A paradox in Geometry concerning the perimeter of an ellipse

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waveplumber

Junior Member level 3 Can someone compute the exact value of the perimeter of an ellipse of semi major axis a = 8.0 m and semi minor axis b = 7.5 m and tell me the results? From my calculations, the answer is less than the perimeter of a circle of radius 7.5 m, which I think does NOT make sense because the later lies inside the boundaries of the ellipse in question.

Thanks

Kral waveplumber,
From "Handbook of Mathematical TYables and Formulas" by Burrington:
p = 2Pi (Sqrt((a^2 +b^2)/2)) (Approximately)
.
p = 4aE (Exactly)
where
E = Complete elliptic integral of the first kind, using
. k = Sqrt(a^2-b^2)/a
. Phi = Arcsin(k)
Regards,
Kral

waveplumber

Junior Member level 3 Thanks Kral. I had used the second formula [p = 4aE(e)] but my worry is that the perimeter of the ellipse (a =8.0, b =7.5) comes out to be less than that of a circle of radius 7.5. This is counter intuitive so I'd like to verify my calculations or have an explanation for the apparently paradoxical situation. Can you look into that?
Thanks

flatulent You can ease your emotions by drawing the ellipse and then us a string to go around the plotted line and then measure the length of the string. Empirical methods sometimes are useful.

waveplumber

Junior Member level 3 No, no no, patulent. I prefer to do this by analytical methods. Thanks though.

Kral waveplumber,
here are my calculations:
k = SQRTSQRT(1-(b^2)/(a^2)) = 0.347985
Phi = ArcSin(k) = 20.36413 degrees
E = 1.522089 (Interpolated value from table)
p = 4*a*E = 48.70683
.
p of circle radius 7.5 = 2*Pi*7.5 = 47.12389
p of circle radius 8.0 = 2*Pi*8 = 50.26548
.
So the caluclate perimeter falls in between the two circle perimeters.
Regards,
Kral

btwang

Newbie level 5 Send via mobile phone wap.edaboard.com

waveplumber

Junior Member level 3 Thanks Kral. I was using Matlab to compute E(e) and I just realized that the parameter that Matlab requires is the modulus of the elipticity e

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