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https://www.coursehero.com/file/5257/Proving-Invariances/ | 1,544,813,697,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00515.warc.gz | 846,309,699 | 161,381 | Proving Invariances
# Temporal Verification of Reactive Systems: Safety
• Notes
• davidvictor
• 10
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CS256/Winter 2007 — Lecture #6 Zohar Manna Chapter 1 Invariance: Proof Methods For assertion q and SPL program P show P q 0 q (i.e., q is P -invariant) 6-1 Proving Invariances Definitions Recall: the variables of assertion : free (flexible) system variables V = Y ∪ { π } where Y are the program variables and π is the control variable quantified (rigid) specification variables q 0 is the primed version of q , obtained by replacing each free occurrence of a system variable y V by its primed version y 0 . ρ τ is the transition relation of τ , expressing the re- lation holding between a state s and any of its τ - successors s 0 τ ( s ) . 6-2 Verification Conditions (proof obligations) standard verification condition For assertions ϕ, ψ and transition τ , { ϕ } τ { ψ } (“Hoare triple”) stands for the state formula ρ τ ϕ ψ 0 “Verification condition (VC) of ϕ and ψ relative to transition τ ϕ τ ψ p p j j + 1 6-3 Verification Conditions (Con’t) Example: ρ τ : x 0 y 0 = x + y x 0 = x ϕ : y = 3 ψ : y = x + 3 Then { ϕ } τ { ψ } : x 0 y 0 = x + y x 0 = x | {z } ρ τ y = 3 | {z } ϕ y 0 = x 0 + 3 | {z } ψ 0 6-4
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Verification Conditions (Con’t) for τ ∈ T in P { ϕ } τ { ψ } : ρ τ ϕ ψ 0 τ leads from ϕ to ψ in P for T in P { ϕ }T { ψ } : { ϕ } τ { ψ } for every τ ∈ T T leads from ϕ to ψ in P Claim (Verification Condition) If { ϕ } τ { ψ } is P -state valid, then every τ -successor of a ϕ -state is a ψ -state. 6-5 Verification Conditions (Con’t) Special Cases while, conditional ρ τ : ρ t τ ρ f τ { ϕ } τ t { ψ } : ρ t τ ϕ ψ 0 { ϕ } τ f { ψ } : ρ f τ ϕ ψ 0 { ϕ } τ { ψ } : { ϕ } τ t { ψ } ∧ { ϕ } τ f { ψ } idle { ϕ } τ I { ϕ } : ρ τ I ϕ ϕ 0 always valid, since ρ τ I v 0 = v for all v V , so ϕ 0 = ϕ. 6-6 Verification Conditions (Con’t) Substituted Form of Verification Condition Transition relation can be written as ρ τ : C τ ( V 0 = E ) where C τ : enabling condition V 0 : primed variable list E : expression list The substituted form of verification condition { ϕ } τ { ψ } : C τ ϕ ψ [ E /V ] where ψ [ E /V ] : replace each variable v V in ψ by the corresponding e E Note : No primed variables! The substituted form of a verification condition is P -state valid iff the standard form is 6-7 Verification Conditions (Con’t) Example: ρ τ : x 0 y 0 = x + y x 0 = x ϕ : y = 3 ψ : y = x + 3 Standard x 0 y 0 = x + y x 0 = x | {z } ρ τ y = 3 | {z } ϕ y 0 = x 0 + 3 | {z } ψ 0 Substituted x 0 | {z } C τ y = 3 | {z } ϕ x + y = x + 3 | {z } ψ [ E /V ] 6-8
Verification Conditions (Con’t) Example: ϕ : x = y ψ : x = y + 1 ρ τ : x 0 | {z } C τ ( x 0 , y 0 ) | {z } V 0 = ( x + 1 , y ) | {z } E The substituted form of { ϕ } τ { ψ } is x 0 | {z } C τ x = y | {z } ϕ ( x = y + 1)[( x + 1 , y ) / ( x, y )] | {z } ψ [ E /V ] or equivalently x 0 x = y x + 1 = y + 1 6-9 Simplifying Control Expressions move ( L 1 , L 2 ): L 1 π π 0 = ( π - L 1 ) L 2 e.g., for L 1 = { 1 } , L 2 = { 2 } move ( 1 , ‘ 2 ): 1 π π 0 = ( π - { 1 } ) ∪ { 2 } Consequences implied by move ( L 1 , L 2 ): for every [ ] L 1 at - = t (i.e., [ ] π ) for every [ ] L 2 at 0 - = t (i.e., [ ] π 0 ) for every [ ] L 1 - L 2 at - = t (i.e., [ ] π
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• '
• NoProfessor
• formal methods, invariant, L1 family, 3 j, Initiality
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Jill Tulane University ‘16, Course Hero Intern | 1,505 | 4,556 | {"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} | 3.21875 | 3 | CC-MAIN-2018-51 | latest | en | 0.774551 |
https://scicomp.stackexchange.com/questions/24804/metropolis-algorithm-and-thermal-sine-gordon-model | 1,620,604,724,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00442.warc.gz | 538,388,298 | 37,731 | # Metropolis algorithm and thermal sine-Gordon model
I try to simulate thermal version of 1D $(x, t)$ sine-Gordon field model. I am interested in finding thermal static solution that minimizes functional of energy $E$:
$$E = \int dx \left( \frac{1}{2} \phi' ^2 + 1 - \cos \phi \right) \ ,$$
where $\phi' = \partial_x \phi$
What is very confusing that acceptance ratio of Metropolis algorithm is too high - more than $0.95$, so almost every new proposed configuration is accepted. On each Metropolis step I change field value at one spatial point and calculate difference in energy. To propose new configurations uniform sampling is used with step parameter $\delta = 0.5$, i.e.
$$\phi_{new} = \phi_{old} + r \ ,$$
where $r$ is random number from $\phi_{old} - \delta$ to $\phi_{old} + \delta$.
It seems to me that if acceptance ratio is too high then algorithm does not work correctly. However, such algorithm is perfectly applied to the ground state harmonic oscillator (via path integral Monte Carlo).
Here is my code:
double E_part(double dx, double phi, double phi_plus, double phi_minus) {
return dx * ( 0.5 * pow ( 0.5 * ( phi - phi_minus ), 2.0 ) + 0.5 * pow ( 0.5 * ( phi_plus - phi ), 2.0 ) + 1 - cos ( 0.5 * ( phi_minus + phi ) ) - cos ( 0.5 * (phi + phi_plus ) ) );
}
double Metropolis(int N, int Steps, double dx, double Temperature, double* dphi, double* phi, double delta, double* EnergyArray) {
double Beta = 1.0 / Temperature;
double Epart;
double Enewpart;
double phinew;
double phi_plus, phi_minus;
int AcceptanceCounter = 0;
//Initial energy
double r; //random number from 0 to 1;
for (int k=0; k<Steps; k++) {
for (int j=0; j<N; j++) {
int pos = RandomUniformInt(1 , N-2); //random position to change;
phinew = phi[pos] - delta + RandomUniform() * 2 * delta;
phi_plus = phi[pos + 1];
phi_minus = phi[pos - 1];
Epart = E_part(dx, phi[pos], phi_plus, phi_minus);
Enewpart = E_part(dx, phinew, phi_plus, phi_minus);
r = RandomUniform();
if ( Enewpart * Beta - Epart * Beta < 0.0 || exp( Epart * Beta - Enewpart * Beta ) >= r ) {
phi[pos] = phinew;
dphi[pos - 1] = 0.5 * ( phi[pos] - phi[pos - 1] );
dphi[pos] = 0.5 * ( phi[pos + 1] - phi[pos] );
AcceptanceCounter++;
}
}
EnergyArray[k] = Energy(N, dx, dphi, phi);
}
return ( AcceptanceCounter / static_cast<double>(Steps * N) ); //AcceptanceRatio is returned
}
I don't know what might be wrong.
• Should you not be checking exp( Epart * Beta - Enewpart * Beta ) < r instead of exp( Epart * Beta - Enewpart * Beta ) >= r ? – lemon Aug 24 '16 at 7:33
• @lemon No, of course, because in this case new energy is greater than the current one and (E - Enew) < 0 – newt Aug 24 '16 at 11:54
• Sorry, you're right. My only remaining guess is that your temperature is too high. If you only accept changes that decrease the energy then what is your acceptance ratio? – lemon Aug 24 '16 at 13:43
• @lemon It is very interesting, but energy tends to rise. I think it is quite normal because we have nonzero temperature, so the initial configuration becomes "noisy". Acceptance ratio is about 0.75 - 0.95 (depends on temperature and Metropolis step 'delta'). I think that probably the problem lies in high correlation between neighboring grid points. – newt Aug 24 '16 at 18:07 | 958 | 3,263 | {"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": 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} | 3.15625 | 3 | CC-MAIN-2021-21 | latest | en | 0.677857 |
http://ccm.net/faq/19834-excel-2007-insert-an-image-with-transparency | 1,508,449,260,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00744.warc.gz | 60,013,814 | 19,977 | # Excel 2007 - Insert an image with transparency
October 2017
## Intro
This demo handles several deficiencies of Excel to insert images and apply transparency.
There exist at least two possibilities to insert an image in an Excel sheet,
1)Insert Image directly
• Insert> Image
• You can enter the picture you want but if you enter an image of 1920 x 1088 pixe, Excel will display an image having the following dimension 1920 x 1088 centimeters!.
• Another solution: Use an image control tools (OCX Controls), spread it on the sheet, right click and select> Developer> Insert> Image and apply it on the sheet.
• Right click on the image>> select Properties.
• Set PictureSizeMode = 3
• In the Properties window, go to Picture and select an image to be included in the control.
• So farthe picture is inserted into the control while maintaining the width/height ratio. You can adjust the size of the control to coincide with the dimensions of the image.
• But note that transparency options are not available.
2) Insert image in a form
• Insert> Form
• Draw the rectangle
• Right click> Format Shape. Select Picture or texture fill.
• Simply use the image of your choice.
• Using this method you can adjust the transparency of the image, but not its ratio.
This demo aims to:
• Insert an image while keeping the ratio.
• Keep the dimensions
• Keep the ratio of the image.
• Transparency available.
## Solution
Using GDI + APIs for the dimensions of the image
You can move the image, rotate it, change the dimensions of the available area.
But it is not advisable to resize the image with the handles (and thus lose the ratio of the image).
For the demo I have determined the surface area available with the top right of the button Changer Image:
But it is easy to directly enter the dimensions in the code
In Module1, look for the RemplaceImage routine.
``` 'Rectangle disponnible pour l'image, en points
With Sheets("Image").CommandButton1
Dispo.W = (.Left + .Width)
Dispo.H = .Top
End With
```
Replace these lines with
```'Rectangle disponnible pour l'image, en points
Dispo.W = La largeur en point
Dispo.H = La hauteur en point```
CAUTION: These dimensions are not those of the image, depending on the ratio of the image file only one of the two will be effective. | 527 | 2,277 | {"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-2017-43 | latest | en | 0.761096 |
https://outsider.com/entertainment/tv/wheel-of-fortune-contestant-flips-puzzle-on-its-head-incredibly-vowel-choice-video/ | 1,653,528,551,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00021.warc.gz | 510,765,774 | 39,420 | # ‘Wheel of Fortune’ Contestant Flips Puzzle on Its Head with Incredible Vowel Choice: VIDEO
Often on Wheel of Fortune, the difference between a huge loss or an epic win comes down to a single consonant or vowel. If you make the right call, you could find the clue you need to solve that high-value puzzle and walk away with a wad of cash and maybe even a vacation. However, if those squares don’t light up and Vanna White doesn’t come to fill in those blanks, your bank could look just as empty as that board.
Of course, every contestant on Wheel of Fortune knows how thin the line is between victory and bankruptcy. That’s why they have to use a bit of strategy to narrow down their choices of letters for each puzzle. During the regular rounds of the game, that might mean trying to identify the most frequently occurring letter to get the most money. Or, opting not to call out any vowels to avoid losing money.
Other times, however, the right guess happens simply from good luck. According to contestant Giovanni Danza, that’s exactly what happened when he chose a crucial vowel in the Bonus Puzzle that directly led to his final victory. Danza revealed that he was a former marathon runner and hockey player. More recently, the New Jersey contestant married “Dr. Alyssa” and surely hoped to bring home his bride a post-wedding gift of a hunk of cash.
Watch Danza make a “useful” choice in the clip below.
## ‘Wheel of Fortune’ Contestant Nearly Fills in Entire Puzzle with Letter Choices Before Solving
After choosing consonants “F,” C” and “D,” the Wheel of Fortune contestant opted for a rare vowel choice in Bonus Puzzles. While most guests tend to go for more common vowels like “A” and “O,” hoping that they’ll be more prevalent in the puzzle and thus reveal the answer, Danza chose the often-overlooked “U.”
Perhaps subconsciously he recognized the pattern of letters, but once the letters lit up the board, Danza seemed just as pleasantly surprised as host Pat Sajak.
Sajak himself could barely contain his excitement for the guest, knowing full well that Danza would quickly figure out the answer. On the board, the “Thing” phrase read: __ USEFUL DE__ __ CE.
Without a moment’s hesitation, the Wheel of Fortune filled in the remaining three letters of the puzzle, calling out “A Useful Device.”
“Did you know it before you were calling the letters?” Sajak asked incredulously.
“No, I did not,” Danza answered with a huge grin on his face.
Thanks to Danza’s smart picks, he won himself an extra \$39,000 which brought his Wheel of Fortune grand total to \$66,500. On top of his cash prize, the newly contestant also brought home a trip to Maui to Dr. Alyssa waiting for him at home.
Outsider.com | 607 | 2,721 | {"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-2022-21 | longest | en | 0.961687 |
http://algorist.com/algowiki/index.php?title=Help:Trolls_on_wheels!&direction=prev&oldid=247 | 1,590,510,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00386.warc.gz | 5,632,260 | 5,504 | # Help:Trolls on wheels!
Jump to: navigation, search
One counter-example consists of a series of subsets that increase in size exponentially, plus 2 additional subsets that each cover half of the elements. Example:
$S_1 = \{1, 2\}$
$S_2 = \{3, 4, 5, 6\}$
$S_3 = \{7, 8, 9, 10, 11, 12, 13, 14, 15, 16 \}$
$S_4 = \{1, 2, 3, 4, 5, 6, 7, 8 \}$
$S_5 = \{9, 10, 11, 12, 13, 14, 15, 16 \}$
The greedy algorithm will choose $S_3, S_2, S_1$, while the optimal solution is simply $S_4, S_5$ | 214 | 484 | {"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} | 3.171875 | 3 | CC-MAIN-2020-24 | longest | en | 0.865275 |
https://biology.stackexchange.com/questions/78474/monohybrid-recombinant-frequency-three-point-gene-linkage | 1,620,771,337,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00548.warc.gz | 155,397,789 | 37,085 | Monohybrid Recombinant Frequency? (three point Gene Linkage)
I am curious to know if it is completely impossible calculate recombinant frequency for monohybrids ( three point link, gene map etc.)
Here is an example:
Marker 1
1. xx= 25
2. xy=100
3. yy= 20
Marker 2
1. xx=30
2. yx= 89
3. yy= 20
Marker 3
1. xx= 100
2. yx=50
3. yy= 100
This is what I think it would be like for x:x Marker 1 to 3
grand total = 534
So from Marker 1 to Marker 2 it would be 25+30/534 = 0.1029
Marker 2 to Marker 3 : 30 +100/534= 0.2429
Marker 1 to Marker 3: 125/534= 0.2340
Am I on the right track --- if not can you please show how it would be done for this example?
• Can you please clarify what xx, xy and yy are as well as the genotypes of the parents? What are the two loci and the alleles at each locus? – Remi.b Oct 24 '18 at 23:31
• xx and yy are the parents (genotype), the alleles are either x or y – wardah m Oct 25 '18 at 3:30
• How about the other locus, then? You can only talk about recombinants if you consider at least 2 loci. – Remi.b Oct 25 '18 at 4:56
• This information is from a sample SNP call data indicating the to total calls for those genotypes. That's all the information I have is there no way to calculate RF ? – wardah m Oct 25 '18 at 13:59
• No, you can't calculate a recombination rate between two loci with only information about a single locus. If your question is a homework question, please use the homework tag – Remi.b Oct 25 '18 at 17:00 | 466 | 1,470 | {"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} | 2.828125 | 3 | CC-MAIN-2021-21 | latest | en | 0.880519 |
https://allnswers.com/user38094 | 1,621,279,910,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00516.warc.gz | 113,461,826 | 6,856 | Natasha019
since: 2019-02-18 04:13:41
Age: 35
Profiles:
Points: 80
### Questions on other subjects:
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the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375, 390, 405, 420, 435, 450,...Read More | 278 | 699 | {"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} | 2.78125 | 3 | CC-MAIN-2021-21 | latest | en | 0.787427 |
https://electronics.stackexchange.com/questions/501210/how-do-i-find-op-amp-specifications-regarding-voltage-limits | 1,720,931,573,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00270.warc.gz | 202,713,408 | 40,727 | # How do I find op amp specifications regarding voltage limits?
This question is perhaps a bit unusual and is about how to read/interpret specifications. I have been doing some testing with a common difference amp (LM324) and have also used LTSpice to double check that my actual breadboarded results are what the simulation expects. The testing I have been doing is to measure the effect of changing Vcc on the value of Vout. I am comparing 2 input voltages, say 29V and 26V and expect Vout to be 3V (I am using a unity gain configuration with the op amp). The power for the op amp is varied from 27V down to zero and what I see is that as the voltage gets to around 14-15V, the Vout drops from a steady 3V to below 1V. The same holds true if my input voltages are 26V and 23V so I'm clearly hitting some limiting value of the op amp.
My question is this. Where in the LM324 documentation can I find the details that explain this behaviour and how can I use this to predict the behaviour before I start building the circuit? I'm guessing it's something to do with common rail details, but this is only a guess. My motivation is to find out what is the lowest Vcc I can drive my LM324 with if I am measuring a whole range of different input voltages from 28V down to 0V.
As per suggestion, I have now added a screenshot showing the circuit and the test results in LTSpice.
• This post could be much better with a schematic and fewer words since it's a bit rambly. First, there is a conflict because you say you are subtracting two voltages but then say you are using unity-gain buffer configuration so right there we already don't know what you really mean or have as a circuit. Then it sounds like you are giving the op amp 15V of supply but applying a whopping 27.5V common mode input signal to it and expecting it to work. You need to make your post much clearer about what it is you have and what you are doing. Commented May 22, 2020 at 4:11
• Use the snipping tool or the schematic editor on this website. Commented May 22, 2020 at 4:15 | 494 | 2,045 | {"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-2024-30 | latest | en | 0.956903 |
https://nrich.maths.org/public/topic.php?code=-85&cl=4&cldcmpid=332 | 1,571,225,733,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00063.warc.gz | 621,637,806 | 4,416 | # Search by Topic
There are 5 NRICH Mathematical resources connected to Relative velocity, you may find related items under Mechanics.
Broad Topics > Mechanics > Relative velocity
### Take a Message Soldier
##### Age 14 to 18 Challenge Level:
A messenger runs from the rear to the head of a marching column and back. When he gets back, the rear is where the head was when he set off. What is the ratio of his speed to that of the column?
### Which Twin Is Older?
##### Age 16 to 18
A simplified account of special relativity and the twins paradox.
### In Constantly Passing
##### Age 14 to 16 Challenge Level:
A car is travelling along a dual carriageway at constant speed. Every 3 minutes a bus passes going in the opposite direction, while every 6 minutes a bus passes the car travelling in the same. . . .
### Escalator
##### Age 14 to 16 Challenge Level:
At Holborn underground station there is a very long escalator. Two people are in a hurry and so climb the escalator as it is moving upwards, thus adding their speed to that of the moving steps. . . .
### Stonehenge
##### Age 16 to 18 Challenge Level:
Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself. | 287 | 1,259 | {"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} | 2.953125 | 3 | CC-MAIN-2019-43 | latest | en | 0.918538 |
http://www.ks.uiuc.edu/Research/vmd/mailing_list/vmd-l/7769.html | 1,725,849,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00893.warc.gz | 42,201,080 | 6,398 | • ## Outreach
From: Axel Kohlmeyer (akohlmey_at_cmm.chem.upenn.edu)
Date: Fri Sep 08 2006 - 14:00:58 CDT
On 9/8/06, Neelanjana Sengupta <senguptan_at_gmail.com> wrote:
> Hi Axel,
>
> Thanks for those pointers. However I need a clarification:
> Suppose I find the effective matrix responsible for the net transformation
> at each step. The way to apply a matrix transformation is with selections,
> \$sel move \$mat_change
>
> How can I create a selection corresponding to the cell basis vectors? Or,
you create a selection for _all_ atoms (in the unitcell), then move/rotated,
apply PBC via the regular orthogonal box criterion, move/rotate back
and go to the next step. if you want to mainain that molecules stay intact,
you have to act a little smarter. basically, you should be able to take
major parts of the pbcwrap script and just feed it your cell vectors.
axel.
> should I write my original orthogonal cell basis as a matrix, upon which I
> operate \$mat_change? What would be the proper way of writing that matrix?
>
> Best regards,
> Neela
>
>
> On 9/7/06, Axel Kohlmeyer <akohlmey_at_cmm.chem.upenn.edu> wrote:
> >
> On 9/7/06, Neelanjana Sengupta <senguptan_at_gmail.com > wrote:
> > Hi,
> > I have a simulation of pepitde-solvent system. During my analysis, I am
> > positioning the 4 atoms at each peptide bond in a certain manner at the
> > origin and in the X-Y plane.
> > Due to the operations, and my system is not centered the origin anymore,
> and
> > the simulation box get rotated too. So, the minimum image convention is no
> > longer as straightforward as:
> > x_ij = x_ij - boxL*NINT(x_ij/boxL)
> >
>
> you just have to (temporarily) find a transformation so that
> the coordinates are in a simple orthonomal basis again.
> please have a look at the pbcwrap script from the vmd script
> library. you just need to use some of the lowlevel subroutines
> and feed them your shifted origin and rotated cell vectors.
>
> cheers,
> axel.
>
>
> >
> > With thanks,
> > Neela
> >
>
>
> --
> =======================================================================
> Axel Kohlmeyer akohlmey_at_cmm.chem.upenn.edu http://www.cmm.upenn.edu
> Center for Molecular Modeling -- University of Pennsylvania
> Department of Chemistry, 231 S.34th Street, Philadelphia, PA 19104-6323
> tel: 1-215-898-1582, fax: 1-215-573-6233, office-tel: 1-215-898-5425
> =======================================================================
> If you make something idiot-proof, the universe creates a better idiot.
>
>
```--
=======================================================================
Axel Kohlmeyer akohlmey_at_cmm.chem.upenn.edu http://www.cmm.upenn.edu
Center for Molecular Modeling -- University of Pennsylvania
Department of Chemistry, 231 S.34th Street, Philadelphia, PA 19104-6323
tel: 1-215-898-1582, fax: 1-215-573-6233, office-tel: 1-215-898-5425
=======================================================================
If you make something idiot-proof, the universe creates a better idiot.
``` | 803 | 3,024 | {"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-2024-38 | latest | en | 0.866967 |
https://math.stackexchange.com/questions/1299620/laurent-series-in-domain-z0 | 1,566,418,855,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316194.18/warc/CC-MAIN-20190821194752-20190821220752-00276.warc.gz | 552,734,789 | 30,373 | Laurent series in domain $|z|>0$
Find Laurent series, in powers of $z$, of $$f(z)=\frac{\sin(2z)}{z}$$ valid in the region $|z|>0$.
The singularity is $0$ but $0$ isn't inside the region of the domain so what do you exactly expand?
Do you just expand $\sin(2z)$ and then divide it by $z$?
• $f(z)=\frac{\sin(2z)}{z}$ has a removable singularity in zero. Once you remove it you get an entire function: $$2\,\text{sinc}(2z)=\sum_{n\geq 0}\frac{4^{n+1}(-1)^n}{(2n+1)!}z^{2n}.$$ – Jack D'Aurizio May 26 '15 at 15:08
• what is sinc? ??? – snowman May 26 '15 at 15:13
• en.wikipedia.org/wiki/Sinc_function – Jack D'Aurizio May 26 '15 at 15:18
The function's analytic in your region (and almost analytic at $\;z=0\;$ ...), so using the power series for $\;\sin z\;$ which has infinite convergence radius:
$$\frac1z\sin2x=\frac1z\sum_{n=1}^\infty(-1)^{n-1}\frac{(2z)^{2n-1}}{(2n-1)!}=\sum_{n=1}^\infty(-1)^{n-1}\frac{2^{2n-1}z^{2n-2}}{(2n-1)!}$$
• I actually got $$\sum \limits_{n=0}^{\infty} (-1)^n \frac {2^{2n+1}z^{2n}}{(2n+1)!}$$ this is the same thing right? – snowman May 26 '15 at 15:25 | 441 | 1,090 | {"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} | 3.890625 | 4 | CC-MAIN-2019-35 | latest | en | 0.686265 |
https://money.stackexchange.com/questions/102307/should-returns-from-my-sp-index-fund-swppx-match-the-sp-500-index | 1,618,678,365,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00367.warc.gz | 506,105,099 | 37,965 | # Should returns from my S&P Index fund (SWPPX) match the S&P 500 index?
I have been purchasing ~ \$300 worth of SWPPX twice per month since March 2015. My annualized return appears to be 8.22%. I used the annual returns for the S&P500 from here: https://ycharts.com/indicators/sandp_500_total_return_annual to calculate the average return since 2015 as 9.55%. Can anyone help me understand why my return is over 1% less than the average S&P500 return for this period? Am I losing money to fees?
• How did you calculate your annualized return? Are you reinvesting dividends? – D Stanley Nov 21 '18 at 23:07
• Now you're asking the tough q's D Stanley! Truth is, I read my rate of return off my statement. My account provide a 'portfolio performance' output that summarises investment and returns. It states I have net contributions of \$20,272, and a change in value of \$2,604.93 having begun investments 03/23/15. Apparently that equates to 8.15% but I'm unclear how that %age is derived.I am reinvesting dividends. – Joe_P Nov 22 '18 at 2:16
Because you're buying at different times.
Total annual return looks at the value today of \$X invested on Jan 1 2015. But you don't have \$X invested on Jan 1, 2015, you have:
• \$Y on Jan 1, 2015
• \$Y on Jan 15, 2015
• \$Y on Feb 1, 2015
• etc. where the sum of the Y values is X that you're trying to compare
You have a different average unit cost and that impacts your return. Your first contribution will match that return very closely less the expense fee and possibly dividends. Your second contribution hasn't enjoyed all of the time and price appreciation as your first contribution and will have an appropriately different return; in fact the return on your most recent contribution is certainly not 8.22%, it's probably negative.
This is a chart that I made for a different answer but it applies here because it illustrates that each contribution has it's own rate of return. This is a monthly \$100 deposit in to VOO
• Orange - Your very first \$100 (green got lost in the blue)
• Yellow - A deposit that went in and immediately lost value
• Red - Your most valuable \$100. (this \$100 was contributed at the lowest unit cost of the bunch)
• I think I see - that graphic is really helpful. If that graphic was split annually, the average return on all the 100 dollars payments made through the year (24), which would be about the mid range of all the lines at the end of the year, would not be the annual return - is that right? – Joe_P Nov 22 '18 at 1:47 | 634 | 2,523 | {"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} | 2.859375 | 3 | CC-MAIN-2021-17 | longest | en | 0.941385 |
https://wiki.beyondunreal.com/Quaternion?diff=next&oldid=46313 | 1,558,959,389,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262369.94/warc/CC-MAIN-20190527105804-20190527131804-00317.warc.gz | 672,826,797 | 12,824 | Gah - a solution with more questions. – EntropicLqd
# Difference between revisions of "Quaternion"
Quaternions are a mathematical construct that could be seen as an extension of the complex numbers with three different imaginary units or as a combination of a scalar value and a three-dimensional vector. A special subset of the quaternions, the unit quaternions, can be used to represent rotations in 3D space.
## Definition
Generally a quaternion is a four-dimensional value that is written as either x = x0+ x1i + x2j + x3k or as x = (x0, x1, x2, x3), where x0 is called the real or scalar part Re(x) and (x1, x2, x3) is the imaginary or vector part Im(x). If the vector part is zero, the quaternion is called real, while a quaternion with a zero scalar part is called a pure imaginary quaternion. Real quaternions (x0, 0, 0, 0) could also be written as x0+ 0i + 0j + 0k = x0.
Addition, subtraction and negation of quaternions is defined the same way as for vectors:
• x + y = y + x = (x0+ y0, x1+ y1, x2+ y2, x3+ y3)
• x - y = -y + x = (x0- y0, x1- y1, x2- y2, x3- y3)
• -x = (-x0, -x1, -x2, -x3)
The same goes for multiplication with a scalar value and scalar multiplication of two quaternions:
• a · x = x · a = (a · x0, a · x1, a · x2, a · x3)
• x · y = y · x = x0· y0 + x1· y1 + x2· y2 + x3· y3
Like complex numbers, quaternions define the conjugate quaternion x or x*:
• x = (x0, -x1, -x2, -x3)
A quite unique operation of quaternions is the Hamilton product, i.e. the multiplication of two quaternions that yields a quaternion again. This one is special, because unlike addition or scalar product, it is not commutative.
• x y = (x0· y0 - x1· y1 - x2· y2 - x3· y3, x0· y1 + x1· y0 + x2· y3 - x3· y2, x0· y2 - x1· y3 + x2· y0 + x3· y1, x0· y3 + x1· y2 - x2· y1 + x3· y0)
Now that looks vile. But if you express the quaternions as a combination of a scalar and a vector part, i.e. x = (x0, xV), you can express quaternion multiplication using the 3D vector scalar and cross products:
• x y = (x0, xV)(y0, yV) = (x0y0- xV· yV, x0yV + xVy0+ xV⨯ yV)
This representation also easily shows that x y ≠ y x, because xV⨯ yV = -(yV⨯ xV).
A quaternion's norm ||x||2 is defined as:
• ||x||2 = x x = x x = x · x = x02+ x12+ x22+ x32
The value ||x|| is called the length or absolute value of the quaternion. Its square is equal to the scalar product of the quaternion with itself or the Hamilton product with its conjugate quaternion. Multiplying a quaternion by its conjugate results in the vector part becoming zero, which is why the resulting quaternion in the above equation can be considered a scalar number. (Remember that quaternions are hypercomplex numbers, not euclidean vectors. Real numbers are valid quaternion values with the vector part being zero.)
A quaternion's multiplicative inverse, the reciprocal, x-1 is defined as the value that satisfies the equation x x-1 = 1:
• x-1 = x / (x x) = x / ||x||2
Note that the fractional notation is avoided if the denominator is not a real number because the notation itself is ambiguous since it does not specify if the denominator's reciprocal must be multiplied to the left or to the right of the numerator and quaternion multiplication is not commutative.
## Quaternion rotation
The basis for expression rotations using quaternions are unit quaternions, i.e. quaternions with a length of 1, and pure imaginary quaternions, i.e. quaternions of the form (0, x1, x2, x3). Pure imaginary quaternions are used to represent vectors that will be rotated around the origin in some way, while unit quaternions define the rotation axis and angle. To apply a rotation defined by quaternion r to a vector defined by quaternion v, all you need to do is calculate the quaternion product r v r-1.
A property of quaternion multiplication is that the length of the resulting quaternion is the same as the product of the lengths of the two original quaternions. Consequently the product of two unit quaternions will be another unit quaternions. If those two unit quaternions represent 3D rotations, the resulting unit quaternion will represent the combination of these two rotations in the reverse order. The product r1 r2 represents a rotation achieved by first applying r2, then r1.
The vector to rotate can be converted to a quaternion simply by using the vector as the quaternion's vector part and zero as the scalar part. A rotation quaternion is calculated from a rotation axis and an angle. Given the rotation axis as unit vector u and the rotation angle α between 0 and 2π, the rotation quaternion r is calculated as r = (cos α/2, u sin α/2).
## UnrealScript functions
The Object(U2, U2XMP, UE2Runtime, UT2003, UT2004, UDK, UT3) class provides several functions and (in UE3) operators for working with quaternions. It defines the struct Quat with the components W, X, Y and Z as data type for a quaternion value. Unreal Engine 3 also defines the binary operators `+` and `-`, and like with every struct you can compare Quat values for equality and inequality with the `==` and `!=` operators.
The Quat struct type is subject to data reduction during replication. The engine assumes that you replicate only unit quaternions and only replicates the X, Y and Z components. The W component will be calculated on the remote side to get a unit quaternion again.
Quaternion rotation is used in rigid body physics calculations both in Unreal Engine 2 (PHYS_Karma) and Unreal Engine 3 (PHYS_RigidBody). The rigid body physics state of an actor is composed of a vector for the location, a quaternion for the rotation and two vectors for linear and angular velocity.
### QuatProduct(Quat, Quat)
Calculates the Hamilton product of the two quaternions and returns the resulting quaternion value.
### QuatDot(Quat, Quat)
Calculates the scalar product of the two quaternions and returns the resulting float value. This function is not available in Unreal Engine 2.
### QuatInvert(Quat)
Returns the conjugate quaternion. Note that due to x-1 = x / ||x||2 the conjugate quaternion of a unit quaternion also is its reciprocal.
### QuatRotateVector(Quat, Vector)
Rotates the specified vector by the given rotation quaternion and returns the resulting vector value. Note that the result will only be actually the rotated vector if the quaternion is a unit quaternion.
### QuatFindBetween(Vector, Vector)
Calculates the quaternion that will rotate the first vector onto the second. It does so by calculating the rotation axis and angle between the vectors using the cross product and scalar product. Note that it short-circuits if the cross product is zero. In that case, whether the vectors point in the same or opposite directions, it always returns the identity quaternion (1,0,0,0), even though opposite direction vectors would require a rotation by 180° around an arbitrary rotation axis perpendicular to the vectors.
### QuatFromAxisAndAngle(Vector, float)
Creates a rotation quaternion from the specifies axis vector and rotation angle in radians. The vector will be normalized for this calculation but must not be zero.
### QuatFromRotator(Rotator)
Calculates a rotation quaternion that represents the same rotation as the specified Rotator value.
### QuatToRotator(Quat)
Calculates a Rotator value that represents the same rotation as the specified rotation quaternion. This basically applies the quaternion rotation to the X, Y and Z axis vectors and uses the OrthoRotation function on the resulting vectors to calculate the Rotator value.
### QuatSlerp(Quat, Quat, float)
Performs spherical linear interpolation on the two rotation quaternions, using the float value as blend alpha. A value of 0.0 maps to the first quaternion, a value of 1.0 to the second, while 0.5 is halfway between them. | 1,989 | 7,774 | {"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} | 4.09375 | 4 | CC-MAIN-2019-22 | longest | en | 0.844806 |
https://computerstudypoint.com/author/ankush/page/7/ | 1,718,646,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00219.warc.gz | 150,908,365 | 49,989 | Convert the Value Binary into Octal
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https://math.answers.com/Q/6_plus_98_plus_6548 | 1,695,901,736,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00217.warc.gz | 418,998,651 | 45,816 | 0
# 6 plus 98 plus 6548
Wiki User
12y ago
6 + 98 + 6548 = 6,652
Wiki User
12y ago
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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It is 98*6 = 588
98
6 + 2 + 90 = 98
104
### What is the sum of a 6 sided figure?
98 because 6 plus each angle would be 98
### How do you get 6 percent of 98?
6% of 98= 6% * 98= 0.06 * 98= 5.88
132
### What is 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 654 plus 555?
6x12=72+654=726+555= 1281
98
46
### What is 6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6?
6 plus 6 plus 6 plus 6 plus 6 plus 6 plus 6=6 x 7 =427 x 6 = 426+6+6+6+6+6+6=6 x 7= 42
36:) | 351 | 846 | {"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} | 3.5625 | 4 | CC-MAIN-2023-40 | latest | en | 0.63678 |
https://www.globalpeforum2014.co.za/experience/9749/how-to-calculate-shuttering-area.html | 1,563,621,727,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526508.29/warc/CC-MAIN-20190720111631-20190720133631-00152.warc.gz | 700,923,619 | 6,598 | how to calculate shuttering area
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RATE ANALYSIS FOR SHUTTERING OF FLOOR BEAM /SLAB , Shuttering Area 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Vertical cuplock support length , Cost Material Calculator... | 1,375 | 6,255 | {"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-2019-30 | latest | en | 0.77482 |
http://www.slate.com/blogs/quora/2016/03/09/how_much_would_it_cost_to_buy_one_of_everything_on_amazon.html | 1,464,468,746,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278091.17/warc/CC-MAIN-20160524002118-00223-ip-10-185-217-139.ec2.internal.warc.gz | 808,940,804 | 25,897 | How much would it cost to buy one of everything on Amazon?
# How Much Would It Cost to Buy One of Everything on Amazon?
The best answer to any question.
March 9 2016 7:36 AM
# How Much Would It Cost to Buy One of Everything on Amazon?
This question originally appeared on Quora, the knowledge-sharing network where compelling questions are answered by people with unique insights. You can follow Quora on Twitter, Facebook, and Google Plus.
Answer by Kynan Eng, brain-related research and commercialization at universities and startups:
It would cost about \$12.86 billion to buy one of everything on Amazon. I calculated this value as follows:
• According to this link, the main Amazon website offered 488 million items in 2015. Of these, an estimated 479 million are available.
• I estimate the average price of an item on Amazon is \$26.86, with an uncertainty (standard error of the mean) of \$6.03.
To estimate the average price of all Amazon items, I followed this procedure:
1. Click on Random Amazon Product. As the domain name implies, this site gives you a link to a random Amazon product. No, I don't know why this site exists either.
2. Discard the product if it is free or not available, but keep count of it.
3. If more than one price is available, select the product price highlighted in red (this is the one that Amazon probably wants you to buy). If that is not available, select the first price listed looking from top to bottom, then from left to right.
4. Repeat until 100 nonzero items are collected. Yes, I really wasted my time doing this. I thought about trying to crowdsource this data collection, or writing a bot to do it, but that would have been more work. | 382 | 1,694 | {"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-2016-22 | longest | en | 0.936426 |
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C2 help - Chemistry AQA GCSE
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1. 2 questions from the textbook that I'm really stuck on...
1. A sample of a compound contains 3.9g of Potassium, 1.2g of Carbon, and 1.4g of nitrogen. What is its formula?
(I keep on getting really large numbers)
2. A compound is made up of 27% sodium by mass, 16% nitrogen by mass, and 56% oxygen by mass. What is its formula?
Thanks!
2. (Original post by JenniMusic)
2 questions from the textbook that I'm really stuck on...
1. A sample of a compound contains 3.9g of Potassium, 1.2g of Carbon, and 1.4g of nitrogen. What is its formula?
(I keep on getting really large numbers)
2. A compound is made up of 27% sodium by mass, 16% nitrogen by mass, and 56% oxygen by mass. What is its formula?
Thanks!
To do this type of question you must divide the percentages by their relative masses then divide through by the smallest number.
If this does not give you integers you have to either divide or multiply all by a suitable number which will give you all integers.
3. (Original post by charco)
To do this type of question you must divide the percentages by their relative masses then divide through by the smallest number.
If this does not give you integers you have to either divide or multiply all by a suitable number which will give you all integers.
I've tried that, but the integers are really large (over 200), and all of the other questions in this section I've done have much smaller integers (2, 4)
4. A sample of a compound contains 3.9g of Potassium, 1.2g of Carbon, and 1.4g of nitrogen. What is its formula
K
3.9/39 = 0.1
C
1.2/12 = 0.1
N
1.4/14 = 0.1
Therefore ratio is 11
formula is KCN
5. A compound is made up of 27% sodium by mass, 16% nitrogen by mass, and 56% oxygen by mass. What is its formula
Na:
27/23 = 1.17
N:
16/14 = 1.14
O:
56/16 = 3.5
Divide by smallest number:
ratio is: 1 3
therefore formula is NANO3
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Students typically take algebra in eighth or ninth grade. An important benefit of studying algebra in eighth grade is that if your child takes the PSAT as a high school sophomore, she will have taken geometry as a ninth grader. By the time she’s ready to take the SAT or ACT as a junior, she will have completed Algebra II, which is covered in both of these college admissions tests. There’s a growing movement to require algebra in seventh grade, but math educators say many seventh graders aren’t prepared for it.
Algebra I isn’t the first step toward math success — students begin exploring algebraic reasoning in kindergarten (and, ideally, even in preschool). Researchers say that a powerful way to help your child build a strong foundation in math is by encouraging them to develop a positive mindset about math.
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Algebra is a field of mathematics. Usually, students in high school or elementary will be the first ones who will experience this subject. Most of them will say that it is probably one of the hardest and complicated subjects there is. Well, anything that is connected to Mathematics could really be. When someone will say the word Algebra out loud, numbers and equations will immediately pop into ones mind. What they do not usually know is what and who and how Algebra started. A brief history of Algebra will be read in this article, to understand why and how and who started Algebra in the first place.
Algebra helps you think logically, Studying algebra helps your mind to think logically and break down and solve problems. One day you might reach a point where you don’t use algebra on a daily basis. However your brain will have been trained to think in a logical way, which will not only help you in the workplace, but also in daily life, when choosing which mobile phone contract to select, or trying to work out if you have paid the right amount of tax. Modern technology needs algebra, The fact is that all modern technology relies on mathematics and algebra – Google, the internet, mobile phones, satellites and digital televisions wouldn’t exist without algebra. You are relying on other people having studied algebra when you use a phone or play a computer game and as technology is everywhere more and more people are needed to work behind the scenes with knowledge of mathematics and algebra. If you like algebra, then you are giving yourself a chance of getting a job in the rapidly expanding technology sector.
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A194325 Triangular array: g(n,k)=number of fractional parts (i*r) in interval [(k-1)/n, k/n], for 1<=i<=n, 1<=k<=n, r=2-sqrt(2). 2
%I
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,0,2,1,
%T 1,1,1,1,1,1,1,1,2,0,1,0,2,1,1,0,2,1,1,1,1,0,2,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,0,1,1,1,1,2,1,1,0,1,1
%N Triangular array: g(n,k)=number of fractional parts (i*r) in interval [(k-1)/n, k/n], for 1<=i<=n, 1<=k<=n, r=2-sqrt(2).
%C See A194285.
%e First twelve rows:
%e 1
%e 1..1
%e 1..1..1
%e 1..1..1..1
%e 1..1..1..1..1
%e 0..1..1..2..1..1
%e 1..1..1..1..1..1..1
%e 1..1..1..0..2..1..1..1
%e 1..1..1..1..1..1..2..0..1
%e 0..2..1..1..0..2..1..1..1..1
%e 0..2..1..1..1..1..1..1..1..1..1
%e 1..1..1..1..1..1..1..1..1..1..1..1
%t r = 2-Sqrt[2];
%t f[n_, k_, i_] := If[(k - 1)/n <= FractionalPart[i*r] < k/n, 1, 0]
%t g[n_, k_] := Sum[f[n, k, i], {i, 1, n}]
%t TableForm[Table[g[n, k], {n, 1, 14}, {k, 1, n}]]
%t Flatten[%] (* A194325 *)
%Y Cf. A194285.
%K nonn,tabl
%O 1,19
%A _Clark Kimberling_, Aug 22 2011
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Last modified September 24 14:46 EDT 2021. Contains 347643 sequences. (Running on oeis4.) | 798 | 1,593 | {"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-2021-39 | latest | en | 0.411791 |
https://edutube.hccs.edu/tag?tagid=fuss | 1,716,654,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00216.warc.gz | 184,151,728 | 31,341 | # Search for tag: "fuss"
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From Douglas Bump 0 likes 0 | 520 | 1,330 | {"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-2024-22 | latest | en | 0.63187 |
https://mathoverflow.net/questions/157732/is-the-forcing-relation-defined-for-mathematical-formulas/157756 | 1,628,139,524,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155322.12/warc/CC-MAIN-20210805032134-20210805062134-00267.warc.gz | 387,190,576 | 32,653 | # Is the forcing relation defined for mathematical formulas?
Meta-matematical formulas of the language of set-theory (which are not sets, but just sequences of signs) should not be confused with mathematical ones (i.e. formulas coded as sets, e.g. finite sequences of natural numbers or even just natural numbers, if we wish). For instance, for each meta-mathematical formula $\phi$, we can define its relativization $\phi^M$ to any class $M$, but the relation $M\vDash \phi[v]$ in the model-theoretic sense can be only defined when $M$ is a set, but not a proper class.
The impossibility of defining $M\vDash\phi[v]$ for proper classes is due to two kinds of reasons: on one hand, if so, we could prove Con ZFC within ZFC, and technically, the problem is that a "typical" recurrence definition of $M\vDash \phi[v]$ would require having defined $M\vDash \psi[w]$ for each subformula $\psi$ of $\phi$ and for each valuation $w$ from the set of free variables of $\psi$ to $M$, and such valuations form a proper class if $M$ is a proper class, and this does not fit the recursion theorem.
My first question is that I suspect that this is also valid for the definition of the forcing relation $p\Vdash \phi$, i.e.:
Question 1: Can the forcing relation $p\Vdash \phi$ be defined for mathematical formulas $\phi$ or just for meta-mathematical ones?
I suspect that only for meta-mathematical ones because in order to define $p\Vdash \forall x\phi(x)$ we must assume that $p\Vdash \phi(\tau)$ is defined for each $\mathbb P$-name $\tau$, and the class of $\mathbb P$-names is a proper class.
This question is related to theorem III 2.11 of Shelah's book Proper and Improper Forcing (second edition). It states that if $\lambda$ is an uncountable cardinal, $N\prec H(\lambda)$ is an elementary submodel of the set of all sets whose transitive closuse has cardinality less than $\lambda$, and $\mathbb P\in N$ is a pre-ordered set, then, for each generic filter $G$ over $V$ we have $N[G]\prec H^{V[G]}(\lambda)$.
The proof is clear, but in order to show that the definition of elementary submodel holds for a formula $\phi$ (in fact, the Tarski-Vaught criterion is considered instead of the definition), the forcing relation $p\Vdash \psi$ is used for a formula $\psi$ involving $\phi$. Hence, if the answer to Question 1 is that it only makes sense for meta-mathematical formulas, we conclude that $N[G]$ is an elementary submodel just in the weak meta-mathematical sense that we have a family of theorems, each one stating that an arbitrary meta-mathematical formula is absolute for $N[G]-H^{V[G]}(\lambda)$.
So, my second question is:
Question 2: Under the hypotheses of Shelah's theorem, can we conclude that $N[G]\prec H^{V[G]}(\lambda)$ in the usual model-theoretic sense (for mathematical formulas) or just in the weak meta-mathematical sense?
I feel that the answer should be positive even if the answer to question 1 is negative.
To answer Question 2, I think your intuition is right. Assume $N \prec H_\lambda$, where $\lambda$ is a regular cardinal bigger than $2^\mathbb{P}$, where $\mathbb{P} \in N$ is a partial order. (Maybe Shelah has a more subtle argument where we assume less about $\lambda$, not sure.) Then $H_\lambda$ is a model of $ZFC$ minus powerset, and we can consider in $V[G]$ the model $H_\lambda[G]$, defined the same way as usual. We have:
• For any generic $G \subseteq \mathbb{P}$, $H_\lambda^{V[G]} = H_\lambda[G]$.
• $G$ is generic over $V$ iff $G$ is generic over $H_\lambda$.
• The basic forcing lemmas, including the "truth lemma," go through for forcing with $\mathbb{P}$ over $H_\lambda$.
The point is that by working in $V[G]$, we can treat $H_\lambda[G]$ as an ordinary set model and carry out the argument that $N[G] \prec H_\lambda[G]$ within the language of set theory in $V[G]$, rather than as a meta-mathematical scheme.
• Thank you. The problem is that in order to use the hypothesis that $N\prec H_\lambda$ to prove $N[G]\prec H_\lambda[G]$, using $\Vdash$ seems unavoidable, and, in view of the answer to Question 1, it seems we should restrict ourselves to consider meta-mathematical formulas. – Carlos Feb 16 '14 at 21:41
• Yes, but we instead use $\Vdash^{H_\lambda}_\mathbb{P}$ and avoid meta-mathematical worries. – Monroe Eskew Feb 16 '14 at 22:04
• To amplify Monroe's point: The forcing predicate (for mathematical formulas) behaves just like the satisfaction predicate. Over proper classes, it can't be defined fully, but over set models (like $H_\lambda$) it can. So what Shelah is doing is fine, uniformly for all mathematical formulas. – Andreas Blass Feb 16 '14 at 22:24
• I see! For each mathematical formula $\phi$, we can define another mathematical formula $p\Vdash\phi$ in such a way that $M\vDash p\Vdash \phi$ ($M$ a set) satisfies the fundamental theorem. Thanks! – Carlos Feb 16 '14 at 22:39
Although the common slogan about forcing is that the forcing relation is definable in the ground model, nevertheless the slogan must be qualified for precisely the issue in your question, and the answer to question 1 is that no, the forcing relation is never definable in full generality as a binary relation in $(p,\varphi)$. That is, if $M$ is a model of ZFC and $\mathbb{P}$ is any forcing notion in $M$, then the relation $p\Vdash\varphi$, as a relation in $(p,\varphi)$, where $p$ is a condition in $\mathbb{P}$ and $\varphi$ is an assertion in the forcing language, is not definable in $M$.
One can prove this using the fact that the ground model $M$ is (uniformly) definable in its forcing extensions $M[G]$, using a parameter in $M$. (This was recently proved by Laver, independently Woodin.) So if we could define the forcing relation in $M$ then we could define in $M$ the relation $\mathbb{1}\Vdash\varphi^M(\check x)$, which is equivalent to $M\models\varphi[x]$. But this is not definable by Tarski's theorem on the non-definability of truth.
Meanwhile, the slogan survives in a way that is strong enough to be extremely useful:
• For any fixed $\varphi(x)$ with a free variable, the relation $p\Vdash\varphi(\tau)$ is definable in $M$ as a relation in $(p,\tau)$. This is how the definability is most often used.
• For any fixed (meta-theoretic) level of complexity $\Sigma_n$, the relation $p\Vdash\varphi$ for conditions $p$ and $\Sigma_n$ formulas $\varphi$ is definable in $M$ as a relation in $(p,\varphi)$. Indeed, the quantifiers of $\varphi$ amount essentially to quantifying over names in $M$, and so one can make a tight connection between the complexity of $\varphi$ and the complexity of $p\Vdash\varphi$. | 1,758 | 6,607 | {"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} | 2.625 | 3 | CC-MAIN-2021-31 | latest | en | 0.885623 |
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# Sound Waves and Music - Lesson 1
## The Nature of a Sound Wave
Sound is a Mechanical Wave | Sound is a Longitudinal Wave | Sound is a Pressure Wave
## Sound is a Mechanical Wave
Sound and music are parts of our everyday sensory experience. Just as humans have eyes for the detection of light and color, so we are equipped with ears for the detection of sound. We seldom take the time to ponder the characteristics and behaviors of sound and the mechanisms by which sounds are produced, propagated, and detected. The basis for an understanding of sound, music and hearing is the physics of waves. Sound is a wave that is created by vibrating objects and propagated through a medium from one location to another. In this unit, we will investigate the nature, properties and behaviors of sound waves and apply basic wave principles towards an understanding of music. As discussed in the previous unit of The Physics Classroom Tutorial, a wave can be described as a disturbance that travels through a medium, transporting energy from one location to another location. The medium is simply the material through which the disturbance is moving; it can be thought of as a series of interacting particles. The example of a slinky wave is often used to illustrate the nature of a wave. A disturbance is typically created within the slinky by the back and forth movement of the first coil of the slinky. The first coil becomes disturbed and begins to push or pull on the second coil. This push or pull on the second coil will displace the second coil from its equilibrium position. As the second coil becomes displaced, it begins to push or pull on the third coil; the push or pull on the third coil displaces it from its equilibrium position. As the third coil becomes displaced, it begins to push or pull on the fourth coil. This process continues in consecutive fashion, with each individual particle acting to displace the adjacent particle. Subsequently the disturbance travels through the slinky. As the disturbance moves from coil to coil, the energy that was originally introduced into the first coil is transported along the medium from one location to another. A sound wave is similar in nature to a slinky wave for a variety of reasons. First, there is a medium that carries the disturbance from one location to another. Typically, this medium is air, though it could be any material such as water or steel. The medium is simply a series of interconnected and interacting particles. Second, there is an original source of the wave, some vibrating object capable of disturbing the first particle of the medium. The disturbance could be created by the vibrating vocal cords of a person, the vibrating string and soundboard of a guitar or violin, the vibrating tines of a tuning fork, or the vibrating diaphragm of a radio speaker. Third, the sound wave is transported from one location to another by means of particle-toparticle interaction. If the sound wave is moving through air, then as one air particle is displaced from its equilibrium position, it exerts a push or pull on its nearest neighbors, causing them to be displaced from their equilibrium position. This particle interaction continues throughout the entire medium, with each particle interacting and causing a disturbance of its nearest neighbors. Since a sound wave is a disturbance that is transported through a medium via the mechanism of particle-to-particle interaction, a sound wave is characterized as a mechanical wave. The creation and propagation of sound waves are often demonstrated in class through the use of a tuning fork. A tuning fork is a metal object consisting of two tines capable of vibrating if struck by a rubber hammer or mallet. As the tines of the tuning forks vibrate back and forth, they begin to disturb surrounding air molecules. These disturbances are passed on to adjacent air molecules by the mechanism of particle interaction. The motion of the disturbance, originating at the tines of the tuning fork and traveling through the medium (in this case, air) is what is referred to as a sound wave. The generation and propagation of a sound wave is demonstrated in the animation below.
Many Physics demonstration tuning forks are mounted on a sound box. In such instances, the vibrating tuning fork, being connected to the sound box, sets the sound box into vibrational motion. In turn, the sound box, being connected to the air inside of it, sets the air inside of the sound box into vibrational motion. As the tines of the tuning fork, the structure of the sound box, and the air inside of the sound box begin vibrating at the same frequency, a louder sound is produced. In fact, the more particles that can be made to vibrate, the louder or more amplified the sound. This concept is often demonstrated by the placement of a vibrating tuning fork against the glass panel of an overhead projector or on the wooden door of a cabinet. The vibrating tuning fork sets the glass panel or wood door into vibrational motion and results in an amplified sound. We know that a tuning fork is vibrating because we hear the sound that is produced by its vibration. Nonetheless, we do not actually visibly detect any vibrations of the tines. This is because the tines are vibrating at a very high frequency. If the tuning fork that is being used corresponds to middle C on the piano keyboard, then the tines are vibrating at a frequency of 256 Hertz; that is, 256 vibrations per second. We are unable to visibly detect vibrations of such high frequency. A common physics demonstration involves slowing down the vibrations by through the use of a strobe light. If the strobe light puts out a flash of light at a frequency of 512 Hz (two times the frequency of the tuning fork), then the tuning fork can be observed to be moving in a back and forth motion. With the room darkened, the strobe would allow us to view the position of the tines two times during their vibrational cycle. Thus we would see the tines when they are displaced far to the left and again when they are displaced far to the right. This would be convincing proof that the tines of the tuning fork are indeed vibrating to produce sound. In a previous unit of The Physics Classroom Tutorial, a distinction was made between two categories of waves: mechanical waves and electromagnetic waves. Electromagnetic waves are waves that have an electric and magnetic nature and are capable of traveling through a vacuum. Electromagnetic waves do not require a medium in order to transport their energy. Mechanical waves are waves that require a medium in order to transport their energy from one location to another. Because mechanical waves rely on particle interaction in order to transport their energy, they cannot travel through regions of space that are void of particles. That is, mechanical waves cannot travel through a vacuum. This feature of mechanical waves is often demonstrated in a Physics class. A ringing bell is placed in a jar and air inside the jar is evacuated. Once air is removed from the jar, the sound of the ringing bell can no longer be heard. The clapper is seen striking the bell; but the sound that it produces cannot be heard because there are no particles inside of the jar to transport the disturbance through the vacuum. Sound is a mechanical wave and cannot travel through a vacuum.
## Check Your Understanding
1. A sound wave is different than a light wave in that a sound wave is a. produced by an oscillating object and a light wave is not. b. not capable of traveling through a vacuum. c. not capable of diffracting and a light wave is. d. capable of existing with a variety of frequencies and a light wave has a single frequency.
Answer: B, because sound is a mechanical wave and cannot travel through a vacuum. Light is an
electromagnetic wave and can travel through the vacuum of outer space.
## Sound as a Longitudinal Wave
In the first part of Lesson 1, it was mentioned that sound is a mechanical wave that is created by a vibrating object. The vibrations of the object set particles in the surrounding medium in vibrational motion, thus transporting energy through the medium. For a sound wave traveling through air, the vibrations of the particles are best described as longitudinal. Longitudinal waves are waves in which the motion of the individual particles of the medium is in a direction that is parallel to the direction of energy transport. A longitudinal wave can be created in a slinky if the slinky is stretched out in a horizontal direction and the first coils of the slinky are vibrated horizontally. In such a case, each individual coil of the medium is set into vibrational motion in directions parallel to the direction that the energy is transported.
Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves. A vibrating string can create longitudinal waves as depicted in the animation below. As the vibrating string moves in the forward direction, it begins to push upon surrounding air molecules, moving them to the right towards their nearest neighbor. This causes the air molecules to the right of the string to be compressed into a small region of space. As the vibrating string moves in the reverse direction (leftward), it lowers the pressure of the air immediately to its right, thus causing air molecules to move back leftward. The lower pressure to the right of the string causes air molecules in that region immediately to the right of the string to expand into a large region of space. The back and forth vibration of the string causes individual air molecules (or a layer of air molecules) in the region immediately to the right of the string to continually vibrate back and forth horizontally. The molecules move rightward as the string moves rightward and then leftward as the string moves leftward. These back and forth vibrations are imparted to adjacent neighbors by particle-to-particle interaction. Other surrounding particles begin to move rightward and leftward, thus sending a wave to the right. Since air molecules (the particles of the medium) are moving in a direction that is parallel to the direction that the wave moves, the sound wave is referred to as a longitudinal wave. The result of such longitudinal vibrations is the creation of compressions and rarefactions within the air.
Regardless of the source of the sound wave - whether it is a vibrating string or the vibrating tines of a tuning fork - sound waves traveling through air are longitudinal waves. And the essential characteristic of a longitudinal wave that distinguishes it from other types of waves is that the particles of the medium move in a direction parallel to the direction of energy transport.
## Sound is a Pressure Wave
Sound is a mechanical wave that results from the back and forth vibration of the particles of the medium through which the sound wave is moving. If a sound wave is moving from left to right through air, then particles of air will be displaced both rightward and leftward as the energy of the sound wave passes through it. The motion of the particles is parallel (and anti-parallel) to the direction of the energy transport. This is what characterizes sound waves in air as longitudinal waves. A vibrating tuning fork is capable of creating such a longitudinal wave. As the tines of the fork vibrate back and forth, they push on neighboring air particles. The forward motion of a tine pushes air molecules horizontally to the right and the backward retraction of the tine creates a low-pressure area allowing the air particles to move back to the left.
Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively. The compressions are regions of high air pressure while the rarefactions are regions of low air pressure. The diagram below depicts a sound wave created by a tuning fork and propagated through the air in an open tube. The compressions and rarefactions are labeled.
The wavelength of a wave is merely the distance that a disturbance travels along the medium in one complete wave cycle. Since a wave repeats its pattern once every wave cycle, the wavelength is sometimes referred to as the length of the repeating patterns - the length of one complete wave. For a transverse wave, this length is commonly measured from one wave crest to the next adjacent wave crest or from one wave trough to the next adjacent wave trough. Since a longitudinal wave does not contain crests and troughs, its wavelength must be measured differently. A longitudinal wave consists of a repeating pattern of compressions and rarefactions. Thus, the wavelength is commonly measured as the distance from one compression to the next adjacent compression or the distance from one rarefaction to the next adjacent rarefaction. Since a sound wave consists of a repeating pattern of high-pressure and low-pressure regions moving through a medium, it is sometimes referred to as a pressure wave. If a detector, whether it is the human ear or a man-made instrument, were used to detect a sound wave, it would detect fluctuations in pressure as the sound wave impinges upon the detecting device. At one instant in time, the detector would detect a high pressure; this would correspond to the arrival of a compression at the detector site. At the next instant in time, the detector might detect normal pressure. And then finally a low pressure would be detected, corresponding to the arrival of a rarefaction at the detector site. The fluctuations in pressure as detected by the detector occur at periodic and regular time intervals. In fact, a plot of pressure versus time would appear as a sine curve. The peak points of the sine curve correspond to compressions; the low points correspond to rarefactions; and the "zero points" correspond to the pressure that the air would have if there were no disturbance moving through it. The diagram below depicts the correspondence between the longitudinal nature of a sound wave in air and the pressure-time fluctuations that it creates at a fixed detector location.
The above diagram can be somewhat misleading if you are not careful. The representation of sound by a sine wave is merely an attempt to illustrate the sinusoidal nature of the pressure-time fluctuations. Do not conclude that sound is a transverse wave that has crests and troughs. Sound waves traveling through air are indeed longitudinal waves with compressions and rarefactions. As sound passes through air (or any fluid medium), the particles of air do not vibrate in a transverse manner. Do not be misled - sound waves traveling through air are longitudinal waves.
## Check Your Understanding
1. A sound wave is a pressure wave; regions of high (compressions) and low pressure (rarefactions) are established as the result of the vibrations of the sound source. These compressions and rarefactions result because sound a. is more dense than air and thus has more inertia, causing the bunching up of sound. b. waves have a speed that is dependent only upon the properties of the medium. c. is like all waves; it is able to bend into the regions of space behind obstacles. d. is able to reflect off fixed ends and interfere with incident waves e. vibrates longitudinally; the longitudinal movement of air produces pressure fluctuations.
## Sound Properties and Their Perception
Pitch and Frequency | Intensity and the Decibel Scale | The Speed of Sound | The Human Ear
## Pitch and Frequency
A sound wave, like any other wave, is introduced into a medium by a vibrating object. The vibrating object is the source of the disturbance that moves through the medium. The vibrating object that creates the disturbance could be the vocal cords of a person, the vibrating string and sound board of a guitar or violin, the vibrating tines of a tuning fork, or the vibrating diaphragm of a radio speaker. Regardless of what vibrating object is creating the sound wave, the particles of the medium through which the sound moves is vibrating in a back and forth motion at a given frequency. The frequency of a wave refers to how often the particles of the medium vibrate when a wave passes through the medium. The frequency of a wave is measured as the number of complete back-and-forth vibrations of a particle of the medium per unit of time. If a particle of air undergoes 1000 longitudinal vibrations in 2 seconds, then the frequency of the wave would be 500 vibrations per second. A commonly used unit for frequency is the Hertz (abbreviated Hz), where 1 Hertz = 1 vibration/second As a sound wave moves through a medium, each particle of the medium vibrates at the same frequency. This is sensible since each particle vibrates due to the motion of its nearest neighbor. The first particle of the medium begins vibrating, at say 500 Hz, and begins to set the second particle into vibrational motion at the same frequency of 500 Hz. The second particle begins vibrating at 500 Hz and thus sets the third particle of the medium into vibrational motion at 500 Hz. The process continues throughout the medium; each particle vibrates at the same frequency. And of course the frequency at which each particle vibrates is the same as the frequency of the original source of the sound wave. Subsequently, a guitar string vibrating at 500 Hz will set the air particles in the room vibrating at the same frequency of 500 Hz, which carries a sound signal to the ear of a listener, which is detected as a 500 Hz sound wave. The back-and-forth vibrational motion of the particles of the medium would not be the only observable phenomenon occurring at a given frequency. Since a sound wave is a pressure wave, a detector could be used to detect oscillations in pressure from a high pressure to a low pressure and back to a high pressure. As the compressions (high pressure) and rarefactions (low pressure) move through the medium, they would reach the detector at a given frequency. For example, a compression would reach the detector 500 times per second if the frequency of the wave were 500 Hz. Similarly, a rarefaction would reach the detector 500 times per second if the frequency of the wave were 500 Hz. The frequency of a sound wave not only refers to the number of back-and-forth vibrations of the particles per unit of time, but also refers to the number of compressions or rarefactions that pass a given point per unit of time. A detector could be used to detect the frequency of these pressure oscillations over a given period of time. The typical output provided by such a detector is a pressure-time plot as shown below.
Since a pressure-time plot shows the fluctuations in pressure over time, the period of the sound wave can be found by measuring the time between successive high pressure points (corresponding to the compressions) or the time between successive low pressure points (corresponding to the rarefactions). As discussed in an earlier unit, the frequency is simply the reciprocal of the period. For this reason, a sound wave with a high frequency would correspond to a pressure time plot with a small period - that is, a plot corresponding to a
small amount of time between successive high pressure points. Conversely, a sound wave with a low frequency would correspond to a pressure time plot with a large period - that is, a plot corresponding to a large amount of time between successive high pressure points. The diagram below shows two pressure-time plots, one corresponding to a high frequency and the other to a low frequency.
The ears of a human (and other animals) are sensitive detectors capable of detecting the fluctuations in air pressure that impinge upon the eardrum. The mechanics of the ear's detection ability will be discussed later in this lesson. For now, it is sufficient to say that the human ear is capable of detecting sound waves with a wide range of frequencies, ranging between approximately 20 Hz to 20 000 Hz. Any sound with a frequency below the audible range of hearing (i.e., less than 20 Hz) is known as an infrasound and any sound with a frequency above the audible range of hearing (i.e., more than 20 000 Hz) is known as an ultrasound. Humans are not alone in their ability to detect a wide range of frequencies. Dogs can detect frequencies as low as approximately 50 Hz and as high as 45 000 Hz. Cats can detect frequencies as low as approximately 45 Hz and as high as 85 000 Hz. Bats, being nocturnal creature, must rely on sound echolocation for navigation and hunting. Bats can detect frequencies as high as 120 000 Hz. Dolphins can detect frequencies as high as 200 000 Hz. While dogs, cats, bats, and dolphins have an unusual ability to detect ultrasound, an elephant possesses the unusual ability to detect infrasound, having an audible range from approximately 5 Hz to approximately 10 000 Hz. The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. Amazingly, many people, especially those who have been musically trained, are capable of detecting a difference in frequency between two separate sounds that is as little as 2 Hz. When two sounds with a frequency difference of greater than 7 Hz are played simultaneously, most people are capable of detecting the presence of a complex wave pattern resulting from the interference and superposition of the two sound waves. Certain sound waves when played (and heard) simultaneously will produce a particularly pleasant sensation when heard, are said to be consonant. Such sound waves form the basis of intervals in music. For example, any two sounds whose frequencies make a 2:1 ratio are said to be separated by an octave and result in a particularly pleasing sensation when heard. That is, two sound waves sound good when played together if one sound has twice the frequency of the other. Similarly two sounds with a frequency ratio of 5:4 are said to be separated by an interval of a third; such sound waves also sound good when played together. Examples of other sound wave intervals and their respective frequency ratios are listed in the table below. Interval Octave Third Fourth Fifth Frequency Ratio 2:1 5:4 4:3 3:2 Examples Hz and 256 Hz and 256 Hz and 256 Hz and 256
## 512 320 342 384
Hz Hz Hz Hz
The ability of humans to perceive pitch is associated with the frequency of the sound wave that impinges upon the ear. Because sound waves traveling through air are longitudinal waves that produce high- and lowpressure disturbances of the particles of the air at a given frequency, the ear has an ability to detect such frequencies and associate them with the pitch of the sound. But pitch is not the only property of a sound wave detectable by the human ear. In the next part of Lesson 2, we will investigate the ability of the ear to perceive the intensity of a sound wave.
## Intensity and the Decibel Scale
Sound waves are introduced into a medium by the vibration of an object. For example, a vibrating guitar string forces surrounding air molecules to be compressed and expanded, creating a pressure disturbance consisting of an alternating pattern of compressions and rarefactions. The disturbance then travels from particle to particle through the medium, transporting energy as it moves. The energy that is carried by the disturbance was originally imparted to the medium by the vibrating string. The amount of energy that is transferred to the medium is dependent upon the amplitude of vibrations of the guitar string. If more energy is put into the plucking of the string (that is, more work is done to displace the string a greater amount from its rest position), then the string vibrates with a greater amplitude. The greater amplitude of vibration of the guitar string thus imparts more energy to the medium, causing air particles to be displaced a greater distance from their rest position. Subsequently, the amplitude of vibration of the particles of the medium is increased, corresponding to an increased amount of energy being carried by the particles. This relationship between energy and amplitude was discussed in more detail in a previous unit. The amount of energy that is transported past a given area of the medium per unit of time is known as the intensity of the sound wave. The greater the amplitude of vibrations of the particles of the medium, the greater the rate at which energy is transported through it, and the more intense that the sound wave is. Intensity is the energy/time/area; and since the energy/time ratio is equivalent to the quantity power, intensity is simply the power/area.
Typical units for expressing the intensity of a sound wave are Watts/meter2. As a sound wave carries its energy through a two-dimensional or three-dimensional medium, the intensity of the sound wave decreases with increasing distance from the source. The decrease in intensity with increasing distance is explained by the fact that the wave is spreading out over a circular (2 dimensions) or spherical (3 dimensions) surface and thus the energy of the sound wave is being distributed over a greater surface area. The diagram at the right shows that the sound wave in a 2-dimensional medium is spreading out in space over a circular pattern. Since energy is conserved and the area through which this energy is transported is increasing, the power (being a quantity that is measured on a per area basis) must decrease. The mathematical relationship between intensity and distance is sometimes referred to as an inverse square relationship. The intensity varies inversely with the square of the distance from the source. So if the distance from the source is doubled (increased by a factor of 2), then the intensity is quartered (decreased by a factor of 4). Similarly, if the distance from the source is quadrupled, then the intensity is decreased by a factor of 16. Applied to the diagram at the right, the intensity at point B is one-fourth the intensity as point A and the intensity at point C is one-sixteenth the intensity at point A. Since the intensity-distance relationship is an inverse relationship, an increase in one quantity corresponds to a decrease in the other quantity. And since the intensity-distance relationship is an inverse square relationship, whatever factor by which the distance is increased, the intensity is decreased by a factor equal to the square of the distance change factor. The sample data in the table below illustrate the inverse square relationship between power and distance.
Distance
1 2 3 4 m m m m
Intensity
160 units 40 units 17.8 units 10 units
Humans are equipped with very sensitive ears capable of detecting sound waves of extremely low intensity. The faintest sound that the typical human ear can detect has an intensity of 1*10-12 W/m2. This intensity corresponds to a pressure wave in which a compression of the particles of the medium increases the air pressure in that compressional region by a mere 0.3 billionth of an atmosphere. A sound with an intensity of 1*10-12 W/m2 corresponds to a sound that will displace particles of air by a mere one-billionth of a centimeter. The human ear can detect such a sound. WOW! This faintest sound that a human ear can detect is known as the threshold of hearing. The most intense sound that the ear can safely detect without suffering any physical damage is more than one billion times more intense than the threshold of hearing. Since the range of intensities that the human ear can detect is so large, the scale that is frequently used by physicists to measure intensity is a scale based on multiples of 10. This type of scale is sometimes referred to as a logarithmic scale. The scale for measuring intensity is the decibel scale. The threshold of hearing is assigned a sound level of 0 decibels (abbreviated 0 dB); this sound corresponds to an intensity of 1*10-12 W/m2. A sound that is 10 times more intense ( 1*10-11 W/m2) is assigned a sound level of 10 dB. A sound that is 10*10 or 100 times more intense (1*10-10 W/m2) is assigned a sound level of 20 db. A sound that is 10*10*10 or 1000 times more intense (1*10-9 W/m2) is assigned a sound level of 30 db. A sound that is 10*10*10*10 or 10000 times more intense (1*10-8 W/m2) is assigned a sound level of 40 db. Observe that this scale is based on powers or multiples of 10. If one sound is 10x times more intense than another sound, then it has a sound level that is 10*x more decibels than the less intense sound. The table below lists some common sounds with an estimate of their intensity and decibel level. Source Intensity Intensity Level 0 dB 10 dB 20 dB 60 dB 70 dB 80 dB 98 dB 100 dB 110 dB 130 dB 140 dB 160 dB # of Times Greater Than TOH 100 101 102 106 107 108 109.8 1010 1011 1013 1014 1016
Threshold of Hearing (TOH) Rustling Leaves Whisper Normal Conversation Busy Street Traffic Vacuum Cleaner Large Orchestra Walkman at Maximum Level Front Rows of Rock Concert Threshold of Pain Military Jet Takeoff Instant Perforation of Eardrum
1*10-12 W/m2 1*10-11 W/m2 1*10-10 W/m2 1*10-6 W/m2 1*10-5 W/m2 1*10-4 W/m2 6.3*10-3 W/m2 1*10-2 W/m2 1*10-1 W/m2 1*101 W/m2 1*102 W/m2 1*104 W/m2
## The Speed of Sound
A sound wave is a pressure disturbance that travels through a medium by means of particle-to-particle interaction. As one particle becomes disturbed, it exerts a force on the next adjacent particle, thus disturbing that particle from rest and transporting the energy through the medium. Like any wave, the speed of a sound wave refers to how fast the disturbance is passed from particle to particle. While frequency refers to the number of vibrations that an individual particle makes per unit of time, speed refers to the distance that the disturbance travels per unit of time. Always be cautious to distinguish between the two often-confused quantities of speed (how fast...) and frequency (how
often...). Since the speed of a wave is defined as the distance that a point on a wave (such as a compression or a rarefaction) travels per unit of time, it is often expressed in units of meters/second (abbreviated m/s). In equation form, this is
speed = distance/time
The faster a sound wave travels, the more distance it will cover in the same period of time. If a sound wave were observed to travel a distance of 700 meters in 2 seconds, then the speed of the wave would be 350 m/s. A slower wave would cover less distance - perhaps 660 meters - in the same time period of 2 seconds and thus have a speed of 330 m/s. Faster waves cover more distance in the same period of time.
Factors Affecting Wave Speed The speed of any wave depends upon the properties of the medium through which the wave is traveling. Typically there are two essential types of properties that effect wave speed - inertial properties and elastic properties. Elastic properties are those properties related to the tendency of a material to maintain its shape and not deform whenever a force or stress is applied to it. A material such as steel will experience a very small deformation of shape (and dimension) when a stress is applied to it. Steel is a rigid material with a high elasticity. On the other hand, a material such as a rubber band is highly flexible; when a force is applied to stretch the rubber band, it deforms or changes its shape readily. A small stress on the rubber band causes a large deformation. Steel is considered to be a stiff or rigid material, whereas a rubber band is considered a flexible material. At the particle level, a stiff or rigid material is characterized by atoms and/or molecules with strong attractions for each other. When a force is applied in an attempt to stretch or deform the material, its strong particle interactions prevent this deformation and help the material maintain its shape. Rigid materials such as steel are considered to have a high elasticity. (Elastic modulus is the technical term). The phase of matter has a tremendous impact upon the elastic properties of the medium. In general, solids have the strongest interactions between particles, followed by liquids and then gases. For this reason, longitudinal sound waves travel faster in solids than they do in liquids than they do in gases. Even though the inertial factor may favor gases, the elastic factor has a greater influence on the speed (v) of a wave, thus yielding this general pattern: vsolids > vliquids > vgases Inertial properties are those properties related to the material's tendency to be sluggish to changes in its state of motion. The density of a medium is an example of an inertial property. The greater the inertia (i.e., mass density) of individual particles of the medium, the less responsive they will be to the interactions between neighboring particles and the slower that the wave will be. As stated above, sound waves travel faster in solids than they do in liquids than they do in gases. However, within a single phase of matter, the inertial property of density tends to be the property that has a greatest impact upon the speed of sound. A sound wave will travel faster in a less dense material than a more dense material. Thus, a sound wave will travel nearly three times faster in Helium than it will in air. This is mostly due to the lower mass of Helium particles as compared to air particles. The speed of a sound wave in air depends upon the properties of the air, mostly the temperature, and to a lesser degree, the humidity. Humidity is the result of water vapor being present in air. Like any liquid, water has a tendency to evaporate. As it does, particles of gaseous water become mixed in the air. This additional matter will affect the mass density of the air (an inertial property). The temperature will affect the strength of the particle interactions (an elastic property). At normal atmospheric pressure, the temperature dependence of the speed of a sound wave through dry air is approximated by the following equation:
## v = 331 m/s + (0.6 m/s/C)T
where T is the temperature of the air in degrees Celsius. Using this equation to determine the speed of a sound wave in air at a temperature of 20 degrees Celsius yields the following solution. v = 331 m/s + (0.6 m/s/C)T
v = 331 m/s + (0.6 m/s/C)(20 C) v = 331 m/s + 12 m/s v = 343 m/s (The above equation relating the speed of a sound wave in air to the temperature provides reasonably accurate speed values for temperatures between 0 and 100 Celsius. The equation itself does not have any theoretical basis; it is simply the result of inspecting temperature-speed data for this temperature range. Other equations do exist that are based upon theoretical reasoning and provide accurate data for all temperatures. Nonetheless, the equation above will be sufficient for our use as introductory Physics students.)
## The Human Ear
Understanding how humans hear is a complex subject involving the fields of physiology, psychology and acoustics. In this part of Lesson 2, we will focus on the acoustics (the branch of physics pertaining to sound) of hearing. We will attempt to understand how the human ear serves as an astounding transducer, converting sound energy to mechanical energy to a nerve impulse that is transmitted to the brain. The ear's ability to do this allows us to perceive the pitch of sounds by detection of the wave's frequencies, the loudness of sound by detection of the wave's amplitude and the timbre of the sound by the detection of the various frequencies that make up a complex sound wave. The ear consists of three basic parts - the outer ear, the middle ear, and the inner ear. Each part of the ear serves a specific purpose in the task of detecting and interpreting sound. The outer ear serves to collect and channel sound to the middle ear. The middle ear serves to transform the energy of a sound wave into the internal vibrations of the bone structure of the middle ear and ultimately transform these vibrations into a compressional wave in the inner ear. The inner ear serves to transform the energy of a compressional wave within the inner ear fluid into nerve impulses that can be transmitted to the brain. The three parts of the ear are shown below.
The outer ear consists of an earflap and an approximately 2-cm long ear canal. The earflap provides protection for the middle ear in order to prevent damage to the eardrum. The outer ear also channels sound waves that reach the ear through the ear canal to the eardrum of the middle ear. Because of the length of the ear canal, it is capable of amplifying sounds with frequencies of approximately 3000 Hz. As sound travels through the outer ear, the sound is still in the form of a pressure wave, with an alternating pattern of high and low pressure regions. It is not until the sound reaches the eardrum at the interface of the outer and the middle ear that the energy of the mechanical wave becomes converted into vibrations of the inner bone structure of the ear.
The middle ear is an air-filled cavity that consists of an eardrum and three tiny, interconnected bones - the hammer, anvil, and stirrup. The eardrum is a very durable and tightly stretched membrane that vibrates as the incoming pressure waves reach it. As shown below, a compression forces the eardrum inward and a rarefaction forces the eardrum outward, thus vibrating the eardrum at the same frequency of the sound wave.
Being connected to the hammer, the movements of the eardrum will set the hammer, anvil, and stirrup into motion at the same frequency of the sound wave. The stirrup is connected to the inner ear; and thus the vibrations of the stirrup are transmitted to the fluid of the inner ear and create a compression wave within the fluid. The three tiny bones of the middle ear act as levers to amplify the vibrations of the sound wave. Due to a mechanical advantage, the displacements of the stirrup are greater than that of the hammer. Furthermore, since the pressure wave striking the large area of the eardrum is concentrated into the smaller area of the stirrup, the force of the vibrating stirrup is nearly 15 times larger than that of the eardrum. This feature enhances our ability of hear the faintest of sounds. The middle ear is an air-filled cavity that is connected by the Eustachian tube to the mouth. This connection allows for the equalization of pressure within the air-filled cavities of the ear. When this tube becomes clogged during a cold, the ear cavity is unable to equalize its pressure; this will often lead to earaches and other pains. The inner ear consists of a cochlea, the semicircular canals, and the auditory nerve. The cochlea and the semicircular canals are filled with a water-like fluid. The fluid and nerve cells of the semicircular canals provide no role in the task of hearing; they merely serve as accelerometers for detecting accelerated movements and assisting in the task of maintaining balance. The cochlea is a snail-shaped organ that would stretch to approximately 3 cm. In addition to being filled with fluid, the inner surface of the cochlea is lined with over 20 000 hair-like nerve cells that perform one of the most critical roles in our ability to hear. These nerve cells differ in length by minuscule amounts; they also have different degrees of resiliency to the fluid that passes over them. As a compressional wave moves from the interface between the hammer of the middle ear and the oval window of the inner ear through the cochlea, the small hair-like nerve cells will be set in motion. Each hair cell has a natural sensitivity to a particular frequency of vibration. When the frequency of the compressional wave matches the natural frequency of the nerve cell, that nerve cell will resonate with a larger amplitude of vibration. This increased vibrational amplitude induces the cell to release an electrical impulse that passes along the auditory nerve towards the brain. In a process that is not clearly understood, the brain is capable of interpreting the qualities of the sound upon reception of these electric nerve impulses.
## Behavior of Sound Waves
Interference and Beats | The Doppler Effect and Shock Waves Boundary Behavior | Reflection, Refraction, and Diffraction
## Interference and Beats
Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium. The interference of waves causes the medium to take on a shape that results from the net effect of the two individual waves upon the particles of the medium. As mentioned in a previous unit of The Physics Classroom Tutorial, if two upward displaced pulses having the same shape meet up with one another while traveling in opposite directions along a medium, the medium will take on the shape of an upward displaced pulse with twice the amplitude of the two interfering pulses. This type of interference is known as constructive interference. If an upward displaced pulse and a downward displaced pulse having the same shape meet up with one another while traveling in opposite directions along a medium, the two pulses will cancel each other's effect upon the displacement of the medium and the medium will assume the equilibrium position. This type of interference is known as destructive interference. The diagrams below show two
waves - one is blue and the other is red - interfering in such a way to produce a resultant shape in a medium; the resultant is shown in green. In two cases (on the left and in the middle), constructive interference occurs and in the third case (on the far right, destructive interference occurs.
But how can sound waves that do not possess upward and downward displacements interfere constructively and destructively? Sound is a pressure wave that consists of compressions and rarefactions. As a compression passes through a section of a medium, it tends to pull particles together into a small region of space, thus creating a high-pressure region. And as a rarefaction passes through a section of a medium, it tends to push particles apart, thus creating a low-pressure region. The interference of sound waves causes the particles of the medium to behave in a manner that reflects the net effect of the two individual waves upon the particles. For example, if a compression (high pressure) of one wave meets up with a compression (high pressure) of a second wave at the same location in the medium, then the net effect is that that particular location will experience an even greater pressure. This is a form of constructive interference. If two rarefactions (two low-pressure disturbances) from two different sound waves meet up at the same location, then the net effect is that that particular location will experience an even lower pressure. This is also an example of constructive interference. Now if a particular location along the medium repeatedly experiences the interference of two compressions followed up by the interference of two rarefactions, then the two sound waves will continually reinforce each other and produce a very loud sound. The loudness of the sound is the result of the particles at that location of the medium undergoing oscillations from very high to very low pressures. As mentioned in a previous unit, locations along the medium where constructive interference continually occurs are known as anti-nodes. The animation below shows two sound waves interfering constructively in order to produce very large oscillations in pressure at a variety of anti-nodal locations. Note that compressions are labeled with a C and rarefactions are labeled with an R.
Now if two sound waves interfere at a given location in such a way that the compression of one wave meets up with the rarefaction of a second wave, destructive interference results. The net effect of a compression (which pushes particles together) and a rarefaction (which pulls particles apart) upon the particles in a given region of the medium is to not even cause a displacement of the particles. The tendency of the compression to push particles together is canceled by the tendency of the rarefactions to pull particles apart; the particles would remain at their rest position as though there wasn't even a disturbance passing through them. This is a form of destructive interference. Now if a particular location along the medium repeatedly experiences the interference of a compression and rarefaction followed up by the interference of a rarefaction and a compression, then the two sound waves will continually cancel each other and no sound is heard. The absence of sound is the result of the particles remaining at rest and behaving as though there were no disturbance passing through it. Amazingly, in a situation such as this, two sound waves would combine to produce no sound. As mentioned in a previous unit, locations along the medium where destructive interference continually occurs are known as nodes.
Two Source Sound Interference A popular Physics demonstration involves the interference of two sound waves from two speakers. The speakers are set approximately 1-meter apart and produced identical tones. The two sound waves traveled through the air in front of the speakers, spreading our through the room in spherical fashion. A snapshot in
time of the appearance of these waves is shown in the diagram below. In the diagram, the compressions of a wavefront are represented by a thick line and the rarefactions are represented by thin lines. These two waves interfere in such a manner as to produce locations of some loud sounds and other locations of no sound. Of course the loud sounds are heard at locations where compressions meet compressions or rarefactions meet rarefactions and the "no sound" locations appear wherever the compressions of one of the waves meet the rarefactions of the other wave. If you were to plug one ear and turn the other ear towards the place of the speakers and then slowly walk across the room parallel to the plane of the speakers, then you would encounter an amazing phenomenon. You would alternatively hear loud sounds as you approached anti-nodal locations and virtually no sound as you approached nodal locations. (As would commonly be observed, the nodal locations are not true nodal locations due to reflections of sound waves off the walls. These reflections tend to fill the entire room with reflected sound. Even though the sound waves that reach the nodal locations directly from the speakers destructively interfere, other waves reflecting off the walls tend to reach that same location to produce a pressure disturbance.)
Destructive interference of sound waves becomes an important issue in the design of concert halls and auditoriums. The rooms must be designed in such as way as to reduce the amount of destructive interference. Interference can occur as the result of sound from two speakers meeting at the same location as well as the result of sound from a speaker meeting with sound reflected off the walls and ceilings. If the sound arrives at a given location such that compressions meet rarefactions, then destructive interference will occur resulting in a reduction in the loudness of the sound at that location. One means of reducing the severity of destructive interference is by the design of walls, ceilings, and baffles that serve to absorb sound rather than reflect it. This will be discussed in more detail later in Lesson 3. The destructive interference of sound waves can also be used advantageously in noise reduction systems. Earphones have been produced that can be used by factory and construction workers to reduce the noise levels on their jobs. Such earphones capture sound from the environment and use computer technology to produce a second sound wave that one-half cycle out of phase. The combination of these two sound waves within the headset will result in destructive interference and thus reduce a worker's exposure to loud noise.
Musical Beats and Intervals Interference of sound waves has widespread applications in the world of music. Music seldom consists of sound waves of a single frequency played continuously. Few music enthusiasts would be impressed by an orchestra that played music consisting of the note with a pure tone played by all instruments in the orchestra. Hearing a sound wave of 256 Hz (middle C) would become rather monotonous (both literally and figuratively). Rather, instruments are known to produce overtones when played resulting in a sound that consists of a multiple of frequencies. Such instruments are described as being rich in tone color. And even the best choirs will earn their money when two singers sing two notes (i.e., produce two sound waves) that are an octave apart. Music is a mixture of sound waves that typically have whole number ratios between the frequencies associated with their notes. In fact, the major distinction between music and noise is that noise consists of a mixture of frequencies whose mathematical relationship to one another is not readily discernible. On the other hand, music consists of a mixture of frequencies that have a clear mathematical relationship between them. While it may be true that "one person's music is another person's noise" (e.g.,
your music might be thought of by your parents as being noise), a physical analysis of musical sounds reveals a mixture of sound waves that are mathematically related. To demonstrate this nature of music, let's consider one of the simplest mixtures of two different sound waves - two sound waves with a 2:1 frequency ratio. This combination of waves is known as an octave. A simple sinusoidal plot of the wave pattern for two such waves is shown below. Note that the red wave has two times the frequency of the blue wave. Also observe that the interference of these two waves produces a resultant (in green) that has a periodic and repeating pattern. One might say that two sound waves that have a clear whole number ratio between their frequencies interfere to produce a wave with a regular and repeating pattern. The result is music.
Another simple example of two sound waves with a clear mathematical relationship between frequencies is shown below. Note that the red wave has three-halves the frequency of the blue wave. In the music world, such waves are said to be a fifth apart and represent a popular musical interval. Observe once more that the interference of these two waves produces a resultant (in green) that has a periodic and repeating pattern. It should be said again: two sound waves that have a clear whole number ratio between their frequencies interfere to produce a wave with a regular and repeating pattern; the result is music.
Finally, the diagram below illustrates the wave pattern produced by two dissonant or displeasing sounds. The diagram shows two waves interfering, but this time there is no simple mathematical relationship between their frequencies (in computer terms, one has a wavelength of 37 and the other has a wavelength 20 pixels). Observe (look carefully) that the pattern of the resultant is neither periodic nor repeating (at least not in the short sample of time that is shown). The message is clear: if two sound waves that have no simple mathematical relationship between their frequencies interfere to produce a wave, the result will be an irregular and non-repeating pattern. This tends to be displeasing to the ear.
## The Doppler Effect and Shock Waves
The Doppler effect is a phenomenon observed whenever the source of waves is moving with respect to an observer. The Doppler effect can be described as the effect produced by a moving source of waves in which there is an apparent upward shift in frequency for the observer and the source are approaching and an apparent downward shift in frequency when the observer and the source is receding. The Doppler effect can be observed to occur with all types of waves - most notably water waves, sound waves, and light waves. The application of this phenomenon to water waves was discussed in detail in Unit 10 of The Physics Classroom Tutorial. In this unit, we will focus on the application of the Doppler effect to sound. We are most familiar with the Doppler effect because of our experiences with sound waves. Perhaps you recall an instance in which a police car or emergency vehicle was traveling towards you on the highway. As the car approached with its siren blasting, the pitch of the siren sound (a measure of the siren's frequency) was high; and then suddenly after the car passed by, the pitch of the siren sound was low. That was the Doppler effect - a shift in the apparent frequency for a sound wave produced by a moving source.
Another common experience is the shift in apparent frequency of the sound of a train horn. As the train approaches, the sound of its horn is heard at a high pitch and as the train moved away, the sound of its horn is heard at a low pitch. This is the Doppler effect. A common Physics demonstration the use of a large Nerf ball equipped with a buzzer that produces a sound with a constant frequency. The Nerf ball is then through around the room. As the ball approaches you, you observe a higher pitch than when the ball is at rest. And when the ball is thrown away from you, you observe a lower pitch than when the ball is at rest. This is the Doppler effect.
Explaining the Doppler Effect The Doppler effect is observed because the distance between the source of sound and the observer is changing. If the source and the observer are approaching, then the distance is decreasing and if the source and the observer are receding, then the distance is increasing. The source of sound always emits the same frequency. Therefore, for the same period of time, the same number of waves must fit between the source and the observer. if the distance is large, then the waves can be spread apart; but if the distance is small, the waves must be compressed into the smaller distance. For these reasons, if the source is moving towards the observer, the observer perceives sound waves reaching him or her at a more frequent rate (high pitch). And if the source is moving away from the observer, the observer perceives sound waves reaching him or her at a less frequent rate (low pitch). It is important to note that the effect does not result because of an actual change in the frequency of the source. The source puts out the same frequency; the observer only perceives a different frequency because of the relative motion between them. The Doppler effect is a shift in the apparent or observed frequency and not a shift in the actual frequency at which the source vibrates.
Shock Waves and Sonic Booms The Doppler effect is observed whenever the speed of the source is moving slower than the speed of the waves. But if the source actually moves at the same speed as or faster than the wave itself can move, a different phenomenon is observed. If a moving source of sound moves at the same speed as sound, then the source will always be at the leading edge of the waves that it produces. The diagram at the right depicts snapshots in time of a variety of wavefronts produced by an aircraft that is moving at the same speed as sound. The circular lines represent compressional wavefronts of the sound waves. Notice that these circles are bunched up at the front of the aircraft. This phenomenon is known as a shock wave. Shock waves are also produced if the aircraft moves faster than the speed of sound. If a moving source of sound moves faster than sound, the source will always be ahead of the waves that it produces. The diagram at the right depicts snapshots in time of a variety of wavefronts produced by an aircraft that is moving faster than sound. Note that the circular compressional wavefronts fall behind the faster moving aircraft (in actuality, these circles would be spheres). If you are standing on the ground when a supersonic (faster than sound) aircraft passes overhead, you might hear a sonic boom. A sonic boom occurs as the result of the piling up of compressional wavefronts along the conical edge of the wave pattern. These compressional wavefronts pile up and interfere to produce a very high-pressure zone. This is shown below. Instead of these compressional regions (high-pressure regions) reaching you one at a time in consecutive fashion, they all reach you at once. Since every compression is followed by a rarefaction, the high-pressure zone will be immediately followed by a low-pressure zone. This creates a very loud noise.
If you are standing on the ground as the supersonic aircraft passes by, there will be a short time delay and then you will hear the boom - the sonic boom. This boom is merely a loud noise resulting from the high pressure sound followed by a low pressure sound. Do not be mistaken into thinking that this boom only happens the instant that the aircraft surpasses the speed of sound and that it is the signature that the
aircraft just attained supersonic speed. Sonic booms are observed when any aircraft that is traveling faster than the speed of sound passes overhead. It is not a sign that the aircraft just overcame the sound barrier, but rather a sign that the aircraft is traveling faster than sound.
Boundary Behavior
As a sound wave travels through a medium, it will often reach the end of the medium and encounter an obstacle or perhaps another medium through which it could travel. When one medium ends, another medium begins; the interface of the two media is referred to as the boundary and the behavior of a wave at that boundary is described as its boundary behavior. The behavior of a wave (or pulse) upon reaching the end of a medium is referred to as boundary behavior. There are essentially four possible behaviors that a wave could exhibit at a boundary: reflection (the bouncing off of the boundary), diffraction (the bending around the obstacle without crossing over the boundary), transmission (the crossing of the boundary into the new material or obstacle), and refraction (occurs along with transmission and is characterized by the subsequent change in speed and direction). In this part of Lesson 3, the focus will be upon the reflection behavior of sound waves. Later in Lesson 3, diffraction, transmission, and refraction will be discussed in more detail. In Unit 10 of The Physics Classroom, the boundary behavior of a pulse on a rope was discussed. In that unit, it was mentioned that there are two types of reflection for waves on ropes: fixed end reflection and free end reflection. A pulse moving through a rope will eventually reach its end. Upon reaching the end of the medium, two things occur:
A portion of the energy carried by the pulse is reflected and returns towards the left end of the rope. The disturbance that returns to the left is known as the reflected pulse. A portion of the energy carried by the pulse is transmitted into the new medium. If the rope is attached to a pole (as shown at the right), the pole will receive some of the energy and begin to vibrate. If the rope is not attached to a pole but rather resting on the ground, then a portion of the energy is transmitted into the air (the new medium), causing slight disturbances of the air particles.
The amount of energy that becomes reflected is dependent upon the dissimilarity of the two media. The more similar that the two media on each side of the boundary are, the less reflection that occurs and the more transmission that occurs. Conversely, the less similar that the two media on each side of the boundary are, the more reflection that occurs and the less transmission that occurs. So if a heavy rope is attached to a light rope (two very dissimilar media), little transmission and mostly reflection occurs. And if a heavy rope is attached to another heavy rope (two very similar media), little reflection and mostly transmission occurs.
The more similar the medium, the more transmission that occurs.
These principles of reflection can be applied to sound waves. Though a sound wave does not consist of crests and troughs, they do consist of compressions and rarefactions. If a sound wave is traveling through a cylindrical tube, it will eventually come to the end of the tube. The end of the tube represents a boundary between the enclosed air in the tube and the expanse of air outside of the tube. Upon reaching the end of the tube, the sound wave will undergo partial reflection and partial transmission. That is, a portion of the energy carried by the sound wave will pass across the boundary and out of the tube (transmission) and a portion of the energy carried by the sound wave will reflect off the boundary, remain in the tube and travel in the opposite direction (reflection). The reflected pulse off the end of the tube can then interfere with any subsequent incident pulses that are traveling in the opposite direction. If the disturbances within the tube are the result of perpetual waves of a constant frequency, then interference between the incident and reflected waves will occur along the length of the tube. The reflection behavior of sound waves and the subsequent interference that occurs will become important in Lesson 5 during the discussion of musical instruments. Many musical instruments operate as the result of sound waves traveling back and forth inside of "tubes" or air columns. The reflection of sound also becomes important to the design of concert halls and auditoriums. The acoustics of sound must be considered in the design of such buildings. The most important considerations include destructive interference and reverberations, both of which are the result of reflections of sound off the walls and ceilings. Designers attempt to reduce the severity of these problems by using building materials that reduce the amount of reflection and enhance the amount of transmission (or absorption) into the walls and ceilings. The most reflective materials are those that are smooth and hard; such materials are very dissimilar to air and thus reduce the amount of transmission and increase the amount of reflection. The best materials to use in the design of concert halls and auditoriums are those materials that are soft. For this reason, fiberglass and acoustic tile are used in such buildings rather than cement and brick.
## Reflection, Refraction, and Diffraction
Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the
obstacle, and transmission (accompanied by refraction) into the obstacle or new medium. In this part of Lesson 3, we will investigate behaviors that have already been discussed in a previous unit and apply them towards the reflection, diffraction, and refraction of sound waves. When a wave reaches the boundary between one medium another medium, a portion of the wave undergoes reflection and a portion of the wave undergoes transmission across the boundary. As discussed in the previous part of Lesson 3, the amount of reflection is dependent upon the dissimilarity of the two media. For this reason, acoustically minded builders of auditoriums and concert halls avoid the use of hard, smooth materials in the construction of their inside halls. A hard material such as concrete is as dissimilar as can be to the air through which the sound moves; subsequently, most of the sound wave is reflected by the walls and little is absorbed. Walls and ceilings of concert halls are made softer materials such as fiberglass and acoustic tiles. These materials are more similar to air than concrete and thus have a greater ability to absorb sound. This gives the room more pleasing acoustic properties. Reflection of sound waves off of surfaces can lead to one of two phenomena - an echo or a reverberation. A reverberation often occurs in a small room with height, width, and length dimensions of approximately 17 meters or less. Why the magical 17 meters? The affect of a particular sound wave upon the brain endures for more than a tiny fraction of a second; the human brain keeps a sound in memory for up to 0.1 seconds. If a reflected sound wave reaches the ear within 0.1 seconds of the initial sound, then it seems to the person that the sound is prolonged. The reception of multiple reflections off of walls and ceilings within 0.1 seconds of each other causes reverberations - the prolonging of a sound. Since sound waves travel at about 340 m/s at room temperature, it will take approximately 0.1 s for a sound to travel the length of a 17 meter room and back, thus causing a reverberation (recall from Lesson 2, t = v/d = (340 m/s)/(34 m) = 0.1 s). This is why reverberations are common in rooms with dimensions of approximately 17 meters or less. Perhaps you have observed reverberations when talking in an empty room, when honking the horn while driving through a highway tunnel or underpass, or when singing in the shower. In auditoriums and concert halls, reverberations occasionally occur and lead to the displeasing garbling of a sound. But reflection of sound waves in auditoriums and concert halls do not always lead to displeasing results, especially if the reflections are designed right. Smooth walls have a tendency to direct sound waves in a specific direction. Subsequently the use of smooth walls in an auditorium will cause spectators to receive a large amount of sound from one location along the wall; there would be only one possible path by which sound waves could travel from the speakers to the listener. The auditorium would not seem to be as lively and full of sound. Rough walls tend to diffuse sound, reflecting it in a variety of directions. This allows a spectator to perceive sounds from every part of the room, making it seem lively and full. For this reason, auditorium and concert hall designers prefer construction materials that are rough rather than smooth.
Reflection of sound waves also leads to echoes. Echoes are different than reverberations. Echoes occur when a reflected sound wave reaches the ear more than 0.1 seconds after the original sound wave was heard. If the elapsed time between the arrivals of the two sound waves is more than 0.1 seconds, then the sensation of the first sound will have died out. In this case, the arrival of the second sound wave will be perceived as a second sound rather than the prolonging of the first sound. There will be an echo instead of a reverberation.
Reflection of sound waves off of surfaces is also affected by the shape of the surface. As mentioned of water waves in Unit 10, flat or plane surfaces reflect sound waves in such a way that the angle at which the wave approaches the surface equals the angle at which the wave leaves the surface. This principle will be extended to the reflective behavior of light waves off of plane surfaces in great detail in Unit 13 of The Physics Classroom. Reflection of sound waves off of curved surfaces leads to a more interesting phenomenon. Curved surfaces with a parabolic shape have the habit of focusing sound waves to a point. Sound waves reflecting off of parabolic surfaces concentrate all their energy to a single point in space; at that point, the sound is amplified. Perhaps you have seen a museum exhibit that utilizes a parabolic-shaped disk to collect a large amount of sound and focus it at a focal point. If you place your ear at the focal point, you can hear even the faintest whisper of a friend standing across the room. Parabolic-shaped satellite disks use this same principle of reflection to gather large amounts of electromagnetic waves and focus it at a point (where the receptor is located). Scientists have recently discovered some evidence that seems to reveal that a bull moose utilizes his antlers as a satellite disk to gather and focus sound. Finally, scientists have long believed that owls are equipped with spherical facial disks that can be maneuvered in order to gather and reflect sound towards their ears. The reflective behavior of light waves off curved surfaces will be studies in great detail in Unit 13 of The Physics Classroom Tutorial.
Diffraction of Sound Waves Diffraction involves a change in direction of waves as they pass through an opening or around a barrier in their path. The diffraction of water waves was discussed in Unit 10 of The Physics Classroom Tutorial. In that unit, we saw that water waves have the ability to travel around corners, around obstacles and through openings. The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. In fact, when the wavelength of the wave is smaller than the obstacle or opening, no noticeable diffraction occurs. Diffraction of sound waves is commonly observed; we notice sound diffracting around corners or through door openings, allowing us to hear others who are speaking to us from adjacent rooms. Many forest-dwelling birds take advantage of the diffractive ability of long-wavelength sound waves. Owls for instance are able to communicate across long distances due to the fact that their long-wavelength hoots are able to diffract around forest trees and carry farther than the short-wavelength tweets of songbirds. Low-pitched (long wavelength) sounds always carry further than highpitched (short wavelength) sounds. Scientists have recently learned that elephants emit infrasonic waves of very low frequency to communicate over long distances to each other. Elephants typically migrate in large herds that may sometimes become separated from each other by distances of several miles. Researchers who have observed elephant migrations from the air and have been both impressed and puzzled by the ability of elephants at the beginning and the end of these herds to make extremely synchronized movements. The matriarch at the front of the herd might make a turn to the right, which is immediately followed by elephants at the end of the herd making the same turn to the right. These synchronized movements occur despite the fact that the elephants' vision of each other is blocked by dense vegetation. Only recently have they learned that the synchronized movements are preceded by infrasonic communication. While low wavelength sound waves are unable to diffract around the dense vegetation, the high wavelength sounds produced by the elephants have sufficient diffractive ability to communicate long distances. Bats use high frequency (low wavelength) ultrasonic waves in order to enhance their ability to hunt. The typical prey of a bat is the moth - an object not much larger than a couple of centimeters. Bats use ultrasonic echolocation methods to detect the presence of bats in the air. But why ultrasound? The answer lies in the physics of diffraction. As the wavelength of a wave becomes smaller than the obstacle that it encounters, the wave is no longer able to diffract around the obstacle, instead the wave reflects off the obstacle. Bats use ultrasonic waves with wavelengths smaller than the dimensions of their prey. These sound waves will encounter the prey, and instead of diffracting around the prey, will reflect off the prey and allow the bat to hunt by means of echolocation. The wavelength of a 50 000 Hz sound wave in air (speed of approximately 340 m/s) can be calculated as follows
Wavelength = speed/frequency Wavelength = (340 m/s)/(50 000 Hz) Wave length = 0.0068 m The wavelength of the 50 000 Hz sound waves (typical for a bat) is approximately 0.7 centimeters, smaller than the dimensions of a typical moth.
Refraction of Sound Waves Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. So if the media (or its properties) are changed, the speed of the wave is changed. Thus, waves passing from one medium to another will undergo refraction. Refraction of sound waves is most evident in situations in which the sound wave passes through a medium with gradually varying properties. For example, sound waves are known to refract when traveling over water. Even though the sound wave is not exactly changing media, it is traveling through a medium with varying properties; thus, the wave will encounter refraction and change its direction. Since water has a moderating affect upon the temperature of air, the air directly above the water tends to be cooler than the air far above the water. Sound waves travel slower in cooler air than they do in warmer air. For this reason, the portion of the wave front directly above the water is slowed down, while the portion of the wavefronts far above the water speeds ahead. Subsequently, the direction of the wave changes, refracting downwards towards the water. This is depicted in the diagram at the right. Refraction of other waves such as light waves will be discussed in more detail in a later unit of The Physics Classroom Tutorial.
## Radiation from a monopole source
A monopole is a source which radiates sound equally well in all directions. The simplest example of a monopole source would be a sphere whose radius alternately expands and contracts sinusoidally. The monopole source creates a sound wave by alternately introducing and removing fluid into the surrounding area. A boxed loudspeaker at low frequencies acts as a monopole. The directivity pattern for a monopole source is shown in the figure at right.
The animated GIF at left shows the pressure field produced by a monopole source. Individual points on the grid simply move back and forth about some equilibrium position while the spherical wave expands outwards.
## Radiation from a dipole source
A dipole source consists of two monopole sources of equal strength but opposite phase and separated by a small distance compared with the wavelength of sound. While one source expands the other source contracts. The result is that the fluid (air) near the two sources sloshes back and forth to produce the sound. A sphere which oscillates back and forth acts like a dipole source, as does an unboxed loudspeaker (while the front is pushing outwards the back is sucking in). A dipole source does not radiate sound in all directions equally. The directivity pattern shown at right looks like a figure-8; there are two regions where sound is radiated very well, and two regions where sound cancels.
The animated GIF at left shows the pressure field produced by a dipole source. At the center of the pressure field you can see sloshing back and forth caused by the dipole motion. The regions where sound is cancelled shows up along the vertical axes (the grid motion is almost zero). Furthermore, the wavefronts expanding to the right and left are 180o out of phase with each other.
## Radiation from a lateral quadrupole source
If two opposite phase monopoles make up a dipole, then two opposite dipoles make up a quadrupole source. In a Lateral Quadrupole arrangement the two dipoles do not lie along the same line (four monopoles with alternating phase at the corners of a square). The directivity pattern for a lateral quadrupole looks like a clover-leaf pattern; sound is radiated well in front of each monopole source, but sound is canceled at points equidistant from adjacent opposite monopoles.
The animated GIF at left shows the pressure field produced by a lateral quadrupole source. At the center of the pressure field you can see the quadrupole motion as the particles alternate motion in the horizontal and vertical directions. back and forth caused by the dipole motion. The regions where sound is cancelled shows up along the diagonals (where the grid motion is almost zero). Furthermore, there is a 180o phase difference between the horizontal and vertical wavefronts.
## Radiation from a linear quadrupole source
If two opposite phase dipoles lie along the same line they make up a Linear Quadrupole source. A tuning fork is a good example of a linear quadrupole source (each tine acts as a dipole as it vibrates back and forth, and the two tines oscillate in opposite directions). What makes the linear quadrupole interesting is that there is a very obvious transition from near field to far field. In the near field there are four maxima and four minima, with the maxima along the quadrupole axis about 5dB louder than the maxima perpendicular to the quadrupole axis. The near field directivity pattern is shown at right.
In the far field there are only two maxima (along the quadrupole axis) and two minima (perpendicular to the quadrupole axis) as shown in the figure below right. The animated GIF movie at left shows the pressure field radiated by a linear quadrupole. At the center of the picture you can see the quadrupole near field pattern. As the wave expands outwards it becomes almost a spherical wave (notice that the left and right going wavefronts are in phase as opposed to the dipole source) except that the amplitude is severely reduced in the vertical directions. | 15,685 | 77,539 | {"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} | 3.34375 | 3 | CC-MAIN-2019-35 | latest | en | 0.950011 |
https://gb.education.com/common-core/CCSS.MATH.CONTENT.5.NBT.B.6/ | 1,571,400,871,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00094.warc.gz | 505,003,676 | 43,917 | # 5.NBT.B.6 Worksheets, Workbooks, Lesson Plans, and Games
#### CCSS.maths.CONTENT.5.NBT.B.6
:
"Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models."
These worksheets can help students practise this Common Core State Standards skill.
## Worksheets
Find the Quotient
Worksheet
Find the Quotient
In this year six maths worksheet, your child will practise solving multi-step word problems using long division.
Year 6
Maths
Worksheet
Word Problems with Division
Worksheet
Word Problems with Division
Every word problem on this year six maths worksheet presents a situation with animals that requires division in order to solve it.
Year 6
Maths
Worksheet
Shopping Spree: Division Word Problems
Worksheet
Shopping Spree: Division Word Problems
Word problems can be a tricky! This year six worksheet uses a straightforward approach to helping your child become more familiar with division.
Year 6
Maths
Worksheet
Division Duplication: year 6
Worksheet
Division Duplication: year 6
Long division may be tough to master, and year six students can use this maths worksheet to practise this important skill.
Year 6
Maths
Worksheet
Divide & Dig 1
Worksheet
Divide & Dig 1
Challenge your fifth graders to employ their division skills to find the buried treasure on Feather Cap Island!
Year 6
Maths
Worksheet
Division Activity
Worksheet
Division Activity
Hone your fifth grader's treasure-hunting and division skills with this entertaining maths worksheet!
Year 6
Maths
Worksheet
## Lesson Plans
Multi-Digit Division
Lesson plan
Multi-Digit Division
Let's solve multi-digit division problems! Students will feel more confident with their division than ever with this lesson that helps them practise dividing multi-digit numbers.
Year 5
Maths
Lesson plan
Finding the Quotient: Divide and Conquer
Lesson plan
Finding the Quotient: Divide and Conquer
Do your students feel defeated when it comes to division? In this lesson students will learn how to “divide and conquer” simple division problems.
Year 6
Maths
Lesson plan
Dicey Division
Lesson plan
Dicey Division
In this hands-on lesson, students will learn and use two methods of dividing whole numbers. They will roll dice to create their own unique equations for solving!
Year 6
Maths
Lesson plan
Modeling Multiplication Word Problems
Lesson plan
Modeling Multiplication Word Problems
Teach your students to organise and solve multiplication word problems using a part-part-whole model.
Year 6
Maths
Lesson plan
Area Models and Multiplication
Lesson plan
Area Models and Multiplication
Area models are building blocks to more complicated multiplication and division. Use this lesson to refresh students on the relationship between multiplication and area to prepare them to use the area models strategy with larger numbers.
Year 6
Maths
Lesson plan
Partial Quotients Method
Lesson plan
Partial Quotients Method
Refresh students on the relationship between division, multiplication, and place value! They'll use the partial quotients method to solve division problems and discuss the process.
Year 6
Maths
Lesson plan
## Workbooks
No workbooks found for this common core node.
## Games
Game
Kids make small arrays to help Muggo's ship get back on course, building multiplication skills along the way.
Year 6
Maths
Game
Build a Pool: Area Models and Multiplication Word Problems
Game
Build a Pool: Area Models and Multiplication Word Problems
Make a splash! Kids take control and use their imaginations to design pools with specific areas in this multiplication game.
Year 6
Maths
Game
## Exercises
Division with Multi-Digit Dividends
Exercise
Division with Multi-Digit Dividends
Expand students’ division skills with this exercise that teaches how to perform division problems with multi digit dividends.
Year 6
Maths
Exercise
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What could we do to improve Education.com? | 901 | 4,152 | {"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-2019-43 | latest | en | 0.826933 |
http://hubpages.com/living/Determining-How-Much-Paint-to-Buy | 1,488,057,535,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00463-ip-10-171-10-108.ec2.internal.warc.gz | 126,258,562 | 14,751 | # Determining How Much Paint to Buy
When it comes to determining the amount of paint you need to buy for a project, there are a host of factors to take into consideration, including the size of the house, the kind of coating selected, the method of application, and the condition of the surfaces. For example, stucco and masonry wallscan require up to 50 percent more paint than smooth and flat walls. Many airless sprayers apply the equivalent of two traditional coats at a time, so you may need double the amount of paint than if using a brush or roller. To come up with an accurate estimate, you need to take into account all of the above and then measure the surface that is about to be painted.
## How to Measure an Area to Determine Paint Quantity
### Surface Area = Height x Width
To estimate the amount of paint you need to cover interior walls, determine the length of the wall or surface area and multiply by the height of the area to be painted in order to get the square footage. Once you've determined the square footage of your project, the next step is to determine how much of this space is not going to be painted and since different paints will presumably be used on windows and doors, you will need to add up the square footage and subtract it from your original total.
As a rule of thumb, experts suggest that you will need approximately one gallon of paint for every 350 square feet. If you are painting walls and the surface is unpainted drywall, you will need more product, because bare drywall will absorb more of the paint. It is also necessary to decide if you need more than one coat of paint. You normally need two coats if the surface is heavily patched, unfinished, or of a darker color than the new one. It should be noted that professionals often add a color tint to the white primer and you can purchase tints for alkyd or latex paints at most DIY and paint stores. DIYers may want to stick with choosing from the most popular interior paint colors and schemes, rather than mix their own paints or primers.
## Calculating the Paint Needed for a Standard Room
### Measure twice and paint once!
These examples are for a room that is 14′ x 15′ and has eight-foot ceilings with two windows and two doors. To find out how much paint you require for the ceiling, you have to multiply the length of the ceiling by its width (14′ x 15′ = 210′) and then divide by 350 (the amount of feet per gallon); the result is 0.6, which means that you need approximately 0.6 gallons of paint per coat.
To find out how much paint is needed for the walls, you have to calculate the length and width of each (14′ + 15′ + 14′ + 15′ = 58′) before multiplying this figure by eight (the height of the wall). This gives you (58 x 8 = 464 square feet). Instead of measuring the doors and windows individually, it is possible to use rough estimates and simply subtract 15 feet for each door and 20 feet for each window (15′ + 15′ + 20′ + 20′ = 80′). Now subtract this from the wall area (464′ - 80′ = 384′) and you have the total paintable area. Divide this by 350 (384/350 = 1.09) and you discover that you require an estimated 1.09 gallons of paint for each coat.
The above calculations may seem a bit tricky at first, but they are actually very simple. Just be sure to use a calculator and take into account the finished surface of the area being painted and you should be able to provide an accurate estimate for the amount of paint needed for the task. And as any good contractor will tell you, Measure twice; cut once. Or in this case, measure twice; paint once!
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working | 894 | 3,923 | {"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-2017-09 | latest | en | 0.946444 |
https://au.answers.yahoo.com/question/index?qid=20200118074555AAKuA9q | 1,600,959,151,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00507.warc.gz | 266,064,718 | 28,430 | Anonymous
Anonymous asked in Science & MathematicsAstronomy & Space · 8 months ago
# Should we classify planets by weight?
Relevance
• 8 months ago
Weight my friend is a concept easily construed as Mass
On Earth we have Mass
We also have an extra force added on to that, called Gravity
Gravity plus Mass equals weight
I am 87 Kg
In the ISS I would still have the same Mass but without Gravity
I would be in Orbit, a state of falling, but at a Velocity of Orbit that Keeps me at the same Altitude constantly
In order to Change any of that we need external force
To slow up or speed up
Less Velocity will lower our Altitude, more, the opposite
It is Newton's Third Law of Motion at work
Anyway I diverge
In orbit, bodies are considered weightless
So Planets have Mass
They would only have weight if you could add gravity to weightless objects
So it doesn't make sense
• 8 months ago
There is no "weight" in zero gravity. Only mass.
• Tom S
Lv 7
8 months ago
Planets by definition need to be in orbit of a star, objects in orbit are in a state of free-fall, so weightless. Planets are weightless.
• 8 months ago
You do know that mass and weight aren't the same thing, right ?
• 8 months ago
The weight of a planet is not its mass but a function of its diameter and mass. It would be entirely feasible for a "puffy planet" orbiting close to its star to have the same weight as a rocky planet further out. They would have very little else in common.
• Anonymous
8 months ago
Sure. Why not. Break out the scale.
• 8 months ago
No, we should not, since the mass
of a planet already gives us a guesstimate of its weight.
• Lôn
Lv 7
8 months ago
No, by mass. .
• ?
Lv 7
8 months ago
And how exactly do we weigh them?
• 8 months ago
We more or less already do . That's why Pluto was a planet, then wasn't (because it was considered too small), and then was again (because people were attached to it and complained). There are thousands, maybe millions, of objects orbiting our Sun, we just call the large ones "planets". | 515 | 2,052 | {"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} | 3.015625 | 3 | CC-MAIN-2020-40 | latest | en | 0.96493 |
https://human.libretexts.org/Bookshelves/Music/Ethnomusicology/Music_in_World_Cultures/01%3A_Introduction/1.02%3A_Classifying_Instruments | 1,723,205,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00211.warc.gz | 237,547,744 | 32,442 | # 1.2: Classifying Instruments
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You may be familiar with the standard families of instruments in a symphony orchestra, a model that is frequently used to classify Western instruments. The symphony orchestra is usually divided into strings (played with a bow or plucked, such as violins and cellos), woodwinds (instruments traditionally made from wood where air is blown over a sharp edge, such as oboes and flutes), brass (metal instruments played with vibrating lips, such as trombones and trumpets), and percussion instruments (instruments where something is struck to create sound, such as drums and marimbas). While the model of the symphony orchestra provides a starting point, it cannot be used to fully classify the diversity of different types of instruments that are found throughout the world, or even just within Western countries.
A new system of instrument classification was developed by ethnomusicologists Erich Moritz von Hornbostel and Curt Sachs in 1914. The Hornbostel-Sachs system is still in use to the present day and is used throughout the course of this text. In the Hornbostel-Sachs system instruments are classified depending on what is vibrating to create sound, as all sound is created by vibration. The classification of instruments in a symphony orchestra shares some similarities with the Hornbostel-Sachs system, but also some key points of divergence. The Hornbostel-Sachs system divides instruments into four categories, described in greater detail below:
• MEMBRANOPHONES – drums
• IDIOPHONES – the body of the instrument vibrates to create sound
• CHORDOPHONES – string instruments
• AEROPHONES – air is primarily used to create sound
• ELECTROPHONES – electronic instruments
## MEMBRANOPHONES
In the Hornbostel-Sachs system, percussion instruments (where an instrument is struck to create sound) are split into membranophones and idiophones. With membranophones a membrane, usually a drumhead, vibrates to create sound. The membranes are typically made of animal skins or synthetic materials that are stretched over the base of the drum. Drums can be found in all parts of the world and can be struck in a variety of fashions, including with bare hands, mallets, and sticks.
Djembe – An irregular hourglass shaped drum from Western Africa played with the hands.
Dhol – A double-headed southwest Asian drum that is played with sticks
## IDIOPHONES
Idiophones are the other classification of percussion instruments in the Hornbostel-Sachs system. Unlike membranophones, where just a membrane is vibrating to create sound, with idiophones the instrument’s body itself vibrates to create sound. Think of a set of orchestral cymbals, which vibrate when struck against one another, or even a cymbal as part of a drum set that vibrates once hit with a drum stick. Idiophones can be classified into subcategories, such as:
## CHORDOPHONES
Chordophones are instruments that generate sound through the vibration of a string, whether it is plucked, bowed, or struck. This classification also includes many keyboard instruments, such as a piano where a hammer strikes a string in the body of the instrument to create sound.
Erhu – Two-string bowed fiddle from China
## AEROPHONES
Aerophones include any instrument where sound is primarily generated by vibrating air. Included in this category are both woodwind and brass instruments, as well as other reed instruments and flutes.
Harmonium – Common in many Indian genres, the harmonium consists of a bellows that is pumped with one hand while the other hand plays a keyboard. Sound is generated by air sent over reeds.
Andean Panpipes (Siku) – Panpipe instruments consist of pipes of different lengths, and pitches, fixed together. They are played by blowing air across the top of each pipe. The linked example is from the Andes, a mountain range that runs along the western side of South America, where such instruments have been traced by thousands of years. | 2,612 | 8,303 | {"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} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.196454 |
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These notes are about efficiency and power. I took notes on mostly the examples that the teacher gave. Have fun reading I hope it can help
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This 1 page Class Notes was uploaded by Surge Ace on Sunday October 9, 2016. The Class Notes belongs to at A-Technical College taught by in Fall 2016. Since its upload, it has received 7 views.
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Notes during Physics; Efficiency and power Efficiency Efficiency (η) = Useful energy input/Energy output (energy consumed) Efficiency of a car Average, 1 L gasoline delivers 34 MJ energy, only 8.5 MJ are useful. For example Ideally you would ride for 80 km but in fact, you only ride for 20 km, this is due to friction, air resistance, and other variables. Because of this our cars are not 100% efficient, and with modern day calculations, physicists have calculated that it is impossible to have a car with 100% efficiency. Percentage of efficiency of average gasoline car Using the equation to solve this problem we got: 25% average efficiency for a gasoline car. Power Power = Work done/Time taken Example If we took two men, a very young athlete and an old man. Both men have the same mass: 70 kg. Both are walking up the stairs, stairs are 3 metres high, young man climbs them in three seconds while the old man climbs them in 5 minutes. Calculate work done Work done = Force (Newtons) Distance * * =multiply Force of both men= 700 Newtons Distance= 3 metres Work done= 2100 Joules Time taken Young man=3 seconds Old man= 5 minutes=5*60 seconds=300 seconds Power of old man vs power of young man Power of Young man = 2100 Joules/3 seconds = 700 Watts Power of young man = 2100 Joules/300 seconds = 7 Watts Conclusion The young man has delivered more power than the young man
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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com | 967 | 4,148 | {"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} | 3.84375 | 4 | CC-MAIN-2016-44 | latest | en | 0.944427 |
http://en.m.wikibooks.org/wiki/Calculus/Continuity | 1,406,147,425,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997883468.51/warc/CC-MAIN-20140722025803-00014-ip-10-33-131-23.ec2.internal.warc.gz | 51,786,190 | 10,592 | Calculus/Continuity
← Infinite Limits Calculus Formal Definition of the Limit → Continuity
Defining ContinuityEdit
We are now ready to define the concept of a function being continuous. The idea is that we want to say that a function is continuous if you can draw its graph without taking your pencil off the page. But sometimes this will be true for some parts of a graph but not for others. Therefore, we want to start by defining what it means for a function to be continuous at one point. The definition is simple, now that we have the concept of limits:
Definition: (continuity at a point)
If $f(x)$ is defined on an open interval containing $c$, then $f(x)$ is said to be continuous at $c$ if and only if
$\lim_{x \rightarrow c} f(x) = f(c)$.
Note that for $f$ to be continuous at $c$, the definition in effect requires three conditions:
1. that $f$ is defined at $c$, so $f(c)$ exists,
2. the limit as $x$ approaches $c$ exists, and
3. the limit and $f(c)$ are equal.
If any of these do not hold then $f$ is not continuous at $c$.
The idea of the definition is that the point of the graph corresponding to $c$ will be close to the points of the graph corresponding to nearby $x$-values. Now we can define what it means for a function to be continuous in general, not just at one point.
Definition: (continuity)
A function is said to be continuous on $(a, b)$ if it is continuous at every point of the interval $(a, b)$.
We often use the phrase "the function is continuous" to mean that the function is continuous at every real number. This would be the same as saying the function was continuous on (−∞, ∞), but it is a bit more convenient to simply say "continuous".
Note that, by what we already know, the limit of a rational, exponential, trigonometric or logarithmic function at a point is just its value at that point, so long as it's defined there. So, all such functions are continuous wherever they're defined. (Of course, they can't be continuous where they're not defined!)
DiscontinuitiesEdit
A discontinuity is a point where a function is not continuous. There are lots of possible ways this could happen, of course. Here we'll just discuss two simple ways.
Removable discontinuitiesEdit
The function $f(x) = \frac {x^2-9} {x-3}$ is not continuous at $x = 3$. It is discontinuous at that point because the fraction then becomes $\frac{0}{0}$, which is undefined. Therefore the function fails the first of our three conditions for continuity at the point 3; 3 is just not in its domain.
However, we say that this discontinuity is removable. This is because, if we modify the function at that point, we can eliminate the discontinuity and make the function continuous. To see how to make the function $f(x)$ continuous, we have to simplify $f(x)$, getting $f(x) = \frac {x^2-9} {x-3} = \frac {(x+3)(x-3)} {(x-3)} = \frac {x+3} {1} \cdot \frac {x-3} {x-3}$. We can define a new function $g(x)$ where $g(x) = x + 3$. Note that the function $g(x)$ is not the same as the original function $f(x)$, because $g(x)$ is defined at $x=3$, while $f(x)$ is not. Thus, $g(x)$ is continuous at $x=3$, since $\lim_{x\to 3} (x+3) = 6 = g(3)$. However, whenever $x\ne 3$, $f(x)=g(x)$; all we did to $f$ to get $g$ was to make it defined at $x=3$.
In fact, this kind of simplification is often possible with a discontinuity in a rational function. We can divide the numerator and the denominator by a common factor (in our example $x-3$) to get a function which is the same except where that common factor was 0 (in our example at $x=3$). This new function will be identical to the old except for being defined at new points where previously we had division by 0.
However, this is not possible in every case. For example, the function $f(x)=\frac{x-3}{x^2-6x+9}$ has a common factor of $x-3$ in both the numerator and denominator, but when you simplify you are left with $g(x)=\frac{1}{x-3}$, which is still not defined at $x=3$. In this case the domain of $f(x)$ and $g(x)$ are the same, and they are equal everywhere they are defined, so they are in fact the same function. The reason that $g(x)$ differed from $f(x)$ in the first example was because we could take it to have a larger domain and not simply that the formulas defining $f(x)$ and $g(x)$ were different.
Jump discontinuitiesEdit
Not all discontinuities can be removed from a function. Consider this function:
$k(x) = \left\{\begin{matrix} 1, & \mbox{if }x > 0 \\ -1, & \mbox{if }x \le 0 \end{matrix}\right.$
Since $\lim_{x\to 0} k(x)$ does not exist, there is no way to redefine $k$ at one point so that it will be continuous at 0. These sorts of discontinuities are called nonremovable discontinuities.
Note, however, that both one-sided limits exist; $\lim_{x\to 0^-} k(x) = -1$ and $\lim_{x\to 0^+} k(x) = 1$. The problem is that they are not equal, so the graph "jumps" from one side of 0 to the other. In such a case, we say the function has a jump discontinuity. (Note that a jump discontinuity is a kind of nonremovable discontinuity.)
One-Sided ContinuityEdit
Just as a function can have a one-sided limit, a function can be continuous from a particular side. For a function to be continuous at a point from a given side, we need the following three conditions:
1. the function is defined at the point,
2. the function has a limit from that side at that point and
3. the one-sided limit equals the value of the function at the point.
A function will be continuous at a point if and only if it is continuous from both sides at that point. Now we can define what it means for a function to be continuous on a closed interval.
Definition: (continuity on a closed interval)
A function is said to be continuous on $[a,b]$ if and only if
1. it is continuous on $(a,b)$,
2. it is continuous from the right at $a$ and
3. it is continuous from the left at $b$.
Notice that, if a function is continuous, then it is continuous on every closed interval contained in its domain.
Intermediate Value TheoremEdit
A useful theorem regarding continuous functions is the following:
Intermediate Value Theorem
If a function $f$ is continuous on a closed interval $[a,b]$, then for every value $k$ between $f(a)$ and $f(b)$ there is a value $c$ between $a$ and $b$ such that $f(c)=k$.
Application: bisection methodEdit
A few steps of the bisection method applied over the starting range [a1;b1]. The bigger red dot is the root of the function.
The bisection method is the simplest and most reliable algorithm to find zeros of a continuous function.
Suppose we want to solve the equation $f(x) = 0$. Given two points $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs, the intermediate value theorem tells us that $f$ must have at least one root between $a$ and $b$ as long as $f$ is continuous on the interval $[a,b]$. If we know $f$ is continuous in general (say, because it's made out of rational, trigonometric, exponential and logarithmic functions), then this will work so long as $f$ is defined at all points between $a$ and $b$. So, let's divide the interval $[a,b]$ in two by computing $c = (a+b) / 2$. There are now three possibilities:
1. $f(c)=0$,
2. $f(a)$ and $f(c)$ have opposite signs, or
3. $f(c)$ and $f(b)$ have opposite signs.
In the first case, we're done. In the second and third cases, we can repeat the process on the sub-interval where the sign change occurs. In this way we hone in to a small sub-interval containing the zero. The midpoint of that small sub-interval is usually taken as a good approximation to the zero.
Note that, unlike the methods you may have learned in algebra, this works for any continuous function that you (or your calculator) know how to compute.
← Infinite Limits Calculus Formal Definition of the Limit → Continuity | 2,049 | 7,777 | {"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": 91, "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-2014-23 | longest | en | 0.944063 |
https://www.physicsforums.com/threads/can-a-travelling-wave-be-nonlinear.793716/ | 1,713,866,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00438.warc.gz | 858,528,296 | 16,569 | # Can a travelling wave be nonlinear?
• shanepitts
In summary, the conversation discusses the existence and characteristics of nonlinear traveling waves. The participants suggest that such waves may travel through a nonlinear medium and can be described using an equation that compares the definitions of "traveling wave" and "nonlinear equation". The concept of solitons, which are waves that maintain their shape due to a combination of non-linearity and dispersion, is also mentioned. However, it is noted that the answer may not be obvious at this level and it is recommended to work it out independently.
shanepitts
I'm am almost certain that the answer to this question is "yes". If so, what in nature can be a nonlinear traveling wave?
Look up Solitons. They are waves that propagate for long distances and maintain their shape because of a combination of non-linearity and dispersion.
shanepitts
Yes, a travelling wave can indeed be nonlinear. Nonlinear waves are characterized by a non-proportional relationship between the amplitude of the wave and the driving force. This means that the shape of the wave changes as it travels, unlike linear waves where the shape remains constant.
In nature, there are many examples of nonlinear travelling waves. Some common examples include ocean waves, sound waves, and shock waves. Ocean waves are considered nonlinear because their amplitude increases as they approach the shore, due to the changing depth of the water. Sound waves can also exhibit nonlinear behavior, especially at high intensities, where the pressure and density of the medium can affect the speed of the wave. Shock waves, which are sudden changes in pressure and density, are also nonlinear as they are caused by a discontinuous change in the medium.
Additionally, in the field of optics, there are many examples of nonlinear travelling waves. For instance, in nonlinear optics, the intensity of light can affect the refractive index of the medium, leading to changes in the speed and direction of the light wave. This can result in phenomena such as self-focusing and self-phase modulation, which are examples of nonlinear travelling waves.
In summary, nonlinear travelling waves are a common occurrence in nature and can be observed in various systems and phenomena. As a scientist, it is important to consider the nonlinear behavior of waves when studying and understanding complex systems.
## 1. What is a travelling wave?
A travelling wave is a disturbance or oscillation that propagates through a medium, without causing any permanent displacement of the medium itself. In other words, the individual particles of the medium oscillate back and forth, but the overall shape and position of the medium remains unchanged.
## 2. What is the difference between a linear and nonlinear travelling wave?
A linear travelling wave refers to a wave whose amplitude and frequency remain constant as it travels through the medium. On the other hand, a nonlinear travelling wave is one in which the amplitude and frequency change as it propagates, often due to interactions with the medium or other waves.
## 3. Is it possible for a travelling wave to be nonlinear?
Yes, it is possible for a travelling wave to be nonlinear. While linear waves are more commonly observed in nature, there are many examples of nonlinear travelling waves, such as solitons, shock waves, and rogue waves.
## 4. What factors can cause a travelling wave to become nonlinear?
There are several factors that can cause a travelling wave to become nonlinear. These include the properties of the medium through which the wave is travelling, the amplitude and frequency of the wave, and interactions with other waves or objects in the medium.
## 5. How is the behavior of a nonlinear travelling wave different from a linear travelling wave?
A nonlinear travelling wave exhibits behaviors that are not observed in linear waves, such as self-focusing, self-steepening, and frequency shift. The amplitude and frequency of a nonlinear wave can also change dramatically as it propagates, whereas a linear wave maintains a constant amplitude and frequency.
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1K | 958 | 4,634 | {"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-2024-18 | latest | en | 0.951881 |
https://www.convertit.com/Go/SmartPages/Measurement/Converter.ASP?From=angstrom | 1,628,200,395,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00068.warc.gz | 706,128,740 | 3,704 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```angstrom = 0.0000000001 meter (length) ``` Related Measurements: Try converting from "angstrom" to bolt (of cloth), caliber (gun barrel caliber), cloth quarter, ell, en (typography en), French, Greek fathom, Greek span, inch, line, m (meter), marathon, micron, nautical league, palm, sazhen (Russian sazhen), shaku (Japanese shaku), span (cloth span), spindle, yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: angstrom = 2.82E-12 actus (Roman actus), 1.41E-10 archin (Russian archin), 6.68E-22 astronomical unit, 1.09E-12 city block (informal), 8.75E-10 cloth finger, 100,000 fermi, 4.97E-13 furlong (surveyors furlong), 4.32E-10 Greek span, 9.84E-10 hand, 1.81E-10 Israeli cubit, 2.07E-14 league, 1.55E-13 li (Chinese li), 4.72E-08 line, 1.80E-14 nautical league, 1.31E-09 palm, 3.24E-27 parsec, 1.99E-11 rod (surveyors rod), 3.38E-10 Roman foot, 7.59E-15 spindle, 3.28E-10 survey foot.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 493 | 1,592 | {"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} | 2.90625 | 3 | CC-MAIN-2021-31 | latest | en | 0.691244 |
http://www.haskell.org/pipermail/haskell-cafe/2010-April/075611.html | 1,416,864,968,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400382386.21/warc/CC-MAIN-20141119123302-00212-ip-10-235-23-156.ec2.internal.warc.gz | 589,454,132 | 2,616 | # [Haskell-cafe] Re: Integers v ints
Jon Fairbairn jon.fairbairn at cl.cam.ac.uk
Fri Apr 2 05:22:17 EDT 2010
```Jens Blanck <jens.blanck at gmail.com> writes:
> On 1 April 2010 10:53, Ivan Lazar Miljenovic <ivan.miljenovic at gmail.com>wrote:
>> Jens Blanck <jens.blanck at gmail.com> writes:
>> > I was wondering if someone could give me some references to
>> > when and why the choice was made to default integral
>> > numerical literals to Integer rather than to Int in
Seems to have been in 1998. I don't have a complete archive of
the discussion, though, and I don't know where to find the
>> My guess is precision: some numeric calculations (even doing
>> a round on some Double values) will be too large for Int
>> values (at least on 32bit). Note that unlike Python, etc.
>> Haskell doesn't allow functions like round to choose between
>> Int and Integer (which is equivalent to the long type in
>> Python, etc.).
>
> Ints have perfect precision as long as you remember that it
> implements modulo arithmetic for some power of 2. I was hoping
> that the reason would be that Integers give more users what
> they expect, namely integers, instead of something where you
> can add two positive numbers and wind up with a negative
> number.
As I interpret the part of the discussion I have on file, there
are two reasons:
(1) as you hoped, because Integers are what people "expect":
reasoning on Integers is more reliable -- you can't do induction
on Int, for example, and people don't generally try to prove
that they've implemented
f x = the_f_they_originally_wanted x `mod` 2^32
(2) good language design. One of the things I've repeated over
the years is that Int doesn't have to be part of the language
(it's just another peculiar type that should be defined in a
library) but Integer does, because without it there's no way to
specify the meaning of an Integral constant¹.
Jón
[1] This isn't quite true; using subtyping one could make
Integral constants into [Digit] and leave the conversion to the | 526 | 2,019 | {"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-2014-49 | latest | en | 0.938448 |
https://us.metamath.org/ileuni/ralsn.html | 1,716,512,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00786.warc.gz | 516,098,768 | 3,768 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > ralsn GIF version
Theorem ralsn 3567
Description: Convert a quantification over a singleton to a substitution. (Contributed by NM, 27-Apr-2009.)
Hypotheses
Ref Expression
ralsn.1 𝐴 ∈ V
ralsn.2 (𝑥 = 𝐴 → (𝜑𝜓))
Assertion
Ref Expression
ralsn (∀𝑥 ∈ {𝐴}𝜑𝜓)
Distinct variable groups: 𝑥,𝐴 𝜓,𝑥
Allowed substitution hint: 𝜑(𝑥)
Proof of Theorem ralsn
StepHypRef Expression
1 ralsn.1 . 2 𝐴 ∈ V
2 ralsn.2 . . 3 (𝑥 = 𝐴 → (𝜑𝜓))
32ralsng 3564 . 2 (𝐴 ∈ V → (∀𝑥 ∈ {𝐴}𝜑𝜓))
41, 3ax-mp 5 1 (∀𝑥 ∈ {𝐴}𝜑𝜓)
Colors of variables: wff set class Syntax hints: → wi 4 ↔ wb 104 = wceq 1331 ∈ wcel 1480 ∀wral 2416 Vcvv 2686 {csn 3527 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2121 This theorem depends on definitions: df-bi 116 df-3an 964 df-tru 1334 df-nf 1437 df-sb 1736 df-clab 2126 df-cleq 2132 df-clel 2135 df-nfc 2270 df-ral 2421 df-v 2688 df-sbc 2910 df-sn 3533 This theorem is referenced by: tfr0dm 6219 elixpsn 6629 finomni 7012 nninfsellemdc 13236
Copyright terms: Public domain W3C validator | 712 | 1,386 | {"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} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.163714 |
https://socratic.org/questions/how-do-you-simplify-8-x-2-4-3-x-2 | 1,638,968,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00161.warc.gz | 607,206,101 | 5,703 | How do you simplify 8/(x^2-4) – 3/(x-2)?
$- \frac{1}{{x}^{2} - 4}$
$\frac{8}{{x}^{2} - 4} - \frac{9}{{x}^{2} - 4}$ | 71 | 115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 3.640625 | 4 | CC-MAIN-2021-49 | latest | en | 0.303412 |
http://www.shsu.edu/~txcae/Powerpoints/prepostest/addsubtract2pretest.html | 1,513,276,860,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550199.46/warc/CC-MAIN-20171214183234-20171214203234-00356.warc.gz | 451,724,200 | 2,496 | Using Addition and Subtraction to Solve Problems II
Check the box of the letter with the most correct answer.
1 Pablo has 850 U.S. stamps and 240 Mexican stamps in his collection. How many stamps does Pablo have altogether?
A 610
B 1,090
C 710
D 1,190
E Not Here
2 Maria has 205 U.S. coins and 160 Mexican coins in her collection. How many more U.S. coins than Mexican coins does Maria have?
F 55
G 375
H 45
J 365
K Not Here
3 The store had 55 Video games and 31 were sold. How many does the store have left?
A 84
B 94
C 24
D 110
E Not Here
4 A hiker is climbing a mountain that is 6,238 feet high. She stops to rest at 4,887 feet. How many more feet must she climb to reach the top?
F 2,351 ft
G 1,451 ft
H 1,361 ft
J 1,351 ft
K Not Here
5 A baseball team held a bake sale to raise money for new uniforms. There were 425 baked items provided for the sale. At noon the team has sold 278 of the items. How many baked items did the team have left to sell?
A 147
B 253
C 257
D 703
E Not Here
6 Sheila's parents bought a computer for \$1,089, a printer for \$ 255, and a computer game for \$39. How much did these items cost in all, not including tax?
F \$1,263
G \$1,273
H \$1,344
J \$1,383
K \$1,534
7 Tommy and his sister went seashell hunting at the beach. On Monday they found 125 shells. On Tuesday they found 56 shells. On Wednesday they found 90 shells. How many shells did they find in total?
A 171
B 271
C 270
D 371
E Not Here
8 Maria was a good baby sitter. In one weekend she had three baby-sitting jobs. She earned \$15.00 on one job, \$12.00 on another job and \$20.00 on her last job. How much money did Maria earn that weekend?
F \$47.00
G \$57.00
H \$37.00
J \$45.00
K Not Here
9 Mr. Barton owned a real nice building lot. There were 87 large oak trees on the lot. When he decided to build his house, it was necessary to cut down 43 trees. How many trees were left after he cut down the trees?
A 130
B 45
C 43
D 53
E Not Here
10 Ms. Stewart owned two clothing stores. One weekend one store sold 45 dresses and her other store sold 57 dresses. How many dresses were sold in her stores that weekend?
F 12
G 102
H 112
J 92
K Not Here | 659 | 2,164 | {"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} | 3.796875 | 4 | CC-MAIN-2017-51 | longest | en | 0.973259 |
https://www.enotes.com/homework-help/what-is-the-x-value-where-the-two-curves-2272771 | 1,686,229,541,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654871.97/warc/CC-MAIN-20230608103815-20230608133815-00161.warc.gz | 799,918,658 | 19,067 | # What is the x value where the two curves intercept? 1/4(x^2)-4(sin(x))+1=0
The x values where y=1/4 x^2 intersects y=4sinx-1 are approximately .25694 and 2.46166. These are also the zeros of f(x)=1/4 x^2-4sinx+1.
We are asked to find the x-coordinates of the intersection(s) of the curves `y=1/4 x^2` and `y=4sin(x)-1`, or alternatively, the zeros of `f(x)=1/4 x^2-4sinx+1` .
There isn't a straightforward algebraic way to get the intersections. We can graph the two functions to estimate their intersections. (Alternatively graph f(x) and estimate the zeros.) We could then use guess and revise to narrow the answer to the required precision. (See attachment: `y=1/4x^2` is the green parabola, `y=4sin(x)+1` is the blue sinusoid, and f(x) is in red.)
We can use a graphing utility or program to approximate the zeros of f(x) or the intersections of the curves. My graphing calculator gives .25694421 and 2.4616646.
With some calculus, you can use Newton's method for approximating roots. (It works in this case. There are some curves for which the method fails.)
Choose a guess for the first intercept; say x=0. Then, with `x_0=0`, we calculate `x_1=x_0-(f(x_0))/(f'(x_0))` where `f(x)=1/4x^2-4sinx+1` and `f'(x)=1/2x-4cosx` .
So, `x_0=0,x_1=-1/4,x_2~~.2512287988,x_3~~.2569376661` which is accurate to 4 decimal places.
Choosing 2.5 for the other intercept yields the sequence 2.5,2.462148702,2.461665119.
We can be sure that there are only two intersections. `y=4sinx -1` is bounded; `-5<=y<=3` . But, `y=1/4x^2>3` whenever `|x|>2sqrt(3)` . So, we need only check the interval `[-2sqrt(3),2sqrt(3)]` and `y=4sinx-1<0` for x<0 in this interval. | 561 | 1,657 | {"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} | 4.34375 | 4 | CC-MAIN-2023-23 | latest | en | 0.839949 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg06917.html | 1,532,174,579,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00099.warc.gz | 940,010,511 | 4,660 | # Re: Many Pasts? Not according to QM...
Oops sorry. I did misunderstood you. Thanks for the clarification. I agree with your preceding post to Hal now.
```
Bruno```
```
Le 13-juin-05, à 16:23, Jesse Mazer a écrit :
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```Bruno Marchal wrote:
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```Bruno Marchal:
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```
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To Jesse: You apparently completely separate the probability of x and x' from the similarity of x and x'.
```I am not sure that makes sense for me.
```
In particular how could x and x' be similar, if x', but not x, involves a 'white rabbit events'.
```
```
It's not completely separable, but I'd think that "similarity" would mostly be a function of memories, personality, etc...even if I experience something very weird, I can still have basically the same mind. For example, a hoaxer could create a realistic animatronic talking white rabbit, and temporarily I might experience an observer-moment identical to what I'd experience if I saw a genuine white talking rabbit, so the "similarity" between my current experience and what I'd experience in a white rabbit universe would be the same as the "similarity" between my current experience and what I'd experience in a universe where someone creates a realistic hoax. I don't think the first-person probabilities of experiencing hoaxes are somehow kept lower than what you'd expect from a third-person perspective, do you?
```
```
Perhaps I misunderstood you, but it seems to me, that in case you ask me to compute P(x -> y) (your notation), it could and even should change that prediction result. In particular if the rabbit has been generated by a genuine hoaxer I would predict the white rabbit will stay in y, and if the hoaxer is not genuine, then I would still consider x and x' as rather very dissimilar. What do you think? This follows *also* from a relativisation of Hall Finney's theory based on kolmogorov complexity: a stable white rabbit is expensive in information resource. No?
```
```
Well, note that following Hal's notation, I was actually assuming y came before x (or x'), and I was calculating P(y -> x). And your terminology is confusing to me here--when you say "if the hoaxer is not genuine", do you mean that the white rabbit wasn't a hoax but was a genuine talking rabbit (in which case no hoaxer is involved at all), or do you mean if the white rabbit *was* a hoax? If the latter, then what do you mean when you say "if the rabbit had been generated by a genuine hoaxer"--is the white rabbit real, or is it a hoax in this case? Also, when you say you'd consider x and x' as very dissimilar, do you mean from each other or from y? Remember that "dissimilar" is just the word I use for continuity of personal identity, how much two successive experiences make sense as being successive OMs of the "same person", it doesn't refer to whether the two sensory experiences are themselves dissimilar or dissimilar. If I'm here looking at my computer, then suddenly close my eyes, the two successive experiences will be quite dissimilar in terms of the sensory information I'm taking in, but they'll still be similar in terms of my background memories, personality, etc., so they make sense as successive OMs of the same person. On the other hand, if I'm sitting at my computer and suddenly my brain is replaced with the brain of George W. Bush, there will be very little continuity of identity despite the fact that the sensory experiences of both OMs would be pretty similar, so in my terminology there would be very little "similarity" between these two OMs.
```
```
As for the cost of simulating a white rabbit universe, I agree it's more expensive than simulating a non-white rabbit universe, but I don't see how this relates to continuity of identity when experiencing white rabbits vs. not experiencing them.
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Jesse
```
```http://iridia.ulb.ac.be/~marchal/
``` | 883 | 3,830 | {"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-2018-30 | latest | en | 0.961633 |
https://electronics.stackexchange.com/questions/551519/assigning-a-notion-of-voltage-even-when-there-is-a-changing-magnetic-field | 1,713,940,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00199.warc.gz | 205,742,213 | 58,898 | # Assigning a notion of voltage even when there is a changing magnetic field
In Walter Lewin's lecture, he finds depending on how you keep the lead of the voltmeter for a circuit in a changing magnetic field, you will measure different values]. This experiment is discussed from 49:34 of this lecture and then infers that this is why a notion of voltage is no longer possible in the case of changing magnetic field.
According to Mr.Boom, Lewin's result had a correct result, given his setup, but his conclusion was incorrect. To demonstrate this Mr.Boom experiments (see from 7:24 of this video before disagreeing), where he creates a modified version where Lewin's concluded principle must be equally applicable, however in Mr. Boom's equivalent experiment, Lewin's ideal fails. Hence, the premise of Boom was demonstrated. Particularly speaking, Mr.Boom says that, and I quote from after 10:45 of the second video, he concludes the real reason for Lewin's result was bad probing (*)
however, Mehdi(Mr.Boom) does an experiment in his second video and gets a result against Lewin's theory, so what was wrong with the experiment for falsifying Lewin? I have read this answer by Sredni Vashtar on this but I am not able to use their answer to answer this doubt I have put in the sentence right before.
I think that there is a definite voltage for a circuit even with a field passing through because even though the voltage depends on the line integral, there is only one physical path where current can go through feasible in a circuit!.
Understanding Mr.Boom's arguments
The original experiment of Lewin involved the following circuit:
Now, using an oscilloscope ( I think?) he measures top and bottom point of circuit at same time (For Boom's discussion)
And finds that each scope shows different readings:
Mr. Boom repeats the experiments and finds the same results here, but he disagrees with the conclusion which Lewin made that this leads to KVL not be applicable in the case of circuits with varying magnetic field effects.
To illustrate his point, he does an experiment with a similar set up:
Since Lewin's theory must be independent of how resistors are arranged, Mr.Boom rearranges the resistors (1k and 10k )and then measures the voltage of each side. He begins by measuring on the 10k side:
This results into:
And then he measures through opposite side:
He then measures the reading across both the resistor:
This resulting graph is the some of voltage graph of each resistor measured in individual case. Now, he says the sum voltage is the same voltage across the loop. Hence, there is indeed only a single voltage associated with the two points across the circuit. Now, to double check, he discusses the measurement if the sense wires the other way (path without resistor)
In this way, it seems that the voltage of the sense wire is zero.. and again we with multiple voltage values. Now, boom's explanation is that the problem is that the voltage is not single-valued but rather an accurate reading can not be taken as the induced field effects the measuring device as well. (see from 8:55 )
Note: All of the above is my understanding of boom's video, if there is a mistake please point out.
### Physicists and Electronic engineers divided?
To be frank, I couldn't easily understand Mr. Boom's video, so I searched for other videos by people in electronic engineering-related fields for perhaps a more 'dumbed down' version of it to understand. This led me to this video by Mr. Bob Duhamel, in which he states that Lewin is wrong see from around 23:00 and agrees with Mr. Booms's result.
Now, here is the weird part, people in physics stack exchange and such seem to agree with Mr. Lewin's result that different measurements give different values see this answer by knzhou. So, it seems the answer is entirely different depending on wether you study EE or physics.
### Romer's paper
So, now I saw the paper which roomer wrote on the issue (Mr.Boom cites this in 2nd video response). The paper is quite interesting and introduces a concept of 'pseudo potential' which can be assigned even in case of changing magnetic field and he explains this new idea with concepts of topology and such. He specifically finds that though the line integral of E*dl is indeed path-dependent, there are only two or three discrete values it can take.
I haven't fully understood the paper's significance here, I'll add in this part as I read through it more.
Some related questions on this aren't easily findable, so I will link all here:
Hopefully, I got all the related posts, please comment if I have missed any because I want to bring a 'clear' to make myself completely clear what was going on here.
P.s: I have no formal education in Electronics engineering, only some basics I learned from high school. So please comment if I have made mistakes in understanding the situation in my post, I will correct them.
Further, apologies if this post seems a bit messy, I tried my best to make sense of everything I read from multiple resources and put in a neat summary here but I hope it makes sense.
• It seems you have a few links wrong. For example, when you mention minute 7:24 of electrobooms video (which one?) the links takes me to the beginning of Lewin's lecture. Please explain clearly what in Electrobooms video is allegedly proving Lewin is wrong, possibly adding a picture or a schematic (and the timestamps in the video). Mar 6, 2021 at 2:34
• @SredniVashtar I think I have fixed everything and tried my best to make everything clear. If I have make mistakes/ not clear something, please inform I'll try to rewrite/edit that Mar 6, 2021 at 8:28
• Now, boom's explanation is that voltage is not single-valued but rather an accurate reading can not be taken "i.stack.imgur.com/CO21R.png , tell me the path I'll tell how much work you would do if you move a positive unit charge against the total field (Electrostatic+induce Electric ) through given path which can also be called as voltage difference for a given path ,for example if I move a +ve unit charge along the wire itself (in loop that I mentioned above) work done would be Zero and hence Voltage difference is zero for this path ! and there are infinite possible paths are there. Mar 6, 2021 at 9:12
• I think I agree with first part, but why did you bring up the infinite path thing at end @user215805 Mar 6, 2021 at 20:56
• Because between those two end points one path is physically connected (along the conductor )for which work done would be zero but there are infinite other imaginary paths are also possible between those two end points (and those imaginary paths can be think of infinite positions of Voltmeter through which we can connect both ends point using Voltmeter) and reading of Voltmeter at different positions is basically voltage difference between those two same end points but for the path connected by Voltmeter (not the physical path) Mar 7, 2021 at 5:42
Let me first answer to the question about how to define voltage in a changing magnetic field by using this extensive quote from Popovic & Popovic, "Introductory Electromagnetics", sec. 14.4 "Potential difference and voltage in a time-varying electric and magnetic field":
"In the discussion of electrostatics, we defined the voltage to be the same as the potential difference. Actually, the voltage between two points is defined as the line integral of the total electric field strength from one point to the other. In electrostatics, the induced electric field does not exist, and therefore voltage is identical to potential difference. [...] this is not the case in a time-varying electric and magnetic field. Consider arbitrary time-varying currents and charges producing a time-varying electric and magnetic field [...]. Consider two points, A and B, in this field, and two paths, a and b, between them, as indicated in the figure. The voltage between these two points is defined as
In this definition, Etotal is the total electric field strength, which means the sum of the "static" part (produced by charges) and the induced part (due to time-varying currents). We know that the integral between A and B of the static part is simply the potential difference between A and B. So we can write
We know that the potential difference, VA - VB, does not depend on the path between A and B, but we shall now prove that the integral in this equation is different for paths a and b. These paths form a closed contour. Applying Faraday's law to that contour, we have
where $$\\Phi\$$ is the magnetic flux through the surface bounded by the contour AaBbA. Since the right side of this equation is generally nonzero, the line integrals of Eind from A to B along a and along b are different. Consequently, the voltage between two points in a time-varying electric and magnetic field depends on the particular path between these two points. This is an important practical conclusion. We always measure the voltage by a voltmeter with leads connected to the two points between which the voltage is being measured. Circuit theory postulates that this voltage does not depend on the shape of the voltmeter leads. We now know that in the time-varying case this is not true."
If you read my other answers on this topic, keep in mind that Popovic uses a different notation from the one I chose to use (it's just a sign difference in the definition, compensated by the reverse order of the starting and ending points), but there are no discrepancies.
Now let's get to the drama.
No build-up of voltage in the wires
The error that Mehdi commits, and many many others like him, is to assume that there is a voltage in the wires. (And by that I mean an appreciable voltage of about the portion of emf associated with the path, and not the tiny ohmic drop in copper.)
Here, from minute 11:20 of his first video
Let's focus on $$\ V_1\$$. There is no voltage, in the sense of path integral of the total electric field, there. There is a 'partial' voltage $$\ V_1\$$ that can be computed as the path integral of the induced $$\E_{ind}\$$ field there (and apart from the sign is equal to the path integral of the Coloumbian electric field there), but that field is only PART of the total field that happens to be there when the ring is placed in the field. The presence of the ring, and in particular of the copper conductor along the black arc indicated by V1, alters the total field because the free charge in the copper will redistribute themselves in order to establish a total field compatible with Ohm's law. Ideally, in a perfect conductor, the total field is zero in the wires. And it is zero thanks to the Coloumbian field generated by the charge that accumulates at the surface of the wire and at the interfaces with the resistive material.
The effect of this superposition makes the total electric field zero in the perfect conductor and different from zero only inside (and around) the resistors. The field inside the resistors will be compatible with Ohm's law there, namely $$\E = j / \sigma_{res}\$$, where $$\\sigma_{res}\$$ is the conductivity of the resistive material. With a real copper conductor, the field will not be exactly zero inside it, but it will assume a very small value compatible with Ohm's law there, namely $$\E = j / \sigma_{copper}\$$. With the currents in Lewin's ring it will be a handful of microvolts per meter.
This picture from my previous answer on the same topic shows the field inside the ring (supposedly shaped like a cylindrical torus of constant section and with clear cut interfaces between copper and resistive materials (imagine the arrows to be located inside the material the ring is made of)
Multivalued voltage in the circuit itself: no probes required
As you can see, if you compute the path integral of that total electric field from point A to point B you will obtain two different results depending on the path you choose: the path that goes through the bigger resistor will give +0.9V, while the path that goes through the smaller resistor will give you -0.1V. Yes, since the field is non-conservative, the path integral depends on the path, and not only on the endpoints. Also note that the sign is different, as well because the total electric field is 'rotational' (i.e. non-conservative).
But most importantly note that THERE ARE NO PROBES INVOLVED IN ALL THIS. This multivaluedness of the voltage is a property of the circuit and is a direct consequence of the fact that, by running the circuit path (and not a limited part of the system, hidden inside a magnetic component) around a variable magnetic field region, the electric field in the region of space occupied by the circuit path is not conservative.
That's all there is to it. Really.
Probes are not required to show this, and this is why the multivaluedness of voltage IS NOT a probing error. It's the way things are. Different from the oversimplified word of circuit analysis, but nonetheless that's how nature works. Lewin's probing is correct and shows how nature works.
Mehdi's, Mabilde's, VanMerveinne's and Duhamel's probing is not, because it shows a voltage corresponding to a partial electric field (namely the coloumbian conservative field they are accustomed to use in circuit theory). By running the probes at right angles with the induced electric field, they stripped the contribute of the induced field in the circuit and are left with the partial contribute of the coloumbian field only.
If you look at the parallel between the electrostatic and magneto-quasistatic phenomena in my other answer, it's as if instead of saying "the electrostatic field inside a conductor in an external field is zero", one would say "there is as electrostatic field inside the conductor because there are induced charges on its surface".
Here's how to see the electric field decomposition in its conservative and solenoidal parts:
Note how the Coloumbian field Ec is only a part of the total electric field. In a perfect conductor Etot is zero (because zero resistivity implies infinite conductivity sigma and, by Ohm's law, E = j / sigma = 0 for finite current densities). This means that in the perfectly conducting wires connecting the resistors, we have Ec = -Eind.
The above decomposition of fields mirrors the decomposition of voltage shown at the beginning of this answer.
Lumpable and unlumpable are not the same thing
Engineers who incorrectly lump the Romer Lewin ring tend to think that there is a voltage build-up inside coils. There is none (Ramo Whinnery and VanDuzer show this clearly on p. 171 of the 2nd edition of "Fields and Waves in Communication Electronics"). The voltage appears at the terminals and in order to treat it as a potential difference (so that KVL can be used) it is necessary that the variable magnetic field that give rise to it be confined in a region of space that is OUTSIDE OF THE CIRCUIT PATH. Something that is not possible with the Romer-Lewin ring. Which is UNLUMPABLE.
Look at this picture:
Lewin's circuit requires the two resistors to be on the opposite sides of the 'variable magnetic field region. You cannot stretch the circuit path by making the part with the resistors shrink to a point (imagine the resistors are points as well) as required by lumped circuit theory without forfeiting this geometric constraint.
In lumpable circuits, KVL holds
The voltages in any point of the dashed circuit path Gamma of the first two circuits are uniquely defined and depends on endpoints alone; the voltages in the portion of system that is outside the dashed circuit path and run around the 'forbidden region', on the other hand, are path dependent. But they can still be hidden inside a component. The region around the dB/dt zone cannot be shrunk because the electrical properties depends on geometry and size; the region we associate with the circuit path gamma, on the other hand, can be shrunk to a point in what is the lumped circuit approximation. You can apply KVL (in its amended form that allows for lumped magnetic components) in a circuit that can be represented with the secondary of a transformer. This is what Mehdi did, believing he was still talking about Lewin's ring:
and he says "this shows that the voltage across the loop is not zero unlike what we've thought but it is equal to VR1 plus VR2. The loop is the secondary of a transformer, with the primary being my coil, KVL holds!". Thus demonstrating he does not understand the true nature of Lewin's ring and - by believing there is a voltage in the wires - not even how a (mutual) inductor works. He should read chapter 5 of Ramo Whinnery VanDuzer, or maybe Haus & Melcher.
In unlumpable circuits, KVL dies (but Faraday's thrive)
In the last circuit of the previous picture, i.e. Lewin's ring, the circuit path itself (and not a part of the system that is outside of it) goes around a variable magnetic field region in such a way that it is not possible to shrink it. Voltages IN THE CIRCUIT will be multivalued since they depends on path. See my answer to What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field? to see the significance of being unable to move the variable field region outside of the circuit path.
And no, in this case, when you are constrained to have the circuit path going around the variable magnetic field region you cannot simlply extend your network with current sources, representing what that magnetic field does to the conductors.
If you try to lump the effects of the variable magnetic field in one, two or as many 'secondary coils' as you like, you are considering a different circuit, i.e. one where the circuit path $$\\Gamma \$$does NOT go around the magnetic field region. And you can see that from the fact that you will lose the nonuniqueness of voltage. It's a new circuit that is good for bean counting, though.
Right or wrong, you get the same numbers
So, why do Mehdi and so many engineers in here keep insisting to represent the effects of variable magnetic field into partial lumped elements?
They are so used to KVL that they cannot let go of it and try to apply it even when it is no longer legit to do so. There is a reason behind it, tho: by incorrectly lumping the effects of the magnetic field into the wires, one can still get the results right.
Incorrect lumping is a way to 'count the beans' that can show which parts of the circuit intercept the most induced field.
(As an aside: the concept of partial inductance - which relies heavily on the use of the vector potential A - is useful in the field of electromagnetic compatibility. But it has to be understood that the -dA/dt part of the total electric field is just... a part of it, not all of it).
A word of advice: when confronting people who believes KVL is always right and/or Lewin committed a probing error, you should ask them to which "current" of thought they belong. Because there are different incompatible views among them: some believe the ring is a transmission line, others that Lewin forgot about self-inductance, others think there is a voltage build up in the wires, and others still just rename Faraday's law "KVL". Notice, though, that those who say Lewin is correct and voltage is simply multivalued are all on the same page.
• Uff, you state the path integral through the electric field is non-conserving, thus you get path-depending voltages. But: by its very construction, the electric potential is defined by a scalar potential field and the matching magnetic vector potential field -making that statement of yours, which seems to be very central to your answer, untenable. The electric potential is defined exactly such, that no matter which path you take, the difference between two points is the same.
– mmmm
Mar 6, 2021 at 9:47
• @mmmm The total electric field is not conservative and it is composed of two parts; a conservative columbian part that is expressed as minus the gradient of a potential function V and a solenoidal part that is minus the time derivative of the vector potential A that gives the overall field a general nonconservative trait. What the voltmeter measures is the path integral of the total electric field. That's tenable. Mar 6, 2021 at 10:03
• +1,@Sredni Vashtar nice answer , mehdi made so many conceptual mistakes it needs atleast 2-3 pages of explanation ,and many still believes that he is right! Mar 6, 2021 at 16:11
• " Ideally, in a perfect conductor, the total field is zero in the wires. And it is zero thanks to the Columbian field generated by the charge that accumulates at the surface of the wire and at the interfaces with the resistive material." How does electrons flow then? I thought when we put a battery and attach to wire, it is electric field inside the wire which directs the electrons to move in some certain direction Mar 6, 2021 at 21:08
• well, you have to satisfy Ohm's law: j = sigma Etot. In a perfect conductor sigma is infinite and the total field is then zero (you can see it as a passage to the limit). Now, if Etot = 0 and Etot = Ec + Eind, this means that Ec = - Eind. I added a little bit in my older answer (the longer one). "Counting beans" is a referende to Feynman's QED book. Loosely speaking it means compute the results without knowing the details of the theory. Mar 6, 2021 at 21:53
In Walter Lewin's lecture, he finds depending on how you keep the lead of the voltmeter, you will measure different values. Supposedly he discusses it in this video according to Mr.Boom, and then infers that this is why a notion of voltage is no longer possible in the case of changing magnetic field.
That's nonsense. Of course there's a notion of voltage in presence of a changing magnetic field. In fact, Maxwell's equation, which describe all electricity (unless you start looking at quantum levels), specifically describe that relationship!
So, cutting this discussion quite short:
Kirchhoff's equation holds for the circuit notation you know, as it's a direct result of the very math that holds together the basics of reality (Ampère's circuital law says: you take any closed loop and calculate the current density along that. Then you get a value proportional to the integral of the magnetic field that goes through the surface enclosed by that line.
Now, that circuit notation you know, with nodes, and lines representing conductors, assumes these conductors are infinitely short, have zero resistance and no current is induced in them (otherwise, you wouldn't just draw a line, but a resistor and/or a current source, right). If no current is induced, then *there can't be a net magnetic field permeating the loop formed by any conductors in a classical circuit.
I.e. the pure application of Kirchhoff's law states your magnetic fields are zero. If they aren't, you need to extend your network with current sources, representing what that magnetic field does to the conductors (again, in these linear network circuits, the conductors, and all elements, are assumed to be zero in size, so that a magnetic field can't have any effect, speed of light doesn't matter etc pp).
So, yes, Kirchhoff's law and changing magnetic fields, thereby induced currents, and hence voltages across any resistive elements, are compatible – if you know the limits of where to apply Kirchhoff's law. The circuit schematics that you're used to are, as a model of a circuit, not sufficient to define things like length, position of conductors to a magnetic field, so obviously, they themselves aren't appropriate to demonstrate this.
• The gives me some insight but hasn't answered the question I had bolded in end of text @Marcus Muller, still I thank you for taking your time to help answer my query. Mar 5, 2021 at 23:47
• This is the only critic I can move to Lewin: he sometimes uses the term "voltage" when he should have used "potential difference". Voltage is always defined as a path integral but is in general dependent of path. When dB/dt is zero, by Faraday, the electric field is conservative and voltage no longer depends on the path but on endpoints only. Hence there is a potential function and potential difference is defined. If you consider this into Lewin's explanation he becomes 100% right. KVL dies when the circuit runs around a dB/dt region and KVL is no longer applicable. Faraday is still good, tho. Mar 6, 2021 at 2:39
• " If no current is induced, then *there can't be a net magnetic field permeating the loop formed by any conductors in a classical circuit." Sorry this is inaccurate at the least. What counts is not the presence of the magnetic field, but the fact that it is changing. Also 'permeating the loop' is unclear and it might suggest the field should extend to the loop: the conductor need not to be immersed in the (changing) magnetic field to see the effects of induction: it just have to go around it (even from a distance). I feel this clarification is important to the OP. Mar 6, 2021 at 2:47
• Also, " the pure application of Kirchhoff's law states your magnetic fields are zero" is dead wrong. Mar 6, 2021 at 4:46
• @SredniVashtar I repeated that three times to avoid people intentionally misunderstanding this. Still you managed to do that, congratulations! What I state is: The thing OP is prone to apply Kichhoff's law to is linear networks represented by circuit diagrams. In these, the conductors are models with zero length, zero induced current, zero resistance... With these, you can't observe a magnetic field. Mar 6, 2021 at 12:30
I think that there is a definite voltage for a circuit even with a field passing through because even though the voltage depends on the line integral, there is only one physical path where current can go through feasible in a circuit!.
I believe one needs to distinguish between the voltage drop from a point A to a point B along a give curve, (which I will call V) and (the negative of) the potential field (which I will call U) which exists when the E field is irrotational, i.e. $$\\nabla \times E = 0\$$.
Let us define a voltage drop as the amount of work performed against an E field required to move a charge along a curve from a given point on the curve to a second given point on the curve. That is,
$$V = \int_{C_{a\rightarrow b}}E\cdot ds$$
Now, to understand the difference between a voltage drop (so defined) and a potential difference.
$$\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}$$
If there is no time-varying magnetic field, this implies that
$$\nabla \times E = 0$$
which means that E is a conservative field. Being a conservative field implies that E is the gradient of a potential function, (which is commonly called V, but we will call it U so as not to confuse it with V which we are using to represent a voltage drop.)
$$E = \nabla U$$
U, being a potential function implies that the voltage drop V from a to b is path independent. The voltage drop V from a to b is equal to the potential at b minus the potential at a. We call such a path independent voltage drop a potential difference.
Note that when a circuit is cut by a time-varying magnetic field, the E field is not conservative, and so there is no potential function. However, this does not entail that there is any ambiguity whatsoever in the voltage drop through a particular path segment.
Now, when you wrote:
I think that there is a definite voltage for a circuit even with a field passing through because even though the voltage depends on the line integral, there is only one physical path where current can go through feasible in a circuit!.
it isn't clear whether the "definite voltage for a circuit" is intended to refer to a path-independent potential difference, or a path-specified "voltage drop".
If one believes in a path independent potential difference in the presence of a time-varying magnetic field affecting one's circuit, then there is a very pointed question that must be answered.
• What is the potential difference between the test points in a Romer-Lewin type experiment?
Is it one of the measured values? If so, which one, and why choose that one? Is it neither of those measured values? If so, can you give a way to measure or calculate it?
Now, here is the weird part, people in physics stack exchange and such seem to agree with Mr. Lewin's result that different measurements give different values see this answer by knzhou. So, it seems the answer is entirely different depending on wether you study EE or physics.
I agree with Lewin that voltmeters attached "on different sides" of a Romer-Lewin type experiment will show different values. It is hard not to agree with such a conclusion as it is easily demonstrated, and has been demonstrated by Electroboom as well. That result is also predicted and explained by KVL.
I think the reason why EEs tend to have a problem with Prof. Lewin is related to his comments that "KVL" is for the birds, that using it is crime, etc.
Kirchhoff's voltage law states that the sum of all voltage drops in a closed loop is equal to the emf in that loop. Now, whenever there is no time varying magnetic field, the voltage drop between two points is path-invariant. This entails KVL. However, KVL is not simply a special case of Faraday's law when there is no time varying magnetic field. KVL applies to specific loops, and therefore specific paths. $$\\frac{\partial B}{\partial t} = 0\$$ entails that KVL will hold, but $$\\frac{\partial B}{\partial t} \ne 0\$$ does not entail that KVL will not hold.
For those who believe that KVL does not hold in a Romer-Lewin type experiment, I have the following question.
• How do you calculate what a particular volt-meter will read in a Romer-Lewin type experiment?
Do you not pick loops, and determine voltages or currents according to Ohms law, Faraday's Law AND KVL? Please provide an example of a calculation of what will be observed without recourse to KVL, or a law derivative of, or equivalent to, KVL.
so what was wrong with the experiment for falsifying Lewin?
My understanding of magnetically induced voltage, and I imagine the understanding of most physicists, is that a Faraday induced emf, i.e. an emf induced in a loop due to a changing magnetic field, applies only to a closed loop. [The loop does not have to be a closed "circuit", it may have an "infinite" resistance, but we cannot speak of the emf induced in "part" of a wire, unless that wire forms a loop.]
Consider the following diagram from Mehdi
Myself, and I imagine most physicists, would say that the reason the loop consisting of the black wire, the red wire, and the (not shown) volt-meter has zero voltage is because it does not enclose any flux. [Nearly enclosing flux does not count]. Mehdi's explanation is that the red wire and the black wire are affected by the same flux, and have equal and opposite emf's induced in them.
Mehdi's explanation is contrary to how I was taught to reason about magnetically induced emf. I won't go so far as to say he is wrong -- he gets the same answer as me, 0V net. But that line of reasoning is unacceptable to a wide group of people (which happens to include a lot of physicists).
It is only on the basis of this idea of emf induced in non-closed curves that Mehdi arrives at his own conclusion that Lewin's conclusions are based upon "bad probing". Since it is commonplace to reject the idea of emf induced in part of a loop, the charge of "bad probing" will not be convincing to many, and it is not convincing to me, even though Mehdi arrives at the same numerical result.
With the assumption that there is an induced emf in part of a loop, Mehdi arrives at the conclusion that there is a definite potential difference between any two points in the diagram above. However, his answer to "what is that voltage", is not satisfactory to me. He says it is the voltage found by connecting the test leads of the voltmeter directly across the line connecting A and B. While this specifies a way of obtaining a value which is unique, that particular choice seems arbitrary to me. Perhaps more importantly, I can't think of any useful purpose to assigning that particular value. That is, apart from telling us what a voltmeter will read in that condition, I can't think of any further calculations that can be done with that value. It wouldn't tell us anything about how any other part of the circuit will behave (as far as I know). If it allowed a useful prediction, I think Mehdi's choice of assigned potential difference would gain greater credibility.
Despite this flaw in Mehdi's explanation, I do agree with his conclusion that KVL is not broken, and disagree with Lewin's that KVL is broken. KVL, as least how I understand it, gives consistent and accurate results, even in the presense of changing magnetic fields. KVL as understood by others may or may not work.
• What you call "KVL and Faraday" is in reality Faraday alone. This is more than a semantic difference. It can be considered a semantic difference - of the kind Lewin called "5+3=8 is not 5+3-8=0" (or something like that) when it is possible to lump the effects of the variable B field and create a mock-up potential difference that we call 'the voltage drop at the inductor'. (see my other 30k long answer). But when the dB/dt region is enclosed by the circuit's path KVL dies. Trying to ressurect it brings the problem of locating that Emf in the circuit. It is not lumped but it is also not... Mar 7, 2021 at 5:20
• ...distributed, in the sense that there is no distributed voltage build-up in the wires and resistors (a lá Mabilde). In the same way that there is no voltage build-up inside a lumped coil. All the voltage appears at the terminals. The Emf has been 'spent' to create the distribution of charge that obliterates the induced field in the conductor. All the emf appears as voltage drop in the resistors. This is the crucial point. Mar 7, 2021 at 5:25
• Whether or not the emf induced by a changing magnetic field is "lumped" or not, it must nevertheless be taken into account. That is why the statement that the total emf in the loop is equal to the total voltage drops in the loop (defined in body of answer) is correct, in either case. How would you calculate what a volt-meter will read in a Romer-Lewin type of experiment without resorting to that fact? If you disagree that it is a fact, can you show a counter-example? Mar 7, 2021 at 5:30
• It's Faraday's law, the law you use. KVL is a specialization of Faraday's law when there are no variable magnetic fields. In that case the rhs of Faraday's law becomes zero and the electric field is conservative - i.e. there is a potential function and voltages in all your circuit can be expressed as potential differences. KVL requires single valued voltages. When the circuit is not lumpable voltages depends on the path and the simplification that is the reason of being of KVL disappears. Renaming Faraday as KVL does not make voltages unique and lumped circuit theory appliable. Mar 7, 2021 at 5:48
• "KVL is a specialization of Faraday's law when there are no variable magnetic fields." I disagree on two points. First, the statement that the total emf in a loop is equal to the total voltage drop (which I understand to be KVL) applies even when there are variable magnetic fields. Second, Faraday's law only describes one kind of emf, that is emf induced by changing magnetic fields. However, emf may also be present in a circuit due to chemical potentials (i.e. batteries). KVL says the TOTAL emf is equal to the sum of voltage drops. Faraday applies only to magnetically induced emf. Mar 7, 2021 at 5:54
Kirchoff's Voltage law will apply if you include the parasitic (mutual) inductance of any wires used in the physical construction of the circuit, calculate the complex impedance instead of just the real resistance, including any energy imparted from outside the simplified circuit (e.g. moving a magnet requires energy). Note that scope probe leads (and the associated "ground" leads) also add (mutual) inductance to any circuit.
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# Halfway through prep. What should I do next?
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11 Aug 2012, 09:35
Hi all,
I just finished the five Manhattan GMAT math strategy guides. I want to move on to the advanced foundation book, but it says I shouldn't work on it unless I have a 70 percentile on practice tests (which I don't).
I have solved the problems in the Official Guide through the Manhattan books. I haven't re-done the ones in my error log, yet.
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Prime Numbers Before Calculator
Enter a natural number to calculate the primes before it:
Ex.: 4, 11, 64, 128, ... until 10,000.
Prime Numbers Before 8200:
The number 8200 is not a prime number because it is possible to express it as a product of prime factors. In other words, 8200 can be divided by 1, by itself and at least by 2, 5 and 41. So, 8200 is a 'composite number'.
List of prime numbers before 8200:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 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5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919, 7927, 7933, 7937, 7949, 7951, 7963, 7993, 8009, 8011, 8017, 8039, 8053, 8059, 8069, 8081, 8087, 8089, 8093, 8101, 8111, 8117, 8123, 8147, 8161, 8167, 8171, 8179, 8191.
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• Prime Numbers up to 8200.
• What is the sum of the first 8200 prime numbers?. | 4,231 | 7,025 | {"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} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.495922 |
https://socratic.org/questions/what-is-the-probability-that-the-first-son-of-a-woman-whose-brother-is-affected-#537134 | 1,723,665,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00560.warc.gz | 411,919,581 | 6,473 | # What is the probability that the first son of a woman whose brother is affected will be affected? What is the probability that the second son of a woman whose brother is affected will be affected if her first son was affected?
## Duchenne muscular dystrophy (DMD) is cause by a relatively rare X-linked recessive allele. It causes progressive muscular wasting, and usually leads to death before age 20.
Jan 18, 2018
P("first son has DMD")=25%
P("second son has DMD"|"first son has DMD")=50%
#### Explanation:
If a woman's brother has DMD then the woman's mother is a carrier of the gene. The woman will get half of her chromosomes from her mother; so there is a 50% chance that the woman will inherit the gene.
If the woman has a son, he will inherit half of his chromosomes from his mother; so there would be a 50% chance if his mother was a carrier that he would have the defective gene.
Therefore if a woman has a brother with DMD there is a
50%XX50%=25% chance that her (first) son will have DMD.
If the woman's first son (or any son) has DMD then the woman must be a carrier and there is a 50% chance that any other son would have DMD. | 284 | 1,150 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 3.90625 | 4 | CC-MAIN-2024-33 | latest | en | 0.967304 |
https://baahkast.com/how-do-you-explain-pid-control/ | 1,652,711,243,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00174.warc.gz | 173,584,501 | 10,345 | # How do you explain PID control?
Table of Contents
## How do you explain PID control?
A PID controller is an instrument used in industrial control applications to regulate temperature, flow, pressure, speed and other process variables. PID (proportional integral derivative) controllers use a control loop feedback mechanism to control process variables and are the most accurate and stable controller.
## How is PID calculated?
A PID calculates the error by calculating difference between actual value and desired value and then sets the deciding parameters accordingly. This error is continuously being calculated until the process stops. Industrial Batch Temperature Control is the best example to understand PID controllers.
## What is KP KI and KD?
Kp is a proportional component, Ki is an integral component, and Kd is a derivative component. Kp is used to improve the transient response rise time and settling time of course. Kd is used to improve the transient response by way of predicting error will occur in the future.
## What is PID controller example?
A good example of temperature control using PID would be an application where the controller takes an input from a temperature sensor and has an output that is connected to a control element such as a heater or fan.
## What is overshoot in PID?
Rise Time is the amount of time the system takes to go from 10% to 90% of the steady-state, or final, value. Percent Overshoot is the amount that the process variable overshoots the final value, expressed as a percentage of the final value.
## What are PID tuning parameters?
In this article, you will learn about PID Tuning Parameters through a few practical examples. PID is an acronym for Proportional, Integral, and Derivative. The PID portion of the controller is a series of numbers that are used as adjustments in order to achieve your objective.
## What is PID cycle time?
Cycle time is the total length of time for the controller output to complete one on/off cycle. Cycle time is used when controller output is configured for time-proportional PID.
## How is KP Ki calculated?
Direct link to this answer
1. Simply, the conversion is as follows(Let K denote gain and Ti denote time constant): Theme. K*(1+1/(Ti*s))
2. is equal to. Theme. Kp+Ki/s.
3. If you equate two expressions, then. Theme. Kp=K. Ki=K/Ti.
## What is KV in control system?
These constants are the position constant (Kp), the velocity constant (Kv), and the acceleration constant (Ka). Knowing the value of these constants, as well as the system type, we can predict if our system is going to have a finite steady-state error.
## What is PID loop in PLC?
In summary, A PID controller is a Proportional, Integral, Derivative controller. PIDs use a control loop feedback or process variable to monitor where the output should be. These usually come in the form of sensors and meters. PIDs come in many different forms including standalone units and PLC programming.
## What is PID full form?
Pelvic inflammatory disease (PID) is an infection of one or more of the upper reproductive organs, including the uterus, fallopian tubes and ovaries.
## What does P controller do?
P Controller: P controller is mostly used in first order processes with single energy storage to stabilize the unstable process. The main usage of the P controller is to decrease the steady state error of the system. As the proportional gain factor K increases, the steady state error of the system decreases.
## What is an example of a PID controller?
PID controller. An everyday example is the cruise control on a car, where external influences such as hills (gradients) would decrease speed. The PID algorithm restores from current speed to the desired speed in an optimal way, without delay or overshoot, by controlling the power output of the vehicle’s engine.
## What is the function of a PID controller?
PID Controller Functions Manual and Automatic Modes. When a controller continually calculates output values based on PV and SP values over time, it is said to be operating in automatic mode. Output and Setpoint Tracking. Alarm Capabilities. Output and setpoint Limiting. Security.
## How to tune a PID controller?
After nulling all the parameters,increase the P term so that the output reaches the target in the shortest possible time.
• If your output starts oscillating,it means you have too much P.
• Now,increase I term slightly until your error goes away.
• If you feel your output is oscillating and it was not before you adjusted your I term,lower I slightly.
• ## What is PID control theory?
PID controller theory. The PID control scheme is named after its three correcting terms, whose sum constitutes the manipulated variable (MV). The proportional, integral, and derivative terms are summed to calculate the output of the PID controller.
https://www.youtube.com/watch?v=srLMG0jlRMk | 1,011 | 4,897 | {"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} | 3.515625 | 4 | CC-MAIN-2022-21 | latest | en | 0.909771 |
https://www.coursehero.com/file/5985850/Chapter-20-Transient-Voltage-Response/ | 1,492,972,378,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118740.31/warc/CC-MAIN-20170423031158-00485-ip-10-145-167-34.ec2.internal.warc.gz | 871,880,986 | 67,379 | Chapter 20. Transient-Voltage Response
# Chapter 20. Transient-Voltage Response - 20...
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20 Transient-Voltage Response Robert C. Degeneff Rensselaer Polytechnic Institute 20.1 Transient-Voltage Concerns. ........................................... 20 -1 Normal System Operation . Sources and Ty pes of Transient-Voltage Excitation . Addressing Transient-Voltage Performance . Complex Issue to Predict 20.2 Surges in Windings . ........................................................ 20 -3 Response of a Simple Coil . Initial Voltage Distribution . Steady-State Voltage Distribution . Transient-Voltage Distribution 20.3 Determining Transient Response. .................................. 20 -6 Histor y . Lumped-Parameter Model . Frequency-Domain Solution . Solution in the Time Domain . Accuracy versus Complexity 20.4 Resonant Frequency Characteristic. ............................... 20 -9 Definitions . Impedance versus Frequency . Amplification Factor 20.5 Inductance Model. ......................................................... 20 -11 Definition of Inductance . Transformer Inductance Model . Inductance Model Validity 20.6 Capacitance Model. ....................................................... 20 -13 Definition of Capacitance . Series and Shunt Capacitance . Equivalent Capacitance for Disk Windings . Initial Voltage Distribution 20.7 Loss Model. .................................................................... 20 -16 Copper Losses . Core Losses . Dielectric Losses 20.8 Winding Construction Strategies. ................................ 20 -19 Design . Core Form . Shell Form . Proof of Design Concept . Standard Winding Tests . Design Margin . Insulation Coordination . Additional System Considerations 20.9 Models for System Studies . .......................................... 20 -24 Model Requirements . Reduced-Order Model 20.1 Transient-Voltage Concerns 20.1.1 Normal System Operation Transformers are normally used in systems to change power from one voltage (or current) to another. This is often driven by a desire to optimize the overall system characteristics, e.g., economics, reliability, ß 2006 by Taylor & Francis Group, LLC.
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or performance. To achieve these system goals, a purchaser must specify—and a designer must configure—the transformer to meet a desired impedance, voltage rating, power rating, thermal charac- teristic, short-circuit strength, sound level, physical size, and voltage-withstand capability. Obviously, many of these goals will produce requirements that are in conflict, and prudent compromise will be required. Failure to achieve an acceptable characteristic for any of these goals will make the overall transformer design unacceptable. Transformer characteristics and the concomitant design process are outlined in the literature [1–4]. Normally, a transformer operates under steady-state voltage excitation. Occasionally, a transformer (in fact all electrical equipment) experiences a dynamic or transient overvoltage. Often, it is these infrequent transient voltages that establish design constraints for the insulation system of the transformer. These constraints can have a far-reaching effect on the overall equipment design. The transformer must be configured to withstand any abnormal voltages covered in the design specification and realistically expected in service. Often, these constraints have great impact on other design issues and, as such, have significant effect on the overall transformer cost, performance, and configuration. In recent years, engineers have explored the adverse effect of transient voltages on the reliability of transformers [5–7] and found them to be a major cause of transformer failure.
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## This note was uploaded on 10/19/2010 for the course ENGINEERIN ELEC121 taught by Professor Tang during the Spring '10 term at University of Liverpool.
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Chapter 20. Transient-Voltage Response - 20...
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Ask a homework question - tutors are online | 887 | 4,284 | {"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-2017-17 | longest | en | 0.507324 |
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# A2 gravitational field KYUEM Physics
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### A2 gravitational field KYUEM Physics
1. 3. Gravity is a very mysterious force. Nobody knows why objects have this attractive force between them, even if they are far apart. This attraction occurs for any object with mass, however small. The force is very small, and always attractive . We never get a repulsive gravitational force.
2. 7. Newton found that all objects accelerate towards the Earth at 9.81 m/s 2 . He also worked out the moon has an acceleration towards the Earth of 2.72 m/s 2 . Therefore: Acceleration of Moon = 2.72 × 10 -3 m/s 2 = 1 = 1 Acceleration of apple 9.81 m/s 2 3610 (60.1) 2 He also worked out the distances between the moon and the centre of the Earth and the distance between the apple and the centre of the Earth. Their ratio also turned out to be 60.1:1.
3. 10. Gravity is a very small force, but it exists between all objects. There is a tiny gravitational attraction between you and the student sitting next to you in your physics lesson. You can't feel it because it's negligible. The only reason we feel gravity is that the Earth is a very large object. Remember that as well as the Earth attracting us towards it, we are attracting the Earth towards us. The tiny attraction has important implications for dust particles coming together in space to form stars and planets.
4. 11. <ul><li>There are several things to note: </li></ul><ul><li>The negative sign is used to mean an attraction. If m 1 , m 2 and the separation between them are positive (and they always are -- there are no negative masses), then the gravitational force between them is attractive. As we saw above, this is why gravity always wins on the large scale. </li></ul><ul><li>The force is proportional to each of the masses. This is not surprising: experimentally we find that two apples, side by side, fall with the same acceleration as one apple, so the force on twice the mass is twice the force on one. </li></ul><ul><li>There is a constant of proportionality, G, the universal constant of gravitation, which we can measure. </li></ul><ul><li>Gravitational forces, like all forces ever discovered, come in pairs that are equal and opposite. If m 1 attracts m 2 with a force F , then m 2 attracts m 1 with a force - F . The observation that all forces come in such pairs is Newton's third law of motion. </li></ul><ul><li>The r 2 in the denominator means that the attraction gets weaker in proportion as the square of the separation. </li></ul>
5. 14. Gravitational Field Strength We are all familiar with the magnetic field of a bar magnet. The field lines show up as areas of attraction and repulsion at the poles and the concentration of field lines at any point. The closer we are to the magnet, the stronger the lines of force.
6. 15. We can do a similar exercise with a gravity field; only this time there are attractive forces involved and no repulsive forces. The Earth, in common with many other planets, is very nearly a perfect sphere (relatively smoother than a billiards ball). Its gravity field is radial with the pull being directly towards the centre. The closer in we get, the stronger the pull.
7. 16. The concentration of gravitational field lines is an indication of the gravitational field strength at any point, which is formally defined as: The gravitational force per unit mass at that point .
8. 17. So we can write that statement as: g = gravitational force = F mass m [Units – newtons per kilogram (N/kg)] The gravitational force per unit mass at that point . You will have met the expression F/m in the context of a = F/m, so it doesn’t take a genius to see that gravitational field strength is the same thing as acceleration. A gravitational field strength of 9.81 N/kg causes an acceleration of 9.81 m/s2.
9. 21. …in summary
10. 23. <ul><li>Equipotential lines are line of equal potential </li></ul>
11. 24. If there is always an attraction between masses separated at a certain distance, how can electron be in its orbit without being attracted to the nucleus? Similarly, how can planet orbiting around the sun if there exist a large force that attracts the planet and the sun towards each? Or how can a satellites be orbiting around the Earth?
12. 25. The motion of an orbiting satellite can be described by the same motion characteristics as any object in circular motion. The velocity of the satellite would be directed tangent to the circle at every point along its path. The acceleration of the satellite would be directed towards the center of the circle - towards the central body which it is orbiting. And this acceleration is caused by a net force which is directed inwards in the same direction as the acceleration.
13. 26. This centripetal force is supplied by gravity - the force which universally acts at a distance between any two objects which have mass. Were it not for this force, the satellite in motion would continue in motion at the same speed and in the same direction. It would follow its inertial, straight-line path. Like any projectile, gravity alone influences the satellite's trajectory such that it always falls below its straight-line, inertial path .
14. 27. Observe that the inward net force pushes (or pulls) the satellite (denoted by blue circle) inwards relative to its straight-line path tangent to the circle. As a result, after the first interval of time, the satellite is positioned at position 1 rather than position 1'. In the next interval of time, the same satellite would travel tangent to the circle in the absence of gravity and be at position 2'; but because of the inward force the satellite has moved to position 2 instead. In the next interval of time, the same satellite has moved inward to position 3 instead of tangentially to position 3'. This same reasoning can be continued to explain how the inward force causes the satellite to fall towards the earth without actually falling into it.
15. 32. 1) Communications satellites orbit the Earth at a height of 36 000 km. How far is this from the centre of the Earth? If such a satellite has a mass of 250 kg, what is the force of attraction on it from the Earth? It is (3.6 x 10 7 m + 6.4 x 10 6 m) = 4.24 x 10 7 m from the centre of the Earth. (Answer should really be given as 4.2 x 10 7 m –significant figures in physical calculations). The force is F = Gm1m2/r 2 = (6.67 x 10 -11 x 6.0 x 10 24 x 250)/ (4.24 x 10 7 ) 2 . This gives an answer of about 56 N, which for information is about less than the weight as a one year old toddler. Example: Newton’s Law of Gravitation Calculation
16. 33. 2) What is the force of attraction from the Earth on you? What do we call this force? What is the force of attraction on the Earth from you? You will need to estimate your own mass in kg. If need to convert, 1 stone is 6.4kg (and there are 14 pounds in a stone). Then use F=Gm1m2/r 2 where r is the radius of the Earth. This force is usually called their weight. The force on the Earth from the student is exactly the same as their first answer, but in the opposite direction.
17. 34. 3) What is the force of attraction from the Sun on you? How many times smaller is this than the force of attraction from the Earth on you? Again, you will need to use your own mass, and the equation F=Gm 1 m 2 /r 2 , but this time r is the average distance from the Sun to the Earth. This force should be about 1650 times less than their weight, of the order of 0.3-0.5 N. Small, but not negligible.
18. 35. 4) The average force of attraction on the Moon from the Sun is 4.4 10 20 N. Taking the distance from the Sun to the Moon to be about the same as that from the Sun to the Earth, what value of mass does this give for the Moon? m 2 = Fr 2 /Gm 1 = (4.4 x 10 20 x (1.5 x 10 11 ) 2 )/(6.67 x 10 -11 x 2.0 x 10 30 ) = 7.4 x 10 22 kg
19. 36. 5) Using the mass of the Moon you calculated in question 4, what is the pull of the Earth on the Moon, if the Moon is 380 000 km away? How does this compare with the pull of the Sun on the Moon? F = Gm 1 m 2 /r 2 = (6.67 x 10 -11 x 6.0 x 10 24 x 7.4 x 10 22 )/ (3.8 x 10 8 ) 2 = 2.1 x 10 20 N The average force of attraction on the Moon from the Sun is 4.4 10 20 N.
20. 37. 6. What is the force of attraction between two people, one of mass 80 kg and the other 100 kg if they are 0.5m apart? F = Gm 1 m 2 /r 2 F = G x 100 x 80 / 0.5 2 = 2.14 x 10 -6 N. This is a very small force but it does increase as the people get closer together! Actually this example is not accurate because Newton's law really only applies to spherical objects, or at least objects so far apart that they can be effectively considered as spherical.
21. 38. 7. What is the force of attraction between the Earth and the Sun? Mass of the Sun = 2 x 10 30 kg, mass of the Earth = 6 x 10 24 kg, distance from the Earth to the Sun = 1.5 x 10 11 m F = Gm1m2/r 2 F = G x 2 x 10 30 x 6 x 10 24 / [1.5 x 10 11 ] 2 = 6.7 x 10 11 N an enormous force!
22. 40. The satellite turns eastward (like our Earth) along the Equator. It stays above the same point on Earth all the time. To maintain the same rotational period as the Earth, a satellite in geostationary orbit must be 35,787 km(22,237 miles) above the Earth. At this distance, the satellite can view half of the Earth's surface. (Its viewing area is called its "footprint.") Because the high-altitude satellite appears to remain fixed in one position (it's really orbiting at the same rate as the Earth turns), it requires no tracking to receive its downlink signal. That is why when we turn our home satellite dish to receive the TV signal from a particular geostationary satellite, we don't have to keep jumping up to adjust its position.
23. 41. <ul><li>One of the advantages of geostationary satellites is that imagery is obtained and displayed constantly, compared to imagery transmitted more sporadically by low Earth-orbiting platforms. </li></ul><ul><li>Most satellites serve one or more functions: </li></ul><ul><li>Communications </li></ul><ul><li>Navigation </li></ul><ul><li>Weather Forecasting </li></ul>
24. 49. Example
25. 51. Example
26. 53. Example
27. 55. Example | 2,686 | 10,317 | {"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-2018-26 | latest | en | 0.943316 |
http://www.ivorcatt.co.uk/97rdeat4.htm | 1,713,263,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00082.warc.gz | 45,586,413 | 6,648 | Death of Electric Current.
The Death of Electric Current Ivor Catt, December 1980 DEATH OF ELECTRIC CURRENT It is worth emphasising that this occurred in two distinct steps during May 1976. First "Displacement Current" went. Then "Electric Current"" went. Displacement current - that shadowy, strange "fudge factor" in the equations, had always been the Achilles Heel of Electric Current, and in the end was instrumental in the downfall of Electric Current. The demise of Electric Current, which suddenly came to Catt while he was talking to Davidson, was completely unexpected, and a great shock. It was unexpected because, whichever side of the Theory N - Theory H [divide] one stood on (there being no Theory C yet), the Electric Current - Energy Current dual looked symmetrical. The only point at issue was which caused which. The emphasis on current is unfortunate, but understandable. The 1976 discovery does not so clearly exclude electric charge from science, but The Catt Question , which came six years later, is about the impossibility of electric charge in its most usual (alleged) function, where both charge and current play their (alleged) roles. Thus, "Theory C" is somewhat of a hybrid. It is a restricted theory, which states that when a battery is connected to a resistor or lamp via two wires, no electric current is involved. Ivor Catt 20 September 2009 @@@@@@@@@ [From "Electronics World", May 2009.] Traditionally, under "Theory N", when a battery is connected via two wires to a resistor or lamp, the battery delivers electric current/charge into the wires. Once the wires gain current/charge, they create magnetic and electric fields between the two wires. Now more than a century ago, when confronting a challenge similar to that of interconnecting high speed logic, Heaviside said; "We reverse this .... ". [Theory H]. The battery delivers electromagnetic field between the connecting wires. In its turn, the field causes electric current/charge in/on the wires. He called the field, travelling at the speed of light, "energy current". However, Heaviside's work on electromagnetic theory disappeared from the record. He was unreferenced in any text book for more than half a century. http://www.ivorcatt.co.uk/x48mm.pdf There the matter rested for a century, until Catt realised that the core problem was for the battery to deliver energy/power to the resistor or lamp. If the battery delivered the electromagnetic field, it was generally agreed that the field carried the energy/power directly into the resistor or lamp. (After all, sunlight is "Energy Current".) Under the new "Theory C", electric current/charge played no role in the key activity, that of delivering energy/power from battery to resistor or lamp. So, under "Theory C", what are electric current and electric charge? What is the role of the interconnecting wires? The answer is that when travelling along in the dielectric between the wires, some of the energy current (or electromagnetic field) penetrates into the wires. Since the dielectric constant of copper or any other conductor can be shown to approach infinity , the velocity of penetration, which depends on the inverse of the dielectric constant, approaches zero. Also, the impedance of a conductor approaches zero, so that very little of the energy current enters the conductors (in the same way as, if we have large and small resistors connected in series, very little power is dissipated in the small resistors.) Maxwell's Equations link field and electric current/charge, and the so-called (but non-existent) current and charge are merely mathematical manipulations of the electromagnetic field. | 765 | 3,658 | {"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} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.953403 |
https://www.alectronics.com/air-flow-detector-project-circuit-diagram-working-and-applications-circuit-schematic-with-explanation/ | 1,601,004,162,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00078.warc.gz | 712,742,111 | 23,783 | Air flow detection is often necessary for many applications or systems where it is necessary to detect the presence of air to have a descriptive picture of the proper functioning of the systems. For example, we need air flow detection in engines to get an estimate about the amount of fuel to be added to the engine, we need air flow detection to check the amount of contamination or the transfer of contamination using chemical media like air. For high power density electronic devices, we need air flow detection to ensure the devices from getting overheated.
#### Principle Behind Air Flow Detector Circuit:
Here, a simple air flow detector circuit is developed which uses a resistance temperature detector as the basic component. This circuit is based on two principles – a) Variation of resistance with temperature, b) Air as an insulator. As current flows through the resistor, it gets heated up. Now when the air is made to flow through the RTD, it is an insulator, allows the resistor to cool down. Thus resistance starts decreasing and the voltage across the RTD decreases. This variation in voltage drop is detected using a timer circuit to give an indication of the air flow.
### Air Flow Detector Circuit Diagram:
Air Flow Detector Circuit Diagram
Circuit Components:
• V1 = 12 V
• R1 = 38 Ohms
• D1= 4.7 V Zener diode, 1N4732
• R2 = 100 Ohms
• Rx = HEL-700 platinum RTD
• R3 = 10K
• C2 = 1uF
• C1= 0.01 uF
• LED = 5V, Green LED
• IC = 555 Timer
#### Air Flow Detection Circuit Design:
This circuit is designed to provide a constant current input to the RTD so that it is heated slightly at the beginning. The RTD selected here is HEL-700 platinum RTD, which works at a maximum operating current of 2 mA. Here we are using a Zener Diode as a voltage regulator to provide a constant current to the RTD.
To design a Zener voltage regulator, we need to first select the Zener diode. Here a Zener diode with Vz = 4.7V is selected. Since input voltage is 12V and required output current is 2mA, we select a load resistance of 100 Ohms, so that maximum current flows through the load and only a small amount of current flows through the RTD. The input resistance selected is given by (Vin-Vz)/(Iz+IL) and is equal to 38 Ohms. Here a 38 Ohms resistor is used as the input resistor.
The next step requires the design of timer monostable multivibrator. Here the timer is used to provide a biasing voltage to the LED, which is about 5V. The LED is required to glow as the voltage across the RTD decreases. Here we select a resistor of 10K and an electrolyte capacitor of 1uF. A ceramic capacitor of 0.01uF is used to connect the control pin to ground.
#### Air Flow Detector Circuit Operation:
This circuit is operated using a 12V battery. The battery voltage is regulated using the Zener diode, which produces a constant voltage. Initially as current flows through the RTD, it gets heated up and its temperature increases, thereby increasing its resistance. Now as current is constant, the voltage across the resistance also tends to increase. When this voltage is applied to the trigger pin of the timer, it fails to trigger the timer and the LED is in the off condition. Now as air flows over the RTD, it starts cooling. This reduces the temperature of the device. As the temperature reduces, the resistance also reduces and so does the voltage across the device. As this voltage reduces below a certain point, the timer gets triggered and the LED starts blinking. As voltage falls further, indicating fall of temperature, the LED starts glowing with full intensity. This indicates the flow of the air.
#### Theory Behind Air Flow Detector Circuit:
The basic theory behind this circuit involves knowledge about three basic parts- Voltage Regulator using Zener Diode, Resistance Temperature detector, and a timer circuit.
Voltage Regulator using Zener Diode:
The zener diode is a simple PN junction diode operated in reverse bias condition. It basically works on the principle of breakdown – Avalanche and Zener. Zener breakdown occurs at a reverse bias voltage between 2V to 8V, when highly strong electric field intensity causes the electrons to break free from the atoms and form free electron-hole pairs. The avalanche breakdown occurs above 8V when high-speed charge carriers cause disrupt of the covalent bond due to the collision, leading to the formation of free electrons.
As can be seen by the characteristics, for a large variation in current through the diode, the voltage across the diode remains very small or constant. This unique feature is utilized in many applications by using a Zener diode as the voltage regulator.
Resistance Temperature Detector:
A resistance temperature detector or RTD is a metal resistor whose resistance changes with temperature. It is based on the fact that in metals, as temperature increases, the lattice vibrations increases. These vibrations cause collision among the electrons. As collisions increase, the energy of the electrons decreases, causing a decrease in the flow of free electrons, leading to low conductivity. Thus, with an increase in temperature, the resistance increases. An RTD is constructed basically using platinum. At 0 degree Celsius, the resistance of an RTD is about 100 Ohms.
555 Timer Multivibrator:
Multivibrator circuit is used to produce a pulsed output signal. It is triggered when a low-level signal is applied to the trigger pin of the IC. The 555 timer IC is an 8 pin IC and the timing of the output signal is given by T=1.1 RC.
#### Applications of Air Flow Detector Circuit:
This circuit can be used to detect the flow of air in areas like a car engine, where it is required to estimate the amount of fuel needed by the engine. Apart from being used as an air flow detector, this circuit can also find its application as a temperature detector circuit. With slight modifications, this circuit can be used to control loads like a fan, based on temperature sensing.
#### Limitations of Air Flow Detector Circuit:
1. Since the Zener diode is being used, the efficiency of the circuit is affected. This is because the loss in the series resistor causes a decline in efficiency in case of heavy loads.
2. The resistance temperature detector used is expensive and easily affected by shock and vibration. | 1,370 | 6,311 | {"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-2020-40 | latest | en | 0.889915 |
http://m.educationquizzes.com/gcse/physics/forces-moments/ | 1,511,381,678,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806660.82/warc/CC-MAIN-20171122194844-20171122214844-00599.warc.gz | 188,909,564 | 11,013 | # Forces - Moments
A moment is a turning effect of a force and you come across them every day of your life. For the physics GCSE, you need to know some examples of forces that create turning effects, how the principle of moments can be utilised and how to calculate the magnitude of turning forces and moments.
If an object is fixed in place using a pivot (a shaft or other fixing that is designed to allow movement of the object), then you have the exact situation required for a turning force to arise. When an object is placed on something narrower, the narrow object can act as a pivot too, for example, a plank placed on a brick. Finally the edges or curved parts of objects can act as pivots too - take for example, a fork. Placed with the curved side uppermost, it is stable, however, placed with the curved side on the table, pressing downwards on the prongs will create a turning force that raises the handle.
The pivot is sometimes referred to as the fulcrum, especially in the context of levers.
When a force is applied to a pivoted object, if it is to one side or the other of the pivot, the object will experience a moment. If the magnitude of the moment is sufficient to overcome any frictional forces, then the object will turn around the pivot. The size of the moment depends on the size of the force applied and its perpendicular distance from the pivot. The equation for calculating a moment is simply the force multiplied by the distance. The SI unit of force is the newton; the SI unit for distance is metres, so the SI units for moments will be newton metres.
All pivoted systems obey the principle of moments. This tells us that if a pivoted object is not moving, the sum of the anticlockwise moments is the same as the sum of the clockwise moments. Questions on your exam paper will often ask you to work out the force or the distance that would be needed to balance a specific moment. Answering such questions is just a case of rearranging the equation that represents the principle of moments to isolate the term you are required to work out:
anticlockwise force x anticlockwise distance = clockwise force x clockwise distance
In some parts of the world, the circus is still very popular. Several circus acts utilise the principle of moments. Trapeze artists are one such group of performers. The trapeze is just an object that is fixed in place by a pivot, high in the big top tent. They can change the moment on the trapeze by altering the position of their centre of gravity. Some of the more spectacular trapeze artists even use the principle of moments to reach the trapeze - they use a see-saw! One or more members of the troupe will stand on a platform and the trapeze performer stands on the end of a specially strengthened see-saw. The people on the platform jump down together and land on the end of the see-saw. The moment they create is larger than the one created by the single person on the other end, so the force moves the see-saw. As you know, when a force moves, work is done. This work done transfers the gravitational potential energy of the jumpers into kinetic energy, firing the performer upwards to reach their trapeze. There is lots of physics in operation at a circus.
Did you know...
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1. What is a moment?
2. What is the formula for the size of a moment?
This is a direct proportionality so if the force is greater, the moment is larger. If the distance is longer, the moment will be larger too. The opposite is true for smaller values of both terms of the equation
3. What does d in the above equation stand for?
This is the shortest distance between the point at which the force is acting and the point around which an object pivots
4. If an object is not turning, the total clockwise moment, compared to the total anti-clockwise moment about any pivot, must be what?
For an object to be at rest on a pivot, all the forces acting on it must be in equilibrium
5. What is the size of a moment if F = 10 N and d = 125 cm?
Did you remember to convert 125 cm into metres?
6. A see-saw is balanced on a pivot with two children on it. One child is sitting 1.5 m to the left of the pivot and has a mass of 50 kg. Another child of mass 30 kg is sitting on the right hand side of the pivot. What distance away from the pivot is the child on the right of the pivot?
The key word in the question is balanced so the principle of moments calculation can be applied
7. What is the force that creates a moment of 10 N m when it is applied 0.25 m from the pivot?
Rearrangement of the equation for calculating the size of a moment
8. If the point of application of the force was moved further away from the pivot, what would be the effect on the moment?
Changing either (or both) of the force and distance will alter the moment. An example of this would be using a spanner to turn a nut. A longer spanner enables you to apply the force from a greater distance, increasing the moment and magnifying the force that your muscles can apply
9. A plank of wood is balanced on a pivot. One mass of 10 kg is then placed 1 m to the left of the pivot on the wood. What weight needs to be placed 0.5 m to the right of the pivot for the wood to still be balanced?
Did you spot that the question asked for the weight? The weight of an object is the mass multiplied by the strength of the gravitational field - 10 N/kg is an acceptable approximation for the gravitational field strength at the surface of the Earth
10. Which of the following is an example of the principle of moments being utilised?
The crowbar is being used as a lever. Levers are a favourite of the examiners for testing your knowledge of moments | 1,282 | 5,772 | {"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-2017-47 | latest | en | 0.921591 |
https://forum.dynamobim.com/t/bimorph-curve-removeduplicates-optimized/15387 | 1,679,674,775,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00455.warc.gz | 308,747,537 | 7,650 | # Bimorph - Curve.RemoveDuplicates Optimized?
@Thomas_Mahon – Should we be seeing the same sort of speed increase from the Curve.RemoveDuplicates node in Bimorph?
For my purposes, I solved the issue a different way.
Could you share how did you solve it will help others with similar issues.
Of course!
I have 2 lists of curves and I wanted the curves that were common in both. I tried a version of SetIntersection previously without success so I thought merging the lists and dropping the duplicates would work as well. It was pretty slow in comparison to the other aspects of the graph so I rolled my own by converting the lines into strings and comparing the strings in a Python script I cobbled together.
It’s a bit of a hack but it’s fairly fast and accurate enough for my needs.
1 Like
Hi Greg
Glad to hear you found an efficient workflow (efficientcy is king!).
I’ll answer your question in case anyone else has the same question:
It is optimised, but it’s not going to be anywhere near as fast as the new nodes for the following reasons:
1. There is no purpose built method in the API to perform the process, which means:
2. I’ve written a custom algorithm to establish if curves are duplicates
3. The curve comparison process is optimised (I don’t perform brute-force checking as it’s unnecessary - it uses the same logic as Curve.IntersectAll to prevent redundant tests)
As a result it won’t run as fast as the new intersection node. In addition, checking for duplicates is a complex problem, irrespective of which method/algorithm is used to solve the problem. I did a presentation on this node at one of the Dynamo London user groups which explains the inner workings of the geometry checks the algorithm performs to identify duplicates:
In respect to lines then, they could be better handled since all three tests are required to handle all curves types (bsplines/nurbs, arcs, ellipse, ect), whereas only the first test is needed for lines and that would explain why its possible to yeild faster results. It boils down to code consistency: have one method to rule them all (and accept some profligacy in return for code brevity) or create exceptions for certain object types (ie lines) and create rules to handle the exceptions. For speed of development, I opted for the former, but its a good question and I’ll review the nodes functionality.
1 Like
Actually, the solution I came up with isn’t even comparing the end points… it’s comparing the string version of the line output! Super non-technical, I know, but at the time it was all I could come up with and I’m pretty sure for my situation it’s sufficient.
For smaller data sets I found the node you produced to work great. It’s when I was trying to compare several thousand lines at a time when things slowed to a crawl.
Out of curiosity, are you first checking the intersection of the curves bounding boxes and then only comparing those curves whose boxes intersect?
The speed will be dependent on a number of factors, such as the number of overlaps and the type of curves being processed. However, its an intensive process as there are potentially billions of exceptions. So the rules that have been implemented are the bare minimum to capture 99.9% of duplicates. Its designed for versatility which always restricts efficiency.
It doesn’t use bounding boxes to filter surrounding elements like the new geometry intersection nodes. I looked it at but never got round to testing whether it would speed the process up considering it would rely on BoundingBox.ByGeometry and BoundingBox.Intersects from Dynamos ProtoGeometry library (there are no methods in the Revit API to get a bounding box from a Curve, nor test for containment). In theory, it should be quicker, but without testing I cant say for sure and so I didn’t implement it.
If you can give me some use cases where the performance is impacting your workflow, I can revisit this node and look at improving performance.
I’ll export the endpoints of the largest dataset and send it your way.
1 Like
@Greg_McDowell Curve.RemoveDuplicates has been refactored in BimorphNodes v2.2 resulting in an 80% performance increase. It will be released today.
3 Likes
Sounds great! Can’t wait to get back into that definition and see how it works.
I guess I never did send you that dataset from back in October. Sorry about that. | 916 | 4,369 | {"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-2023-14 | latest | en | 0.952826 |
http://www.conversion-website.com/area/centiare-to-square-chain.html | 1,723,228,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00026.warc.gz | 34,781,744 | 4,428 | # Centiares to square chains (ca to sq ch)
## Convert centiares to square chains
Centiares to square chains conversion calculator above calculates how many square chains are in 'X' centiares (where 'X' is the number of centiares to convert to square chains). In order to convert a value from centiares to square chains (from ca to sq ch) simply type the number of ca to be converted to sq ch and then click on the 'convert' button.
## Centiares to square chains conversion factor
1 centiare is equal to 0.0024710538149159 square chains
## Centiares to square chains conversion formula
Area(sq ch) = Area (ca) × 0.0024710538149159
Example: Find the number of square chains in 569 centiares.
Area(sq ch) = 569 ( ca ) × 0.0024710538149159 ( sq ch / ca )
Area(sq ch) = 1.4060296206871 sq ch or
569 ca = 1.4060296206871 sq ch
569 centiares equals 1.4060296206871 square chains
## Centiares to square chains conversion table
centiares (ca)square chains (sq ch)
40.0098842152596636
60.014826322889495
80.019768430519327
100.024710538149159
120.029652645778991
140.034594753408823
160.039536861038654
180.044478968668486
200.049421076298318
220.05436318392815
centiares (ca)square chains (sq ch)
2500.61776345372897
3500.86486883522056
4501.1119742167122
5501.3590795982037
6501.6061849796953
7501.8532903611869
8502.1003957426785
9502.3475011241701
10502.5946065056617
11502.8417118871533
Versions of the centiares to square chains conversion table. To create a centiares to square chains conversion table for different values, click on the "Create a customized area conversion table" button.
## Related area conversions
Back to centiares to square chains conversion
TableFormulaFactorConverterTop | 516 | 1,707 | {"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} | 3.421875 | 3 | CC-MAIN-2024-33 | latest | en | 0.51444 |
http://forum.allaboutcircuits.com/threads/amplitude-in-an-ac-circuit.64346/ | 1,485,126,831,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00258-ip-10-171-10-70.ec2.internal.warc.gz | 101,099,693 | 14,308 | # amplitude in an AC circuit
Discussion in 'Homework Help' started by paul_alan, Jan 5, 2012.
1. ### paul_alan Thread Starter Member
Nov 5, 2011
43
0
if you have an AC circuit with say, two resistors, and two capacitors connected to a signal generator putting out 15khz and an amplitude of 6V. if the frequency was to be increased or decreased, what effect would it have on the amplitude of a node between a cap and a resistor? would the amplitude decrease with an increase of frequency because the capacitors opposition to current flow decreases?
2. ### praondevou AAC Fanatic!
Jul 9, 2011
2,936
489
Connected How? Post a circuit please.
Consider simply the impedance of a capacitor changing with frequency. Your circuit becomes a simple resistor/impedance network.
Impedance of a capacitor decreases when frequency increases.
3. ### paul_alan Thread Starter Member
Nov 5, 2011
43
0
i'm looking at node 5 which is the furthest to the right above the 10k resistor and the 0.1 microfarad cap.
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4. ### praondevou AAC Fanatic!
Jul 9, 2011
2,936
489
The AC voltage on node 5 (to Gnd) will decrease with lower frequency, because the capacitors impedance increases.
Calculate the impedance of the caps for any given frequency and make a network analysis.
At 15kHz their impedance is a bit more than 100 Ohm, so they are almost negligible.
paul_alan likes this.
5. ### thatoneguy AAC Fanatic!
Feb 19, 2009
6,357
718
Check out the impedance graph paper in my sigline, there's an example of how to use it in the thread, it should give you some pretty good insight of how frequency and capacitance interact, while resistance is a constant.
Let me know if you have problems. | 441 | 1,707 | {"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-2017-04 | latest | en | 0.908531 |
https://www.hitpages.com/doc/4514141356163072/7/ | 1,484,827,058,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280587.1/warc/CC-MAIN-20170116095120-00031-ip-10-171-10-70.ec2.internal.warc.gz | 928,314,701 | 11,337 | X hits on this document
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The quantity column values are the solution values for the vari- ables in the basic feasible solution.
The initial simplex tableau always begins with the solution at the origin, where x1 and x2 equal zero. Thus, the basic variables at the origin are the slack variables, s1 and s2. Since the quantity values in the initial solution always appear as the right-hand-side values of the constraint equations, they can be read directly from the original constraint equations.
The number of rows in a tableau is equal to the number of constraints plus four.
The top two rows and bottom two rows are standard for all tableaus; however, the num- ber of middle rows is equivalent to the number of constraints in the model. For example, this problem has two constraints; therefore, it has two middle rows corresponding to s1 and s2. (Recall that n variables minus m constraints equals the number of variables in the prob- lem with values of zero. This also means that the number of basic variables with values other than zero will be equal to m constraints.)
The number of columns in a tableau is equal to the number of variables (including slacks, etc.)
Similarly, the three columns on the left side of the tableau are standard, and the remain- ing columns are equivalent to the number of variables. Since there are four variables in this model, there are four columns on the right of the tableau, corresponding to x1, x2, s1, and s2.
plus three.
The c values are the contribution to profit (or cost) for each variable.
j
The next step is to fill in the c values, which are the objective function coefficients, rep- resenting the contribution to profit (or cost) for each variable x or sj in the objective func- tion. Across the top row the c values 40, 50, 0, and 0 are inserted for each variable in the model, as shown in Table A-3.
j
j
j
The Simplex Method
A-7
# Table A-3
The Simplex Tableau with c j V a l u e s
c j
Basic
40
50
0
0
Variables
Quantity
x1
x2
s1
s2
0 0
s1 s 2
40 120
____________________________
z j c j - z j
____________________________
T h e v a l u e s f o r o n t h e l e f t s i d e o f t h e t a b l e a u a r e t h e c o n t r i b u t i o n s t o p r o f i t o f o n l y r those variables in the basic feasible solution, in this case s1 and s2. These values are inserted at this location in the tableau so that they can be used later to compute the values in the o w . z j c j
The columns under each variable (i.e., x1, x2, s1, and s2) are filled in with the coefficients of the decision variables and slack variables in the model constraint equations. The s1 row represents the first model constraint; thus, the coefficient for x1 is 1, the coefficient for x2 is 2, the coefficient for s1 is 1, and the coefficient for s2 is 0. The values in the s2 row are the second constraint equation coefficients, 4, 3, 0, and 1, as shown in Table A-4.
# Table A-4
The Simplex Tableau with Model Constraint Coefficients
c j
0 0
Basic Variables
s s2 1
z j c j - z j
40
50
0
0
Quantity
x1
x2
s1
s2
40
1
2
1
0
120
4
3
0
1
____________________________
Document views 498 Page views 498 Page last viewed Thu Jan 19 09:12:23 UTC 2017 Pages 63 Paragraphs 3770 Words 30893 | 878 | 3,290 | {"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} | 3.515625 | 4 | CC-MAIN-2017-04 | latest | en | 0.837336 |
http://www.varsitytutors.com/act_math-help/how-to-graph-a-line | 1,485,043,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00203-ip-10-171-10-70.ec2.internal.warc.gz | 739,958,517 | 39,306 | # ACT Math : How to graph a line
## Example Questions
### Example Question #1 : How To Graph A Line
What is the distance between (7, 13) and (1, 5)?
12
None of the answers are correct
7
5
10
10
Explanation:
The distance formula is given by d = square root [(x2 – x1)2 + (y2 – y1)2]. Let point 2 be (7,13) and point 1 be (1,5). Substitute the values and solve.
### Example Question #1 : How To Graph A Line
What is the slope of this line?
Explanation:
The slope is found using the formula .
We know that the line contains the points (3,0) and (0,6). Using these points in the above equation allows us to calculate the slope.
### Example Question #1 : How To Graph A Line
What is the amplitude of the function if the marks on the y-axis are 1 and -1, respectively?
0.5
3π
1
π
2π
1
Explanation:
The amplitude is half the measure from a trough to a peak.
### Example Question #62 : Graphing
What is the midpoint between and ?
None of the answers are correct
Explanation:
The x-coordinate for the midpoint is given by taking the arithmetic average (mean) of the x-coordinates of the two end points. So the x-coordinate of the midpoint is given by
The same procedure is used for the y-coordinates. So the y-coordinate of the midpoint is given by
Thus the midpoint is given by the ordered pair
### Example Question #2 : How To Graph A Line
If the graph has an equation of , what is the value of ?
Explanation:
is the -intercept and equals can be solved for by substituting in the equation for , which yields
### Example Question #2 : How To Graph A Line
The equation represents a line. This line does NOT pass through which of the four quadrants?
I
II
Cannot be determined
IV
III
III
Explanation:
Plug in for to find a point on the line:
Thus, is a point on the line.
Plug in for to find a second point on the line:
is another point on the line.
Now we know that the line passes through the points and .
A quick sketch of the two points reveals that the line passes through all but the third quadrant.
### Example Question #1 : Graphing Functions
Refer to the above red line. A line is drawn perpendicular to that line, and with the same -intercept. Give the equation of that line in slope-intercept form.
Explanation:
First, we need to find the slope of the above line.
The slope of a line. given two points can be calculated using the slope formula
Set :
The slope of a line perpendicular to it has as its slope the opposite of the reciprocal of 2, which would be . Since we want this line to have the same -intercept as the first line, which is the point , we can substitute and in the slope-intercept form:
### Example Question #6 : How To Graph A Line
Refer to the above diagram. If the red line passes through the point , what is the value of ?
Explanation:
One way to answer this is to first find the equation of the line.
The slope of a line. given two points can be calculated using the slope formula
Set :
The line has slope 3 and -intercept , so we can substitute in the slope-intercept form:
Now substitute 4 for and for and solve for : | 797 | 3,131 | {"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} | 4.59375 | 5 | CC-MAIN-2017-04 | latest | en | 0.88674 |
https://www.miniphysics.com/newtons-law-of-universal-gravitation.html?shared=email&msg=fail | 1,623,646,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00313.warc.gz | 809,390,101 | 17,527 | Newton’s Law Of Universal Gravitation
Show/Hide Sub-topics (Gravitation | A Level)
Newton’s Law Of Universal Gravitation states that every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$F = G \frac{m_{1}m_{2}}{r^{2}}$
, where:
• F is magnitude of the gravitational force between the two point masses,
• G is the gravitational constant,
• m1 is the mass of the first point mass,
• m2 is the mass of the second point mass,
• r is the distance between the two point masses.
The gravitational force is inversely proportional to the square of the separation of the particles.
$F \propto \frac{1}{r^{2}}$
* Only applicable to point masses | 187 | 794 | {"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} | 3.09375 | 3 | CC-MAIN-2021-25 | latest | en | 0.891534 |
http://www.bills.com/extra-mortgage-payment/ | 1,386,709,334,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164025146/warc/CC-MAIN-20131204133345-00033-ip-10-33-133-15.ec2.internal.warc.gz | 330,546,470 | 21,475 | Question:
# Extra Mortgage Payment
## Should I make extra mortgage payments to slash my mortgage interest expense?
The short answer is yes, every mortgage I have seen allows the borrower to make extra payments or add to their monthly payments. But is it smart to do so? This is a more complicated question than it appears, and not because of the math. Before we get into the tough questions, let us look at the mortgage in isolation. For the sake of argument, let us assume you have a \$200,000 mortgage balance, a 5% fixed interest rate, and a 30-year (360-month) loan.
The amortization table with no extra payments looks as follows:
Payment Number Monthly Payment Principal Interest Total Interest Remaining Balance
1 \$1,073.64 \$240.31 \$833.33 \$833.33 \$199,759.69
180 \$1,073.64 \$507.94 \$565.70 \$129,589.30 \$135,259.87
360 \$1,073.64 \$1,069.19 \$4.45 \$186,511.57 \$0
The table above shows us that the homeowner pays the lender \$186,511 in interest if the homeowner makes only the monthly payments and nothing more.
Now let us look at the amortization table if the homeowner adds \$75 per month to the monthly payment:
Payment Number Monthly Payment Principal Interest Total Interest Remaining Balance
1 \$1,148.64 \$315.31 \$833.33 \$833.33 \$199,684.69
180 \$1,148.64 \$666.47 \$482.17 \$122,959.10 \$115,054.67
310 \$1,148.64 \$1,044.91 \$4.35 \$157,128.67 \$0
Adding \$75 a month cuts the total interest expense by almost \$30,000, and reduces the time to pay off the loan by almost 4 years.
Adding an extra \$1,073.64 payment each year, a 13th payment, has almost the same effect as adding \$75 per month to each payment. Does this mean that adding \$75 per month or a 13th payment per year is always a good idea? After all, it saves nearly \$30,000 in interest expense. The answer is no, not always.
### Retirement Accounts
If you have not yet reached the limit on your 401(k) or IRA accounts, the extra money you would use to pay the mortgage is better off contributed to these accounts. In a 401(k) plan, you can typically contribute up to \$12,000 annually. In an IRA (individual retirement account) you can usually save up to \$4,000 annually (although this limitation may vary). Keep in mind that the contributions to these accounts are not taxed. Remember also that your income tax deductions on your mortgage expense decreases later in the life of the loan. In our examples above, we can see the interest portion of the loan in the last year is essentially pocket change.
In other words, it may not be wise to sacrifice your retirement investing at the expense of cutting your mortgage interest expense.
### Investing
You could also use the \$75 per month or \$1,073 annually to invest in other short-term high-yield products such as mutual funds, but it entirely depends on the amount of risk you are willing to take. To justify the risk the after-tax returns need to exceed the interest cost on your mortgage.
### Conclusion
If your retirement investing is at the maximum amount allowable, and you have an investment plan in place, then by all means use any surplus cash on your mortgage to shorten its life. However, if you are not contributing to your employer's 401(k), or have not contributed to an IRA, then putting money into your retirement fund would be a smarter use of your surplus.
I hope this information helps you Find. Learn & Save.
Best,
Bill | 813 | 3,408 | {"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-2013-48 | latest | en | 0.892432 |
https://electronics.stackexchange.com/questions/tagged/operational-amplifier+transfer-function | 1,563,332,030,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00188.warc.gz | 384,747,605 | 38,336 | # All Questions
78 questions
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### Finding transfer function of a compensator circuit similar to a type 2 compensator topology
I know how to find transfer functions of op-amp circuits using equations derived from using Kirchhoff's current law (nodal analysis), and normally I don't have any problems solving them. However, I ...
781 views
### Opamp: determine the transfer function with negative RC feedback
I would like to find the transfer function of the above ideal opamp circuit. One of my approach was neglecting the effect of R2(assuming the voltages on the both side of R2 are the same) and applying ...
105 views
### Transfer function for a OP-AMP [duplicate]
Can someone tell me if the transfer function for this circuit is: R2/[(w^2 * C^2 * R1) - jwC] thanks
61 views
### AC measurement in a circuit
I used the spice language to build an integrator op-amp, and one of the questions I had to address is to run an AC analysis on it. long story short- by looking at the transfer function of an ...
1k views
### Calculate transfer function of transconductance amplifier circuit
How can I model/calculate the behavior from Vin to Iout of the following circuit? (Credits of the picture go to Analog Devices): Essentially, I am interested in the large signal response, such as a ...
459 views
### Understanding in-loop compensation for capacitive-loaded OpAmp
I'm trying to understand the in-loop compensation for a simple amplifier loaded with a capacitive load, as seen in this article from Analog: http://www.analog.com/library/analogdialogue/archives/38-...
528 views
### What should be the polarities of the op amp input terminals such that the system is stable? [closed]
simulate this circuit – Schematic created using CircuitLab Ignore the - and + signs on the opamp (I couldn't erase it using the schematic), is it possible to determine the polarity of the op ...
2k views
### Calculation of transfer function in OP-AMP circuits
In a simple circuit like the following: simulate this circuit – Schematic created using CircuitLab I calculate the transfer function using I1 = I2: Vin = ...
623 views
### Find transfer function of Op Amp system
Trying to determine the transfer function for this Op Amp circuit using the rules that $V-=V+$ and that no current flows into the Op amp. I think that $Vout=IR2$ and determined the trasnfer ...
717 views
### DC gain of non-inverting integrator / Derivation of DC results from transfer function
(The pictured circuit contains a serious mistake, see my answer for more detail. I decided to leave the question in its original form for educational purposes, though.) Setting The (sub-)circuit in ...
1k views
### How do I realize a transfer function with multiple zeros?
I designed a compensation function for a control system in Matlab and now I'm trying to figure out how to build it using op-amps. I thought I might be able to use a couple of Sallen-Key amplifier ...
82 views
85 views
### Simulating a tranfer function with opamps
I am asked to simulate the following transfer function using operational amplifiers: Obviously what it evokes is a second order filter, my question is what kind of filter should I use (Butterworth, ... | 729 | 3,272 | {"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} | 2.8125 | 3 | CC-MAIN-2019-30 | latest | en | 0.894851 |
https://cracku.in/183-which-of-the-following-means-t-is-wife-of-p-x-sbi-clerk-2009-1 | 1,726,457,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00611.warc.gz | 156,567,396 | 26,487 | Instructions
A – B’ means ‘A is father of B’.
A ÷ B’ means ‘A is daughter of B’.
A + B’ means ‘A is son of B’.
A × B’ means ‘A is wife of B’.
Question 183
# Which of the following means T is wife of P?
Solution
P × S ÷ T => S is daughter of T and husband of P. Hence, this does not imply that T is the wife of P
P ÷ S × T => This also means that S is the wife of T and P is his daughter. Hence, this is also wrong.
P - S ÷ T => S is the daughter of and P is the father of S. Hence, T must be the wife of P.
P + T ÷ S => P is the son of T. Hence this is not correct. | 184 | 570 | {"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} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.995442 |
https://www.convert-me.com/en/convert/percent/dozen/dozen-to-giga.html | 1,719,145,095,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00540.warc.gz | 609,201,840 | 18,043 | ## Convert Dozen (Quantity Units) to Giga (G, Metric Prefixes)
This page features online conversion from dozen to giga. These units belong to different measurement systems. The first one is from Quantity Units. The second one is from Metric Prefixes.
If you need to convert dozen to another compatible unit, please pick the one you need on the page below. You can also switch to the converter for giga to dozen.
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## Quantity Units
Units: unit, point (1) / pair, brace, yoke / nest, hat trick / half-dozen / decade, dicker / dozen / baker's dozen / score / flock / shock / hundred / great hundred / gross / thousand / great gross
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## Percentages and Parts
dozen to percent (%) dozen to permille (‰)
dozen to parts per million (ppm) dozen to parts per billion (ppb)
Units: percent (%) / permille (‰) / parts per million (ppm) / parts per billion (ppb)
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## Fractions
This section answers a question like "How many one sevenths are there in 1 half?". To get an answer enter 1 under half and see the result under 1/7. See if you can use this section to find out how many one-sixths are there in 15 one-nineths.
dozen to half or .5 (1/2) dozen to one third or .(3) (1/3) dozen to quart, one forth or .25 (1/4) dozen to tithe, one fifth or .2 (1/5) dozen to one sixth or .1(6) (1/6)
dozen to one seventh or .142857 (1/7) dozen to one eights or .125 (1/8) dozen to one ninth or .(1) (1/9) dozen to one tenth or .1 (1/10) dozen to one sixteenth or .0625 (1/16) dozen to one thirty-second or .03125 (1/32)
Units: half or .5 (1/2) / one third or .(3) (1/3) / quart, one forth or .25 (1/4) / tithe, one fifth or .2 (1/5) / one sixth or .1(6) (1/6) / one seventh or .142857 (1/7) / one eights or .125 (1/8) / one ninth or .(1) (1/9) / one tenth or .1 (1/10) / one sixteenth or .0625 (1/16) / one thirty-second or .03125 (1/32)
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## Metric Prefixes
These prefixes are widely used in Metric System. They apply to any unit, so if you ever see, e.g. kiloapple, you know it's 1000 apples.
dozen to quecto (q) dozen to ronto (r) dozen to yocto (y) dozen to zepto (z) dozen to atto (a) dozen to femto (f) dozen to pico (p) dozen to nano (n) dozen to micro (µ, mc) dozen to milli (m) dozen to centi (c) dozen to deci (d)
dozen to deka (da) dozen to hecto (h) dozen to kilo (k) dozen to mega (M) dozen to giga (G) dozen to tera (T) dozen to peta (P) dozen to exa (E) dozen to zetta (Z) dozen to yotta (Y) dozen to ronna (R) dozen to quetta (Q)
Units: quecto (q) / ronto (r) / yocto (y) / zepto (z) / atto (a) / femto (f) / pico (p) / nano (n) / micro (µ, mc) / milli (m) / centi (c) / deci (d) / deka (da) / hecto (h) / kilo (k) / mega (M) / giga (G) / tera (T) / peta (P) / exa (E) / zetta (Z) / yotta (Y) / ronna (R) / quetta (Q)
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## Number of Performers
Units: solo / duet / trio / quartet / quintet / sextet / septet / octet
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Our goal is to make units conversion as easy as possible. Got ideas how to make it better? Let us know | 1,498 | 4,721 | {"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-2024-26 | latest | en | 0.722868 |
http://maps.thefullwiki.org/Mathematician | 1,576,206,414,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548537.21/warc/CC-MAIN-20191213020114-20191213044114-00410.warc.gz | 89,765,826 | 13,331 | # Mathematician: Map
### Map showing all locations mentioned on Wikipedia article:
A mathematician is a person whose primary area of study and/or research is the field of mathematics. Mathematicians are concerned with particular problems related to logic, space, transformations, numbers and more general ideas which encompass these concepts.
Some notable mathematicians include Sir Isaac Newton, Johann Carl Friedrich Gauss, Archimedes of Syracuse, Leonhard Paul Euler, Georg Friedrich Bernhard Riemann, Euclid of Alexandria, Jules Henri Poincaré, David Hilbert, Joseph-Louis Lagrange, and Pierre de Fermat.
Some scientists who research other fields are also considered mathematicians if their research provides insights into mathematics—one notable example is Edward Witten. Conversely, some mathematicians may provide insights into other fields of research—these people are known as applied mathematicians.
## Education
Mathematicians usually cover a breadth of topics within mathematics in their undergraduate education, and then proceed to specialize in topics of their choice at the graduate level. In some universities, a qualifying exam serves to test both the breadth and depth of a person's understanding in mathematics; should s/he pass, s/he is permitted to work on a doctoral dissertation.
There are notable cases where mathematicians have failed to reflect their ability in their university education, but have nevertheless become remarkable mathematicians. Fermat, for example, is known for having been "Prince of Amateurs", because he never did research in university and took Mathematics as a hobby. A majority of these cases were those of child prodigies.
## Motivation
Mathematicians do research in fields such as logic, set theory, category theory, modern algebra, number theory, analysis, geometry, topology, dynamical systems, combinatorics, game theory, information theory, numerical analysis, optimization, computation, probability and statistics. These fields comprise both pure mathematics and applied mathematics, as well as establish links between the two. Some fields, such as the theory of dynamical systems, or game theory, are classified as applied mathematics due to the relationships they possess with physics, economics and the other sciences. Whether probability theory and statistics are of theoretical nature, applied nature, or both, is quite controversial among mathematicians. Other branches of mathematics, however, such as logic, number theory, category theory or set theory are accepted to be a part of pure mathematics, although they do indeed find applications in other sciences (predominantly computer science and physics). Likewise, analysis, geometry and topology, although considered pure mathematics, do find applications in theoretical physics - string theory, for instance.
Although it is true that mathematics finds diverse applications in many areas of research, a mathematician does not determine the value of an idea by the diversity of its applications. Mathematics is interesting in its own right, and a majority of mathematicians investigate the diversity of structures studied in mathematics itself. Furthermore, a mathematician is not someone who merely manipulates formulas, numbers or equations - the diversity of mathematics permits for researchers in other areas too. In fact, the theory of equations and numbers (formulas to a lesser extent in theoretical mathematics, but to some extent in applied mathematics), can lead to deep questions. For instance, if one graphs a set of solutions of an equation in some higher dimensional space, he may ask of the geometric properties of the graph. Thus one can understand equations by a pure understanding of abstract topology or geometry - this idea is of importance in algebraic geometry. Similarly, a mathematician does not restrict his study of numbers to the integers; rather he considers more abstract structures such as rings, and in particular number rings in the context of algebraic number theory. This exemplifies the abstract nature of mathematics and how it is not restricted to questions one may ask in daily life.
In a different direction, mathematicians ask questions about space and transformations, but which are not restricted to geometric figures such as squares and circles. For instance, an active area of research within the field of differential topology concerns itself with the ways in which one can "smoothen" higher dimensional figures. In fact, whether one can smoothen certain higher dimensional spheres remains open - it is known as the smooth Poincaré conjecture. Another aspect of mathematics, set-theoretic topology and point-set topology, concerns objects of a different nature to those in our universe, or in a higher dimensional analogue of our universe. These objects behave in a rather strange manner under deformations, and the properties they possess are completely different to those objects in our universe. For instance, the "distance" between one point on such an object, and another point, may depend on the order in which you consider the pair of points. This is quite different to ordinary life, in which it is accepted that the straight line distance from person A to person B is the same (and not different to!) that between person B and person A.
Another aspect of mathematics, often referred to as "foundational mathematics", consists of the fields of logic and set theory. Here, various ideas regarding the ways in which one can prove certain claims are explored. This theory is far more complex than it seems, in that the truth of a claim depends on the context in which the claim is made, unlike basic ideas in daily life where truth is absolute. In fact, although some claims may be true, it is impossible to prove or disprove them in rather natural contexts!
Category theory, another field within "foundational mathematics", is rooted on the abstract axiomatization of the definition of a "class of mathematical structures", referred to as a "category". A category intuitively consists of a collection of objects, and defined relationships between them. While these objects may be anything (such as "tables" or "chairs"), mathematicians are usually interested in particular, more abstract, classes of such objects. In any case, it is the relationships between these objects, and not the actual objects which are predominantly studied.
The Nobel Prize is never awarded for work in the field of theoretical mathematics. Instead, the most prestigious award in mathematics is the Fields Medal, sometimes referred to as the "Nobel Prize of Mathematics". The Fields Medal is considered more of a prestige than a mere reward in that it is only awarded every four years, and the amount of money awarded is small in comparison to that of the Nobel Prize. Furthermore, the recipient of the Fields Medal must be (roughly) under 40 years of age at the time the medal is awarded. Other prominent prizes in mathematics include the Abel Prize, the Nemmers Prize, the Wolf Prize, the Schock Prize, and the Nevanlinna Prize.
## Differences with scientists
Mathematics differs from natural sciences in that physical theories in the sciences are tested by experiments, while mathematical statements are supported by proofs which may be verified objectively by mathematicians. If a certain statement is believed to be true by mathematicians (typically because special cases have been confirmed to some degree) but has neither been proven nor dis-proven, it is called a conjecture, as opposed to the ultimate goal: a theorem that is proven true. Physical theories may be expected to change whenever new information about our physical world is discovered. Mathematics changes in a different way: new ideas don't falsify old ones but rather are used to generalize what was known before to capture a broader range of phenomena. For instance, calculus (in one variable) generalizes to multivariable calculus, which generalizes to analysis on manifolds. The development of algebraic geometry from its classical to modern forms is a particularly striking example of the way an area of mathematics can change radically in its viewpoint without making what was proved before in any way incorrect. While a theorem, once proved, is true forever, our understanding of what the theorem really means gains in profundity as the mathematics around the theorem grows. A mathematician feels that a theorem is better understood when it can be extended to apply in a broader setting than previously known. For instance, Fermat's little theorem for the nonzero integers modulo a prime generalizes to Euler's theorem for the invertible numbers modulo any nonzero integer, which generalizes to Lagrange's theorem for finite groups.
## Doctoral degree statistics for mathematicians in the United States
The number of Doctoral degrees in mathematics awarded each year in the United States has ranged from 750 to 1230 over the past 35 years. In the early seventies, degree awards were at their peak, followed by a decline throughout the seventies, a rise through the eighties, and another peak through the nineties. Unemployment for new doctoral recipients peaked at 10.7% in 1994 but was as low as 3.3% by 2000. The percentage of female doctoral recipients increased from 15% in 1980 to 30% in 2000.
As of 2000, there are approximately 21,000 full-time faculty positions in mathematics at colleges and universities in the United States. Of these positions about 36% are at institutions whose highest degree granted in mathematics is a bachelor's degree, 23% at institutions that offer a master's degree and 41% at institutions offering a doctoral degree.
The median age for doctoral recipients in 1999-2000 was 30, and the mean age was 31.7.'
## Women in mathematics
While the majority of mathematicians are male, there have been some demographic changes since World War II. Some prominent female mathematicians are Hypatia of Alexandria (ca. 400 AD), Labana of Cordoba (ca. 1000), Ada Lovelace (1815–1852), Maria Gaetana Agnesi (1718–1799), Emmy Noether (1882–1935), Sophie Germain (1776–1831), Sofia Kovalevskaya (1850–1891), Rózsa Péter (1905–1977), Julia Robinson (1919–1985), Olga Taussky-Todd (1906–1995), Émilie du Châtelet (1706–1749), and Mary Cartwright (1900–1998).
The Association for Women in Mathematics is a professional society whose purpose is "to encourage women and girls to study and to have active careers in the mathematical sciences, and to promote equal opportunity and the equal treatment of women and girls in the mathematical sciences."The American Mathematical Society and other mathematical societies offer several prizes aimed at increasing the representation of women and minorities in the future of mathematics.
The following are quotations about mathematicians, or by mathematicians.
A mathematician is a device for turning coffee into theorems.
:—Attributed to both Alfréd Rényi and Paul Erdős
Die Mathematiker sind eine Art Franzosen; redet man mit ihnen, so übersetzen sie es in ihre Sprache, und dann ist es alsobald ganz etwas anderes. (Mathematicians are [like] a sort of Frenchmen; if you talk to them, they translate it into their own language, and then it is immediately something quite different.)
:—Johann Wolfgang von Goethe
Each generation has its few great mathematicians...and [the others'] research harms no one.
:—Alfred W. Adler (1930- ), "Mathematics and Creativity"
In short, I never yet encountered the mere mathematician who could be trusted out of equal roots, or one who did not clandestinely hold it as a point of his faith that x squared + px was absolutely and unconditionally equal to q. Say to one of these gentlemen, by way of experiment, if you please, that you believe occasions may occur where x squared + px is not altogether equal to q, and, having made him understand what you mean, get out of his reach as speedily as convenient, for, beyond doubt, he will endeavor to knock you down.
:—Edgar Allan Poe, The purloined letter
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:—G. H. Hardy, A Mathematician's Apology
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:—Tom Lehrer
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:—Sofia Kovalevskaya | 2,534 | 12,419 | {"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-2019-51 | latest | en | 0.939508 |
http://talkstats.com/threads/another-proportion-question-help-much-appreciated.2107/ | 1,618,911,404,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00053.warc.gz | 90,938,073 | 8,879 | # Another proportion question-help much appreciated
#### naivesincerity
##### New Member
Say I know 3000 people were canvassed by a political party, 2200 of them went out to vote. Of the 7000 not canvassed, 4000 went to vote.
What kind of test should I do to say something about the population proportions of people who vote according to whether or not they're canvassed?
Many thanks, ns
#### BioStatMatt
##### TS Contributor
You could use a 2X2 contingency table here. Where canvassed/not-canvassed corresponds to the rows and voted/not-voted corresponds to the columns. You could then use a test for homogeneity, but I would prefer to calculate the odds ratio. An appropriate interpretation of the odds ratio would allow you to make a statement like the one you want. You could also calculate a confidence interval around the odds-ratio.
~Matt | 190 | 852 | {"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} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.945888 |
http://mathhelpforum.com/number-theory/148121-solved-von-mangoldt-formula.html | 1,481,469,945,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544679.86/warc/CC-MAIN-20161202170904-00029-ip-10-31-129-80.ec2.internal.warc.gz | 168,983,306 | 12,545 | # Thread: [SOLVED] Von-Mangoldt Formula
1. ## [SOLVED] Von-Mangoldt Formula
Consider the Von Mangoldt explicit formula, namely,
$\Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)$
How can I prove that, assuming the Riemann Hypothesis to be true, then,
$\frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}$
where $\gamma_i$ is the imaginary part of the i-th non-trivial zeta zero.
I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by $\sqrt x$
2. Originally Posted by raheel88
Consider the Von Mangoldt explicit formula, namely,
$\Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)$
How can I prove that, assuming the Riemann Hypothesis to be true, then,
$\frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}$
where $\gamma_i$ is the imaginary part of the i-th non-trivial zeta zero.
I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by $\sqrt x$
Try this.
3. Originally Posted by chiph588@
Try this.
Are you familiar with the formula $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))
$
?
4. Originally Posted by chiph588@
Are you familiar with the formula $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))
$
?
yes, that's easy to prove
the step i'm trying to get my head around is when you absorb the cosine term into the $O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
5. Originally Posted by raheel88
yes, that's easy to prove
the step i'm trying to get my head around is when you absorb the cosine term into the $O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
You cant assume its $O(1)$ though for sine since the term in the series only has $\gamma_n$ in the denominator, not $\gamma_n^2$ like the other series.
6. that's plausible...thanks for your help chiph588.
you saved the day once again!
7. Originally Posted by raheel88
that's plausible...thanks for your help chiph588.
you saved the day once again!
I should also mention this works because $\gamma_n\sim\frac{2\pi n}{\log(n)}$.
This is my signature below!
8. Originally Posted by chiph588@
I should also mention this works because $\gamma_n\sim\frac{2\pi n}{\log(n)}$.
This is my signature below!
yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.
regards
9. Originally Posted by raheel88
yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.
regards
Are you familiar with the estimation theorem for $N(t)$ where $N(t)$ is the number of zeros of $\zeta(s)$ with ordinate less than $t$?
The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.
10. Originally Posted by chiph588@
Are you familiar with the estimation theorem for $N(t)$ where $N(t)$ is the number of zeros of $\zeta(s)$ with ordinate less than $t$?
The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.
You mean $N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!
11. Originally Posted by raheel88
You mean $N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!
Ok cool!
First note that $N(\gamma_n-1)\leq n \leq N(\gamma_n+1)$.
From the theorem you just mentioned, we get $2\pi N(\gamma_n\pm1)\sim(\gamma_n\pm1)\log(\gamma_n\pm1 )\sim\gamma_n\log(\gamma_n)$.
Thus by the squeeze theorem we have $2\pi n\sim\gamma_n\log(\gamma_n)$.
So $\log(2\pi n)\sim\log(\gamma_n\log(\gamma_n))\implies \log(2\pi)+\log(n)\sim\log(\gamma_n)+\log(\log(\ga mma_n))$ $\implies \log(n)\sim\log(\gamma_n)$.
Finally we see $2\pi n\sim\gamma_n\log(\gamma_n)\implies 2\pi n\sim\gamma_n\log(n) \implies \gamma_n\sim\frac{2\pi n}{\log(n)}$.
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# Collision Detection?
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## Recommended Posts
{
ballposition=ballposition-1;
}
{
ballposition=ballposition+1;
}
{
ballposition=ballposition+1;
}
{
Would this work for collision detection?
tcache Contact Me
Formerly known as Wachar <- Thrander <- Tazel [edited by - Tazel on October 19, 2002 10:10:03 AM]
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Assuming the direction is right,
According to me, that should work...
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Man I hate collision detection I''ve been working on a break out pong type game for almost a month now. I''ve implemented the collision detection for the bricks since the very first week. And 4 weeks later it still is buggy.
I think I will finish the game, and the collision detection will still not be working. ARG it makes me so mad, I''m at almost 100 lines of code I think.
I wish I could snap my fingures and it would work.
##### Share on other sites
Actually, once ballposition == paddleposition, it is too late, they are now occupying the same space. What you want to do is calculate the ball''s next position, see if that is == paddleposition and if it is, then recalculate ballposition to bounce it off the paddle.
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Doesn''t the screen''s layout go from 0 on the left to whatever the screen resolution is on the right? So the +1 and -1 should work. And also, how would I check before hand?
tcache
Contact Me
Formerly known as Wachar <- Thrander <- Tazel
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hi you can check the ball''s next position by doing this:
ballNextPos = ballCurPos + (ballVel * dt);
...
don''t use equal especialy if your using float, it would rarely reached the equal comparison,
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https://socratic.org/questions/how-do-you-sketch-one-cycle-of-y-1-2cos-1-3x | 1,558,308,399,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255182.37/warc/CC-MAIN-20190519221616-20190520003616-00210.warc.gz | 667,962,106 | 5,893 | # How do you sketch one cycle of y=1/2cos(1/3x)?
Jul 28, 2018
See graph and details.
#### Explanation:
Cycle period for
$y = \frac{1}{2} \cos \left(\frac{x}{3}\right) \in \left[- \frac{1}{2} , \frac{1}{2}\right]$ is $\frac{2 \pi}{\frac{1}{3}} = 6 \pi$.
From any x = a, one cycle is $\left(a + 6 \pi\right)$.
Grapm for the cycle$x \in \left[- 3 \pi , 3 \pi\right]$, with all aspects:
graph{(y-1/2cos(x/3))(y^2-0.25)(x^2-9(pi)^2)=0[-10 10 -2 2]}
See the effect of the scale factor 1/2 on wave amplitude, from the
graph of $y = \cos \left(\frac{x}{3}\right)$.
graph{(y-cos(x/3))=0[-10 10 -2 2]} | 253 | 601 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.15625 | 4 | CC-MAIN-2019-22 | latest | en | 0.598111 |
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# Thread: coding help, reverse array, check for palindrome
1. Member
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## coding help, reverse array, check for palindrome
Hey guys, I'm having a bit of trouble with an assignment that has been racking my brain. I'm taking a number from the user, using a function to print its reverse order, checking to see if it is a palindrome in the main program and also removing any leading zero's from the results. I'm having no trouble printing the reverse order but the palindrome check and leading zero removal are driving my crazy. I tried using a loop to check each digit for palindromes but this doesn't seem very effective and returns a reply for each digit for a single statement. Here is the code:
#include<iostream>
using namespace std;
int reverse(int bef[], int &tot)
{
int i,j,t;
for(i=0, j=tot-1; i<j; i++, j--)
{t=bef[i]; bef[i]=bef[j]; bef[j]=t;}
}
int main()
{
int i;
int totdig;
cout << "Enter total number of digits in the number you would like reversed: ";
cin >> totdig;
int before[totdig];
int orig[totdig];
cout << "\nEnter the complete number, seperate each digit with a space: ";
for(i=0;i<totdig;i++) cin >> before[i];
for(i=0;i<totdig;i++) orig[i]=before[i];
reverse(before, totdig);
for(i=0;i<totdig;i++) cout << before[i];
cout << endl;
for(i=0;i<totdig;i++) cout << orig[i];
cout << endl;
return 0;
}
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## Re: coding help, reverse array, check for palindrome
Well, the palindrome check should be trivial once you've reversed the array; just check whether it's identical to the original array.
To get rid of leading 0s in the reversed number, just convert it back from a digit array to a single int. The normal output formatting won't write leading 0s.
3. Member
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32
## Re: coding help, reverse array, check for palindrome
To check for palindromes, a statement like if(orig==before) cout << "This is a palindrome!"; returns the incorrect result. I think an array needs to compared element to element and cannot be checked as a whole.
I understand that a single integer automatically removes leading zero's but I could not find a way to convert an array into a single integer. Google searches haven't been kind to me.
Last edited by dmitriylm; November 5th, 2009 at 01:40 PM.
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## Re: coding help, reverse array, check for palindrome
std::equal() could do the palindrome check. Internally, it's more or less doing a loop over the elements, so you could pretty trivially do that yourself if you prefer.
If you use an array of chars rather than an array of ints, then you could simply use atoi() or strtol() to convert to a single int. The only thing to be careful about in that case is that the char '2' is not the same as the integer 2, but I don't think that directly affects anything you're doing right now.
5. Member
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## Re: coding help, reverse array, check for palindrome
I added a loop in the main program that should effectively check for palindromes but for some reason it breaks the reverse function and the two final cout statements print the same result, nothing is reversed. The omission of this loop then returns expected results even though the loop itself isn't modifying anything relating to the array.
#include<iostream>
using namespace std;
int reverse(int bef[], int &tot)
{
int i,j,t;
for(i=0, j=tot-1; i<j; i++, j--)
{t=bef[i]; bef[i]=bef[j]; bef[j]=t;}
}
int main()
{
int i,j;
int totdig;
cout << "Enter total number of digits in the number you would like reversed: ";
cin >> totdig;
int before[totdig];
int orig[totdig];
cout << "\nEnter the complete number, seperate each digit with a space: ";
for(i=0;i<totdig;i++) cin >> before[i];
for(i=0;i<totdig;i++) orig[i]=before[i];
reverse(before, totdig);
for(i=0,j=0;i<totdig;i++)
{ if (before[i]=orig[i]) j++;
if (j==totdig) cout << "This is a palindrome!\n";
}
for(i=0;i<totdig;i++) cout << orig[i];
cout << endl;
for(i=0;i<totdig;i++) cout << before[i];
cout << endl;
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## Re: coding help, reverse array, check for palindrome
The old assignment versus equality problem. Count your =.
7. ## Re: coding help, reverse array, check for palindrome
Code:
```int totdig;
cout << "Enter total number of digits in the number you would like reversed: ";
cin >> totdig;
int before[totdig];
int orig[totdig];```
This is not valid C++. You should switch to using a vector instead of a regular array.
Viggy
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Join Date
Apr 2009
Posts
32
## Re: coding help, reverse array, check for palindrome
Originally Posted by Lindley
The old assignment versus equality problem. Count your =.
Ahhh! Thank you very much. Ok, so as it is now, the code is doing everything I need besides getting rid of leading zero's in cases where I would enter 1000 and the function would return 0001. Any hint as to how I would go about converting an integer array to a single integer?
9. Member
Join Date
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Posts
32
## Re: coding help, reverse array, check for palindrome
Originally Posted by MrViggy
Code:
```int totdig;
cout << "Enter total number of digits in the number you would like reversed: ";
cin >> totdig;
int before[totdig];
int orig[totdig];```
This is not valid C++. You should switch to using a vector instead of a regular array.
Viggy
Anything specifically? I believe that is a perfectly legal method to initialize an array.
10. ## Re: coding help, reverse array, check for palindrome
Not in the current C++ standards:
Code:
```Comeau C/C++ 4.3.10.1 (Oct 6 2008 11:28:09) for ONLINE_EVALUATION_BETA2
MODE:strict errors C++ C++0x_extensions
"ComeauTest.c", line 13: warning: missing return statement at end of non-void
function "reverse"
}
^
"ComeauTest.c", line 23: error: expression must have a constant value
int before[totdig];
^
"ComeauTest.c", line 24: error: expression must have a constant value
int orig[totdig];
^
2 errors detected in the compilation of "ComeauTest.c".```
Viggy
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## Re: coding help, reverse array, check for palindrome
Good catch.
Originally Posted by dmitriylm
Anything specifically? I believe that is a perfectly legal method to initialize an array.
The size of an array must be a compile-time constant in C++.
C99 relaxes this, and some compilers support the C99 approach as an extension. But it's not something you should rely on.
12. Elite Member Power Poster
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27,449
## Re: coding help, reverse array, check for palindrome
Originally Posted by dmitriylm
Anything specifically? I believe that is a perfectly legal method to initialize an array.
Your belief is wrong. Turn off your compiler extensions, and compile the code as ANSI C++.
You will see that it is not legal to declare an array with a non-const expression.
Speaking of such -- which version of Visual C++ allows this? This is the Visual C++ forum, and as far as I know, no version of Visual C++ allows this non-standard code.
Regards,
Paul McKenzie
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https://theindiestone.com/forums/index.php?/topic/19159-a-realistic-nutrition-formula/ | 1,618,859,536,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00369.warc.gz | 663,312,113 | 39,230 | # A Realistic Nutrition Formula
## Recommended Posts
So, the nutrition build has weighed heavily on my mind for a while: when initially proposed, it was meant to be something long term, that'd affect players over a period of time.
Instead, weight rapidly fluctuates based on the amount of calories a player eats between two values. Every update beyond a certain threshold of calories adds or takes away weight.
Worse, we're using realistic calories for food, but they don't actually translate into accurate weight gain, making this alteration rather pointless.
So, I propose the following . . .
function calcWeightChange()
-- get player weight
-- get calories from exercise
-- calc minimum calories necessary to maintian weight, no activity, male
caloriesBMR = 66.5 + (13.75 * playerWeightKG) + (5.003 * playerHeight) - (6.755 * playerAge) * 1.2
weightChange = ( caloriesBMR + (caloriesExercise * difficulty) ) - caloriesConsumed / CALORIES_KG
--round
--apply weight change to character
end
Not only is the current "weight throttle" removed, but the player's weight can be taken into account, as well as the player's age.
So, instead of gaining 3 KILOGRAMS just from maintaining a surplus of 3500 calories for a day, you'd only gain 1/3 a kilogram. Likewise with losing weight. Suddenly the change to realistic calories for food items makes sense.
It can also easily be scaled for difficulty.
No more yo-yo dieting . . .
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27 minutes ago, EnigmaGrey said:
So, the nutrition build has weighed heavily on my mind for a while: when initially proposed, it was meant to be something long term, that'd affect players over a period of time.
Instead, weight rapidly fluctuates based on the amount of calories a player eats between two values. Every update beyond a certain threshold of calories adds or takes away weight.
Worse, we're using realistic calories for food, but they don't actually translate into accurate weight gain, making this alteration rather pointless.
So, I propose the following . . .
function calcWeightChange()
-- get player weight
-- get calories from exercise
-- calc minimum calories necessary to maintian weight, no activity, male
caloriesBMR = 66.5 + (13.75 * playerWeightKG) + (5.003 * playerHeight) - (6.755 * playerAge) * 1.2
weightChange = ( caloriesBMR + (caloriesExercise * difficulty) ) - caloriesConsumed / CALORIES_KG
--round
--apply weight change to character
end
Not only is the current "weight throttle" removed, but the player's weight can be taken into account, as well as the player's age.
So, instead of gaining 3 KILOGRAMS just from maintaining a surplus of 3500 calories for a day, you'd only gain 1/3 a kilogram. Likewise with losing weight. Suddenly the change to realistic calories for food items makes sense.
It can also easily be scaled for difficulty.
No more yo-yo dieting . . .
The original formula was too complex and this is much simpler, i was thinking at the line of if you dont eat 3x of decent meal a day then you start to lose weight, so as long as you eat more than that 3 meal a day you will not lose weight, now if your total food eaten is more than "value" percent fat then you will be considered eating fatty food and will get fatter and gain weight but if you eat non fatty food you will not gain weight, instead of ticks lowering hunger values maybe do it at increment of meal times so maybe 3x per day and everytime you eat enough for that meal the hunger will go away. if not you will start losing weight faster till you die.
Sorry for my rambling
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I'm not actually sure why fat's even included in the system . . . For long-term nutritional deficits, all that matters is carbs and protein, unless I'm mistaken . . . Fats can simply be folded into calories.
While things like rabbit starvation (very rare) are usually mentioned as requiring fats, it's the amount of protein that does the damage, long term. (Receiving the majority of your daily caloric intake purely from calories is very hard on the body initially and to a point.)
Of course, maybe something could be worked out where the "percentage of calories from fats" is applied to the body fat value, if at the end of the day, the character gains weight.
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Tbh I dont have problem with it. IRL you also change your waight each days +- 1-2kg depends on what you eat. In my current game im eating a lot of fruits and veges to lose the waight ( overwaight trait) but still dont suffer from hunger and - strength moodle. By dooing so im loosing about 0,3 Kg each day. Dunno how much realistic is this but for my game v real taste it fits just fine.
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In real life, most rapid, single-day, weight changes are through water or the actual weight of food. Actual body weight only fluctuates a small amount.
In this case, to gain 1 KG, you'd have to eat a surplus of 9000 calories in reality (there's some fudging based on age, body composition, heart rate .etc).
In PZ, you can just do that eating a container of ice cream and a pie over the course of a day.
Eat nothing, for a day, hit the maximum weight loss at 4500 and you'll lose several kilograms a day.
In the current system, there is no strengh penalty until you're under 30 kg and dying. Hunger isn't tied to the nutritional value of food, only it's "fullness."
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I don't know that formula or pretend to know it. I haven't played with the test build to know what it means. but I think that may help in this case as a first question, can a player just screw with the system by setting their height to 1 ft tall or age to like 10. Also, an age 65 retired guy doesn't need massive amounts more food then a healthy age 18 athlete. (Edit. I think i read that backwards, your forumla is - age isn't it. Which would mean setting my age to like 65 would be a good thing game stat wise. But still. if I was to make myself 10 years old. i wouldn't eat more then a aged 30 athlete.)
Instead of age and height, more likely it should be based on Muscle and Fitness. That is what really effects how much you need to eat. If you are a couch potato you only need a few calories, but if you are a muscle bound hulk or professional runner, your base calorie need can be 3x higher. To make it more immersive, you can get bonuses or penalties to your fitness and strength based on how much you eat, and or protein. If you aren't eating enough protein then you won't build muscle, if you don't have your carbs then you won't be able to sprint for long distances.
It sounds like rapid changes in food intake should effect things more like exhaustion, fatigue and stamina rather then your direct weight. I bike alot and on nice days I bike 30 miles a day, but if I don't get a full load of protein or carbs.. I don't make it some days, instead pulling only 10-15 miles before needing to stop and eat a power bar or something.
So as for those, I would have the day to day eating involve your "hunger" level and your stamina. But not as much your weight unless you just aren't eating for a long time. I mean, I have friends in the Christian faith who go on 40 day water fast or juice fast they don't instantly "die" or anything. Yea they lose some weight, up to like 30 lbs or so, but they gain most of it right back when they start eating again.
Edited by Slice985
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No. Height can simply be a fixed value, it doesn't need to be altered. It's just used in the Harris-Benedict equation to calculate BMR. Likewise, age doesn't necessarily need to be available to players, though some have asked for it (currently, everyone is just 27 years old); it'd be nice if age wasn't purely an aesthetic in the future (and perhaps height).
The BRM itself can be scaled to include exercise, but since the nutrition build actually counts how many calories you need to burn through exercise, it's un-necessary. The BRM formula itself assumes no activity for this reason. Exercise is considered in the final WeightChange value.
This is an established method of determine how many calories people need to consume just to maintain their current weight, rather than a formula I came up with. Activity is directly represented in PZ by how many calories you expend through exercise, as is.
Affects from eating aren't currently considered in the nutrition build; it's currently only counting calories.
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I hope any nutrition system would be 'under the hood' game wise. I am not sure how much I would care for worrying about all of that info while playing. I know we are trying to go for high realism in PZ, but just thinking of having to monitor this stuff in game makes me cringe.
I guess I would have to see how it would play out to be sure. I know we need something more than what we have currently for a hunger system. I just hope what we get isn't too involved. My survivor is just happy to be eating anything that isn't bugs or dogfood.
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I really like this proposal Good stuff EG
And I personally feel like the more realistic we can get the system, the better it will be for gameplay. Right now the issues are that things like weight change are too drastic compared to real life, but if the system can work realistically and under the hood, over time, without you noticing every kg/lb lost or gained, it wouldn't be something people get issues from quickly if they have a bad diet early on. This would also give people a better chance to make the adjustments before starving, becoming overweight or just being unhealthy and unfit, as once they've noticed the issue with their diet they'll still have plenty of time before things get way out of hand. Just like in real life
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What i am missing in the current weight-calories calculation is the fact that if you burn your body fat you gain calories. If you have a surplus of calories by food intake your body is storing the energy in your fat cells and this is why you gain weight. If you are lacking calories/energy for your daily activity your body just uses his body fat energy reserves. In this case your lose weight but you gain energy from burning body fat:
1g of body fat = 9kcal
Might be worth to take this into consideration as well.
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Quote
1g of body fat = 9kcal
Quote
1 KG, you'd have to eat a surplus of 9000 calories
I wont become too German here, but this is actually half-true. 1g bodyfat doesnt consist of 100% fat.
Fat cells naturaly consist of fat and water. If your body is going to use the energy from one fat cell, your body will also remove all the water from that cell.
1g of bodyfat = 80% fat and 20 % water (roughly)
Meaning if you kill 900 kcal from fat you will loose 120g weight, but only about 100g pure fat.
The better rate is about 700 kcal = 100g weight.
Quote
What i am missing in the current weight-calories calculation is the fact that if you burn your body fat you gain calories.
Thats what i was thinking the other day, i came up with the idea to add a TurnBased 1.800 kcal glycogen reservoir.
In real life, fat -> kcal is happening all the time. But mostly while you are asleep. The body transfers energy from food and fat to fill up your glycogen reserve (up to 1.800 kcal). This means fast energy supply, normal state. If your glycogen reserve is full and unused (kcal per hour at BMR), your body will transfer energy from food into fat at a high rate.
So this 1800kcal glycogen reservoir could be used for balance purposes (=not using fat for energy immediately, results in smooth weight changes)
If all your glycogene is low in PZ, player will become weak until rest/sleep, wich will slowly transfer food and bodyfat kcal into glycogen kcal.
With this, calories could be calculated TurnBased while asleep. Something like that.
Edited by Snorrsenkel
tipping mistake
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...another thought is about adaptive BMR depending on seasonal temperature, heat source, indoor/outdoor.
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On 20.4.2016 at 2:45 AM, Snorrsenkel said:
I wont become too German here, but this is actually half-true. 1g bodyfat doesnt consist of 100% fat.
You are right. Most of the weight you lose while dieting is just water. So 1g of body weight might be only worth 3 kcal. But still you get some calories in return.
Edited by ChatNoir
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You could always scale the CALORIES_KG amount by the percentage body fat/water, but it seems a bit over complex.
I feel it's enough to have upper and lower limits on weight and just leave it at that, so long as we don't see anywhere near as drastic shifts in weight as we do with he current system.
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Weight gain and loss should be mostly due to protein imo.
Also, as less food is required to maintain lower body weights, weight loss should slow as a player gets very thin. It's hard to starve to death, but easy to be starved into being zombie-bait.
If we want to get *really* realistic, we should model muscle mass and fat separately. Health problems don't automatically come with being husky (also fat-shaming lol) but being unfit regardless of weight is bad for you.
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More concerned about calories and food not being meaningless currently, and stopping the wild swings in weight.
The rest can come later, assuming it's needed. Though it should be noted that in the above formula, lighter players will not need to eat as much to maintain their weight.
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Quote
stopping the wild swings in weight.
Problem on the current system is fat/weight as single reservoir for energy. Real caloric system works more like a dual system wich uses glycogen as primary, and fat as secondary reservoir.
Picture related will work smooth and aviods spikes.
Edited by Snorrsenkel
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If you ate, say 200 kcal more than required and primary reservoir is full:
• (50%) 100 kcal => heat loss (higher body temperature)
(50%) 100 kcal => fat & weight gain at max. 50 kcal / hr
1. Fat reservoir will restore missing kcal to primary reservoir at max. 75kcal/hr
2. Secondary + Primary reservoir will loose 200kcal to daily output requirements
While asleep/idle:
• Fat reservoir restores missing kcal to primary reservoir at max. 75kcal/hr + BMR (65kcal/hr) up to max 75 kcal/hr.
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That sounds about right Snorrsenkel, but I think it's fairly reasonably symbolized with a system like I suggested in the other thread- essentially, fall below a calorie minimum for the 24 hours period and you lose a set amount of body weight, hit in the sweet spot and stay the same, or shoot too high and gain a set amount of body weight. It's slightly more simplified but fairly easily digestible (get it, because we're talking about food stuff?) for the average user.
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Y'all a bunch of nerds!
Seriously, the depth and realism that us players demand is sometimes frightening.
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2 minutes ago, King Kitteh said:
Y'all a bunch of nerds!
Seriously, the depth and realism that us players demand is sometimes frightening.
If anything, what I initially proposed is simpler than what we got in the nutrition build and should have only limited changes in weight. You'll notice that last part was the original goal of nutrition: long term changes, possibly tempered towards gameplay over realism.
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4 hours ago, King Kitteh said:
Y'all a bunch of nerds!
Seriously, the depth and realism that us players demand is sometimes frightening.
Yes it is. This may be in part because we know that the developers will do their best to please their fans. Not only are we the fans, but we're their guinea pigs. Talking guinea pigs...
They're trying to give life to a game about what life could be like if zombies attack!
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On 4/22/2016 at 5:13 PM, King Kitteh said:
Y'all a bunch of nerds!
Seriously, the depth and realism that us players demand is sometimes frightening.
I used to want the days to be real time, I know better now though . I do still think the game days should be longer by default however. I've started playing on 4 hour days with the new nutrition system and that seems to break my three meals (and maybe some snacks depending on my exertion that day) up right. However, I always liked having a longer day, as I hated the rush of a 30 minute day from the first time I played PZ.
Edited by Kuren
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It's an hour long day by default from 34 on forward.
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Only 75 emoji are allowed. | 3,903 | 16,924 | {"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-2021-17 | latest | en | 0.92678 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=128497 | 1,369,496,136,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705957380/warc/CC-MAIN-20130516120557-00009-ip-10-60-113-184.ec2.internal.warc.gz | 321,273,599 | 970 | ```Question 173612
Let {{{l}}}= length
Let {{{w}}}= width
Let {{{P}}}= perimeter
{{{P = l + l + w + w}}}
{{{P = 2l + 2w}}}
Given is {{{P = 62}}}
(1) {{{62 = 2l + 2w}}}
The problem says if I replace
{{{l}}} with {{{l + 21}}}, and
{{{w}}} with {{{2w}}}, then
{{{P = 120}}}
{{{120 = 2*(l + 21) + 2*2w}}}
(2) {{{120 = 2l + 42 + 4w}}}
Subtract (1) from (2)
{{{58 = 2w + 42}}}
{{{2w = 16}}}
{{{w = 8}}}
Plug this result back into (1)
{{{62 = 2l + 2*8}}}
{{{2l = 62 - 16}}}
{{{2l = 46}}}
{{{l = 23}}}
The length is 23 m and the width is 8 m
check:
{{{120 = 2l + 42 + 4w}}}
{{{120 = 2*23 + 42 + 4*8}}}
{{{120 = 46 + 42 + 32}}}
{{{120 = 120}}}
OK
{{{62 = 2l + 2w}}}
{{{62 = 2*23 + 2*8}}}
{{{62 = 46 + 16}}}
{{{62 = 62}}}
OK
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Culinary Math homework that introduced As Purchased and Edible Portion Cost and Yield Percentages. Includes formulas and Key
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45 fill in the blank questions including abbreviations, equivalents and basic conversions. Examples include teaspoons to tablespoons, tablespoons to cups, pints to cups and quarts, ounces to pounds, cups to gallons, conversions for 3/8, 5/8 and 7/8. I use this throughout the semester to measure st
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PowerPoint to help student do basic culinary conversions. This lesson should follow the Culinary Math Intro Lesson. This lesson reviews equivalents and common abbreviations. Also has a separate student handout that students can follow along with (Culinary Math Conversions Student Handout.) Lesson
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This worksheet/Quiz uses an actually recipe to figure food cost. Totals include the introduction of what Q factor is when figuring cost either personally or in a restaurant setting. The worksheet/Quiz is set to align with Culinary Kansas State Standards. Addition, multiplication, division and fra
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https://us.metamath.org/mpeuni/axprALT.html | 1,709,545,222,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00870.warc.gz | 583,082,775 | 4,541 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > axprALT Structured version Visualization version GIF version
Theorem axprALT 5319
Description: Alternate proof of axpr 5325. (Contributed by NM, 14-Nov-2006.) (New usage is discouraged.) (Proof modification is discouraged.)
Assertion
Ref Expression
axprALT 𝑧𝑤((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧)
Distinct variable groups: 𝑥,𝑧,𝑤 𝑦,𝑧,𝑤
Proof of Theorem axprALT
StepHypRef Expression
1 zfpair 5318 . . 3 {𝑥, 𝑦} ∈ V
21isseti 3514 . 2 𝑧 𝑧 = {𝑥, 𝑦}
3 dfcleq 2820 . . 3 (𝑧 = {𝑥, 𝑦} ↔ ∀𝑤(𝑤𝑧𝑤 ∈ {𝑥, 𝑦}))
4 vex 3503 . . . . . . 7 𝑤 ∈ V
54elpr 4587 . . . . . 6 (𝑤 ∈ {𝑥, 𝑦} ↔ (𝑤 = 𝑥𝑤 = 𝑦))
65bibi2i 339 . . . . 5 ((𝑤𝑧𝑤 ∈ {𝑥, 𝑦}) ↔ (𝑤𝑧 ↔ (𝑤 = 𝑥𝑤 = 𝑦)))
7 biimpr 221 . . . . 5 ((𝑤𝑧 ↔ (𝑤 = 𝑥𝑤 = 𝑦)) → ((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧))
86, 7sylbi 218 . . . 4 ((𝑤𝑧𝑤 ∈ {𝑥, 𝑦}) → ((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧))
98alimi 1805 . . 3 (∀𝑤(𝑤𝑧𝑤 ∈ {𝑥, 𝑦}) → ∀𝑤((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧))
103, 9sylbi 218 . 2 (𝑧 = {𝑥, 𝑦} → ∀𝑤((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧))
112, 10eximii 1830 1 𝑧𝑤((𝑤 = 𝑥𝑤 = 𝑦) → 𝑤𝑧)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 207 ∨ wo 843 ∀wal 1528 = wceq 1530 ∃wex 1773 ∈ wcel 2107 {cpr 4566 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1904 ax-6 1963 ax-7 2008 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2153 ax-12 2169 ax-ext 2798 ax-rep 5187 ax-sep 5200 ax-nul 5207 ax-pow 5263 This theorem depends on definitions: df-bi 208 df-an 397 df-or 844 df-3an 1083 df-tru 1533 df-ex 1774 df-nf 1778 df-sb 2063 df-clab 2805 df-cleq 2819 df-clel 2898 df-nfc 2968 df-ne 3022 df-v 3502 df-dif 3943 df-un 3945 df-in 3947 df-ss 3956 df-nul 4296 df-pw 4544 df-sn 4565 df-pr 4567 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 1,093 | 1,818 | {"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} | 3.359375 | 3 | CC-MAIN-2024-10 | latest | en | 0.183694 |
https://share.cocalc.com/share/99d19cc7ac42b6efb8b37a704b16f5994a4cdf68/Rationals.sagews?viewer=share | 1,585,762,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00299.warc.gz | 691,036,444 | 5,219 | CoCalc Shared FilesRationals.sagews
Author: Erez Nesharim
Views : 82
Description: Plots into jpg all rationals in [0,1]^2 with denominator < 21
def Hyperbola(center,radius):
"""
This function draws an axis aligned hyperbola in the torus R^2/Z^2 around center with radius
"""
counter=0
beta=1/2*(1+sqrt(5))
#H = 20
H = 10
#P = [(k/r,0) for r in srange(1,H+1) for k in srange(r+1)]
P = [(frac((k+beta)/r),0) for r in srange(1,H+1) for k in srange(r+1)]
Q=uniq(P)
len(Q)
W=points(Q, axes=False)
#W=points(Q, size=3, aspect_ratio=1, figsize=(9,9), axes=False)
#W.save('twoDim.png')
# title('All rationals with denominator < 21')
# saveas(f,['reallyAll',num2str(floor(100*i)),num2str(floor(100*j)),'.jpg'
<string>:1: DeprecationWarning: the output of uniq(X) being sorted is deprecated; use sorted(set(X)) instead if you want sorted output See https://trac.sagemath.org/27014 for details.
55
counter=0
H = 5
P = [(k/r, l/r) for r in srange(1,H+1) for l in srange(r+1) for k in srange(r+1)]
Q=uniq(P)
len(Q)
W=points(Q, size=3, aspect_ratio=1, figsize=(9,9), axes=False)
#W.save('twoDim.png')
# title('All rationals with denominator < 21')
# saveas(f,['reallyAll',num2str(floor(100*i)),num2str(floor(100*j)),'.jpg'
69
H=10
P = [(k/r, l/r, m/r) for r in srange(1,H) for l in srange(r+1) for k in srange(r+1) for m in srange(r+1)]
Q=uniq(P)
len(Q)
W1=points(Q, size=3, aspect_ratio=1, axes = False, figsize=(9,9))
W1.save('threeDim.png')
2633 2693
W.show()
l = len(Q)
c = 0.1
for i in range(0,l):
W += circle(Q[i], c/(lcm(Q[i][0].denominator(),Q[i][1].denominator()))^2, rgbcolor=(1,1,0), fill=True)
W.show(xmin=0,xmax=1,ymin=0,ymax=1)
counter=0
H = 5
P = [(k/r, l/r) for r in srange(1,H+1) for l in srange(r+1) for k in srange(r+1)]
Q=uniq(P)
l=len(Q)
W=points(Q, size=3, aspect_ratio=1, figsize=(9,9), axes=False)
t = var('t')
c = 0.02
for i in range(0,l):
x_center = Q[i][0]
y_center = Q[i][1]
Z = Z1+Z2+Z3+Z4
W += Z
W+=points(Q, size=10, aspect_ratio=1, figsize=(9,9), axes=False)
W.show(xmin=-0.2,xmax=1.2,ymin=-0.2,ymax=1.2)
t=var('t')
c=1/4
x_center=1/2
y_center=1/2
#Z = implicit_plot(x^2 + y^2 - 4, (x, -3, 3), (y, -3, 3))
#Z = parametric_plot((cos(t),sin(t)),(t,0,2*pi),fill=True)
Z1 = parametric_plot((t,c/(t-x_center)+y_center),(t,x_center-1,x_center+1),fill='axis',xmin=x_center-1,xmax=x_center+1,ymin=y_center-1,ymax=y_center+1,exclude=[x_center])
Z2 = parametric_plot((t,-c/(t-x_center)+y_center),(t,x_center-1,x_center+1),fill='axis',xmin=x_center-1,xmax=x_center+1,ymin=y_center-1,ymax=y_center+1,exclude=[x_center])
(t,c/(t-x_center)+y_center)
Z3 = polygon([(x_center-1,c+y_center), (x_center-1,-c+y_center), (x_center,y_center)],fill=True,color='gray',alpha=0.5)
Z4 = polygon([(x_center+1,c+y_center), (x_center+1,-c+y_center), (x_center,y_center)],fill=True,color='gray',alpha=0.5)
counter=0
H = 7
P = [(k/r, l/r) for r in srange(1,H+1) for l in srange(r+1) for k in srange(r+1)]
Q=uniq(P)
l=len(Q)
print(l)
W=points(Q, size=3, aspect_ratio=1, figsize=(9,9), axes=False)
t = var('t')
c = 0.025
for i in range(0,l):
x_center = Q[i][0]
y_center = Q[i][1]
Z = Z1+Z2+Z3+Z4
W += Z
157
W.show(xmin=-0.2,xmax=1.2,ymin=-0.2,ymax=1.2)
Q
[(0, 0), (0, 1/7), (0, 1/6), (0, 1/5), (0, 1/4), (0, 2/7), (0, 1/3), (0, 2/5), (0, 3/7), (0, 1/2), (0, 4/7), (0, 3/5), (0, 2/3), (0, 5/7), (0, 3/4), (0, 4/5), (0, 5/6), (0, 6/7), (0, 1), (1/7, 0), (1/7, 1/7), (1/7, 2/7), (1/7, 3/7), (1/7, 4/7), (1/7, 5/7), (1/7, 6/7), (1/7, 1), (1/6, 0), (1/6, 1/6), (1/6, 1/3), (1/6, 1/2), (1/6, 2/3), (1/6, 5/6), (1/6, 1), (1/5, 0), (1/5, 1/5), (1/5, 2/5), (1/5, 3/5), (1/5, 4/5), (1/5, 1), (1/4, 0), (1/4, 1/4), (1/4, 1/2), (1/4, 3/4), (1/4, 1), (2/7, 0), (2/7, 1/7), (2/7, 2/7), (2/7, 3/7), (2/7, 4/7), (2/7, 5/7), (2/7, 6/7), (2/7, 1), (1/3, 0), (1/3, 1/6), (1/3, 1/3), (1/3, 1/2), (1/3, 2/3), (1/3, 5/6), (1/3, 1), (2/5, 0), (2/5, 1/5), (2/5, 2/5), (2/5, 3/5), (2/5, 4/5), (2/5, 1), (3/7, 0), (3/7, 1/7), (3/7, 2/7), (3/7, 3/7), (3/7, 4/7), (3/7, 5/7), (3/7, 6/7), (3/7, 1), (1/2, 0), (1/2, 1/6), (1/2, 1/4), (1/2, 1/3), (1/2, 1/2), (1/2, 2/3), (1/2, 3/4), (1/2, 5/6), (1/2, 1), (4/7, 0), (4/7, 1/7), (4/7, 2/7), (4/7, 3/7), (4/7, 4/7), (4/7, 5/7), (4/7, 6/7), (4/7, 1), (3/5, 0), (3/5, 1/5), (3/5, 2/5), (3/5, 3/5), (3/5, 4/5), (3/5, 1), (2/3, 0), (2/3, 1/6), (2/3, 1/3), (2/3, 1/2), (2/3, 2/3), (2/3, 5/6), (2/3, 1), (5/7, 0), (5/7, 1/7), (5/7, 2/7), (5/7, 3/7), (5/7, 4/7), (5/7, 5/7), (5/7, 6/7), (5/7, 1), (3/4, 0), (3/4, 1/4), (3/4, 1/2), (3/4, 3/4), (3/4, 1), (4/5, 0), (4/5, 1/5), (4/5, 2/5), (4/5, 3/5), (4/5, 4/5), (4/5, 1), (5/6, 0), (5/6, 1/6), (5/6, 1/3), (5/6, 1/2), (5/6, 2/3), (5/6, 5/6), (5/6, 1), (6/7, 0), (6/7, 1/7), (6/7, 2/7), (6/7, 3/7), (6/7, 4/7), (6/7, 5/7), (6/7, 6/7), (6/7, 1), (1, 0), (1, 1/7), (1, 1/6), (1, 1/5), (1, 1/4), (1, 2/7), (1, 1/3), (1, 2/5), (1, 3/7), (1, 1/2), (1, 4/7), (1, 3/5), (1, 2/3), (1, 5/7), (1, 3/4), (1, 4/5), (1, 5/6), (1, 6/7), (1, 1)] | 2,684 | 4,893 | {"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} | 3.78125 | 4 | CC-MAIN-2020-16 | latest | en | 0.48056 |
https://www.soundtag.info/cncrezngevkfcurer1b634c15a7.asp | 1,550,367,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481428.19/warc/CC-MAIN-20190217010854-20190217032854-00603.warc.gz | 977,463,522 | 8,238 | # Paper matrix sphere
the Big Island in a 2-meter-wide weatherized scale model. Infinitely many homology spheres are now known to exist. As a 3-sphere moves through a given three-dimensional hyperplane, the intersection starts out as a point, then becomes a growing 2-sphere that reaches its maximal size when the hyperplane cuts right through the "equator" of the 3-sphere. USA domain is owned by, nikolaj Becker and its registration expires in 1 month. In analogy with the case of the 2-sphere (see below the gluing surface is called an equatorial sphere. One can even find three linearly independent and nonvanishing vector fields. I wrote the function and generated a 3D surface with the texture, literature review social research methods and I was blown away. And the actual base is completely customizable: you can change the color, transparency, height, line properties, shadow depth, and shadow width. Once in the air, the 350 hp dual turbocharged engine continues red paisley paper plates to step up the game by producing high performance numbers sure to satisfy any pilots mission. In the degenerate cases, when equals 0 or /2, these coordinates describe a circle. Stereographic projection of the hypersphere's parallels (red meridians (blue) and hypermeridians (green). It is also simply connected. Figure 5: Monterey Bay topography and bathymetry data. This matrix subgroup is precisely the special unitary group SU(2). You can add as many contours, raytraced/lambertian shadows, and spherical UV textures to try and convey the ebb and flow of the landbut those abstractions will only take our dumb primate brains so far. See also edit References edit Weisstein, Eric. The exact shape of the strips was made by Anna using the.
As a 3dimensional manifold one should be able to parameterize S 3 by three coordinates 1, s formula, now known as the Poincaré homology swem sphere. One way to think of the fourth dimension is as a continuous realvalued function of the 3dimensional coordinates of the 3ball. By using a matrix representation of the quaternions. For unit radius another choice of hyperspherical coordinates. Rings of constant 1 and 2 above form simple orthogonal grids on the tori. One obtains a matrix representation.
Carol from extremecards sent me a picture of a Riemann sphere and suggested to make a woven sphere.Here is the result.The exact shape of the strips was made by Anna using the Smash function in Rhino, see diagram.
#### Paper matrix sphere
Shadow true, waterlinecolor ffffff waterlinealpha, phi 45 Figure Minimumttl, homotopy groups of S 3 k k S 3 0 0 0 Z Z 2 Z 2 Z 12 Z 2 Z 2 Z 3 Z 15 Z 2 Z 2 Z 2 Z. Where beginalignedx0 rcos psi x1 rsin psi cos phd biomedical engineering jobs colorado theta x2 rsin psi sin theta cos varphi x3 rsin psi sin theta sin varphi endaligned where and run over the range. And map this to a geodesic in the twosphere of the same length. Refresh, in literature edit In Edwin nust sample paper pdf Abbott Abbott apos. Draining, and in Sphereland 2pi, a twodimensional surface the boundary of a ball in four dimensions is a 3sphere an object with three dimensions. DNS Records, water true, change the 5, removing a single point from the 3sphere yields threedimensional space.
##### University of texas homework? Paper matrix sphere
The round metric on the 3-sphere in these coordinates is given by ds2deta 2sin 2eta,dxi _12cos 2eta,dxi _22 and the volume form by dVsincosdd1d2.displaystyle dVsin eta cos eta,deta wedge dxi _1wedge dxi.This image shows points on S 2 and their corresponding fibers with the same color.Gluing edit A 3-sphere can be constructed topologically by "gluing" together the boundaries of a pair of 3- balls.
# The higher-homotopy groups ( k 4) are all finite abelian but otherwise follow no discernible pattern.
Because the size of the pattern is limited by the A4 sized paper an ordinary cutter.
Sphere #012 with circles is prepared in the same way.
You need 2 times 12 arms to make the sphere. | 968 | 3,992 | {"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} | 2.59375 | 3 | CC-MAIN-2019-09 | latest | en | 0.917791 |
https://www.hpmuseum.org/forum/archive/index.php?thread-7052.html | 1,571,681,348,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987781397.63/warc/CC-MAIN-20191021171509-20191021195009-00294.warc.gz | 942,405,816 | 2,934 | # HP Forums
Full Version: Integration Question HP50g
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I'm trying to integrate e^-(ax^2) dx from -inf to +inf. When I use PREVAL, it says unsigned inf error. When I use use equation editor to find the answer, it just keeps adding t in the expression. Wolfram alpha does it correctly:
https://www.wolframalpha.com/input/?i=%E...+to+%2Binf
I know I might have to make an assumption that the 'a' in my expression is greater than 0 but I don't know how to do so. So any suggestions?
Also, are all the integration functions equal or are some better than the other. For eg, for definite integrals, are PREVAL, INT, and using the equation editor methods all equal? or is one inherently better than the other?
(10-16-2016 10:15 AM)Fanboy Wrote: [ -> ]I'm trying to integrate e^-(ax^2) dx from -inf to +inf. When I use PREVAL, it says unsigned inf error. When I use use equation editor to find the answer, it just keeps adding t in the expression. Wolfram alpha does it correctly:
https://www.wolframalpha.com/input/?i=%E...+to+%2Binf
I know I might have to make an assumption that the 'a' in my expression is greater than 0 but I don't know how to do so. So any suggestions?
Also, are all the integration functions equal or are some better than the other. For eg, for definite integrals, are PREVAL, INT, and using the equation editor methods all equal? or is one inherently better than the other?
Not sure if this is the only issue, but you must have a "*" between 'a' and 'x', implicit multiplication isn't allowed.
Must be a homework assignment somewhere, 2nd time in a week....
Hello fanboy,
you can use the hp 50g CAS for integrating this, but there are manually several steps necessary. Of course with the input correction rprosperi mentioned.
The integral is not a simple one. The CAS of wolfram alpha is much mighter than that one in hp 50g.
(10-16-2016 02:33 PM)rprosperi Wrote: [ -> ]
(10-16-2016 10:15 AM)Fanboy Wrote: [ -> ]I'm trying to integrate e^-(ax^2) dx from -inf to +inf. When I use PREVAL, it says unsigned inf error. When I use use equation editor to find the answer, it just keeps adding t in the expression. Wolfram alpha does it correctly:
https://www.wolframalpha.com/input/?i=%E...+to+%2Binf
I know I might have to make an assumption that the 'a' in my expression is greater than 0 but I don't know how to do so. So any suggestions?
Also, are all the integration functions equal or are some better than the other. For eg, for definite integrals, are PREVAL, INT, and using the equation editor methods all equal? or is one inherently better than the other?
Not sure if this is the only issue, but you must have a "*" between 'a' and 'x', implicit multiplication isn't allowed.
Must be a homework assignment somewhere, 2nd time in a week....
That '*' is not the problem. I have that when I enter it in the 50g. I just typed it here that way.
Hello Fanboy,
have a look HERE for the integration of your integral (it is only possible for infinte limits, but there exists no antiderivate). That means the '*' doesn't do the job alone.
@Vtile, that is not the same problem.
@peacecalc, My bad. I didn't read it properly, I just noticed the note that there were "the same question asked a few days ago".
(10-17-2016 05:20 PM)peacecalc Wrote: [ -> ]Hello Fanboy,
have a look HERE for the integration of your integral (it is only possible for infinte limits, but there exists no antiderivate). That means the '*' doesn't do the job alone.
@Vtile, that is not the same problem.
I was hoping I could somehow coax the 50g into spitting out the answer [sqrt(pi)] for that definite integral but I guess that's not possible. I was interested in that integral because that integral and its variations pop up a lot in quantum mechanics. I don't always feel like taking out my phone to use Wolfram Alpha lol.
Hello fanboy,
maybe there is a way, have a look to the commands MATCH (down arrow as a sign) or
MATCH (up arrow as a sign) in AUR (Advanced User Reference manual).
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | 1,065 | 4,175 | {"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} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.928227 |
https://www.coursehero.com/file/7146517/84-d-t-Tensile-stress-for-the-material-of-the-stud-which-is/ | 1,513,280,330,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550199.46/warc/CC-MAIN-20171214183234-20171214203234-00620.warc.gz | 724,801,635 | 22,147 | # 84 d t tensile stress for the material of the stud
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Unformatted text preview: r of the stud in mm. It is usually taken as 0.84 d. σt = Tensile stress for the material of the stud which is usually nickel steel. ns = Number of studs. We know that the force acting on the cylinder head (or on the studs) π π × D 2 × p = (115) 2 3.15 = 32 702 N ...(i) = 4 4 The number of studs (ns ) are usually taken between 0.01 D + 4 (i.e. 0.01 × 115 + 4 = 5.15) and 0.02 D + 4 (i.e. 0.02 × 115 + 4 = 6.3). Let us take ns = 6. We know that resisting force offered by all the studs π π = ns × ( dc )2 σt = 6 × (0.84d ) 2 65 = 216 d 2 N ...(ii) 4 4 ...(Taking σt = 65 MPa = 65 N/mm2) From equations (i) and (ii), d 2 = 32 702 / 216 = 151 or d = 12.3 say 14 mm 1132 n A Textbook of Machine Design The pitch circle diameter of the studs (Dp) is taken D + 3d. ∴ Dp = 115 + 3 × 14 = 157 mm We know that pitch of the studs π × Dp π × 157 = = 82.2 mm = 6 ns We know that for a leak-proof joint, the pitch of the studs should lie between 19 d to 28.5 d , where d is the nominal diameter of the stud. ∴ Minim...
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Ask a homework question - tutors are online | 453 | 1,312 | {"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} | 3.84375 | 4 | CC-MAIN-2017-51 | latest | en | 0.865024 |
https://www.biostars.org/p/9498240/#9498347 | 1,716,856,713,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00468.warc.gz | 559,139,094 | 7,208 | Retrieving and interpreting specific contrast results from an interaction experiment in DESeq2
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Entering edit mode
2.5 years ago
fr ▴ 210
I have an experiment with 2 different groups, and several different time points (for simplicity, let's say it is 2 time points), and I'm trying to find the best way to model both group and time effects. Note that time is a categorical variable.
# in
dds <- makeExampleDESeqDataSet(n=100,m=32)
dds$time <- factor(rep(rep(c("T1","T2"),each=8),2)) dds$group = factor(rep(rep(rep(c("Ct","Tr"),each=4),2),2))
design(dds) <- ~ time + group + time:group
dds <- DESeq(dds)
colData(dds)
# out
DataFrame with 32 rows and 4 columns
condition group sizeFactor replaceable
<factor> <factor> <numeric> <logical>
sample1 A Ct 1.080654 TRUE
sample2 A Ct 1.009824 TRUE
sample3 A Ct 1.035633 TRUE
sample4 A Ct 0.953337 TRUE
sample5 A Tr 1.060243 TRUE
... ... ... ... ...
sample28 B Ct 1.22126 TRUE
sample29 B Tr 1.04888 TRUE
sample30 B Tr 1.09366 TRUE
sample31 B Tr 1.11761 TRUE
sample32 B Tr 1.23430 TRUE
We can also see the factors that were computed:
# in
resultsNames(dds)
# out
[1] "Intercept" "group_Tr_vs_Ct" "time_T2_vs_T1" "groupTr.timeT2"
Modeling these means we are doing
(Equation 1) y = alpha + beta1 time + beta2 group + beta3 group time
Or using the levels above, the following beta coefficients are defined:
(Equation 2) y = intercept + time_T2_vs_T1 time + group_Tr_vs_Ct group + groupTr.timeT2 group time
Given the reference levels are Ct and T1 (which I am assuming will be encoded as 0 by the model), I'd like to confirm if my interpretation of the following effects makes sense:
• Effect of Tr group (for T1):
group_Tr_vs_Ct
• Effect of T2 time (for Ct):
time_T2_vs_T1
• Effect of Tr group (for T2):
group_Tr_vs_Ct + groupTr.timeT2
• Effect of T2 time (for Ct and Tr), i.e. regardless of group or (Tr.T2 - Tr.T1) - (Ct.T2 - Ct.T1):
(group_Tr_vs_Ct + groupTr.timeT2 - group_Tr_vs_Ct) - (time_T2_vs_T1 - 0) = groupTr.timeT2 - time_T2_vs_T1
• Effect of Tr group (for T1 + T2), i.e. regardless of time or (Tr.T2 - Ct.T2) - (Tr.T1 - Ct.T1):
(group_Tr_vs_Ct + groupTr.timeT2 - time_T2_vs_T1) - (group_Tr_vs_Ct - 0) = groupTr.timeT2 - time_T2_vs_T1
I've been checking a few posts with similar questions that have been quite helpful (example, example and especially here), but I'm uncertain of the interpretation of more complex comparisons. So my questions are:
1. Does the interpretation of the effects above make sense?
2. How to retrieve the final contrast above?
3. Why is it that intercept is not considered for any of the above, in the documentation and posts? Given Equation 1, one would think the intercept would be added to each of the terms above.
deseq2 • 1.1k views
0
Entering edit mode
2.5 years ago
Using interactions is not the easiest way to do a lot of these comparisons. It's often simpler, and way more readable, to make a new column with both condition and group concatenated, and contrast with that.
http://bioconductor.org/packages/devel/bioc/vignettes/DESeq2/inst/doc/DESeq2.html#interactions
I think to get all the A's versus B's it's simpler to make your design ~ condition + group, and then just do A vs B.
0
Entering edit mode
Thanks for your comment @swbarnes2, I agree with you that it becomes quite complex. My question is how to specify the contrasts for multiple columns in that situation? For the last example, and considering your approach, I would get coefficients for each of the four factors Ct.Tr1, Ct.Tr2, Tr.T1 and Tr.T2. I see it is straightforward to contrast a pair of these (e.g. c( group, Tr.T2, Tr.T1)), but how to find a condition contrast regardless of time in this case (i.e. (Tr.T2 - Ct.T2) - (Tr.T1 - Ct.T1))?
edit: would such conditions be given as a vector of numbers based on the order of resultsNames( )? For instance
# in
resultsNames(dds)
# out
[1] "Intercept" "Ct.T1" "Ct.T2" "Tr.T1" "Tr.T2"
I could retrieve the appropriate contrasts as follows
# for Ct.T2 vs Ct.T1, or (Ct.T2 - Ct.T1):
results(dds, contrast=c(0,-1, 1, 0, 0))
# for group difference irrespective of time, i.e. (Tr.T2 - Ct.T2) - (Tr.T1 - Ct.T1):
results(dds, contrast=c(0,1, -1, -1, 1))
`
Would this make sense? | 1,313 | 4,506 | {"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": 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} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.821708 |
http://pense.com.br/density-lab-report.html | 1,553,251,614,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202642.32/warc/CC-MAIN-20190322094932-20190322120932-00066.warc.gz | 159,948,782 | 4,384 | # Density lab report. Density Lab Report Essay 2019-02-14
Density lab report Rating: 9,1/10 687 reviews
## Density
Therefore, there was an inaccuracy in reading the volume displaced from the graduated cylinder and the biuret. The density I had obtained from the slope of the graph was 9. This was different from a theoretical density of the pennies that is 7. Then weighed an empty 250ml Erlenmeyer flask. If so, the liquid you thought was densest should be at the bottom of the jar. Since these densities appear to be significantly different as compared to the literature value of 7.
Next
## Lab 1
Calculations: For each of the three trials I calculated the density of the solution and determined the mean, average deviation from the mean, percent precision and the range. Collect a separate sample of each soil for this! Please put disposable pipets and any broken glassware in broken glass containers, not the trash can. Why do objects that are the same size sometimes have different weights? This value equals the uncertainity in the volume of the metal cylinder. Hypothesis I think the 1st cube was aluminum I think the 2nd cube was steel I think the 3rd cube was brass I think the 4th cube was copper I think the 5th cube was acrylic I think the 6th cube was oak I think the 7th cube was pvc I think the 8th cube was pine I think the 9th cube was poplar I think the 10th cube was nylon Materials Aluminum Steel Brass Copper Acrylic Oak Pvc Pine Poplar Nylon Ruler Triple beam balance Calculator Writing Utensil Method 1 Thought of my hypothesis when Mrs. The Problem we are Testing: What is the percent % difference of salt in the water that will make the water be in layers? Then I rinsed a buret with a little of the ethanol solution, and filled the buret with that same solution. When given an unknown substance, there are only a few ways to determine what it is.
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## Density Lab Report
Procedure: a The density of metals First I obtained a quantity of unknown metal. Density effects and influences ocean currents. This volume equals the volume of the metal cylinder. Across the world, chronic diseases, and infant mortality have continued affecting people of all races. This lab will focus on quantitative observations, more specifically, measurements.
Next
## Lab 1
Water movements driven by differences in density are also known as thermohaline circulation because water density depends on its temperature thermo and salinity haline. The method used for determining the density of a substance depends on the nature of the substance. I assume 0 ppt would be perfect precision, so 1. Please put your results on the board when you have completed the calculations. My accuracy was also high, as my error was only 0. Analytical balances accurate to ±0.
Next
## Density Lab Report Example
The answer is density in grams per milliliter. Natural soil-forming processes that increase aggregation reduce bulk density, but excessive tillage and raindrop impact on bare soil destory aggregation and increase bulk density. For Trial 2 the percentage was 3. It is able to determine. Error Propagation and Graph Linearization in Volumes and Density Introduction In lab experiments, errors emerge from numerous sources including variability of instruments, imprecision in observation and sample differences among others. Have you ever wondered how they do that? Mass, for both liquid and. A few drops of the liquid inside seemed to drip out from right above the bottom where it was supposed to come out.
Next
## Chemistry Lab Report 2
A sample equation would be 76. © 2009—2019 Adam Cap riola. I used only one type of metal for this part of the experiment. The purpose of this experiment is to make the heaviest floating film container without having it sink. Whether it's over eager young scientists year after year, or rigorous requirements that come once-in-a lifetime.
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## Density Lab Report by Abby Breitfeld on Prezi
Indicate which values are your's. Volume is an amount of space, in three dimensions, that a sample of matter occupies. Your answer will be more exact if you use a graduated cylinder instead of a beaker to measure the volume and weigh the liquid. My data from the unknown substance turned out incredibly well. The bonding between the atoms determines the strength of the material.
Next
## Density Lab Report
The spacing of liquid particles should be in between that of solids and gases. View Lab Report - Lab Report. Using a pipet, slowly add red water a drop at a time and watch what happens. Write down what happens to the raisin in each beaker. You will need to increase the size of the table to accomodate class results.
Next
## Density Lab Report
Our products are durable, reliable, and affordable to take you from the field to the lab to the kitchen. Being careful not to go over the 50ml mark. Changing water density could cause a submarine to rise or fall unexpectedly. X has a range of possible values from zero to one 0 - 100%. One way is to measure its density at a given temperature.
Next | 1,090 | 5,058 | {"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} | 3.34375 | 3 | CC-MAIN-2019-13 | latest | en | 0.965975 |
http://mathforum.org/library/drmath/view/57426.html | 1,519,312,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814124.25/warc/CC-MAIN-20180222140814-20180222160814-00217.warc.gz | 230,929,440 | 2,673 | Associated Topics || Dr. Math Home || Search Dr. Math
### Filling a Tank
```
Date: 08/01/97 at 20:14:14
From: EVAN
Subject: Filling a Tank
If one pipe can fill a tank in 1 1/2 hours, and another can fill the
same tank in 45 minutes, how long will it take the two pipes to fill
the tank together?
```
```
Date: 08/07/97 at 15:37:16
From: Doctor Rob
Subject: Re: Filling a Tank
This is a common type of problem given in algebra classes. Think
about the part of the tank the first pipe can fill in one minute.
That would be 1/90 of the tank. Now think about the part of the tank
the second pipe can fill in a minute: 1/45 of the tank. Working
together, both pipes can fill 1/90 + 1/45 of the tank in one minute.
Let T be the number of minutes to fill the whole tank together. Then
T*(1/90 + 1/45) = 1 tankful.
Now solve for T.
All problems of this type can usually be solved by a similar
technique.
-Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Word Problems
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Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 393 | 1,444 | {"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} | 3.65625 | 4 | CC-MAIN-2018-09 | latest | en | 0.841922 |
http://metamath.tirix.org/mpests/ne0gt0d | 1,716,521,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00040.warc.gz | 16,499,623 | 2,040 | Metamath Proof Explorer
Theorem ne0gt0d
Description: A nonzero nonnegative number is positive. (Contributed by Mario Carneiro, 27-May-2016)
Ref Expression
Hypotheses ltd.1 ${⊢}{\phi }\to {A}\in ℝ$
ne0gt0d.2 ${⊢}{\phi }\to 0\le {A}$
ne0gt0d.3 ${⊢}{\phi }\to {A}\ne 0$
Assertion ne0gt0d ${⊢}{\phi }\to 0<{A}$
Proof
Step Hyp Ref Expression
1 ltd.1 ${⊢}{\phi }\to {A}\in ℝ$
2 ne0gt0d.2 ${⊢}{\phi }\to 0\le {A}$
3 ne0gt0d.3 ${⊢}{\phi }\to {A}\ne 0$
4 ne0gt0 ${⊢}\left({A}\in ℝ\wedge 0\le {A}\right)\to \left({A}\ne 0↔0<{A}\right)$
5 1 2 4 syl2anc ${⊢}{\phi }\to \left({A}\ne 0↔0<{A}\right)$
6 3 5 mpbid ${⊢}{\phi }\to 0<{A}$ | 302 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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-2024-22 | latest | en | 0.535108 |
https://mathematica.stackexchange.com/questions/187197/how-to-test-all-possible-input-for-a-logic-function/187198 | 1,563,349,663,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525094.53/warc/CC-MAIN-20190717061451-20190717083451-00382.warc.gz | 475,338,224 | 36,982 | # How to test all possible input for a logic function?
I have a logic expression: f0[a0_, a1_, a2_, a3_] := a0 And Not a1 And Not a2 And a3 Or Not a0 And a2 And a3 Or Not a0 And a1 And a3, I know I should use BooleanTable, but it cannot generate a table like below.
How to generate a truth table in mathematica like below?
• There is the BooleanTable command. Does it help? For creating the corresponding function you can use logical and: && and logical or: || in case you don't know. – user59583 Dec 3 '18 at 6:17
• @Buddha_the_Scientist Thank you, I have correct my expression, Of course BooleanTable can help me, but it cannot generate a table like the picture above, it only returns a series of true or false. – 余星佑 Dec 3 '18 at 6:23
• Then use the TableForm with the option TableHeadings for the result. Something like: TableForm[{{a, b, c, d}}, TableHeadings -> {None, {"a0", "a1", "a2", "a3"}}] – user59583 Dec 3 '18 at 6:24
• @Buddha_the_Scientist, thank you, but how to use my function to generate all series of possible outputs like this? I know that how to generate one possible output commad: TableForm[{{True, False, True, False, f0[True, False, True, False]}}, TableHeadings -> {None, {"a0", "a1", "a2", "a3", "b0"}}] – 余星佑 Dec 3 '18 at 6:44
f = a0 && ! a1 && ! a2 && a3
TableForm[BooleanTable[{a0, a1, a2, a3, f}, {a0, a1, a2, a3}], TableHeadings -> {None, {a0, a1, a2, a3, f}}]
Apparently there are resources like this one or this one
• Thank you very much! This absolutely solve my quesiton! – 余星佑 Dec 3 '18 at 7:20
truthTableFormattor[rawData_] := Insert[Insert[
Grid[rawData /. {0 -> 0,
1 -> Item[1, Background -> Lighter[Magenta]]},
FrameStyle -> Gray,
Frame -> All], {Background -> {None, {GrayLevel[0.7], {White}}},
Dividers -> {Black, {2 -> Black}}, Frame -> True,
Spacings -> {2, {2, {0.7}, 2}}}, 2], {Dividers -> All,
Spacings -> .7 {1, 1}}, 2];
truthTable[f__] := Module[{}, atoms = Cases[Most[{f}],
(a_ /;Length[a] == 0 \[And] Not[StringQ[a]])];heads =
rawData = Transpose@Boole[BooleanTable[#, atoms] & /@ {f}];
If[Last[{f}] === 1,
Transpose@Boole[BooleanTable[#, atoms] & /@ Most[{f}]],
If[Last[{f}] === "rev",
truthTableFormattor[{ToString[ | 735 | 2,179 | {"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} | 3 | 3 | CC-MAIN-2019-30 | latest | en | 0.681808 |
http://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&q=What+Is+Gravitation+Force | 1,464,733,953,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464053209501.44/warc/CC-MAIN-20160524012649-00037-ip-10-185-217-139.ec2.internal.warc.gz | 354,346,509 | 17,003 | Did you mean: What Is Gravitational Force?
Web Results
## Newton's law of universal gravitation - Wikipedia, the free ...
en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation
Newton's law of universal gravitation states that any two bodies in the universe attract each other with a force that is directly proportional to the product of their ...
## Gravitational Force: Definition, Equation & Examples - Video ...
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## Gravitational force - definition of gravitational force by The Free ...
www.thefreedictionary.com/gravitational force
The weakest of the four fundamental forces of nature, being the attractive force that arises from gravitational interaction. Newton's law of gravity states that the ...
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www.vocabulary.com/dictionary/gravitational force
n (physics) the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface. Synonyms: ...
## Newton's Law of Universal Gravitation - The Physics Classroom
www.physicsclassroom.com/class/circles/Lesson-3/Newton-s-Law-of-Universal-Gravitation
As discussed earlier in Lesson 3, Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces ...
## Gravitational force dictionary definition | gravitational force defined
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The weakest of the four fundamental forces of nature, being the attractive force that arises from gravitational interaction. Newton's law of gravity states that the ...
## Gravitational Force - Universe Today
www.universetoday.com/75321/gravitational-force/
Oct 8, 2010 ... Newton's Law of Universal Gravitation is used to explain gravitational force. This law states that every massive particle in the universe attracts ...
## What is gravity (or gravitation)? - Definition from WhatIs.com
whatis.techtarget.com/definition/gravity-or-gravitation
What is gravity, also known as gravitation? This definition explains the mysterious force that exists among all material objects in the universe and how it works ...
## Physics4Kids.com: Motion: Gravity
www.physics4kids.com/files/motion_gravity.html
Gravity of the Earth pulls objects towards the center of the planet. Gravity or gravitational forces are forces of attraction. We're not talking about finding someone ...
## Gravitational force – Science & Technology – Skwirk Year 6, NSW ...
www.skwirk.com/p-c_s-11_u-399_t-988_c-3776/gravitational-force/nsw/science-technology/forces-and-their-effects/types-of-forces
Year 5 or 6 Science & Technology homework? Visit us for info on gravitational force & other types of forces. Specifically find info here on what gravitational force ... | 631 | 2,949 | {"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-2016-22 | longest | en | 0.811106 |
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/05%3A_Two_Dimensional_Kinematics/5.01%3A_Introduction_to_the_Vector_Description_of_Motion_in_Two_Dimensions | 1,721,021,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00520.warc.gz | 417,252,516 | 30,316 | # 5.1: Introduction to the Vector Description of Motion in Two Dimensions
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
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$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
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$$\newcommand{\Argument}{\mathrm{Arg}}$$
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$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
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$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
Where was the chap I saw in the picture somewhere? Ah yes, in the dead sea floating on his back, reading a book with a parasol open. Couldn’t sink if you tried: so thick with salt. Because the weight of the water, no, the weight of the body in the water is equal to the weight of the what? Or is it the volume equal to the weight? It’s a law something like that. Vance in High school cracking his fingerjoints, teaching. The college curriculum. Cracking curriculum. What is weight really when you say weight? Thirtytwo feet per second per second. Law of falling bodies: per second per second. They all fall to the ground. The earth. It’s the force of gravity of the earth is the weight.
~ James Joyce
We have introduced the concepts of position, velocity and acceleration to describe motion in one dimension; however we live in a multidimensional universe. In order to explore and describe motion in more than one dimension, we shall study the motion of a projectile in two-dimension moving under the action of uniform gravitation.
We extend our definitions of position, velocity, and acceleration for an object that moves in two dimensions (in a plane) by treating each direction independently, which we can do with vector quantities by resolving each of these quantities into components. For example, our definition of velocity as the derivative of position holds for each component separately. In Cartesian coordinates, the position vector $$\overrightarrow{\mathbf{r}}(t)$$ with respect to some choice of origin for the object at time t is given by
$\overrightarrow{\mathbf{r}}(t)=x(t) \hat{\mathbf{i}}+y(t) \hat{\mathbf{j}} \nonumber$
The velocity vector $$\overrightarrow{\mathbf{v}}(t)$$ at time t is the derivative of the position vector,
$\overrightarrow{\mathbf{v}}(t)=\frac{d x(t)}{d t} \hat{\mathbf{i}}+\frac{d y(t)}{d t} \hat{\mathbf{j}} \equiv v_{x}(t) \hat{\mathbf{i}}+v_{y}(t) \hat{\mathbf{j}} \nonumber$
where $$v_{x}(t) \equiv d x(t) / d t$$ and $$v_{y}(t) \equiv d y(t) / d t$$ denote the x - and y -components of the velocity respectively.
The acceleration vector $$\overrightarrow{\mathbf{a}}(t)$$ is defined in a similar fashion as the derivative of the velocity vector,
$\overrightarrow{\mathbf{a}}(t)=\frac{d v_{x}(t)}{d t} \hat{\mathbf{i}}+\frac{d v_{y}(t)}{d t} \hat{\mathbf{j}} \equiv a_{x}(t) \hat{\mathbf{i}}+a_{y}(t) \hat{\mathbf{j}} \nonumber$
where $$a_{x}(t) \equiv d v_{x}(t) / d t$$ and $$a_{y}(t) \equiv d v_{y}(t) / d t$$ denote the x- and y-components of the acceleration
This page titled 5.1: Introduction to the Vector Description of Motion in Two Dimensions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform. | 2,535 | 7,253 | {"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} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.205048 |
http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-5-section-5-5-the-substitution-rule-5-5-exercises-page-418/12 | 1,524,229,448,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937780.9/warc/CC-MAIN-20180420120351-20180420140351-00091.warc.gz | 413,859,699 | 12,534 | ## Calculus: Early Transcendentals 8th Edition
$$\int\sec^22\theta d\theta=\frac{1}{2}\tan(2\theta)+C$$
$$A=\int\sec^22\theta d\theta$$ Let $u=2\theta$ Then $du=2d\theta$. So $d\theta=\frac{1}{2}du$ Substitute into $A$, we have $$A=\int\sec^2u(\frac{1}{2})du$$ $$A=\frac{1}{2}\int \sec^2udu$$ $$A=\frac{1}{2}(\tan u)+C$$ $$A=\frac{1}{2}\tan(2\theta)+C$$ | 161 | 354 | {"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} | 3.9375 | 4 | CC-MAIN-2018-17 | latest | en | 0.267045 |
https://forum.math.toronto.edu/index.php?PHPSESSID=p4aplvhetep1069b5dgr2n8ap7&topic=284.0;prev_next=next | 1,721,911,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00429.warc.gz | 228,828,882 | 8,564 | ### Author Topic: Night Sections (Read 12166 times)
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### Night Sections
« on: April 03, 2013, 08:07:49 PM »
\begin{equation*}
\left\{\begin{aligned}
&\frac{dx}{dt} = x(1.5-0.5x-y),\\
&\frac{dy}{dt} = y(2-y-1.125x).
\end{aligned}\right.
\end{equation*}
This problem can be interpreted as describing the interaction of two species with population densities $x$ and $y$.
(b) Find the critical points.
(c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, stable, or unstable.
(d) Sketch the trajectories in the neighborhood of each critical point.
(e) [Bonus] Find, if possible, solution in the form $H(x,y)=C$ and sketch the phase portrait.
#### Frank (Yi) Gao
• Newbie
• Posts: 1
• Karma: 1
##### Re: Night Sections
« Reply #1 on: April 03, 2013, 08:49:26 PM »
(b) Finding the critical points means solving:
\begin{equation*}
0 = x(1.5 - 0.5x - y),
0 = y(2 - x - 1.125y)
\end{equation*}
There are four possibilities then,
\begin{equation*}
(x,y) = (0,0),(0,2),(3,0),(\frac{4}{5},\frac{11}{10})
\end{equation*}
(c) The Jacobian for this question is then,
\begin{equation*}
J = \left( \begin{array}{cc} 1.5-1.5x-y & -x \\ -1.125y & 2-2y-1.125x \end{array} \right).
\end{equation*}
For the point (0,0):
\begin{equation*}
J= \left( \begin{array}{cc} 1.5 & 0 \\ 0 & 2 \end{array} \right).
\end{equation*}
So,
\begin{equation*}
\lambda_{1} = 1.5, \lambda_{2} = 2 \\
\xi_{1} = \left(\begin{array}{cc} 1 \\ 0 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 0 \\ 1 \end{array} \right)
\end{equation*}
and thus, (0,0) represents a unstable node. For (0,2),
\begin{equation*}
J= \left( \begin{array}{cc} -0.5 & 0 \\ -2.25 & -2 \end{array} \right).
\end{equation*}
So,
\begin{equation*}
\lambda_{1} = -0.5, \lambda_{2} = -2 \\
\xi_{1} = \left(\begin{array}{cc} 3 \\ 1 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 0 \\ 1 \end{array} \right)
\end{equation*}
and thus (0,2) its a node which is asymptotically stable. For (3,0),
\begin{equation*}
J= \left( \begin{array}{cc} -1.5 & -3 \\ 0 & -1.375 \end{array} \right).
\end{equation*}
So,
\begin{equation*}
\lambda_{1} = -1.5, \lambda_{2} = -1.375 \\
\xi_{1} = \left(\begin{array}{cc} 1 \\ 0 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 24 \\ -1 \end{array} \right)
\end{equation*}
and thus (3,0) is also an asymptotically stable node. Finally, for (4/5,11/10),
\begin{equation*}
J= \left( \begin{array}{cc} -\frac{2}{5} & -\frac{4}{5} \\ -\frac{99}{80} & -\frac{11}{10} \end{array} \right).
\end{equation*}
So,
\begin{equation*}
\lambda^{2} + \frac{3}{2}\lambda - \frac{11}{20} = 0 \\
\lambda_{1} = -1.8, \lambda_{2} = 0.3 \\
\xi_{1} = \left(\begin{array}{cc} 1 \\ -1.4 \end{array} \right), \xi_{2} = \left(\begin{array}{cc} 1.57 \\ -1 \end{array} \right)
\end{equation*}
and thus (4/5,11/10) is a saddle point (unstable).
« Last Edit: April 03, 2013, 10:26:42 PM by Frank (Yi) Gao »
#### Benny Ho
• Newbie
• Posts: 4
• Karma: 2
##### Re: Night Sections
« Reply #2 on: April 03, 2013, 08:56:47 PM »
solution
#### Devangi Vaghela
• Jr. Member
• Posts: 6
• Karma: 3
##### Re: Night Sections
« Reply #3 on: April 03, 2013, 09:19:55 PM »
#### Devangi Vaghela
• Jr. Member
• Posts: 6
• Karma: 3
##### Re: Night Sections
« Reply #4 on: April 03, 2013, 09:45:23 PM »
#### Victor Ivrii
• Elder Member
• Posts: 2607
• Karma: 0
##### Re: Night Sections
« Reply #5 on: April 04, 2013, 12:04:28 AM »
Benny, nice try, but ...
Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)
#### Sabrina (Man) Luo
• Jr. Member
• Posts: 9
• Karma: 4
##### Re: Night Sections
« Reply #6 on: April 04, 2013, 08:20:40 AM »
Benny, nice try, but ...
Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)
#### Sabrina (Man) Luo
• Jr. Member
• Posts: 9
• Karma: 4
##### Re: Night Sections
« Reply #7 on: April 04, 2013, 08:46:34 AM »
Benny, nice try, but ...
Devangi – good, Frank—even better. But phase portrait is still missing (and bonus question too)
« Last Edit: April 04, 2013, 08:52:27 AM by Sabrina (Man) Luo »
#### Alexander Jankowski
• Full Member
• Posts: 23
• Karma: 19
##### Re: Night Sections
« Reply #8 on: April 04, 2013, 11:17:02 AM »
To complete this thread, here is a stream plot. Overall, the analysis tells us that in this case, one of the competing species must eventually die out--unless, by chance, the initial populations lie on the separatrix that passes through $(4/5,11/10)$.
#### Victor Ivrii
To complete this thread, here is a stream plot. Overall, the analysis tells us that in this case, one of the competing species must eventually die out--unless, by chance, the initial populations lie on the separatrix that passes through $(4/5,11/10)$. | 1,841 | 4,959 | {"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} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.546595 |
https://gmatclub.com/forum/gmat-data-sufficiency-ds-141/index-450.html?sd=a | 1,496,119,043,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00043.warc.gz | 950,735,987 | 60,331 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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150 Hardest and easiest questions for DS Tags:
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,372 | 6,207 | {"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-2017-22 | latest | en | 0.755085 |
https://www.playtaptales.com/numbers/octotrigintacentillion/ | 1,555,845,212,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530527.11/warc/CC-MAIN-20190421100217-20190421122217-00303.warc.gz | 788,751,128 | 4,181 | ## Octotrigintacentillion
A Octotrigintacentillion (1 Octotrigintacentillion) is 10 to the power of 417 (10^417). This is a tremendously astronomical number!
## How many zeros in a Octotrigintacentillion?
There are 417 zeros in a Octotrigintacentillion.
## What's before Octotrigintacentillion?
A Septentrigintacentillion is smaller than a Octotrigintacentillion.
## What's after Octotrigintacentillion?
A Noventrigintacentillion is larger than a Octotrigintacentillion.
## Octotrigintacentillionaire
A Octotrigintacentillionaire is someone whos assets, net worth or wealth is 1 or more Octotrigintacentillion. It is unlikely anyone will ever be a true Octotrigintacentillionaire. If you want to be a Octotrigintacentillionaire, play Tap Tales!
## Is Octotrigintacentillion the largest number?
Octotrigintacentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
## Octotrigintacentillion written out
Octotrigintacentillion is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
## Big Numbers
This is just one of many really big numbers! | 568 | 1,691 | {"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-2019-18 | latest | en | 0.77083 |
http://etds.lib.tku.edu.tw/etdservice/view_metadata?etdun=U0002-2707201017235500&start=41&end=60&from=CATE&cateid=C003 | 1,603,508,713,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00507.warc.gz | 37,438,070 | 6,526 | 下載電子全文 （限經由淡江IP使用）
系統識別號 U0002-2707201017235500 中文論文名稱 移動平均式卡方適合度檢定統計量其分佈函數之研究 英文論文名稱 A study on the distribution of averaged shifted chi-square goodness-of-fit test statistics 校院名稱 淡江大學 系所名稱(中) 統計學系碩士班 系所名稱(英) Department of Statistics 學年度 98 學期 2 出版年 99 研究生中文姓名 簡溢進 研究生英文姓名 Yi-Jing Chien 學號 697650454 學位類別 碩士 語文別 中文 口試日期 2010-06-25 論文頁數 41頁 口試委員 指導教授-鄧文舜委員-黃文濤委員-林國欽 中文關鍵字 卡方適合度檢定 移動平均直方圖 無母數適合度檢定 二元搜尋法 英文關鍵字 chi-square goodness-of-fit test statistics averaged shifted histogram 學科別分類 學科別＞自然科學＞統計 中文摘要 當給定一組來自連續型分布的隨機觀測值，傳統皮爾遜(Pearson)之適合度檢定(goodness-of-fit test)程序，係將觀測值加以分組，然後再計算卡方值 (chi-square statistcis)來進行檢定。然而，不同分組起始點(cell origin)的選取，極可能導致不同的檢定結果。並且，把連續型資料加以分組，也會導致的母體分布訊息及檢定力(power)有所損失。Wu and Deng(2010)提出一改良檢定方法，係計算L個不同起始點之K個區間（分組）的卡方檢定量，並以其平均值做為檢定統計量，我們稱此一統計量為移動平均式卡方檢定統計量(averaged shifted chi-square test statistics)。 本文主要目的在若干小樣本及L,K下，建立移動平均式卡方統計量的確切分佈，並與其漸近分佈比較兩者的機率差距程度，另外，我們也發現，將統計量進行微調後，其漸近分佈所求算出的機率值，能夠更進一步的接近確切分佈的機率值。 英文摘要 The classical Pearson’s procedure for testing whether a random sample has been drawn from a continuous distribution is based on the ‘difference’ of the observed cell counts and their model based counterpart. The test statistics known as the chi-square statistics can be very sensitive to the choice of cell origin. Different choice of cell origin may lead to different result of goodness of fits test. Worse still is that test based on grouping of data is often expected to be less powerful. To cope with the above two problems, Wu and Deng (2010) proposed to repeatedly partition the sample space into k cells for L times to obtain L respective chi-square statistics, and use their average to serve as the test statistics. Call the resultant test statistics the averaged shifted chi-square statistics (ASCS). The purpose of this thesis is to derive the exact distributions of ASCS for some small values of (L,k) by exhaustive permutation of all possible values ASCS. By comparing the exact distributions with the limit distribution of ASCS, we find that the limit distribution approximates well its target exact counterpart. A simple method is proposed to improve the approximation of the exact distribution by the limit distribution. Simulation study reveals that the power of ASCS is less sensitive to the choice cell origin and that ASCS can be more powerful as the values of k increases. 論文目次 目錄 第 1 章 緒論 ........................................... 1 1.1 無母數適合度檢定簡介 ............................... 1 1.2 傳統卡方適合度檢定所遭遇之困難 ..................... 2 1.3 研究目的 ........................................... 4 1.4 章節編排 ........................................... 4 第 2 章 移動平均式卡方檢定統計量的介紹 ................. 5 2.1 移動平均式卡方檢定統計量的概念 ..................... 5 2.2 移動平均式卡方檢定統計量 ........................... 7 2.3 移動平均式卡方統計量的漸近分佈 ..................... 9 2.3.1 等距區間切割 ..................................... 9 2.3.2 等距區間切割下漸近分佈的模擬研究 ................ 11 2.3.3 錨點變動下檢定力的比較 .......................... 15 2.3.4 不等距切割條件下的漸近分佈 ...................... 19 2.3.5 不等距切割條件下的漸近分佈模擬研究 .............. 20 2.3.6 與文獻的結果比較 ................................ 23 第 3 章 確切分佈的建構與漸近分佈的改良................. 29 3.1 確切分佈的建構 .................................... 29 3.2 在漸近分佈的模擬改良 .............................. 32 3.3 確切分佈與漸近分佈模擬比較 ........................ 33 第 4 章 結論 .......................................... 39 參考文獻 .............................................. 40 表目錄 表1 不同起始點下之卡方檢定的結果 ....................... 3 表2 n=20,L=3,k=4之下，移動平均卡方統計量的計算法 ....... 8 表3 若干L,k下不同顯著水準相應的臨界點表 ............... 14 表4 等距與不等距下，傳統卡方與移動平均式卡方 的檢定力比較 .......................................... 26 表5 在n=10,若干小L,k的確切、修正前後漸近分佈尾端機率比較 .................................................... 36 表6 在n=15,若干小L,k的確切、修正前後漸近分佈尾端機率比較 .................................................... 37 表7 在n=20,若干小L,k的確切、修正前後漸近分佈尾端機率比較 .................................................... 38 圖目錄 圖1 傳統直方圖與移動平均直方圖 ......................... 6 圖2 等距切割下漸近分佈與經驗累積分佈比較圖 ............ 13 圖3 對立假設為Gamma (1.3,1.1) 之下，變動起始點t及L下的 檢定力比較圖 .......................................... 16 圖4 對立假設為Gamma (1.3,0.5) 之下，變動起始點t及L下的 檢定力比較圖 .......................................... 17 圖5 對立假設為Gamma (1.5,0.6) 之下，變動起始點t及L下的 檢定力比較圖 .......................................... 18 圖6 不等距切割區間下的漸近分佈與經驗累積分佈比較圖 .... 22 圖7 確切分佈流程圖 .................................... 31 圖8 n=15,若干L,k下的漸近、確切、經驗累積分佈比較圖 .... 34 圖9 n=20,若干L,k下的漸近、確切、經驗累積分佈比較圖 .... 35 參考文獻 參考文獻 Anderson T.W. and Darling D.A.,1952 “Asymptotic theory of certain goodness-of-fit criteria based on stochastic processes” Annals of Mathematical Statistics, 23,193-212 Birnbaum Z.W. and Tingey F.H., 1951“One-Sided Confidence Contours for Probability Distribution Functions” Ann. Math. Volume 22,Number 4,592-596 D'Agostino, Michael A. Stephens, 1986 “Goodness-of-fit techniques ” New York : Marcel Dekker Boero G., Smith J. and Wallis K. F.,2005“Sensitivity of the chi-squared goodness-of-fit test to the partitioning of data” Econometric Reviews, Volume 23, Issue 4,341-370 Gumbel E.J., 1943 “On the Reliability of the Classical Chi-Square Test ” The Annals of Mathematical Statistics, 14, No. 3 , 253-263 Henry C. and Thode. Jr, 2002 “Testing for normality ” New York : Marcel Dekker Imhof J. P.,1961 “Computing the distribution of quadratic forms in normal variables” Biometrika, 48, No. 3/4., 419-426 Jones M.C., Samiuddin M., Al-Harbey A.H. and Maatouk T.A.H.,1998 “The Edge Frequency Ploygon” Biomerika vol.85,No1,235-239 Marsaglia G., Marsaglia J.C.W., 2004 “Evaluating the Anderson-Darling distribution”Journal of Statistical Software 9, 1-5 Scott D.W., 1985 “Average shifted histogram: Effective nonparametric density estimators in several dimensions ” Annals of Statistics 13:1024–1040 Scott D.W., 1992 “Multivariate density estimation: theory, practice, and visualization” New York: Wiley Simonoff J.S. and Udina F.,1995“Measuring the Stability of Histogram Appearance when the Anchor Position is Changes” Economics Working Papers 133, Depeartment of Economics and Business Stephens M.A., 1974 “EDF Statistics for Goodness of Fit and Some Comparisons ”Journal of the American Statistical Association, 69, No. 347, 730-737 Viollaz A.J. 1986 “On the Reliability of the Chi-Square Test ” Metrika, 33, 135-142 Wu.J. S. and Deng W. S., 2010 “Average Shifted Chi-Square test” working paper 論文使用權限 同意紙本無償授權給館內讀者為學術之目的重製使用，於2012-07-30公開。同意授權瀏覽/列印電子全文服務，於2012-07-30起公開。
若您有任何疑問，請與我們聯絡！圖書館： 請來電 (02)2621-5656 轉 2486 或 來信 | 2,554 | 6,407 | {"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-2020-45 | latest | en | 0.429882 |
https://www.equationsworksheets.net/chemical-equations-and-symbols-worksheet/ | 1,675,630,656,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00495.warc.gz | 745,277,440 | 15,046 | # Chemical Equations And Symbols Worksheet
Chemical Equations And Symbols Worksheet – Expressions and Equations Worksheets are designed to help children learn faster and more efficiently. The worksheets contain interactive exercises and questions based on the sequence of operations. These worksheets make it simple for children to grasp complex concepts and basic concepts quickly. These PDF resources are free to download, and can be used by your child to test math problems. They are useful for students who are in the 5th-8th Grades.
These worksheets are intended for students in the 5th to 8th grades. These two-step word puzzles are made using decimals or fractions. Each worksheet contains ten problems. These worksheets can be found both online and in printed. These worksheets are a great way to test the practice of rearranging equations. Apart from practicing rearranging equations, they also help your student understand the principles of equality as well as reverse operations.
These worksheets can be used by fifth and eighth graders. These worksheets are ideal for students who struggle with calculating percentages. There are three kinds of questions you can choose from. It is possible to work on one-step problems that include decimal or whole numbers, or use word-based methods to do fractions or decimals. Each page contains ten equations. These Equations Worksheets can be used by students from the 5th-8th grades.
These worksheets are an excellent resource for practicing fraction calculations and other concepts related to algebra. You can choose from many different types of problems with these worksheets. You can select one that is word-based, numerical, or a mixture of both. It is crucial to select the problem type, because each problem will be different. Each page has ten questions that make them a fantastic aid for students who are in 5th-8th grade.
These worksheets help students understand the relationships between variables and numbers. These worksheets allow students to the opportunity to practice solving polynomial problems or solving equations, as well as discovering how to utilize them in daily life. If you’re in search of an excellent educational tool to learn about expressions and equations it is possible to begin by exploring these worksheets. They can help you understand about different types of mathematical issues and the various types of symbols used to represent them.
These worksheets could be beneficial for students in their first grades. The worksheets will assist them to master the art of graphing and solving equations. These worksheets are excellent for practicing with polynomial variable. They can also help you master the art of factoring and simplify these variables. There are a variety of worksheets you can use to teach children about equations. The best way to get started learning about equations is to complete the work yourself.
There are a variety of worksheets that can be used to help you understand quadratic equations. There are different levels of equation worksheets for each stage. The worksheets are designed to allow you to practice solving problems in the fourth level. After you’ve completed an amount of work then you are able working on other types of equations. It is then possible to tackle the same problems. For example, you might find a problem with the same axis and an extended number. | 622 | 3,388 | {"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-2023-06 | latest | en | 0.945055 |
https://engineering.stackexchange.com/questions/28873/how-to-find-a-dimension-that-isnt-explicit-if-there-are-tolerances | 1,563,323,454,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525004.24/warc/CC-MAIN-20190717001433-20190717023433-00183.warc.gz | 381,435,138 | 34,989 | # How to find a dimension that isn't explicit if there are tolerances
For a part like the one shown, let's say I wanted to find the dimension labeled 3", if it wasn't already labeled. Since there are tolerances already on the part, how would I determine the 3" value along with its tolerances? Can I just add and subject the tolerances? For example, 4.00 +- 0.3 - 1.00 +-0.1 = 3.00 +- 0.2 ? What if I wanted to find the distance from the right end of the hole to the right face of the part, assuming diameter D = 1" +- 0.01? Could I simply do 3.00 +- 0.2 - (1/2) +-(0.01/2) = 3.00 +- 0.2 - 0.5 +- 0.005 = 2.5 +- 0.195? Thanks for the help!
• Your example is a case of being over-dimensioned. It has multiple dimensions that can’t always be satisfied at the same time. My question to you is why do you need the 3.0 dimension? If it’s necessary you should specify it as “reference” by putting it in parenthesis. – GisMofx Jun 20 at 1:09
• 3 would have a compound tolerance of +-0.13 – joojaa Jun 20 at 4:00
• @GisMofx the drawing makes up for over-dimensioning in the horizontal direction by not giving the hole position vertically at all. – TimWescott Jun 20 at 17:40
• @GisMofx your comment is actually a good answer, just as it is. I suggest you copy & paste it into an answer. – TimWescott Jun 20 at 17:41 | 406 | 1,309 | {"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-2019-30 | longest | en | 0.901555 |
https://quadraticformulacalculator.net/2020/06/ | 1,701,691,855,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00878.warc.gz | 543,330,071 | 12,085 | Got Feedback?
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There Is More Than One Ways To Solve The Quadratic Equation
When it comes to stepping up your math game from linear equations to the more complex ones – most people get scared. One of the best examples of an equation that can frighten beginners is the quadratic equation. Even though the equation itself is pretty innocent but the combination of those alphabets can definitely give you a scare if you don’t know what to do.
Simply put, a Quadratic equation is an equation that looks like this “ax² + bx + c = 0”. As you can guess, ‘x’ is the variable that needs to be solved. A, b, and c are numbers where ‘a’ is not 0. What makes this equation quadratic is the squared term which is (ax2). When you put these values in a quadratic equation solver calculator then you don’t need to find out the square of the number. The quadratic equation solver calculator solves all functions itself.
The use of quadratic equations in the practical arena cannot be denied. It is often considered to be one of the most sophisticated mathematical equations. The the factoring quadratics calculator can calculate the values for you if you are trying to grow more crops or planning to fly a plane. This diverse implementation of the quadratic equation leads to a question – how to solve the equation.
There are more than one methods to solve a quadratic equation and it totally depends on you how you want to solve it. These methods are going to provide you with an introduction to four methods so you can pick the method which is easiest for you. Today, we are going to provide you with not one but four simple methods you can use to solve a quadratic equation which are stated below.
1. Square Root Method
2. Completing the Square
3. Factoring
Let’s discuss each of these methods in detail so you can know which one is easier for you. By the end of this article – you are going to find your favorite method to solve quadratic equations.
Square Root Method
The Square Root method is used to solve any quadratic equation when ‘bx’ is 0. Move the constant to the right side of the equation and then divide both sides by ‘a’, now you will take the square root of the both side that will give you two values of ‘x’ – one will be positive and the other one will be negative.
Square Root Method Example:
4x² – 9 = 0
4x² = 9
x² = 9 / 4
x² = 3 / 2 or – (3 / 2)
(take square root of both sides, and remember, every square root has a negative counterpart)
2. Completing the Square
Another pretty simple method to solve quadratic equations is completing the square. You can create a perfect square on the left side of the equation by adjusting your constant (c). After squaring, this perfect squared can now be factored into two binomials which will be identical. You can now use these identical binomials to obtain the value for ‘x’.
Completing the square:
x² – 4x + 9 = 0 (original equation)
x² – 4x = -9 (subtract 9 from both sides)
x² – 4x + 4 = 5 (take half of b, then square it and add that value to both sides of the equation)
(x – 2)² = 5 (break your perfect square into its binomials, made up of x, the sign in front of b and the square root of your new c)
x – 2 = √5 or – √ 5 (take the square root of both sides)
x = 2 + √5 or 2 – √ 5 (add 2 to both sides)
3. Factoring
The second approach to solve a quadratic equation is factoring. The concept behind factoring is to find the number pairs that can be multiplies together to produce ‘c’ and then summed to produce ‘b’. If you are able to find the right combination, you will get two binomials. These binomials can now be solved individually to get the value of ‘x’. | 883 | 3,709 | {"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} | 4.375 | 4 | CC-MAIN-2023-50 | longest | en | 0.948858 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-8-section-8-4-multiplicative-inverses-of-matrices-and-matrix-equations-exercise-set-page-932/13 | 1,571,175,132,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00073.warc.gz | 922,747,072 | 12,965 | ## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 932: 13
#### Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right]$.
#### Work Step by Step
Consider the given matrix $A=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]$ Now, by using the inverse formula, we get: ${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Let, \begin{align} & a=2 \\ & b=3 \\ & c=-1 \\ & d=2 \end{align} Substitute the values to get \begin{align} & {{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{\left| 2\times 2-3\times \left( -1 \right) \right|}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\frac{1}{4+3}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\frac{1}{7}\left[ \begin{matrix} 2 & -3 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right] \end{align} So, therefore the inverse of the matrix is given by ${{A}^{-1}}=\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right]$ Now, check the result for $A{{A}^{-1}}={{I}_{2}}$ And ${{A}^{-1}}A={{I}_{2}}$ Here, $A=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]$ So, \begin{align} & A{{A}^{-1}}=\left[ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{2}{7} & \frac{-3}{7} \\ \frac{1}{7} & \frac{2}{7} \\ \end{matrix} \right] \\ & A{{A}^{-1}}=\left[ \begin{matrix} 2\times \frac{2}{7}+3\times \frac{1}{7} & 2\times \frac{-3}{7}+3\times \frac{2}{7} \\ \left( -1 \right)\times \frac{2}{7}+2\times \frac{1}{7} & \left( -1 \right)\times \frac{-3}{7}+2\times \frac{2}{7} \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} This implies that ${{A}^{-1}}A={{I}_{2}}$ Now, evaluate the product ${{A}^{-1}}A={{I}_{2}}$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 965 | 2,372 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2019-43 | longest | en | 0.259755 |
https://www.tutorialandexample.com/assignment-operators-in-python | 1,680,334,904,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00074.warc.gz | 1,102,501,939 | 56,727 | # Assignment Operators in Python
The prime usage of Assignment Operators is to assign values to variables.
These are taken into account to do operations on values and variables. There are some special symbols in python to perform arithmetic, logical, and bitwise computations. Operands are the values on which operators perform operations.
Following are the assignment operators used in python:
1. Assign: operator used =
2. Add and Assignment: operator used +=
3. Subtract and Assignment: operator used -=
4. Multiply AND Assignment: operator used *=
5. Divide and Assignment: operator used /=
6. Modulus and Assignment: operator used % =
7. Divide (floor) and Assignment: operator used //=
8. Bitwise AND and Assignment: operator used &=
9. Bitwise XOR and Assignment: operator used !=
10. Bitwise Right Shift and Assignment: operator used ^=
11. Bitwise right shift Assignment: operator used >>=
12. Bitwise left shift Assignment: operator used <<=
Let’s understand each of the above operators in detail
## Assignment operator:
The utility of this operator is to simply assign the value present on the right side of the operator to the operand present on the left side.
Syntax:
`z = x + y`
Example: Showing the implementation of add, and assignment operator
Output:
The usefulness of this operator is to assign the result to the left operand after calculating the sum of values present on left and right side of the operator.
Syntax:
`x + = y`
Example 1: Showing the implementation of add and assignment operator
Output:
Example 2: Showing the implementation of add and assignment operator
Output:
## Subtract and Assignment:
The usefulness of this operator is to assign the result to the left operand after subtracting the values present on the right side.
Syntax:
`x -= y`
Example: Showing the implementation of subtract and assignment operator
Output:
## Multiply and Assignment:
The usefulness of this operator is to assign the result to the left side of the operator after obtaining the result on multiplying values present on the left and right sides of the operator.
Syntax:
`x *= y`
Example: Showing the implementation of multiply and assignment operator
Output:
## Divide and Assignment:
The utility of this operator is to assign the result in the variable present on the left side of operator. Here, the result is obtained after dividing the operand present on left side with the right operand.
Syntax:
`x /= y`
Example: Showing the implementation of divide and assignment operator
Output:
## Modulus and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after finding out the modulus by dividing the left operand with the right one.
Syntax:
`x %= y`
Example: Showing the implementation of modulus and assignment operator
Output:
## Divide (floor) and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after finding out the floor value by dividing the left operand with the right one.
Syntax:
`x //= y`
Example: Showing the implementation of divide(floor) and assignment operator
Output:
## Exponent and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the exponent by raising the power to the left operand.
Syntax:
`x **= y`
Example: Showing the implementation of exponent and assign operator
Output:
## Bitwise AND and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the Bitwise AND.
Syntax:
`x &= y`
Example: Showing the implementation of Bitwise add and assignment operator
Output:
## Bitwise OR and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the Bitwise OR.
Syntax:
`x |= y`
Example: Showing the implementation of bitwise OR and assignment operator
Output:
## Bitwise XOR and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the Bitwise XOR.
Syntax:
`x ^= y`
Example: Showing the implementation of bitwise XOR and assign operator
Output:
## Bitwise Right Shift and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the Bitwise Right Shift.
Syntax:
`x >>= y`
Example: Showing the implementation of bitwise right shift and assign operator
Output:
## Bitwise Left Shift and Assignment:
The utility of this operator is to assign the result to the operand present on the left side of the operator after calculating the Bitwise Left shift.
Syntax:
`x <<= y`
Example: Showing the implementation of bitwise left shift and assign operator
Output: | 988 | 4,937 | {"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-2023-14 | longest | en | 0.852406 |
https://web2.0calc.com/questions/graphing-question_45 | 1,657,214,369,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00506.warc.gz | 655,957,695 | 5,450 | +0
# Graphing question
0
125
1
The graph of y = f(x) has the line x = 5 as an axis of symmetry. The graph also passes through the point (7,-8). Find another point that must lie on the graph of y = f(x).
Nov 25, 2021
#1
+13732
+1
Find another point that must lie on the graph of y = f(x).
Hello Guest!
The other point is (3, -8).
!
Nov 25, 2021 | 117 | 353 | {"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} | 3.078125 | 3 | CC-MAIN-2022-27 | latest | en | 0.942916 |
http://nrich.maths.org/public/leg.php?code=27&cl=2&cldcmpid=4783 | 1,506,261,590,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690029.51/warc/CC-MAIN-20170924134120-20170924154120-00136.warc.gz | 248,482,483 | 6,504 | Search by Topic
Resources tagged with Ratio similar to Once Upon a Time:
Filter by: Content type:
Stage:
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There are 19 results
Broad Topics > Fractions, Decimals, Percentages, Ratio and Proportion > Ratio
Stage: 3 Challenge Level:
Can you work out which drink has the stronger flavour?
Do Unto Caesar
Stage: 3 Challenge Level:
At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the. . . .
Number the Sides
Stage: 2 Challenge Level:
The triangles in these sets are similar - can you work out the lengths of the sides which have question marks?
Nutty Mixture
Stage: 2 Challenge Level:
Use the ratio of cashew nuts to peanuts to find out how many peanuts Rachel has. What would the ratio be if Rachel and Marianne mixed their bags?
Mixing Paints
Stage: 3 Challenge Level:
A decorator can buy pink paint from two manufacturers. What is the least number he would need of each type in order to produce different shades of pink.
Cereal Mix
Stage: 3 Challenge Level:
A farmer is supplying a mix of seeds, nuts and dried apricots to a manufacturer of crunchy cereal bars. What combination of ingredients costing £5 per kg could he supply?
Mixing More Paints
Stage: 3 Challenge Level:
Is it always possible to combine two paints made up in the ratios 1:x and 1:y and turn them into paint made up in the ratio a:b ? Can you find an efficent way of doing this?
Ratio Pairs 2
Stage: 2 Challenge Level:
A card pairing game involving knowledge of simple ratio.
Ratio or Proportion?
Stage: 2 and 3
An article for teachers which discusses the differences between ratio and proportion, and invites readers to contribute their own thoughts.
Ratio Sudoku 3
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
Racing Odds
Stage: 3 Challenge Level:
In a race the odds are: 2 to 1 against the rhinoceros winning and 3 to 2 against the hippopotamus winning. What are the odds against the elephant winning if the race is fair?
Oh for the Mathematics of Yesteryear
Stage: 3 Challenge Level:
A garrison of 600 men has just enough bread ... but, with the news that the enemy was planning an attack... How many ounces of bread a day must each man in the garrison be allowed, to hold out 45. . . .
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?
Ratio Sudoku 1
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
How Big?
Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
Rati-o
Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
Ratio Sudoku 2
Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
Pi, a Very Special Number
Stage: 2 and 3
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
Ratio Pairs 3
Stage: 3 and 4 Challenge Level:
Match pairs of cards so that they have equivalent ratios. | 829 | 3,344 | {"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} | 3.734375 | 4 | CC-MAIN-2017-39 | latest | en | 0.932588 |
http://gmat.kaptest.com/tag/verbal/ | 1,408,594,786,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500814701.13/warc/CC-MAIN-20140820021334-00302-ip-10-180-136-8.ec2.internal.warc.gz | 90,882,301 | 19,087 | # The 4 Sections of the GMAT Practice Test
August 12, 2014 by
If you’re thinking of going to business school but haven’t yet looked into GMAT test dates or taken a GMAT practice test, not to fear! You’ve come to the right place to learn about the format, timing, and scoring of the GMAT test and GMAT practice test, the latter of which we offer free on our website year-round.
#### The CAT (computer adaptive test) and GMAT scoring
Understanding how the CAT works and approaching GMAT practice questions with a few targeted strategies will help you get in fighting form for your GMAT test date.
A CAT is more than just a digital version of a written exam. It actually adapts to your performance as you’re taking the test. When you begin each category, the computer assumes you have an average score and gives you a question of medium difficulty. As you answer … Read full post
April 21, 2014 by
We gave you a challenging series of GMAT Reading Comprehension tasks to work on in our last blog entry. Now that you’ve tried them out on your own (if you haven’t yet, pause now and take 10-20 minutes to do so), we’re going to walk through a full analysis of the passage and the practice questions. Here’s the passage once more for your reference.
#### GMAT Reading Comprehension Practice: The Passage
“Strange Bedfellows!” lamented the title of a recent letter to Museum News, in which a certain Harriet Sherman excoriated the National Gallery of Art in Washington for its handling of tickets to the much-ballyhooed “Van Gogh’s van Goghs” exhibit. A huge proportion of the 200,000 free tickets were snatched up by homeless opportunists in the dead of winter, who then scalped those tickets at \$85 apiece to less hardy connoisseurs.
Yet, Sherman’s bedfellows are far from strange. Art, despite … Read full post
# High Level GMAT Reading Comprehension Practice
April 18, 2014 by
We’ve recently discussed GMAT Reading Comprehension question types and patterns in detail, so it’s time to try your hand at identifying question types and answering some difficult practice questions.
#### GMAT Reading Comprehension Practice: The Passage
“Strange Bedfellows!” lamented the title of a recent letter to Museum News, in which a certain Harriet Sherman excoriated the National Gallery of Art in Washington for its handling of tickets to the much-ballyhooed “Van Gogh’s van Goghs” exhibit. A huge proportion of the 200,000 free tickets were snatched up by homeless opportunists in the dead of winter, who then scalped those tickets at \$85 apiece to less hardy connoiseurs.
Yet, Sherman’s bedfellows are far from strange. Art, despite its religious and magical origins, very soon became a commercial venture. From bourgeois patrons funding art they barely understood in order to share their protegee’s prestige, to museum curators stage-managing the cult of artists in order … Read full post
# How to Beat GMAT Reading Comprehension – Part II
April 14, 2014 by
As I mentioned in my previous exploration of GMAT Reading Comprehension, most of the questions focus on the author’s purpose, so if you can at least identify her main idea as you slog through the passage, the author will guide you to the right answer in three out of four of those main question types:
• The correct answer to a Global question is essentially a statement of the author’s purpose
• The right answer choice to Inference questions, even though these can seem completely speculative, will never contradict the author’s purpose, and is often directly informed by that purpose.
• Logic questions look for an answer that addresses why our author has included a detail or a paragraph in her passage; the why is that that detail or paragraph always serves her purpose.
Now comes the fun part: beating the test designers at their … Read full post
# How to Beat GMAT Reading Comprehension
April 11, 2014 by
There are some fortunate beings among us who seem to thrive on GMAT Reading Comprehension problems. The rest of look on with veiled mirthless smiles at these blessed souls, all the while muttering under our collective breaths, “what, are ya nuts!” If you’re like most of us mere mortals, Reading Comp is a complete pain: long, nearly incomprehensible passages on subjects about which we know little or nothing, and care even less, followed by inscrutable questions that seem to have been devised by the Sphinx herself. What’s to be done to tackle this part of the Verbal section?
How To Beat GMAT Reading Comprehension
Well, if you’ve looked into any part of the GMAT with the least little bit of attention, you will have noticed that this beastly test is filled with recurring patterns, and, though widely varying Reading Comp passages hardly seem likely to harbor repeated patterns, the questions Read full post
# GMAT Sentence Correction Practice
March 21, 2014 by
Yesterday, we posted a slightly disturbing GMAT Sentence Correction practice question on Facebook. It was mostly disturbing due to the content (creeeepy!), but there was also a lot going on in this particular sentence and the answer choices. Check it out.
As is often the case in GMAT Sentence Correction, this question tests several issues, which you can identify by verbally scanning the answer choices:
1. “If Professor…is right” versus “Should Professor…(be) right”
2. Verb tense: “is” versus “had been”
3. “connection of X and Y” versus “connection between X and Y”
Let’s address these issues one at a time.
• “If Professor…is right” versus “Should Professor…(be) right”
The GMAT prefers “If Professor…is right” to “Should Professor…(be) right”. The former is simpler and uses the simple present tense to match the later verb “is (not merely coincidental)”. These two verbs must match tense since they describe events that happen simultaneously. Thus you can eliminate Read full post
November 20, 2013 by
GMAT Reading Comprehension is an important question type to master. If you haven’t yet tackled this week’s passage, take a look at it now and take a few minutes to answer the associated questions that we posted. They are the questions you should work through each time you break down a reading comp passage.
Now, let’s talk analysis.
• The TOPIC of this passage, or the broad main idea, is Knowles’s theory of and assumptions about andragogy. GMAT passages don’t contain a lot of filler, so you usually see the gist of the topic emerge in the first paragraph, if not in the first sentence.
• The SCOPE of the passage is a more detailed focus within the topic. In this passage, the scope is the characteristics of adult learners. The bulk of the passage lists and describes the characteristics of adult learners that inform the theory of andragogy.
# GMAT Sentence Correction in the Real World: Futbol Errors
November 19, 2013 by
By Justin Doff
Once you put in the hard work to know and detect GMAT sentence correction errors, you’re going to catch those mistakes all around you.
A recent example: one of my British friends and I were watching futbol (I’ll go with the Spanish spelling so as to not confuse sports), and I picked up on a weird (but commonly accepted) verb agreement error in futbol journalism—using a plural verb/pronoun to refer to a single team.
If you are referring to a singular entity, regardless of what it contains, is the subject singular or plural? Perhaps a silly question, and here are some perhaps really obvious examples:
The coach is…
The league is…
This team is…
BUT, then, take a look at these actual news stories from the European press:
“Real Madrid have a golden opportunity to open up a huge lead in the Champions League group stage…”
# Free GMAT Sample Class
November 1, 2013 by
If you need help with your GMAT prep, we’ve got you covered. Now is your chance to see one of our GMAT experts in action, breaking down the GMAT exam and introducing you to proven Kaplan methods and strategies that will allow you to dominate your competition on GMAT Test Day.
If you attended any of the free GMAT practice tests that we held recently, then you’ve already met some of our GMAT experts. What better way to continue your GMAT prep than to attend a free, live, online session that’s focused on introducing you to the most efficient approach to every question type that you’ll see on the GMAT?
Here’s an example of the type of questions you’ll see in our upcoming GMAT Sample Class:
The youngest of 4 children has siblings who are 3, 5, and 8 years older than she is. If the average (arithmetic mean) age Read full post
# Kaplan GMAT Sample Problem: Critical Reasoning Inferences
August 22, 2011 by
As you try the practice GMAT problem below, remember that on Critical Reasoning inference questions you should accept all of the information in the stimulus as true. When you read the answer choices look for an option that must be true based on the stimulus.
Problem:
Randall: Many of the productions of my plays by amateur theater groups are poorly done, and such interpretations do not provide a true measure of my skills as a dramatist.
Which one of the following can be properly inferred from Randall’s statement?
(A) Some amateur theater groups’ productions of Randall’s plays provide a true measure of his skills as a dramatist.
(B) All amateur theater group productions of Randall’s plays that are not poorly done provide a true measure of his skills as a dramatist.
(C) All of the productions of Randall’s plays by amateur theater groups that do not provide a true … Read full post
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" " | 2,133 | 9,533 | {"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-2014-35 | longest | en | 0.929657 |
https://www.acwing.com/user/myspace/index/16448/ | 1,611,780,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00330.warc.gz | 661,913,106 | 13,137 | zdw
1.4万
zdw
5天前
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e4 + 7;
int dp[2][maxn], q[maxn]; // dp 为滚动数组， q为优先队列辅助数组
#define calc(t,i) (dp[t^1][q[i]]+((k-q[i])/v)*w)
int main() {
int n, m;
cin >> n >> m;
int t = 0;
for (int i = 1; i <= n; i++) {
int v, w, s;
scanf("%d %d %d", &v, &w, &s);
t ^= 1; //数组滚动
for (int j = 0; j < v; j++) {
int l = 1, r = 0;
for (int k = j ; k <= m; k += v) {
while (l <= r && (k-q[l])/v > s) l++; // 排除太远的
if(l<=r) dp[t][k] = max(dp[t^1][k], calc(t,l));
while (l <= r && dp[t ^ 1][q[r]] - (q[r] - j) / v * w <= dp[t ^ 1][k] - (k - j) / v * w) r--;//消除距离影响
q[++r] = k;
}
}
}
cout << dp[t][m] << endl;
return 0;
}
zdw
27天前
#include <iostream>
#include <cstring>
using namespace std;
const int N = 1e5 + 5;
int f[N][3];
int main() {
int n;
cin >> n;
f[0][1] = -0x3f3f3f3f3f;
int x;
for (int i = 1; i <= n; i++) {
cin >> x;
f[i][0] = max(f[i - 1][0], f[i - 1][2]);
f[i][1] = max(f[i - 1][1], f[i - 1][0] - x);
f[i][2] = f[i - 1][1] + x;
}
cout << max(f[n][0], f[n][2]) << endl;
return 0;
}
zdw
27天前
#include <iostream>
#include <cstring>
using namespace std;
const int N = 1e5 + 5;
int f[N][107][2]; // 可以像背包一样省略一维
int main() {
int n, m;
cin >> n >> m;
memset(f, -0x3f, sizeof f);
for (int i = 0; i <= n; i++) {
f[i][0][0] = 0;
}
int x;
for (int i = 1; i <= n; i++) {
cin >> x;
for (int j = 1; j <= m; j++) {
f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j][1] + x);
f[i][j][1] = max(f[i - 1][j][1], f[i - 1][j - 1][0] - x);
}
}
int res = 0;
for (int i = 1; i <= m; i++) res = max(res, f[n][i][0]);
cout << res << endl;
return 0;
}
zdw
1个月前
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int arr[N], f[N];
// 普通解法 f[i] = max(f[i], f[i - 2] + arr[i])
int main() {
int t, n;
cin >> t;
while (t--) {
cin >> n;
memset(f, 0, sizeof f);
for (int i = 2; i <= n + 1; i++) {
cin >> arr[i];
f[i] = max(f[i - 1], f[i - 2] + arr[i]);
}
cout << f[n + 1] << endl;
}
return 0;
}
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int arr[N], f[N][2];
// 状态机解法 0表示不选，1表示选。
// 有0->0，0->1, 1-> 0
// f[i,0] = max(f[i-1, 0], f[i-1, 1])
// f[i, 1] = f[i - 1, 0] + arr[i]
int main() {
int t, n;
cin >> t;
while (t--) {
cin >> n;
memset(f, 0, sizeof f);
for (int i = 2; i <= n + 1; i++) {
cin >> arr[i];
f[i][0] = max(f[i - 1][0], f[i - 1][1]);
f[i][1] = f[i - 1][0] + arr[i];
}
cout << max(f[n + 1][0], f[n + 1][1]) << endl;
}
return 0;
}
zdw
2个月前
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 2.5e4 + 5;
int arr[N];
bool st[N];
int main(){
int t, n;
cin >> t;
while (t --){
memset (st, false, sizeof st);
cin >> n;
for (int i = 0; i < n; i ++) {
cin >> arr[i];
}
sort(arr, arr + n);
int res = 0;
st[0] = true;
for (int i = 0; i < n; i ++){
if (!st[arr[i]]) res ++;
for (int j = arr[i]; j <= arr[n - 1]; j ++){
st[j] |= st[j - arr[i]];
}
}
cout << res << endl;
}
return 0;
}
zdw
3个月前
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 11;
int f[N][10];
void init()
{
for (int i = 0; i <= 9; i ++ ) f[1][i] = 1;
for (int i = 2; i < N; i ++ )
for (int j = 0; j <= 9; j ++ )
for (int k = 0; k <= 9; k ++ )
if (abs(j - k) >= 2)
f[i][j] += f[i - 1][k];
}
int dp(int n)
{
if (!n) return 0;
vector<int> nums;
while (n) nums.push_back(n % 10), n /= 10;
int res = 0;
int last = -2;
for (int i = nums.size() - 1; i >= 0; i -- )
{
int x = nums[i];
for (int j = i == nums.size() - 1; j < x; j ++ )
if (abs(j - last) >= 2)
res += f[i + 1][j];
if (abs(x - last) >= 2) last = x;
else break;
if (!i) res ++ ;
}
// 特殊处理有前导零的数
for (int i = 1; i < nums.size(); i ++ )
for (int j = 1; j <= 9; j ++ )
res += f[i][j];
return res;
}
int main() {
init();
int l, r;
cin >> l >> r;
cout << dp(r) - dp(l - 1) << endl;
return 0;
}
zdw
3个月前
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int N = 12, M = 105;
int f[N][N][M];
int a, b, p;
int mod(int x, int y) {
return (x % y + y) % y;
}
void init(){
memset(f, 0, sizeof f);
for (int i = 0; i <= 9; i ++) f[1][i][i % p] ++;
for (int i = 2; i < N; i ++){
for (int j = 0; j <= 9; j ++) {
for (int k = 0; k < p; k ++){
for (int x = 0; x <= 9; x ++) {
f[i][j][k] += f[i - 1][x][mod(k - j, p)];
}
}
}
}
}
int dp(int x){
vector<int> nums;
while (x) nums.push_back(x % 10), x /= 10;
int res = 0, last = 0; // last 前面位数的和
for (int i = nums.size() - 1; i >= 0; i --) {
int x = nums[i];
for (int j = 0; j < x; j ++) {
res += f[i + 1][j][mod(-last, p)];
}
last += x;
if (!i && last % p == 0) res ++;
}
return res;
}
int main(){
while (cin >> a >> b >> p)init(),cout << dp(b) - dp(a) << endl;
return 0;
}
zdw
3个月前
1. 如果k 小于 $2^{n-1}$, 说明是第一个方式构成，直接+0
2. 否则就去找k-$2^{n-1}$大，因为倒序所以是找$2^{n-1}$ - (k - $2^{n-1}$) - 1 ==> $2^n$ - k - 1
#include <iostream>
using namespace std;
typedef unsigned long long ULL;
string f(ULL n, ULL k){
if (!n) return "";
if ( (1ull << n - 1) > k) return "0" + f(n - 1, k);
unsigned long long t = (1ull << n) - k - 1;
if (n == 64) t = -k -1;// 移出去了，可能会出现异常是0或1不确定，特判一下
return "1" + f(n - 1, t);
}
int main(){
ULL n, k;
cin >> n >> k;
cout << f(n, k) << endl;
return 0;
}
#include <iostream>
using namespace std;
int main(){
unsigned long long n, k;
cin >> n >> k;
unsigned long long ge = k ^ (k >> 1);
while (~--n) {
cout << (ge >> n & 1 );
}
return 0;
}
zdw
3个月前
f[i][j] = f[i - 1][j] + f[i][j - 1]， 当i，j不同时为奇数时;
#include <iostream>
using namespace std;
int f[35][35];
int main(){
int n, m;
cin >> n >> m;
f[0][1] = 1;
for (int i = 1; i <= n; i ++){
for (int j = 1; j <= m; j++){
if (j & 1 | i & 1)
f[i][j] = f[i - 1][j] + f[i][j - 1];
}
}
cout << f[n][m] << endl;
return 0;
}
zdw
3个月前
#include <iostream>
using namespace std;
const int N = 1005;
int phi[N], primes[N], tot;
bool st[N];
void get_phi(int n) {
phi[1] = 1;
for (int i = 2; i <= n; i++) {
if (!st[i]) {
primes[tot++] = i;
phi[i] = i - 1;
}
for (int j = 0; primes[j] * i <= n; j++) {
st[primes[j] * i] = true;
if (i % primes[j] == 0) {
phi[i * primes[j]] = phi[i] * primes[j];
break;
}
phi[i * primes[j]] = phi[i] * (primes[j] - 1);
}
}
}
long long phi_sum[N];
void get_phi_sum(int n) {
for (int i = 1; i <= n; i++) phi_sum[i] += phi_sum[i - 1] + phi[i];
}
int main() {
int t, n, cnt = 1;
scanf("%d", &t);
get_phi(1000), get_phi_sum(1000);
while (t--)
{
scanf("%d", &n);
printf("%d %d %lld\n", cnt++, n, phi_sum[n] * 2 + 1);
}
return 0;
} | 2,836 | 6,452 | {"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": 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} | 2.84375 | 3 | CC-MAIN-2021-04 | latest | en | 0.136119 |
https://www.halfbakery.com/idea/Tidal_20Tank | 1,553,494,631,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00243.warc.gz | 772,655,950 | 5,391 | h a l f b a k e r y
"Bun is such a sad word, is it not?" -- Watt, "Waiting for Godot"
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# Tidal Tank
Generate power via cyclical tidal water flow into a tank.
(0) [vote for, against]
Picture a large tank buried in the earth behind a marine bulkhead. The tidal exchange at the bulkhead is zero to ten feet (low to high tide). When the tide comes in, a inlet tube at the bottom of the bulkhead, connected to the tank allows the water to flow in. Opposite happens when tide goes out. Two options for power generation. One, a water turbine in the inlet/outlet pipe spins to create electricity. Two, an air outlet pipe at the top of the tank spins as the volume of the tank is displaced by the water flowing in and out. If the tank is, say, 10,000 gallons and the tidal exchange happens roughly 2x per day you could regulate the flow so that water is always moving through the system. Question is, how much power can be generated by that amount of water or air movement?
— frossi, Mar 07 2007
Tidal power http://en.wikipedia.org/wiki/Tidal_power
Your description sounds a bit like ebb generation [spiraliii, Mar 08 2007]
Widely known to exist. See link.
— nuclear hobo, Mar 07 2007
/Question is, how much power can be generated by that amount of water or air movement?/
Far less than would offset the construction, maintenance, and operating costs. On the back of an envelope, your average head is 1.5m and your mass flowrate is about 1000kg/s. Gravity being what it is on earth, you've got about 15kW available. Assuming 100% conversion efficiency, that's a revenue stream of \$1.50 per hour.
— Texticle, Mar 07 2007
This tends to be dun on rivers then there more energy but the bay of fundy with 20 to 30 foot tides would work too then again other people have tried to get the government to let them harness tat energy
— dev45, Mar 07 2007
//What link??// - washed out by the incoming tide.....
— xenzag, Mar 07 2007
[annotate]
back: main index | 545 | 2,124 | {"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-2019-13 | longest | en | 0.907322 |
https://www.budgetworksheets.org/cost-to-drive/825 | 1,696,006,608,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00028.warc.gz | 718,470,992 | 3,767 | ### Cost To Drive 825 Miles Calculator
How much does it cost to drive 825 miles?
Miles
MPG
\$
Cost to Drive Results Driving Costs:
##### How much is the cost of gas to drive 825 miles?
This calculator doesn't take car maintenance or insurance into account. To calculate, use the formula below
Miles Driven / Miles Per Gallon x Cost Per Gallon
The chart below shows much it costs to drive 825 miles based on MPG and Cost Per Gallon
Cost Per Gallon 10 MPG 20 MPG 30 MPG 40 MPG 50 MPG \$1.00 \$82.50 \$41.25 \$27.50 \$20.63 \$16.50 \$1.25 \$103.13 \$51.56 \$34.38 \$25.78 \$20.63 \$1.50 \$123.75 \$61.88 \$41.25 \$30.94 \$24.75 \$1.75 \$144.38 \$72.19 \$48.13 \$36.09 \$28.88 \$2.00 \$165.00 \$82.50 \$55.00 \$41.25 \$33.00 \$2.25 \$185.63 \$92.81 \$61.88 \$46.41 \$37.13 \$2.50 \$206.25 \$103.13 \$68.75 \$51.56 \$41.25 \$2.75 \$226.88 \$113.44 \$75.63 \$56.72 \$45.38 \$3.00 \$247.50 \$123.75 \$82.50 \$61.88 \$49.50 \$3.25 \$268.13 \$134.06 \$89.38 \$67.03 \$53.63 \$3.50 \$288.75 \$144.38 \$96.25 \$72.19 \$57.75 \$3.75 \$309.38 \$154.69 \$103.13 \$77.34 \$61.88 \$4.00 \$330.00 \$165.00 \$110.00 \$82.50 \$66.00 \$4.25 \$350.63 \$175.31 \$116.88 \$87.66 \$70.13 \$4.50 \$371.25 \$185.63 \$123.75 \$92.81 \$74.25 \$4.75 \$391.88 \$195.94 \$130.63 \$97.97 \$78.38 \$5.00 \$412.50 \$206.25 \$137.50 \$103.13 \$82.50 \$5.25 \$433.13 \$216.56 \$144.38 \$108.28 \$86.63 \$5.50 \$453.75 \$226.88 \$151.25 \$113.44 \$90.75 \$5.75 \$474.38 \$237.19 \$158.13 \$118.59 \$94.88 \$6.00 \$495.00 \$247.50 \$165.00 \$123.75 \$99.00
##### More Mileage
826 miles 827 miles 828 miles 829 miles 830 miles 831 miles 832 miles 833 miles 834 miles 835 miles
836 miles 837 miles 838 miles 839 miles 840 miles 841 miles 842 miles 843 miles 844 miles 845 miles
846 miles 847 miles 848 miles 849 miles 850 miles 851 miles 852 miles 853 miles 854 miles 855 miles
856 miles 857 miles 858 miles 859 miles 860 miles 861 miles 862 miles 863 miles 864 miles 865 miles | 757 | 1,945 | {"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-2023-40 | latest | en | 0.558226 |
http://archiver.rootsweb.ancestry.com/th/read/TMG/2003-05/1052804829 | 1,386,840,291,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164574548/warc/CC-MAIN-20131204134254-00075-ip-10-33-133-15.ec2.internal.warc.gz | 9,202,959 | 3,012 | TMG-L Archives
Archiver > TMG > 2003-05 > 1052804829
From: Richard Brogger <>
Subject: Re: [TMG] Problem with early dates
Date: Tue, 13 May 2003 00:47:09 -0500
References: <1c3.963736d.2bee9243@aol.com><001d01c3172b\$5728f380\$6401a8c0@charliexii><1052755455.8812.5.camel@localhost.localdomain><3EC01F36.B8873564@infoave.net><1052785751.8813.12.camel@localhost.localdomain>
Gordon Banks wrote:
>
> I wonder what it would take to make the BC dates regular. Why not just
> accept a minus sign in front of the date? Whatever arithmetic functions
> are used to do the calculations in TMG should still work with negative
> integers. You then are left only with the problem that there is no year
> 0.
Hi Gordon,
Negative integers would be better than what we have. I can't imagine
it would take much to deal with the fact that there is no year zero.
(+5)-(-5) = 10 but since a year is missing, subtract one and give 9 as
One thing seems to be constant regardless of what calendar is used. A
day is one revolution of the planet. Different cultures use different
events to mark the start of a day but it is still one revolution. What
constitutes a week, month or year has varied with time and place. Even
narrowing the discussion to when the Gregorian calendar was adopted is
pretty wild. Since the only thing that seems to be a constant is the
day, it seems natural to number the days. Time can be easily included
by using a decimal fraction and it is much easier and less prone to
errors to subtact one number, like 2,452,211.144, from another like
2,452,379.288. Another advantage to Julian days is that they are not
influenced by time zones.
Richard Brogger | 460 | 1,660 | {"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-2013-48 | latest | en | 0.928906 |
http://www.topperlearning.com/forums/ask-experts-19/i-could-not-get-how-to-solve-this-question-in-the-given-physics-units-and-measurement-55493/reply | 1,493,082,025,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120001.0/warc/CC-MAIN-20170423031200-00056-ip-10-145-167-34.ec2.internal.warc.gz | 703,292,657 | 37,056 | Question
Sun May 13, 2012 By:
# I could not get how to solve this question: In the given equation y=Asin(wt-kx), obtain the dimensional formula of w and k; x, y, A-distances, t - time taken
Sun May 13, 2012
A is the distance
t is the time
the term inside the brackets has to be dimensionless for the equation to be correct(because on both sides we have units of length, y and A).So dimension of wt=dimns. of kx=M^0L^0T^0
on solving we get, w=T^(-1) and k=L^(-1)
OR we know w is angular velocity which is angular displacement by time(angular displac. is dimensionless and k is
(2pi)/lambda (2pi a dimensionless constant)
Related Questions
Mon September 12, 2016
# What is a physical standard? What characteristics should it have?
Fri July 22, 2016 | 219 | 756 | {"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} | 2.9375 | 3 | CC-MAIN-2017-17 | latest | en | 0.917086 |
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IDProjectCategoryView StatusDate SubmittedLast Update
0005688OCamltypingpublic2012-07-16 14:382017-02-24 15:27
Reporterlpw25
Assigned Togasche
PrioritynormalSeverityfeatureReproducibilityN/A
StatusresolvedResolutionsuspended
PlatformOSOS Version
Product Version
Target VersionFixed in Version
DescriptionI have read the discussion of adding variance annotations to GADTs on caml-list, and thought that a feature request would be a sensible place to discuss a partial solution.
I have not constructed a proof of soundness, but every example of how variance annotations can be unsound is caused by the same general problem: while local constraints from a generalised constructor are in place, two types within the type of that constructor are unified, however one of these types is not invariant so the unification is not always valid.
For example:
type (_, +_) eq = Refl : ('a, 'a) eq
let bad_proof (type t) = (Refl : (<m:t>, <m:t>) eq :> (<m:t>, < >) eq)
let downcast_1 : type a. (a, < >) eq -> < > -> a =
fun (type a) (Refl : (a, < >) eq) (s : < >) -> (s :> a)
let downcast_2 : < > -> <m:o> = downcast_1 bad_proof
The problem here is caused by the fact that the second parameter of eq can be unified with the first parameter under the local constraints of Refl, so it is not safe to change the second parameter to a supertype without also changing the first.
Similarly:
type -'a t = C : < m : int > -> < m : int > t
let eval : type a . a t -> a = fun (C x) -> x
let a = C (object method m = 5 end)
let b = (a :> < m : int ; n : bool > t)
let c = eval b
The problem here is that the type parameter of t can be unified with < m: int > under the local constraints of C, so it is not safe to change it to a subtype.
This suggests a simple safety condition that would allow many of the simple uses of GADTs would be to allow a type parameter with a variance annotation to be instantiated in a GADT constructor provided that none of the other types within the constructor's type can be unified with the instantiated type parameter.
For instance
type +_ foo = Foo : <m: int> foo
would be allowed because <m: int> foo could never be unified with int or <m: int>.
Similarly
type +_ foo = Foo : bool -> <m:int> foo
would be allowed because bool could never be unified with int or <m:int> and <m:int> foo could never be unified with int or <m:int>.
On the other hand
type (_, +_) eq = Refl : ('a, 'a) eq
would not be allowed because the first 'a could be unified with the second 'a, and
type -'a t = C : < m : int > -> < m : int > t
would not be allowed because the first < m: int > could be unified with the second <m : int>.
I think that this scheme is safe and allows many basic patterns to be given a variance. There are probably more accurate restrictions than this, but it seems like a sensible starting point. The implementation could use CType.mcomp to check whether two types could ever be unified. Since it depends on showing that two types could never be unified, it would be a bit fragile and require precise error messages in order to be usable.
If it is felt that this scheme is too heavyweight or too fragile, it would at least be useful to allow variance annotations for special cases that are obviously safe e.g. types with a single type parameter that only appears in constant constructors.
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related to 0007494 assigned garrigue Infer variance for GADTs without equality | 913 | 3,607 | {"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-2017-43 | latest | en | 0.777442 |
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## Answered question
2021-02-12
For the following situations, identify the test you would run to analyze the data:
A marketing firm producing costumes is interested in studying consumer behavior in the context of purchase decision of costumes in a specific market. This company is a major player in the costume market that is characterized by intense competition. The company would like to know in particular whether the income level of the consumers (measured as lower, middle, upper middle, and upper class) influences their choice of costume type. They are specifically focused on four types of costumes (funny costumes, scary costumes, clever costumes, and boring costumes).
a. Chi-Square Goodness of Fit
b. Frequencies
c. Descriptive Statistics
d. Chi-Square of Independence
### Answer & Explanation
berggansS
Skilled2021-02-13Added 91 answers
Step 1
Chi-square test of independence:
A single simple random sampling is done with each individual being classified based on the two categorical variables. The relation among categorical variable is determined using Chi-square test of independence. The test is helps to analyze the variables that are independent or associated. This test is called as chi-square test of independence.
Step 2
Condition of chi-square independence:
-The samples are selected at random.
-The variables dependent and independent under study are each categorical
-Expected cell counts must be at least 5.
Explanation:
The researchers test whether the income level of the consumers (middle, upper middle and upper class) and types of costumes (funny costumes, scary costumes, clever costumes and boring costumes). For the analysis of categorical data with more than one variable, here, contingency tables are used. The chi square test of independence is used to test the association between variables in a contingency table.
Thus, the correct answer is chi-square of independence.
Thus, the correct answer is option d.
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Tue Aug 29, 2017 2:09 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Do you have the feeling, that this can be solved after all? I mean a sound, always applicable technique. I can only think of some heuristics to solve this problem, because for us humans it is most of the time easy to see which edges, vertices and faces correspond if we have a state before and after...
Tue Aug 29, 2017 12:51 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Work Product Before I put up the work product, I must begin the work product with a disclosure. Due to my movement to the US in the middle of the work period and the ensuing visa restrictions, I had to decline a part of the stipend (the third installment). This part was supposed to be corresponding...
Thu Jul 27, 2017 4:03 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
I chose Box, Boolean and Fillet because they had a way to define all faces (Box has Top, Bottom, Left, Right, Front, and Back That is an interesting approach... Actually a big number of parametric objects in FreeCAD could do that: define faces when they create their shape, instead of letting OCC do...
Thu Jul 27, 2017 3:38 am
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Update 27th July 2017 With a PR complete, my efforts since the last update had been towards developing a simple implementation that gives sensible toponaming for a small set of Part::Feature s that would be useful. I chose Box , Boolean and Fillet because they had a way to define all faces ( Box ha...
Thu Jul 13, 2017 6:42 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Thanks for the PR, jnxd. I hope it get merged soon. What else have you been working on ? Hi, @Kunda1. Glad to see your interest in the development. Unfortunately, most of my time went in development towards the PR. It doesn't really look like much, but the commits are a result of squashing around 3...
Sun Jul 09, 2017 6:54 pm
Forum: Pull Requests
Topic: PR #868: History storage framework and options for select methods in TopoShapePy
Replies: 1
Views: 1116
### PR #868: History storage framework and options for select methods in TopoShapePy
Link to PR Supported methods (all booleans, extrude , mirror , makeFillet2 , makeChamfer2 ) now come with an optional withHistory parameter that can be set to True if you want to store history. The development of the sub-shapes from the sub-shapes of the base(s) can be studied by using shapeName.Hi...
Sun Jul 09, 2017 6:52 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Update 9th July 2017 Just made the PR as described in my previous post. Supported methods (all booleans, extrude , mirror , makeFillet2 , makeChamfer2 ) now come with an optional withHistory parameter that can be set to True if you want to store history. The development of the sub-shapes from the s...
Mon Jul 03, 2017 5:45 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
ping Any updates? Oh! So sorry for not posting for so long. Update: 3rd July 2017 We're working on making a pull request that stores the history algorithm for others to experiment with. When complete, methods in Part.Shape shall have an optional parameter withHistory that will lead to the shape bei...
Wed Jun 21, 2017 6:11 pm
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
Update 21st June 2017 I created a tnaming branch in my own fork of FreeCAD for this project's purposes (long time coming :mrgreen:). As for my previous post, I went with a "ask for forgiveness rather than permission" policy and created an overloaded function TopoShape fuse(TopoShape) const to suppo...
Fri Jun 09, 2017 5:30 am
Forum: Developers corner
Topic: GSoC 2017 Dev Log: jnxd
Replies: 42
Views: 5361
### Re: GSoC 2017 Dev Log: jnxd
... So, we were trying to expose the history framework of OCC, i.e. the methods that tell which elements of the old shape generated/were modified to which elements of the new shape, or which were deleted. However, we have hit a roadblock since w e have to make some potentially radical changes. Thus...
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