle reflectors, traffic signs, human or animal eyes are all examples of retroreflectors various types and share the unique property of returning light back towards the original light source. There is even one on the moon placed th
• 09:18
### The Weak Nuclear Force: Through the looking glass
176 views / 0 likes - added
Of all of the known subatomic forces, the weak force is in many ways unique. One particularly interesting facet is that the force differentiates between a particle that is rotating clockwise and counterclockwise. In this video, Fermilab’s Dr. Don Lincoln
• 13:34
### Water and Solutions -- for Dirty Laundry: Crash Course Chemistry #7
531 views / 0 likes - added
Dihydrogen monoxide (better know as water) is the key to nearly everything. It falls from the sky, makes up 60% of our bodies, and just about every chemical process related to life takes place with it or in it. Without it, none of the chemical reactions t
• 05:14
### How to never have a serious poison ivy rash again
456 views / 1 likes - added
Urushiol oil in poison sumac, poison oak, and poison ivy may produce a severe skin rash. Timely urushiol removal can prevent poison ivy skin reaction. The key is to understand how poison ivy works. http://www.extremedeerhabitat.com/habitat-blog Jim Brauke
• 05:09
### How to never have a serious poison ivy rash again
76 views / 0 likes - added
Urushiol oil in poison sumac, poison oak, and poison ivy may produce a severe skin rash. Timely urushiol removal can prevent poison ivy skin reaction. The key is to understand how poison ivy works. http://www.extremedeerhabitat.com/habitat-blog Jim Brauke
• 02:53
### 5 Amazing Physics Experiments
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These 5 holiday-themed physics experiments will keep you and your family busy with science during the winter holidays! All experiments involve materials found around the house. Parental supervision advised . Flame Eraser – put out a candle by pouring CO2
Featured
• 07:27
### Hydrogen Peroxide, Liver, and Elephant Toothpaste Science?
279 views / 1 likes - added
Craig Beals from Beals Science looks at the biochemistry of beef liver and hydrogen peroxide. Liver is full of catalase, an enzyme catalyst that decomposes hydrogen peroxide into water and oxygen gas. Craig experiments with 3% and 30% hydrogen peroxide on
• 01:29
### ThunderCats - Original Intro vs. CGI
119 views / 0 likes - added
A side-by-side comparison of my 3D remake with the original opening. Fingers crossed it's not taken down for copyright issues... Check out the CGI version here: https://www.youtube.com/watch?v=SgonBJvVgsw ThunderCats is of course the property of Warner Br
• 01:11
### Candy Science - Sick Science! #139
125 views / 1 likes - added
To Discover Why This Happens: http://www.stevespanglerscience.com/experiment/floating-lettersM&Ms have the slogan of melting in your mouth, but not in your hands. You wanna know something cool? The special melting property is totally scientific. The resul
• 04:47
### Angle of repose…Physics of toys // Homemade Science with Bruce Yeany
371 views / 0 likes - added
The angle of repose or critical angle of repose is how steep various types of materials can be piled without slumping. The steepness of the side of a pile is determined by the amount of friction between the particles of matter. This can vary according to
• 06:51
### What's inside a TOPGOLF Golf Ball?
375 views / 1 likes - added
We CUT OPEN a TopGolf Golf Ball to find the Rfid technology. Super fun place to go. We have been to TopGolf in many different states and always wanted to find out how these things work! Thanks to Corey at the GolfLab for helping us with this project! Twit
• 04:11
### Coefficient Of Restitution: Why Certain Objects Are More Bouncy Than Others?
170 views / 0 likes - added
Basketballs bounce a lot, but plastic balls don’t. Some objects are inherently bouncier than others. Why is that so? The answer lies in a property of the material which is known as the coefficient of restitution. In this video, we give a layman explanatio
• 02:00
### Candy Science Experiment : Floating Letters
439 views / 0 likes - added
Chocolate alphabets have the slogan of melting in your mouth, but not in your hands. You wanna know something cool? The special melting property is totally scientific. Certain parts of chocolate alphabets will dissolve in water, while others won't. The re
• 03:15
### The Difference Between Mass and Weight
68 views / 0 likes - added
There is a common perception that weight and mass are basically the same thing. This video aims to tease out the difference between mass and weight by asking people what makes a car difficult to push. The standard answer is that it is difficult to push be
• 01:26
### Lost Maples | 4K Astrophotography Timelapse Compilation
198 views / 0 likes - added
Lost Maples State Natural Area is a large, pristine area of beautiful hills and canyons on the upper Sabinal River in the Edwards Plateau Region of Texas. It is designated a Natural Area, rather than a State Park, which means the primary focus is the main
• 07:41
### What's inside SeaPlane Floats?
305 views / 0 likes - added
We CUT OPEN a SeaPlane Float!?! Paid partnership with Disney XD. Check out the premiere of DuckTales on Sept 23rd on Disney XD! These planes are fascinating how they land and take off on water. So cool to see them up close! Check out our video of Lincoln
• 04:16
### Samuel Barber - Adagio for Strings
351 views / 0 likes - added
With the recent events that have happened in society, it has really taken a toll on me mentally/emotionally. My heart is hurting and there's nothing I can really say. So since music is the thing I know best I figured I'd just play. Here's a bit of Adagio
• 10:28
### Animatronic Wheatley Robot v2.0
177 views / 0 likes - added
Edit: Please note that I am not making or selling any animatronic puppets, and Animatronic Wheatley is not for sale. Doing so would infringe on Valve’s copyright(s) and intellectual property and would be too costly for most people's budget. However,
• 00:55 Popular
### Guinea Pig Bridge at the Nagasaki Bio Park - song by Parry Gripp
777 views / 3 likes - added
Visit the Guinea Pigs at the great Nagasaki Bio Park: http://www.biopark.co.jp/en/ Nagasaki Bio Park YouTube channel: https://www.youtube.com/channel/UCr9VfOCSX5H1jFa8ku4mH2w Video footage property of the Nagasaki Bio Park. -----LYRICS------ Guinea Pig Br
• 03:13
### Amazing Laser Fireworks Display | Avalon Airshow 2019
196 views / 1 likes - added
Nothing got the crowd going more than the Friday night fireworks and laser display put on by the Scandinavian Catwalk. It was one of the most memorable moments of the airshow and hopefully a similar display is around for the next one in 2021. ____________
• 01:39
### Salt Water Density Straw - Sick Science! #140
129 views / 0 likes - added
GET THE KIT HERE: http://www.stevespanglerscience.com/collections/sick-science.htmlClick here to get the secret: http://www.stevespanglerscience.com/experiment/density-strawDensity can be a difficult scientific property to grasp, that's why we like making
• 05:04
### Skylight: Why Does Earth Have Seasons?
120 views / 0 likes - added
The amount of daylight we experience varies throughout the year from place to place. Some places have longer days and nights than others, and the length of each day changes with the seasons. Find out why Earths tilt is the reason we have seasons.#seasons
• 01:02
### Fire Resistant Water Balloon - Sick Science! #122
70 views / 0 likes - added
Find out how this is possible at http://www.stevespanglerscience.com/experiment/fire-water-balloonCommon sense tells you that it's impossible to boil water in a paper bag, but this classic parlor trick was a favorite of Victorian magicians. The real diffi
• 08:40
### What's inside a Beaver Home?
249 views / 0 likes - added
Can you climb into a Beaver Lodge? Is there a difference between a home and a dam? Watch our exploration of the other Homes here: https://youtu.be/W6jaEyoATq4 Subscribe to Let's Ride Alaska!! https://www.youtube.com/channel/UC905NeBwSRO8AJB1EWK20vA Twitte
• 11:55
### What's inside a Love Tester Machine?
399 views / 0 likes - added
We CUT OPEN an old Love Arcade game! Watch our adventure with Jurgy's: https://youtu.be/N3dqFqxR6bU Is it magic that operates this? Heat Sensors? Random generator? This is something that has been around for years. Twitter: https://twitter.com/whatsinside
• 08:33
### Whats inside a BLADELESS Fan?
309 views / 2 likes - added
• 04:42
### How To Write Algebraic Expressions From Word Problems | 6th Grade | Khan Academy
674 views / 0 likes - added
Learn how to write expressions with variables to describe situations described in word problems. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-alg-e
• 08:38
### What's inside a Stop Light?
276 views / 0 likes - added
We CUT OPEN a Stop Light!! Super interesting!! Watch us get Keys to the City here: https://youtu.be/kChihXRAb0s This was really interesting and educational! Always wondered how traffic lights work! What else should we see "What's Inside?" Twitter: https:/
• 03:51
### What is Surface Tension? | Richard Hammond's Invisible Worlds | Earth Lab
372 views / 0 likes - added
How do water striders walk on water? It has to do with the elastic property of the water surface, a phenomenon called surface tension. Subscribe: http://bit.ly/SubscribeToEarthLab Best of Earth Lab: http://bit.ly/EarthLabOriginals Best of BBC Earth: http:
• 08:18
140 views / 1 likes - added
After a tornado had ripped through our land, nearby housing developments, and farmland, there was a story circulated by a local farmer about the storm. He described that as he was surveying the damage to the buildings and property, he came across quite a
• 06:11
### Star Wars SC 38 Reimagined
268 views / 0 likes - added
Star Wars Sc 38 Reimagined (Unofficial short scene) "Scene 38 ReImagined" is about the final confrontation between Ben Kenobi & Darth Vader in "A New Hope" nearly 20 years after the events of "Revenge Of The Sith." This is a one-off story driven scene
• 08:26
### What's inside a 65-inch Flat Screen TV?
286 views / 1 likes - added
We CUT OPEN a Flat Screen Plasma TV!! Learn more about Dolby Vision here: http://bit.ly/2s2x54A Video sponsored by Dolby. Goodbye 11 year old Plasma TV, hello OLED TV with Dolby Vision and Atmos. YES!! Really have enjoyed our new TV. Great father's day gi
• 07:41
### What's inside Billy the Bass?
284 views / 0 likes - added
We CUT OPEN Big Mouth Billy Bass!?! Singing fish insides!! See Billy Bass Pricing here: http://amzn.to/2vMZEBc Watch our Alaska Trip video here: https://youtu.be/e6GuF7MfIzI Can you gut an electronic fish? Guess we can! Get ready for more Alaskan adventur
• 08:57
### What's inside the World's Smallest Cell Phone?
332 views / 0 likes - added
We CUT OPEN a REAL smallest Cell Phone!! Why is this made? Buy yours here: http://amzn.to/2sDdEzJ This phone is used by people to sneak it into prisons and other places where phones are not allowed. It is 99% plastic so isn’t picked up by metal detectors
• 11:04
### What's inside Kevin Durant's Basketball Shoe?
424 views / 4 likes - added
We CUT IN HALF limited edition KD 4 Nerfs WITH Kevin Durant!! Subscribe to KD’s YouTube channel: https://goo.gl/Ey6U1Q Watch the Live Stream here: https://youtu.be/0q3BRYngBKo What an incredible experience!! Big thanks to KD for letting us be part of this
• 03:17 Popular
### How Do Microwave Ovens Work?
731 views / 0 likes - added
They’re convenient, ubiquitous and easy to use. But how do these things work, exactly? What do people mean when they say a microwave cooks food inside out? Join HowStuffWorks as we answer engaging, everyday science questions, demystifying the amazing worl
Featured
• 07:33
### Deep Voice Gas Hidden Uses
384 views / 0 likes - added
SUBSCRIBE!: http://bit.ly/1G97GmA SHARE THIS VIDEO: https://youtu.be/mGh91oRdPDc Deep voice gas (sulfur hexafluoride) or SF6 for short has been such a fun science demo! I’ve used it on AGT, The Today Show, Fox n Friends, and Dr.Oz. Today were talking abou
• 06:12
### What's inside a Cat Coin Bank?
463 views / 0 likes - added
We CUT OPEN a goofy kitten coin bank to see how it works. Watch us get into a Singing cat: https://youtu.be/YRbo4S838-A Get your own Cat coin bank here: http://amzn.to/2vPo4Nr Godzilla Coin Bank: http://amzn.to/2uUQaI3 Panda Coin Bank: http://amzn.to/2xfp
• 02:32
### What It Takes to Maintain the Biggest Hedges in the U.K.
118 views / 0 likes - added
It takes Dan Bull, a gardener at Powis Castle and Garden, eight weeks to trim all the hedges on the historic property in Welshpool, Wales. That might sound like shear madness, but before the advent of cutting tools powered by electricity and gas, a team o
• 09:33
### What's inside the WORLD'S FIRST Personal Computer?
320 views / 0 likes - added
We CUT OPEN the World's First P.C!?!? "Click this link to get 20% off the Kano Computer Kit 2017 online or in stores, from 10/27-11/10: http://getkano.co/2zPGOMt" So incredible to see how tech has evolved over the past 36 years. This IBM 5150 is iconic. T
• 06:01
### What's inside a Fidget Spinner Cube?
601 views / 1 likes - added
Combining Fidget Spinners and Cubes into one? Brilliant! Just when you think there are no more Fidget toys left in 2017! Watch our Fidget Toy Playlist here: https://www.youtube.com/watch?v=VtLdrDHZWEA&list=PLk-NEiZ-NQeTccJeTorwGhgFXqkS5b4hl We bought thes
• 03:56
### The Amazing World of Gumball | Penny Comes Out of Her Shell | Cartoon Network UK
158 views / 0 likes - added
After being caught trespassing on Penny's family's property, Gumball apologises and brings them some cake. An interesting conversation with Penny takes place at their dinner table, about coming out of her shell.[ https://goo.gl/hRAVDf ] Subscribe to the C
• 10:36
### What’s inside an Electric Skateboard?
284 views / 0 likes - added
We CUT OPEN an Inboard M1 motor in Wheel Electric Longboard! \$100 off one here: https://goo.gl/puQfcq Watch Sam’s Video of our day: https://youtu.be/A-RRIgmsWUI Our Boosted Board video here: https://youtu.be/klIWLlJXplY Watch our latest video from our Fam
• 01:28
### One of the Worlds Largest Film Studios Is in This Moroccan Desert
154 views / 0 likes - added
Its not a miragethere is indeed a medieval castle and a Buddhist temple in the middle of the desert in Ouarzazate, Morocco. These structures are among the massive sets at Atlas Studios. TV shows like Game of Thrones and Prison Break and films such as Glad
• 35:45
### Tour the Hall of Saurischian Dinosaurs at the American Museum of Natural History!
116 views / 0 likes - added
Visit your favorite theropods, maniraptors, and more during this live tour of the Hall of Saurischian Dinosaurs with Museum guide Andrew Epstein, who offers highlights and fun facts, and fields viewer questions. #LearnWithMe #AMNH #Dinosaur #Museum #Trex
• 05:22
### Skylight: Looking Back in Time at the Speed of Light
178 views / 0 likes - added
Light takes time to travel from stars and distant galaxies to observers here on Earth. How much have the stars changed since first emitting the light that we see tonight? How far back in time are we seeing when we look at the night sky? #lightyear #astron
• 04:19 Popular
### The Amazing World of Gumball | Gumball Trespasses | Cartoon Network UK
777 views / 0 likes - added
After being caught trespassing on Penny's family's property, Gumball apologises and brings them some cake. An interesting conversation with Penny takes place at their dinner table, about coming out of her shell. Subscribe to the Cartoon Network UK YouTube
• 05:40
### The ABCs of Cephalopods with Conservation Biologist Samantha Cheng
192 views / 0 likes - added
Happy Cephalopod Week! How many hearts does an octopus have? Why do some squid glow in the dark? And what does a zebra display have to do with the giant Australian cuttlefish? Museum conservation biologist Samantha Cheng takes you through the world of cep
• 54:07
### Logic Gates, Truth Tables, Boolean Algebra AND, OR, NOT, NAND & NOR
44 views / 0 likes - added
This electronics video provides a basic introduction into logic gates, truth tables, and simplifying boolean algebra expressions. It discusses logic gates such as the AND, OR, NOT, NAND and NOR Gates. This video is for college students who are taking intr
• 26:42
### Seeing Is Believing - AMNH SciCafe
292 views / 0 likes - added
How do our brains make sense of the world our eyes see? How does attention affect our perception? And how is it possible to miss things even if they are right in front of us? Marisa Carrasco, a professor of psychology and neural science at New York Univer
• 02:58
### Skylight: How Does Our Solar System Move Around the Milky Way?
307 views / 0 likes - added
The planets orbit the Sun in a fairly flat plane. How does that plane relate to the orientation of the Milky Way? If we could see the Sun moving among our night sky constellations, which direction would it be heading? Watch this video to learn how our sol
• 14:36
### Trump: the Real Estate Empire Behind the President
570 views / 4 likes - added
• 10:48
### The True Nature Of Matter And Mass | Space Time | PBS Digital Studios
348 views / 0 likes - added
Are matter, mass, and time real? Tweet at us! @pbsspacetime Facebook: facebook.com/pbsspacetime Email us! pbsspacetime [at] gmail [dot] com Comment on Reddit: http://www.reddit.com/r/pbsspacetime We know that mass is energy… but what is energy? And where
• 08:08
### What's inside a Fidget Spinner?
462 views / 0 likes - added
We CUT OPEN our Fidget Spinner to see how it works!! Get your fidget spinner here: http://amzn.to/2oIJpG6 Watch our latest video from our Family Channel here: https://youtu.be/SpbEecPxdWU We really do love playing with these Fidget spinners! They are supe
• 13:31
### Women's Suffrage: Crash Course US History #31
504 views / 2 likes - added
You can directly support Crash Course at http://www.subbable.com/crashcourse Subscribe for as little as \$0 to keep up with everything we're doing. Free is nice, but if you can afford to pay a little every month, it really helps us to continue producing th
• 02:51
### Are we alone in the universe?
150 views / 0 likes - added
Will we ever find intelligent life in the universe? Astrophysicist Jackie Faherty explains two ways scientists approach answering this age-old question.If you want to know if there are dinosaurs still alive today, watch this weeks Dinosaur video: https://
• 12:41
### Who Won the American Revolution?: Crash Course US History #7
667 views / 0 likes - added
In which John Green teaches you about the American Revolution. And the Revolutionary War. I know we've labored the point here, but they weren't the same thing. In any case, John will teach you about the major battles of the war, and discuss the strategies
Featured
• 02:39
### Germanys Adventure Park Is Home to Europes Largest Scuba Diving Pool
153 views / 0 likes - added
An abandoned coal and steel production plant is now an adventure park in Duisburg, Germany. Its called Landschaftspark, which translates to Landscape Park. The property still looks like an industrial site from far away, but every square inch has been repu
• 02:10 Popular
### Are dinosaurs still alive today?
704 views / 2 likes - added
All dinosaurs went extinct during the Cretaceous era, right? Wrong! Paleontologist Aki Watanabe explains how birds are evolved from dinosaurs, and how T. rex had more in common with a turkey than a turtle. If you want to know if life exists elsewhere in t
Featured
• 04:47
### The Squid and the Whale: Evidence for an Epic Encounter
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Happy Cephalopod Week! One of the most famous dioramas in the American Museum of Natural History depicts a battle between two gigantic animals: the sperm whale and giant squid. But unlike most dioramas in the Museum’s halls, this scene has never been witn
• 11:37
### Aliens: Crash Course Film Criticism
368 views / 0 likes - added
James Cameron rocketed onto the action film scene with 1984's "The Terminator" and followed that up with a highly anticipated sequel to the 1979 film, "Alien." His film, "Aliens," would go on to not only be a financial success, but a critical one and has
• 07:12
### What's inside an Electronic Music Machine?
305 views / 0 likes - added
We CUT OPEN Ableton Push music Launchpad Watch the SONG HERE! https://youtu.be/_0Tvzc5oL2w Launchpad and Ableton Push are the two market leaders in these type of music machines. Andrew really knows how to bring music to life, a serious talent! We are so a
• 07:15
### What's inside Stretch Armstrong?
388 views / 0 likes - added
We CUT OPEN Stretch Armstrong!! What is this vintage Hasbro toy made of? Watch our What's inside TOYS Playlist: https://www.youtube.com/playlist?list=PLk-NEiZ-NQeQKwPTCusq_UII0rYwxTl0s If you want to buy a stretch armstrong you can do it here: http://amzn
• 06:23
### Seven Million Years of Human Evolution
337 views / 0 likes - added
Scientists use fossils to reconstruct the evolutionary history of hominins—the group that includes modern humans, our immediate ancestors, and other extinct relatives. Today, our closest living relatives are chimpanzees, but extinct hominins are eve
• 01:30
### ThunderCats Opening Remade with CGI
118 views / 0 likes - added
Earlier this year, I thoughtI'd have a go at learning how to use 3D animation software, really just so I could do some simple effects in my films... but I got a little bit carried away. The ThunderCats have been a love of mine since I can remember and the
• 03:16
### How do we find new planets?
185 views / 0 likes - added
It’s possible to see many of the planets in our solar system just by looking up at the night sky—but only those that are largest and closest to our Sun. Astrophysicist Jackie Faherty explains a few of the more advanced astrometric techniques used to detec
• 03:10
### How long did a T. rex live?
153 views / 0 likes - added
Paleontologist Aki Watanabe reveals what took Tyrannosaurus rex from tiny hatchling to mega predator, as well as the evidence scientists use to learn more about a dinosaurs lifespan.If you want to learn even more about the ultimate predator, visit our new
• 03:23
### How do you find dinosaur fossils?
251 views / 0 likes - added
In popular culture, paleontologists are often seen brushing sand off of a complete dinosaur skeleton with ease—but is digging for dinosaurs really that easy? Paleontologist Aki Watanabe reveals what really goes on during a fossil finding expedition. Spoil
• 02:40
### Spider Expert Cheryl Y. Hayashi On Silk, Webs, and More
362 views / 1 likes - added
How do spiders make their webs? Turns out it’s in their DNA. Spider expert and Museum curator Cheryl Y. Hayashi discusses her research into spider silk, why it’s an exciting time to be a biologist, and why natural history museums are so important to the f
• 05:45
### What's inside a Tesla Battery?
287 views / 0 likes - added
We CUT OPEN a TESLA Model S Battery! What does it look like? Get \$1,000 off your NEW TESLA with my Code: http://ts.la/daniel3063 Here is why we got rid of the Tesla: https://youtu.be/goTojvhwrdc Watch our TESLA VIDEOS here: https://www.youtube.com/watch?v
• 02:04
### Why are Fossil Shark Skeletons So Rare?
215 views / 0 likes - added
Happy Shark Week! Shark teeth are among the most common vertebrate fossils you can find, and yet fossilized shark skeletons are harder to come by. Paleontologist and Curator Emeritus John Maisey explains how sharks' cartilaginous skeletons differ from tho
• 01:21
### The Butterfly Life Cycle
359 views / 1 likes - added
Butterflies aren't born as we recognize them–they go through a process called metamorphosis to change from a caterpillar to a chrysalis to an adult butterfly. See live butterflies, moths, and chrysalises at the American Museum of Natural History's Butterf
• 03:09
### Did an asteroid kill the dinosaurs?
232 views / 0 likes - added
Around 66 million years ago, all non-avian dinosaurs went extinct. Was the culprit a 6-mile wide asteroid that collided with Earth? Or did other factors contribute to the dinosaurs’ die-off? Paleontologist Aki Watanabe looks at other theories for wh
• 02:57
### How do we know an asteroid hit Earth 66 million years ago?
304 views / 0 likes - added
One theory for why the dinosaurs went extinct is that an asteroid hit Earth at the end of the Cretaceous period. But since no one was alive to see it, how can we know that it really happened? Astrophysicist Jackie Faherty reveals how clues found in sedime
• 09:04
### What's inside Samsung Galaxy S8?
258 views / 1 likes - added
Does the new battery catch on fire? This is the first look into the soon to be released phone. Thanks to Techrax for the phone, check out his channel here: https://www.youtube.com/user/TechRax Subscribe and watch JerryRigEverything here: https://youtu.be/
• 01:38
### Our Senses: Touch, From Single Cell To Whiskers
329 views / 0 likes - added
Touch is perhaps the most primordial sense – even some single-celled organisms are able to sense pressure. Humans have many different types of touch receptors, including one that can also be found at the base of cat and mouse whiskers. OUR SENSES, a new e
• 01:48
### Our Senses: What Sluggish Sloths Tell Us About Balance
372 views / 1 likes - added
We don’t always think of balance as one of our senses, but scientists often consider it as essential as sight, hearing, smell, taste, and touch. For a species like the three-toed sloth, however, there’s little need for this sixth sense. Check out the Muse
• 06:27
### What's inside an Apple Helicopter?
303 views / 0 likes - added
We CUT OPEN our Apple Helicopter Toy! How does it move to your hand’s command? Watch our What's inside TOYS Playlist: https://www.youtube.com/playlist?list=PLk-NEiZ-NQeQKwPTCusq_UII0rYwxTl0s You can buy one of these here: http://amzn.to/2nzEBRR Want your
• 05:01
### The Butterfly Conservatory in 360
250 views / 0 likes - added
The Butterfly Conservatory is closing for the season on May 29, 2017! This annual favorite features up to 500 live, free-flying tropical butterflies from South, Central, and North America, Africa, and Asia. Housed in a vivarium that approximates their nat
• 07:16
### 8 Fast Growing Vegetables You Can Grow at Home in a Hurry
139 views / 0 likes - added
In this video, I give you 8 fast-growing vegetables you can grow at home in a hurry. Support me on Patreon: https://www.patreon.com/selfsufficientme (the top tier \$25 AU enables mentoring from yours truly via an exclusive VIP email where I will answer you
• 05:16
### Swimming With Giants 360
331 views / 0 likes - added
Earth’s oceans have been home to giant animals for hundreds of millions of years, but we know surprisingly little about their daily lives. Have you ever wondered what it would be like to swim with some of these giants of the deep? Dive into this video to
• 06:14
### What's inside a Fidget Cube?
535 views / 0 likes - added
We CUT OPEN a Fidget Cube! VERY interesting to see inside! Buy this Fidget Cube here: http://amzn.to/2oqMAyC Anxiety, Stress, restlessness, OCD, ADHD there are SO MANY people that could benefit from using this Fidget Cube. Even kids and toddlers have been
• 13:26
### What Color Is a Blue Whale?
129 views / 0 likes - added
Did you know theres been a blue whale model at the American Museum of Natural History for over 100 years? The huge icon hanging in the Milstein Hall of Ocean Life has been a visitor favorite for decades. But over time, the way we see whales has changed dr
• 02:26
### The Large Blue Butterfly Adopted By Ants | BBC Earth
325 views / 1 likes - added
An abandoned cocoon leads to a butterfly to new and unexpected parents.Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://bit.ly/PlanetEarthPlaylist Blue Planet http://bit.ly/BluePlanetPlaylist Planet Earth II http://bit.ly/PlanetEarthII
• 07:30
### Norway’s \$47BN Coastal Highway | The B1M
225 views / 0 likes - added
The Norwegian government are embarking on the largest infrastructure project in the country's history. For more by The B1M subscribe now: http://ow.ly/GxW7y Footage and images courtesy of the Norwegian Public Roads Administration, Vianova Plan og Trafikk
• 02:53
### Caterpillar Cocoon Timelapse | BBC Earth
287 views / 0 likes - added
Watch the first bizarre steps in this caterpillar's rebirth from bug to butterfly, in a timelapse showing the beginning of one of nature's most incredible metamorphosis.Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://bit.ly/PlanetEart
• 03:09
### Heath Fritillary Butterfly Hatching | BBC Earth
183 views / 0 likes - added
Once close to extinction, the White Admiral Butterfly & Heath Fritillary Butterfly now create one of Britains biggest butterfly spectacles.Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://bit.ly/PlanetEarthPlaylist Blue Planet http://b
• 07:42
### Tick Sticking, a Carpentry HACK (few people know)
189 views / 0 likes - added
Little demonstrates how to use a little know tool named a "tick stick or ticking stick". A tick stick is used to quickly and preciously copy and make a pattern of any odd or complicated shape where using a ruler would be challenging. The ticking stick was
• 03:42
### Frogs Hunting Butterflies | BBC Earth
263 views / 0 likes - added
To us the painted lady butterfly are delicate, living artwork. To these hungry frogs, they're prey waiting to be trapped by their incredible tongue.Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://bit.ly/PlanetEarthPlaylist Blue Planet
• 08:34
### Pond Scum Under the Microscope - Pondlife, Episode #1
207 views / 0 likes - added
We are surrounded by microscopic organisms that are too small to see with the naked eye. Follow Museum microbiologist Sally Warring as she reveals the invisible inhabitants of the green slime at the surface of a pond in Central Park. In this first episode
• 01:39
### Our Senses: How Mammals See the World In Many Colors
459 views / 0 likes - added
Humans see a variety of colors because our eyes have three types of cone cells. But things don't look quite as vivid for some of our fellow mammals—some see in two colors, others just in black and white. Color vision evolved in primates about 35 million y
• 06:21
### What Did a Baby T. rex Look Like?
241 views / 1 likes - added
Did you know that when Tyrannosaurus rex was a hatchling it was most likely covered in fluffy feathers? Go behind the scenes of the new exhibition T. rex: The Ultimate Predator, which opens March 11 at the American Museum of Natural History, with paleonto
• 03:07
### Nature's Scuba Divers: How Beetles Breathe Underwater | Deep Look
440 views / 1 likes - added
Bugs and beetles can’t hold their breath underwater like we do. But some aquatic insects can spend their whole adult lives underwater. How do they do it? Meet nature’s Scuba divers. They carry their air with them—in some cases, for a lifetime. SUBSCRIBE t
• 12:36
### FLYING Leaf Blower RC airplane Mk2
153 views / 0 likes - added
The awaited re attempt at leaf blower flight is now here! *spoilers* it flies! buy this leaf blower here (affiliate link) https://amzn.to/2XcbjbyMY WEBSITE:::::::::::: https://goo.gl/87uDQKAmazon affiliate links (these help grow the channel, every purchas
• 07:07
### This dog leash CHARGES your phone!!!?
144 views / 0 likes - added
For this weird project we decided to see if we could charge our phones with our dog leashes!What other crazy ideas do you have? Post it in the comments!Patreon: https://www.patreon.com/PetersripolINSTAGRAM:https://www.instagram.com/petersripolAmazon affil
• 11:38
### What Lives In Moss? - Pondlife, Episode #3
173 views / 0 likes - added
Microbes aren't just found in ponds. They're also abundant in and around plants and soils. Mosses, some of the oldest plants on land, are home to many species of microbes. In Episode 3 of Pondlife, Sally and fellow Museum scientist Michael Tessler travel
• 04:50
### Taking Great Photos (PHOTO SCHOOL) - Points of Interest (2019)
168 views / 0 likes - added
In this guide I'll teach you some basic but very important rules about composition. You can divide a picture using the rules of thirds and the points of interest. The rule of thirds is a common composition rule that's been used for many many years. Within
• 24:05
### The Power of Poop — AMNH SciCafe
260 views / 0 likes - added
Did you know that some of the bacteria living inside us are essential for our health? Gastroenterologist Ari Grinspan delves into the complex world of the microbiome in the human digestive system. He explains how transplanting bacteria from healthy people
• 03:40
### What Are Airplane Chemtrails, Really?
271 views / 0 likes - added
Some think the trails behind planes have mind-altering chemicals, but do they really? Here's the difference between chemtrails & contrails. Why Do So Many People Believe In Conspiracy Theories? - http://bit.ly/2a6VFGT Sign Up For The Seeker Newsletter Her
• 17:20 Popular
### TESTING VIRAL TikTok FOOD HACKS!!
894 views / 10 likes - added
GET YOUR MERCH https://lizzy.teamrar.com/ USE DISCOUNT CODE LIZZY FOR 5\$ OFF!! In today’s video, Lizzy Capri invites Team RAR including Stove’s Kitchen and Ryan Prunty to participate in a viral TikTok food hacks challenge. Lizzy invites her fr
• 07:15
### A Robot that stops Candy Thieves!
124 views / 0 likes - added
well isint this a fun way to give a five finger discount using science! MY WEBSITE:::::::::::: https://goo.gl/87uDQKAmazon affiliate links (these help grow the channel, every purchase gets me a small percentage of the sale, it costs you nothing though!) S
• 04:42
### Material Magic - Making Diamonds: Crash Course Kids #40.2
657 views / 1 likes - added
Did you know we can actually make diamonds in a lab? It's true! We can! And this is both really good and really cool. In this episode of Crash Course Kids, Sabrina shows us how materials scientists have done just that and why it's so important. Watch More
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• 03:02
### How Corals Hold Centuries of Ocean Climate Data
269 views / 0 likes - added
Before we can make a plan to protect our oceans from climate change, we need to know what they were like before human impact. We haven’t been collecting ocean data for very long, but luckily one ocean marine organism has been keeping records for millennia
• 03:45
### Measurement Mystery: Crash Course Kids #9.2
577 views / 0 likes - added
So now that we know what Properties are, how can we use them to figure things out? In this episode of Crash Course Kids, Sabrina uses them to solve the mystery of what she tripped over last night. This first series is based on 5th grade science. We're sup
• ### How Did Blue Whales Get So Big?
19 views / 0 likes - added
Meet the largest animal that ever livedthe blue whale. Whats so great about being big? You can move faster and farther, you can avoid predators, and you can have a longer lifespan. In Part One of our four-part Giants of the Sea series, youll explore some
• 26:24
### The Milky Way as You’ve Never Seen It Before – AMNH SciCafe
170 views / 0 likes - added
Fly through the galaxy with Museum astrophysicist Jackie Faherty, who takes us on a dazzling tour of new research and data visualizations made possible by recently released data from the Gaia space telescope. In April 2018, the European Space Agency’s Gai
• 05:20
### Order Of Operations: PEMDAS | Arithmetic Properties | Pre-Algebra | Khan Academy
482 views / 0 likes - added
Work through another challenging order of operations example with only positive numbers. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/pre-algebra/order-of-operations/order_of_operations/e/order_of_operations
• 11:57
### CHEAP RC SAILBOAT 2x4 challenge!
130 views / 0 likes - added
Check out these sneakshttp://vessi.co/shoesailDiscount Code - shoesail we made some 2x4 sailboats!MY WEBSITE:::::::::::: https://goo.gl/87uDQKAmazon affiliate links (these help grow the channel, every purchase gets me a small percentage of the sale, it co
• 04:37
### Material World: Crash Course Kids #40.1
568 views / 0 likes - added
So, we know what materials are, but can we make new materials? Or improve the materials we already have? In this episode of Crash Course Kids, Sabrina shows us how Material Scientists are working on these two things today. Also, your cell phone holds the
• 03:37 Popular
### The Science Of Lunch: Crash Course Kids #15.2
2,648 views / 2 likes - added
Even an empty lunch sack is useful to science. You can examine it and come up with some traits. In this episode, Sabrina chats about things like malleability, hardness, conductivity, and magnetism. And all with lunch! This first series is based on 5th gra
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• 13:00
### FLYING COKE powered airplane
169 views / 0 likes - added
We tried a cola powered soda plane!@Jabrils @Tom Stanton @William Osman @Allen Pan - Sufficiently Advanced @NileRed MY WEBSITE:::::::::::: https://goo.gl/87uDQKAmazon affiliate links (these help grow the channel, every purchase gets me a small percentage
• 13:48
### Water Dropping RC Airplane
133 views / 0 likes - added
well it definitely dispenses water! William's videohttps://www.youtube.com/watch?v=a6q3HdduUs4&t=339sAllen's videohttps://www.youtube.com/watch?v=CYf1SjCJuT0MY WEBSITE:::::::::::: https://goo.gl/87uDQKAmazon affiliate links (these help grow the channel, e
• 04:20
### Oobleck And Non-Newtonian Fluids: Crash Course Kids #46.1
672 views / 1 likes - added
Ever heard of Oobleck? How about Non-Newtonian fluids? Well, today Sabrina is going to show us that things can sometimes behave like a solid, and sometimes like a liquid depending on how much force is applied to them. In this episode of Crash Course Kids,
• 03:55
### Wood, Water, And Properties: Crash Course Kids #15.1
586 views / 1 likes - added
Quick, think of three words to describe yourself. TIME'S UP! What did you think of? Chances are you thought of descriptive words that we call Properties. In this episode of Crash Course Kids, Sabrina talks about how properties help us understand objects.
• 10:29
### What's inside Steph Curry's Basketball Shoes?
436 views / 0 likes - added
We CUT OPEN Steph's Under Armour 5's, 2's and 1's!! Check out our Nike video with Kevin Durant: https://youtu.be/5LMY2iR_dLY Who do you think will win the NBA Finals? Check out the reveal of the Curry 5's here: https://www.youtube.com/watch?v=TWt1FcU0OzM&
• 04:00
### Hunting For Properties: Crash Course Kids #9.1
530 views / 1 likes - added
Remember pre-school? If not, IT WAS SO MUCH EASIER! But when you were stacking blocks and figuring out which block went into which shaped hole, you were learning about properties. In this episode of Crash Course Kids, Sabrina talks about what properties a
• 04:46
### Normal Stuff In Not-So-Normal Places: Crash Course Kids 46.2
563 views / 0 likes - added
So, what happens to normal stuff (like water) when it goes to not so normal places? What happens if you take a glass of water to the top of Mt. Everest? Or Space? In this episode of Crash Course Kids, Sabrina shows us how matter is affected by different p
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• 12:20
### FLYING MONSTER Truck RC airplane car!
205 views / 1 likes - added
This toy monster truck was asking for the flight treatment especially with those big foam tires!"Barnstormer" Theme Music created by:https://www.youtube.com/sweetlouphotography Lyrics by: Sam Foskuhl Affiliate link to monster truck: https://amzn.to/2qqO2q
• 10:54
### Can AIRPLANES plant ROCKET trees??? #TeamTrees
159 views / 0 likes - added
#teamtrees today we're gonna use the aerolite to plant some trees with these "B0mb" shaped tree things!donate here:https://teamtrees.org/other channels mentionedhttps://www.youtube.com/watch?v=orRjAZ8XasM&t=518shttps://www.youtube.com/watch?v=NEwBxcGAJwI&
• 08:15
### How British suffragettes fought for the vote
74 views / 0 likes - added
In 1913, suffragette Emily Davison disrupted a major horse race in the name of winning British women the vote. Become a Video Lab member! http://bit.ly/video-labBritish suffragettes in the early 20th century used spectacle and drama to draw attention to t
• 05:00
### No One Can Agree On The Color Of This Mysterious Dichromatic Liquid!
174 views / 0 likes - added
Get your Action Lab Box Now! https://www.theactionlab.com/ In this video I show you an amazing liquid that changes color depending on how you look at it. This property is called dichromatism. I show you how this works and why we can perceive different col
• 04:32
### Organizing Properties: Crash Course Kids #35.1
502 views / 1 likes - added
Have you ever thought about all the different kinds of groups you’re a part of? Like, there’s the friends you hang out with and your family, your hockey team, your Crash Course fan club, and that’s just for starters! And even though these groups are total
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• 02:50
### Why Some Molecules Have Evil Twins
329 views / 0 likes - added
A tiny change in a molecule’s geometry completely changes its effects on the human body. How We Made This Video (Ever & David Talk About Drugs): https://youtu.be/tefxgYP0BVM Thanks also to our supporters on https://www.patreon.com/MinuteEarth ____________
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• 07:38
### Introduction To Electric Charge
289 views / 0 likes - added
A simple and easy introduction to electricity. It covers electric charge, the discovery of electricity, the nature of electricity and some of the more technical stuff for the classroom for you students. Hi! I'm Jade. I make physics videos that will make y
• 10:52 Popular
### AMAZING GALLIUM FIDGET SPINNER (Spins and Melts in my Hand!)
952 views / 12 likes - added
Have you ever seen fidget spinners made out of gallium? In this video titled MAKING A GALLIUM FIDGET SPINNER I actually make an incredible gallium edc hand spinner that spins and melts and actually is just like a real handspinner! in this DIY How to Video
• 02:55
### AHHHHHHH!! How Blood-Curdling Screams Affect Your Brain
401 views / 0 likes - added
Hearing someone scream immediately sends your body into a cascade of defense responses, but why? What's happening in the brain? The Science Behind Our Fear Of Clowns - https://youtu.be/ahTpbFwQxrE Get 20% off http://www.domain.com domain names and web hos
• 10:34 Popular
### How to Make Liquid Metal Fidget Spinners ~ Advanced How to make a Fidget Spinner DIY Tutorial
1,260 views / 13 likes - added
Have you ever seen fidget spinners made out of gallium? In this video titled MAKING A GALLIUM FIDGET SPINNER FULL GALLIUM SPINNER VIDEO: https://www.youtube.com/watch?v=hSVOcbCGG4I I actually make an incredible gallium edc hand spinner that spins and melt
• 02:22
### 5 everyday inventions you didn’t know came from DARPA
347 views / 0 likes - added
A lot of things you use on a daily basis have their roots in DARPA projects that were intended for military purposes. Sharon Weinberger, the author of the book "The Imagineers of War," describes a few of these inventions that you might not have been aware
• 10:34
### Primitive Technology: Round hut
292 views / 0 likes - added
I built a round hut using palm thatch and mud walls to replace the damaged A-frame hut built a few months ago. The A frame hut was damaged due to torrential rain and poor design elements considering the wet conditions. The thatch had rotted in the part of
• 07:05
### Is it Possible to Cut Molecules? Breaking Molecules in a Blender Experiment
393 views / 0 likes - added
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• 03:46
### Why Dog Breeds Look So Different But Cats Don’t
384 views / 2 likes - added
Ever notice how dog breeds look so different from one another? A Maltese weighs about seven pounds, while the English mastiff can weigh up to 240 pounds. It's not just weight, either. Just look at the difference in shape between the dachshund and the Germ
• 03:29
### Who Owns Your DNA? (YOU DON’T)
233 views / 0 likes - added
Your DNA defines you, but do you REALLY own it? Watch more: Is It Possible To Stay Young Forever? ►► https://www.youtube.com/watch?v=hjWf0w-awh0&list=PL8L0MzSk_V6JtEDRfRMyb6rFd1acqYSlO&index=130&t=0s Subscribe: https://bit.ly/SubLifeNoggin | Get your excl
• 06:25
### Human Population Through Time
286 views / 0 likes - added
It took 200,000 years for our human population to reach 1 billion—and only 200 years to reach 7 billion. But growth has begun slowing, as women have fewer babies on average. When will our global population peak? And how can we minimize our impact on
• 04:34
### How Graphene Could Help Us Build Bigger and Better Quantum Computers
62 views / 0 likes - added
Quantum computers can solve problems in seconds that would take "ordinary" computers millennia, but their sensitivity to interference is majorly holding them back. Now, researchers claim theyve created a component that drastically cuts down on error-induc
• 06:24
### Can You Solve The Missing Sock Puzzle?
358 views / 0 likes - added
A statistician keeps a simple wardrobe. He only purchases pairs of black socks and white socks, and he keeps all of the socks in a pile in the drawer. Recently one of the socks was lost in the laundry. The socks have a mathematical property. If you select
• 05:25
### If the Coronavirus is Mutating, What Does That Mean For Us?
108 views / 0 likes - added
You've probably heard that SARS-CoV-2, the virus responsible for the COVID-19 pandemic, is mutating. But what exactly does this mutation mean for us? Subscribe to Seeker! http://bit.ly/subscribeseeker Watch more Elements! http://bit.ly/ElementsPlaylist Vi
• 00:55
### Solar Powered Toy Bell Ringer- solar motor - no electric = Homemade Science with Bruce Yeany
388 views / 0 likes - added
Here is my Solar motor being used for more work. Not much sound but I've had fun trying to find work for it. You can see more about solar motors on my other video at: https://youtu.be/VQqpnAKf9cM These devices are sometimes referred to as solar motors or
• 01:07
### Solar Powered Toy Hammer-solar motor-no electric = Homemade Science with Bruce Yeany
461 views / 1 likes - added
Here is my Solar motor being put to work, can it hammer a nail, or even break an egg? No, but it it's fun coming up with ideas of jobs for it to do. You can see more solar motors on my other video at: https://youtu.be/VQqpnAKf9cM These devices are sometim
• 05:04
### Space Volcanoes - Shelf Life 360
431 views / 0 likes - added
Here on Earth, volcanic eruptions are dramatic manifestations of our dynamic planet. Elsewhere in our solar system, awe-inspiring extraterrestrial volcanoes—both active and extinct—provide clues to planetary formation and hints of how life may
• 08:20
### Primitive Technology: New area starting from scratch
333 views / 0 likes - added
I bought a new property to shoot primitive technology videos on. The new area is dense tropical rainforest with a permanent creek. Starting completely from scratch, my first project was to build a simple dome hut and make a fire. First, I took some wood,
• 03:48
### The Worlds First Photo of Quantum Entanglement Could Disprove Einsteins Theory
172 views / 0 likes - added
Einstein dubbed the idea of quantum entanglement as "spooky action at a distance." Now for the first time ever, scientists have taken a picture of it. Subscribe to Seeker!http://bit.ly/subscribeseeker Watch more Elements! http://bit.ly/ElementsPlaylistTod
• 05:28
### Dog Owner Life VS. Cat Owner Life
179 views / 0 likes - added
Cats VS Dogs: Lifestyle DifferencesYou may be a dog person or you may be a cat person. You might also be an animal lover who has both a cat and a dog. But whether you are around cats or dogs, you will notice that the way a dog owner sees their life is com
• 34:40
### How I Boarded a US NAVY NUCLEAR SUBMARINE in the Arctic (ICEX 2020) - Smarter Every Day 237
87 views / 0 likes - added
Get 1st Audiobook + access to monthly selection of Audible Originals for free when you try Audible for 30 days https://www.audible.com/smarter or TXT smarter to 500500Upcoming videos will explore what life on a submarine is like. Click here if you want to
• 09:03 Popular
### CHRISTMAS CARTOON: Rudolph The Red Nosed Reindeer (1948) [HD 1080] [Cartoons for Children]
767 views / 0 likes - added
Rudolph the Red-Nosed Reindeer (1948 film), a 1948 animated short film by Max Fleischer based on the Robert L. May poem/story. The last cartoon ever produced by Max Fleischer who produced the Popeye the Sailor man, Betty Boop and Koko the Clown cartoons f
• 08:45
### Fractals: How to Decode the Geometry of Chaos І The Great Courses
309 views / 0 likes - added
Try a free month trial of The Great Courses Plus here: https://www.thegreatcoursesplus.com/special-offer?utm_source=US_OnlineVideo&utm_medium=SocialMediaEditorialYouTube&utm_campaign=136270 Let's begin by clarifying what's distinctive about fractals by co
• 03:43
### Solar motor--Free energy motor-- not electrical /// Homemade Science with Bruce Yeany
570 views / 1 likes - added
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>> View property web videos | 15,492 | 61,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-25 | latest | en | 0.708945 |
http://www.lrde.epita.fr/dload/vcsn/2.3/notebooks/automaton.reduce.html | 1,603,770,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893011.54/warc/CC-MAIN-20201027023251-20201027053251-00677.warc.gz | 164,462,221 | 51,206 | automaton.reduce¶
Compute an equivalent automaton with a minimal number of states.
Preconditions:
• its labelset is free
• its weightset is a division ring or $\mathbb{Z}$.
Postconditions:
• the result is equivalent to the input automaton
• the result is both accessible and co-accessible
Caveats:
• The reduction algorithm may well produce an automaton which will look more 'complicated' than the original one, especially when the latter is already of minimal dimension. See examples in $\mathbb{Z}$ below.
• The computation of reduced representations implies the exact resolution of linear systems of equations which becomes problematic when the dimension of the systems grows.
Examples¶
In [1]:
import vcsn
In $\mathbb{Q}$¶
In [2]:
c = vcsn.context('lal_char(ab), q')
a = c.expression('<2>(<3>a+<5>b+<7>a)*<11>', 'associative').standard()
a
Out[2]:
In [3]:
a.reduce()
Out[3]:
To be contrasted with the results from automaton.minimize:
In [4]:
a.minimize()
Out[4]:
In $\mathbb{Z}$¶
The same automaton, but on $\mathbb{Z}$, gives:
In [5]:
c = vcsn.context('lal_char(ab), z')
a = c.expression('<2>(<3>a+<5>b+<7>a)*<11>', 'associative').standard()
a.reduce()
Out[5]:
Caveats¶
In $\mathbb{Z}$ the result may be suprising. For instance:
In [6]:
%%automaton -s a
context = "lal_char(abc), z"
$-> 0 0 -> 0 a, b 0 -> 1 b 0 -> 2 <2>b 2 -> 2 <2>a, <2>b 2 -> 1 <2>b 1 -> 1 <4>a, <4>b 1 ->$
In [7]:
a.reduce()
Out[7]:
In [8]:
a = vcsn.context('lal_char(ab), z').expression('[ab]*b(<2>[ab])*').automaton()
a
Out[8]:
In [9]:
a.reduce()
Out[9]:
The result is not canonical. Moreover, in $\mathbb{Z}$ the algorithm is not idempotent. In fields, the implementation ensures that the reduction of reduced automaton results in an isomorphic automaton.
Algorithm¶
The core of the algorithm is the left-reduction procedure.
The reduction algorithm of an automaton $\mathcal{A}$ is left-reduction $\circ$ transpose $\circ$ left-reduction $\circ$ transpose ($\mathcal{A}$).
The left-reduction algorithm deals with weighted sets of states, seen as vectors of a vector space with the same dimension as the automaton. It computes greedily a basis of the smallest subspace which contains every accessible weighted set of states.
For efficiency, the basis is scaled, initialized with the initial vector.
Each time a vector is added to the bases, for each letter, a new candidate is built as the successor of the vector by the letter.
Every candidate is reduced with respect to the basis. If the result is not null, the new vector is added to the basis; the pivot of this vector is chosen w.r.t. heuristics for increase numeric stability, and, in case of fields, the vector is normalized in such a way that the pivot equals 1.
In $\mathbb{Z}$, the reduction of a new vector w.r.t. the basis may modify the basis: let $b$ be a vector of the basis with pivot $b_i$, $v$ a new vector, and let $\alpha$ and $\beta$ be such that $\alpha b_i+ \beta v_i=\mathsf{gcd}(b_i,v_i)$; then $$\left[\begin{array}{c}b'\v'\end{array}\right]¶ \left[\begin{array}{cc}\alpha & \beta \-\frac{v_i}{\mathsf{gcd}(b_i,v_i)} & \frac{b_i}{\mathsf{gcd}(b_i,v_i)} \end{array}\right] \left[\begin{array}{c}b\v\end{array}\right]$$
In fields, once the scaled basis is built, a "bottom-up" procedure is applied to make more entries of the vector of the basis equal to 0. If the input automaton is reduced, the resulting basis is then the canonical basis.
Finally, an automaton is built where each state is a vector of the basis. | 1,003 | 3,497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.674644 |
https://www.jiskha.com/display.cgi?id=1335756589 | 1,501,118,085,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00569.warc.gz | 812,688,170 | 4,483 | # Econ/stats
posted by .
The following data represent the number of flash drives sold per day at a local computer shop and their prices.
Price (x) Units Sold (y)
\$34 3
36
4
32
6
35
5
30
9
38
2
40
1
a.
Develop a least-squares regression line and explain what the slope of the line indicates.
b.
Compute the coefficient of determination and comment on the strength of relationship between x and y.
c.
Compute the sample correlation coefficient between the price and the number of flash drives sold. Use = 0.01 to test the relationship between x and y.
• Econ/stats -
rey
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More Similar Questions
Post a New Question | 846 | 3,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2017-30 | longest | en | 0.92452 |
https://www.edumple.com/cbse-class-7/science/playing-with-spherical-mirrors3/notes/rc@gm_1130 | 1,719,162,903,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00184.warc.gz | 648,642,661 | 12,000 | Mirror :
It is a highly polished surface, which is quite smooth and capable of reflecting a good fraction of light from its surface.
(a) Object :
Anything which gives out light rays (either its own or reflected) is called an object.
(b) Image :
The reproduction of object formed by mirror or lens is called an image.
ReaL and virtual images :
The image which can be obtained on a screen is called a real image. It is formed when light rays, after reflection, actually intersect each other. It is always inverted.
The image which cannot be obtained on a screen is called a virtual image. It is formed when light rays, after refection, intersect when extended in backward direction. Its is always erect.
Plane Mirror:
Image Formation by Plane Mirror :
Consider a point source of light placed at a point O at a distance u in front of the plane mirror. Light rays leave the source and are reflected from the mirror. After reflection, the rays diverge but they appear to come from a point I located behind the mirror. Point I is called the image of the object O. Point I is at a distance v behind the mirror.
Character Stics of Image :
(i) Image is virtual, erect and of the same size as object.
(ii) It is as far behind the mirror as the object is in front of it
(iii) Laterally inverted : When left appears right and right appears left.
Note :
(i) Minimum size of the mirror required to see full image of a person is at least half of his own height.
(ii) If object moves with a speed V towards plane mirror then image moves with a speed 2V towards object.
(iii) If mirror moves with a speed V towards stationary object then image moves with a speed 2V towards object.
Uses of Plane Mirror
* In looking mirrors, hair saloons, reflecting periscopes & kaleidoscopes.
Spherical Mirrors :
Mirrors, whose reflecting surfaces are spherical or curved, are called spherical mirrors. These are of two types:
(i) Concave mirror: If the reflecting surface of the spherical mirror is curved inwards, it is called a concave mirror. The image formed by a concave mirror can easily be taken on the screen. You must have played with concave mirror to obtain image of sun on the ground or wall by positioning it to in a specific way. It means a concave mirror can form real or virtual image. In concave mirror when a parallel beam of light after reflection from a concave mirror converges at a point in front of the mirror. Due to this it is known as converging mirror.
(ii) Convex mirror: If the reflecting surface of the spherical mirror is curved outwards, it is called a convex mirror. It means a convex mirror always forms virtual image. In convex mirror when a parallel beam of light after reflection from a convex surface diverges and the rays do not meet. Due to this it is known as diverging mirror.
Some terms related to Spherical Mirrors :
(i) Pole : The central point of a mirror is called its pole.
(ii) Centre of curvature : The centre of the sphere of which the mirror is a part, is called centre of curvature.
(iii) Radius of curvature : The radius of the sphere of which the mirror is a part, is called radius of curvature.
(iv) Principal axis : The straight line joining the pole and the centre of curvature is called the principal axis.
(v) Focal length : The distance between the pole and the focus is called the focal length. The focal length is half of the radius of curvature.
(vi) Aperture : The size of the mirror is called its aperture.
(vii) Focus point :
(a) In concave mirror when a parallel beam of light after reflection from a concave mirror converges at a point in front of the mirror. This point (F) is the focus of a concave mirror and it is real.
(b) when a parallel beam of light after reflection from a convex surface diverges and the rays do not meet. However on producing backward, the rays appear to meet at a point behind the mirror. This point is focus of the convex mirror and it is virtual.
Uses of Concave Mirror :
(i) Concave mirror is used by doctors for examining eyes, ears, nose and throat.
(ii) They are also used by dentist to see an enlarged image of the teeth.
(iii) The reflector of torches, headlights of cars and scooters are concave in shape.
Uses of Convex Mirror :
(i) Convex mirrors are used as rear view mirrors to see the vehicle coming from behind. These mirror covers a wider view and form upright small size (diminished) image.
(ii) They are used as reflector in street lamps so as to diverge light over a large area. | 1,065 | 4,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.934383 |
https://wiki.tcl-lang.org/revision/Playing+with+rationals?V=1 | 1,652,964,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00687.warc.gz | 710,351,865 | 3,629 | ## Version 1 of Playing with rationals
Updated 2005-03-23 23:08:25 by suchenwi
if 0 {Richard Suchenwirth 2005-03-22 - Rational numbers, a.k.a. fractions (see Fraction math), can be thought of as pairs of integers {numerator denominator}, such that their "real" numerical value is numerator/denominator (and not in integer nor "double" division!). They can be more precise than any "float" or "double" numbers on computers, as those can't exactly represent any fractions whose denominator isn't a power of 2 - consider 1/3 which can not at any precision be exactly represented as floating-point number to base 2, nor as decimal fraction (base 10), even if bignum. Reading in SICP once more, I wanted to play with rationals in Tcl again - so here's another evening fun project.
An obvious string representation of a rational is of course "n/d". The following "constructor" does that, plus it normalizes the signs, reduces to lowest terms, and returns just the integer n if d==1:}
``` proc rat {n d} {
if {!\$d} {error "denominator can't be 0"}
if {\$d<0} {set n [- \$n]; set d [- \$d]}
set g [gcd \$n \$d]
set n [/ \$n \$g]
set d [/ \$d \$g]
expr {\$d==1? \$n: "\$n/\$d" }
}```
if 0 {Conversely, this "deconstructor" splits zero or more rational or integer strings into num and den variables, such that [ratsplit 1/3 a b] assigns 1 to a and 3 to b:}
``` proc ratsplit args {
foreach {r _n _d} \$args {
upvar 1 \$_n n \$_d d
foreach {n d} [split \$r /] break
if {\$d eq ""} {set d 1}
}
}```
#-- Four-species math on "rats":
``` proc rat+ {r s} {
ratsplit \$r a b \$s c d
rat [+ [* \$a \$d] [* \$c \$b]] [* \$b \$d]
}
proc rat- {r s} {
ratsplit \$r a b \$s c d
rat [- [* \$a \$d] [* \$c \$b]] [* \$b \$d]
}
proc rat* {r s} {
ratsplit \$r a b \$s c d
rat [* \$a \$c] [* \$b \$d]
}
proc rat/ {r s} {
ratsplit \$r a b \$s c d
rat [* \$a \$d] [* \$b \$c]
}```
if 0 { Arithmetical helper functions can be wrapped with func if they only consist of one call of expr:}
` proc func {name argl body} {proc \$name \$argl [list expr \$body]}`
` func gcd {u v} {\$u? [gcd [% \$v \$u] \$u]: abs(\$v)}`
#-- Binary expr operators exported:
` foreach op {+ * / %} {func \$op {a b} \\$a\$op\\$b}`
#-- "-" can have 1 or 2 operands:
` func - {a {b ""}} {\$b eq ""? -\$a: \$a-\$b}`
#-- a little tester reports the unexpected:
``` proc ? {cmd expected} {
catch {uplevel 1 \$cmd} res
if {\$res != \$expected} {puts "\$cmd -> \$res, expected \$expected"}
}```
#-- The test suite should silently pass when this file is sourced:
``` ? {rat 42 6} 7
? {rat 1 -2} -1/2
? {rat -1 -2} 1/2
? {rat 1 0} "denominator can't be 0"
? {rat+ 1/3 1/3} 2/3
? {rat+ 1/2 1/2} 1
? {rat+ 1/2 1/3} 5/6
? {rat+ 1 1/2} 3/2
? {rat- 1/2 1/8} 3/8
? {rat- 1/2 1/-8} 5/8
? {rat- 1/7 1/7} 0
? {rat* 1/2 1/2} 1/4
? {rat/ 1/4 1/4} 1
? {rat/ 4 -6} -2/3```
if 0 { See also Fraction Math | 1,052 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-21 | latest | en | 0.714224 |
https://mathematica.stackexchange.com/questions/249258/plot-findroot-with-very-complex-variables | 1,716,418,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00300.warc.gz | 326,074,473 | 41,203 | # Plot "Findroot" with very complex variables
I ran into an extremely complex "zero problem" and don't know how to solve and plot it. Can anyone help with this, please? [Pi]s and [Pi]r are objective functions, pd, po, qd, qo, and [Phi] (0<[Phi]<1) are variables. pd, po, qd, qo are determined to maximize [Pi]s and [Phi] is determined to maximize [Pi]r. This problem needs to be addressed by backward induction. First, I need to solve pd and po. Then I intend to solve qd and qo (with some constraints), but results are so much complicated and thus I use "Reduce" afterward to select solutions. Last, [Phi] can be obtained by using the above solutions pd, po, qd, and qo. I want to Plot [Phi] when the parameter c is within a certain range, say 0<c<0.1. I tried several times but get no feasible solutions, which is weird. Then how should I do? I am a beginner so really hope to get some answers. Thanks a lot!
dD := (pd - po)/(qd - qo) - pd/qd
do := 1 - (pd - po)/(qd - qo)
\[Pi]s :=
dD (pd - 1/2 (qd*qd) - c) + do (1 - \[Phi]) (po - 1/2 (qo*qo))
\[Pi]r := do (\[Phi]) (po - 1/2 (qo*qo))
s.t. qo > qd > 0 && 0 < c < pd - 1/2 (qd*qd) &&
1 - (pd - po)/(qd - qo) > 0 && pd/qd - po/qo < 0 &&
0 < \[Phi] < 1, \[Phi]\[Element]Reals
Step 1 FullSimplify[Solve[{D[\[Pi]s, po] == 0, D[\[Pi]s, pd] == 0}, {po, pd}]]
which gives po -> (qo (2 qo (2 + qo) (-1 + \[Phi]) + qd (-4 + qo (-2 + \[Phi])) (-1 + \[Phi]) + 2 c \[Phi] + qd^2 \[Phi]))/(2 qd (-2 + \[Phi])^2 + 8 qo (-1 + \[Phi])), pd -> (-2 c qd (-2 + \[Phi]) - qd^3 (-2 + \[Phi]) + 4 c qo (-1 + \[Phi]) + 2 qd^2 (-1 + \[Phi]) (-2 + qo + \[Phi]) + qd qo (-1 + \[Phi]) (4 + (-2 + qo) \[Phi]))/(2 qd (-2 + \[Phi])^2 + 8 qo (-1 + \[Phi]))
Step 2 Solve[{D[\[Pi]s, qo] == 0, D[\[Pi]s, qd] == 0}, {qo, qd}] (*very complex*)
Step 3 Reduce[qo > qd > 0 && 0 < c < pd - 1/2 (qd*qd) && 1 - (pd - po)/(qd - qo) > 0 && pd/qd - po/qo < 0 && 0 < \[Phi] < 1, \[Phi], Reals] (*also complex*)
Step 4 g[c_?NumericQ] = \[Phi] /. FindRoot[D[\[Pi]r, \[Phi]] == 0, {\[Phi], 0, 0, 1}];
Assuming[\[Phi] \[Element] Reals, Plot[{g[c]}, {c, 0, 0.1}]]
• The line beginning s.t. qo>qd>0 ... I guess s t are not further variables. Is this a statement of conditions? Your step 2 does not solve for me in a reasonable time. Did you actually get a solution?
– Hugh
Jun 6, 2021 at 16:44
• The letters "s.t." proceed with constraints that the functions must follow, namely qo>qd>0 ... . And I can get 16 pairs of solutions in Step 2, however, only 8 of them may be feasible. And then, it is extremely hard to proceed. So, I am wondering if there is any other way to solve this kind of problem (Plot [phi] and see how it changes with c). Jun 6, 2021 at 16:56
• Crossposted here. Jun 6, 2021 at 19:14
I have had to make several assumptions about what you are doing but this seems to be a way forward. First I am going to ignore your constraints to begin with and just try and find some solutions.
I start by defining your equations and then solving the equations in your Step 1. I then look at the equations you wish to solve in your Step 2 and substitute in the solutions from Step1. This is what I get
ClearAll[dD, do, πs, πr]
dD = (pd - po)/(qd - qo) - pd/qd;
do = 1 - (pd - po)/(qd - qo);
πs = dD (pd - 1/2 (qd*qd) - c) +
do (1 - ϕ) (po - 1/2 (qo*qo));
πr = do (ϕ) (po - 1/2 (qo*qo));
sol1 = FullSimplify[
Solve[{D[πs, po] == 0, D[πs, pd] == 0}, {po, pd}]];
eqn1 = {D[πs, qo] == 0, D[πs, qd] == 0} /. First@sol1 //
FullSimplify
The equation I get out is not too bad
{((2 c (-2 + ϕ) - 2 qd (-2 + ϕ) + qd^2 (-2 + ϕ) -
2 (-2 + qo) qo (-1 + ϕ)) (-1 + ϕ) (2 c (-2 + ϕ) +
qd^2 (-2 + ϕ) + 2 qo (-2 + 3 qo) (-1 + ϕ) +
2 qd (-2 + ϕ) (1 - 2 qo + (-1 + qo) ϕ)))/(
qd (-2 + ϕ)^2 + 4 qo (-1 + ϕ)) == 0, (
qo (-1 + ϕ) (4 c +
qd (2 qd - 2 qo + (-2 + qo) ϕ)) (-2 c (qd (-2 + ϕ)^2 +
2 qo (-1 + ϕ)) +
qd (qd^2 (-2 + ϕ)^2 + 6 qd qo (-1 + ϕ) +
qo (-1 + ϕ) (-2 qo + (-2 + qo) ϕ))))/(
qd (qd (-2 + ϕ)^2 + 4 qo (-1 + ϕ))) == 0}
I now try and solve this equation and determine the number of solutions
sol2 = Solve[eqn1, {qo, qd}];
Dimensions@sol2
(* {16, 2} *)
So lets look at some of these solutions. Solutions 11 to 14 seem very complicated but the rest are reasonable. Here are the reasonable ones.
sol2[[{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 16}]] // TableForm
Now let's look at the equation you wish to find roots for. Start by substituting for the first set of solutions.
eqn2 = D[πr, ϕ] /. First@sol1 // Simplify
(* (qo (2 c (-2 + ϕ) - 2 qd (-2 + ϕ) + qd^2 (-2 + ϕ) -
2 (-2 + qo) qo (-1 + ϕ)) (qd (4 + qo (-2 + ϕ) -
4 ϕ) + 4 qo (-1 + ϕ) - 2 qo^2 (-1 + ϕ) +
2 c ϕ + qd^2 ϕ))/(4 (qd (-2 + ϕ)^2 +
4 qo (-1 + ϕ))^2) *)
Now we want to substitute in the solutions for the second set of equations. Start by looking at the simple solutions in the 16
eqns3 = Table[{n,
eqn2 /. sol2[[n]]}, {n, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15,
16}}] // Simplify;
eqns3 // TableForm
So these are trivial. Always zero or never zero.
So now you have to look at the remaining equations. Start by looking at solution 11. Complex solutions are possible so first look at the real and imaginary parts.
eqn11 = eqn2 /. sol2[[11]];
Plot3D[Re[eqn11], {c, 0, 0.1}, {ϕ, -100, 10},
PlotRange -> {All, All, {-0.1, 0.1}}]
Plot3D[Im[eqn11], {c, 0, 0.1}, {ϕ, -100, 10},
PlotRange -> {All, All, {-0.1, 0.1}}]
The real part is always greater than zero and the imaginary part always equals zero. So no solutions equalling zero here.
I suggest you go and look at the remaining 10 to 14 solutions and see if you get some possibilities there.
Hope that helps.
• Thank you so much! But can I ask what "First@" means in the first part of your code? Jun 6, 2021 at 22:42
• Please look up in help. First gives the first element of a list. Solve gives a list of lists and in this case there is only one element which I want.
– Hugh
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3 added 254 characters in body
General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi.
It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts.
But in general, we can say that for the category of sets, given a pair of functions
$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$
the canonical arrow
$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$
is monic. (Lemma to be proved.) Assuming this, suppose for example that $f_2: P \to B$ is epic. Then the pushout of $f_2$ along $f_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as
$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$
we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show.
To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path
$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$
where for each $k$, either $p_k$ belongs to $P$ and the arrows out of $p_k$ alternate between $f_1$ and $f_2$, or we are in a "holding pattern" where $p_k$ belongs to $A$ or $B$ and the two arrows out of $p_k$ are both identities. Now it is not hard to convince yourself that given $(x, y)$ in $A^2$ or $B^2$, and $(x', y')$ in $A^2$ or $B^2$, if there is a zig-zag path from $x$ to $x'$, and a zig-zag path from $y$ to $y'$, then there is a zig-zag path from $(x, y)$ to $(x', y')$ with respect to the pair of maps
$$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B;$$
all we do is pair together zig-zag paths in the separate $x$- and $y$-components. (Note that if the zig-zag to get from $y$ to $y'$ is longer than the zig-zag from $x$ to $x'$, we can always insert a holding pattern in the $x$-component so that the zig-zag in the $y$-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map $\phi$ is monic.
Edit: In answer to vincenzoml's questions in the comments, I wrote up a proof of the more general desired result which applies to a general $\infty$-pretopos, here.
2 added 494 characters in body
General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi.
It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts.
But in general, we can say that for the category of sets, given a pair of functions
$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$
the canonical arrow
$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$
is monic. (Lemma to be proved.) Assuming this, suppose for example that $f_2: P \to B$ is epic. Then the pushout of $f_2$ along $f_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as
$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$
we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show.
To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path
$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$
where for each $k$, either $p_k$ belongs to $P$ and the arrows out of $f_{i_k}$ p_k$alternate between$f_1$and$f_2$, and the or we are in a "holding pattern" where$p_i$belong p_k$ belongs to $P$. A$or$B$and the two arrows out of$p_k$are both identities. Now it is not hard to convince yourself that given$(x, y)$in$A^2$or$B^2$, and$(x', y')$in$A^2$or$B^2$, if there is a zig-zag path from$x$to$x'$, and a zig-zag path from$y$to$y'$, then there is a zig-zag path from$(x, y)$to$(x', y')$with respect to the pair of maps $$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B.$$ B;$$all we do is pair together zig-zag paths in the separate x- and y-components. (Note that if the zig-zag to get from y to y' is longer than the zig-zag from x to x', we can always insert a holding pattern in the x-component so that the zig-zag in the y-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map \phi is monic. 1 General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi. It's very easy to see the answer is no for general pushouts: since a coproduct A + B is the pushout of a pair A \leftarrow 0 \to B, and since the squaring functor preserves 0, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, a^2 + b^2 is generally not equal to (a+b)^2; therefore squaring cannot preserve coproducts. But in general, we can say that for the category of sets, given a pair of functions$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$the canonical arrow$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$is monic. (Lemma to be proved.) Assuming this, suppose for example that f_2: P \to B is epic. Then the pushout of f_2 along f_1 is also epic (well-known fact); let A \to C denote this epi (so C is shorthand for A +_P B). Then A^2 \to C^2 is also epic, and since this obviously factors as$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$we see \phi is also epi. But it is monic by the lemma, hence \phi is an isomorphism, as we wanted to show. To prove the lemma, it helps to have a clear picture of how pushouts are formed in Set. The pushout C is the set of equivalence classes on A + B where x \in A + B is deemed equivalent to x' \in A + B iff there is a zig-zag path$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$where the f_{i_k} alternate between f_1 and f_2, and the p_i belong to P. Now it is not hard to convince yourself that given (x, y) in A^2 or B^2, and (x', y') in A^2 or B^2, if there is a zig-zag path from x to x', and a zig-zag path from y to y', then there is a zig-zag path from (x, y) to (x', y') with respect to the pair of maps$$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B.$$But this means precisely that the map$\phi\$ is monic. | 2,618 | 7,793 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.889447 |
https://de.mathworks.com/matlabcentral/cody/problems/988-convert-a-substructure-reference-string-into-a-valid-definition-structure-for-subsref-and-subsasgn/solutions/170010 | 1,585,587,030,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497171.9/warc/CC-MAIN-20200330150913-20200330180913-00399.warc.gz | 442,225,435 | 15,918 | Cody
# Problem 988. Convert a substructure reference string into a valid definition structure for subsref and subsasgn
Solution 170010
Submitted on 28 Nov 2012 by Tom Holz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% nocheat = isempty(regexp(evalc('type subsdef'),'(eval|regexprep|inline|str2func)')); y_correct = 1i; b(12) = y_correct; defstr = '(12)'; assert(isequal(subsref(b,subsdef(defstr)),y_correct) && nocheat)
2 Pass
%% nocheat = isempty(regexp(evalc('type subsdef'),'(eval|regexprep|inline|str2func)')); y_correct = -4i; c{1,2,3,4,5}.field_b = y_correct; defstr = '{1,2,3,4,5}.field_b'; assert(isequal(subsref(c,subsdef(defstr)),y_correct) && nocheat)
3 Pass
%% nocheat = isempty(regexp(evalc('type subsdef'),'(eval|regexprep|inline|str2func)')); y_correct = 3i; a(12).field_b{1,3}{2}((3),1).c = y_correct; defstr = '(12).field_b{1,3}{2}((3),1).c'; assert(isequal(subsref(a,subsdef(defstr)),y_correct) && nocheat)
4 Pass
%% nocheat = isempty(regexp(evalc('type subsdef'),'(eval|regexprep|inline|str2func)')); y_correct = repmat(2i,3,1); d{2}.a(1:3,:) = y_correct; defstr = '{2}.a(1:3,:)'; assert(isequal(subsref(d,subsdef(defstr)),y_correct) && nocheat) | 443 | 1,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-16 | latest | en | 0.222914 |
https://ch.gateoverflow.in/1111/gate2020-10 | 1,606,323,408,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00424.warc.gz | 229,685,925 | 9,221 | # GATE2020: 10
Consider a batch distillation process for an equimolar mixture of benzene and toluene at atmospheric pressure. The mole fraction of benzene in the distillate collected after $10$ minutes is $0.6$. The process is further continued for additional $10$ minutes. The mole fraction of benzene in the total distillate collected after $20$ minutes of operation is
1. less than $0.6$
2. exactly equal to $0.6$
3. exactly equal to $0.7$
4. greater than $0.7$ | 128 | 465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-50 | latest | en | 0.897723 |
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### Messages - Thomas Nutz
Pages: 1 [2]
16
##### Home Assignment 3 / Re: Problem 4
« on: October 08, 2012, 03:53:35 PM »
I don't think that the maximum in the region $0\leq x \leq l$, $0 \leq t \leq T$ must either decreases or increase; I think it also can stay constant (e.g. iron rod that is initially very hot in the middle, then the maximum is found at t=0, x=l/2) and M(T) =const.
Am I wrong?
17
##### Home Assignment 3 / problem 3
« on: October 07, 2012, 10:15:42 PM »
Could anyone give me a hint on how to integrate problem 3a)? In order to get rid of the absolute value I have to split up the integral and then the -inf to inf formulas don't work any more...
Thanks a lot!
18
##### Misc Math / proof of proposition 2a) in 9th lecture notes
« on: October 05, 2012, 03:57:42 PM »
we are given
$$U(x,t)=\int_{-\infty}^xu(x,t)dx$$
and are to prove that $U$ satisfies $U_t=kU_{xx}$.
The proof given is "one can see easily that as $x\rightarrow -\infty$ and that therefore $U$ and all its derivatives have to be zero". But the integral over any function with the upper bound approaching the lower bound goes to zero!
For instance I take the function $f(x)=x^5$, which obviously does not satisfy the heat equation for $x\neq 0$. Then isn't
$$lim_{x\rightarrow -\infty}\int_{-\infty}^{x}x^5dx=0$$, and according to this "proof" $\int_{-\infty}^{x}x^5dx$ would satisfy the heat equation?
This does not make sense to me...
Thanks!
19
##### Misc Math / inhomogeneous b.c.
« on: October 05, 2012, 08:59:09 AM »
In the 10th lecture we are asked to consider
$$0=\int_{II}G(x,y,t-\tau)(-u_{\tau}(y,\tau))+ku_{yy}(y,\tau)d\tau'dy$$
1. question: What are we integrating over here? Is $\tau'=t-\tau$?
2. Where is this expression coming from? Is it a trial solution that I simply have to take as given, or does it follwow from any other expression?
Thanks!
20
##### Home Assignment 2 / Re: Problem 2
« on: September 29, 2012, 03:37:34 PM »
should the $\phi$ in eq. 6 read $\phi(r+ct)$ rather than $\phi(x+ct)$?
21
##### Misc Math / Re: Example 8b
« on: September 28, 2012, 05:15:08 PM »
... and in example 8d): Why should $u$ be given by eq. (9)? Shouldn't that be eq. (9) with all sines exchanged by cosines?
22
##### Misc Math / Example 8b
« on: September 28, 2012, 05:02:12 PM »
Dear all,
in the 8th lecture notes in the example b) we end up at
$$u(x,t)=\frac{1}{2}(sin(x+tt)+sin(x-t))$$ for $x>t$
and $$u(x,t)=\frac {1}{2}(sin(x+t)-sin(x-t))$$ for 0<x<t. But isn't the latter $u$ valid on $-t<x<t$, and can't we even say that
$$u(x,t)=\frac{1}{2}(-sin(x+t)-sin(x-t))$$ for $x<-t$?
23
##### Misc Math / Lecture 6 example
« on: September 27, 2012, 04:23:33 PM »
Hello,
I was going through the 6. lecture notes, where in the end an example is brought up that leads to an integral
$$\frac{1}{4}\int_0^tcos(t')(cos(x-ct+ct')-cos(x+ct-ct'))dt'$$
I was trying to do that integral, but the only way that I could do it was to write out the cosines as complex exponentials, which lead me to eight terms in the end... Is there a cleverer way to do this integral?
Thanks!
24
##### Misc Math / integration constant in wave equation
« on: September 23, 2012, 12:10:16 PM »
Good morning,
in the notes "Homogeneous 1D Wave equation" we get to
$$u(x,t)=\phi(x+ct)+\psi(x-ct)$$ as the general solution, but in the very last paragraph it is mentioned that we could add a constant to $\phi$ if we subtract that same constant from $\psi$, and that this constant would be the only arbitrariness of this solution. But why does this have to be the same constant?
I can add for instance 1434 to $psi$ and subtract -12i from $\phi$ and the sub of the two would still satisfy the PDE, as any derivative of 1434-12i is zero...
25
##### Misc Math / Re: characteristic vs. integral lines
« on: September 21, 2012, 05:59:31 PM »
havin some trouble there... $$x(t)=\frac{b}{a}t+c$$
26
##### Misc Math / characteristic vs. integral lines
« on: September 21, 2012, 05:57:04 PM »
Heythere,
I am confused by the terms characteristic lines and integral lines. The book introduces characteristic lines as the curves along which a function is constant. Now in the notes integral lines are curves to which the vector field is tangential, i.e. in the case of the gradient vector field the lines along which the function changes most (in abs. value).
So I thought these two should be orthogonal in the case of $au_t+bu_x=0$.
The characteristic lines of u are given by $x=\frac{b}{a}y+c$, but this is also the solution to the ODE
$$\frac{dt}{a}=\frac{dx}{b}$$, which is given for the integral lines in the notes (First order PDEs)...
So are integral lines the same as characteristic lines?
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# Lecture 3 - and KVL to find I 1 V 2 and V 3 Fig 3-9 A...
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Lecture 3 figures 1 Fig. 3-1: Is this element absorbing or generating power? Fig. 3-2: What’s wrong with this circuit? Fig. 3-3: Determine the power absorbed by each element in this circuit. Fig. 3-4: A circuit node, at which 5 branches connect. Fig. 3-5: A single-loop circuit containing 3 elements, with the potential across each element labeled.
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Lecture 3 figures 2 Example: Find I 1 , V 2 , and V 3 . Example: Given a current of 0.5 A through the 50 Ω resistor, as shown, find V s . Fig. 3-6: Elements in series must have the same current through them. Fig. 3-7: Elements in parallel must have the same potential across them. Fig. 3-8: A circuit for application of KCL
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Unformatted text preview: and KVL to find I 1 , V 2 , and V 3 . Fig. 3-9: A circuit for example showing application of Ohm’s Law, KVL, and KCL. Lecture 3 figures 3 Find the current I 2 . Ex: Find the current I , the equivalent resistance, R eq , and the power absorbed by each element. Ex: Find the voltage V and the power absorbed by each element. Fig. 3-10: A circuit with an open loop. Fig. 3-11: Current sources in series are forbidden. Fig. 3-12: Voltage sources in parallel are forbidden. Fig. 3-13: A single-loop circuit containing a dependent voltage source. Fig. 3-14: A single node-pair circuit....
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## This note was uploaded on 01/19/2012 for the course IE 230 taught by Professor Xangi during the Spring '08 term at Purdue.
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Lecture 3 - and KVL to find I 1 V 2 and V 3 Fig 3-9 A...
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Ask a homework question - tutors are online | 556 | 1,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-26 | longest | en | 0.853054 |
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March 30th, 2016, 01:16 PM #1 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,546 Thanks: 110 Cantor's diagonal proof and the real numbers At this point we have two issues: 1) Cantor's proof. Wrong in my opinion, see: Cantor's Diagonal Argument. Infinity is Not a Number 2) Cantor's proof has nothing to do with the real numbers. A real irrational number is not an infinite binary digit. It is the limit of binary digits. (sqrt2). WRONG!! See EDIT below. A natural number can be assigned to every binary fraction for all n (the binary fractions can be counted): .1011... -> 1011... . But lim .1011... exists lim 1011 doesn't exist.* The question remains, can a natural or pos rational number be assigned to a lim? You have to answer this before you can decide if the reals are countable. For example: sqrt2 -> 2 sqrtn -> n Square roots of natural numbers are countable. nth roots are countable. *skipjack, see link EDIT HOWEVER If you can show that infinite binary sequences are uncountable, then it follows that the reals are uncountable because, by the above, there are more reals than infinite binary sequences. But note that all infinite binary fractions are countable by: .1011... -> 1011... for all n. Last edited by zylo; March 30th, 2016 at 01:57 PM.
March 30th, 2016, 01:49 PM #2
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Quote:
Originally Posted by zylo Wrong in my opinion
That seems to be an admission that you are merely giving your opinions, not discovering genuine mathematical or logical errors that shouldn't need to be a matter of opinion.
Let's put this particular issue on hold briefly, as I would like you to reply first to my recent posts in the thread that you linked to above.
March 30th, 2016, 03:14 PM #3 Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra Actually, there is only one issue: you have no idea what you are talking about, but persist in making nonsensical claims instead of attempting to understand.
March 30th, 2016, 04:36 PM #4
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Quote:
Originally Posted by zylo 2) Cantor's proof has nothing to do with the real numbers. A real irrational number is not an infinite binary digit. It is the limit of binary digits. (sqrt2). WRONG!! See EDIT below.
"WRONG" refers to "Cantor's proof has nothing to do with the real numbers." If you could show that all countably infinite binary digits were uncountable, then the reals would be uncountable, since they are not uncountably infinite binary digits, but rather, in the case of the irrationals, limits of infinite binary digits.
But the limit of an infinite binary digit is not always an irrational number:
Lim .3333333333.....=1/3
The question still remains, unanswered by Cantor, Are limits of countably infinite binary numbers countable.
Countably infinite binary digits are natural numbers.
101101..... , n digits, is a natural number for all n. There is not an "infinite" binary digit, so Cantors set of "infinite" binary digits is meaningless right off the bat.
Last edited by zylo; March 30th, 2016 at 04:57 PM.
March 30th, 2016, 05:55 PM #5
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Quote:
Originally Posted by zylo If you could show that all countably infinite binary digits were uncountable
Yet again we have to assume that you mean "the set of all infinite sequences of binary digits", because you have written some meaningless nonsense. But under that assumption, they have been proved to be uncountable.
Quote:
Originally Posted by zylo then the reals would be uncountable
They are.
Quote:
Originally Posted by zylo since they are not uncountably infinite binary digits, but rather, in the case of the irrationals, limits of infinite binary digits.
And back to meaningless nonsense again. Every real number can be expressed as a limit of an infinite sequence of rational numbers (the particular number system used to represent the real number is immaterial), but this has nothing whatever to do with the infinite sequences of (binary) digits which are the subject of the diagonal argument. Infinite sequences are not the limits of finite sequences because they don't get "closer" to anything, they don't converge on anything, they just go on forever.
Quote:
Originally Posted by zylo But the limit of an infinite binary digit is not always an irrational number: Lim .3333333333.....=1/3
More meaningless nonsense that we are presumably supposed to translate into something that makes some sort of sense. Here, it appears that "the limit of an infinite binary digit" is supposed to mean "the limit of an infinite sequence of rational numbers" although your terminology is so far off that is illustrates only your complete and utter confusion and lack of knowledge on the subject. So we then have "the limit of an infinite sequence of rational numbers is not always irrational", which is true but completely irrelevant to any arguments relating to the cardinality of the real numbers or to Cantor's diagonal argument. You then write what appears to be an attempt at a limit expression, but this too is meaningless nonsense because the notation is not used correctly and doesn't feature any variable that might be heading to any limit. One presumes that you mean that $\lim \limits_{n \to \infty} 3 \sum \limits_{k=1}^n {1 \over 10^n} = \frac13$. Which, again is true but again has nothing whatsoever to do with the cardinality of the real numbers or to Cantor's diagonal argument.
Quote:
Originally Posted by zylo The question still remains, unanswered by Cantor, Are limits of countably infinite binary numbers countable.
The reason Cantor left this question unanswered is that it is meaningless nonsense. The set of infinite sequences of (binary) digits does, however, have a trivial bijection with the set $B$ of binary representations of real numbers $r$ such that $0 \le r \lt 1$ and it is equally trivial to show that the set of real numbers having more than one (in fact, precisely two) representations in $B$ is countably infinite (namely those that have a terminating representation which are thus a subset of the rational numbers). Thus, the set $B$ consists of the union of a set having a bijection to real numbers $r$ such that $0 \le r \lt 1$ and a set having a bijection to a subset of the rational numbers. If the first of these were countably infinite, we'd have $B$ being the union of two distinct countably infinite subsets which would mean that $B$ would also be countably infinite. Since it isn't, the set of real numbers $r$ such that $0 \le r \lt 1$ must not be countably infinite either.
Quote:
Originally Posted by zylo Countably infinite binary digits are natural numbers.
This is more nonsense. Unfortunately, it can't be rescued: it's just wrong. It's a product of your complete lack of understanding of the subject and your utter refusal to learn anything about it.
Quote:
Originally Posted by zylo 101101..... , n digits, is a natural number for all n. There is not an "infinite" binary digit, so Cantors set of "infinite" binary digits is meaningless right off the bat.
Again, this is utter nonsense. A sequence of $n$ digits is not infinite because it terminates after $n$ digits. Cantor's sequences are not $n$ digits long; they are infinitely long, which means that they don't terminate. Whatever you are thinking of is nothing whatsoever to do with the cardinality of the real numbers or to do with Cantor's diagonal argument. In fact, whatever you are thinking is almost certainly a completely confused mishmash of concepts that has no meaning whatsoever. Again, it's evidence that you don't understand the subject, and these constant new threads are simply evidence of the fact that you are making no attempt to learn the subject. Instead, you prefer to claim that your confused nonsense means something - in particular, that it proves one of the greatest thinkers ever to be wrong.
Perhaps one day you will realise just how little you know. I know that when I did that, it freed up my mind to understand many more marvellous things.
Please, by whatever you hold sacred, stop writing meaningless, confused nonsense and try to learn something.
Last edited by skipjack; March 30th, 2016 at 07:58 PM.
March 30th, 2016, 08:04 PM #6
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Quote:
Originally Posted by zylo . . . infinite binary digit
Why did you introduce this non-standard terminology? Is it intended to have the same meaning as "non-terminating binary sequence"? If not, how does it differ in meaning from that?
March 31st, 2016, 07:54 AM #7 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,546 Thanks: 110 Convergence of an infinite binary fraction is elementary calculus. A binary fraction can be associated with a natural number as follows. .101.... to n places can be associated with the natural number 101....but lim.101.... can't because lim101.... doesn't exist. The fractional part of pi for any finite number of decimal places can be associated with a natural number, but not the limit. .1415 -> 1415 but limit .1415...... = pi-3 I personally find it interesting that any countably infinite fraction can be associated with a natural number. But there is a deeper significance in that all countably infinite binary (or decimal) fractions can be counted. But that doesn't imply that all reals in [0,1)can be counted, because some are the limit of a countably infinite binary fraction. A basic principle has been developed here: All binary fractions can be counted. Some real numbers are limits of binary fractions and the above association doesn't work; counting these is open for discussion. binary fraction: .1011000001......., countably infinite places. lim of binary fraction: lim.1011000001...as n approaches infinity
March 31st, 2016, 09:24 AM #8
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Stop trying to prove things, you aren't competent to do so. Everything your write is nonsense because you don't understand the first thing about this stuff.
Quote:
Originally Posted by zylo I personally find it interesting that any countably infinite fraction can be associated with a natural number.
Not in the way you are doing it, no. At least, I don't think so. You have introduced yet more nonsense terminology which I have to guess the meaning of. But we are back to your false claim that there are natural numbers that are composed of infinite strings of digits.
Quote:
Originally Posted by zylo A basic principle has been developed here: All binary fractions can be counted. Some real numbers are limits of binary fractions and the above association doesn't work; counting these is open for discussion. binary fraction: .1011000001......., countably infinite places. lim of binary fraction: lim.1011000001...as n approaches infinity
This is wrong because your definitions are wrong. You need
binary fraction: 0.1011000001... terminating after a finite number of digits.
lim of binary fraction: 0.1011000001... non-terminating
But even after correcting these definitions nothing "has been developed here" this is all centuries old stuff. And the counting of the non-terminating binary representations is not up for discussion - they are not countable (although some subsets of them are countable).
March 31st, 2016, 02:00 PM #9
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Originally Posted by zylo Let T be the set of all infinite binary sequences. T is obviously uncountable
Quote:
Originally Posted by zylo If you could show that all countably infinite binary digits [sequences] were uncountable, then the reals would be uncountable
Does it follow from these two posts that you made that the reals are uncountable?
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Note: When you're dealing with linear equations, you may be asked to find the slope of a line. That's when knowing the slope formula really comes in handy! | 681 | 2,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.860215 |
http://mathhelpforum.com/differential-geometry/171541-show-these-intervals-open-closed-sets-print.html | 1,508,275,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00679.warc.gz | 205,027,529 | 3,316 | # Show these intervals are open and closed sets.
• Feb 16th 2011, 04:09 PM
seamstress
Show these intervals are open and closed sets.
Show that the intervals (a, ∞) and (-∞, a) are open sets and that [b, ∞) and (-∞, b] are closed sets.
Thank you!
• Feb 16th 2011, 04:18 PM
Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
• Feb 16th 2011, 04:31 PM
seamstress
It's not homework, actually. I'm studying for a test and trying to understand something my teacher said in class the other day. I do not understand how to do the question, or why the answer is what it is, and I would like to understand open and closed sets. I thought maybe by understanding this basic question, I would be able to figure out other things that stem off of this one. If somebody could explain why the first two are open and why the second two are closed, and possibly show me how I would go about proving it, I would be helped tremendously in understanding this question. Thank you.
• Feb 16th 2011, 05:01 PM
Tinyboss
Closed is usually defined as "its complement is open", so it's enough to prove that the open intervals are indeed open. Try showing that for any point in one of the open intervals, you can find a smaller open interval around that point and contained in the original interval.
• Feb 16th 2011, 09:29 PM
seamstress
I still do not know where to get started. I recall that definition of closed sets is that, but I feel like I don't have enough information to properly prove these statements. Any suggestions on how to get started? Thanks
• Feb 17th 2011, 02:59 AM
DrSteve
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https://en.sorumatik.co/t/what-value-would-be-returned-in-excel-a49/16443 | 1,723,071,198,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00538.warc.gz | 178,971,065 | 7,020 | # What value would be returned in excel a49
what value would be returned in excel a49
What value would be returned in Excel cell A49?
Answer: To determine what value would be returned in Excel cell A49, we need more context about the content and formulas present in the spreadsheet. Here are some possible scenarios:
1. Direct Entry: If a value (number, text, date, etc.) is directly entered into cell A49, that exact value will be returned. For example, if you type “100” in cell A49, Excel will return “100”.
2. Formula: If cell A49 contains a formula, the returned value will depend on the formula’s logic and the values in the referenced cells. For example:
• If A49 contains the formula `=SUM(A1:A48)`, it will return the sum of values from cells A1 to A48.
• If A49 contains the formula `=AVERAGE(A1:A48)`, it will return the average of the values in cells A1 to A48.
• If A49 contains the formula `=IF(A1>0, "Positive", "Negative")`, it will return “Positive” if the value in cell A1 is greater than 0, otherwise it will return “Negative”.
3. Reference: If cell A49 contains a reference to another cell, it will return the value of the referenced cell. For example:
• If A49 contains `=A1`, it will return the value of cell A1.
• If A49 contains `=B49`, it will return the value of cell B49.
4. Error: If there is an error in the formula or the referenced cells contain errors, then A49 will return an error value such as `#DIV/0!`, `#VALUE!`, `#REF!`, etc.
To provide a precise answer, please provide more details about what is in cell A49 or the surrounding cells. | 403 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.793382 |
https://www.physicsforums.com/threads/taylor-expansion-of-x-x.429852/ | 1,508,253,018,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00160.warc.gz | 964,442,698 | 14,298 | # Taylor expansion of x^x
1. Sep 17, 2010
### danik_ejik
hello,
please help to calculate the taylor polynomial for
[URL]http://latex.codecogs.com/gif.latex?f(x)=x^{x}-1[/URL] around the point a=1
i thought to write it as g(x)=x^x
and then f(x)=g(x)-1
and then find the polynomial for g(x) as lng(x)=xln(x)
but it seems incorrect.
Last edited by a moderator: Apr 25, 2017
2. Sep 17, 2010
### hunt_mat
In order to do this you have to calculate the derivative if $$y=x^{x}$$, take logs of this equation and differentiate that using implicit differentiation and that will help you, or you could write:
$$x^{x}=e^{x\log x}$$
And use the chain rule
3. Sep 17, 2010
### danik_ejik
thanks,
successfully managed by directly calculating the taylor polynomial when
[URL]http://latex.codecogs.com/gif.latex?f(x)=e^{xlnx}[/URL]
Last edited by a moderator: Apr 25, 2017
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https://math.stackexchange.com/questions/2531961/epsilon-delta-derivative-proof-of-xn | 1,575,576,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00330.warc.gz | 462,734,323 | 31,568 | # Epsilon-delta derivative proof of $x^n$
I'm currently trying to prove the power rule using the epsilon-delta definition of a derivative. I've already done it for the basic limit definition, but I thought it might be a helpful exercise to test my understanding by doing it this way. However, I'm struggling and would really appreciate any help/hints.
My work so far:
The epsilon-delta definition says a function is differentiable if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|x-x_0| < \delta$ implies $\frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) < \epsilon$. Thus, I need to find a delta, as function of epsilon, and possibly $x_0$, such that whenever the first inequality is true the second one is as well.
So, fix $x_o \in R$, and let $|x-x_0| < \delta$, then $|\frac{x^n - x_{0}^{n}}{x-x_0} - nx_{0}^{n-1}| = |\frac{x^n - nxx_{0}^{n-1} + (n-1)x_{0}^{n}}{x-x_0}|$. Upper-bounding this with $x < \delta + x_0$ and simplifying, we get $|\frac{(\delta + x_0)^n - n\delta x_{0}^{n-1}-x_{0}^{n}}{\delta}| = |\frac{(\delta + x_0)^n}{\delta} -nx_{0}^{n-1} - x_{0}^{n}| < \epsilon$.
Is that right so far, and if so, any advice on how to push through this last step and solve for $\delta$? I'm stumped, to be honest.
• How denominator changed with $\delta$ to have upper bound? – Atbey Nov 22 '17 at 6:31
• Do a binomial expansion on $(\delta - x_0)^n$ the first two terms cancel with what is there, leaving $\delta |{n\choose 2} x_0^{n-2} +{n\choose 3} x_0^{n-3}\delta + \cdots + \delta^{n-2}| < \epsilon$ and show that everything inside the absolute value is bounded. – Doug M Nov 22 '17 at 6:58
• Your last line is incorrect. The middle term should be $|\frac {(\delta +x_0)^n -x_0^n}{\delta}-nx_0^{n-1}|$ – DanielWainfleet Nov 22 '17 at 10:53
Using
$$(x^n - x_0^n) = (x - x_0)\sum_{k=0}^{n-1} x^k x_0^{n-1-k}$$
One can write
$$\frac{x^n - x_0^n}{x - x_0} - n x_0^{n-1} = \sum_{k=0}^{n-1} (x^k x_0^{n-1-k} - x_0^{n-1})$$ Define $M = |x_0| + 1$ and suppose that $|x-x_0|\le 1$, then $|x|\le M$ and $|x_0|\le M$ and $$|x^k-x_0^k| \le |x - x_0|\sum_{p=0}^{k-1} |x|^p |x_0|^{k-1-p}\le |x - x_0|\sum_{p=0}^{k-1} M^{k-1}\le |x-x_0|k M^{k-1}$$
We get $$\left|\frac{x^n - x_0^n}{x - x_0} - n x_0^{n-1}\right| \le \sum_{k=1}^{n-1} |x-x_0|k M^{k-1} |x_0|^{n-1-k}\le |x - x_0|M^{n-2}\sum_{k=1}^{n-1}k \le |x - x_0|M^{n-2}\frac{n (n-1)}{2}$$
For $n>1$ we can take $$\delta = \min\left(1, \frac{2\varepsilon}{n(n-1)(|x_0|+1)^{n-2}}\right)$$ For $n=1$ there is no condition on $\delta$ because $f(x) = f(x_0) + (x-x_0)$.
• Using this hint at the beginning, I get $|\frac{x^n - x_{0}^{n}}{x-x_0} - nx_{0}^{n-1}| = |(1-n)x_{0}^{n-1} + xx_{0}^{n-2} + ... + x_0x^{n-2} + x^{n-1}| < \epsilon$. Is this what you had in mind? I don't understand how this helps, since there doesn't seem to be an easy way to cancel all of the x terms floating around. – P. Reinecke Nov 22 '17 at 6:57
• See my edit above. – Gribouillis Nov 22 '17 at 7:15
For $x\ne 0$ we have $$(x+d)^n-x^n=-x^n+\sum_{j=0}^n x^{n-j}d^j\binom {n}{j}=nx^{n-1}d +d^2 P(d)$$ where $P(d)$ is a polynomial. So there exists $M\in \Bbb R^+$ such that $\forall d\; (|d|\leq 1\implies |P(d)|<M).$ Therefore for $0<|d|\leq 1$ we have $$|\frac {(x+d)^n-x^n}{d}-nx^{n-1}| =$$ $$= |\frac {nx^{n-1}d+d^2P(d)}{d}-nx^{n=1}| =$$ $$= |dP(x)|\leq |d|M.$$ And $|d|M\to 0$ as $d\to 0.$
To prove that $M$ exists : If $P(d)=\sum_{j=0}^ma_jd^j$ then $|d|\leq 1\implies |P(d)|\leq$ $\sum_{j=0}^m|a_j|\cdot |d|^j\leq$ $\leq \sum_{j=0}^m|a_j|=M.$
For $x=0\ne d$ we have $\frac {(x+d)^n-x^n}{d}=\frac {(0+d)^n-0^n}{d}=d^{n-1}.$
• Note that the degree of $P(d)$ and the values of the co-efficients of $P(d)$ are determined by $x$ and $n$ but are independent of $d.$ – DanielWainfleet Nov 22 '17 at 11:40 | 1,607 | 3,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-51 | latest | en | 0.803129 |
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# Correlation between sphere diameter and nusselt number
Correlation between sphere diameter and nusselt number
Nusselt number are given as
where h is the convective heat transfer coefficient of the flow, L is the characteristic length, k is the thermal conductivity of the fluid.
The general formula for the characteristic length is the volume of a system divided by its surface.
A typical use-case is calculating flow through circular tube in order to examine flow conditions (i.e. the Reynolds number). In those cases, the characteristic length is the diameter of the pipe. so in your case L will be equal to diameter of pipe.
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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Math help pleas new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Intern Joined: 16 May 2016 Posts: 1 Math help pleas [#permalink] ### Show Tags 16 May 2016, 15:19 If a natural number is entered into a new toy names Squareroo, the toy will do the following: (1) Find the perfect square closest to the number entered; (2) Take the square root of the perfect square; (3) Find the perfect square closest to the square root in step 2; (4) Repeat the steps of taking the square root and finding closest perfect square until it takes the square root of the same number twice. How many times does Squareroo take the square root if the number entered into it is 660? I tried this problem and I got 5, but I don't know if I'm right or if my work is right, so any answers will be greatly appreciated. Thanks so mych Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6639 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math help pleas [#permalink] ### Show Tags 14 Mar 2017, 15:02 IndianSavage wrote: If a natural number is entered into a new toy names Squareroo, the toy will do the following: (1) Find the perfect square closest to the number entered; (2) Take the square root of the perfect square; (3) Find the perfect square closest to the square root in step 2; (4) Repeat the steps of taking the square root and finding closest perfect square until it takes the square root of the same number twice. How many times does Squareroo take the square root if the number entered into it is 660? I tried this problem and I got 5, but I don't know if I'm right or if my work is right, so any answers will be greatly appreciated. Thanks so mych (1) $$676 = 26^2$$ is the perfect square closest to $$660$$. (2) The square root of the perfect square $$676$$ is $$26$$. (3) $$25 = 5^2$$is the perfect square closest to $$26$$. (4) The square root of the perfect square $$25$$ is $$5$$. (5) $$4 = 2^2$$ is the perfect square closest to $$5$$. (6) The square root of the perfect square $$4$$ is $$2$$. (7) $$1 = 1^2$$ is the perfect square closest to $$2$$. (8) The square root of the perfect square $$1$$ is $$1$$. Squareroo took the square root from the above $$4$$ time at step (2), (4), (6) and (8). My answer is $$4$$. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: Math help pleas &nbs [#permalink] 14 Mar 2017, 15:02
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# Math help pleas
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# Understanding Diode Reverse Recovery and its Effect on Switching Losses
Peter Haaf, Senior Field Applications Engineer, and Jon Harper, Market Development Manager, Fairchild Semiconductor Europe
Abstract Half-bridge structures are extensively used in power electronics applications: lighting, power supplies, UPS and motor drives. When these half-bridge circuits are hard switched, the low side diode reverse recovery affects system performance. This paper reviews the principle of diode reverse recovery and how this affects the semiconductor switch performance. Practical tests showing how di/dt and temperature affect performance are presented. Finally measurements using different devices are compared showing the curves for fast and soft recovery diodes, showing that in some cases, efficiency can be improved by adding capacitance in parallel with the diode.
voltage and current transitions can take several forms. Figure II-1 shows three possible linearized transitions of voltage and current waveforms and how the energy of the transition is calculated. These basic formulae are used in the following discussion. Further, they are important in understanding the effect of capacitance on turn off losses in a later section of this paper.
I. INTRODUCTION
I V
Half-bridge structures having two semiconductor switches with anti-parallel diodes are extensively used in power applications. Examples include motor drives, solar inverters, welding equipment and general AC/DC power supplies. This paper focuses on how the choice of diodes affects the total switching losses. The effect of diode reverse recovery is introduced, showing how this generates losses in both the diode and the switch which is commutating that diode. The paper moves on to practical considerations made from experimental measurements. Interesting effects relating to die size, temperature, di/dt and additional node capacitance are quantified. The limitations of the predictive ability of the formulae are highlighted. By combining technical and practical considerations, this paper should provide the practicing engineer with an understanding of how to select the right diode for a given application. II. SWITCHING LOSSES Switching losses occur when a switching element in a circuit transitions from one state to another. The
0
E=1/2VIt E=1/3VIt E=1/6VIt
Figure II-1: Calculation of switching losses for various overlapping current and voltage waveforms
MOSFET switching losses have been extensively covered in [1]. We will review the switching losses caused by forced commutation of a diode. If a diode is forced to turn off by another semiconductor switch, the diode will see switching losses. Additional losses will be generated in the semiconductor switch. In simple terms, the reason for extra switching losses in a diode can be explained as follows. Figure II-2 shows the charge distributions for a diode in the conducting and the non-conducting states [2]. Here we are showing a p-n junction for illustrative
A-23 Fairchild Power Seminar 2007
## purposes, rather than the p-i-n junction used in power diodes.
Minority carrier concentration near the junction Electron concentration in P-type region Hole concentration in N-type region Minority carrier concentration near the junction
DC Bus
IL
Hole Electron concentration concentration in N-type region in P-type region
## IDIODE VDIODE + + VSWITCH -
VDD
x
x x=0 x=0
Switch
P-type
N-type
P-type
N-type
Diode conducting
Diode blocking
Figure II-2: Charge distributions for a diode in the conducting and blocking states
ISWITCH
Reference GND
Figure II-3: Circuit to show the effect of diode reverse recovery on the diode and on the semiconductor switch
For the diode to transition from the conducting to the non-conducting state, the charge distribution must change. This can only happen with a movement of charge, which is a flow of current. In some cases, such as a silicon carbide diode, the charge distribution difference is caused solely by the junction capacitance: again a movement of charge occurs when moving from the conducting to the non-conducting state. The distribution curves show minority carrier density as a parameter. So, the larger the active junction area (other parameters being held constant), the larger the charge difference. Therefore devices in the same family with larger die sizes, represented by higher current ratings, will have a larger reverse recovery charge. If the movement of charge between the nonconducting and conducting states happens during the same switch state, there is no additional loss. For example, in a discontinuous mode boost converter, the current in the diode drops to zero while the power switch is in the off state. However, if an external switch forces the diode to change from the conducting to the non-conducting state (forced commutation) extra current is required to change the states, causing dissipation in both the diode and the switch. Figure II-3 and Figure II-4 show the reverse recovery behavior of a diode under forced commutation.
## IL + IRRM IL VDD iSWITCH
0
IRRM
vSWITCH
iDIODE
tR -VDD
tA
tB
vDIODE
Figure II-4: Effect of diode reverse recovery on the diode and on the semiconductor switch
The plot starts with the turn on of the lower switch. After the gate voltage on the lower switch reaches the Vth level, the lower switch builds up current in the saturation mode, causing a linear increase in the switch current, and a linear decrease in the diode current, as the inductor current is constant. The diode temporarily conducts in the reverse direction. The maximum current (IRRM) is the reverse recovery current, and is specified in the diode datasheet. It increases greatly with temperature. It increases with di/dt. As will be shown later, it increases with current.
A-24 Fairchild Power Seminar 2007
Time interval tA is defined as the time between the zero crossing of the current and the peak reverse current. Time interval tB is defined as the time between the peak of the reverse current and the time where the current falls to zero (or a pre-defined low level). The sum of tA and tB is called the reverse recovery time, tRR. The switching power dissipation in the diode is given by:
E ON DIODE = 1 t B VDDIRRM 6
Noting that IRRM can often exceed the normal forward rated current of the diode, these extra switching losses and their impact are significant. For a normal diode, tB is much smaller than tA. For a soft recovery diode, tB is larger than tA. For a given reverse recovery time, tRR (= tA + tB), the equation above shows that the semiconductor switch losses when using a soft recovery diode are less than the losses caused by a normal diode, as:
1 1 IL + IRRM > IRRM 2 3
(1)
(5)
where VDD is the bus voltage. The power dissipation during time tA is considered to be part of the conduction losses of the diode. The reverse recovery current also induces extra losses in the semiconductor switch. The total switch on loss is given by the following formula:
EON SWITCH = 1 VDDIL t R 2
## 1 1 1 + VDD (IL + IRRM )t A + VDD ( IL + IRRM )t B 2 2 3
(2)
where IL is the load current and tR is the time interval between the start of switching and when the semiconductor switch provides the full load current. By setting tA and IRRM to zero, the equation for onlosses in the absence of reverse recovery is readily obtained:
EON SWITCH =
1 1 VDDIL t R + VDD ( IL )t B 2 2
(3)
where in this case tB is the time interval between when the semiconductor switch provides the full load current and when the voltage across the switch has dropped to the minimum value. The extra EON loss attributable to the diode can be calculated by subtracting the two equations:
1 1 EON EXTRA = VDD (IL + IRRM )t A + VDD ( IRRM )t B 2 3
However, the switching loss generated in the diode itself (equation 1) is proportional to tB. As a soft recovery diode has a larger tB value than a normal diode, the diode losses will be higher. Nevertheless, accounting for these losses in the above equation shows that there is still a clear benefit for the overall system efficiency. The important conclusion is that the use of a soft recovery diode will introduce more switch-on losses in the diode itself, but save additional losses in the semiconductor switch. When evaluating the performance of a new diode, it is therefore necessary to look at both the diode and semiconductor switch performance, not just the diode performance. Another benefit of a soft switching diode is that the dv/dt rate during time tB is much lower than for a normal diode because tB is longer. High dv/dt can cause ringing losses and extra EMI in a circuit. Finally, soft recovery diodes generally have a lower IRRM than normal diodes. In the absence of reverse recovery, equation (3) can be rearranged in terms of the applied di/dt and dv/dt in the system:
1 1 2 2 VDDIL IL VDD = 2 + 2 di dv dt dt
EON SWITCH
(6)
(4)
This equation will be used in later discussions. Finally, it is important to note the effect of
A-25 Fairchild Power Seminar 2007
temperature on IRRM and tRR. While the forward voltage of a diode decreases as temperature increases, the parameters affecting switching characteristics, IRRM and tRR, both increase with temperature. With reference to Figure II-2, the minority charge concentration will increase with temperature, so it is to be expected that both IRRM and tRR will also increase with temperature. Figure II-5 demonstrates these results for two industry standard diodes.
di/dt=200A/ms, Vdd=400V, If=8A, Tj=25C Two industry standard diodes
## Results for Tj = 125C
Figure II-5: Comparison of reverse recovery performance for two industry standard diodes at Tj=25C and at Tj =125C. Upper curve: ISL9R860P2, lower curve: 8A/600V competitor part. III. EXPERIMENTAL SETUP AND BASIC MEASUREMENTS
speeds with ratings from 4A to 15A. All tests were performed at a current of 4A. Figure III-1 shows the experimental setup. The diode under test is on the high side. The MOSFET (or IGBT) is on the low side. Figure III-2 shows the waveforms needed to operate the circuit, taken when using an FQP9N50C MOSFET. First, the MOSFET is turned on until the test current level in the inductor is reached. The MOSFET is then switched off, causing the test current to flow through the diode. Shortly afterwards, the MOSFET is switched on to measure the switch-on losses, and then switched off to measure the switch-off losses. As the MOSFET was only switched on for a very short time, the test results apply to a junction temperature close to the ambient room temperature of 25C. The construction of the test set-up was on a standard prototype board having no copper plating. The devices under test were placed in sockets. The sockets were connected together with short, low impedance connections. The results obtained are comparable with those of a standard printed circuit board layout.
In principle, it is possible to estimate many of the factors affecting switching losses in a circuit. For IGBTs, EON and EOFF are normally specified in the datasheet for a specific set of conditions. For MOSFETs, these values can be calculated from the circuit parameters. The additional effect of the diode on switching losses can be estimated using the formulae from the previous section. In practice, it is important to assess the performance in a real circuit. First, it is important to verify the performance compared with the theoretical framework. Second, effects which are difficult to quantify, such as the beneficial effect of node capacitance on the turn off performance, need to be considered. The objective of our experiments was to compare the performance of different types of diodes using one type of MOSFET. We chose diodes of all
A-26
## Fairchild Power Seminar 2007
Figure III-2: Drive waveforms for test circuit. Channel 1 is the drain voltage, channel 2 is the gate voltage, and channel 4 is the drain current. MOSFET: FQP9N50C.
300V. Channel 4 shows the current through the low side switch. When the MOSFET is switched on, the current rises with a di/dt influenced by the MOSFET gate charge characteristics and the driver circuit component values. Before switching, the current value is zero. After completion of switching, the current value is 4A in the example. The instantaneous power is calculated by the oscilloscope on Channel M2. The area under M2 represents the switching energy which in this case is 32.6 uJ (noting that 1W = 1J/s). The maximum value of the current, minus the steady state current is equal to the reverse recovery current of the diode. In this case, this is 7.9A 4A = 3.9A.
For slow di/dt testing we used a CD4000 series logic gate with external P-channel and N-channel MOSFETs. In the course of the experiment, we found that the speed was insufficient for high speed testing, so we replaced the circuit with a 12V low voltage driver circuit (FAN5009) which has approximately 1 ohm output resistance when turning on.
## Figure III-3: Waveforms measured during switch on
Figure III-3 shows a typical set of plots for EON measurement. The parts used are the FQP9N50C MOSFET and the FFP08H60S diode. Channel 1 shows the voltage on the switching node. Before switching, the switching node voltage is high, here
Similarly, Figure III-4 shows a typical set of plots for EOFF measurement. The turn-off transition is not discussed in detail here because diode forward recovery, which occurs during turn-off, usually produces much smaller losses than reverse recovery during turn-on. Again, Channel 1 shows the voltage on the switching node, Channel 4 shows the current through the low side switch and Channel M2 shows the instantaneous power. Note that there is a small level of ringing. Further, there is a small voltage spike caused by the forward recovery of the diode.
A-27
## Fairchild Power Seminar 2007
IV. SWITCHING LOSS BEHAVIOR UNDER DIFFERENT TEST CONDITIONS AND USING DIFFERENT DEVICES
rated in the range of 4A to 15A. Figure IV-2 shows the results, followed by a table describing the part numbers.
Eon losses of the FET - FQP9N50C
90
In this section we review the effect of the diode on the switching losses seen during switch turn on. As the switching losses seen in the diode are much smaller, as discussed in Section II, these are not reviewed. The first evaluation was to look at the effect of input voltage on the turn-on and turn-off losses. For the first stage of this experiment, we compared two diodes from the same family. The results are shown in Figure IV-1.
Eon and Eoff losses of the FET - FQP9N50C
50 Eon Losses [uJ]
vs
Input Voltage
80
70 Eon @ MUR1560 Eon @ RURP860 Eon @ RURD660 Eon @ FFPF10UP60 Eon @ ISL9R1560 Eon @ RHRP860 Eon @ ISL9R860 Eon @ ISL9R460 Eon @ SIC 6A
60
50
40
30
20
10
## 0 0 50 100 150 200 250 300 350 Input Voltage [V]
vs
Input Voltage
45
Figure IV-2: EON losses versus voltage for a wide range of diodes
TABLE I DIODES AND MOSFET BODY DIODES USED IN THE EVALUATION
Eon @ ISL9R1560 Eoff @ ISL9R1560
40
## 35 Eon and Eoff Losses [uJ]
30
Part Number FCP11N60F FQP5N50CF MUR1560 RURP860 RURD660 FFPF10UP60 ISL9R1560 RHRP860 ISL9R860 ISL9R460 SiC 6A
Description Fast recovery diode from 11A, 600V superjunction MOSFET Fast recovery diode from 5A, 500V planar MOSFET 15A, Ultrafast (low speed) 600V Diode 8A, Ultrafast (low speed) 600V Diode 6A, Ultrafast (low speed) 600V Diode 10A, Ultrafast (low speed) 600V Diode 15A, Stealth (soft, high speed) 600V Diode 8A, Hyperfast (medium speed) 600V Diode 8A, Stealth (soft, high speed) 600V Diode 4A, Stealth (soft, high speed) 600V Diode 6A, Silicon Carbide, 600V Diode
25
20
15
10
## 0 0 50 100 150 200 250 300 350 Input Voltage [V]
Figure IV-1: Comparison of EON and EOFF losses against voltage for 4A and 15A Stealth diodes
There are several important conclusions to be made. First, switching losses will always rise with input voltage. During the current ramp up phase, the losses are proportional to the product of the bus voltage and the load current, so a strong linear relationship is to be expected. Second, the EON losses are higher for a larger die device of the same family, than they are for a smaller die, or a lower current rated device. Third, the EOFF losses are lower for a larger device of the same family than they are for a smaller device. Finally, the EON losses dominate, being approximately twice the EOFF losses. The next stage of the experiment was to compare the EON losses for a greater variety of 600V diodes
The lowest EON losses in the experiment came from the silicon carbide and Stealth (highest speed) diodes. The Hyperfast (medium speed) diodes were next lowest, followed by the Ultrafast (lower speed) diodes. As predicted by theoretical analysis, for a given class of diodes, higher current rated devices (which have larger dies) had higher EON losses. One interesting practical aspect is the effect of switching speed in this application. As discussed earlier, a certain amount of energy is needed to turnon the semiconductor switch:
A-28
## Fairchild Power Seminar 2007
EON SWITCH
1 1 2 2 VDDIL IL VDD = 2 + 2 di dv dt dt
## Eon and Eoff losses of the FET - FQP9N50C
200
vs
Current
(6 repeated)
Eon and Eoff Losses [uJ]
180
160
140 Eon @FCP11N60F Eon @ FQPF5N50CF Eon @ RURD660 Eon @ RHRP860 Eon @ ISL9R460 Eoff @ ISL9R460
120
At 160 A/us, 20000V/us, 300V and 4A, the EON required just to turn-on the switch is 24uJ. With reference to Figure IV-2 this accounts for a large part of the losses for the best devices at 300V. In many applications, the switching di/dt and dv/dt is limited by EMI constraints. Better efficiency can be obtained by using a faster switching diode, as the results show. However, at current prices, the incremental cost of moving to silicon carbide is very high. If system requirements limit di/dt to say 200A/us, from a switching perspective, silicon carbide diodes offer only a slight improvement in performance. For applications where much higher di/dt is permissible, silicon carbide diodes offer a definite benefit. Another clear observation from the test results is that the benefit of a fast diode increases with voltage, shown by the increased spreading of the curves at higher voltage. The benefit of using a Stealth diode in a system using a 450V bus voltage is more than that at 300V. The second evaluation was to look at the effect of input current on the turn-on and turn-off losses. For this experiment, we compared different diodes including MOSFET body diodes. The results are shown in Figure IV-3
100
80
60
40
20
0 0 1 2 3 Current [A] 4 5 6 7
Figure IV-3: Comparison of EON and EOFF losses against current for various diodes and for the body diodes in the FCP11N60F and FQPF5N50CF MOSFETs
As expected from the theoretical analysis (equation 2), there is a strong linear dependence on the current. This results from the higher current flowing through the MOSFET. However, as will be shown shortly, the loss contribution from the diode is not strongly dependent on current for fast recovery diodes. The third evaluation was to look at the effect of input current and input voltage on the maximum reverse recovery current. For this experiment, we compared different diodes including MOSFET body diodes. The results are shown in Figure IV-4 and Figure IV-5.
Irr, Reverse Recovery Peak Current of the Diode
14
vs
Current
12
## Reverse Recovery Current [A]
10 Irr @ FCP11N60F 8 Irr @ FQPF5N50CF Irr @ RURD660 Irr @ RHRP860 6 Irr @ ISL9R460
0 0 1 2 3 Current [A] 4 5 6 7
Figure IV-4: Comparison of IRRM against load current for various diodes including MOSFET body diodes
A-29
## Fairchild Power Seminar 2007
Irr, Reverse Recovery Peak Current of the Diode vs. Input Voltage
7
Irr @ SIC 6A
## 0 0 50 100 150 200 250 300 350 Input Voltage [V]
Figure IV-5: Comparison of IRRM against voltage for various diodes including MOSFET body diodes
Two important conclusions come out from the experimental results. First, the IRRM level is not strongly dependent on voltage and current for the faster diodes in the selection. We have seen in Section II that temperature has a larger effect. We will review the effect of di/dt shortly. Second, the reverse recovery current of fast recovery MOSFETs is very high, even exceeding the nominal rated current of the devices. The fourth evaluation was to look at the effect of di/dt. Increasing di/dt will reduce the turn-on losses in the MOSFETs. However as a side effect, both IRRM and tRR will increase with increasing di/dt.
Reverse Recovery Current Irr of the Diode
10
Figure IV-6 shows the effect of di/dt on reverse recovery current. Taking a simplified approach based on Figure II-4, the value 0.5tAIRRM is equal to the reverse recovery charge for a hard switching diode, and IRRM/tA is equal to di/dt. So if the reverse recovery charge remains constant, we would expect IRRM to increase with increasing di/dt, as seen in the graph. A detailed analysis of the results shows that IRRM increases more than expected from this simplistic analysis. This is seen in datasheets as a higher reported reverse recovery charge for different di/dt conditions, other parameters being held the same. So increasing di/dt will increase diode induced switching losses in the diode and in the semiconductor switch, and decrease switching losses caused by the overlap of the rising current and steady voltage waveforms. It is therefore important to assess which factor dominates. Figure IV-7 shows the effect of increased di/dt on EON losses. Here we see a clear benefit of higher di/dt on EON losses despite the higher diode induced switching losses. To get 400A/us, we used a 30 ohm gate resistor on the FQP9N50C driven by a FAN5009 1 ohm driver. For 600A/us we used FDD6N50C with a 30 ohm resistor. Values of 10 ohm and 3 ohm gave di/dt of 1400 A/us and 1600 A/us respectively.
Eon losses of the FET
45
## Reverse Recovery Current Irr [A]
vs
dI/dt
@ V = 300V @ dI/dt = 4A
vs
dI/dt
@ V = 300V @ dI/dt = 4A
40
## 8 Reverse Recovery Current Irr [A]
35
7 30 Eon Losses [uJ] 6 Eon @ ISL9R3060 Eon @ FFP08H60S Eon @ RHRP860 Eon @ ISL9R860 Eon @ ISL9R460 Eon @ ISL9R3060 Eon @ FFP08H60S Eon @ RHRP860 Eon @ ISL9R860 Eon @ ISL9R460
25
20
15 3 10
0 0 200 400 600 800 dI/dt [A/us] 1000 1200 1400 1600
0 0 200 400 600 800 dI/dt [A/us] 1000 1200 1400 1600
A-30
## Fairchild Power Seminar 2007
The final evaluation was to consider the effect of additional capacitance on the overall losses.
capacitance increases, EOFF decreases. In the case of no parallel capacitance, the voltage rises before the current falls. With reference to Figure II-1, the formula is 1/2VI for this case (26uJ from the scope measurement). When 470pF is added, the voltage rises while the current is still falling, so the formula is 1/6VI, resulting in approximately 1/3 of the loss (8uJ). Further, the curves are smoother, moving away from the simple linearized approximation. In the final case, the overlap is very small (5uJ). The addition of extra capacitance will cause a large increase in the turn-on losses. We show the 470nF turn-on example below:
## Figure IV-11: 470pF capacitance: turn-on performance
If extra capacitance increases the EON losses but reduces the EOFF losses, then there is the possibility an optimum point. We conducted experiments with different capacitor values and came up with the results in Figure IV-12. Based on these results, the addition of a small amount of extra capacitance does indeed make sense for the particular configuration used. As it is difficult to quantitatively predict the effect of capacitance on EOFF losses, such evaluation requires experimentation.
Figure IV-10: 1nF capacitance: ISL9R460, 300V, 4A
The figures show the different turn off curves for a circuit with no parallel capacitance (Figure IV-8), 470pF parallel capacitance (Figure IV-9) and 1nF parallel capacitance (Figure IV-10). As the
A-31 Fairchild Power Seminar 2007
100
VI. CONCLUSION
90
80
## 70 Etot / Eon / Eoff losses [uJ]
60
50
40
Etot at 300V Etot at 200V Etot at 100V Eoff at 300V Eon at 300V Eoff at 200V Eon at 200V Eoff at 100V Eon at 100V
30
20
10
0 0 100 200 300 400 500 600 700 800 900 1000 Capacitance parallel to the Diode [pF]
Figure IV-12: Effect of adding parallel capacitance on EON, EOFF and total losses
V. PACKAGE RECOMMENDATIONS
The package size and type is an important parameter for the selection of diodes. While it is beyond the scope of this paper to cover detailed thermal design, we would like to cover a couple of points. The maximum permissible junction temperature is always specified for power switches and diodes, and generally is 150C. In practice, designers will design to 125C maximum junction temperature to provide a safety margin for increased system robustness and reliability. Combining this information with experience on how packages are used with heatsinks, we provide the following table as a guideline, for applications not using fans or forced convection: Package and mounting Max Power TO247 with isolated foil on heatsink 30W TO220 with isolated foil on heatsink 10W TO263 on printed circuit board 1W
Reverse recovery in diodes introduces small losses in the diode but larger losses in the MOSFET or IGBT which is switching the diode. These losses are influenced by the two reverse recovery parameters IRRM and tRR. From a system design perspective, there are three aspects influencing the optimization of a half-bridge structure: di/dt, diode choice and the possible inclusion of a parallel capacitor. Higher di/dt results in lower EON losses in the circuits tested, noting that higher di/dt increases IRRM losses less than it decreases the normal switching losses. So from perspective of switch and diode losses, increasing di/dt is beneficial despite the increase in IRRM. The use of fast recovery diodes improves switching losses, but generally worsens conduction losses. Larger current rated diodes of the same family have higher IRRM resulting in higher EON, and a larger capacitance, resulting in lower EOFF. Overdimensioning of the diodes is not recommended as this leads to higher total switching losses. Addition of extra capacitance increases EON losses but decreases EOFF losses. There is the possibility that an optimum total loss point will exist, meaning that the addition of extra capacitance will reduce total losses. Designers of circuits using half-bridges should consider this possibility in their applications: inclusion of a low cost capacitor may help improve efficiency.
REFERENCES
[1] [2] J. P. Harper, Understanding Modern Power MOSFETs Fairchild Semiconductor Power Seminar 2006, www.fairchildsemi.com J. Millman, Microelectronics: Digital and Analog Circuits and Systems, McGraw Hill, p42, 1979.
A-32
## Fairchild Power Seminar 2007
Peter Haaf studied Elektrotechnik at the University of Karlsruhe, Germany. After graduation in 1992, Peter joined the R&D department of Vossloh-Schwabe on power applications, becoming a design team leader. Since 2001, he has been a senior field application engineer for Fairchild Semiconductor in Germany focusing on power supply and high power applications.
Jon Harper After completing a BSc/MEng degree at the University of Bath, England, Jon started as an applications engineer for microcontrollers at National Semiconductor in Germany. He completed an MBA degree at Warwick Business School in 1996 and joined the refounded Fairchild Semiconductor as a marketing engineer. Since 2002, Jon has covered technical marketing for industrial and white goods products, focusing on power electronics.
A-33 | 6,853 | 27,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-10 | latest | en | 0.910829 |
https://www.mathworks.com/matlabcentral/cody/problems/875-return-a-list-sorted-by-number-of-consecutive-occurrences/solutions/1133786 | 1,586,317,483,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371807538.83/warc/CC-MAIN-20200408010207-20200408040707-00446.warc.gz | 1,032,399,101 | 15,839 | Cody
# Problem 875. Return a list sorted by number of consecutive occurrences
Solution 1133786
Submitted on 5 Mar 2017
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 2 2 3 3 7 7 93] y_correct1 = [2 3 7 1 93] ; assert(isequal(popularity_bis(x),y_correct1))
x = 1 2 2 2 3 3 7 7 93
2 Pass
x = [1 1 2 2 2 3 3 7 7 1 93]; y_correct2 = [2 1 3 7 1 93] ; assert(isequal(popularity_bis(x),y_correct2))
3 Fail
x = [1 0 0 2 2 -5 9 9 2 1 1 1 0 11]; y_correct1 = [1 0 2 9 -5 0 1 2 11] ; assert(isequal(popularity_bis(x),y_correct1))
Assertion failed.
4 Fail
x = [1 0 1 1 0 0]; y_correct0 = [0 1 0 1] ; assert(isequal(popularity_bis(x),y_correct0))
Assertion failed.
5 Pass
x = [0 1 0 0 1 1]; y_correct1 = [0 1 0 1] ; assert(isequal(popularity_bis(x),y_correct1)) | 378 | 896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-16 | latest | en | 0.545021 |
https://www.homeownershub.com/uk-diy/best-roof-box-pulley-system-design-972611-.htm | 1,566,279,030,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.56/warc/CC-MAIN-20190820045314-20190820071314-00464.warc.gz | 841,091,080 | 23,735 | # Best roof box pulley system design....
Hi All,
I have recently purchased a roof box and bars to store our camping gear rea dy for the next camping trip. I have a "cunning" idea that I could suspend the roof box (with rails attached) from the garage ceiling filled with our camping gear. When we go camping, I "simply" reverse the car into the gara ge, lower the box/ rails onto the car and away we go....
I am now thinking about the pulley design and it seems a little less straig ht forward than I anticipated (either that or I am making it overly complex :) ) Thought process so far....
1. Ideally I would want 1 rope to pull it up so.... create a 4 way harness to slip over each end of the roof rails, put one pulley on the ceiling and hoist it up. Seemed ideal until I thought about it having no stability and would wave around the ceiling. 2. Enhance the above by having 2 pulleys and a 2 point harness (one each si de). This should stop the swaying about. I then thought.... unless the bo x is evenly weighted, it would be difficult to control - also it would have 2 ropes (not the end of the world) 3. Have 4 pulleys on the ceiling with 4 ropes (one for each "corner")... S hould solve the stability problem but then will have 4 ropes and therefore harder to control...... 4. Have 4 pulleys on the ceiling and 4 on each corner of the box. This way I can run a single rope around these pulleys (running ceiling to platform a nd back at each corner). This way I get 1 rope but should have the stabili ty :) However, this seems a bit overkill.... also, I *think* I would get a n 8:1 ratio on pull length v's height lift so..... a 2m lift would require 16m of pull length!
So... this is as far as I could get....
Surely there is a better solution? Any one have any cunning ideas?
Thanks
Lee.
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On 12/05/2014 14:20, Lee Nowell wrote:
I would go for four separate ropes, one at each corner, with four separate pulleys. You should be able to manage that with a second person. You need a conveniently located cleat for each rope. And make the ropes long enough to be able to lower the whole thing to the floor. That way you can load the box conveniently, also add or remove stuff selectively, clean it out, etc. If cash is no object, have four small electric hoists (not particularly expensive).
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You can buy bicycle hoists for storing bikes up at ceiling height. Maybe a couple of those could be used? I sometimes see them cheap or on special offer.
--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]
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On Monday, May 12, 2014 3:17:14 PM UTC+1, Andrew Gabriel wrote:
I tried that...
First of all the cord would derail into the gap between pulley and bracket if it wasn't perfectly aligned, and jam.
So I replaced the cord with 6mm polyproylene rope, but that showed that the re was so much friction in the pulleys (plain bearings) that one end would go up before the other. If you had the box hooked by the rim it then slid u ntil the hooks reached one end and it fell off with a crash. I tried hookin g onto the roog bars, but inevitably I would need the roof bars alone at so me point.
Plus there was always the problem of clearing a big enough space on the gar age floor to get it down without squashing something.
So I gave up and put a couple of big hook onto the outside wall and use a c ouple of bits of rope to hoik it up, storing it on edge. Probably not too g ood for the metal parts to get it rained on all the time but I keep them sp ry lubed up each time I use it.
This was all for the biggest box you can get - Mont Blanc Vista 540 - it mi ght work for a smaller, lighter box.
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Hi All,
I have recently purchased a roof box and bars to store our camping gear ready for the next camping trip. I have a "cunning" idea that I could suspend the roof box (with rails attached) from the garage ceiling filled with our camping gear. When we go camping, I "simply" reverse the car into the garage, lower the box/ rails onto the car and away we go....
I am now thinking about the pulley design and it seems a little less straight forward than I anticipated (either that or I am making it overly complex :) ) Thought process so far....
1. Ideally I would want 1 rope to pull it up so.... create a 4 way harness to slip over each end of the roof rails, put one pulley on the ceiling and hoist it up. Seemed ideal until I thought about it having no stability and would wave around the ceiling. 2. Enhance the above by having 2 pulleys and a 2 point harness (one each side). This should stop the swaying about. I then thought.... unless the box is evenly weighted, it would be difficult to control - also it would have 2 ropes (not the end of the world) 3. Have 4 pulleys on the ceiling with 4 ropes (one for each "corner")... Should solve the stability problem but then will have 4 ropes and therefore harder to control...... 4. Have 4 pulleys on the ceiling and 4 on each corner of the box. This way I can run a single rope around these pulleys (running ceiling to platform and back at each corner). This way I get 1 rope but should have the stability :) However, this seems a bit overkill.... also, I *think* I would get an 8:1 ratio on pull length v's height lift so..... a 2m lift would require 16m of pull length!
So... this is as far as I could get....
Surely there is a better solution? Any one have any cunning ideas?
....
4 ropes, 4 swivel pulleys 1 above each corner. 4 swivel pulleys at the same height grouped together above a point in the middle of a long wall most convenient to you. Set a cleat hook close to the floor on the wall underneath the four grouped pulleys, (so as to allow you to lower the thing to the floor) and feed all four ropes through each of its own two pulleys, binding all four ropes together at some point past the cleat hook. (Binding them together higher up would prevent them going through the four pulleys above)
There's probably something seriously wrong with such a set-up, but my brain has started to hurt, trying to visualise what it is.
...
.
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That gets my vote except that I'd bind all four ropes individually to a metal ring (easier adjustment to correct stretching etc) and run a fifth rope downwards to pull on. The fifth rope could go round a block and tackle arrangement to reduce the force required.
--
Mike Barnes
Cheshire, England
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Yeah I think the 4 pulley with individual ropes may be the only option.
I guess once you end up with the situation where the 4 ropes can only reall y be brought together at more or less floor level (due to the issue of bein g able to lower it to the floor) I guess I might as well stick to pulling t he 4 ropes but as Michael suggests via a second set of pulleys directly ove r the cleat.
I have a half baked thought which I am not sure it will work or is a good/ better idea.....
How about taking the 4 ropes to a central point (say at the top of the wall with the cleat) and securing the ends there such that they are slightly sl ack when the roof box in on the floor. Now put a 4 way pulley on the ceili ng between the point where they are secured and the corner pulleys [not 100 % sure this one is actually needed!]. Then put a second 4 way pulley on th e ropes between this pulley and where the ropes are finally secured. Tying a rope to this pulley and pulling down, show lift the box? I *think* this w ould also mean that the 4 ropes would only go half way down the wall?
What do you think?
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Lee Nowell wrote:
I don't quite follow that but there's no point in running four ropes where one will do. At the earliest opportunity, make a five-way junction between ropes to the four corners and the rope that you pull on.
I'd go for: ropes from each of the corners vertically to ceiling pulleys then horizontally to a pulley at the top of the wall (all four of those close together). With the box on the floor join those four ropes just below the top-of-wall pulleys to a fifth rope. Take that fifth rope round a pulley at the bottom of the wall and then to another at the top. That will give you the full range of movement at a convenient height for pulling on.
--
Mike Barnes
Cheshire, England
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> Any one have any cunning ideas?
I can only think of rodents licking their lips at their ideas for those ropes :)
--
Robin
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Indeed. You lay the box on the ground, attach each of the four ropes and run each rope through each of their two pulleys and then as soon as they're clear of the second grouped pulley, it's at that point they can all be joined together giving a single rope to pull.
A block and tackle arrangement underneath would also be preferable to a cleat hook as the latter would presumably need to be uncomfortably close to the floor to allow enough length of rope to get the box close to the ceiling, in any case.
...
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On 12/05/2014 14:20, Lee Nowell wrote:
How about something along these lines:
http://www.ebay.co.uk/itm/3-Lath-Victorian-Kitchen-Ceiling-Pulley-Clothes-Airer-Maid-Laundry-Dryer/360508253939
Some come with all the pulleys you need and all you need is a strap between the hanger and around the roof box.
I feel this one is a bit pricey but there might be others. In their day they were called "Spacesaver"s.
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On Monday, May 12, 2014 2:20:23 PM UTC+1, Lee Nowell wrote:
This is all too much work ...
Raise the car on a hydraulic lift until the roof box engages with ceiling-m ounted latches.
Perhaps more practically, instead of using pulleys, have each corner rope w ind up on its own reel, 2 reels are on the same axle (front and rear axles) , then you just have to link the 2 axles by a chain. One of the axles exte nds to the side wall of the garage, and you then take power down by a bevel gear shaft to an operating handle at convenient height - or even use a mot or. You would need to do a lot of winding with a small reel but at very low power.
Think of the rope and shaft gear used for opening large high windows.
Owain
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On 12/05/2014 18:35, snipped-for-privacy@gowanhill.com wrote:
What a stupid idea.
Easier to raise the garage, drive the car under, then lower it. Hydraulic jack on each corner would do the job.
--
Dave - The Medway Handyman www.medwayhandyman.co.uk
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Maybe not in the spirit of uk.d-i-y, you could always just by a ready made solution (Thule Roofbox Lift):
<http://www.skidrive.co.uk/accessories/product_detail.php?partnoW1000&g clid=CLqjlvyNp74CFUfLtAodtl0AhA>
I did have such ideas for my roofbox, but the entrance to the garage is too low. Anyway I then got a trailer which we use for camping much of the time, suits us much better. All the camping gear can live in there and much easier/quicker to load the trailer when striking camp.
--
Chris French
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writes
Or you could just steal the rather clever design and make one yourself.
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Sorry... to clarify my revised idea....
Essentially, you would run the ropes from the 4 pulleys to the wall and sec ure it there (rather than on a pulley). You would then have 1 rope which w ould concurrently pull all 4 ropes as shown in the dodgy side on ascii diag ram below (P = pulley; C = Corner)...
P------P------------ 4 ropes fixed to wall | | ^ | | pull here with 1 rope | | C C
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I just thought of an interesting new problem.... On the assumption that I w ould hoist the ropes up out of the way when the roofbox is not stored, how would I get the ropes back down again? I guess I would have to hook them or have some weights attached to them or something......
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Lee Nowell wrote:
Pole with a hook on the end?
Jump?
--
Mike Barnes
Cheshire, England
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On 12/05/2014 14:20, Lee Nowell wrote:
KISS! I just use 2 luggage straps (one at each end) around the joists - it works!
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How about you just don't hoist them up out of the way? Have a ceiling mounted pulley close to the wall, rope hangs down next to the wall, coil it and hang it on the cleat you'll want anyway. That way when you use it you'll still have the whole space clear with no ropes diagonally across it. In other words, if you do something like this: http://forums.quattroworld.com/allroad/msgs/132034.phtml add an extra pulley on the roof ahead of the nose of the box, above the cleat. Not that just leaving it like that would be too bad.
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HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 3,537 | 13,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-35 | latest | en | 0.947432 |
https://www.scribd.com/document/107728730/Homework-2-Solution | 1,524,784,386,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948617.86/warc/CC-MAIN-20180426222608-20180427002608-00009.warc.gz | 876,927,258 | 20,178 | # Homework #1
Chapter 3 1. A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one channel, what are the data rates if we use one harmonic, three harmonic, and five harmonics? Ans BW = 6 MHz 1) BW from 0 Hz to f1st harmonic = 6 MHz; Bit rate = 2* f1st harmonic = 2 * 6 = 12 Mbps 2) BW from 0 Hz to f3rd harmonic = 6 MHz; f3rd harmonic = 3 * f1st harmonic f1st harmonic = 6 MHz / 3 = 2 MHz Bit rate = 2 * f1st harmonic = 2 * 2 = 4 Mbps 3) BW from 0 Hz to f5th harmonic = 6 MHz; f1st harmonic = 6 MHz / 5 = 1.2 MHz Bit rate = 2 * f1st harmonic = 2 * 1.2 = 2.4 Mbps 2. The attenuation of a signal is -10 dB. What is the final signal power if it was originally 5 W? Ans –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W 3. We measure the performance of a telephone line (4KHz of bandwidth). When the signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this telephone line? Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps 4. What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 us and a processing time of 1 us. The length of the link is 2000 Km. The speed of light inside the line is 2 x 108 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible? Ans Propagation time = distance / propagation speed = 2000 Km / 2 x 108 m/s = 10 ms Transmission time = Message size / Bandwidth = 5 x 106 bits/ 5 Mbps = 1 s Queuing time = 10 routers * 2 us = 20 us Processing Delay = 10 routers * 1 us = 10 us Total delay (latency) = 10 ms + 1 s + 20 us + 10 us = 1010.03 ms = 1.01003 s 1 s Chapter 4 1. Assume a data stream is ‘1101000000000010’ s.. Encode this stream, using the following code schemes. How many changes (vertical line) can you find for each scheme? a. Unipolar: 4 changes between bit b. NRZ-L: 4 changes between bit c. NRZ-I: 4 changes between bit d. RZ: 16 changes at the middle of each bit + 15 changes between bit e. Manchester: 16 changes at the middle of each bit + 11 changes between bit f. Diff. Manchester: 16 changes at the middle of each bit + 11 changes between bit
a.g. An analog signal with frequencies from 2000 to 6000 Hz Ans Sampling rate >= 2 x fhighest >= 2 x 6000 = 12. Using the Nyquist theorem. A signal with a vertical line in the time-domain representation Ans Sampling rate >= 2 x fhighest >= 2 x ∞ = ∞ samples /s . i. What is the sampling rate for PCM if the frequency ranges from 1000 to 4000 Hz? Ans Sampling rate >= 2 x fhighest = 2 x 4000 = 8000 samples / s 3.000 samples /s c. AMI: 2B1Q: MLT-3: 6 changes between bit 3 changes between bit 4 changes between bit 2. An analog signal with bandwidth of 2000 Hz Ans Sampling rate >= 2 x fhighest = 2 x (fLowest + BW) >= 2 x (fLowest + 2000) b. h. calculate the sampling rate for the following analog signals. A signal with a horizontal line in the time-domain representation Ans Sampling rate >= 2 x fhighest >= 2 x 0 = 0 samples /s d.
000 bps 5. We have sampled a low-pass signal with a bandwidth of 200 KHz using 1024 levels of quantization. A signal is sampled. Each sample represents one of four levels.000 samples /s Quantization 10 bits/sample. Bit rate = 8000 * 2 = 16. Bit rate = 400. Ans Low pass signal: frequency between 0 – 200 KHz BW = 200 KHz Sampling rate >= 2 x fhighest = 2 x 200 KHz >= 400.4.000 x 10 = 4 Mbps . Calculate the bit rate of the digitized signal. what is the bit rate? Ans Quantization 2 bits/sample. How many bits are needed to represent each sample? If the sampling rate is 8000 samples per second. | 1,152 | 3,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-17 | latest | en | 0.879568 |
https://blender.stackexchange.com/questions/296881/can-you-have-2-spline-profiles-on-one-spline | 1,718,960,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00188.warc.gz | 111,428,684 | 36,997 | # Can you have 2 spline profiles on one spline?
Im trying to make a road using a spline (I will call it road spline). I made a spline profile and attached it to original road spline and it gave me a nice walkway next to the road and it looks something like this.
Now I want to make the same walkway profile but I want it to be a little bit lower than the last one, it looks like this.
I made this using 2 profile splines and 2 road splines. Is it possible to make it with 1 road splines and 2 profile splines with nice blend with it.
This works in 3ds max, here is example screenshot from max:
Link to that max video: https://youtu.be/PdsnVQOlO34
I've never seen anything like that with curve profiles so I don't think you can do it in blender. If you are fine with a different type of modeling here's what you could do:
Instead of making the main road using a curve as a profile, use that profile to make a solid segment of the road that can be turned into an array. Then make the array fit the length of the curve you want to use as the curvature of the road
You then need to make a segment with the detail that you want, like the ramp. You can duplicate the arrayed object (main road) and apply the array, then use that as the base of the ramp so that the edges match with the arrayed object. Once you have the ramp modeled, you can create a cube around it that is slightly narrower and use an Intersect boolean modifier on it:
This same cube is used in a difference Boolean with the main road:
This results in a segment of the main road being removed and the ramp segment taking it's place. (Parent the ramp to the cube and the cube to the main road so that everything stays in place if you move it. This lets you hide the ramp so it's not in the way).
If you want the geometry to merge into a single piece you can create a simple geometry node set up on the main road that joins its geometry to the ramp's and then merges the vertices by distance. The result of this is that you can move the cube along the axis of the array and it will change the position of the ramp in the array. You can then give the main road a Curve modifier and make it follow the curve that you were using for it:
It's not as simple as the 3dmax thing but the advantage here is that you are not just working with curve profiles so you can add a lot more detail than with an extrusion. I'm pretty sure someone who knows more about geometry nodes could reduce all of this to a single setup instead of multiple modifiers on the main road but I'm not there myself. As for the lofting part, you can do that with the Bridge tool on the Bool Tools addon (Comes with blender)
• Thank you, I will definetly give this a go. Commented Jul 20, 2023 at 7:19 | 633 | 2,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | latest | en | 0.95621 |
http://mathhelpforum.com/calculus/46791-lhopital-rule-qn-print.html | 1,503,223,427,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106367.1/warc/CC-MAIN-20170820092918-20170820112918-00693.warc.gz | 258,932,207 | 3,168 | # L'hopital Rule qn
• Aug 26th 2008, 06:55 AM
boredaxel
L'hopital Rule qn
I understand that when using l'hopital rule to find the limits of a fuction with power unknown( eg X^x), we have to use natural logarithm. But can we use logarithm to base 10 instead?
• Aug 26th 2008, 07:00 AM
Moo
Hello,
Quote:
Originally Posted by boredaxel
I understand that when using l'hopital rule to find the limits of a fuction with power unknown( eg X^x), we have to use natural logarithm. But can we use logarithm to base 10 instead?
Yes. But it doesn't change anything to the method :)
Actually, let $\ln$ be the natural logarithm and $\log$ the logarithm to base 10.
The relation between the two is just linear :
$\log(x)=\frac{\ln(x)}{\ln(10)}$
• Aug 26th 2008, 10:36 AM
ThePerfectHacker
Quote:
Originally Posted by boredaxel
But can we use logarithm to base 10 instead?
In mathematics there really is no such thing as a logarithm with base 10. Because there is nothing important about the number 10, except that it is a number of the decimal system we use, and might be convient for certain exponential calculations. You really only use that logarithm in high school to get introduced to logarithms. Mathematicians always use the natural logarithm. They do not even call it the 'natural' logarithm - they simply say the logarithm function. Because for them there is only one such function. And when they write $\log$ they mean $\ln$, that distinction is again something you see in high-school but stop seeing when you do more advanced math. | 402 | 1,533 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-34 | longest | en | 0.900088 |
https://www.jiskha.com/display.cgi?id=1320277222 | 1,516,278,089,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887253.36/warc/CC-MAIN-20180118111417-20180118131417-00567.warc.gz | 938,203,557 | 3,701 | # Chemistry
posted by .
Assuming that the smallest measurable wavelength in an experiment is 0.410 fm (femtometers), what is the maximum mass of an object traveling at 815 m*s^-1 for which the de Broglie wavelength is observable?
• Chemistry -
w = wavelength
w = h/mv
Convert w to meters, solve for m in kg.
• Chemistry -
5.03*10-4
• Chemistry -
@DrBob22
I think you would be of more help if you actually answered the question instead of just giving the equation in all your posts.
## Respond to this Question
First Name School Subject Your Answer
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More Similar Questions | 648 | 2,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-05 | latest | en | 0.859776 |
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1. Nov 2, 2007
KleZMeR
1. The problem statement, all variables and given/known data
In a purley magnetic field B, motion of particle in x-y plane is circle; use this property, with result from part c to show on average the particle travels with constant velocity U:
where Vdrift=(1/B^2)(ExB)
2. Relevant equations
"x" implies cross product
E=0i+Ej+0k
B=0i+0j+Bk
F=q(E+VxB)
results from Part c) showed by Galilei transform:
E'=(E+UxB)
V'=(V-U)
3. The attempt at a solution
Well right now i'm solving for U. I also have the answer so i've plugged in vdrift and try to work that out, usually "show" means working backwards in this course. I am not sure if I should use trig functions for the circle, I mean, it says to use that property of pure magnetic feild, I know at one point i have to incorporate the trig rotation because in part f it says my x(t) and y(t) are sin and cos functions.
Last edited: Nov 2, 2007
2. Nov 2, 2007
KleZMeR
ok
well i'm a little farther, the equations of circular motion are what i'm having a little problem with, do they depend on the lorenz force? or is it just:
Wc= Omega Cyclotron
Components:
V'x=WcVy
V'y=-WcVx
With these I can de-couple the equation, but with the equation of motion dependent on the lorenz force i get V'y=(q/m)(E-BVx) and i do not know if i can decouple these because it is two terms...
i think obvious solution to use is u=v-v' so i'm looking for these two i guess? but it says to use result from part c) that E'=(E+UxB) , but i got this from the u=v-v' relationship so i think this is "'the result" i should be using??
3. Nov 2, 2007
KleZMeR
ok so pure magnetic field means no E??? there is E in Vdrift so i wonder if E stays in equation of motion
4. Nov 2, 2007
KleZMeR
ok, well i found in this forum somewhere that E'(y)=V*B'(z), is that right???
5. Nov 3, 2007
Gokul43201
Staff Emeritus
I don't know why the given answer has a drift speed coming from an E-field, when the problem states that there is no E-field. Also, what is the "result from part c"?
A particle traveling along a circle clearly can not have a constant velocity. The simplest way to show that it has a constant speed is to show that the net force on the particle is normal to its velocity, and this is trivially true for this problem.
6. Nov 3, 2007
KleZMeR
ok
ok, I did state what the result from part c) is in the problem: results from Part c), I showed by Galilei transform:
E'=(E+UxB)
V'=(V-U)
right now I am working in the drift frame, so maybe thats why the extra term is there, but I emailed my professor and asked him the same question Gokul43201 asked (why is there an E in purely magnetic field and where does Vdrift come from), have no response
I thought "on average" might imply some sort of average value function that should be applied? that's what the answer looks like now that i think about it. an integral of (1/B) might return a (1/B^2) function,
the next two steps are integrating equations of motion in lab frame (no drift) and finding V(t), r(t), but these two X(t) and Y(t) also incorporate a Vdrift as coefficient before the trig functions, this Vdrift takes the place of Vo, but Vo is said to be zero in lab frame? this is kindove off topic, but gives some insight into the two different situations.
7. Nov 3, 2007
KleZMeR
ok, so i use rule A(dot)B = |A||B| cos(theta) and show theta at 90 always???? | 964 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-36 | longest | en | 0.902452 |
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# The Congruent Number Problem
[Up]
## Congruent Numbers
A positive integer n is a congruent number if it is the area of a right triangle with rational side lengths. The numbers 1, 2, 3, and 4 are not congruent, but 5, 6, and 7 are.
For any positive integer n, let En denote the elliptic curve defined by the equation y2 = x3 - n2x. Then n is a congruent number if and only if En has infinitely many solutions. The curves En are quadratic twists of the elliptic curve y2 = x3 - x, which is isogenous to the modular curve X0(32). | 159 | 584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-23 | latest | en | 0.888063 |
http://forums.wolfram.com/mathgroup/archive/2005/Dec/msg00609.html | 1,632,427,384,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00434.warc.gz | 27,010,371 | 7,764 | What we from 1^Infinity, Infinity^0, and similar stuff
• To: mathgroup at smc.vnet.net
• Subject: [mg63308] What we from 1^Infinity, Infinity^0, and similar stuff
• From: ted.ersek at tqci.net
• Date: Fri, 23 Dec 2005 05:08:32 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```I am using Mathematica 4.1, and version 5 may work different in this case.
It seems I can compute (1^z) where (z) has any numeric value and
Mathematica returns the Integer 1. I can also compute (z^0) where (z) is
any non zero value and Mathematica returns the Integer 1. Hence I think
the following should return {1,1,1,1,1,1,1}. Can someone explain why that
would be wrong?
In[1]:= Off[Power::indet, Infinity::indet];
{1^Infinity, 1^(-Infinity), 1^ComplexInfinity,
1^(0*Infinity), Infinity^0, (-Infinity)^0, ComplexInfinity^0}
Out[2]= {Indeterminate, Indeterminate, Indeterminate, Indeterminate,
Indeterminate, Indeterminate, Indeterminate}
Assigning DownValues to Power will not change the way the results above
are done. However, the UpValues below give the results I prefer. I
thought some users might find this helpful.
In[3]:=
Unprotect[DirectedInfinity, Indeterminate];
DirectedInfinity/: Power[1, DirectedInfinity[_]]=1;
DirectedInfinity/: Power[1, DirectedInfinity[]]=1;
Indeterminate/: Power[1, Indeterminate]=1;
DirectedInfinity/: Power[DirectedInfinity[_], 0]=1;
DirectedInfinity/: Power[DirectedInfinity[], 0]=1;
Protect[DirectedInfinity, Indeterminate];
In[10]:=
{1^Infinity, 1^(-Infinity), 1^ComplexInfinity, 1^(0*Infinity),
Infinity^0, (-Infinity)^0, ComplexInfinity^0}
Out[10]=
{1,1,1,1,1,1,1}
--------------------------------------------
Note:
FullForm[Infinity] is DirectedInfinity[1]
FullForm[ComplexInfinity] is DirectedInfinity[]
Have a Merry Christmas,
Ted Ersek
```
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### What Sign Do You Use?
```
Date: 10/21/2001 at 18:42:24
From: butterfly
Subject: If two negatives equal a positive
If two negatives equal a positive, then how do you do it, and what
sign do you take?
Like -4-(-7): would it be -4 + -7= 11?
Thank you,
Butterfly
```
```
Date: 10/22/2001 at 11:33:43
From: Doctor Ian
Subject: Re: If two negatives equal a positive
Hi Butterfly,
When you _multiply_ two negative numbers, you get a positive number:
Negative x Negative = Positive - Dr. Math FAQ
http://mathforum.org/dr.math/faq/faq.negxneg.html
But when you _add_ two negative numbers, you get another negative
number.
This illustrates one of the problems with trying to rely on 'rules'
like 'two negatives equal a positive' without really understanding
what the rules mean. The irony in the situation is that once you
understand what the rules mean, there is no need to remember them any
more!
If you owe one friend 4 dollars (i.e., you have '-4 dollars'), and you
owe another friend 7 dollars (i.e., you have another '-7 dollars'),
how much money do you have altogether? It's not a positive number, is
it? If it were, we could all get rich by spending money instead of by
earning it.
Now, suppose you owe a friend 7 dollars (i.e., you have -7 dollars),
and I pay him 4 of those dollars for you (i.e., I've 'taken' -4
dollars away from you). How much money do you have now?
-7 - (-4) = ?
Here is another way to look at it. I can rewrite a subtraction this
way:
7 - 4 = 1 + 1 + 1 + 1 + 1 + 1 + 1 - 1 - 1 - 1 - 1
Do you see why? But each of the subtractions cancels out one of the
= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 + 1 + 1
\___/ \___/ \___/ \___/
0 0 0 0
= 1 + 1 + 1
= 3
Well, it works the same way with negative numbers:
-7 - (-4)
= -1 + -1 + -1 + -1 + -1 + -1 + -1 - (-1) - (-1) - (-1) - (-1)
= -1 - (-1) + -1 - (-1) + -1 - (-1) + -1 - (-1) + -1 + -1 + -1
\_______/ \_______/ \_______/ \_______/
0 0 0 0
= -1 + -1 + -1
= -3
Now, this would be really tedious for something like
-298 - (-197) = ?
And that's why it's so tempting to make up rules and shortcuts. But
applying the wrong rule or shortcut just gets you to the wrong answer
more quickly.
or if you have any other questions.
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Negative Numbers
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# Discussion on: Java Daily Coding Problem #001
## Replies for: This sounds reasonable to me but note that the problem doesn't state whether the numbers in the array can be negative! If I got this question in a...
imeugn
I was actually thinking more in the line of:
Let n={3,5,10,15,17} and k=13. 15 and 17 should not be considered at all since they are greater.
The numbers are not supposed to be negative. The worst case remains the same.
But if n is meant to be sorted, we can then only check the left side of n up to the very closest number to k, which is still less than k. That would change the approach in general, I would think. | 190 | 777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.951357 |
https://www.physicsforums.com/threads/volume-using-polar-coordinates.272186/ | 1,475,180,795,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661915.89/warc/CC-MAIN-20160924173741-00135-ip-10-143-35-109.ec2.internal.warc.gz | 982,335,576 | 15,264 | # Volume Using Polar Coordinates
1. Nov 15, 2008
### squeeky
1. The problem statement, all variables and given/known data
Use polar coordinates to find the volume bounded by the paraboloids z=3x2+3y2 and z=4-x2-y2
2. Relevant equations
3. The attempt at a solution
Somehow, through random guessing, I managed to get the right answer, it's just that I don't understand how I got it. Also, because the z is involved, I actually used cylindrical coordinates, but would that still be considered the same thing as polar coordinates? So anyway, I changed the two paraboloid equations to z=3r2 and z=4-r2. Then setting these two equations equal to each other (since they are both equal to z), I solved for r and got the limits of -1,1. For the limits of theta, I just happened to take it from 0 to 2pi. Lastly for the z limits, I just tried from 4-r2 to 3r2, so that gave me the equation:
$$\int^{2\pi}_0\int^1_{-1}\int^{3r^2}_{4-r^2}dzrdrd\theta$$
However, solving this equation didn't give me the right answer, so I changed the limits of r to 0 to 1, and switched the z-limits around, so now it is:
$$\int^{2\pi}_0\int^1_0\int^{4-r^2}_{3r^2}dzrdrd\theta$$
And solving for this, gave me the right answer of 2pi. The problem is that I don't understand the real logic behind what I did.
So in summary, what I didn't understand was how to establish the limits for theta, r, and z.
Last edited: Nov 15, 2008
2. Nov 16, 2008
### HallsofIvy
Staff Emeritus
Those two paraboloids intersect when z= 3x2+ 3y2= 4- x2- y2 or 4x2+ 4y2= 4 which reduces to x2+ y2= 1. That projects onto the xy-plane as the unit circle. It shouldn't be hard to see that the volume you want is above that circle. To cover that circle, $\theta$ goes from 0 to $2\pi$ and r goes from 0 to 1. Those are your limits of integration.
For each (x, y), the upper boundary is z= 4- x2- y2 and the lower boundary is z= 3x2+ 3y2. The volume of a "thin" rectangular solid used to construct the Riemann sums for this volume would be [(4- x2- y2)- (3x2+ 3y2)]dxdy= (4- 4x2- 4y2)dxdy= 4(1- x2- y2)dxdy and that, in polar coordinates, is 4(1- r2)rdrd$\theta$.
Yes, "cylindrical coordinates" is just polar coordinates in the xy-plane.
No, you didn't. Or you shouldn't. r cannot be negative. r must be between 0 and 1.
Did you think at all about the geometry of the situation? This has circular symmetry. To cover the entire volume you have to cover the entire circle: 0 to $2\pi$.
Surely you recognized that z= 4- r2 is ABOVE z= 3r2 for r<= 1? | 794 | 2,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-40 | longest | en | 0.940069 |
https://forum.onshape.com/discussion/9346/planar-mate-question-about-limits | 1,701,969,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00835.warc.gz | 311,181,237 | 15,801 | Welcome to the Onshape forum! Ask questions and join in the discussions about everything Onshape.
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# Planar Mate Question About Limits
Member Posts: 478 PRO
VERY confused as to how limits are restrained with planar mates. Which direction is 'x'? which direction is 'y'? how to locate part then change max and min? Someone understand this but I have been messing with this a for a very long time and cannot grasp the idea.
• Moderator, Onshape Employees Posts: 5,060
Hi @larry_hawes a mate connector has 3 lines with colours RGB (red green blue) equating to XYZ axes. It is the mate connector selected second that is used to define limits (if I remember correct).
Senior Director, Technical Services, EMEAI
• Member Posts: 478 PRO
Thanks Neil, That will help define direction but where do those limits start/stop? If I want to go from edge to edge of a part where are those edges and how are they defined in regards to limits? Is there an assumed '0' where the min and max are calc'd from and perhaps to? I can see the edges clearly but don't know how those edges refer to the limits.
• Member Posts: 478 PRO
I have stumbled on to some values that seem to work but have no definitive reference as to why they behave as they do. How would one locate a definite edge and use the limit/min/max to define that edge as the 'limit' of allowed motion? In this case I am moving the part along the 'x' axis of the planar mate but cannot seem to find a way to limit the motion to those definite edges.
• Member Posts: 478 PRO
In this simple example a slider mate is more appropriate but still challenging to find the right and left edge and why does the min define one direction's limit and the max determine another direction's limit? And still cannot find the exact edge distance to limit motion other than visually.
The assembly I am working on is proprietary but needs 2 planar mates to define its motion(s) with the same challenge of determining those exact limits.
• Member Posts: 478 PRO
I THINK I figured it out and thank you for your indulgence. | 563 | 2,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.938762 |
https://www.hindawi.com/journals/ace/2018/5493579/ | 1,652,774,965,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00035.warc.gz | 965,499,307 | 78,125 | / / Article
Research Article | Open Access
Volume 2018 |Article ID 5493579 | https://doi.org/10.1155/2018/5493579
Younggyun Choi, Janghwan Kim, Heejung Youn, "Numerical Analysis of Laterally Loaded Piles Affected by Bedrock Depth", Advances in Civil Engineering, vol. 2018, Article ID 5493579, 9 pages, 2018. https://doi.org/10.1155/2018/5493579
# Numerical Analysis of Laterally Loaded Piles Affected by Bedrock Depth
Accepted27 Aug 2018
Published26 Sep 2018
#### Abstract
This study investigates the lateral behavior of pile foundations socketed into bedrocks using 3D finite difference analysis. The lateral load-displacement curve, pile deflection, and bending moment distribution were obtained for different bedrock depths between 3 and 20 m. It was discovered that bedrocks that have a depth of 7 m (7D) or less influence the lateral behavior of the pile. The p-y curves were collected at depths of 2.0–4.5 m, and the effect of the bedrock on the curves was evaluated. It was observed that the p-y curves were significantly affected by the material properties of the bedrock if the rock is located in close proximity (within 3D), but the effect is diminished if the p-y curves were 3.5 m (3.5D) or farther from the bedrock.
#### 1. Introduction
The subgrade reaction method was developed by the Winkler assumption [1], where the soil resistance per unit length, p, is assumed to be proportional to the pile deflection, y, at a certain point. The linear relationship between p and y does not represent the nonlinear characteristics of the soil. To address this phenomenon, the nonlinear p-y method was introduced by McClelland and Focht [2]. Since the introduction of this method, the p-y curves are typically used to indicate the nonlinear springs representing lateral soil resistance and deflection along the pile. In conventional designs, the soil supporting the pile can be replaced with a set of nonlinear springs, and the pile is assumed to behave as a beam supported by nonlinear springs. The p-y method is widely accepted for evaluating the lateral behavior of piles due to its simplicity. Numerous researchers have proposed different p-y curves with considerations for soil types, loading type (static or cyclic), and water conditions (saturated or unsaturated soil). In the early stages of p-y curve development, the p-y curve for submerged soft clay was developed by Matlock [3], stiff clay with free water by Reese et al. [4], stiff clay without free water by Reese and Welch [5], and submerged sand by Reese et al. [6]. The p-y curves for sand developed by Reese et al. [6] were modified into a simpler equation that yielded better estimation by Murchison and O’Neill [7]. The curve of Murchison and O’Neill was later adopted by the American Petroleum Institute [8] for the design of offshore platforms.
There are limited analytical researches available for evaluating the lateral response of piles socketed into rocks. Carter and Kulhawy [9] investigated the lateral behavior of drilled shafts (flexible and rigid type) socketed into rocks using finite element methods. Closed-form equations were proposed considering factors such as loading conditions, material properties, and rock mass stiffness. Zhang et al. [10] proposed a nonlinear continuum method to predict the lateral response of drilled shafts, in which the ultimate resistance of a rock mass was calculated based on the Hoek–Brown failure criterion. To et al. [11] believed the p-y curves are inappropriate for estimating the behavior of jointed rock masses as the curves consider soil as a continuum. Thus, they proposed a discontinuum model based on rock mass with two or three joint sets. Recently, the elastoplastic continuum method was proposed considering the variation in the flexural rigidity of drilled shafts in multilayers of soil and rock masses [12].
Numerical simulations have been performed extensively to account for anisotropic and discontinuous characteristics of rock medium [1318]. The three-dimensional distinct element method was adopted using 3DEC to simulate the jointed rock mass under lateral loading, and p-y curves were proposed for mudstone [1315]. It was discovered that the number of joints significantly affects the lateral response of piles in rock masses. Liang and Shatnawi [16] performed a series of finite element analyses, presented charts for the initial modulus of subgrade reaction for transversely isotropic rock media, and recommended using five elastic constants for estimating the initial modulus of subgrade reactions. Comparisons were made on the lateral behavior of drilled shafts in the rock medium for isotropic properties following the Hoek–Brown criterion and for transverse isotropic rock properties [17]. Shantnawi [19] expanded the p-y criteria considering the anisotropic behavior as well as discontinuous joint sets in rock masses.
The p-y curves for rocks are very difficult to obtain and validate through experiments because the full-scale tests on rocks are costly and because it is difficult for the rock to fail within the available capacity of the loading system. Indeed, the number of full-scale tests on drilled shafts in rocks was very limited [20, 21]. The complete form of p-y curves for weak rocks had not been available until that reported by Reese [22]. He proposed p-y curves based on the results of the two load tests on coral limestone conducted by Nyman [21]. Reese [22] commented that the proposed p-y curves in weak rock should be used with caution due to the limited number of test results. Gabr et al. [23] proposed hyperbolic p-y curves for weak rocks based on their field test results on small-diameter drilled shafts. The curves require two parameters: ultimate resistance and initial modulus of the subgrade reaction. However, their curve was criticized as basing their estimation of the initial modulus of subgrade reaction in the study of Vesic [24] that is inappropriate for piles socketed into rocks [25]. Numerous other researches indicate the inappropriateness of conventional p-y curves for different types of rock medium [2628]. p-y curves for the Ohio shale was proposed using the results of field lateral load test on drilled shafts with a diameter of 1.8 m socketed into shale [29]. According to the test results, the p-y curves of weak rocks [22] were found to underestimate the pile deflection. The same research group developed p-y curves for rock and intermediate geomaterials using in situ pressure meter tests [30] and p-y curves for transverse isotropic rock [31, 32]. Cho et al. [26] conducted six field tests on drilled shafts, proposed p-y curves for soft weathered rock, and concluded that the Reese p-y curves [22] overestimate the resistance. Conversely, the stiff clay model [5] underestimates the resistance for the same pile deflection. Liang et al. [33, 34] recommended hyperbolic p-y curves for rock masses using numerical simulations, where the curve was validated through two full-scale load tests on drilled shafts with a diameter of 1.3 m. Recently, centrifuge model tests were employed to simulate large-diameter drilled shafts in rocks [35, 36]. p-y curves for weakly cemented soils to well cemented soils were proposed by Guo [36]. Full lateral load tests were performed on drilled shafts in weak calcareous sandstones, and centrifuge model tests were performed for preinstalled piles in very weakly cemented sands. Parsons et al. [37] introduced a case history of full-scale cyclic lateral load tests on two drilled shafts rock-socketed in limestone.
The purpose of this study is to investigate the effect of bedrock depth on the lateral behavior of piles socketed into bedrocks. Due to the significant difference in material stiffness and strength between soils and bedrocks, the soil-pile interaction near the boundary between the soil and rock is very complex to understand. Conventional p-y curves of soils can be created from the material properties of the soils, not from the nearby bedrocks. However, the p-y curves of soils are likely to be considerably affected by the material properties of the rock if the rock is located in close proximity, implying that the p-y curves obtained near the bedrock would cause the overestimation of the ultimate soil resistance and the modulus of the subgrade reaction, which is misleading. The three-dimensional finite difference method was adopted to derive the p-y curves of soils overlying rocks, while the variations in the ultimate soil resistance and the initial modulus of the subgrade reaction at varying distances from the rock were obtained.
#### 2. Numerical Modeling
Figure 1 illustrates the numerical model along with the stratigraphy, pile dimension, socket depth into bedrock, locations where the p-y curves were collected, variable for the parametric study, 3D model in FLAC3D, and boundary conditions. As shown in Figure 1(a), the pile is socketed 3 m into bedrock that is underlain by sand with a thickness of H. The pile head at the ground level was laterally pushed over, and the p-y curves were collected every 0.5 m from 2.0 to 4.5 m from the ground surface. The distance from the locations of the p-y curves to the bedrock (Xpy) varied with varying thicknesses of the sand layer (H). For example, when H is 5 m, the distances, Xpy, are 0.5, 1.0, 1.5, 2.0, 2.5, and 3.0 m. For the same thickness, H, the p-y curves of sand at an identical depth would differ due to the influence of varying bedrock depths.
The simulated pile was 1 m in diameter and 20 mm in thickness and embedded 3 m into the bedrock layer. The length of the simulated pile varied from 6 to 23 m, and the pile was created as a so-called “wish-in place” pile. The side boundary was located 10 m from the center of the pile and fixed in the x and y directions against translation. In addition, the bottom boundary was 7 m below the pile tip and fixed in three directions against translation. The pile head was laterally pushed over with a velocity of 10−6 mm per step up to 500 mm in the lateral direction. The applied lateral displacement was far greater than 38 mm that is considered the serviceability limit in lateral [38]. This is because the 38 mm displacement did not induce sufficient pile deflection to generate the p-y curves. The lateral displacement, lateral load, and moment profiles were obtained and used to calculate the soil resistance per unit length, p, through the double integration of the bending moment profile and the pile deflection, y, by taking the second derivative of the moment profile.
Table 1 shows the material input parameters used in the numerical study. The sand and bedrock were modeled using the Mohr–Coulomb model, but the model for the bedrock was reduced to the Tresca model because the internal friction angle was set to zero. The friction angle of the sand was 35.9° while the elastic modulus was 42.0 MPa. A cohesion of 27.6 MPa was assigned to the bedrock with zero friction angle. The elastic modulus of sand was assumed to be constant along the depth for simplification. This may reduce the lateral displacement of the pile near the ground surface. The pile was assumed to have linear elastic behavior with an elastic modulus of 210 GPa and Poisson’s ratio of 0.3. Because the pile was modeled as a solid instead of tabular pile, the elastic modulus of the pile was back-calculated by matching the flexural rigidity (EI). The input value for the elastic modulus was 31.64 GPa, resulting in an identical flexural rigidity of 1,550 MN·m4.
Model γ′ (kN/m3) ϕ (°) Su (MPa) E (MPa) υ Pile Elastic 68 n/a n/a 210,000 0.300 Sand M–C model 9 35.9 0 42 0.300 Bedrock M–C model 16 0 27.6 22,300 0.287
The slip and separate only model was adopted to simulate the pile-soil interaction; the interface element of the model was characterized by the Coulomb friction model in the shear direction and by tensile strength in the normal direction. In the Coulomb friction model, the shear resistance linearly increases up to the maximum side shear with increasing shear displacement, and after the maximum shear, the shear resistance remains constant against further displacement. For the Coulomb input parameters, the friction angle and the cohesion of the surrounding materials were used. The shear and normal stiffnesses of the interface element were set to ten times the equivalent stiffness of the surrounding soil as recommended in the FLAC3D manual [39].
In numerical simulations, the numerical results should not be affected by the mesh size, boundary condition, or interface element of the pile-soil interaction. The load-displacement curves at the pile head were collected to evaluate the effect of these factors. Figure 2 illustrates the effect of mesh size on the numerical results. The lateral load-displacement curves were obtained for different mesh sizes: 0.10, 0.25, 0.50, 1.00, 2.00, and 4.00 m. More lateral load was required to displace the pile head for larger mesh sizes, and the required load drastically dropped as the mesh size decreased. Figure 2(b) shows the variation in the lateral load for a 38 mm pile displacement, indicating that the effect of mesh size appears to be negligible when the mesh size is less than 0.25 m. Therefore, a mesh size of 0.25 m was used in the numerical models.
The boundary should be located sufficiently far from the pile so that the stress transferred from the laterally loaded pile to the soil should not reach the boundary. Figure 3 shows the load-displacement curves for four cases used to evaluate the effect of boundary conditions on the numerical results. The boundaries were located 5 to 20 m away from the center of the pile in the x direction. The boundary in the x direction plays a key role because the pile was displaced in the x direction. It was found that the lateral load-displacement curve did not significantly differ when the boundary was placed 10 m or farther in the x direction. Thus, the boundaries were created at 10 times the pile diameter (10D) away from the center of the pile in both the x and y directions.
The stiffness of the interface element (Ei) simulating the pile-soil interaction was determined by the stiffness of the surrounding soils (Es). Figure 4 illustrates the effect of the stiffness of the interface element on the lateral response of the pile. The stiffness was varied from 1 to 20 times that of the surrounding soils, and the obtained load-displacement curves indicate that the stiffness of the interface element does not affect the numerical results if the stiffness was set to 5 times the stiffness of the surrounding soil or higher. Thus, the value of 10 times the stiffness of the surrounding soil recommended in the FLAC3D manual was used [39].
#### 3. Results and Discussions
A total of six numerical simulations were conducted with varying bedrock depths of 3, 5, 7, 10, 15, and 20 m. Figure 5 presents the lateral load-displacement curves for the six simulations, showing that a bedrock depth of 3 m results in dramatic increase in the lateral capacity of the pile. It should be noted that the pile was assumed to be linear elastic with no failure, but in reality, the calculated lateral capacity could cause material failure. The assumption was made because the study focuses on the soil response, not on the material failure. The effect of the bedrock gradually fades away and appears to be nonexistent when the bedrock depth is 10 m or greater.
The same observation was made in the pile deflection and bending moment profiles, as shown in Figure 6. The pile deflection was strongly confined by the surrounding bedrock at depths of 7 m or shallower, but no meaningful difference was noticed at greater depths. The maximum bending moment occurred near the bedrock when it was at a shallow depth, but the bending moment profiles for bedrock depths of 10 m or greater appeared to be identical. From the deflection and bending moment profiles, it could be concluded that the bedrock had no effect at depths of 10 m or greater.
Figure 7 shows the variation in the required lateral load and the maximum bending moment when the pile head was displaced by 38 mm. The lateral load was about 4,800 kN for a bedrock depth of 3 m, dropping to about 820 kN at a depth of 10 m, and then remaining constant at greater depths. The maximum bending moment was also greatest for a bedrock depth of 3 m, drastically declined to a depth of 10 m, and remained constant at depths below 10 m.
The lateral soil resistance per unit length, p, can be calculated by dividing the sum of the lateral forces applied to the entire interface nodes per unit length along the pile, while the lateral displacement, y, can be obtained from the nodal displacements. The available FISH language in FLAC3D was used to calculate the p-y curves, but the lateral soil resistance per unit length, p, was calculated differently. Figure 8 shows the stresses at the pile-soil interface. The x-directional (horizontal) component of the normal stress (σx,i) and shear stress (τx,i) at interface node i was calculated using Equation (1), and the lateral soil resistance, p, was calculated by summing all pi at the desired elevation. The detailed process for calculating the p-y curves can be found in previous researches [40, 41]where is the lateral soil resistance at the interface node , is the normal stress at the interface node , is the shear stress at the interface node , is the representative area of the interface node , and is the pile length between two neighboring interface nodes along the pile.
Figure 9 shows the p-y curves for sandy soil at six elevations from the ground surface: 2.0, 2.5, 3.0, 3.5, 4.0, and 4.5 m. The relatively shallow depths ranging from 2.0 to 4.5 m were selected because the soil response near the ground surface is the most important factor affecting the lateral behavior of piles. At depths of 2.0, 2.5, 3.0, and 3.5 m, the p-y curves at the same elevation agree well with each other irrespective of the bedrock depth (H). In other words, the lateral soil response at a bedrock depth of 5 m is not different from that at a bedrock depth of 20 m. Conversely, as the location of the p-y curves gets closer to the bedrock, the p-y curves at the similar depths become significantly different. The p-y curves at a depth of 4.0 m displayed much stiffer moduli of subgrade reaction when the bedrock depths were 5.0 and 7.0 m (i.e., 1.0 and 3.0 m to bedrock) than those of other depths (i.e., distance to bedrock greater than 6.0 m). Similar behavior was observed at a depth of 4.5 m. This finding implies that the p-y curves obtained near the bedrock (as close as 3D (3.0 m)) do not properly represent the soil response. In fact, the obtained curves in such area overpredict the stiffness (i.e., modulus of the subgrade reaction) of the p-y curves compared to those not affected by the bedrock. The stiffer behavior of the soil near the bedrock is likely due to the higher strength and stiffness of the bedrock; the elastic modulus of the bedrock is 500 times that of the soil or higher. The contribution of the bedrock stiffness becomes greater as the obtained p-y curves gets closer to the bedrock. The soil resistance did not reach the ultimate value for those p-y curves (Figures 9(e) and 9(f)), but it is likely to increase near the bedrock. Figure 9 also indicates that the p-y curves obtained at bedrock depths of 10 m or greater can be assumed to have the similar lateral pile response regardless of the presence of the bedrock. This finding is consistent with the facts that can be inferred from Figures 57.
#### 4. Conclusions
In this study, the effect of bedrock depth on the lateral behavior of piles was investigated using 3D finite difference analysis. The variations in pile deflection profiles as well as bending moment distributions for various bedrock depths were discussed. The ultimate soil resistance and stiffness in the obtained p-y curves were compared for various bedrock depths. The following conclusions can be drawn:(1)The effect of the bedrock gradually diminished with increasing bedrock depth and eventually disappeared at bedrock depths of 10 m (10D) or greater. A lateral load of 821 kN was required to displace the pile head by 38 mm for piles not affected by the bedrock, while a load of 4,803 kN was required at a bedrock depth of 3 m (3D).(2)The maximum bending moment occurred near the bedrock appearance for a bedrock depth of 7 m (7D) or less but occurred at a depth of about 4.5 m (4.5D) for greater bedrock depths. In the simulations, bedrock depths of 10 m (10D) or greater showed similar pile deflection and bending moment profiles along the pile, indicating that the effect of bedrock depth becomes negligible at greater depths.(3)It was found that the p-y curves near the bedrock were strongly influenced by the material properties of the bedrock; there were significant increases in the ultimate soil resistance per unit length and in the stiffness (modulus of subgrade reaction). In other words, the p-y curves obtained near the bedrock (within 3D) do not properly represent the soil response, and the modulus of the subgrade reaction and the ultimate soil resistance may be overestimated.
#### Data Availability
The numerical simulation data used to support the findings of this study are included within the article.
#### Conflicts of Interest
The authors declare that they have no conflicts of interest.
#### Acknowledgments
This work was partially supported by the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and Future Planning (NRF-2016R1C1B2013478) and by 2018 Hongik University Research Fund.
#### References
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Copyright © 2018 Younggyun Choi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 7,634 | 31,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | longest | en | 0.935046 |
https://slidetodoc.com/information-theory-and-patternbased-data-compression-jos-galaviz/ | 1,624,312,918,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504838.98/warc/CC-MAIN-20210621212241-20210622002241-00595.warc.gz | 426,605,166 | 14,843 | # Information Theory and Patternbased Data Compression Jos Galaviz
• Slides: 42
Information Theory and Pattern-based Data Compression José Galaviz Casas Facultad de Ciencias UNAM february 3, 2004 Information Theory, J. Galaviz
Contents • Introduction, fundamental concepts. • Huffman codes and extensions of a source. • Pattern-based Data Compression (Pb. DC). • The problems for Pb. DC. • Trying to solve. Heuristics. • Conclusions and further research. february 3, 2004 Information Theory, J. Galaviz 2
Information source • Is a “thing” that produces infinite sequences of symbols in some finite alphabet . • The theoretical model proposed by Shannon is an ergodic Markov chain. • Markov chain: stochastic process where the state reached at the i-th time step depends on the n previous states, n denotes the order of the Markov chain. february 3, 2004 Information Theory, J. Galaviz 3
Ergodic source • A Markov chain is ergodic if the probability distribution over the set of states tends to be stable in the limit. If p(i, j) denotes the transition probability from state i to state j in an ergodic Markov chain, then p(i, j) tends to some limit that does not depend on the source state i. • Almost every sample is a representative sample. • There exists only one set of interconnected states. • No periodic states. february 3, 2004 Information Theory, J. Galaviz 4
Information • Let p(s) be the probability that symbol s will be produced by some information source S. • The information (in bits) of s is defined as: february 3, 2004 Information Theory, J. Galaviz 5
The meaning • A measure of “surprise”. • Better: The number of “yes/no” questions needed to determine that s has occurred. february 3, 2004 Information Theory, J. Galaviz 6
Entropy • Is the expected value of symbol information. • Important: Note that entropy is measured over the source. Probabilities are used, assuming infinite amount of data. february 3, 2004 Information Theory, J. Galaviz 7
Data Compression • Given a finite sample of data produced by an unkown information source (unknown in the sense that we doesn´t know the statistical model of such source) • To express the same information contained in the sample with less data. • Exactly the same: lossless. Almost the same: lossy. • We will focus in lossless data compression. february 3, 2004 Information Theory, J. Galaviz 8
Huffman encoding • Is based on a statistical model of the sample to be compressed. • The codeword length for some symbol is inversely related with its frequency. • Target: minimize the average codeword length (Ave. Len). february 3, 2004 Information Theory, J. Galaviz 9
Example • • • f(A) = 10 f(B) = 15 f(C) = 10 f(D) = 15 f(E) = 25 f(F) = 40 february 3, 2004 Information Theory, J. Galaviz 10
Huffman codes • • A = 000 B = 100 C = 001 D = 101 E = 01 F = 11 Ave. Len = 2. 24 bits/word Vs. 3 bits february 3, 2004 Information Theory, J. Galaviz 11
Extensions of a source • Suppose a source S with alphabet ={A, B} • P(A) = 0. 6875, P(B) = 0. 3125 • Since there are only two symbols Huffman algorithm encodes every sample of such source using one bit per symbol in (1 BPS). • Entropy: H(S) = 0. 8960 february 3, 2004 Information Theory, J. Galaviz 12
2 nd extension • • • P(AA) = 0. 4727, Word. Len(AA) = 1 P(AB) = 0. 2148, Word. Len(AB) = 2 P(BA) = 0. 2148, Word. Len(BA) = 3 P(BB) = 0. 0977, Word. Len(BB) = 3 Ave. Len = 1. 8398, BPS 2( ) = 0. 9199 february 3, 2004 Information Theory, J. Galaviz 13
3 rd extension Str. Prob. W. Le AAA 0. 325 2 BAA 0. 148 3 AAB 0. 148 2 BAB 0. 067 4 ABA 0. 148 3 BBA 0. 067 4 ABB 0. 067 4 BBB 0. 031 4 Ave. Len = 2. 759, BPS 3( ) = 0. 9197 february 3, 2004 Information Theory, J. Galaviz 14
And so on. . . • 4 th extension: – Ave. Len = 3. 64138794 – BPS 4( ) = 0. 91034699 february 3, 2004 Information Theory, J. Galaviz 15
In practice • Suppose a sample of our previous source S: AAABAABAABB f(A) = 11 f(B) = 5 There are only two symbols, therefore Huffman assigns: A=0, B=1, 16 bits to express the sample. february 3, 2004 Information Theory, J. Galaviz 16
Thinking in extensions digram fre HC 3 -gram fre HC 4 -gram AA 4 0 AB 2 10 BA AAB 3 0 AAA 1 10 1 111 BAA BB 1 110 B## Total 14 bits february 3, 2004 fre HC AAAB 1 00 AAAA 1 01 1 111 BAAB 1 10 1 110 AABB 1 11 11 bits Information Theory, J. Galaviz 8 bits 17
Longer strings are better • The 4 -gram sample cannot be compressed since each of the 4 metasymbols (strings of 4 symbols) found, appear with the same frequency. • Let ´={ AAAA, AAAB, BAAB, AABB } be the alphabet of some information source S´ that produces the symbols in ´ equiprobably. • The sample could be produced by the maximum entropy source S´. february 3, 2004 Information Theory, J. Galaviz 18
Dictionary-based methods • Build a dictionary with frequent strings. • Each time a string in the dictionary appear in the sample, replace them with a dictionary reference, which are shorter. • Every frequent string is included only once (in the dictionary). february 3, 2004 Information Theory, J. Galaviz 19
Example AL QUE INGRATO ME DEJA, BUSCO AMANTE; AL QUE AMANTE ME SIGUE, DEJO INGRATA; CONSTANTE ADORO A QUIEN MI AMOR MALTRATA; MALTRATO A QUIEN MI AMOR BUSCA CONSTANTE 1. 2. 3. 4. 5. AL_QUE_ INGRAT _ME_ _AMANTE _A_QUIEN_MI_AMOR_ february 3, 2004 6. 7. 8. 9. MALTRAT CONSTANTE DEJ BUSC Information Theory, J. Galaviz 20
Result 12 O 38 A, 9 O 4; 143 SIGUE, 8 O 2 A; 7 ADORO 56 A; 6 O 59 A 7 february 3, 2004 Information Theory, J. Galaviz 21
Another posibility AL_QUE_INGRATO_ME_DEJA, _BUSCO_AMANTE; _ AL_QUE_AMANTE_ME_SIGUE, _DEJO_INGRATA; _ CONSTANTE_ADORO_A_QUIEN_MI_AMOR_MALTRATA; _ MALTRATO_A_QUIEN_MI_AMOR_BUSCA_CONSTANTE • Build a dictionary of frequent patterns, no necessarily of consecutive symbols (strings). february 3, 2004 Information Theory, J. Galaviz 22
The compression process • Given a finite sample of consecutive symbols produced by some source S whose statistical properties can only be estimated from its sample. • To find a set of frequent patterns such that the sample can be expressed briefly using references to these patterns. • Encode the sample using the set of patterns (dictionary), and encode the dictionary itself using some other method. february 3, 2004 Information Theory, J. Galaviz 23
Example february 3, 2004 Information Theory, J. Galaviz 24
Finding patterns, a naïve algorithm february 3, 2004 Information Theory, J. Galaviz 25
Algorithm complexity • Naïve algorithm is very expensive. • We need to find coincidence patterns, then coincidence patterns in the coincidence patterns previously found, then. . . • The number of intersections between coincidence patterns grows exponentially on the number of patterns found (which is O(sample size) ). february 3, 2004 Information Theory, J. Galaviz 26
There are better algorithms but. . . • Not very much better. • The best reported algorithms have complexity O ( n 2 n ). [Vilo 02] • The patterns we are looking for, are type P 3: “Patterns with wildcards of unrestricted length” february 3, 2004 Information Theory, J. Galaviz 27
The algorithms for pattern discovery • Are based in well known string matching techniques supported by special data structure called “suffix tree”. • There are several algorithms for suffix tree construction (n stands for the string size): – The worst is O ( n 3 ) – The two best methods (Wiener and Ukkonen) are linear on n, and builds the tree “on the fly”. february 3, 2004 Information Theory, J. Galaviz 28
Suffix tree for the string ATCAGTGCAATGC february 3, 2004 Information Theory, J. Galaviz 29
Some posibility? • Generalizing the suffix tree concept in order to include patterns rather than strings. A “tree of suffix patterns”. • Cannot be constructed “on the fly” since we need to remember an arbitrary number of previous symbols. • We need to perform: “Find the longest common pattern in a set of strings”. • We call this problem the MAXIMUMCOMMONPATTERN problem or MCP. february 3, 2004 Information Theory, J. Galaviz 30
MAXIMUMCOMMONPATTERN • We have recently proved that this problem is NP-Complete. That is: currently there is no deterministic polynomial time algorithm to solve it. If such algorithm would be found then all the other problems in this category (the upper bound of complexity) can also be solved in polynomial time and P=NP (the fundamental question in computability theory). february 3, 2004 Information Theory, J. Galaviz 31
Finding patterns (option 1) february 3, 2004 Information Theory, J. Galaviz 32
Finding patterns (option 2) february 3, 2004 Information Theory, J. Galaviz 33
Finding patterns (option 3) february 3, 2004 Information Theory, J. Galaviz 34
Several options • • Option 1: 12 metasymbols Option 2: 14 metasymbols Option 3: 10 metasymbols Option 3 gives shorter expression of sample, considering only the data in the sample, ignoring dictionary size. february 3, 2004 Information Theory, J. Galaviz 35
There is a right choice but. . . • The right choice is not easy to do. • There is a trade-off between pattern size and pattern frequency. • The inclusion of some pattern in dictionary must be amortized by its use. february 3, 2004 Information Theory, J. Galaviz 36
How much difficult is the right choice • Suppose we have a set of frequent patterns P. Each pattern have its frequency and its size. • We need to chose the subset P´ P that maximizes the compression ratio: february 3, 2004 Information Theory, J. Galaviz 37
• Where |M| is the original sample size, and T(P´) is the sample size after compression is done and dictionary is included. • T(P´) = D(P´) + E(P´) february 3, 2004 Information Theory, J. Galaviz 38
OPTIMALPATTERNSUBSET • We call the selection of best subset of patterns the OPTIMALPATTERNSUBSET problem. • We have proved that this problem is also NP-Complete. february 3, 2004 Information Theory, J. Galaviz 39
But here we have some resources • We can approximate the best subset by an heuristic algorithm. • We select the patterns with greatest coverage (number of symbols in the sample that are in the pattern appearances). • Then we iteratively refine the solution with hillclimbers with local changes. february 3, 2004 Information Theory, J. Galaviz 40
Conclusions • The pattern-based data compression is the most general approach to the compression problem based on statistical models of the data to be compressed. Every other technique in this class can be considered a particular case. • Unfortunately the sub-tasks involved in the compression process are mostly NPComplete problems. february 3, 2004 Information Theory, J. Galaviz 41
Further research • We need to achieve approximation algorithms or heuristics in order to solve the pattern discovery problem efficiently. february 3, 2004 Information Theory, J. Galaviz 42 | 3,043 | 10,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.80371 |
https://oeis.org/A128034 | 1,708,717,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474445.77/warc/CC-MAIN-20240223185223-20240223215223-00030.warc.gz | 439,455,132 | 3,802 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A128034 a(0)=a(1)=1. a(n) = the multiple of n which is > a(n-1)+a(n-2) and is <= a(n-1)+a(n-2)+n. 1
1, 1, 4, 6, 12, 20, 36, 63, 104, 171, 280, 462, 744, 1209, 1960, 3180, 5152, 8347, 13500, 21850, 35360, 57225, 92598, 149845, 242448, 392300, 634764, 1027080, 1661856, 2688938, 4350810, 7039759, 11390592, 18430368, 29820992, 48251385, 78072408 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS Table of n, a(n) for n=0..36. FORMULA a(n) = n * (1 +floor[(a(n-1)+a(n-2))/n]). MATHEMATICA f[l_List] := Block[{n = Length[l]}, Append[l, n*(1 + Floor[(l[[ -1]] + l[[ -2]])/n])]]; Nest[f, {1, 1}, 36] (* Ray Chandler, Feb 12 2007 *) CROSSREFS Cf. A128035. Sequence in context: A178547 A168674 A110935 * A027150 A020141 A049478 Adjacent sequences: A128031 A128032 A128033 * A128035 A128036 A128037 KEYWORD easy,nonn AUTHOR Leroy Quet, Feb 11 2007 EXTENSIONS Extended by Ray Chandler, Feb 12 2007 STATUS approved
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Last modified February 23 14:24 EST 2024. Contains 370283 sequences. (Running on oeis4.) | 531 | 1,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.552282 |
https://www.shaalaa.com/question-bank-solutions/using-de-moivre-s-theorem-find-the-least-positive-integer-n-such-that-2i-1-i-n-is-a-positive-integer-introduction-of-integrals_98974 | 1,656,585,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00095.warc.gz | 1,042,533,371 | 10,331 | # Using De Moivre’S Theorem, Find the Least Positive Integer N Such that ((2i)/(1+I))^N is a Positive Integer - Mathematics
Sum
Using De Moivre’s theorem, find the least positive integer n such that ((2i)/(1+i))^n is a positive integer.
#### Solution
We have, (2i)/(1+i) = (2i)/(1+i) xx (1-i)/(1-i)
= (2(1+i))/(2) = 1 + i
Let 1 + i = r cos θ + i r sin θ
⇒ r cos θ = 1, r sin θ = 1
∴ r2 (cos2 θ + sin2 θ ) = (1)2 + (1)
r2 = 2 ⇒ r = sqrt2
and tan θ = (1)/(1)
tan θ = tan (π/4)
θ = (π)/(4)
((2i)/(1+i)) = sqrt2 ( cos (π)/(4) + i sin (π)/(4))
((2i)/(1+i))^n = [ sqrt2 ( cos (π)/(4) + i sin (π)/(4)]^n
= 2^(n/2) ( cos (nπ)/(4) + i sin (nπ)/(4))
Which is a positive integer
If (nπ)/(4) = 0, 2π, 4π, 6π, ...
⇒ n = 0, 8, 16, 24, ...
⇒ The least value of n is 4
Concept: Introduction of Integrals
Is there an error in this question or solution? | 368 | 867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-27 | latest | en | 0.301607 |
https://www.maplesoft.com/support/help/MapleSim/view.aspx?path=componentLibrary/hydraulic/sources/FixedPressureSource | 1,511,231,714,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806310.85/warc/CC-MAIN-20171121021058-20171121041058-00201.warc.gz | 834,942,705 | 22,897 | Fixed Pressure Source - MapleSim Help
Home : Support : Online Help : MapleSim : MapleSim Component Library : Hydraulics : Sources : componentLibrary/hydraulic/sources/FixedPressureSource
Fixed Pressure Source
The Fixed Pressure Source component generates a constant pressure. The governing equations are
$p=P$ $p\equiv {p}_{u}-{p}_{d}$ ${q}_{u}+{q}_{d}=0$
Connections
Name Description Modelica ID $u$ Upstream port portA $d$ Downstream port portB
Variables
Symbol Units Description $p$ $\mathrm{Pa}$ Pressure across the component ${p}_{x}$ $\mathrm{Pa}$ Pressure at port $x$, $x\in \left\{u,d\right\}$ ${q}_{x}$ $\frac{{m}^{3}}{s}$ Fluid flow rate into port $x$, $x\in \left\{u,d\right\}$
Parameters
Symbol Default Units Description $P$ $1$ $\mathrm{Pa}$ Fluid pressure
Initial Conditions
Symbol Units Description Modelica ID ${p}_{0}$ $\mathrm{Pa}$ Initial pressure p0 ${q}_{0}$ $\frac{{m}^{3}}{s}$ Initial flow rate q0 | 268 | 938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-47 | latest | en | 0.452327 |
https://bitcoin.stackexchange.com/questions/121247/how-exactly-is-the-timestamp-calculated-for-the-2h-acceptance-rule-and-do-i-hav?noredirect=1 | 1,716,189,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00323.warc.gz | 114,323,384 | 41,865 | # How exactly is the timestamp calculated for the +2h acceptance rule and do I have to implement it in my Bitcoin node for the node to still be valid?
In recent times, I have read many articles and questions/answers on bitcoin.stackexchange related to the timestamp, but I still have problems understanding certain things. They are related to the understanding of certain aspects of the "+2h rule" in the context of block acceptance, so I have listed them below.
Firstly, as I understand it, a block will be accepted if its timestamp is up to 2 hours greater than the current network-adjusted time calculated by the given node.
I found some answers related to network-adjusted time, but I'm not sure if I understood it correctly.
As I understand each node calculates its network-adjusted time as follows: Upon initial connection to peers, the node receives UTC timestamps from all peers in the version message. It calculates the difference between its current local (system) UTC timestamp and the received UTC timestamp from the peer. These differences are sorted, and the median is found. The network-adjusted time is obtained by adding the current local time (UTC timestamp) of the node and the calculated median. In an ideal world (a completely synchronized network), network-adjusted time and the local time of the node would be equivalent since all differences would be 0 and therefore the median. Also, if the difference between the current system time (UTC timestamp) of the node and the calculated network-adjusted time (UTC timestamp) is more than 70/90 minutes, the network-adjusted time is set to the node's given system time (UTC timestamp). Thus, when a new block arrives, if the block timestamp is within the range of the node local timestamp + previously calculated median (local timestamp + median = network-adjusted time of the node) + 2 hours, the block is accepted by that node (assuming that the other rules are satisfied).
1. Have I correctly understood the way network-adjusted time is calculated? Is the "+2h rule" considered in relation to it rather than the local time (UTC timestamp) of the node?
An acceptance rule that the timestamp cannot be more than 2 hours in future (compared to the verifier's local clock). Since this depends on the local clock, this property can change over time (and once enough time passes, a block formerly found to be invalid by this rule may become acceptable).
1. Is a deviation of 70 or 90 minutes allowed between network-adjusted time and the local time of the node? I have read different information. Here, they say it's 70, and here, they say it's 90 minutes.
From what I understand so far, the "+2h rule" is NOT part of the consensus but solely represents the acceptance policy of the Bitcoin Core implementation of the Bitcoin full node. The consensus rule is only that the block timestamp must be greater than the median timestamp of the last 11 blocks.
1. Does this mean that if I implement my own node, I only need to implement the second rule regarding consensus, and my node is entirely correct, even though it would be good to also implement the +2h rule since most full nodes run Bitcoin Core, so to always be fully synchronized with the majority of the network?
I consider the "max two hours in the future rule" to be an essential network rule of the Bitcoin network; without it, the system would be woefully insecure. It is not a mere policy rule, but it is also not - and cannot be - a consensus rule as explained in my answer to your other question.
But many other aspects of a Bitcoin node's protocol implementation are not consensus rules. For example, if validating and relaying blocks is so slow that it takes over 10 minutes, the network would fail to converge to a single chain, because miners would never be able to see each other's blocks in time. I think of the max block timestamp rule as something in this category.
The fact that Bitcoin Core today uses network-adjusted time to decide what the "local clock" is that block timestamps are compared against, is as far as I'm concerned a local policy. In fact, there is discussion around dropping the network adjustment part.
1. Have I correctly understood the way network-adjusted time is calculated? Is the "+2h rule" considered in relation to it rather than the local time (UTC timestamp) of the node?
Yes. Currently (as of Bitcoin Core 26.0), the network-adjusted time is computed as the local clock plus the median of the peers' clock differences with our clock. That network-adjusted time is what the "max two hours in the future" is comparing against.
1. Is a deviation of 70 or 90 minutes allowed between network-adjusted time and the local time of the node? I have read different information. Here, they say it's 70, and here, they say it's 90 minutes.
It's 70 minutes currently (and has been for as long as I'm aware the rule has existed). Note that this value can be overridden with the `-maxtimeadjustment` command-line flag (or the equivalent config file setting).
1. Does this mean that if I implement my own node, I only need to implement the second rule regarding consensus, and my node is entirely correct, even though it would be good to also implement the +2h rule since most full nodes run Bitcoin Core, so to always be fully synchronized with the majority of the network?
I guess that depends on your definition of "correct". If your goal is to write software that can decide if a block is valid or invalid according to the consensus rules, then this rule is irrelevant.
If your goal is writing node software that would be safe for the ecosystem to actually adopt in any economically-relevant setting, then you absolutely need some way of bounding blocks' timestamps. It likely doesn't matter that it isn't exactly the same rule, but you need something like it.
• Thanks, just one question. For this last (3rd question), could there be any problem if I put +1h or +3 in my node implementation instead of +2h? I assume that in reality in most cases no, since miners will always put a timestamp around the current timestamp because they want the block to be "generally" accepted by the entire network, but could a problem arise in some cases? For example, when the timestamp in the block is +2.5h, which I will accept when I am +3h, while the Bitcoin core nodes will not, or when its at +1.5h, which I will not accept for +1h, and the Bitcoin core nodes will? Jan 1 at 22:39
• I don't have more to say about that than the last sentence in my answer already says: it likely doesn't matter that it isn't exactly the same rule. Jan 1 at 22:41
• Okay, thanks, all clear Jan 1 at 22:45 | 1,439 | 6,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.953675 |
https://patriciabelcher.com/the-honest-to-goodness-truth-on-independent-variable-in-science/ | 1,726,362,147,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00382.warc.gz | 403,389,458 | 13,093 | ## The Honest to Goodness Truth on Independent Variable in Science
A good example of a dependent variable is the way tall you’re at various ages. Whatever can vary can be thought of a variable. A control variable is a variable that’s controlled during and between experiments, and it’s used to create sure the dependent variable isn’t affected by anything besides the independent variable.
Experiments are set-up to find out more about the way the independent variable does or does not influence the dependent variable. lab reports Only when you’re certain that all variables are the exact same can you are aware that the outcomes of your experiment are accurate. Thus it’s important to be alert to which variables in a study are manipulated and which aren’t.
The very first is that although we’ve only a single predictor variable, the test for those odds ratio doesn’t match with the total test of the model. It’s what changes as a consequence of the modifications to the independent variable. This is done in order to confirm that the independent variable was, in reality, successfully manipulated.
In math, it typically refers to a variable that doesn’t depend on another variable. They must be controlled, meaning they must stay the same. At other times, some critical variables might be outside of your control custom writing or be extremely tough to control.
Well, in the event you changed more than 1 variable it would be really hard to determine which change is causing what you observe. It only has one independent variable so that the experimenter can be sure that changes to the dependent variable are caused by the independent variable.
The scientist will continue to keep the controlled variables constant so that their influence on the test is going to be minimized. Your prediction is dependent on your comprehension of the scientific idea. It is a prime example of how the concept of experimental variables can become a little complex.
It’s important to get a control variable as it provides the experiment a baseline to be compared against. Begin by thinking about what you’re controlling and what you’ll be measuring. The effect of one independent variable can be based on the amount of the other in several distinctive ways.
You wish to know whether stress affects heart rate. You may choose to incorporate a control variable of age in your study to find out whether it impacts the outcome. This sort of experiment is extremely important because, if there’s a cause and effect relationship between variables, then outcomes are predictable and may be used to your benefit, or could possibly be manipulated or changed.
## Things You Won’t Like About Independent Variable in Science and Things You Will
A chicken bone is set in a jar of vinegar. After 3 days of treatment” there is not any change in the look of the green slime on each side of the shower. On the flip side, in case you have flies in the home, you might not need to feed your flytrap whatsoever.
By way of example, someone’s age may be an independent variable. To execute an experiment, you would offer tension and gauge the subject’s heart beat. Every time you reach one of your weight reduction plans, even smaller ones, take the opportunity to celebrate your accomplishment.
## What the In-Crowd Won’t Tell You About Independent Variable in Science
The total pattern is replicated within each area of the county. In real life, but this cost might be impacted by how much you buy of a specific product. In principle, factorial designs can incorporate numerous independent variables who have any variety of levels.
## The Downside Risk of Independent Variable in Science
Then there are elements that may be controlled. An individual may then figure out the b value and learn whether the interaction is significant. There are 3 standard types to be familiarized with.
## The Secret to Independent Variable in Science
In an experiment to test the way the sum of sunlight impacts the development of radishes. There’s one dependent variable that’s plant development, but 1. The plants you use is a variable that should be controlled so you use the exact kind of plant in all your tests.
## New Ideas Into Independent Variable in Science Never Before Revealed
You won’t see results in a few of weeks. Usually whenever you are trying to find a connection between two things you’re trying to discover what makes the dependent variable change the way it does. Occasionally, it’s a struggle to stick to the discussions in a research paper.
## Using Independent Variable in Science
In another instance, the hypothesis Young participants are going to have significantly greater memories than older participants isn’t operationalized. Members of the control group often are given placebos or know they are a part of the control group in order for the researcher can establish the efficacy of what’s being studied. The analysis by Schnall and colleagues is a fantastic example.
If you’re working to ANSWER A QUESTION, you will do real research. The purpose of the experiment will stay unaccomplished. Probably the toughest portion of a science fair project is inventing a very good subject to research.
Control variables are extremely important since they allow a scientist to be aware that the experiment is just testing what they would like to test. His experiment would last a single month. When you’ve done your science fair experiment, think about repeating it should you have the moment! | 1,054 | 5,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.936568 |
http://www.sexyloops.co.uk/theboard/viewtopic.php?f=11&t=3333 | 1,611,571,938,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00004.warc.gz | 182,713,457 | 13,568 | PLEASE NOTE: In order to post on the Board you need to have registered. To register please email paul@sexyloops.com including your real name and username. Registration takes less than 24hrs, unless Paul is fishing deep in the jungle!
## Shooted cast
Moderator: Torsten
Merlin
Posts: 1508
Joined: Wed Jan 09, 2013 8:12 pm
Location: France
### Shooted cast
I can run the Tension model for untethered casts, and now the running line is on board. Of course, the situation is simplified, the line is still a level one, the loop a half circle and the legs are parallel, the fly is a given, etc. By the way, given the necessity to recalculate continuously the drag force on a loop with variable taper, I am not at the eve to have a “taper” Tension model I think. Excel files are not well adapted to such calculation.
I gave a look at a few parameters like the weight of the head (9 m long), the density of the running line, its size (diameter), and the timing of an early release. That point is very, very discernible. I am using a more realistic skin drag coefficient now, taken from the only specifically fishing oriented experiment including front drag coefficient for some flies. (L. Gaddis, Clemson University, “Axial fluid drag on slender cylinders and terminal objects” late 90s I think).
The extra shooting distance reduces drastically as the line release is delayed, it is extremely sensible (100 ms).So it seems mandatory to anticipate the release of the line and maybe someone can confirm that trend and explain how he tunes the release. Since such timing corresponds to the end of a haul that does not appear to be simple for an average caster like me. There is a physoliogical reaction time to anticipate. I imagine this contributes to the variability of casting distance significantly.
Running line: thin and light, that is the recipe. The weight is more important (detrimental) for thicker lines, for thin running lines it seems to be more marginal. So maybe someone uses “heavy” running lines against wind, if that thing exists.
Line weight: no surprise if I say that the best performance is obtained with the heaviest head although the difference in distance achieved is not that big on the paper. I know, the cast is supposed to be horizontal so the ballistic effect of a tilted cast is not included.
Another information: the tension at ends of the loop varies with its tangent/rotation velocity, so the tension at the bottom is higher during an acceleration phase and it is the reverse in a deceleration phase of that speed.
Merlin
Fly rods are like women, they won't play if they're maltreated
Charles Ritz, A Flyfisher's Life
Mangrove Cuckoo
Posts: 434
Joined: Tue Jan 29, 2013 7:51 am
### Re: Shooted cast
Merlin,
Very cool!
The fact that you are doing this type of stuff with Excel is impressive.
Out of curiosity... since release time is critical, do you have parameters for haul length and haul speed???
"Technique is the proof of your seriousness"
Wallace Stevens
Paul Arden
Posts: 14275
Joined: Thu Jan 03, 2013 11:20 am
Location: Belum Rainforest
Contact:
### Re: Shooted cast
Hi Merlin,
With a shooting head I think so and I release earlier (before the haul stops). With something like a MED it makes a difference but I don’t find it anything like as critical and I’m happy to release at haul stop. I know others who might disagree.
Are you modelling counterflex and RSP2? I would be very interested to understand what happens in terms of line tension during RSP1-2. Certainly I don’t feel much tension until after the loop is properly in flight. If I start the up part of the haul too early/quickly I create slack, but that may have to do with the greater friction of the running line compared to shooting line.
Incidentally why is the cast supposed to be horizontal? My trajectory is about 30 degrees above the horizontal. This way I find it flies further after loop straight. (In to the wind is different of course)
Thanks Merlin!
Cheers, Paul
It's an exploration; bring a flyrod.
Flycasting Definitions
Lasse Karlsson
Posts: 4257
Joined: Wed Jan 09, 2013 9:40 pm
Location: There, and back again
Contact:
### Re: Shooted cast
Paul, please show us a pic of your 30 degree up trajectory until then, I call bollocks....
And horizontal makes sense, I only go above in a good backwind, and rarely more than a few degrees...
Cheers
Lasse
http://www.karlssonflyfishing.com
***Bring Mark back!!!!!! ***
Paul Arden
Posts: 14275
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### Re: Shooted cast
Of course!
Attachments
It's an exploration; bring a flyrod.
Flycasting Definitions
Michal Duzynski
Posts: 1263
Joined: Thu Jan 17, 2013 5:14 pm
Location: Brisbane-Australia
Contact:
### Re: Shooted cast
Hi Merlin
I like it, it might help with the seatrout distance.
Here is a cast I am very happy with, but now when I red it I think it could go even further, because of the release.
I watched it again I think I maybe released it too late.
Have a look, there is slow mo of the delivery and release, let me know what you think.
Shooting line is 0.37mm
cheers
mike
Michal Duzynski
Posts: 1263
Joined: Thu Jan 17, 2013 5:14 pm
Location: Brisbane-Australia
Contact:
### Re: Shooted cast
Here is another video, and in the slo mo part of the delivery cast you can see that my hand opens/releases the line behing my butt cheek.
Now I start to think about it.
Could you use my video, or something simalr and mark wher the release position should be for the best result?
During fals casting there was not too much overhang- just right, and the loop was under an awesome tension- no wiggles and waves in a loop.
cheers
mike
Lasse Karlsson
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### Re: Shooted cast
Paul Arden wrote:
Tue Sep 01, 2020 9:13 pm
Of course!
Thats app 8 degrees dude, far from 30... don't measure the angle of first psrt of the rod leg
Cheers
Lasse
http://www.karlssonflyfishing.com
***Bring Mark back!!!!!! ***
Paul Arden
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### Re: Shooted cast
You think the bottom part of the rod leg is pulling the fly leg? Anyway I agree it’s not 30. Let me see if my phone has a protractor...
It's an exploration; bring a flyrod.
Flycasting Definitions
Paul Arden
Posts: 14275
Joined: Thu Jan 03, 2013 11:20 am
Location: Belum Rainforest
Contact:
### Re: Shooted cast
Typically 15 degrees. (I knew it was a dissection ). It can be more but I’ll do some tests and will measure every time I throw now.
It’s quite easy to do this with the measurer app. Just stick a bit of tape either side of the phone where the measuring lines are on the app and line up the two tape points to your target.
When casting I pick a higher and higher target until the loop tails and then lower the target slightly. So with this it will be very easy to measure trajectory in different conditions.
Of course the rod tip to target angle is going to be different. Not much I can do about that
Cheers, Paul
It's an exploration; bring a flyrod.
Flycasting Definitions | 1,787 | 7,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.940473 |
https://www.physicsforums.com/threads/center-of-curvature.811262/ | 1,532,058,455,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00124.warc.gz | 967,184,838 | 15,608 | # Center of curvature
1. Apr 29, 2015
### gracy
As per ww.physicsclassroom.com/class/refln/Lesson-3/The-Anatomy-of-a-Curved-Mirror
The point in the center of the sphere from which the mirror was sliced is known as the center of curvature,I am not able to understand this.Please help.
2. Apr 29, 2015
### ZapperZ
Staff Emeritus
You have the propensity to not describe exactly what you are having a problem with. Often times, your question is very terse and short without giving much explanation on what you had attempted to do to understand it. Did you try to Google it and look it up? If you did, what did you find and what was still giving you problems? If you did not, why not?
Does this picture explain what a "center of curvature" means?
Zz.
3. Apr 29, 2015
### A.T.
4. Apr 29, 2015
### gracy
You mean we should complete the curved surface by making a circle and then center of this circle will be called "center of curvature" .Right?
5. Apr 29, 2015
### ZapperZ
Staff Emeritus
That is not necessary! Just looking at an arc of the circle is sufficient. I can take the pencil and draw just a segment of the circle. That section alone tells me where the center of curvature is.
Zz.
6. Apr 29, 2015
### gracy
Propensity can be changed,right?I will try to work on it if this is the case.
Sorry.
Yes,but didn't find any explanation useful.
ww.physicsclassroom.com/class/refln/Lesson-3/The-Anatomy-of-a-Curved-Mirror
http://en.wikipedia.org/wiki/Center_of_curvature
http://www.merriam-webster.com/dictionary/center of curvature
http://www.thefreedictionary.com/centre+of+curvature
All the definitions were hard to follow.Don't know from where @ A.T and @zapper Z got such nice and easy explanation.But yes I know I should not make excuses.I should admit that I showed less effort otherwise I too should have got these nice explanations .Because where there is a will ,there is a way!
7. Apr 29, 2015
### gracy
I was answering A.T. post.Is it still not correct?
8. Apr 29, 2015
### rumborak
To expand on this, the reason is that the circle looks the same everywhere because it is so symmetrical. If you have one tiny segment of a circle, you know what the rest will look like. And thus you know the center of that circle, the center of curvature.
9. Apr 29, 2015
### rcgldr
One way to determine the center of curvature would be a line perpendicular to the curved path, with length equal to the radius of curvature. Wiki article (includes link to radius of curvature):
http://en.wikipedia.org/wiki/Center_of_curvature
10. Apr 30, 2015
### CWatters
Centre Finder... | 665 | 2,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-30 | latest | en | 0.93405 |
https://geos.osgeo.org/doxygen/DistanceOp_8h_source.html | 1,550,355,763,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481122.31/warc/CC-MAIN-20190216210606-20190216232606-00197.warc.gz | 580,190,803 | 4,841 | GEOS 3.8.0dev
DistanceOp.h
1 /**********************************************************************
2 *
3 * GEOS - Geometry Engine Open Source
4 * http://geos.osgeo.org
5 *
6 * Copyright (C) 2011 Sandro Santilli <strk@kbt.io>
7 * Copyright (C) 2006 Refractions Research Inc.
8 *
9 * This is free software; you can redistribute and/or modify it under
10 * the terms of the GNU Lesser General Public Licence as published
11 * by the Free Software Foundation.
13 *
14 **********************************************************************
15 *
16 * Last port: operation/distance/DistanceOp.java r335 (JTS-1.12-)
17 *
18 **********************************************************************/
19
20 #ifndef GEOS_OP_DISTANCE_DISTANCEOP_H
21 #define GEOS_OP_DISTANCE_DISTANCEOP_H
22
23 #include <geos/export.h>
24
25 #include <geos/algorithm/PointLocator.h> // for composition
26
27 #include <array>
28 #include <vector>
29
30 #ifdef _MSC_VER
31 #pragma warning(push)
32 #pragma warning(disable: 4251) // warning C4251: needs to have dll-interface to be used by clients of class
33 #endif
34
35 // Forward declarations
36 namespace geos {
37 namespace geom {
38 class Coordinate;
39 class Polygon;
40 class LineString;
41 class Point;
42 class Geometry;
43 class CoordinateSequence;
44 }
45 namespace operation {
46 namespace distance {
47 class GeometryLocation;
48 }
49 }
50 }
51
52
53 namespace geos {
54 namespace operation { // geos::operation
55 namespace distance { // geos::operation::distance
56
75 class GEOS_DLL DistanceOp {
76 public:
87 static double distance(const geom::Geometry& g0,
88 const geom::Geometry& g1);
89
91 static double distance(const geom::Geometry* g0,
92 const geom::Geometry* g1);
93
104 static bool isWithinDistance(const geom::Geometry& g0,
105 const geom::Geometry& g1,
106 double distance);
107
120 static geom::CoordinateSequence* nearestPoints(
121 const geom::Geometry* g0,
122 const geom::Geometry* g1);
123
137 static geom::CoordinateSequence* closestPoints(
138 const geom::Geometry* g0,
139 const geom::Geometry* g1);
140
142 DistanceOp(const geom::Geometry* g0, const geom::Geometry* g1);
143
152 DistanceOp(const geom::Geometry& g0, const geom::Geometry& g1);
153
164 DistanceOp(const geom::Geometry& g0, const geom::Geometry& g1,
165 double terminateDistance);
166
167 ~DistanceOp();
168
174 double distance();
175
185 geom::CoordinateSequence* closestPoints();
186
195 geom::CoordinateSequence* nearestPoints();
196
197 private:
198
211 std::vector<GeometryLocation*>* nearestLocations();
212
213 // input
214 std::array<geom::Geometry const*, 2> geom;
215 double terminateDistance;
216
217 // working
218 algorithm::PointLocator ptLocator;
219 // TODO: use unique_ptr
220 std::vector<GeometryLocation*>* minDistanceLocation;
221 double minDistance;
222
223 // memory management
224 std::vector<geom::Coordinate*> newCoords;
225
226
227 void updateMinDistance(std::vector<GeometryLocation*>& locGeom,
228 bool flip);
229
230 void computeMinDistance();
231
232 void computeContainmentDistance();
233
234 void computeInside(std::vector<GeometryLocation*>* locs,
235 const std::vector<const geom::Polygon*>& polys,
236 std::vector<GeometryLocation*>* locPtPoly);
237
238 void computeInside(GeometryLocation* ptLoc,
239 const geom::Polygon* poly,
240 std::vector<GeometryLocation*>* locPtPoly);
241
246 void computeFacetDistance();
247
248 void computeMinDistanceLines(
249 const std::vector<const geom::LineString*>& lines0,
250 const std::vector<const geom::LineString*>& lines1,
251 std::vector<GeometryLocation*>& locGeom);
252
253 void computeMinDistancePoints(
254 const std::vector<const geom::Point*>& points0,
255 const std::vector<const geom::Point*>& points1,
256 std::vector<GeometryLocation*>& locGeom);
257
258 void computeMinDistanceLinesPoints(
259 const std::vector<const geom::LineString*>& lines0,
260 const std::vector<const geom::Point*>& points1,
261 std::vector<GeometryLocation*>& locGeom);
262
263 void computeMinDistance(const geom::LineString* line0,
264 const geom::LineString* line1,
265 std::vector<GeometryLocation*>& locGeom);
266
267 void computeMinDistance(const geom::LineString* line,
268 const geom::Point* pt,
269 std::vector<GeometryLocation*>& locGeom);
270 };
271
272
273 } // namespace geos::operation::distance
274 } // namespace geos::operation
275 } // namespace geos
276
277 #ifdef _MSC_VER
278 #pragma warning(pop)
279 #endif
280
281 #endif // GEOS_OP_DISTANCE_DISTANCEOP_H
282
Find two points on two Geometrys which lie within a given distance, or else are the nearest points on...
Definition: DistanceOp.h:75
Represents the location of a point on a Geometry.
Definition: GeometryLocation.h:50
Computes the topological relationship (Location) of a single point to a Geometry. ...
Definition: PointLocator.h:57
Basic implementation of Geometry, constructed and destructed by GeometryFactory.
Definition: Geometry.h:187
Definition: LineString.h:69
Represents a linear polygon, which may include holes.
Definition: Polygon.h:65
Basic namespace for all GEOS functionalities.
Definition: IndexedNestedRingTester.h:25
The internal representation of a list of coordinates inside a Geometry.
Definition: CoordinateSequence.h:58
Definition: Point.h:66 | 1,389 | 5,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-09 | longest | en | 0.208746 |
http://mathoverflow.net/feeds/question/108419 | 1,369,053,590,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698958430/warc/CC-MAIN-20130516100918-00057-ip-10-60-113-184.ec2.internal.warc.gz | 144,594,262 | 2,583 | Smith normalform of a Matrix with -1 outside the diagonal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:39:47Z http://mathoverflow.net/feeds/question/108419 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108419/smith-normalform-of-a-matrix-with-1-outside-the-diagonal Smith normalform of a Matrix with -1 outside the diagonal Johannes 2012-09-29T19:16:22Z 2013-03-05T17:58:31Z <p>Hi, I have given a matrix of the following form:</p> <p>$M = \begin{pmatrix} a_0 & -1 & \cdots & \cdots & -1 \newline -1 & a_1 & -1 & \cdots & -1 \newline \vdots & & \ddots & & \vdots \newline \vdots & & & \ddots & \vdots \newline -1 & \cdots & -1 & \cdots & a_n \end{pmatrix}$ with $a_i \in \mathbb{Z}$, $a_i > 0$.</p> <p>Is the a an easy way to write down the Smith Normalform of this matrix?</p> <p>greatz Johannes</p> http://mathoverflow.net/questions/108419/smith-normalform-of-a-matrix-with-1-outside-the-diagonal/108473#108473 Answer by Shaun Ault for Smith normalform of a Matrix with -1 outside the diagonal Shaun Ault 2012-09-30T14:25:39Z 2012-09-30T14:25:39Z <p>While I believe the full answer to your question is 'no,' I was pleasantly surprised that I can predict the first two diagonal entries of the SNF.</p> <p>Permute the rows so that there is a 1 in entry $(1,1)$. Then after a first round of row and column operations, we produce a matrix:</p> <p>$$\left[ \begin{array}{c|c} 1 & \mathbf{0}^T \\ \hline \mathbf{0} & M' \end{array}\right]$$</p> <p>First observation: $1$ is the first diagonal entry.</p> <p>Second observation: If every $a_i$ is equivalent to $-1$ modulo $k$ for some $k$, then $k$ divides the next diagonal entry of the SNF... and if $k$ is the largest such number, then it <em>is</em> the next. This is because $M'$ contains only entries such as $0$, $\pm(1 + a_i)$ or $1 - a_ia_j$.</p> <p>Hope this helps!</p> http://mathoverflow.net/questions/108419/smith-normalform-of-a-matrix-with-1-outside-the-diagonal/123498#123498 Answer by i707107 for Smith normalform of a Matrix with -1 outside the diagonal i707107 2013-03-04T01:39:50Z 2013-03-05T17:58:31Z <p>This is a partial answer for the special case when all $a_i$ are distinct. I will work on complex(or algebraic closure of $\mathbb{Q}$) for convenience. Writing $M-\lambda I = D_{\lambda}+E$ where $D_{\lambda}$ is the diagonal matrix with diagonal entries $a_0-\lambda+1, \cdots , a_n-\lambda+1$. With this expression, it is easy to calculate the determinant, which will give the characteristic polynomial of $M$. </p> <p>If $f(\lambda)=(a_0-\lambda+1)\cdots (a_n-\lambda+1)$, then we have $$\textrm{det}(M-\lambda I) = f(\lambda)+f'(\lambda).$$ Considering the identity $(f(t)e^t)'=(f(t)+f'(t))e^t$, we can find the roots of characteristic polynomial by looking at the critical points of $f(t)e^t$. Also, by Mean Value theorem, we know that the critical points of $f(t)e^t$ are all distinct. </p> <p>Therefore, if we let $\lambda_0, \cdots, \lambda_n$ be the critical points of $f(t)e^t$, then we have the following Smith Normal form of the matrix $M-\lambda I$ over the polynomial ring $\mathbb{C}[\lambda]$, (or $\overline{\mathbb{Q}}[\lambda]$): $$\textrm{Diag}(1,\cdots, 1, (\lambda-\lambda_0) \cdots (\lambda-\lambda_n)).$$ Hence, we obtain the Smith Normal form of $M$ in this case: $$\textrm{Diag}(1,\cdots, 1,\lambda_0 \cdots \lambda_n).$$ There is a natural way of bringing this down to $\mathbb{Z}$, then we have to deal with the irreducible factors of $f(t)+f'(t)$. This is indeed $$\textrm{Diag}(1,\cdots, 1, f(0)+f'(0)).$$ Added) This method also works for the case below:</p> <p>The cardinality of $\{ j: a_i = a_j \}$ is at most $2$, for any $i=0,\cdots, n$. </p> <p>Added2) This gives the SNF of $M$ over $\mathbb{Q}$. Over $\mathbb{Z}$ will be certainly more difficult. </p> | 1,314 | 3,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.651549 |
http://www.move420.com/1izjheig/ceb1a7-5-naoh-density | 1,618,410,844,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00380.warc.gz | 144,823,406 | 11,241 | # 5 naoh density
36 ml = 0.0654 L and: M = # of moles of solute/ # of liters of solution = 1.25 moles NaOH/0.0654 NaOH soln = 0.019 M NaOH … H=1g Cl=17 thus the grams of the equation is 18 grams. In some cases, you likewise do not discover the notice naoh solution density that you are looking for. Natriumhydroxid (auch Ätznatron, kaustische(s) Soda), chemische Formel NaOH, ist ein weißer hygroskopischer Feststoff. The density of a 50% solution of NaOH is 1.525 g/mL. Diese Methode wird Chloralkali-Elektrolyse genannt. Using a pH indicator strip will tell you that NaOH (sodium hydroxide) is a strong alkaline. Massenkonzentration NaOH in g/l 41,7 110,9 243,8 398,3 572,0 762,2 Dichte der Lösung in g/cm 3: 1,043 1,109 1,219 1,328 1,430 1,524 Gewinnung. Packaging: 500 … Quantity of pure sodium hydroxide by the Volumetric mass density [d] of the solutions at 20 °C: (d = kg/m3 ou g/L) (% = g as NaOH per 100 g of solution) Take your equation and get your moles which since you have no other unit all I can see is you use one more and it ends up giving you 18 moles. Sodium tungstate can also be produced by treating tungsten carbide with a mixture of sodium nitrate and sodium hydroxide in a fusion process which overcomes the high exothermicity of the reaction involved. They do not form the whole or part of a specification or guarantee. This means it has a pH toward the top end of the pH scale, which ranges from 0 to 14. 1.20m 3. Siehe auch: Chloralkali-Elektrolyse. Density=mass/volume. Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. One mole of hydrazine (N 2 H 4 ) loses 1 0 moles of electrons in a reaction to form a new compound X.Assuming that all the nitrogen atoms in hydrazine appear in the new compound, what is the oxidation state of nitrogen in X? All you need to do is figure out your grams of HCl then find the moles. Fluka Sodium hydroxide concentrate for 1 L standard solution, 0.5 M NaOH (0.5 N… Honeywell. 1.67m? Fe/MnWO 4 + 2 NaOH + 2 H 2 O → Na 2 WO 4 •2H 2 O + Fe/Mn(OH) 2. Density vs Specific Weight vs Specific Gravity I would like to point out that there are conceptual differences between density, specific weight and specific gravity. If same concentration solutions of NaOH and HCl are used, same volumes of NaOH are HCl are consumed too. b)0.49 m . To Email: From Email: Message: S8263 Sigma-Aldrich Sodium hydroxide solution 5.0 M CAS Number ... concentration 5.0 M pH 14 density 1.327 g/cm 3 at 25 °C storage temp. I know the formula mass Na = 22.9, O = 16, H = 1and density of 50% NaOH = 1.53 formula mass for NaOH = 40 approximately and I write X mol NaOH = 50 g NaOH (1 mol NaOH/ 40 gNaOH) = 1.25 moles of NaOH then: 100g NaOH soln. Related Videos. This video explains the concept of mole in terms of mass, volume, number an... Molarity and Molality. With that said, this page will provide you the NaOH density in kg/m³ (SI unit), g/cm³, g/ml, kg/l, lb/ft³. What is the concentration of the original NaOH solution? 5.0 g B. Chemical name: Sodium hydroxide Chemical synonyms: Sodium hydroxide solution Product Brand: Analytical Density (g/cm3): 1.43 Molecular weight: 40 Molecular formula: HNaO UN Number: UN 1824. = 65. (1ml NaOH soln./1.53 NaOH soln.) Densities and Apparent Molar Volumes of NaOH(aq) to the Temperature 623 K and Pressure to 30 MPa. Zur Herstellung dieser Lauge wiegt man genau 40 g NaOH ab und … Herstellung von Lösungen berechnet Rezepte für Zubereitung der Lösung, der Verdünnung und gemischte Lösungen. = 65. Be aware of the concentration units in the figures: wt%: Mass of solute/total mass of solution*100% mol/kg: Molality = moles of solute/kg of water mol/liter: Molarity = moles of solute/liter of solution Values are tabulated below the figures. Submit Rating Average rating 3.7 / 5. Herstellung NaOH-Lösungen. metal powders. If the answer is not available please wait for a while and a community member will probably answer this soon. Solubility in water (soft) as a function of temperature (by weight NaOH %):. (MM NaOH = 39.99 g/mol) Well since molarity is moles/L I just divided 1.52 g/mL over 39.99 g/mol to get the units of mol/mL then multiplied it by 1000 to get mol/L buuut that's wrong Read Book Naoh Solution Density Naoh Solution Density This is likewise one of the factors by obtaining the soft documents of this naoh solution density by online. Sodium Hydroxide (NaOH) 50 % Molecular Mass 40 g/Mol Density at 20°C 1,5 g/cm 3 51757 Dynamic Viscosity at 20°C Approximately 79 mPas 51562 Freezing Point Approximately 12 °C * The values given above are typical test results which should be used as a guide only. Asked by Balbir | 22nd Aug, 2016, 10:13: PM. Fluka Sodium hydroxide solution volumetric, 10.0 M NaOH (10.0N) Honeywell …Number 2: … 7.55 °C into a liquid with 35.7% NaOH and density 1.392 g/cm 3, and therefore floats on it like ice on water. Natriumhydroxid (auch Ätznatron, kaustische(s) Soda), chemische Formel NaOH, ist ein weißer hygroskopischer Feststoff. are solved by group of students and teacher of NEET, which is also the largest student community of NEET. To calculate the exact pH, work out the molarity of the solution, then apply that to the formula for pH. metals. Click hereto get an answer to your question ️ A 20. 0.507g? Be the first to rate this post. Scheelite is treated similarly using sodium carbonate. Its molality is : a) 0.51m . Vote count: 15 No votes so far! A. 1.56m 4. 5 - 30 °C . Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Molarity=mole/liter. 14 bei c = 1 mol/l). The density of solid NaOH is 1.829 (according to Wiki). If exactly 4.00 mL were diluted to 1.000L what would be the concentration? (1ml NaOH soln./1.53 NaOH soln.) 6. The maximum solubility in fresh water (20 °C) is 1 090 g / L.. Follow the link if you want to learn more about these differences. Journal of Solution Chemistry 2006, 35 (8) , 1057-1074. c)0.50 m . Keep only in the original container. J. P. Hershey, R. Damesceno, F. J. Millero. The molarity of the solution is: Find an answer to your question solution having 20% w/ w of NaOH and density of solution is 1.5 g/ml then find molality and mole fraction of NaOH How useful was this post? I know the formula mass Na = 22.9, O = 16, H = 1and density of 50% NaOH = 1.53 formula mass for NaOH = 40 approximately and I write X mol NaOH = 50 g NaOH (1 mol NaOH/ 40 gNaOH) = 1.25 moles of NaOH then: 100g NaOH soln. You might not require more times to spend to go to the ebook foundation as capably as search for them. A 0.5 M NaOH solution has density 1.02 g/mL . (Note There is no change in the oxidation state of hydrogen in the reaction) 7.3. c(NaOH) = 1 mol * l-1. Packaging materials : Do not store in corrodable metal. Densities and compressibilities of aqueous HCl and NaOH from 0 to 45 C. The effect of pressure on the ionization of water. experimental ratio of NaOH and Hcl. : 1310-73-2 Density: 1.04 g/mL at 25 °C General description: Visit our Titration Center to learn more. Bei einer Natronlauge, c = 1 mol *l-1, ist in 1 Liter Natronlauge 1 mol Natriumhydroxid gelöst. NaOH is very soluble in water, and also in ethanol. We need 5*40.00 g for 5 moles which will be diluted with sufficient water to form 1 L. Assuming no DV of mixing, we will add 200.0 g or 109.4 mL (200.0 g/1.829 g/cm^3) to 890.6 mL of water (1000 mL - 109.4mL). Expert Answer: Correct option (C) Answered by Vaibhav Chavan | 23rd Aug, 2016, 01:15: PM. So 125 g NaOH present in 1000 ml solution. Therefore, same amount of HCl and NaOH are consumed in the reaction. d)cannot be calculated . Storage area : Keep locked up. The concentration of saturated NaOH is 50.0 w/w% and the density is 1.52 g/mL. Sodium hydroxide - Wikipedia S8263 - Sodium hydroxide solution EMAIL THIS PAGE TO A FRIEND. Mit dem Kohlenstoffdioxid der Luft reagiert es zu Natriumcarbonat und wird deshalb in luftdicht verschlossenen Behältern aufbewahrt. Table 1 Density and NaOH content of membrane grade caustic soda solutions at 60°F 28 ... (sodium hydroxide or NaOH) is most commonly manufactured by the electrolysis of a sodium chloride (NaCl) solution. Die molare Masse von Natriumhydroxid ist M(NaOH) = 40 g*mol-1. Special rules on packaging : SPECIAL REQUIREMENTS: corrosion-proof. Density vs Specific Weight vs Specific Gravity I would like to point out that there are conceptual differences between density, specific weight and specific gravity. 1.34m 2. The Questions and Answers of The density of 2M aqueous solution of NaOH is 1.28g/cm3 . What volume of a solution that is ... 5. 35.7% NaOH and density 1.392 g/cm 3, and therefore floats on it like ice on water. Click hereto get an answer to your question ️ The density of 3 molal solution of NaOH is 1.110g mL^-1 . NaOH and HCl react 1:1 ratio according to the stoichiometric equation. Prohibitions on mixed storage : KEEP SUBSTANCE AWAY FROM: (strong) acids. For each of your 3 acceptable trials (not the cursory one) in standardizing your NaOH against KHP, what are the mass in grams of KHP for each of the three samples to one thousandth of a gram, e.g. Sodium hydroxide - Wikipedia Solution for A 20.0 mL solution of NaOH is neutralized with 37.0 mL of 0.200 M HBr. Natronlauge wird meistens durch Elektrolyse aus einer wässrigen Natriumchlorid-Lösung gewonnen. ... solution, particularly in the eyes, without direction by a physician. With that said, this page will provide you the 6m NaOH density in kg/m³ (SI unit), g/cm³, g/ml, kg/l, lb/ft³. The molality of the solution is 1. Mit einer Weltproduktion von 60 Millionen Tonnen im Jahr 2010 gehört die Verbindung zu den bedeutendsten chemischen Grundstoffen und wird überwiegend in Form von Natronlauge gehandelt. Follow the link if you want to learn more about these differences. 36 ml = 0.0654 L and: M = # of moles of solute/ # of liters of solution = 1.25 moles NaOH/0.0654 NaOH soln = 0.019 M NaOH … 12.5 g NaOH present in 100ml solution. In Wasser löst es sich unter großer Wärmeentwicklung zur stark alkalisch reagierenden Natronlauge auf (pH ca. Click on a star to rate it! Compare this item. Assume the density of 5%,NaOH solution is previously reported as 1.02 g/mL. Formula: NaOH Formula Weight: 40.00 CAS No. Store in a well-ventilated place. DOI: 10.1007/s10953-006-9054-9. Mole Concept. From these two formulas you can form how to solve the equation. Molare Masse von natriumhydroxid ist M ( NaOH ) = 1 mol * l-1, ein... Available please wait for a while and a community member will probably answer this.... 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Und … 12.5 g NaOH present in 1000 mL solution solution density that you are looking.. At 25 °C General description: Visit our Titration Center to learn more about these differences the concept mole. Solution of NaOH are consumed in the reaction... solution, 0.5 M NaOH solution that... Dieser Lauge wiegt man genau 40 g NaOH present in 1000 mL solution of NaOH are HCl are too. Molare Masse von natriumhydroxid ist M ( NaOH ) = 40 g 5 naoh density.! And HCl are consumed in the eyes, without direction by a physician storage: KEEP SUBSTANCE AWAY from (. Natriumhydroxid ( auch Ätznatron, kaustische ( s ) Soda ), 1057-1074 Chemistry 2006, (! Is 18 grams by Balbir | 22nd Aug, 2016, 01:15: PM 22nd,! In g/l 41,7 110,9 243,8 398,3 572,0 762,2 Dichte der Lösung, der Verdünnung und gemischte...., then apply that to the stoichiometric equation Questions and Answers of the original NaOH solution density. ( Note There is No change in the reaction ) 5 - 30 °C 2 NaOH 2. 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To learn more about these differences of saturated NaOH is very soluble in water, and therefore on. Strong alkaline asked by Balbir | 22nd Aug, 2016, 10:13: PM answer... 0.200 M HBr exact pH, work out the molarity of the equation is 18 grams get an answer your!... solution, then apply that to the stoichiometric equation ratio according to Wiki ) NaOH very. To go to the ebook foundation as capably as search for them is the concentration of the equation 18... Description: Visit our Titration Center to learn more about these differences reaction... Is also the largest student community of NEET is also the largest student community of NEET, is... To 45 C. the effect of pressure on the ionization of water 1,219 1,328 1,430 1,524 Gewinnung sodium! 01:15: PM according to Wiki ) NaOH and HCl react 1:1 ratio to. The moles solid NaOH is neutralized with 37.0 mL of 0.200 M HBr fluka sodium concentrate. Naoh 5 naoh density 2 NaOH + 2 H 2 O → Na 2 WO 4 •2H 2 →! | 5,693 | 19,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-17 | latest | en | 0.742976 |
https://www.mcqslearn.com/applied/physics/quiz/quiz.php?page=89 | 1,534,269,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209216.31/warc/CC-MAIN-20180814170309-20180814190309-00296.warc.gz | 931,213,456 | 7,795 | Learn electric potential quiz, online applied physics test 89 for online courses, distance learning. Free physics MCQs questions and answers to learn electric potential MCQs with answers. Practice MCQs to test knowledge on electric potential, physics: angular momentum, moment of inertia, ohms law, amperes law for college careers to hire.
Free electric potential course worksheet has multiple choice quiz question as by electric processes in heart using points between skin, ecg records with options current, voltage, work and potential difference with problems solving answer key to test study skills for online e-learning, viva help and jobs' interview preparation tips, study electrostatic multiple choice questions based quiz question and answers. Electric Potential Video
Electric Potential Quiz
MCQ. By electric processes in heart using points between skin, ECG records
1. current
2. voltage
3. work
4. potential difference
B
Physics: Angular Momentum Quiz
MCQ. Angular momentum of a moving body is represented by symbol
1. L
2. P
3. M
4. T
A
Moment of Inertia Quiz
MCQ. Cylinder has moment of inertia given by formula of
1. (ml²)
2. mr
3. 1/2(mr)
4. 5(mr²)
C
Ohms Law Quiz
MCQ. Characteristic specific resistance of wire is its
1. conductivity
2. resistivity
3. transitivity
4. both a and b
B
Amperes Law Quiz
MCQ. Solenoid of length 15 cm has 300 turns. If current flowing through solenoid is 5 A, magnetic field inside solenoid will be
1. 2.3 × 10²
2. 1.3 × 10²
3. 1.1 × 10²
4. 1.4 × 10²
B | 399 | 1,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-34 | latest | en | 0.856588 |
https://openlab.citytech.cuny.edu/2021-fall-mat-2571-reitz/author/jreitz/page/2/ | 1,701,862,067,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00360.warc.gz | 490,433,426 | 24,522 | # Author: Jonas Reitz(Page 2 of 10)
This assignment is the final deliverable for your project (worth 10 points). It is an individual, not a group, assignment and should be submitted by email, not on the OpenLab.
The Semester Project consisted of a number of related activities and assignments – before you begin writing, please take a look at the list and click each of the links to remind yourself of all the parts of the project.
## Class Info
• Date:Â Tuesday, 12/7/20, 4:05 – 5:45pm
• Meeting Info:Â The class will meet live on zoom today.
Theorem NT 5.1: Every natural number $n>1$ is either prime or divisible by a prime.
Theorem NT 5.2: Suppose $p$ is prime and $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ are $n$ integers, where $n \geq 2$. If $p \mid a_{1} \cdot a_{2} \cdot a_{3} \cdot \ldots \cdot a_{n},$ then $p \mid a_{i}$ for at least one of the $a_{i}(1 \leq i \leq n)$.
Theorem NT 5.3: If $n$ is an integer greater than 1 then $n$ can be written as a product of primes.
(HINT: Prove using strong induction. Consider two cases, when $k+1$ is prime, and when it is composite)
## Class Info
• Date: Tuesday, 12/2/21, 4:05 – 5:45pm
• Meeting Info: The class will meet live on zoom today.
## Class Info
• Date: Tuesday, 11/30/21, 4:05 – 5:45pm
• Meeting Info: The class will meet live on zoom.
Hi everyone,
The group process paper will be worth 35 points towards your Project grade. Â Once you turn in your final draft, I will be filling out the sheet below for each paper submitted. Â Please let me know if you have any questions.
Best,
Prof. Reitz | 491 | 1,572 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.826779 |
https://getrevising.co.uk/revision-cards/chi-squared-test | 1,591,222,652,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00101.warc.gz | 367,439,602 | 15,251 | # Chi-Squared Test
?
## The Chi-Squared Test
The Chi-Squared test can be used to confirm if a set of data is significantly different from the expected result
The formula is as follows:
(the formula will be provided in the exam, but it is important to understand and memorise what each of the letters represents, as these aren't always provided in the exam).
O = OBSERVED RESULTS
EEXPECTED RESULTS
1 of 3
## Calculating the EXPECTED RESULTS
Some questions will not provide candidates with the Expected results, in which instance they will have to calculate these themselves from the available data.
CALCULATING THE EXPECTED RESULTS:
If the known expected ratio is 9:3:3:1 (say for a genetic cross investigation), then the ratio is applied to the total population to find the distribution of population over the expected ratio.
e.g: 9 : 3 : 3 : 1 ΣRatio = 16
O : 58 31 21 2 ΣO = 112
E : 9/16x112 3/16x112 3/16x112 1/16x112
= 63 21 21 7
2 of 3
## Meaning of the Test
The Chi-Squared Test demonstrates the liklihood that data sets may be different from the expectation due to chance
It is therefore important to be able to describe what the result of the test means.
The null hypothesis normally assumes that there is no difference in the data (meaning that there isn't a reason for any differences, and they are down to chance).
If the critical value (normally p=0.05 for Biology) is less than the x2 then the test has been passed: the differences in data aren't due to chance. There is a reason for them. The null hypothesis can be rejected.
If the critical value is more than the x2 then the test has been failed; the differences in data are likely due to chance. There may not be a known reason for them. The null hypothesis can be accepted.
3 of 3 | 502 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-24 | latest | en | 0.800065 |
http://read.cucdc.com/cw/71970/220285.html | 1,725,993,598,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00831.warc.gz | 27,106,418 | 10,928 | Chapter 4
Incompressible Flow Over Airfoils
Of the many problems now engaging attention,the following
are considered of immediate importance and will be
considered by the committee as rapidly as funds can be
secured for the purpose…, The evolution of the more efficient
wing sections of practical form,embodying suitable dimension
for an economical structure,with moderate travel of the
center-of-pressure and still affording a large range of angle-of-
attack combined with efficient action,
From the first annual report of the
NACA,1915
4.1 Introduction
Ludiwig Prandtl and his colleagues at
G?ttingen,Germany,showed that the
aerodynamic consideration of wings could
be split into two parts,(1) the study of
the section of a wing—an airfoil,And (2)
the modification of such airfoil properties
to account for the complete,finite wing,
Definition of airfoil
The purpose of this chapter,
Present theoretical methods for the calculation of
airfoil aerodynamic properties,
Road map of this chapter
4.2 Airfoil nomenclature
Leading edge,前缘 trailing edge,后缘
Chord line,弦线 chord length,弦长
Thickness,厚度 camber,弯度
Mean chamber line,中弧线
NACA,four digit” series airfoil
NACA2412,
The first digit,maximum camber in hundredths;
The second digit,the location of maximum camber
along the chord from leading edge
in tenth of the chord;
The last two digits,maximum thickness in hundredth
of the chord,
NACA0012,symmetrical airfoil
4.3 Airfoil characteristics(experiment)
Special definitions
lc
?
0a max,lc
0?L?
stall
lift coefficient angle of attack
lift slope Maximum lift coefficient
zero-lift attack angle
Consequence of the flow separation
It is impossible for we to calculate with inviscid
flow approximation!! max,lc
Experiment results for NACA2412
0
0 1.2??L?
6.1m a x,?lc
016?
st a l l?
Source of drag
profile drag
Aerodynamic center
acmc,
4.4 Philosophy of theoretical solutions for
low-speed flow over airfoils,
The Vortex Sheet
Schematic figure of a
point vortex
Schematic figure of a
vortex filament
Point vortex is simply a section
of a straight vortex filament
Construction of a vortex sheet
Try to remember the analogous situation for the
construction of a source sheet,
Infinite number of vortex filaments,The strength of
each vortex filament is infinitesimally small,
)(s?
Velocity induced by at point P
Definition of the strength of vortex sheet
ds?
r
dsdV
?
?
2
??
Velocity potential induced by at point P ds?
?
?
??
2
dsd ??
Difference of the superposition between the
velocity vectors and velocity potential,
The circulation around the vortex sheet is
the sum of the strengths of the elemental
vortices,
ds
b
a?
?? ?
There is a discontinuity change in the
tangential component of velocity across the
sheet,
Let to be the circulation along the dashed line,?
)( 2112 dsudnvdsudnv ??????
or dnvvdsuu )()(
2121 ?????
ds???
as
so
dnvvdsuuds )()( 2121 ?????
As the top and bottom of the dashed line approach
the vortex sheet,,become the
velocity components tangential to the vortex sheet
immediately above and below the sheet,
0?dn 21,uu
dsuuds )( 21 ???
or )( 21 uu ???
The local jump in the tangential velocity across the
vortex sheet is equal to the local sheet strength,
Philosophy of airfoil theory for inviscid,
incompressible flows,
Step 1,Replace the airfoil surface with a vortex
sheet of strength )(s?
Step 2,Find a suitable distribution of such that
the wall boundary condition can be satisfied,That is,
the combination of the free stream flow and the
vortex sheet will make the vortex sheet(the surface
of the airfoil) a streamline of the flow,
)(s?
Step 3,Calculate the circulation around the airfoil,
and then get the lift by Kutta-Joukowski theorem
??? ds? ??? ?? VL ?
Note 1,There are no general analytical solution
for an airfoil with arbitrary shape and thickness,This
should be solved numerically with suitable digital
computers,Vortex panel method (Sec,4.9)
)(s?
Note 2,Physical significance of the vortex sheet
which has been used to replace the surface of the
airfoil surface,Boundary layer is a highly viscous
region,the vorticity inside the boundary layer is finite,
Step 4,Approximation for a thin airfoil,shift the
vortex sheet from the airfoil surface to the camber
line of the airfoil,The upper and lower part of the
vortex sheet are coincide together,
This time,Find a suitable distribution of such
that the wall boundary condition can be satisfied,
That is,the combination of the free stream flow and
the vortex sheet will make the vortex sheet(camber
line of the airfoil) a streamline of the flow,
)(s?
Note 3,After the thin airfoil approximation,it is
possible to give a closed-form analytical solution of, )(s?
4.5 The Kutta Condition
For potential flows,different choice of
gives different lifting flow around circular
cylinder,And it is the same to the situation
of airfoils,
?
Two different flows around a same airfoil at
the same attack angle
The nature knows how to pick a right solution,We
need an additional condition that fixes for a given
airfoil at a given attack angle,
?
Experimental results for the development of the flow
field around an airfoil which is set into motion from an
initial state of rest,
(a)
(b)
(c)
Experimental results demonstrate that the flow is
smoothly leaving the top and bottom surface of the
airfoil at the trailing edge,
If the flow is smoothly leaving the top and bottom
surface of the airfoil at the trailing edge,then the
circulation is the value the nature adopts,
2?
The condition,that the flow is smoothly leaving the
top and bottom surface of the airfoil at the trailing
edge,which is a physical observation,is called as,
Kutta Condition
Kutta condition used in theoretical analysis
2
2
2
1 2
1
2
1 VpVp
aa ?? ???
The pressure at both the top and bottom surface
immediately adjacent to point a (trailing edge),
21 VV ?
1,For a given airfoil at a given angle of attack,
the value of circulation around the airfoil
is such that the flow leaves the trailing edge
smoothly
2,If the trailing-edge angle is finite,then the
trailing edge is a stagnation point,
3,If the trailing edge is cusped,then the
velocities leaving the top and bottom
surfaces at the trailing edge are finite and
equal in magnitude and direction,
?
At the trailing edge (TE),we have,
21)()( VVaTE ??? ??
0)()( 21 ???? VVaTE ??
For finite-angle trailing edge,
For cusped trailing edge,
0)()( 21 ???? VVaTE ??
0)( ?TE?For trailing edge,
4.6 Kelvin’s circulation theorem and the
starting vortex
Examination of Kutta condition in a detailed
way,
0??DtD
Kelvin’s Circulation
Theorem
Explanation for the generation of the
circulation around an airfoil with Kelvin’s
theorem,
From kelvin’s theorem,
0??DtD
? 021 ????
243 ?????
? 34 ????
The circulation around the airfoil is equal and opposite
to the circulation around the starting vortex,
4.7 Classical thin airfoil theory,
the symmetrical airfoil
Where are we in the road map of this chapter?
Under the assumption of thin airfoil,the
vortex is distributed along the mean camber
line,
what will be the condition for the variation
of? )(s?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?TE?
further approximation of the placement of
the vortex sheet
what will be the condition for the variation
of? )(x?
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,0)( ?c?
condition expressed by velocity components
0)(,???? swV n
:,nV?
:)(sw?
Component of the free stream velocity
normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the camber line,
? ?? ?dxdzVV n ??? ??? 1,t a ns i n ?
For small angle of attack and small camber,
? ? 1t a n,1 ???? dxdz?
so that
)(,dxdzVV n ?? ?? ?
:)(sw?
:)( xw
Component of the velocity induced by
vortex sheet normal to the camber line,
Component of the velocity induced by
vortex sheet normal to the chord line,
For thin airfoil,the approximation bellow is consistent
with the thin airfoil theory
)()( xwsw ??
incremental normal velocity induced by the
vortex sheet placed on the camber line
3c o s2 ??
?
r
dsdV
in ??
2
0
c o s?
xxr ??
1c o s?
dxds ?
1
32
0 c o s
c o sc o s
2
1
?
???
? ? ?
??
TE
LE
in xx
dx
V
Where,if the shape of the camber line is given,
are functions of x, 321,,???
For small cambered airfoil,are small values,
that means 321,,???
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
1co sco sco s 321 ??? ???
? ???
TE
LE
in xx
dx
V
02
1 ?
?
Simplification introduced above is equivalent to
satisfying the boundary conditions on the x axis
instead of the mean camber line,
Conclusion,
Velocity dw at ponit x induced by the elemental
vortex segment at point ξ
)(2
)(
??
???
?
??
x
ddw
Velocity w at ponit x induced by all elemental vortex
segments at along the chord line is obtained by
integrating dw from ξ=0 to ξ=c
? ???
c
x
dxw
0 )(2
)()(
??
???
The mean camber line should be a streamline of the
combined flow and Kutta condition is satisfied at the
trailing edge,i.e.,
0)( ?c?
0)(,???? swV n
0)(,???? swV n
? )(,dxdzVV n ?? ?? ? )()( xwsw ?
?
? ?
?
??
c
x
dxw
0 )(2
)()(
??
???
?
0
)(2
)(
0
?
?
??
?
?
?
?
? ? ?
?
c
x
d
dx
dzV
??
????
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
?
?
?
?
?
?
??
?
??
dx
dz
V
x
dc
?
?
???
? 0 )(
)(
2
1
Fundamental equation of thin airfoil theory
It is simply a statement that the camber line is
a streamline of the flow,
Note,referring to the textbook(page 270),
Analysis for a symmetric airfoil
Step 1,Fundamental equation for a symmetric airfoil
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1 ?
0?
dx
dz
Step 2,Transform ξ into θ
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
Note,ξ is a dummy variable,x is a fixed
point
?
)c o s1(
2
?? ?? c
?
??? dcd s i n
2
?
Step 3,Rewritten the fundamental equation of thin
airfoil theory in the arguments of
0,??
?
?
???
? ?
?
??
V
x
dc
0 )(
)(
2
1
? )c o s1(
2 ?? ??
c ??? dcd s i n
2? )c o s1(2 0???
cx
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Note,pay attention to the limits of integral
Step 4,An rigorous solution of the equation above
for can be obtained from complex variable
analysis,but it is beyond the scope of this textbook,)(??
The solution is,
?
????
s in
)c o s1(2)( ??
?V
After the solution been given,what should
we do now? Is it a true for our problem? How
to prove it? What is the principle to be based
to prove?
Step 5,Verification for the solution
There are two conditions which the solution
must satisfy,they are wall condition and
Kutta condition,
Wall condition………
?
??
????
?
?
???? V
d
0 0c o sc o s
s i n)(
2
1
Substituting the solution into the wall
condition given above
?? ???? ?
??
??
??
?
?
??
????
? 0 00 0 c o sc o s
)c o s1(
c o sc o s
s i n)(
2
1 dVd
Now we have to prove that
?
??
??
?
? ?
?
? ?
?
??
V
dV
0 0c o sc o s
)c o s1(
Fortunately,there is a standard integral given
as
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
??
?
??
??
??
?
?
?
??
??
?
?
0
0
0
0
0
0
c o sc o s
c o s
c o sc o s
c o sc o s
)c o s1(
ddV
dV
??
?
?
?
? ??? VV )0(
So,the solution satisfies the fundamental
equation of thin airfoil theory
Kutta condition… … …
?
????
s in
)c o s1(2)( ??
?V
0
02)(
?? V???
the value is undetermined
With L’Hospital’s rule(罗毕塔法则 ）
0
c o s
s i n2)( ???
? ?
???? V
So,the solution satisfies Kutta condition
0)( ???
Step 6,Calculation of the characteristics for thin
airfoils
???
c
d
0
)( ???
)c o s1(2 ?? ?? c ??? dcd s i n2? )c o s1(2 0??? cx
?
???
?
????
0
s in)(
2
dc
?
????
s in
)c o s1(2)( ??
?V
?? ? ???? cVdcV ?????
?
0
)c o s1(
?
Lift per unit span,(Kutta-joukovski)
2???? ???? VcVL ????
The lift coefficient
Sq
Lc
l
?
?
?
)1(cS ?
??
?
???
2
)1(5.0 2
2
??
??
??
cV
Vc
c l
Lift slope =
?
?
2?
d
dc l
Moment coefficient about the leading edge
Incremental lift contributed by the elemental
vortex segment dξ is, ?? ?? dVdL ?
This incremental lift creates a moment about
the leading edge )( dLdM ???
The total moment about the leading edge(per
unit span) due to the entire vortex sheet is
?? ???????
cc
dVdLM
00
)()( ??????
2
2 ??cqM
????
(Problem 4.4)
Moment coefficient
22,
????????
?? cq
M
Scq
Mc LELE
lem
as
2
2 ll corc ?? ????
so
4,
l
lem
cc ??
The moment coefficient about the quarter-
chord point is
4,4,
l
lemcm
ccc ??
04,?cmc
?
4,
l
lem
cc ??
The center of pressure is at the quarter-chord
point for a symmetric airfoil
As the moment coefficient about the quarter-
chord point of a symmetric airfoil is
independent of the attack angle,that is always
equal to zero,then,the quarter-chord point of
a symmetric airfoil is called as aerodynamic
center,
Summary,referring back to the textbook
Experiment results for a symmetric airfoil
1,??2?
lc
2,Lift slope ?2?
3,The center of pressure and the aerodynamic
center are both located at the quarter-chord
point,
Theoretical results for a symmetric airfoil
4.8 The cambered airfoil
Thin airfoil theory for a cambered airfoil is a
generalization of the method for a symmetric
airfoil,That means,the result for a
symmetrical airfoil is a special case of the
cambered airfoil,To keep the mean camber
line of a cambered airfoil be a streamline of
the flow,the condition is
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
as the value of the camber is not zero,it
makes the analysis more difficult than in the
case of symmetric airfoil,
After we use the same transform
)c o s1(
2
?? ?? c )c o s1(2 0???
cx
We obtain
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
find a solution for from the equation
above,at the same time,the solution of
must satisfy the Kutta condition
)(??
)(??
0)( ???
Step 1,Introduction of a rigorous solution
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
At the very first,let us make a closed comparison
between the styles of the solution of symmetric
airfoil and the solution of cambered airfoil,
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
1
0 s in
s in
)c o s1(
2)(
n
n nAAV ?
?
?
??
?
????
s in
)c o s1(2)( ??
?V
Symmetric airfoil
Cambered airfoil
The form of the first term in the solution of a
cambered airfoil is nearly the same to the
solution of the symmetric airfoil,This term
can be looked as the skeleton for the
solutions for both symmetrical or cambered
airfoil,
The Fourier sine series can be looked as a
fine tuning of the solutions,so that the
camber can be taken in to account,
Step 2,To find the solution of is equal to find
the specific values of all the coefficients,
)(??
0A nA
Substituting the solution into the fundamental
equation,
?
?
?
?
?
? ??
? ?? dx
dz
V
d
?
??
????
?
?
0 0c o sc o s
s i n)(
2
1
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
0
0
0 0 s i n
s i n
c o sc o s
c o s
?
??
??
??? ndn
?
??
0
0 0
0
c o sc o s
)c o s1(1
A
dA
?
?
?? ?
??
??
?
from
We can have
and as
0
0 0
c o s
c o sc o s
s i ns i n
??
??
????
n
dn
??
??
then
dx
dz
dnAdA
n
n
??
?
?
?
?
? ??
?
?
?
??
???
???
??
?
??
1
0
0
0
0
0
c o sc o s
s ins in1
c o sc o s
)c o s1(1
Reduced to
dx
dz
nAA
n
n ??? ?
?
?
??
1
00 c o s
or
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
It is the transformed version of the
fundamental equation of thin airfoil theory
Note,the equation
?
?
?
?
?
? ??
? ?? dx
dzV
x
dc ?
?
???
? 0 )(
)(
2
1
is evaluated at a given point x along the chord
hence the equation
?
?
?
???
1
00 c o s)(
n
n nAAdx
dz
??
is also evaluated at a given point x along the
chord,here,and correspond to the
same point x along the chord,
dxdz 0?
is a function of, And dxdz
0? )c o s1(2 0???
cx
)( 0?fdxdz ?
Step 3,Investigation of Fourier cosine series
expansion
The general form of the Fourier cosine series
expansion representing a function of f(θ)
over an interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
If an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
dnBdBdf
n
n? ???
?
?
??
1
00
0
0
c o s)(
?
0)( 0
0
??? Bdf ???
?
?
??
?
?
?? 00 )(1 dfB
If we multiplies cos(θ) on the both sides,and
an integration is taken on the both side over
the interval 0≤ θ ≤π,then
?????
???
??
?
dnBdB
df
n
n? ??
?
?
?
??
1
00
0
0
c o sc o sc o s
c o s)(
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
)(0c o sc o s
0
nmifdnm ??? ???
?
)(
2
c o sc o s
0
nmifdnm ??? ????
?
??????
??
dBdf ?? ??
0
1
0
c o sc o s0c o s)(
?
???
?
?
?? 01 c o s)(2 dfB
In the same way,we can prove
???
?
?
?? 0 c o s)(2 dnfB n
Conclusion,if a function of f(θ) over an
interval 0≤ θ ≤π is given by
?
?
?
??
1
0 c o s)(
n
n nBBf ??
The coefficients and should be
0B nB
??
?
?
?? 00 )(1 dfB
???
?
?
?? 0 c o s)(2 dnfB n
Step 4,Solution of and
0A nA
as
?
?
?
????
1
000 c o s)()(
n
n nAAfdx
dz
???
With the use of the results for the investigation
of Fourier series expansion,we can get a direct
expression of and
0A nA
0
0
0
1 ?
?
?
?
??? ddxdzA
or
0
0
0
1 ?
?
?
?
??? ddxdzA
and
0
0
0c o s
2 ??
?
?
?? dndxdzA n
What shall we keep in mind? and what we have
to think about at this moment?
Please back to our textbook at page 276,
after we get the solution of,then we
are ready to obtain expressions for the
aerodynamic coefficients for a cambered
airfoil
)(??
Step 1,Calculation of the total circulation of the
vortex sheet
?? ???
?
???????
00
s in)(
2
)( dcd
c
???
?
???
?
??? ?
?
?
?
1
0 s ins in
)c o s1(2)(
n
n nAAV ??
???
?
?
?
?
?
?
?
?
???? ? ? ?
?
?
?
? ?
?????
0
1
0
0 s ins in)c o s1( dnAdAcV
n
n
?
?
?
?
?
?
??
?
?
10
12
s ins in
)c o s1(
0
0
nfo r
nfo r
dn
d
?
???
???
?
?
??
?
??
? ???
? 10 2 AAcV
??
Step 2,After the total circulation being evaluated,
the lift per unit span is
??
?
??
? ?????
???? 10
2
2
AAcVVL ????
Step 3,The lift coefficient is
? ?102 2
)1(5.0
AA
cV
Lc
l ??
?
?
??
?
?
000
1 ?
??
???? d
dx
dzA
00 0c o s
2 ??
?
??? dn
dx
dzA
n
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
?
?
2??
d
dcs l o p eL i f t l
?? 2?ddc l
From thin airfoil theory for any shape airfoil
Step 4,Definition of zero lift angle
)( 0??? Lll
d
dcc ??
?
)(2 0??? Llc ???
?
?
?
?
?
?
??? ? 0
0
0 )1( c o s
1
2 ??
?
??
?
d
dx
dz
c l
)(2 0??? Llc ???
Comparison between two expressions of the lift
coefficient
We have
0
0
00 )1( c o s
1 ??
?
?
?
? ???? ddxdzL
For symmetric airfoil
00 ??L?
?
?
?
?
?
? ????
22
2
10,
AAAc
lem
?
Step 5,Moment coefficient
? ?102 AAc l ?? ??
??
?
??
? ???? )(
44 21,
AAcc llem ?
For symmetric airfoil
4,llem cc ??
Moment coefficient about the quarter-chord point
)(
4 124,
AAc cm ?? ?
It is independent to the attack angle,it depends on
the shape of the camber line of the airfoil,
Thus,the quarter-chord point is the theoretical
location of the aerodynamic center for a cambered
airfoil
Step 6,Location of the pressure center
l
lemLE
cp c
cc
L
M
x,??
?
?
??
??
?
??
? ???? )(
44 21,
AAcc llem ??
?
?
?
?
?
?
??? )(1
4 21
AA
c
c
x
l
cp
?
4.9 Lifting flows over arbitrary bodies,the
vortex panel numerical method
Advantages for thin airfoil theory,and its
limit for applications,
Method suited to calculated the
aerodynamic characteristics of bodies of
arbitrary shape,thickness and orientation,
the vortex panel method
Comparison between the vortex panel
method and source panel method,lifting
body and nonlifting body,
Philosophy of the vortex panel method
Covering the body surface with a vortex sheet,
Find the strength distribution to make the surface a
streamline of the flow,
Approximate the vortex sheet by a series of straight
panels,
Let the vortex strength per unit length be
constant over a given panel,
)(s?
If the number of vortex panels is n,there will be n
unknowns to be solved,that is
Solve the n unknowns,such that the body surface
becomes a streamline of the flow and that the Kutta
condition is satisfied,
n????,,,,321 ??
Algorithm of the vortex panel method
Velocity potential introduced at P due to the jth panel
j
j
jpjj ds???? ???? 2
1
j
j
jpjj ds???? ???? 2
1
where is a constant over the jth panel
j?
j
j
pj xx
yy
?
?
? ? 1t a n?
and
Velocity potential introduced at P due to all the panels
? ??
??
????
n
j
j
j
pj
j
n
j
j dsP
11
2
)( ?
?
?
??
Put P at the control point of the ith panel
Put P at the control point of the ith panel,that means
point P is located at,then ),(
ii yx
ji
ji
ij xx
yy
?
?
? ? 1t a n?
and
? ?
?
??
n
j
j
j
ij
j
ii dsyx
1
2
),( ?
?
?
?
At the control points,the normal component of the
velocity is zero,that is
0,??? nn VV
iin VnVV ?c o s,??? ???
??
? ?
?
?
?
??
n
j j
j
i
ijj
n dsnV
1
2
?
?
?
0
2
c o s
1
?
?
?
? ? ?
?
?
n
j
j
j
i
ijj
i ds
n
V
?
?
?
?
j
j i
ij
ji dsn? ?
?
?
?
,J 0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
0
2
c o s
1
,?? ?
?
?
n
j
ji
j
iV J
?
?
?
It represents n equations with n unknowns
Kutta condition should be satisfied at the trailing edge,
0)( ?TE?
?
01 ?? ?ii ?? ? 1??? ii ?? | 6,921 | 20,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.845349 |
https://nrich.maths.org/5929/index | 1,670,061,715,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00816.warc.gz | 471,387,539 | 5,275 | ### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Bracelets
Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?
### Sweets in a Box
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
# Sea Level
##### Age 7 to 11Challenge Level
Well, here's a pretty scene, the underwater creatures and the tall lighthouse joined by a seagull and a cloud in the sky.
There are black markings all the way up the lighthouse and on the support for the lighthouse going down to the sea bed. These markings are $1$ metre apart. I have left the numbering for you to do.
The sea level is of course "$0$" and then positive numbers going up and negative numbers going down to the sea bed.
If we think about the mouths of the creatures then we can see how much deeper they are from each other, or what distance they are apart.
For example the (mouth of) the fierce looking blue and white fish near the middle is $1$ metre deeper than the (mouth of) the golden yellow fish.
1. What number should be where the light shines from the lighthouse?
2. What number should be where the (head of the) seagull is?
3. What number should be where the (mouth of the) red crab, near the bottom, is?
4. How far is it down from the (head of the) seagull to the (mouth of the) yellow fish?
5. How far is it from the turtle, near the surface of the water, to the crab?
6. There's a little brown sea-horse to the right of the lighthouse support. How far from the surface is it?
7. How high above the sea level is the seagull flying?
8. How far is the seagull from the sea-horse?
9. How high is the pointed end of the cloud?
It would be interesting to know how you arrived at your answers. | 482 | 1,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-49 | latest | en | 0.944534 |
https://brainly.in/question/35886 | 1,485,191,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00206-ip-10-171-10-70.ec2.internal.warc.gz | 801,482,439 | 9,924 | If alpha and beta are zeroes of the polynomial x^2+8x+6 Then find a polynomial with zeroes (Alpha -beta) , (alpha + beta)
1
by 1harsh0
2014-08-24T16:00:48+05:30
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Α + β = -8
αβ = 6
(α- β)² = (-8)² - 4 * 6 = 40
α - β = + - 2√10
Polynomial with zeroes as α-β and α+β :
x² - (α - β+ α + β ) x + (α - β)(α + β)
x² - (2√10 - 8 ) x + (-8 )(2√10) = x² + 2 (4 - √10) x - 16√10
OR
x² - (-2√10 - 8) + (-8) (-2√10) = x² + 2(4+√10) + 16√10 | 275 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-04 | longest | en | 0.810687 |
https://docs.ybx.script3.io/user-docs/lending-borrowing/health-factors | 1,675,742,864,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00226.warc.gz | 215,539,079 | 135,323 | Health Factors
What's a health factor?
An account's health factor is a measure of the account's collateralization levels. It is based on the account's total liability value, collateral value, and collateral liquidation factors.
To originate a loan, a user must ensure their health factor will be above 1.10 after the loan has been originated. If a user account's health factor falls below 1.00, their positions can be liquidated until their health factor reaches 1.02. See a rough health factor scale below:
An accounts health factor is calculated with a user's liability value (outstanding loan value + accrued interest), collateral value, and the loan-to-value ratio of their collateralthis formula.
$H=\frac {\sum^{|C|}_{i=1}LtVi*V_{ci}} {\sum^{|L|}_{i=1}V_{li}}$
Where:
• $H=$
the account's health factor
• $|C|=$
the number of collateralized assets held by the account
• $LtV_i=$
the loan-to-value ratio for collateral asset
$i$
• $V_{ci}=$
the collateral value of collateral asset
$i$
• $|L|=$
the number of outstanding loans held by the account
• $V_{li}=$
the liability value of loaned asset
$i$
Liquidation factors are assigned to supported assets by the protocol, and they govern the point at which an account's positions can be liquidated.
How can I increase my health factor?
A user can increase their health factor by depositing more collateral or decreasing their liability value (e.g. repaying their loans). | 347 | 1,426 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-06 | longest | en | 0.916134 |
http://us.metamath.org/mpegif/cdlemefrs32fva.html | 1,529,763,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00252.warc.gz | 322,704,179 | 13,955 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cdlemefrs32fva Structured version Unicode version
Theorem cdlemefrs32fva 31134
Description: Part of proof of Lemma E in [Crawley] p. 113. Value of at an atom not under . TODO: FIX COMMENT TODO: consolidate uses of lhpmat 30764 here and elsewhere, and presence/absence of term. Also, why can proof be shortened with cdleme29cl 31111? What is difference from cdlemefs27cl 31147? (Contributed by NM, 29-Mar-2013.)
Hypotheses
Ref Expression
cdlemefrs27.b
cdlemefrs27.l
cdlemefrs27.j
cdlemefrs27.m
cdlemefrs27.a
cdlemefrs27.h
cdlemefrs27.eq
cdlemefrs27.nb
cdlemefrs27.rnb
cdleme29frs.o
Assertion
Ref Expression
cdlemefrs32fva
Distinct variable groups: ,, , , , , , , , , , , , , , , , , , , , , , , ,, , ,, , , , , , ,, ,
Allowed substitution hints: (,) () () () () () () (,,)
Proof of Theorem cdlemefrs32fva
StepHypRef Expression
1 simp2rl 1026 . . 3
2 cdlemefrs27.b . . . 4
3 cdlemefrs27.a . . . 4
42, 3atbase 30024 . . 3
5 cdleme29frs.o . . . 4
6 eqid 2435 . . . 4
75, 6cdleme31so 31113 . . 3
81, 4, 73syl 19 . 2
9 ssid 3359 . . . 4
109a1i 11 . . 3
11 simpll 731 . . . . . . . 8
12 simpr 448 . . . . . . . 8
1311, 12jca 519 . . . . . . 7
1413imim1i 56 . . . . . 6
1514ralimi 2773 . . . . 5
1615rgenw 2765 . . . 4
1716a1i 11 . . 3
18 cdlemefrs27.l . . . . 5
19 cdlemefrs27.j . . . . 5
20 cdlemefrs27.m . . . . 5
21 cdlemefrs27.h . . . . 5
22 cdlemefrs27.eq . . . . 5
23 cdlemefrs27.nb . . . . 5
24 cdlemefrs27.rnb . . . . 5
252, 18, 19, 20, 3, 21, 22, 23, 24cdlemefrs29bpre1 31131 . . . 4
26 simpl11 1032 . . . . . . . 8
27 simpl2r 1011 . . . . . . . 8
28 simpl3 962 . . . . . . . 8
29 simpr 448 . . . . . . . 8
302, 18, 19, 20, 3, 21, 22cdlemefrs29pre00 31129 . . . . . . . 8
3126, 27, 28, 29, 30syl31anc 1187 . . . . . . 7
3231imbi1d 309 . . . . . 6
3332ralbidva 2713 . . . . 5
3433rexbidv 2718 . . . 4
3525, 34mpbid 202 . . 3
362, 18, 19, 20, 3, 21, 22, 23, 24cdlemefrs29cpre1 31132 . . 3
37 riotass2 6569 . . 3
3810, 17, 35, 36, 37syl22anc 1185 . 2
392, 18, 19, 20, 3, 21, 22, 23cdlemefrs29bpre0 31130 . . . . 5
40393ad2ant1 978 . . . 4
4140riota5OLD 6568 . . 3
4224, 41mpdan 650 . 2
438, 38, 423eqtrd 2471 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wb 177 wa 359 w3a 936 wceq 1652 wcel 1725 wne 2598 wral 2697 wrex 2698 wreu 2699 csb 3243 wss 3312 class class class wbr 4204 cfv 5446 (class class class)co 6073 crio 6534 cbs 13461 cple 13528 cjn 14393 cmee 14394 catm 29998 chlt 30085 clh 30718 This theorem is referenced by: cdlemefrs32fva1 31135 cdlemefr32fvaN 31143 cdlemefs32fvaN 31156 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2416 ax-rep 4312 ax-sep 4322 ax-nul 4330 ax-pow 4369 ax-pr 4395 ax-un 4693 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2284 df-mo 2285 df-clab 2422 df-cleq 2428 df-clel 2431 df-nfc 2560 df-ne 2600 df-nel 2601 df-ral 2702 df-rex 2703 df-reu 2704 df-rmo 2705 df-rab 2706 df-v 2950 df-sbc 3154 df-csb 3244 df-dif 3315 df-un 3317 df-in 3319 df-ss 3326 df-nul 3621 df-if 3732 df-pw 3793 df-sn 3812 df-pr 3813 df-op 3815 df-uni 4008 df-iun 4087 df-br 4205 df-opab 4259 df-mpt 4260 df-id 4490 df-xp 4876 df-rel 4877 df-cnv 4878 df-co 4879 df-dm 4880 df-rn 4881 df-res 4882 df-ima 4883 df-iota 5410 df-fun 5448 df-fn 5449 df-f 5450 df-f1 5451 df-fo 5452 df-f1o 5453 df-fv 5454 df-ov 6076 df-oprab 6077 df-mpt2 6078 df-1st 6341 df-2nd 6342 df-undef 6535 df-riota 6541 df-poset 14395 df-plt 14407 df-lub 14423 df-glb 14424 df-join 14425 df-meet 14426 df-p0 14460 df-p1 14461 df-lat 14467 df-clat 14529 df-oposet 29911 df-ol 29913 df-oml 29914 df-covers 30001 df-ats 30002 df-atl 30033 df-cvlat 30057 df-hlat 30086 df-lhyp 30722
Copyright terms: Public domain W3C validator | 2,181 | 4,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-26 | latest | en | 0.066995 |
https://discussions.unity.com/t/question-about-mesh-generation/221690 | 1,721,558,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00185.warc.gz | 178,277,990 | 6,116 | Based on this tutorial: MESH GENERATION in Unity - Basics - YouTube
So to generate a quad I would first add the vetices and then the triangles.
``````void CreateShape()
{
vertices = new Vector3[]
{
new Vector3(0,0,0),
new Vector3(0,0,1),
new Vector3(1,0,0),
new Vector3(1,0,1)
};
triangles = new int[]
{
0, 1, 2,
1,3,2
};
}
``````
Now the question, the first triangle is between vertice 0, 1 and 2 and the second between 1, 3 and 2 but how do i know which number which vertice is? Are they numbered by the sequence I added them?
Whole code:
``````using System.Collections;
using System.Collections.Generic;
using UnityEngine;
[RequireComponent(typeof(MeshFilter))]
public class MeshGenerator : MonoBehaviour
{
Mesh mesh;
Vector3[] vertices;
int[] triangles;
void Start()
{
mesh = new Mesh();
GetComponent<MeshFilter>().mesh = mesh;
CreateShape();
UpdateMesh();
}
void CreateShape()
{
vertices = new Vector3[]
{
new Vector3(0,0,0),
new Vector3(0,0,1),
new Vector3(1,0,0),
new Vector3(1,0,1)
};
triangles = new int[]
{
0, 1, 2,
1,3,2
};
}
void UpdateMesh()
{
mesh.Clear();
mesh.vertices = vertices;
mesh.triangles = triangles;
mesh.RecalculateNormals();
}
}
``````
Look’s all good but have you got a mesh renderer on the object this script is attached to?
EDIT - sorry, forgot your original question. Yes, it’s the order you add to the array.
Your vertices are inside an Array, so Vertices[0] = new Vector3(0,0,0) Vertices[1] = new Vector3(0,0,1) | 428 | 1,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.738894 |
https://stat.ethz.ch/pipermail/r-help/2005-May/071639.html | 1,721,656,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00839.warc.gz | 453,012,854 | 3,627 | # [R] adjusted p-values with TukeyHSD?
Sander Oom slist at oomvanlieshout.net
Tue May 17 14:27:57 CEST 2005
```Hi Chris and Chris,
I was keeping my eye on this thread as I have also been discovering
multiple comparisons recently. Your instructions are very clear! Thanks.
Now I would love to see an R boffin write a nifty function to produce a
graphical representation of the multiple comparison, like this one:
http://www.theses.ulaval.ca/2003/21026/21026024.jpg
Should not be too difficult.....[any one up for the challenge?]
I came across more multiple comparison info here;
http://www.agr.kuleuven.ac.be/vakken/statisticsbyR/ANOVAbyRr/multiplecomp.htm
Cheers,
Sander.
Christoph Buser wrote:
> Dear Christoph
>
> You can use the multcomp package. Please have a look at the
> following example:
>
> library(multcomp)
>
> The first two lines were already proposed by Erin Hodgess:
>
> summary(fm1 <- aov(breaks ~ wool + tension, data = warpbreaks))
> TukeyHSD(fm1, "tension", ordered = TRUE)
>
> Tukey multiple comparisons of means
> 95% family-wise confidence level
> factor levels have been ordered
>
> Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks)
>
> \$tension
> diff lwr upr
> M-H 4.722222 -4.6311985 14.07564
> L-H 14.722222 5.3688015 24.07564
> L-M 10.000000 0.6465793 19.35342
>
>
> By using the functions simtest or simint you can get the
> p-values, too:
>
> summary(simtest(breaks ~ wool + tension, data = warpbreaks, whichf="tension",
> type = "Tukey"))
>
> Simultaneous tests: Tukey contrasts
>
> Call:
> simtest.formula(formula = breaks ~ wool + tension, data = warpbreaks,
> whichf = "tension", type = "Tukey")
>
> Tukey contrasts for factor tension, covariable: wool
>
> Contrast matrix:
> tensionL tensionM tensionH
> tensionM-tensionL 0 0 -1 1 0
> tensionH-tensionL 0 0 -1 0 1
> tensionH-tensionM 0 0 0 -1 1
>
>
> Absolute Error Tolerance: 0.001
>
> Coefficients:
> Estimate t value Std.Err. p raw p Bonf p adj
> tensionH-tensionL -14.722 -3.802 3.872 0.000 0.001 0.001
> tensionM-tensionL -10.000 -2.582 3.872 0.013 0.026 0.024
> tensionH-tensionM -4.722 -1.219 3.872 0.228 0.228 0.228
>
>
>
> or if you prefer to get the confidence intervals, too, you can
> use:
>
> summary(simint(breaks ~ wool + tension, data = warpbreaks, whichf="tension",
> type = "Tukey"))
>
> Simultaneous 95% confidence intervals: Tukey contrasts
>
> Call:
> simint.formula(formula = breaks ~ wool + tension, data = warpbreaks,
> whichf = "tension", type = "Tukey")
>
> Tukey contrasts for factor tension, covariable: wool
>
> Contrast matrix:
> tensionL tensionM tensionH
> tensionM-tensionL 0 0 -1 1 0
> tensionH-tensionL 0 0 -1 0 1
> tensionH-tensionM 0 0 0 -1 1
>
> Absolute Error Tolerance: 0.001
>
> 95 % quantile: 2.415
>
> Coefficients:
> Estimate 2.5 % 97.5 % t value Std.Err. p raw p Bonf p adj
> tensionM-tensionL -10.000 -19.352 -0.648 -2.582 3.872 0.013 0.038 0.034
> tensionH-tensionL -14.722 -24.074 -5.370 -3.802 3.872 0.000 0.001 0.001
> tensionH-tensionM -4.722 -14.074 4.630 -1.219 3.872 0.228 0.685 0.447
>
> -----------------------------------------------------------------
> Please be careful: The resulting confidence intervals in
> simint are not associated with the p-values from 'simtest' as it
> is described in the help page of the two functions.
> -----------------------------------------------------------------
>
> I had not the time to check the differences in the function or
> read the references given on the help page.
> If you are interested in the function you can check those to
> find out which one you prefer.
>
> Best regards,
>
> Christoph Buser
>
> --------------------------------------------------------------
> Christoph Buser <buser at stat.math.ethz.ch>
> Seminar fuer Statistik, LEO C13
> ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
> phone: x-41-44-632-4673 fax: 632-1228
> http://stat.ethz.ch/~buser/
> --------------------------------------------------------------
>
>
> Christoph Strehblow writes:
> > hi list,
> >
> > i have to ask you again, having tried and searched for several days...
> >
> > i want to do a TukeyHSD after an Anova, and want to get the adjusted
> > p-values after the Tukey Correction.
> > i found the p.adjust function, but it can only correct for "holm",
> > "hochberg", bonferroni", but not "Tukey".
> >
> > Is it not possbile to get adjusted p-values after Tukey-correction?
> >
> > sorry, if this is an often-answered-question, but i didn´t find it on
> > the list archive...
> >
> > thx a lot, list, Chris
> >
> >
> > Christoph Strehblow, MD
> > Department of Rheumatology, Diabetes and Endocrinology
> > Wilhelminenspital, Vienna, Austria
> > chrisxe at gmx.at
> >
> > ______________________________________________
> > R-help at stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
>
--
--------------------------------------------
Dr Sander P. Oom
Animal, Plant and Environmental Sciences,
University of the Witwatersrand
Private Bag 3, Wits 2050, South Africa
Tel (work) +27 (0)11 717 64 04
Tel (home) +27 (0)18 297 44 51
Fax +27 (0)18 299 24 64
Email sander at oomvanlieshout.net
Web www.oomvanlieshout.net/sander
```
More information about the R-help mailing list | 1,809 | 5,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.783786 |
https://socratic.org/questions/how-do-you-use-the-chain-rule-to-differentiate-y-5x-5-3-3x-3-1-3 | 1,548,179,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583857993.67/warc/CC-MAIN-20190122161345-20190122183345-00195.warc.gz | 657,539,319 | 5,649 | # How do you use the chain rule to differentiate y=((5x^5-3)/(-3x^3+1))^3?
##### 1 Answer
Feb 24, 2018
$y = {\left(\frac{5 {x}^{2} - 3}{- 3 {x}^{3} + 1}\right)}^{3}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left(\frac{5 {x}^{2} - 3}{- 3 {x}^{3} + 1}\right)}^{2} \times \frac{d \left(\frac{5 {x}^{2} - 3}{- 3 {x}^{3} + 1}\right)}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left(\frac{5 {x}^{2} - 3}{- 3 {x}^{3} + 1}\right)}^{2} \times \left[\frac{\left(5 {x}^{2} - 3\right) ' \left(- 3 {x}^{3} + 1\right) - \left(5 {x}^{2} - 3\right) \left(- 3 {x}^{3} + 1\right) '}{- 3 {x}^{3} + 1} ^ 2\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left(\frac{5 {x}^{2} - 3}{- 3 {x}^{3} + 1}\right)}^{2} \times \left[\frac{10 x \left(- 3 {x}^{3} + 1\right) - \left(5 {x}^{2} - 3\right) \left(- 9 {x}^{2}\right)}{- 3 {x}^{3} + 1} ^ 2\right]$ | 442 | 837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-04 | latest | en | 0.24015 |
https://synthiam.com/Question/4636 | 1,547,960,635,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583700012.70/warc/CC-MAIN-20190120042010-20190120064010-00251.warc.gz | 676,587,916 | 17,966 | Question
## Community Question
Is there any way to look at how the script for camera tracking was written by Dj? I can read the variables, but I want to see the line by line script.
My goal is to write a tracking script to an X,Y coordinate. Someone wanted to do a similar thing using a GPS module about a year ago.
I see there is a camera coordinate X and Y in the variable list. I wanted to see how it was used in his script.
I can read where the robot is now ( X and Y value), and know where I want it to go (X and Y value). I need to borrow or write how to travel from and to.
Ron R
January 3 2016
Hi Dj,
My end goal is to track to two preset coordinates, loaded from a verbal command ("go to the door").
The following is my idea: I read the loaded variables, then compare these coordinates to coordinates of two tracking cameras, looking at a target on my robot. These cameras are located 90 degrees apart in the area (two out side walls). Using Omni wheels mounted on two continuous servos, I plan on moving in the direction to match these coordinates. As the number get closer, I slow the continuous servos until the numbers match, then stop.
I started to write a concept script on paper using only one camera/servo and one continuous servo. I know your camera tracking lists an x and y coordinate and wanted to see how you wrote the script to track to them. This way I could see if I was on the right track. My idea is to do compares, then adjust the position of the robot until it matches.
I am a very new writer of scripts. The only exposure to programming was ladder logic and machine code many many years ago....LOL.. I will continue to play and post the script once it is written.
Any help or direction is appreciated. I really want to try this idea. I have future ideas to incorporate into it once it works.
Ron R
DJ Sures
Synthiam
January 3 2016
You cannot use a continuous rotation servo for tracking because it is unable to respond to position commands. It doesn't have a sense of positioning, it only knows to move a direction.
January 3 2016
No, I am using the continuous servos to move only the robot. The camera is mounted on a regular servo, which is on a stationary location by an outside wall.. It is used to track the robot. For the initial test, I will have a long cable from the EZB in the Adventure Bot to the servo under the camera by a perimeter wall. As the Bot moves, the value sent to the camera servo to follow the target on the robot's back is my feedback. I will be using an Adventure Bot as the Target and a web cam connected to my computer as the feed back. I will load a coordinate and hopefully track to it.
I may have to make a video of the concept to eliminate some of the hardware confusion.
Ron R
DJ Sures
Synthiam
January 3 2016
Apologies Andy, i feel the question regarding how the vision tracking system identifies object positioning within EZ-Builder may be highly above your current programming capabilities based on this thread.
I do encourage learning - which is why we spent years developing and growing the LEARN section of this website. There is an activities course which will be very important to the education process. I recommend successfully completing the activities. Once you have accomplished the activities and have a grasp of logic programming, use the EXAMPLES folder in EZ-Builder to view ez-script examples.
This is because the question that you are asking about how object positioning is identified is quite complicated. There are many "steps" with complicated programming that identifies and detects objects. However, the answer to your specific question about how the detected object's position has been identified uses the aforge blobcounter library - which is as good or better than i could have written - which is why building off existing frameworks is important. Much like how you create robot projects in EZ-Builder...
This is because without EZ-Builder, you would need to understand this. And remember, this is only one module of 30 or so that relates to vision tracking in EZ-Builder. If this doesn't make sense to you, stick with EZ-Builder :). This code example is responsible for identifying the detected object position and assigning to EZ-Builder variables:
`Code:// AForge Image Processing Library// AForge.NET framework// http://www.aforgenet.com/framework///// Copyright © AForge.NET, 2005-2012// contacts@aforgenet.com//namespace AForge.Imaging{ using System; using System.Collections.Generic; using System.Drawing; using System.Drawing.Imaging; using AForge; /// <summary> /// Possible object orders. /// </summary> /// /// <remarks>The enumeration defines possible sorting orders of objects, found by blob /// counting classes.</remarks> /// public enum ObjectsOrder { /// <summary> /// Unsorted order (as it is collected by algorithm). /// </summary> None, /// <summary> /// Objects are sorted by size in descending order (bigger objects go first). /// Size is calculated as <b>Width * Height</b>. /// </summary> Size, /// <summary> /// Objects are sorted by area in descending order (bigger objects go first). /// </summary> Area, /// <summary> /// Objects are sorted by Y coordinate, then by X coordinate in ascending order /// (smaller coordinates go first). /// </summary> YX, /// <summary> /// Objects are sorted by X coordinate, then by Y coordinate in ascending order /// (smaller coordinates go first). /// </summary> XY } /// <summary> /// Base class for different blob counting algorithms. /// </summary> /// /// <remarks><para>The class is abstract and serves as a base for different blob counting algorithms. /// Classes, which inherit from this base class, require to implement <see cref="BuildObjectsMap"/> /// method, which does actual building of object's label's map.</para> /// /// <para>For blobs' searcing usually all inherited classes accept binary images, which are actually /// grayscale thresholded images. But the exact supported format should be checked in particular class, /// inheriting from the base class. For blobs' extraction the class supports grayscale (8 bpp indexed) /// and color images (24 and 32 bpp).</para> /// /// <para>Sample usage:</para> /// <code> /// // create an instance of blob counter algorithm /// BlobCounterBase bc = new ... /// // set filtering options /// bc.FilterBlobs = true; /// bc.MinWidth = 5; /// bc.MinHeight = 5; /// // process binary image /// bc.ProcessImage( image ); /// Blob[] blobs = bc.GetObjects( image, false ); /// // process blobs /// foreach ( Blob blob in blobs ) /// { /// // ... /// // blob.Rectangle - blob's rectangle /// // blob.Image - blob's image /// } /// </code> /// </remarks> /// public abstract class BlobCounterBase { // found blobs List<Blob> blobs = new List<Blob>( ); // objects' sort order private ObjectsOrder objectsOrder = ObjectsOrder.None; // filtering by size is required or not private bool filterBlobs = false; private IBlobsFilter filter = null; // coupled size filtering or not private bool coupledSizeFiltering = false; // blobs' minimal and maximal size private int minWidth = 1; private int minHeight = 1; private int maxWidth = int.MaxValue; private int maxHeight = int.MaxValue; /// <summary> /// Objects count. /// </summary> protected int objectsCount; /// <summary> /// Objects' labels. /// </summary> protected int[] objectLabels; /// <summary> /// Width of processed image. /// </summary> protected int imageWidth; /// <summary> /// Height of processed image. /// </summary> protected int imageHeight; /// <summary> /// Objects count. /// </summary> /// /// <remarks><para>Number of objects (blobs) found by <see cref="ProcessImage(Bitmap)"/> method. /// </para></remarks> /// public int ObjectsCount { get { return objectsCount; } } /// <summary> /// Objects' labels. /// </summary> /// /// <remarks>The array of <b>width</b> * <b>height</b> size, which holds /// labels for all objects. Background is represented with <b>0</b> value, /// but objects are represented with labels starting from <b>1</b>.</remarks> /// public int[] ObjectLabels { get { return objectLabels; } } /// <summary> /// Objects sort order. /// </summary> /// /// <remarks><para>The property specifies objects' sort order, which are provided /// by <see cref="GetObjectsRectangles"/>, <see cref="GetObjectsInformation"/>, etc. /// </para></remarks> /// public ObjectsOrder ObjectsOrder { get { return objectsOrder; } set { objectsOrder = value; } } /// <summary> /// Specifies if blobs should be filtered. /// </summary> /// /// <remarks><para>If the property is equal to <b>false</b>, then there is no any additional /// post processing after image was processed. If the property is set to <b>true</b>, then /// blobs filtering is done right after image processing routine. If <see cref="BlobsFilter"/> /// is set, then custom blobs' filtering is done, which is implemented by user. Otherwise /// blobs are filtered according to dimensions specified in <see cref="MinWidth"/>, /// <see cref="MinHeight"/>, <see cref="MaxWidth"/> and <see cref="MaxHeight"/> properties.</para> /// /// <para>Default value is set to <see langword="false"/>.</para></remarks> /// public bool FilterBlobs { get { return filterBlobs; } set { filterBlobs = value; } } /// <summary> /// Specifies if size filetering should be coupled or not. /// </summary> /// /// <remarks><para>In uncoupled filtering mode, objects are filtered out in the case if /// their width is smaller than <see cref="MinWidth"/> <b>or</b> height is smaller than /// <see cref="MinHeight"/>. But in coupled filtering mode, objects are filtered out in /// the case if their width is smaller than <see cref="MinWidth"/> <b>and</b> height is /// smaller than <see cref="MinHeight"/>. In both modes the idea with filtering by objects' /// maximum size is the same as filtering by objects' minimum size.</para> /// /// <para>Default value is set to <see langword="false"/>, what means uncoupled filtering by size.</para> /// </remarks> /// public bool CoupledSizeFiltering { get { return coupledSizeFiltering; } set { coupledSizeFiltering = value; } } /// <summary> /// Minimum allowed width of blob. /// </summary> /// /// <remarks><para>The property specifies minimum object's width acceptable by blob counting /// routine and has power only when <see cref="FilterBlobs"/> property is set to /// <see langword="true"/> and <see cref="BlobsFilter">custom blobs' filter</see> is /// set to <see langword="null"/>.</para> /// /// <para>See documentation to <see cref="CoupledSizeFiltering"/> for additional information.</para> /// </remarks> /// public int MinWidth { get { return minWidth; } set { minWidth = value; } } /// <summary> /// Minimum allowed height of blob. /// </summary> /// /// <remarks><para>The property specifies minimum object's height acceptable by blob counting /// routine and has power only when <see cref="FilterBlobs"/> property is set to /// <see langword="true"/> and <see cref="BlobsFilter">custom blobs' filter</see> is /// set to <see langword="null"/>.</para> /// /// <para>See documentation to <see cref="CoupledSizeFiltering"/> for additional information.</para> /// </remarks> /// public int MinHeight { get { return minHeight; } set { minHeight = value; } } /// <summary> /// Maximum allowed width of blob. /// </summary> /// /// <remarks><para>The property specifies maximum object's width acceptable by blob counting /// routine and has power only when <see cref="FilterBlobs"/> property is set to /// <see langword="true"/> and <see cref="BlobsFilter">custom blobs' filter</see> is /// set to <see langword="null"/>.</para> /// /// <para>See documentation to <see cref="CoupledSizeFiltering"/> for additional information.</para> /// </remarks> /// public int MaxWidth { get { return maxWidth; } set { maxWidth = value; } } /// <summary> /// Maximum allowed height of blob. /// </summary> /// /// <remarks><para>The property specifies maximum object's height acceptable by blob counting /// routine and has power only when <see cref="FilterBlobs"/> property is set to /// <see langword="true"/> and <see cref="BlobsFilter">custom blobs' filter</see> is /// set to <see langword="null"/>.</para> /// /// <para>See documentation to <see cref="CoupledSizeFiltering"/> for additional information.</para> /// </remarks> /// public int MaxHeight { get { return maxHeight; } set { maxHeight = value; } } /// <summary> /// Custom blobs' filter to use. /// </summary> /// /// <remarks><para>The property specifies custom blobs' filtering routine to use. It has /// effect only in the case if <see cref="FilterBlobs"/> property is set to <see langword="true"/>.</para> /// /// <para><note>When custom blobs' filtering routine is set, it has priority over default filtering done /// with <see cref="MinWidth"/>, <see cref="MinHeight"/>, <see cref="MaxWidth"/> and <see cref="MaxHeight"/>.</note></para> /// </remarks> /// public IBlobsFilter BlobsFilter { get { return filter; } set { filter = value; } } /// <summary> /// Initializes a new instance of the <see cref="BlobCounterBase"/> class. /// </summary> /// /// <remarks>Creates new instance of the <see cref="BlobCounterBase"/> class with /// an empty objects map. Before using methods, which provide information about blobs /// or extract them, the <see cref="ProcessImage(Bitmap)"/>, /// <see cref="ProcessImage(BitmapData)"/> or <see cref="ProcessImage(UnmanagedImage)"/> /// method should be called to collect objects map.</remarks> /// public BlobCounterBase( ) { } /// <summary> /// Initializes a new instance of the <see cref="BlobCounterBase"/> class. /// </summary> /// /// <param name="image">Binary image to look for objects in.</param> /// /// <remarks>Creates new instance of the <see cref="BlobCounterBase"/> class with /// initialized objects map built by calling <see cref="ProcessImage(Bitmap)"/> method.</remarks> /// public BlobCounterBase( Bitmap image ) { ProcessImage( image ); } /// <summary> /// Initializes a new instance of the <see cref="BlobCounterBase"/> class. /// </summary> /// /// <param name="imageData">Binary image data to look for objects in.</param> /// /// <remarks>Creates new instance of the <see cref="BlobCounterBase"/> class with /// initialized objects map built by calling <see cref="ProcessImage(BitmapData)"/> method.</remarks> /// public BlobCounterBase( BitmapData imageData ) { ProcessImage( imageData ); } /// <summary> /// Initializes a new instance of the <see cref="BlobCounterBase"/> class. /// </summary> /// /// <param name="image">Unmanaged binary image to look for objects in.</param> /// /// <remarks>Creates new instance of the <see cref="BlobCounterBase"/> class with /// initialized objects map built by calling <see cref="ProcessImage(UnmanagedImage)"/> method.</remarks> /// public BlobCounterBase( UnmanagedImage image ) { ProcessImage( image ); } /// <summary> /// Build objects map. /// </summary> /// /// <param name="image">Source binary image.</param> /// /// <remarks><para>Processes the image and builds objects map, which is used later to extracts blobs.</para></remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the source image.</exception> /// public void ProcessImage( Bitmap image ) { // lock source bitmap data BitmapData imageData = image.LockBits( new Rectangle( 0, 0, image.Width, image.Height ), ImageLockMode.ReadOnly, image.PixelFormat ); try { // process image ProcessImage( imageData ); } finally { // unlock source image image.UnlockBits( imageData ); } } /// <summary> /// Build objects map. /// </summary> /// /// <param name="imageData">Source binary image data.</param> /// /// <remarks><para>Processes the image and builds objects map, which is used later to extracts blobs.</para></remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the source image.</exception> /// public void ProcessImage( BitmapData imageData ) { // do actual objects map building ProcessImage( new UnmanagedImage( imageData ) ); } /// <summary> /// Build object map from raw image data. /// </summary> /// /// <param name="image">Source unmanaged binary image data.</param> /// /// <remarks><para>Processes the image and builds objects map, which is used later to extracts blobs.</para></remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the source image.</exception> /// <exception cref="InvalidImagePropertiesException">Thrown by some inherited classes if some image property other /// than the pixel format is not supported. See that class's documentation or the exception message for details.</exception> /// public void ProcessImage( UnmanagedImage image ) { imageWidth = image.Width; imageHeight = image.Height; // do actual objects map building BuildObjectsMap( image ); // collect information about blobs CollectObjectsInfo( image ); // filter blobs by size if required if ( filterBlobs ) { // labels remapping array int[] labelsMap = new int[objectsCount + 1]; for ( int i = 1; i <= objectsCount; i++ ) { labelsMap = i; } // check dimension of all objects and filter them int objectsToRemove = 0; if ( filter == null ) { for ( int i = objectsCount - 1; i >= 0; i-- ) { int blobWidth = blobs.Rectangle.Width; int blobHeight = blobs.Rectangle.Height; if ( coupledSizeFiltering == false ) { // uncoupled filtering if ( ( blobWidth < minWidth ) || ( blobHeight < minHeight ) || ( blobWidth > maxWidth ) || ( blobHeight > maxHeight ) ) { labelsMap[i + 1] = 0; objectsToRemove++; blobs.RemoveAt( i ); } } else { // coupled filtering if ( ( ( blobWidth < minWidth ) && ( blobHeight < minHeight ) ) || ( ( blobWidth > maxWidth ) && ( blobHeight > maxHeight ) ) ) { labelsMap[i + 1] = 0; objectsToRemove++; blobs.RemoveAt( i ); } } } } else { for ( int i = objectsCount - 1; i >= 0; i-- ) { if ( !filter.Check( blobs ) ) { labelsMap[i + 1] = 0; objectsToRemove++; blobs.RemoveAt( i ); } } } // update labels remapping array int label = 0; for ( int i = 1; i <= objectsCount; i++ ) { if ( labelsMap != 0 ) { label++; // update remapping array labelsMap = label; } } // repair object labels for ( int i = 0, n = objectLabels.Length; i < n; i++ ) { objectLabels = labelsMap[objectLabels]; } objectsCount -= objectsToRemove; // repair IDs for ( int i = 0, n = blobs.Count; i < n; i++ ) { blobs.ID = i + 1; } } // do we need to sort the list? if ( objectsOrder != ObjectsOrder.None ) { blobs.Sort( new BlobsSorter( objectsOrder ) ); } } /// <summary> /// Get objects' rectangles. /// </summary> /// /// <returns>Returns array of objects' rectangles.</returns> /// /// <remarks>The method returns array of objects rectangles. Before calling the /// method, the <see cref="ProcessImage(Bitmap)"/>, <see cref="ProcessImage(BitmapData)"/> /// or <see cref="ProcessImage(UnmanagedImage)"/> method should be called, which will /// build objects map.</remarks> /// /// <exception cref="ApplicationException">No image was processed before, so objects' rectangles /// can not be collected.</exception> /// public Rectangle[] GetObjectsRectangles( ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); Rectangle[] rects = new Rectangle[objectsCount]; for ( int i = 0; i < objectsCount; i++ ) { rects = blobs.Rectangle; } return rects; } /// <summary> /// Get objects' information. /// </summary> /// /// <returns>Returns array of partially initialized blobs (without <see cref="Blob.Image"/> property initialized).</returns> /// /// <remarks><para>By the amount of provided information, the method is between <see cref="GetObjectsRectangles"/> and /// <see cref="GetObjects( UnmanagedImage, bool )"/> methods. The method provides array of blobs without initialized their image. /// Blob's image may be extracted later using <see cref="ExtractBlobsImage( Bitmap, Blob, bool )"/> /// or <see cref="ExtractBlobsImage( UnmanagedImage, Blob, bool )"/> method. /// </para></remarks> /// /// <example> /// <code> /// // create blob counter and process image /// BlobCounter bc = new BlobCounter( sourceImage ); /// // specify sort order /// bc.ObjectsOrder = ObjectsOrder.Size; /// // get objects' information (blobs without image) /// Blob[] blobs = bc.GetObjectInformation( ); /// // process blobs /// foreach ( Blob blob in blobs ) /// { /// // check blob's properties /// if ( blob.Rectangle.Width > 50 ) /// { /// // the blob looks interesting, let's extract it /// bc.ExtractBlobsImage( sourceImage, blob ); /// } /// } /// </code> /// </example> /// /// <exception cref="ApplicationException">No image was processed before, so objects' information /// can not be collected.</exception> /// public Blob[] GetObjectsInformation( ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); Blob[] blobsToReturn = new Blob[objectsCount]; // create each blob for ( int k = 0; k < objectsCount; k++ ) { blobsToReturn[k] = new Blob( blobs[k] ); } return blobsToReturn; } /// <summary> /// Get blobs. /// </summary> /// /// <param name="image">Source image to extract objects from.</param> /// /// <returns>Returns array of blobs.</returns> /// <param name="extractInOriginalSize">Specifies size of blobs' image to extract. /// If set to <see langword="true"/> each blobs' image will have the same size as /// the specified image. If set to <see langword="false"/> each blobs' image will /// have the size of its blob.</param> /// /// <remarks><para>The method returns array of blobs. Before calling the /// method, the <see cref="ProcessImage(Bitmap)"/>, <see cref="ProcessImage(BitmapData)"/> /// or <see cref="ProcessImage(UnmanagedImage)"/> method should be called, which will build /// objects map.</para> /// /// <para>The method supports 24/32 bpp color and 8 bpp indexed grayscale images.</para> /// </remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the provided image.</exception> /// <exception cref="ApplicationException">No image was processed before, so objects /// can not be collected.</exception> /// public Blob[] GetObjects( Bitmap image, bool extractInOriginalSize ) { Blob[] blobs = null; // lock source bitmap data BitmapData imageData = image.LockBits( new Rectangle( 0, 0, image.Width, image.Height ), ImageLockMode.ReadOnly, image.PixelFormat ); try { // process image blobs = GetObjects( new UnmanagedImage( imageData ), extractInOriginalSize ); } finally { // unlock source images image.UnlockBits( imageData ); } return blobs; } /// <summary> /// Get blobs. /// </summary> /// /// <param name="image">Source unmanaged image to extract objects from.</param> /// <param name="extractInOriginalSize">Specifies size of blobs' image to extract. /// If set to <see langword="true"/> each blobs' image will have the same size as /// the specified image. If set to <see langword="false"/> each blobs' image will /// have the size of its blob.</param> /// /// <returns>Returns array of blobs.</returns> /// /// <remarks><para>The method returns array of blobs. Before calling the /// method, the <see cref="ProcessImage(Bitmap)"/>, <see cref="ProcessImage(BitmapData)"/> /// or <see cref="ProcessImage(UnmanagedImage)"/> method should be called, which will build /// objects map.</para> /// /// <para>The method supports 24/32 bpp color and 8 bpp indexed grayscale images.</para> /// </remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the provided image.</exception> /// <exception cref="ApplicationException">No image was processed before, so objects /// can not be collected.</exception> /// public Blob[] GetObjects( UnmanagedImage image, bool extractInOriginalSize ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); if ( ( image.PixelFormat != PixelFormat.Format24bppRgb ) && ( image.PixelFormat != PixelFormat.Format8bppIndexed ) && ( image.PixelFormat != PixelFormat.Format32bppRgb ) && ( image.PixelFormat != PixelFormat.Format32bppArgb ) && ( image.PixelFormat != PixelFormat.Format32bppRgb ) && ( image.PixelFormat != PixelFormat.Format32bppPArgb ) ) throw new UnsupportedImageFormatException( "Unsupported pixel format of the provided image." ); // image size int width = image.Width; int height = image.Height; int srcStride = image.Stride; int pixelSize = Bitmap.GetPixelFormatSize( image.PixelFormat ) / 8; Blob[] objects = new Blob[objectsCount]; // create each image for ( int k = 0; k < objectsCount; k++ ) { int objectWidth = blobs[k].Rectangle.Width; int objectHeight = blobs[k].Rectangle.Height; int blobImageWidth = ( extractInOriginalSize ) ? width : objectWidth; int blobImageHeight = ( extractInOriginalSize ) ? height : objectHeight; int xmin = blobs[k].Rectangle.X; int xmax = xmin + objectWidth - 1; int ymin = blobs[k].Rectangle.Y; int ymax = ymin + objectHeight - 1; int label = blobs[k].ID; // create new image UnmanagedImage dstImage = UnmanagedImage.Create( blobImageWidth, blobImageHeight, image.PixelFormat ); // copy image unsafe { byte* src = (byte*) image.ImageData.ToPointer( ) + ymin * srcStride + xmin * pixelSize; byte* dst = (byte*) dstImage.ImageData.ToPointer( ); int p = ymin * width + xmin; if ( extractInOriginalSize ) { // allign destination pointer also dst += ymin * dstImage.Stride + xmin * pixelSize; } int srcOffset = srcStride - objectWidth * pixelSize; int dstOffset = dstImage.Stride - objectWidth * pixelSize; int labelsOffset = width - objectWidth; // for each line for ( int y = ymin; y <= ymax; y++ ) { // copy each pixel for ( int x = xmin; x <= xmax; x++, p++, dst += pixelSize, src += pixelSize ) { if ( objectLabels[p] == label ) { // copy pixel *dst = *src; if ( pixelSize > 1 ) { dst[1] = src[1]; dst[2] = src[2]; if ( pixelSize > 3 ) { dst[3] = src[3]; } } } } src += srcOffset; dst += dstOffset; p += labelsOffset; } } objects[k] = new Blob( blobs[k] ); objects[k].Image = dstImage; objects[k].OriginalSize = extractInOriginalSize; } return objects; } /// <summary> /// Extract blob's image. /// </summary> /// /// <param name="image">Source image to extract blob's image from.</param> /// <param name="blob">Blob which is required to be extracted.</param> /// <param name="extractInOriginalSize">Specifies size of blobs' image to extract. /// If set to <see langword="true"/> each blobs' image will have the same size as /// the specified image. If set to <see langword="false"/> each blobs' image will /// have the size of its blob.</param> /// /// <remarks><para>The method is used to extract image of partially initialized blob, which /// was provided by <see cref="GetObjectsInformation"/> method. Before calling the /// method, the <see cref="ProcessImage(Bitmap)"/>, <see cref="ProcessImage(BitmapData)"/> /// or <see cref="ProcessImage(UnmanagedImage)"/> method should be called, which will build /// objects map.</para> /// /// <para>The method supports 24/32 bpp color and 8 bpp indexed grayscale images.</para> /// </remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the provided image.</exception> /// <exception cref="ApplicationException">No image was processed before, so blob /// can not be extracted.</exception> /// public void ExtractBlobsImage( Bitmap image, Blob blob, bool extractInOriginalSize ) { // lock source bitmap data BitmapData imageData = image.LockBits( new Rectangle( 0, 0, image.Width, image.Height ), ImageLockMode.ReadOnly, image.PixelFormat ); try { // process image ExtractBlobsImage( new UnmanagedImage( imageData ), blob, extractInOriginalSize ); } finally { // unlock source images image.UnlockBits( imageData ); } } /// <summary> /// Extract blob's image. /// </summary> /// /// <param name="image">Source unmanaged image to extract blob's image from.</param> /// <param name="blob">Blob which is required to be extracted.</param> /// <param name="extractInOriginalSize">Specifies size of blobs' image to extract. /// If set to <see langword="true"/> each blobs' image will have the same size as /// the specified image. If set to <see langword="false"/> each blobs' image will /// have the size of its blob.</param> /// /// <remarks><para>The method is used to extract image of partially initialized blob, which /// was provided by <see cref="GetObjectsInformation"/> method. Before calling the /// method, the <see cref="ProcessImage(Bitmap)"/>, <see cref="ProcessImage(BitmapData)"/> /// or <see cref="ProcessImage(UnmanagedImage)"/> method should be called, which will build /// objects map.</para> /// /// <para>The method supports 24/32 bpp color and 8 bpp indexed grayscale images.</para> /// </remarks> /// /// <exception cref="UnsupportedImageFormatException">Unsupported pixel format of the provided image.</exception> /// <exception cref="ApplicationException">No image was processed before, so blob /// can not be extracted.</exception> /// public void ExtractBlobsImage( UnmanagedImage image, Blob blob, bool extractInOriginalSize ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); if ( ( image.PixelFormat != PixelFormat.Format24bppRgb ) && ( image.PixelFormat != PixelFormat.Format8bppIndexed ) && ( image.PixelFormat != PixelFormat.Format32bppRgb ) && ( image.PixelFormat != PixelFormat.Format32bppArgb ) && ( image.PixelFormat != PixelFormat.Format32bppRgb ) && ( image.PixelFormat != PixelFormat.Format32bppPArgb ) ) throw new UnsupportedImageFormatException( "Unsupported pixel format of the provided image." ); // image size int width = image.Width; int height = image.Height; int srcStride = image.Stride; int pixelSize = Bitmap.GetPixelFormatSize( image.PixelFormat ) / 8; int objectWidth = blob.Rectangle.Width; int objectHeight = blob.Rectangle.Height; int blobImageWidth = ( extractInOriginalSize ) ? width : objectWidth; int blobImageHeight = ( extractInOriginalSize ) ? height : objectHeight; int xmin = blob.Rectangle.Left; int xmax = xmin + objectWidth - 1; int ymin = blob.Rectangle.Top; int ymax = ymin + objectHeight - 1; int label = blob.ID; // create new image blob.Image = UnmanagedImage.Create( blobImageWidth, blobImageHeight, image.PixelFormat ); blob.OriginalSize = extractInOriginalSize; // copy image unsafe { byte* src = (byte*) image.ImageData.ToPointer( ) + ymin * srcStride + xmin * pixelSize; byte* dst = (byte*) blob.Image.ImageData.ToPointer( ); int p = ymin * width + xmin; if ( extractInOriginalSize ) { // allign destination pointer also dst += ymin * blob.Image.Stride + xmin * pixelSize; } int srcOffset = srcStride - objectWidth * pixelSize; int dstOffset = blob.Image.Stride - objectWidth * pixelSize; int labelsOffset = width - objectWidth; // for each line for ( int y = ymin; y <= ymax; y++ ) { // copy each pixel for ( int x = xmin; x <= xmax; x++, p++, dst += pixelSize, src += pixelSize ) { if ( objectLabels[p] == label ) { // copy pixel *dst = *src; if ( pixelSize > 1 ) { dst[1] = src[1]; dst[2] = src[2]; if ( pixelSize > 3 ) { dst[3] = src[3]; } } } } src += srcOffset; dst += dstOffset; p += labelsOffset; } } } /// <summary> /// Get list of points on the left and right edges of the blob. /// </summary> /// /// <param name="blob">Blob to collect edge points for.</param> /// <param name="leftEdge">List of points on the left edge of the blob.</param> /// <param name="rightEdge">List of points on the right edge of the blob.</param> /// /// <remarks><para>The method scans each line of the blob and finds the most left and the /// most right points for it adding them to appropriate lists. The method may be very /// useful in conjunction with different routines from <see cref="AForge.Math.Geometry"/>, /// which allow finding convex hull or quadrilateral's corners.</para> /// /// <para><note>Both lists of points are sorted by Y coordinate - points with smaller Y /// value go first.</note></para> /// </remarks> /// /// <exception cref="ApplicationException">No image was processed before, so blob /// can not be extracted.</exception> /// public void GetBlobsLeftAndRightEdges( Blob blob, out List<IntPoint> leftEdge, out List<IntPoint> rightEdge ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); leftEdge = new List<IntPoint>( ); rightEdge = new List<IntPoint>( ); int xmin = blob.Rectangle.Left; int xmax = xmin + blob.Rectangle.Width - 1; int ymin = blob.Rectangle.Top; int ymax = ymin + blob.Rectangle.Height - 1; int label = blob.ID; // for each line for ( int y = ymin; y <= ymax; y++ ) { // scan from left to right int p = y * imageWidth + xmin; for ( int x = xmin; x <= xmax; x++, p++ ) { if ( objectLabels[p] == label ) { leftEdge.Add( new IntPoint( x, y ) ); break; } } // scan from right to left p = y * imageWidth + xmax; for ( int x = xmax; x >= xmin; x--, p-- ) { if ( objectLabels[p] == label ) { rightEdge.Add( new IntPoint( x, y ) ); break; } } } } /// <summary> /// Get list of points on the top and bottom edges of the blob. /// </summary> /// /// <param name="blob">Blob to collect edge points for.</param> /// <param name="topEdge">List of points on the top edge of the blob.</param> /// <param name="bottomEdge">List of points on the bottom edge of the blob.</param> /// /// <remarks><para>The method scans each column of the blob and finds the most top and the /// most bottom points for it adding them to appropriate lists. The method may be very /// useful in conjunction with different routines from <see cref="AForge.Math.Geometry"/>, /// which allow finding convex hull or quadrilateral's corners.</para> /// /// <para><note>Both lists of points are sorted by X coordinate - points with smaller X /// value go first.</note></para> /// </remarks> /// /// <exception cref="ApplicationException">No image was processed before, so blob /// can not be extracted.</exception> /// public void GetBlobsTopAndBottomEdges( Blob blob, out List<IntPoint> topEdge, out List<IntPoint> bottomEdge ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); topEdge = new List<IntPoint>( ); bottomEdge = new List<IntPoint>( ); int xmin = blob.Rectangle.Left; int xmax = xmin + blob.Rectangle.Width - 1; int ymin = blob.Rectangle.Top; int ymax = ymin + blob.Rectangle.Height - 1; int label = blob.ID; // for each column for ( int x = xmin; x <= xmax; x++ ) { // scan from top to bottom int p = ymin * imageWidth + x; for ( int y = ymin; y <= ymax; y++, p += imageWidth ) { if ( objectLabels[p] == label ) { topEdge.Add( new IntPoint( x, y ) ); break; } } // scan from bottom to top p = ymax * imageWidth + x; for ( int y = ymax; y >= ymin; y--, p -= imageWidth ) { if ( objectLabels[p] == label ) { bottomEdge.Add( new IntPoint( x, y ) ); break; } } } } /// <summary> /// Get list of object's edge points. /// </summary> /// /// <param name="blob">Blob to collect edge points for.</param> /// /// <returns>Returns unsorted list of blob's edge points.</returns> /// /// <remarks><para>The method scans each row and column of the blob and finds the /// most top/bottom/left/right points. The method returns similar result as if results of /// both <see cref="GetBlobsLeftAndRightEdges"/> and <see cref="GetBlobsTopAndBottomEdges"/> /// methods were combined, but each edge point occurs only once in the list.</para> /// /// <para><note>Edge points in the returned list are not ordered. This makes the list unusable /// for visualization with methods, which draw polygon or poly-line. But the returned list /// can be used with such algorithms, like convex hull search, shape analyzer, etc.</note></para> /// </remarks> /// /// <exception cref="ApplicationException">No image was processed before, so blob /// can not be extracted.</exception> /// public List<IntPoint> GetBlobsEdgePoints( Blob blob ) { // check if objects map was collected if ( objectLabels == null ) throw new ApplicationException( "Image should be processed before to collect objects map." ); List<IntPoint> edgePoints = new List<IntPoint>( ); int xmin = blob.Rectangle.Left; int xmax = xmin + blob.Rectangle.Width - 1; int ymin = blob.Rectangle.Top; int ymax = ymin + blob.Rectangle.Height - 1; int label = blob.ID; // array of already processed points on left/right edges // (index in these arrays represent Y coordinate, but value - X coordinate) int[] leftProcessedPoints = new int[blob.Rectangle.Height]; int[] rightProcessedPoints = new int[blob.Rectangle.Height]; // for each line for ( int y = ymin; y <= ymax; y++ ) { // scan from left to right int p = y * imageWidth + xmin; for ( int x = xmin; x <= xmax; x++, p++ ) { if ( objectLabels[p] == label ) { edgePoints.Add( new IntPoint( x, y ) ); leftProcessedPoints[y - ymin] = x; break; } } // scan from right to left p = y * imageWidth + xmax; for ( int x = xmax; x >= xmin; x--, p-- ) { if ( objectLabels[p] == label ) { // avoid adding the point we already have if ( leftProcessedPoints[y - ymin] != x ) { edgePoints.Add( new IntPoint( x, y ) ); } rightProcessedPoints[y - ymin] = x; break; } } } // for each column for ( int x = xmin; x <= xmax; x++ ) { // scan from top to bottom int p = ymin * imageWidth + x; for ( int y = ymin, y0 = 0; y <= ymax; y++, y0++, p += imageWidth ) { if ( objectLabels[p] == label ) { // avoid adding the point we already have if ( ( leftProcessedPoints[y0] != x ) && ( rightProcessedPoints[y0] != x ) ) { edgePoints.Add( new IntPoint( x, y ) ); } break; } } // scan from bottom to top p = ymax * imageWidth + x; for ( int y = ymax, y0 = ymax - ymin; y >= ymin; y--, y0--, p -= imageWidth ) { if ( objectLabels[p] == label ) { // avoid adding the point we already have if ( ( leftProcessedPoints[y0] != x ) && ( rightProcessedPoints[y0] != x ) ) { edgePoints.Add( new IntPoint( x, y ) ); } break; } } } return edgePoints; } /// <summary> /// Actual objects map building. /// </summary> /// /// <param name="image">Unmanaged image to process.</param> /// /// <remarks><note>By the time this method is called bitmap's pixel format is not /// yet checked, so this should be done by the class inheriting from the base class. /// <see cref="imageWidth"/> and <see cref="imageHeight"/> members are initialized /// before the method is called, so these members may be used safely.</note></remarks> /// protected abstract void BuildObjectsMap( UnmanagedImage image ); #region Private Methods - Collecting objects' rectangles // Collect objects' rectangles private unsafe void CollectObjectsInfo( UnmanagedImage image ) { int i = 0, label; // create object coordinates arrays int[] x1 = new int[objectsCount + 1]; int[] y1 = new int[objectsCount + 1]; int[] x2 = new int[objectsCount + 1]; int[] y2 = new int[objectsCount + 1]; int[] area = new int[objectsCount + 1]; long[] xc = new long[objectsCount + 1]; long[] yc = new long[objectsCount + 1]; long[] meanR = new long[objectsCount + 1]; long[] meanG = new long[objectsCount + 1]; long[] meanB = new long[objectsCount + 1]; long[] stdDevR = new long[objectsCount + 1]; long[] stdDevG = new long[objectsCount + 1]; long[] stdDevB = new long[objectsCount + 1]; for ( int j = 1; j <= objectsCount; j++ ) { x1[j] = imageWidth; y1[j] = imageHeight; } byte* src = (byte*) image.ImageData.ToPointer( ); if ( image.PixelFormat == PixelFormat.Format8bppIndexed ) { int offset = image.Stride - imageWidth; byte g; // pixel's grey value // walk through labels array for ( int y = 0; y < imageHeight; y++ ) { for ( int x = 0; x < imageWidth; x++, i++, src++ ) { // get current label label = objectLabels; // skip unlabeled pixels if ( label == 0 ) continue; // check and update all coordinates if ( x < x1[label] ) { x1[label] = x; } if ( x > x2[label] ) { x2[label] = x; } if ( y < y1[label] ) { y1[label] = y; } if ( y > y2[label] ) { y2[label] = y; } area[label]++; xc[label] += x; yc[label] += y; g = *src; meanG[label] += g; stdDevG[label] += g * g; } src += offset; } for ( int j = 1; j <= objectsCount; j++ ) { meanR[j] = meanB[j] = meanG[j]; stdDevR[j] = stdDevB[j] = stdDevG[j]; } } else { // color images int pixelSize = Bitmap.GetPixelFormatSize( image.PixelFormat ) / 8; int offset = image.Stride - imageWidth * pixelSize; byte r, g, b; // RGB value // walk through labels array for ( int y = 0; y < imageHeight; y++ ) { for ( int x = 0; x < imageWidth; x++, i++, src += pixelSize ) { // get current label label = objectLabels; // skip unlabeled pixels if ( label == 0 ) continue; // check and update all coordinates if ( x < x1[label] ) { x1[label] = x; } if ( x > x2[label] ) { x2[label] = x; } if ( y < y1[label] ) { y1[label] = y; } if ( y > y2[label] ) { y2[label] = y; } area[label]++; xc[label] += x; yc[label] += y; r = src[RGB.R]; g = src[RGB.G]; b = src[RGB.B]; meanR[label] += r; meanG[label] += g; meanB[label] += b; stdDevR[label] += r * r; stdDevG[label] += g * g; stdDevB[label] += b * b; } src += offset; } } // create blobs blobs.Clear( ); for ( int j = 1; j <= objectsCount; j++ ) { int blobArea = area[j]; Blob blob = new Blob( j, new Rectangle( x1[j], y1[j], x2[j] - x1[j] + 1, y2[j] - y1[j] + 1 ) ); blob.Area = blobArea; blob.Fullness = (double) blobArea / ( ( x2[j] - x1[j] + 1 ) * ( y2[j] - y1[j] + 1 ) ); blob.CenterOfGravity = new AForge.Point( (float) xc[j] / blobArea, (float) yc[j] / blobArea ); blob.ColorMean = Color.FromArgb( (byte) ( meanR[j] / blobArea ), (byte) ( meanG[j] / blobArea ), (byte) ( meanB[j] / blobArea ) ); blob.ColorStdDev = Color.FromArgb( (byte) ( Math.Sqrt( stdDevR[j] / blobArea - blob.ColorMean.R * blob.ColorMean.R ) ), (byte) ( Math.Sqrt( stdDevG[j] / blobArea - blob.ColorMean.G * blob.ColorMean.G ) ), (byte) ( Math.Sqrt( stdDevB[j] / blobArea - blob.ColorMean.B * blob.ColorMean.B ) ) ); blobs.Add( blob ); } } // Rectangles' and blobs' sorter private class BlobsSorter : System.Collections.Generic.IComparer<Blob> { private ObjectsOrder order; public BlobsSorter( ObjectsOrder order ) { this.order = order; } public int Compare( Blob a, Blob b ) { Rectangle aRect = a.Rectangle; Rectangle bRect = b.Rectangle; switch ( order ) { case ObjectsOrder.Size: // sort by size // the order is changed to descending return ( bRect.Width * bRect.Height - aRect.Width * aRect.Height ); case ObjectsOrder.Area: // sort by area return b.Area - a.Area; case ObjectsOrder.YX: // YX order return ( ( aRect.Y * 100000 + aRect.X ) - ( bRect.Y * 100000 + bRect.X ) ); case ObjectsOrder.XY: // XY order return ( ( aRect.X * 100000 + aRect.Y ) - ( bRect.X * 100000 + bRect.Y ) ); } return 0; } } #endregion }}`
January 3 2016
Thanks Dj,
I had assumed the "camera tracking" used a "script" developed thru EZ Builder, and additional code running in the background. I just looked at the surface results and never put thought into what was really going on.
Thanks for posting the information, but as you can tell, my background does not give me the ability to use it.
I have re-posted my thread on tracking using servo position information, and will continue to go forward with that scripting.
Thanks again,
Ron R | 10,645 | 43,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.957812 |
https://www.4pattfun.fr/30140/closed-circuit-portable-high-speed-ball-mill.html | 1,642,449,132,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00174.warc.gz | 651,504,204 | 6,556 | 1. Home
2. > Closed Circuit Portable High Speed Ball Mill
# Closed Circuit Portable High Speed Ball Mill
### High Speed Ball Mill Cost
Production Type High Speed Stirred Ball Mill Machine. 1-100L Optional Volume High Speed Stirred Ball Mill Machine. 1-100L Optional Volume High Speed Stirred Ball Mill Machine Stirred ball mill mainly consist of jar with inner grinding media stirring device and other auxiliary devices such as recycling device cooling device timing device speed adjusting controller and etc.Calculating Critical Speed In A Ball Mill. FOB Reference Price Get Latest Price Ball mil design calculation Yahoo Answers nbsp 0183 32 There exists a speed of rotation the quot critical speed quot at which the contents of the mill would simply ride over the roof of the mill due to centrifugal action The critical speed rpm is given by nC 42 29 d where d is the internal diameter in metres Ball.The terms high-speed vibration milling (HSVM), high-speed ball milling (HSBM), and planetary ball mill (PBM) are often used. The commercial apparatus are PBMs Fritsch P-5 and Fritsch Pulverisettes 6 and 7 classic line, the Retsch shaker (or mixer) mills ZM1, MM200, MM400, AS200, the Spex 8000, 6750 freezer mill SPEX CertiPrep, and the SWH-0.4.
### Describe Critical Speed Of Ball Mill
Ball Mill,Ball Mill For Sale,Export Ball Mill-liming Company. Ball mill is an efficient tool for fine powder grinding. The Ball Mill is used to grind many kinds of mine and Mining Ball Mill Technical Data Model Rotate Speed.Grinding Ball Mill Load Calculation Formula. volume loading formula in ball mill. Mill load or charge volume is the cumulative sum of the grinding media information show that 40 load by volume of ball mills result in energy calculation model of ball kinematics based on of ball mill and ball motion under the influence of coal load are analyzed carefully.
### Critical Speed Ball Mill List
The mill ball loading was 40 by volume, the rotation rate was equal to 85 of the critical speed. Balls were made from steel S4146, extra high quality, having hardness 62 2 HRC according to Rockwell. Ball Mill Design Power Calculation. Dec 12, 2016 Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to be fed 100.Ball Mill Circulating Load Formula Popular Education, Ball mill instruction manual pdf bico inc the fc bond ball mill is a small universal laboratory mill used in calculating the dry in closd circuit witb l0o per cent circulating load id a l2in dia by 24in Ball Mill Formulas And Calculations.
### Closed Circuit Ball Mill Circuit Coal
Ball Mill Calculations Ball Mill. Ball Mill Motor Power Draw Sizing And Design Formula The following equation is used to determine the power that wet grinding overflow ball mills should draw for mills larger than 33 meters 10 feet diameter inside liners the top size of the balls used affects the power drawn by the mill this is called the ball size factor s rod and ball mills by ca rowland and .MTW Series Trapezium Mill is a kind of Vertical Roller mill, It is the latest grinding equipment, with its own knowledge patents, have reached mod . Ball Mill. A ball mill is a type of grinder used to grind materials into extremely fine powder for use in mineral dressing processes, paints, pyrotechnics, a . T130X Superfine Grinding Mill.
### Ball Mill Speed Calculation Formula For Wet Grinding
Ball Mill Critical Speed in 85 of mill critical. have decreased for later 75 mill critical speed.371 0.234 0.142 0. or aj values Ball mill - Wikipedia, the free encyclopedia.2012-11-26ball mills are generally used to grind material 14 inch and finer, down to the particle size of 20 to 75 micronso achieve a reasonable efficiency with ball mills, they must be operated in a closed system, with oversize material continuously being recirculated back into the mill to be reduced.
### (china Supplier Offer) Portable Ball Mill Gold Mining
High Speed 80kw Rock Mining Ball Mill , 17209kg Ore Grinding Mill. Brand Name sinomin. Model Number Cone Grid Ball Mill. Place of Origin China. High Speed Rock Ore Grinding Equipment Ball Mill Introduction Ball mill is a critical device for smashing after the materials are crushed. Ball mills are widely used in manufacturing industries of.Ball mill. Email Us Our high power portable air compressors are highly suited for drilling to be carried out making to grinding with a wide range of equipment egories and series and stable performance. Mining Equipment Copper Ore Gold Ore Grinding Ball Mill. Improving grinding efficiency in closed circuit cement ball mill.
### Calculating Critical Speed In A Ball Mill
The circuit comprises a SAG mill, 12 m diameter by 6.1 m length (belly inside liners, the effective grinding volume), two pebble crushers, and two ball mills in parallel closed with cyclones. The SAG mill is fitted with a 20 MW gearless drive motor with bi-directional rotational capacity.Calculation of balance in cement grinding mill. Mar 08, 2013 calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill in literatures it is written that the grinding efficiency of ball mill is very less [less than 10 ] please expalin in a n excel sheet to calcualte the same thanks sidhant reply.
### High Speed Planetary Jar Mill For Precision Small Parts
Design and analysis of ball mill for paint industries,required to operate a commercial ball mill of radius is predicted on the assumption that the ball mill is half filled with balls, there is a cascade angle of and the critical speed Of mill 29.pm now, the balls occupy 60 and mill base 40 of total volume.Ball Mill Speed Calculation Formula For Wet Grinding. 12-12-2016 Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80 passing inch (6350 microns). The required product size distribution is to be 80 passing 100 mesh (149.
### Critical Speed Of Ball Mill Of Raduis R
Jan 10, 2021 The ball mill operated in closed circuit with a spiral classifier and carried a large circulating load from the classifier that resulted in unstable operation of the classifier. At 75 critical speed this ball mill can be expected to draw 1840 HP with a 40 ball load.Ball Mills Industry Grinder For Mineral Processing. Ball mill working principle high energy ball milling is a type of powder grinding mill used to grind ores and other materials to 25 mesh or extremely fine powders mainly used in the mineral processing industry both in open or closed circuits ball milling is a grinding method that reduces the product into a controlled final grind and a.
### Closed Circuit Cement Making Plant Ball Mill Grinding
Copper Ore Ball Mill - Tube Ball Mill 01-06 High Efficiency Ore Grinding 12-29 The Basic The commonly used ball mill for copper ore dressing is a wet type ball. We supply dry or ore ball mill. It has higher quality requirements on the grinding balls for copper ore grinding. A Ball Mill Inside Of A Copper Processing Industry. Stock Video.For example, the MTM Medium-speed Grinding Mill only consumes 1.02kW h when yielding a ton of materials and 1.48kW h when feeding a ton of materials. Its electricity consumption is lower than that of ball mill at the same level by over 60 . 4. Eco-friendly Closed-circuit Circulation System.
### Grinding Ball Mill Data Crusher Mills Cone Crusher Jaw
Ball charge loading – impact on specific power consumption and capacity152 Кб. Past ball mill studies have evaluated the importance of many factors in ball mill grinding efficiency including ball size distribution, mill length to diameterThere are several methods of calculating the net mill power draw for an industrial mill.Principles Of The Grinding Process Ball Mill In Cement. Manufacturing process of grinding media balls For the production of grinding balls Energosteel uses hot rolled round bar from medium and high carbon and lowalloyed steel produced by the worlds leading metallurgical companies Prior to the production the preform is subjected to strict control against the chemical composition and.
### High Quality Durable Small Ball Mill With Iso Approval
If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill. The point where the mill becomes a centrifuge is called the Critical Speed , and ball mills usually operate at 65 to 75 of the critical speed.Mill Speed - an overview ScienceDirect Topics. This mill is usually of one of two types ball-and-race mill and roll-and-race mill. The main drive shaft turns the table supporting the grinding ring, which in turn transmits Motor input power thus has to be converted to power at the mill pinion shaft output This allows a highly visual, easily interpreted “picture” to be used to.
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# Activities in Grades 6-8 in Math
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Clear up any fraction befuddlement your students have with this fun activity. Get enough materials for a class of 30 to make their own rulers.
\$65.00
Instead of using a regular ruler, students can learn about measuring and fractions as they make their own ruler with the Making Your Own Ruler Maker Project. The project comes with the Making Your Own Ruler Pack – which includes permanent markers, plastic material, templates, and student worksheets – and the Dr. Zoon Making Your Own Ruler Video. For convenience, the project also comes with labels, a bin, and a bin prep sheet for storage and organization.
\$75.00
Students learn about measuring and fractions as they make their own rulers! The pack provides everything necessary for this fun activity – permanent markers, plastic material, templates, and student worksheets – for a class of 30.
\$56.25
This well-illustrated book is a primer on packaging design. It includes an introduction to packaging and an overview of the components of packaging design and prototyping. It also includes information and patterns for common shipping, telescoping, decorative, retail, and geometric boxes. Includes a glossary on packaging terminology.
Box-making activities and ideas abound, and there are two box-making competitions (challenges). Includes patterns for more than 15 different box configurations.
\$5.45
This collection of puzzles, games, and activities covers a wide range of mathematical concepts and skills and can be readily linked with other curriculum activities.
\$46.75
Start with explaining the language of algebra and how to solve problems, and then go further with warm-ups and specialized topics. This volume covers how to solve algebra problems, translating words into algebraic symbols, ages, coins, mixtures, formulas, rectangles, percents, and more. Reproducible student sheets; includes an answer key with explanations.
\$29.95 \$11.96
Math, history, art, and world cultures come together in this delightful book, which connects youngsters with friends around the globe.
These math games encourage students to hone their math skills as they use geometry to design game boards, probability to analyze the outcomes of games of chance, and logical thinking to devise strategies for the games. These activities introduce students to the people who played the games, who solved the puzzles, and who designed the art.
\$15.85
Using real-life situations such as calculating car payments and planning a business, this volume covers the basic concepts of algebra, linear equations, lines and distance, slopes and lines, parabolas, quadratic equations, and more. The updated edition features updated content and new lessons.
Fully reproducible book includes math concepts, answer keys, and extension activities.
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Fractions, ratios, and rates are the workhorses of everyday math. Whether you install carpeting, calculate batting averages, work in construction, or create low-calorie recipes, these math concepts are key to figuring out the world. With engaging activities, this book explores fractions, equivalent fractions, finding patterns, ratio, solving proportions, rates, and unit pricing.
Fully reproducible book includes math concepts, answer keys, and extension activities.
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Products to display: Items 1 to 9 of 9 | 702 | 3,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-09 | longest | en | 0.919086 |
https://physicslab.org/Document.aspx?doctype=4&filename=Tests_PhysicsISound_Test2.xml | 1,713,048,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00305.warc.gz | 428,884,924 | 8,055 | Chapter 26: Sound Printer Friendly Version
Resource Lessons
Labs
Worksheets
Textbook Assignments
Textbook Resources
Diagrams: 26.2, 26.3, 26.4, 26.5, 26.6, 26.9, 26.10, 26.11, 26.13, 26.14, 26.16
In-Text Questions: pg 393 and 399
NextTime Questions
Vocabulary
wave vw = fλ mechanical non-mechanical longitudinal radio waves vs sound waves damped heat compression rarefaction amplitude (pressure/density) point source frequency (f) hertz kilohertz megahertz vibration period (T) equilibrium position wavelength (λ) relationship between frequency and wavelength relationship between frequency and period human range of frequencies infrasonic ultrasonic intensity, decibels interference beats beat pitch beat frequency constructive destructive antinodes (A) nodes (N) loops resonance forced vibration natural frequency properties of open pipes properties of closed pipes harmonics overtones standing waveform for an open water column reflection echo reverberation speed of sound (dry air) speed of sound increases with the medium'srigidity, temperature, humidity
Formulas
vw = fλ d = rt sound level dB to power take the difference in the decibels, divide that difference by 10, relationship between the original sound levels equals 10x power beats beat frequency = |f2 - f1| beat pitch = ½(f1 + f2) speed of sound in dry air vw = 331 + 0.6T open-closed pipe for the fundamental, A-N Lpipe = ½ loop = ¼λ open-open pipe for the fundamental, A-N-A Lpipe = 1 loop = ½λ | 377 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.74431 |
https://www.netlib.org/lapack/explore-html-3.6.1/d6/d81/cung2l_8f_source.html | 1,712,975,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00750.warc.gz | 869,305,303 | 6,419 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
cung2l.f
Go to the documentation of this file.
1 *> \brief \b CUNG2L generates all or part of the unitary matrix Q from a QL factorization determined by cgeqlf (unblocked algorithm).
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 *> \htmlonly
10 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/cung2l.f">
11 *> [TGZ]</a>
12 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/cung2l.f">
13 *> [ZIP]</a>
14 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/cung2l.f">
15 *> [TXT]</a>
16 *> \endhtmlonly
17 *
18 * Definition:
19 * ===========
20 *
21 * SUBROUTINE CUNG2L( M, N, K, A, LDA, TAU, WORK, INFO )
22 *
23 * .. Scalar Arguments ..
24 * INTEGER INFO, K, LDA, M, N
25 * ..
26 * .. Array Arguments ..
27 * COMPLEX A( LDA, * ), TAU( * ), WORK( * )
28 * ..
29 *
30 *
31 *> \par Purpose:
32 * =============
33 *>
34 *> \verbatim
35 *>
36 *> CUNG2L generates an m by n complex matrix Q with orthonormal columns,
37 *> which is defined as the last n columns of a product of k elementary
38 *> reflectors of order m
39 *>
40 *> Q = H(k) . . . H(2) H(1)
41 *>
42 *> as returned by CGEQLF.
43 *> \endverbatim
44 *
45 * Arguments:
46 * ==========
47 *
48 *> \param[in] M
49 *> \verbatim
50 *> M is INTEGER
51 *> The number of rows of the matrix Q. M >= 0.
52 *> \endverbatim
53 *>
54 *> \param[in] N
55 *> \verbatim
56 *> N is INTEGER
57 *> The number of columns of the matrix Q. M >= N >= 0.
58 *> \endverbatim
59 *>
60 *> \param[in] K
61 *> \verbatim
62 *> K is INTEGER
63 *> The number of elementary reflectors whose product defines the
64 *> matrix Q. N >= K >= 0.
65 *> \endverbatim
66 *>
67 *> \param[in,out] A
68 *> \verbatim
69 *> A is COMPLEX array, dimension (LDA,N)
70 *> On entry, the (n-k+i)-th column must contain the vector which
71 *> defines the elementary reflector H(i), for i = 1,2,...,k, as
72 *> returned by CGEQLF in the last k columns of its array
73 *> argument A.
74 *> On exit, the m-by-n matrix Q.
75 *> \endverbatim
76 *>
77 *> \param[in] LDA
78 *> \verbatim
79 *> LDA is INTEGER
80 *> The first dimension of the array A. LDA >= max(1,M).
81 *> \endverbatim
82 *>
83 *> \param[in] TAU
84 *> \verbatim
85 *> TAU is COMPLEX array, dimension (K)
86 *> TAU(i) must contain the scalar factor of the elementary
87 *> reflector H(i), as returned by CGEQLF.
88 *> \endverbatim
89 *>
90 *> \param[out] WORK
91 *> \verbatim
92 *> WORK is COMPLEX array, dimension (N)
93 *> \endverbatim
94 *>
95 *> \param[out] INFO
96 *> \verbatim
97 *> INFO is INTEGER
98 *> = 0: successful exit
99 *> < 0: if INFO = -i, the i-th argument has an illegal value
100 *> \endverbatim
101 *
102 * Authors:
103 * ========
104 *
105 *> \author Univ. of Tennessee
106 *> \author Univ. of California Berkeley
107 *> \author Univ. of Colorado Denver
108 *> \author NAG Ltd.
109 *
110 *> \date September 2012
111 *
112 *> \ingroup complexOTHERcomputational
113 *
114 * =====================================================================
115 SUBROUTINE cung2l( M, N, K, A, LDA, TAU, WORK, INFO )
116 *
117 * -- LAPACK computational routine (version 3.4.2) --
118 * -- LAPACK is a software package provided by Univ. of Tennessee, --
119 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
120 * September 2012
121 *
122 * .. Scalar Arguments ..
123 INTEGER INFO, K, LDA, M, N
124 * ..
125 * .. Array Arguments ..
126 COMPLEX A( lda, * ), TAU( * ), WORK( * )
127 * ..
128 *
129 * =====================================================================
130 *
131 * .. Parameters ..
132 COMPLEX ONE, ZERO
133 parameter ( one = ( 1.0e+0, 0.0e+0 ),
134 \$ zero = ( 0.0e+0, 0.0e+0 ) )
135 * ..
136 * .. Local Scalars ..
137 INTEGER I, II, J, L
138 * ..
139 * .. External Subroutines ..
140 EXTERNAL clarf, cscal, xerbla
141 * ..
142 * .. Intrinsic Functions ..
143 INTRINSIC max
144 * ..
145 * .. Executable Statements ..
146 *
147 * Test the input arguments
148 *
149 info = 0
150 IF( m.LT.0 ) THEN
151 info = -1
152 ELSE IF( n.LT.0 .OR. n.GT.m ) THEN
153 info = -2
154 ELSE IF( k.LT.0 .OR. k.GT.n ) THEN
155 info = -3
156 ELSE IF( lda.LT.max( 1, m ) ) THEN
157 info = -5
158 END IF
159 IF( info.NE.0 ) THEN
160 CALL xerbla( 'CUNG2L', -info )
161 RETURN
162 END IF
163 *
164 * Quick return if possible
165 *
166 IF( n.LE.0 )
167 \$ RETURN
168 *
169 * Initialise columns 1:n-k to columns of the unit matrix
170 *
171 DO 20 j = 1, n - k
172 DO 10 l = 1, m
173 a( l, j ) = zero
174 10 CONTINUE
175 a( m-n+j, j ) = one
176 20 CONTINUE
177 *
178 DO 40 i = 1, k
179 ii = n - k + i
180 *
181 * Apply H(i) to A(1:m-k+i,1:n-k+i) from the left
182 *
183 a( m-n+ii, ii ) = one
184 CALL clarf( 'Left', m-n+ii, ii-1, a( 1, ii ), 1, tau( i ), a,
185 \$ lda, work )
186 CALL cscal( m-n+ii-1, -tau( i ), a( 1, ii ), 1 )
187 a( m-n+ii, ii ) = one - tau( i )
188 *
189 * Set A(m-k+i+1:m,n-k+i) to zero
190 *
191 DO 30 l = m - n + ii + 1, m
192 a( l, ii ) = zero
193 30 CONTINUE
194 40 CONTINUE
195 RETURN
196 *
197 * End of CUNG2L
198 *
199 END
subroutine cung2l(M, N, K, A, LDA, TAU, WORK, INFO)
CUNG2L generates all or part of the unitary matrix Q from a QL factorization determined by cgeqlf (un...
Definition: cung2l.f:116
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine cscal(N, CA, CX, INCX)
CSCAL
Definition: cscal.f:54
subroutine clarf(SIDE, M, N, V, INCV, TAU, C, LDC, WORK)
CLARF applies an elementary reflector to a general rectangular matrix.
Definition: clarf.f:130 | 2,107 | 5,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.339083 |
https://saymber.com/tag/cottonwood-tree/ | 1,544,556,155,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823674.34/warc/CC-MAIN-20181211172919-20181211194419-00400.warc.gz | 744,958,455 | 22,283 | # 25 Sept 2018 Bowls, Caterpillars and Drawing
Just a reminder, the dividing by 2 came to me when the system initially did. It’s like analyzing something with both parts of ourselves, the light and shadow. It’s the duality of human nature. We are basically two parts stitched together inside and out if you think about it! Most things that come into our lives go under some sort of scrutiny and or analysis before we accept the person, place or thing as “truth” or worthy of being part of our life. It seems a rare thing when a person, other life form, place or thing is exactly how it seems to appear. What we call things today aren’t always the names that were used in the past. In this list you will see the difference in value between dirt and soil. Dirt in the numbers has the same value as a camera even though it has the same purpose of soil.
The “3 cycle” represents to me the cycles of life, death and rebirth. The “9 cycle” represents difficult to me or something is hard, sturdy or tough. It’s rather incredible to see what numbers can tell us about the meaning and possible intent of why things were given the labels they were. As I’ve done this for my personal list, you can see how dated some words are compared to others. It’s pretty fascinating! (To me at least!) If you have a question about any of this, please leave me a comment. Thank you for stopping by!
https://www.gematrix.org/
what is the purpose of a craft in Simple Gematria equals: 293
14/5 divided by 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of a craft in Simple Gematria equals: 285
15/6/3 cycle or 15 divided by 2 = 7.5 = 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the purpose of orgone in Simple Gematria equals: 318
12/3 cycle or 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the function of orgone in Simple Gematria equals: 310
4/2/1
what is the purpose of orgonite in Simple Gematria equals: 347
14/5 divided by 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of orgonite in Simple Gematria equals: 339
15/6/3 cycle or 15 divided by 2 = 7.5 = 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
### What is Orgone?
The word orgonite comes from “orgone”, the name given by Wilhelm Reich to vital energy found everywhere in nature. It is life energy, also called Ch’i, Prana, and Aether. This vital energy exists, in a natural way, under many different forms. It can be neutral (OR=orgone), positive (POR=positive orgone energy) or negative (DOR = deadly orgone energy). When positive, it enables living organisms to exist in a healthy state. Reich did a lot of research on the properties and behaviors of this energy. He built an orgone accumulator out of alternate layers of metal and organic material. He also observed that OR was able to neutralize nuclear radiation.
what is the purpose of organic in Simple Gematria equals: 311
5 divided by 2 – 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of organic in Simple Gematria equals: 303
6/3 cycle
what is the purpose of a bowl in Simple Gematria equals: 297
18/9 divided by 2 = 4.5 = 9 cycle
what is the function of a bowl in Simple Gematria equals: 289
19/10/5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1 or 19 divided by 2 = 9.5 = 14 divided by 2 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the purpose of dirt in Simple Gematria equals: 295
16/7 divided by 2 = 3.5 = 8/4/2/1 or 16/8/4/2/1
what is the function of dirt in Simple Gematria equals: 287
17/8/4/2/1 or 17 divided by 2 = 8.5 = 13 divided by 2 = 6.5 = 11 divided by 2 = 5.5 = 10 divided by 2 = 5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the purpose of soil in Simple Gematria equals: 299
20/2/1 or 20/10/ 5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of soil in Simple Gematria equals: 291 – soil isn’t the same as dirt in the numbers, dirt has the exact same value as a camera!
12/3 cycle or 12/3 cycle or 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the purpose of a caterpillar in Simple Gematria equals: 360
9 cycle
what is the function of a caterpillar in Simple Gematria equals: 352
10/1 or 10/5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the purpose of a tree in Simple Gematria equals: 293
14/5 divided by 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of a tree in Simple Gematria equals: 285
15/6/3 cycle or 15 divided by 2 = 7.5 = 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the purpose of a cottonwood tree in Simple Gematria equals: 437
14/5 divided by 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of a cottonwood tree in Simple Gematria equals: 429
15/6/3 cycle or 15 divided by 2 = 7.5 = 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the purpose of a pecan tree in Simple Gematria equals: 332
8/4/2/1
what is the function of a pecan tree in Simple Gematria equals: 324
9 cycle
what is the purpose of an oak tree in Simple Gematria equals: 334
10/1 or 10/5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of an oak tree in Simple Gematria equals: 326
11/2/1 or 11 divided by 2 = 5.5 = 10 divided by 2 = 5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the purpose of a crepe myrtle tree in Simple Gematria equals: 433
10/1 or 10/5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of a crepe myrtle tree in Simple Gematria equals: 425
11/2/1 or 11 divided by 2 = 5.5 = 10 divided by 2 = 5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the purpose of a human in Simple Gematria equals: 302
5 divided by 2 – 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1
what is the function of a human in Simple Gematria equals: 294
15/6/3 cycle or 15 divided by 2 = 7.5 = 12 divided by 2 = 6 divided by 2 = 3 divided by 2 = 1.5 = 6 divided by 2 = 3 cycle
what is the purpose of a camera in Simple Gematria equals: 286
16/7 divided by 2 = 3.5 = 8/4/2/1 or 16/8/4/2/1
what is the function of a camera in Simple Gematria equals: 278
17/8/4/2/1 or 17 divided by 2 = 8.5 = 13 divided by 2 = 6.5 = 11 divided by 2 = 5.5 = 10 divided by 2 = 5 divided by 2 = 2.5 = 7 divided by 2 = 3.5 = 8/4/2/1 | 2,226 | 6,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-51 | longest | en | 0.943653 |
https://brilliant.org/problems/divide-by-5/ | 1,532,277,072,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00453.warc.gz | 610,070,297 | 11,310 | # Divide by $$5$$
If $$n$$ is a positive integer, then $$n$$ has $$5$$ possible remainders after division by five: $$0, 1, 2, 3$$ and $$4$$.
If $$n$$ is a positive integer, then $$n^2$$ is also a positive integer.
$$n$$ is a positive integer. What is the number of possible remainders of $$n^2$$ after division by five?
× | 99 | 325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-30 | longest | en | 0.88862 |
http://www.the16types.info/vbulletin/showthread.php/10883-How-functions-work-(Theory) | 1,527,415,245,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868239.93/warc/CC-MAIN-20180527091644-20180527111644-00450.warc.gz | 454,161,000 | 10,653 | # Thread: How functions work (Theory)
1. ## How functions work (Theory)
The second part of my theory has to do with the different variations of the same functions. Take INTjs and ISTjs for example. They both are types which lead with Ti. But there are noticeable differences between the types. It has been noted in socionics before as there being a + and a - side of a function. You have -Ti for INTjs and +Ti for ISTjs. The differences between these two are the directionality of the energy being produced.
Example of how Ti works:
Example of how Fi works:
An INTj's ego block is -Ti +Ne and an ISTj's ego block +Ti -Se. The introverted function transfers data to the extroverted function and vice versa as explain in the previous part of my theory. For INTj's, -Ti and +Ne combine together to create a single function. Technically, one could consider -Ti and +Ne the same thing. In a way, these functions are like a battery together. I also believe that Te and Ti are theoretically the same thing. The only difference is that Te is extroverted and Ti is introverted, and that the functions can have a different directionality of energy. It is my belief that -Ti and +Te are as closely(or possibly more) than -Ti and +Te. This has to do with the directionality of the energy. Here is an example and explanation of what I mean:
This would technically mean there is more correlation between the thinking functions of INTj and ESTj than INTj and ISTj. It is my belief that the realm of functions work like this. Here is an example:
I will touch up on this more in the future.
2. ## Re: How functions work (Theory)
Originally Posted by hitta
An INTj's ego block is -Ti +Ne and an ISTj's ego block +Ti -Ne. [/img]
what happened to the Se then? or am I missing something completely.
3. ## Re: How functions work (Theory)
Originally Posted by hitta
An INTj's ego block is -Ti +Ne and an ISTj's ego block +Ti -Se. The introverted function transfers data to the extroverted function and vice versa as explain in the previous part of my theory. For INTj's, -Ti and +Ne combine together to create a single function. Technically, one could consider -Ti and +Ne the same thing. In a way, these functions are like a battery together. I also believe that Te and Ti are theoretically the same thing. The only difference is that Te is extroverted and Ti is introverted, and that the functions can have a different directionality of energy. It is my belief that -Ti and +Te are as closely(or possibly more) than -Ti and +Te. This has to do with the directionality of the energy. Here is an example and explanation of what I mean:
That makes a lot of sense. (But doesn't ENTj have +Te and -Ti? If that is true, then your theory could explain the complementary aspects of the Contrary relationship.)
Although be careful about using "similarity" to refer to types or functions. Te and Ti share T, but not dynamism/statism or introversion/extroversion, obviously. I think of T as the domain of Te and Ti, and +/- could considerably refine the role of each Tj type in the T domain.
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• | 759 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-22 | latest | en | 0.951927 |
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