added 13 characters in body

Source Link

edited Jan 10 at 9:37

3.9k
1
6
45

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference (arc length) of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which (the crossed-out one) does not meet the error requirements. The second approximation formula (Equation 5) barely does (verification) for up to \$a=5b\$ (which is also the upper limit of the test cases, so you can use it for your answer).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which (the crossed-out one) does not meet the error requirements. The second approximation formula (Equation 5) barely does (verification) for up to \$a=5b\$ (which is also the upper limit of the test cases, so you can use it for your answer).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference (arc length) of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which (the crossed-out one) does not meet the error requirements. The second approximation formula (Equation 5) barely does (verification) for up to \$a=5b\$ (which is also the upper limit of the test cases, so you can use it for your answer).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Notice removed Reward existing answer by Bubbler

occurred Nov 9, 2020 at 0:29

Bounty Ended with Razetime's answer chosen by Bubbler

occurred Nov 9, 2020 at 0:29

Notice added Reward existing answer by Bubbler

occurred Nov 2, 2020 at 0:23

Bounty Started worth 150 reputation by Bubbler

occurred Nov 2, 2020 at 0:23

Tweeted twitter.com/StackCodeGolf/status/1313675441572642818

occurred Oct 7, 2020 at 3:00

Became Hot Network Question

occurred Sep 29, 2020 at 13:08

added 131 characters in body

Source Link

edited Sep 29, 2020 at 6:39

Bubbler

78.4k
5
149
469

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which (the crossed-out one) does not meet the error requirements. The second approximation formula (the secondEquation 5) barely does (verification) for up to \$a=5b\$, checked using this code(which is also the upper limit of the test cases, so you can use it for your answer).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which does not meet the error requirements (the second barely does for up to \$a=5b\$, checked using this code).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which (the crossed-out one) does not meet the error requirements. The second approximation formula (Equation 5) barely does (verification) for up to \$a=5b\$ (which is also the upper limit of the test cases, so you can use it for your answer).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

Source Link

asked Sep 29, 2020 at 5:03

Bubbler

78.4k
5
149
469

Circumference of an ellipse

Challenge

Unlike the circumference of a circle (which is as simple as \$2\pi r\$), the circumference of an ellipse is hard.

Given the semi-major axis \$a\$ and semi-minor axis \$b\$ of an ellipse (see the image below, from Wikipedia), calculate its circumference.

By definition, you can assume \$0 < b \le a\$ for input values. The output value must be within \$10^{-6}\$ relative error from the expected answer for the given test cases.

Standard code-golf rules apply. The shortest code in bytes wins.

Formulas

Relevant information can be found on Wikipedia and MathWorld. \$C\$ is the value of the circumference; \$e\$ and \$h\$ are helper values. The last two are Ramanujan's approximations, the first of which does not meet the error requirements (the second barely does for up to \$a=5b\$, checked using this code).

$$ \require{enclose} \\ \begin{align} e &= \sqrt{1-\frac{b^2}{a^2}} \\ C &= 4aE(e) = 4a\int^{\pi/2}_{0}{\sqrt{1-e^2 \sin^2 \theta} \;d\theta} \tag{1} \\ C &= 2 \pi a \left(1-\sum^{\infty}_{n=1}{\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{e^{2n}}{2n-1}}\right) \tag{2} \\ h &= \frac{(a-b)^2}{(a+b)^2} \\ C &= \pi (a + b) \left( 1 + \sum^{\infty}_{n=1} { \left( \frac{(2n-1)!!}{2^n n!} \right)^2 \frac{h^n}{(2n-1)^2} } \right) \tag{3} \\ C &= \pi (a + b) \sum^{\infty}_{n=0} { \binom{1/2}{n}^2 h^n } \tag{4} \\ \enclose{horizontalstrike}{C} &\enclose{horizontalstrike}{\approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)} \right)} \\ C &\approx \pi (a+b) \left( 1+ \frac{3h}{10 + \sqrt{4-3h}} \right) \tag{5} \end{align} $$

Test cases

All the values for C (circumference) are calculated using Equation 4 with 1000 terms, and presented with 10 significant figures.

a     b     C
1     1     6.283185307
1.2   1     6.925791195
1.5   1     7.932719795
2     1     9.688448220
3     1     13.36489322
5     1     21.01004454
20    10    96.88448220
123   45    556.6359936

code-golf math geometry

Stack Exchange Network

Return to Question

Challenge

Formulas

Test cases

Challenge

Formulas

Test cases

Challenge

Formulas

Test cases

Challenge

Formulas

Test cases

Challenge

Formulas

Test cases

Challenge

Formulas

Test cases

Circumference of an ellipse

Challenge

Formulas

Test cases