Posts: 2,761
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00 { 48Byte 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(1b^2/a^2)
where
E(x) is the complete elliptic integral of the second kind
a is the semimajor axis and
b is the semiminor axis
Example:
a = 2, b = 1
p = 4*2*E(3/4) = 9.68844822054767619842850319639...
http://www.wolframalpha.com/input/?i=4*2*E%2811%2F4%29
Approximate formula:
p ~ pi*(a + b)*(4/(4h)  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/(4h). The resulting expression is
4/(4h)  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
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I don't have an HP42S, 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*ab*b)/(a*a)*sin^2(t), t, 0, pi/2)
http://www.wolframalpha.com/input/?i=4+*+3+*+integrate+sqrt%281%283*32*2%29%2F%283*3%29*sin%5E2%28t%29%29%2Ct%2C0%2Cpi%2F2
BTW quoted "exact" answer for a=3,b=2 should be 15.8654395893 not 15.8654396893
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Problem is numerical integration is rather slow to full accuracy on the HP42S, 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!
Posts: 2,761
Threads: 100
Joined: Jul 2005
Quote:
I don't have an HP42S, 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*ab*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 { 41Byte 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 
+++++
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Joined: Jul 2005
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 semimajor axis
// b semiminor axis
// e = eccentricity = sqrt(1(b/a)^2)
// p = 4*a*E(e)
// use AGM variant
return 2*_PI*a*agm2(b/a);
}
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Joined: May 2006
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 semimajor axis and
b is the semiminor 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^2b^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.
Posts: 56
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Joined: May 2006
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
Posts: 2,761
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Joined: Jul 2005
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 semimajor axis and
b is the semiminor 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 semiaxis 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.
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Joined: Jul 2005
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 HP42S 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.
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Joined: Nov 2005
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
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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.
Posts: 1,253
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Joined: Nov 2005
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.
Posts: 2,761
Threads: 100
Joined: Jul 2005
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.
Posts: 56
Threads: 10
Joined: May 2006
Quote:
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
I know why. The documentation linked to from Wolfram Alpha is inconsistent with Mathematica's definition.
What Wolfram Alpha's EllipticE(.) takes as an argument is the modulus k *squared*, and *not* k (in other words the eccentricity *squared*).
The convention used by W.Alpha is the one used by the function elliptic_ec(m) in venerable opensource program Maxima as correctly documented at:
http://maxima.sourceforge.net/docs/manual/en/maxima_16.html
Try:
http://maximaonline.org/#?in=4*3*elliptic_ec%20(5.0%2F9)%0A%20%09
Eduardo
Posts: 2,761
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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.
Posts: 1,253
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Joined: Nov 2005
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.
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In your program ELC you use both, the builtin routine AGM and your own program MAG which implements N(x), the modified arithmeticgeometric 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
