00 { 48-Byte Prgm }

01>LBL "P"

02 RCL- ST Y

03 X<>Y

04 RCL+ ST L

05 /

06 LASTX

07 PI

08 *

09 X<>Y

10 X^2

11 STO ST Z

12 4

13 +

14 RCL* ST Z

15 RCL* ST Z

16 -3

17 *

18 256

19 /

20 R^

21 4

22 /

23 1

24 -

25 1/X

26 -

27 *

28 .END.

Exact formula:

p = 4*a*E(1-b^2/a^2)

where

E(x) is the complete elliptic integral of the second kind

a is the semi-major axis and

b is the semi-minor axis

Example:

a = 2, b = 1

p = 4*2*E(3/4) = 9.68844822054767619842850319639...

http://www.wolframalpha.com/input/?i=4*2*E%281-1%2F4%29
Approximate formula:

p ~ pi*(a + b)*(4/(4-h) - 3/256*h^2*(4 + h))

where

h = ((a - b)/(a + b))^2

The approximate formula is a rework of the Infinite Series 2 in this reference, whose terms are a subset of the infinite series SUM(k=0,inf,(h/4)^k), which converges to 4/(4-h). The resulting expression is

4/(4-h) - 3/64*h^2 - 3/256*h^3 - 39/16384*h^4 - 15/65536*h^5 + 185/1048576*h^6 ...

This was done by hand and haven't been doublechecked. Anyway, the approximation formula above uses only terms up to h^3. The percent error ranges from 0% (circle) to about 0.1% in the worst case (two coincidental lines).

+-----+-----+---------------+---------------+

| a | b | approximate | exact |

+-----+-----+---------------+---------------+

| 1 | 0 | 4.00471251023 | 4.00000000000 |

| 2 | 1 | 9.68845167284 | 9.68844822055 |

| 3 | 2 | 15.8654396854 | 15.8654395893 |

| 4 | 3 | 22.1034921699 | 22.1034921607 |

| 5 | 4 | 28.3616678905 | 28.3616678890 |

| 5 | 5 | 31.4159265359 | 31.4159265359 |

+-----+-----+---------------+---------------+

Thanks Eduardo Duenez for his recent post below.

_{Edited to fix a typo per Ernie's observation below}

_{Edited again to correct an error in the parameter of the elliptic function per Eduardo's observatin below}

*Edited: 24 May 2013, 4:44 p.m. after one or more responses were posted*

Problem is numerical integration is rather slow to full accuracy on the HP-42S, but I'll check this later.

Quote:

BTW quoted "exact" answer for a=3,b=2 should be 15.8654395893 not 15.8654396893

Fixed, thanks!

Quote:

I don't have an HP-42S, but you may be able to improve on this by computing the line integral from 0 to pi/2 using the numeric integrator on the 42S
p = 4 * a * integral(sqrt(1-(a*a-b*b)/(a*a)*sin^2(t), t, 0, pi/2)

The drawback is a somewhat long running time, but that's an option.

Regards,

Gerson.

00 { 41-Byte Prgm }

01>LBL "E2"

02 MVAR "A"

03 MVAR "B"

04 MVAR "T"

05 4

06 RCL* "A"

07 1

08 1

09 RCL "B"

10 RCL/ "A"

11 X^2

12 -

13 RCL "T"

14 SIN

15 X^2

16 *

17 -

18 SQRT

19 *

20 .END.

LLIM = 0

ULIM = 1.5707963268 (pi/2)

ACC = 0.0000000010

+-----+-----+---------------+------+

| a | b | integral | t(s) |

+-----+-----+---------------+------+

| 1 | 0 | 3.99999999999 | 44.9 |

| 2 | 1 | 9.68844822056 | 45.6 |

| 3 | 2 | 15.8654395893 | 46.2 |

| 4 | 3 | 22.1034921608 | 45.7 |

| 5 | 4 | 28.3616678889 | 45.8 |

| 5 | 5 | 31.4159265360 | 2.9 |

+-----+-----+---------------+------+

Hi. Good stuff!

Here's another way i found a while back, using a variation of arithmetic and geometric means. The iteration is fairly rapid and small to code.

This program is in C, but i hope you can see the method:

#include <math.h>

#include <stdio.h>
double agm2(double b)

{

double s = (1 + b*b)/2;

double a = 1;

double t = 1;

for (;;)

{

double a1 = (a + b)/2;

if (a == a1) break;

double c = (a - b)/2;

b = sqrt(a*b);

a = a1;

s = s - t*c*c;

t = t * 2;

}

return s/a;

}

#define _PI 3.141592653589793238462643

double ellipse(double a, double b)

{

// perimeter of an ellipse

// a semi-major axis

// b semi-minor axis

// e = eccentricity = sqrt(1-(b/a)^2)

// p = 4*a*E(e)

// use AGM variant

return 2*_PI*a*agm2(b/a);

}

Quote:

Exact formula:

p = 4*a*E((a + b)/(a + b*(b + 1)))

where

E(x) is the complete elliptic integral of the second kind

a is the semi-major axis and

b is the semi-minor axis

The above is not the correct relation between the complete elliptic integral E(x) and the perimeter p(a,b) of the ellipse. The argument "x" (usually denoted k, "the modulus") of E(k) is dimensionless, whereas (a+b)/(a+b(b+1)) is not. The correct relation is

p(a,b) = 4*a*E(c/a)

where a is the major semiaxis and

c = sqrt(a^2-b^2)

is the focal semidistance (so c/a is the eccentricity of the ellipse). Note that c/a is dimensionless.

Eduardo

EDIT: Coincidentally enough, Hugh Steers' C program above (posted as I was writing this, a few minutes earlier) documents the correct relation as well.

*Edited: 24 May 2013, 11:25 a.m. *

The algorithm used in Hugh Steers' C program is

Semjon Adlaj's, upon which my earlier post for the WP34s is based.

(As an academic I feel the urge to always credit original ideas to their source.)

Eduardo

Quote:

Quote:

Exact formula:

p = 4*a*E((a + b)/(a + b*(b + 1)))

where

E(x) is the complete elliptic integral of the second kind

a is the semi-major axis and

b is the semi-minor axis

The above is not the correct relation between the complete elliptic integral E(x) and the perimeter p(a,b) of the ellipse. The argument "x" (usually denoted k, "the modulus") of E(k) is dimensionless, whereas (a+b)/(a+b(b+1)) is not. The correct relation is

p(a,b) = 4*a*E(c/a)

I'd found the formula

p(a,b) = 4*a*E(pi/2,e), where e is the eccentricity, sqrt(1 - (b/a)^2)

However, trying this on WolframAlpha for a = 3 and b =2 I get 14.567698609052450048.. instead of 15.865439589290589791..

http://www.wolframalpha.com/input/?i=4*3*E%28pi%2F2%2Csqrt%281-%282%2F3%29%5E2%29+%29

Oddly enough my wrong parameter works when the difference between a and b is 1 (at least for integer values) and when a = b or one of the semi-axis is 0, as in all of my examples. I should have tried a = 5 and b = 3, for instance, for which it fails.

I wasn't able to find the right WolframAlpha syntax for this function. As soon as I find it, I will correct my original post. Thanks again.

Gerson.

Thanks for the C code, Hugh!

I am not proficient in C, but this appears to be easier to follow than RPL code. Anyway, my goal is a simple and short approximation formula that is accurate enough for practical cases, something that would give an error of a few meters in the length of the Earth's orbit for instance and takes about 20 or so steps on the HP-42S and no more than 1 second to run. Perhaps the approximate formula above meets this goal, but I haven't checked yet. Other approximate formulas are welcome, in case anyone knows about them (I've found only a few -- it appears measuring ellipses is not a popular sport :-)

Gerson.

To add a log to the fire, here´s the program on the 41 + SandMath:

Assumes a in Y and b in X, where ELI2 is the Incomplete Elliptic Integral of the second kind.-

01 LBL "ELIPER"

02 X^2

03 CHS

04 X<>Y

05 STO 05

06 X^2

07 +

08 SQRT

09 RCL 05

10 /

11 90

12 DEG

13 ELI2

14 RCL 05

15 *

16 4

17 *

18 RCL 05

19 *

20 END

For a=3, b=2 I get: p = 14.56769861

in 1.8 seconds on the CL (gotta love this machine :-)

Cheers,

'AM

Hi Ángel,

What is the physical meaning of this result? Not the perimeter, I presume. That should be about 15.9, according to the best approximations I've seen.

Regards,

Gerson.

This is what happens when rushing thru not paying attention to the details:-

I checked the formula and corrected the program accordingly, the result is now as yours:

http://www.wolframalpha.com/input/?i=perimeter+of+ellipse

01 LBL "ELIPER"

02 X<>Y

03 STO 05

04 /

05 X^2

06 CHS

07 1

08 +

09 90

10 DEG

11 LEI2

12 RCL 05

13 ST+ X

14 *

15 ST+ X

16 END

3, ENTER^, 2, XEQ"ELIPER" --> 15,86543959

Cheers,

'AM

*Edited: 24 May 2013, 4:24 p.m. *

Thanks, Ángel!

The parameter required by WolframAlpha is the square of the eccentricity, not the eccentricity. That's what I was doing wrong.

Cheers,

Gerson.

Quote:

What Wolfram Alpha's EllipticE(.) takes as an argument is the modulus k *squared*, and *not* k (in other words the eccentricity *squared*).

Yes, we'd just found that out too (see my reply to Ángel above). Thanks again!

Gerson.

Exactly, so beating this dead horse just once more, this is the correct syntax for WolframAlpha:

4*3*E(pi/2,(1-(2/3)^2) )

[url:http://www.wolframalpha.com/input/?i=4*3*E%28pi%2F2%2C%281-%282%2F3%29^2%29+%29]

Best,

ÁM

*Edited: 25 May 2013, 1:41 a.m. *

In your program *ELC* you use both, the built-in routine *AGM* and your own program *MAG* which implements *N(x)*, the modified arithmetic-geometric mean of 1 and *x* from the paper you mentioned. However Hugh Steers' program uses just one method *agm2* to calculate the perimeter of the ellipse. I see that numerically both methods produce the same result but I don't understand how Hugh's method is related to the paper you cited. Could you explain that?

Kind regards

Thomas