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Perimeter of an elipse - exact formula

  1. Jun 7, 2009 #1
    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:

    [tex]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}[/tex]



    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 :biggrin:
     
    Last edited: Jun 8, 2009
  2. jcsd
  3. Jun 7, 2009 #2

    dx

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    Re: perimeter of an elipse -- exact formula

    Just try it on a circle, a = b = r, with r = 1. You should get 2pi. What does your formula give?
     
    Last edited: Jun 7, 2009
  4. Jun 7, 2009 #3
    Re: perimeter of an elipse -- exact formula

    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.
     
  5. Jun 7, 2009 #4

    dx

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    Re: perimeter of an elipse -- exact formula

    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.
     
    Last edited: Jun 7, 2009
  6. Jun 7, 2009 #5
    Re: perimeter of an elipse -- exact formula


    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:


    [tex]{\frac{\frac{(a^2+b^2)\frac{1}{b}\pi}{4}}[/tex]

    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.


    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.
     
  7. Jun 7, 2009 #6

    dx

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    Re: perimeter of an elipse -- exact formula

    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.
     
    Last edited: Jun 7, 2009
  8. Jun 7, 2009 #7
    Re: perimeter of an elipse -- exact formula

    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....
     
  9. Jun 7, 2009 #8

    dx

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    Re: perimeter of an elipse -- exact formula

    Eccentricity is just e = √(1 - (b/a)2).
     
  10. Jun 8, 2009 #9
    Re: perimeter of an elipse -- exact formula

    you posted this thread twice :surprised, 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 [tex]\frac{2 \pi r}{ \frac{0}{0} + 1 }[/tex], but 0/0 is not supposed to happen?
     
    Last edited by a moderator: Apr 24, 2017
  11. Jun 8, 2009 #10
    Re: perimeter of an elipse -- exact formula

    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...
     
    Last edited: Jun 8, 2009
  12. Jun 8, 2009 #11
    Re: perimeter of an elipse -- exact formula

    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?
     
    Last edited by a moderator: Apr 24, 2017
  13. Jun 8, 2009 #12

    Mute

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    Re: perimeter of an elipse -- exact formula

    How are you getting a 0/0?

    The "offending" term is

    [tex]\frac{\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}})-45}{90-(\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}})-45)}[/tex]

    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.
     
    Last edited by a moderator: Apr 24, 2017
  14. Jun 8, 2009 #13
    Re: perimeter of an elipse -- exact formula

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

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

    Also I originally messed up posting the formula, i've now edited it
     
    Last edited: Jun 8, 2009
  15. Jun 8, 2009 #14

    arildno

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    Re: perimeter of an elipse -- exact formula

    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.
     
  16. Jun 8, 2009 #15
    Re: perimeter of an elipse -- exact formula

    yes, i will
     
    Last edited: Jun 8, 2009
  17. Jun 8, 2009 #16
    Re: perimeter of an elipse -- exact formula

    yup, you're right, my mistake. i wrote it down on paper omitting the brackets.
     
  18. Jun 8, 2009 #17
    Re: perimeter of an elipse -- exact formula

    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 = [tex]\sqrt{a^2+b^2}[/tex]

    Angle CAB =[tex]\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}})[/tex]



    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 = [tex]\sqrt{a^2+b^2}[/tex]

    Angle DAB =[tex]\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}}) - 45[/tex]

    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 = [tex]\frac{1}{2}(270 degrees)[/tex] = 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:

    [tex]\sin(\frac{angle ADB}{side c}}) = \sin(\frac{angle ABD}{segment AF})[/tex]

    AD = [tex]\frac{(side c)(\sin(angle ABD))}{\sin(angle ADB)}[/tex]


    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:

    [tex]\frac{segment AD}{\sqrt{2}}[/tex]

    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= [tex]\frac{2(ED)\pi}{4}[/tex]



    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 =
    [tex]\frac{BD}{AB}x[/tex]

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

    [tex](1+\frac{BD}{AB})x[/tex]

    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 =[tex]\frac{arc AD}{1+\frac{arc BD}{arc AB}}[/tex]

    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?
     
    Last edited: Jun 9, 2009
  19. Jun 9, 2009 #18

    arildno

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    Re: perimeter of an elipse -- exact formula

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

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

    So, unfortunately, you are wrong.
     
  20. Jun 9, 2009 #19
    Re: perimeter of an elipse -- exact formula

    Can you explain?

    I would think that, surely, an elliptical arc could be expressed as part of a circular arc -- is this not true?
     
  21. Jun 9, 2009 #20

    dx

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    Re: perimeter of an elipse -- exact formula

    No, elliptical arcs are not circular arcs.
     
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