# Perimeter of an ellipse

1. Jun 6, 2009

### realitybugll

perimeter of an ellipse -- 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:

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

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

Last edited: Jun 7, 2009
2. Jun 7, 2009

### tiny-tim

Welcome to PF!

Hi realitybugll! Welcome to PF!

For complete info on LaTeX, see http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken] and linked pages

Last edited by a moderator: May 4, 2017
3. Jun 7, 2009

### HallsofIvy

Staff Emeritus
However, I doubt very seriously that you have a formula for perimeter of an ellipse that does not involve an "elliptic" integeral.

I might be wrong: post it here and let's see.

4. Jun 7, 2009

### realitybugll

tiny tim,

Ok thank you, ill write it up now. Im also going to try and draw the proof on cabri -- its a geometry program thing

5. Jun 7, 2009

### realitybugll

the equation above has a small error...i cant edit any more so here's the fix:

exact formula for perimeter of a ellipse

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

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

6. Jun 7, 2009

### Unit

how did you find this nice-looking equation? i would love to see a proof
will your equation converge down to $$2 \pi r$$ if a = b = r, as in the case of a circle?

i just worked it out right now on paper, and i get $$\frac{2 \pi r}{\left( \frac{0}{0} + 1 \right) }$$. i mean, i guess it's acceptable, because it turns into 2 pi r over one, but the 0/0 is sketchy.

i thought the whole area of elliptic integrals was devoted to finding the arc lengths of ellipses... because it's impossible otherwise?

check this up against some good approximations, and see how your answers work out. like $$C \approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)}\right)$$ (from http://en.wikipedia.org/wiki/Ellipse" [Broken])

Last edited by a moderator: May 4, 2017
7. Jun 8, 2009

### realitybugll

yea i'm going to do that now

the proof is pretty cool. I did it on cabri, a geometry program.

Also i'm 16 btw so i don't know a lot of math stuffs...

testing my formula against this (good) approximation formula: $$C \approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)}\right)$$

a=3, b=1

that formula:13.364
my formula:12.870

a=84, b=9

that formula: 341.777
my formula:338.56

a=7, b=2

that formula: 30.499
my formula: 29.499

a=2, b=1

that formula: 9.668
my formula: 9.272

testing my formula against this more accurate formula which uses elliptic integrals:

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

As you can see my formula is more accurate against this more accurate formula (at least for these values of a,b...)

Last edited by a moderator: May 4, 2017
8. Jun 8, 2009

### Unit

hey, don't worry, i'm 16 too

9. Jun 8, 2009

### realitybugll

unit,

I tested it a couple times against that approximation equation...

Also You seem to know a lot for being 16 haha

10. Jun 8, 2009

### uart

Hi realitybugll. Your formula can be simplified a lot, with a bit of algebra it reduces down to :

$$b (\alpha^2+1) (\pi/2 + 2 \cos^{-1}(\frac{\alpha}{\sqrt{1+\alpha^2}}) )$$

Where $\alpha = a/b$

Unfortunately however it isn't even a good approximation, let alone an exact expression, for the perimeter of an ellipse.

The attached image shows the perimeter versus eccentricity for your formula against some other methods (the elliptic integral method being the "exact" value).

#### Attached Files:

• ###### ellipse.png
File size:
21.3 KB
Views:
76
11. Jun 8, 2009

### uart

OK then there's still something wrong with how you have expressed you formula in post #5 then.

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

If you substitute a=7 and b=2 into that you get approx 56.4 (not 29.5), so it seems you still need to post the corrected formula.

12. Jun 8, 2009

### realitybugll

I checked again it is 29.5

you get: $$4{\frac{20.813}{2.822)}$$

Ill check the simplification you made later don't have time atm

Also, i couldn't open the picture, but if you did that, thank you!

13. Jun 8, 2009

### uart

Ok we agree on the numerator but could you show me how you're getting 2.822 for the denominator.

I get :

$$a/\sqrt{a^2+b^2} = 0.9615$$.

$$x = \sin^{-1}(a/\sqrt{a^2+b^2}) = 74.06$$ degrees.

$$\frac{x-45}{90-(x-45)} + 1 = 1.4767$$

So that's 1.4767 for the denominator which gives the overall value of the formula as 83.252/1.4767 = 56.376.

I still think you have posted a different formula than the one you are evaluating.

BTW. If you use my equivalent formula then you'll need to use radians instead of degrees for the inverse sine ok.

14. Jun 8, 2009

### uart

Ok, from your numbers I think I can guess what you're doing differently. You see that convoluted denominator within a denominator that you're using. You obviously aren't including the "-45 in that sub-donominator (denominator within the main denominator). Sorry to say this but it looks like you really need to learn some basic algebra skills there realitybug.

Anyway, if I remove that second "-45" from the sub-denominator then I get the following algebraic simplification of your formula :

$$4b * (\alpha^2 + 1) \, \cos^{-1} \left( \frac{\alpha} {\sqrt{1+\alpha^2}} \right)$$

Where $\alpha=a/b$ as before.

AND remember that you have to use radians for the inverse cosine here.

This is a much better approximation than before, but still inferior to the simpler approximation I gave in the previous attachment.

Last edited: Jun 8, 2009
15. Jun 8, 2009

### realitybugll

uart,

You were right: i was evaluating it with a different formula -- i posted it wrong

this is what i meant:

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

This is different than the formula that you get if you just remove the second -45

can you evaluate this one like you did the other (wrong) one ?

Last edited: Jun 8, 2009
16. Jun 8, 2009

### 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

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:

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

Then for the perimeter of the entire ellipse multiply by 4

I generalized this process to get the formula

Last edited: Jun 9, 2009
17. Jun 9, 2009

### realitybugll

It turns out I didn't know what exactly an ellipse was...this formula is for a "lens"

So i am wrong. But at least now I know the difference...