perimeter of an elipse -- exact formula

**realitybugll** · Jun7-09, 08:28 PM

I found an exact formula for the perimeter of an ellipse in terms of its major and minor axis

a = 1/2(major axis)
b=1/2(minor axis)

my equation for the perimeter of an ellipse:

$LaTeX Code: 4{\\frac{\\frac{(a^2+b^2)\\frac{1}{b}\\pi}{4}}{\\frac{2 ({\\sin^{-1}(\\frac{a}{\\sqrt{a^2+b^2}})-45)}}{90-2{(\\sin^{-1}(\\frac{a}{\\sqrt{a^2+b^2}})-45)}}+1}$

The perimeter of an ellipse with semi-major axis a and eccentricity e is given by 4aE(pi/2,e), where E is the complete elliptic integral of the second kind. This can be calculated to great precision instantly on any mathematics program like mathematica

I tested it against this formula a couple times:

for a =3, b =1

the formula: 12.808
my formula:12.870

for a=7 b=2

the formula:29.462
my formula:29.499

for a=84 b = 9

the formula:339356
my formula:338.56555

on cabri II plus i drew a proof that shows how i got this -- if u want that tell me.

any insight is greatly appreciated

dx · Jun7-09, 08:43 PM

Just try it on a circle, a = b = r, with r = 1. You should get 2pi. What does your formula give?

**realitybugll** · Jun7-09, 10:48 PM

dx,

yea it gives 2pi, i also simplified it quite a bit.

I also tried it against perimeter of ellipse "calculators" online and got similar results -- the formulas they use have up to 11% margin for error though.

dx · Jun7-09, 11:26 PM

Are you sure you didn't just chage it so that it would give the right answer for a circle? (I didn't check your formula btw)

How did you get rid of the sin inverse that you previously had in the numerator? Where are these "11% margin of error" calculators that you mention. Do you have a link? And yes, please show us your proof.

**realitybugll** · Jun7-09, 11:48 PM

Originally Posted by dx

Are you sure you didn't just chage it so that it would give the right answer for a circle? (I didn't check your formula btw)

no i did not. And i don't blame you for not checking it; its kind of a pain. The numerator did originally have inverse sin in it. But it simplyfied on my calculator to:

$LaTeX Code: {\\frac{\\frac{(a^2+b^2)\\frac{1}{b}\\pi}{4}}$

im not sure how it did it but they do equal each other, I checked the previous numerator equation against the new one multiple times.

How did you get rid of the sin inverse that you previously had in the numerator? Where are these "11% margin of error" calculators that you mention. Do you have a link? And yes, please show us your proof.

I found a "perimeter of an ellipse calculator online" that use an approximation formula with 11% margin for error

My formula is not a guess -- the proof is very convincing. I did it on Cabri II plus because its geometric. I can send you the file if you want, you would have to download the trial version though to see it if you don't have it.

dx · Jun8-09, 12:05 AM

Well, the thing is, the perimenter of an ellipse involves an elliptic integral, which cannot be expressed in closed form in terms of elementary functions (which you claim to have done).

Also, I find this "11% margin of error" thing odd. It can easily be calculated to much greater precision.

The perimeter of an ellipse with semi-major axis a and eccentricity e is given by 4aE(pi/2,e), where E is the complete elliptic integral of the second kind. This can be calculated to great precision instantly on any mathematics program like mathematica. See this for example:

http://www79.wolframalpha.com/input/?i=4*1*E(pi%2F2%2C0)

Just substitue whatever a and e you want in the above link and check your formula.

**realitybugll** · Jun8-09, 12:22 AM

sorry i don't know a lot about this stuff.

So my equation uses the 2 radius's of the ellipse to calculate its perimeter

how can i get the eccentricity from the radius's?

so int he equation i put in the largest radius and e?

-- never mind i found what e is in terms of a,b....

dx · Jun8-09, 12:27 AM

eccentricity is just e = √(1 - (b/a)²).

**Unit** · Jun8-09, 01:08 AM

you posted this thread twice

, the other place is here.

with your equation, like i explained in that thread also, letting

results in $LaTeX Code: \\frac{2 \\pi r}{ \\frac{0}{0} + 1 }$ , but 0/0 is not supposed to happen?

**realitybugll** · Jun8-09, 01:09 AM

for a =3, b =1

the link: 12.808
my formula:12.870

for a=7 b=2

the link:29.462
my formula:29.499

for a=84 b = 9

the link:339356
my formula:338.56555

hmm...so i guess my formula is not exact

I'm going to go look back at my proof...

**realitybugll** · Jun8-09, 01:10 AM

Originally Posted by Unit

you posted this thread twice , the other place is here.

with your equation, like i explained in that thread also, letting results in $LaTeX Code: \\frac{2 \\pi r}{ \\frac{0}{0} + 1 }$ , but 0/0 is not supposed to happen?

yeah i know, for it to work 0/0 has to equal 0, not be undefined...

Ive read that there are exceptions where 0/0 can equal 0...so maybe this is one of them?

**Mute** · Jun8-09, 12:06 PM

Originally Posted by Unit

you posted this thread twice , the other place is here.

with your equation, like i explained in that thread also, letting results in $LaTeX Code: \\frac{2 \\pi r}{ \\frac{0}{0} + 1 }$ , but 0/0 is not supposed to happen?

How are you getting a 0/0?

The "offending" term is

$LaTeX Code: \\frac{\\sin^{-1}(\\frac{a}{\\sqrt{a^2+b^2}})-45}{90-(\\sin^{-1}(\\frac{a}{\\sqrt{a^2+b^2}})-45)}$

The parentheses on the bottom cover both the arcsin and the 45 deg: it's not 90 - arcsin - 45. Hence, when a = b, the arcsin gives 45 degrees, so the numerator is zero, and the denominator is 90 deg - (45 - 45). So, you've got 0/90 = 0, so in this particular instance the formula gives the correct result.

I doubt the formula given here is exact, since that would seem to imply there's a closed form expression for the complete elliptic function of the second kind. I suppose it's possible the formula is a decent approximation, though.

**realitybugll** · Jun8-09, 02:06 PM

The parentheses on the bottom cover both the arcsin and the 45 deg: it's not 90 - arcsin - 45. Hence, when a = b, the arcsin gives 45 degrees, so the numerator is zero, and the denominator is 90 deg - (45 - 45). So, you've got 0/90 = 0, so in this particular instance the formula gives the correct result.

yeah your right you do get 0/90 -- i had forgot, i assumed unit was correct

I doubt the formula given here is exact, since that would seem to imply there's a closed form expression for the complete elliptic function of the second kind. I suppose it's possible the formula is a decent approximation, though.

the proof for how i got this formula is very convincing so i have faith in it

Also I originally messed up posting the formula, i've now edited it

**arildno** · Jun8-09, 02:23 PM

Originally Posted by realitybugll

yeah your right you do get 0/90 -- i had forgot that unit was correct

the proof for how i got this formula is very convincing so i have faith in it

Well, we do not have that.

Not the least since it has been proven long ago that the elliptic integral can't generally be expressed as a closed expression of elementary functions.

Which you claim to have done..

So, please post the actual proof.

**realitybugll** · Jun8-09, 02:34 PM

Originally Posted by arildno

So, please post the actual proof.

yes, i will

**Unit** · Jun8-09, 03:21 PM

Originally Posted by Mute

and the denominator is 90 deg - (45 - 45). So, you've got 0/90 = 0, so in this particular instance the formula gives the correct result.

yup, you're right, my mistake. i wrote it down on paper omitting the brackets.