Perimeter of an elipse - exact formula

[realitybugll]

perimeter of an elipse -- exact formula

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:

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

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]

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

 

[realitybugll]

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]

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]

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:


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

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]

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/2,0)

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

 

[realitybugll]

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]

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

 

[Unit]

you posted this thread twice , the other place is https://www.physicsforums.com/showthread.php?t=318458".

with your equation, like i explained in that thread also, letting [tex]a = b = r[/itex] results in $$\frac{2 \pi r}{ \frac{0}{0} + 1 }$$, but 0/0 is not supposed to happen?

 

[realitybugll]

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]

you posted this thread twice , the other place is https://www.physicsforums.com/showthread.php?t=318458".

with your equation, like i explained in that thread also, letting [tex]a = b = r[/itex] results in $$\frac{2 \pi r}{ \frac{0}{0} + 1 }$$, 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]

you posted this thread twice , the other place is https://www.physicsforums.com/showthread.php?t=318458".

with your equation, like i explained in that thread also, letting [tex]a = b = r[/itex] results in $$\frac{2 \pi r}{ \frac{0}{0} + 1 }$$, but 0/0 is not supposed to happen?

How are you getting a 0/0?

The "offending" term is

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

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]

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]

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]

So, please post the actual proof.

yes, i will

 

[Unit]

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.

 

[realitybugll]

This is how i got the formula. Its pretty simple...hopefully I explained it well enough so its followable.



a= 1/2(length of major axis)

b= 1/2(length of minor axis)


1. draw an ellipse and its major and minor axis


2. take the triangle formed by a, b and the right angle between them.

Label its angles A, B , C where A is opposite the length a, and B is opposite b. Let c be the length of the triangles hypotenuse.

We have an SAS right triangle so we can use trig to find the other side/angles

side c = $$\sqrt{a^2+b^2}$$

Angle CAB =$$\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}})$$



3. Draw a ray from angle A that intercepts side a, and makes a 45 degree angle with side b.

Extend the arc of AB until it intersects the ray. We will call this point of intersection D.

Using the segment AD as a hypotenuse draw in the two perpendicular side lengths so you get a right angle. Label this right angle E

So now you should have two triangles, each with an arc.

Because the triangle ADE (the larger one) is a 45-45-90 triangle we can calculate the perimeter of its arc, AD, because its arc is one-fourth the perimeter of the circle with the radius equal to the lengths of thesegments AE and DE (they are the same)

draw the whole circle in if you want



4. finding the length of segment AD (the hypotenuse of the larger triangle EFG)

Draw in segment BD and label so that you form the triangle. We can call the triangle ABD

Solving for the angles/sides of this triangle with trig...

We already know side c = $$\sqrt{a^2+b^2}$$

Angle DAB =$$\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}}) - 45$$

Angle ABD is an inscribed angle and intercepts an arc of 270 degrees (3/4) of the circumference of the circle (we know this because arc AF is 1/4 the circumference of the circle).

By the inscribed angle theorem we can say that
Angle ABD = $$\frac{1}{2}(270 degrees)$$ = 135 degrees


We know two angles so therefore angle ADB= 180-135-angle BAD

Now we can use the law of sines to solve for the side length of segment AD:

$$\sin(\frac{angle ADB}{side c}}) = \sin(\frac{angle ABD}{segment AF})$$

AD = $$\frac{(side c)(\sin(angle ABD))}{\sin(angle ADB)}$$


5. finding the measure of Arc AD

It’s a 45-45-90 triangle so we can find the lengths segments DE and AE by:

$$\frac{segment AD}{\sqrt{2}}$$

Now we can find the arc AD because its 1/4 the circumference of the circle with the radius the lengths of the segments ED or EA

arc AD= $$\frac{2(ED)\pi}{4}$$



6. But we just want the length of arc AB…

Angle CAB is also an inscribed angle so we can say that the measure of its arc BD in degrees is: 2(angle CAB)

Because arc AD is one fourth of a circle we know that it corresponds to 90 degrees.

If we subtract the measure that we have for Arc BD (in degrees) from 90 degrees we get the length of arc AB in degrees:

90 degrees - angle of arc BD = angle of arc AB

Now we can set up a ratio of the degrees of arc BD over arc AB:

if we let arc AB = x (its actual numerical value) then we can say arc BD =
$$\frac{BD}{AB}x$$

therefore the value of the arc AD in term of x (length of arc AB) is:

$$(1+\frac{BD}{AB})x$$

We have already calculated the exact measure of Arc AD so we can divide this by
(1+(BD/AB))x to solve for x which = the measure of Arc AB:

Arc AB =$$\frac{arc AD}{1+\frac{arc BD}{arc AB}}$$

Then to get the perimeter of the entire ellipse multiply by 4


I generalized this process to get the formula

So why wouldn't this yield the exact value of the perimeter of an ellipse -- what am i missing?

 

[arildno]

3. Draw a ray from angle A that intercepts side a, and makes a 45 degree angle with side b.

Extend the arc of AB until it intersects the ray. We will call this point of intersection D.

How do you "extend" arc of AB?
It is not a circular arc anywhere!

So now you should have two triangles, each with an arc.

Because the triangle ADE (the larger one) is a 45-45-90 triangle we can calculate the perimeter of its arc, AD, because its arc is one-fourth the perimeter of the circle with the radius equal to the lengths of thesegments AE and DE (they are the same)

draw the whole circle in if you want

Impossible, since no finite segment of the elliptical arc is a circular arc.

So, unfortunately, you are wrong.

 

[realitybugll]

Impossible, since no finite segment of the elliptical arc is a circular arc.

Can you explain?

I would think that, surely, an elliptical arc could be expressed as part of a circular arc -- is this not true?

 

[dx]

No, elliptical arcs are not circular arcs.

 

[realitybugll]

No, elliptical arcs are not circular arcs.

But i don't get it. It seems to me that circular arcs (segments of a circle) can also be elliptical arcs.

If you take two circles for instance and interlock them you get an ellipse in the middle which is formed form arcs of the circles

 

[m00npirate]

well if you can prove it then you are all set =P

 

[slider142]

But i don't get it. It seems to me that circular arcs (segments of a circle) can also be elliptical arcs.

If you take two circles for instance and interlock them you get an ellipse in the middle which is formed form arcs of the circles

No you do not. You get a lens shape; the curve is not the locus of points whose sum of distances from two points is a constant. Given a fixed constant, this is only true for 4 (and in one case 2) points on opposite sides of the lens.

 

[realitybugll]

Ok well thank you for clearing that up.

So my formula does not work for ellipses then, it works for "lenses" :yuck:

I guess I should of made sure that I knew exactly what an ellipse was first. I get the difference between an ellipse and a lens now...

 

[arildno]

realitybugll:

The circle has the property of constant curvature.
The ellipse has a non-constant curvature everywhere.

 

[pershi]

hey there!
greetings from Romania

so, i was searching on the internet for a formula to determinate an ellipse length and i found out tht it is complicated stuff
from what i saw on your replyes on the 1'st guy's request i could only figure out only some of your calculations, so it made me wonder how old are these guys who really know theyr mathematics?
i'm asking this because i heard that is a difference in age between our cowntryes, in terms of studyes
i'll be waiting 4 your answer and maybe we'll talk some more

PS if i made some mistakes up there please let it slide :D

 

[Dwight Gill]

The figures given are hardly "exact." A 3 to 1 relationship between the major and minor axis, should result in a perimeter of 13.365. 7 to 2 should be 30.502, and 84 to 9 should be 342.036 (rounding to 3 decimals for consistency).

Dwight Gill