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Perimeter of an ellipse

  1. Jun 6, 2009 #1
    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:

    [tex]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}}[/tex]




    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 7, 2009
  2. jcsd
  3. Jun 7, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi realitybugll! Welcome to PF! :smile:

    For complete info on LaTeX, see http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000 [Broken] and linked pages :wink:
     
    Last edited by a moderator: May 4, 2017
  4. Jun 7, 2009 #3

    HallsofIvy

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    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.
     
  5. Jun 7, 2009 #4
    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
     
  6. Jun 7, 2009 #5
    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)

    [tex]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}}[/tex]
     
  7. Jun 7, 2009 #6
    how did you find this nice-looking equation? i would love to see a proof :smile:
    will your equation converge down to [tex]2 \pi r [/tex] if a = b = r, as in the case of a circle?

    i just worked it out right now on paper, and i get [tex]\frac{2 \pi r}{\left( \frac{0}{0} + 1 \right) }[/tex]. 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 [tex]C \approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)}\right) [/tex] (from http://en.wikipedia.org/wiki/Ellipse" [Broken])
     
    Last edited by a moderator: May 4, 2017
  8. Jun 8, 2009 #7
    yea i'm going to do that now :smile:

    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: [tex]C \approx \pi \left( 3(a+b) - \sqrt{(3a+b)(a+3b)}\right) [/tex]

    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 :smile: (at least for these values of a,b...)
     
    Last edited by a moderator: May 4, 2017
  9. Jun 8, 2009 #8
    hey, don't worry, i'm 16 too :smile:
     
  10. Jun 8, 2009 #9
    unit,

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

    Also You seem to know a lot for being 16 haha
     
  11. Jun 8, 2009 #10

    uart

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

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

    Where [itex]\alpha = a/b[/itex]


    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:

  12. Jun 8, 2009 #11

    uart

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

    [tex]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}}[/tex]

    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.
     
  13. Jun 8, 2009 #12
    I checked again it is 29.5

    you get: [tex]4{\frac{20.813}{2.822)}[/tex]

    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!
     
  14. Jun 8, 2009 #13

    uart

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

    I get :

    [tex] a/\sqrt{a^2+b^2} = 0.9615 [/tex].

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

    [tex] \frac{x-45}{90-(x-45)} + 1 = 1.4767 [/tex]

    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.
     
  15. Jun 8, 2009 #14

    uart

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    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 :

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

    Where [itex]\alpha=a/b[/itex] 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
  16. Jun 8, 2009 #15
    uart,

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

    this is what i meant:

    [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]

    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 :smile:?
     
    Last edited: Jun 8, 2009
  17. Jun 8, 2009 #16
    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:

    numerical value of Arc AB =[tex]\frac{arc AD}{1+\frac{arc BD}{arc AB}}[/tex]

    Then for the perimeter of the entire ellipse multiply by 4



    I generalized this process to get the formula
     
    Last edited: Jun 9, 2009
  18. Jun 9, 2009 #17
    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...
     
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