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Department of Science, Math, & Technology

E = mcï¿½
Constructing an ellipse from a circle
Posted: Posted August 16th by EN
 Here's a way to cut up a circle to make an ellipse. I don't know if it's an actual formal ellipse, but it looks close enough. Can you find the area of it?
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Posted August 17th by Xhin
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Xhin
Ground's what's around

Please bear with me since I don't know the math syntax here.

(pi)*a*b

where a and b are the two axes of the ellipse. We will define a as the longer axis and b as the shorter axis for this explanation.

To find a, we realize that it spans two of the radius of the smaller circle, or 2*.5r = r; to get the rest of a, we need to find the hypotenuse of the triangle formed if you draw a straight line from corner to corner of one of those sections. The triangle formed will be a 45-45-90 triangle, which will yield a hypotenuse of sqrt(2) * leg length, or in this case sqrt(2)*.5r = r(sqrt(2))/2. This yields a = r((1+sqrt(2))/2.

Axis b is somewhat more complicated: we know again that we start with r given that the inner circle is half the radius of the outer circle, but the middle bit is a bit tricky. To find it, we need to focus on one of the smaller sections (e.g. Section 3), and the aforementioned triangle. We know that the hypotenuse is r*sqrt(2).2; to get the remaining bit, the distance between the hypotenuse and the edge of the circle, we subtract that from .5r:

(r/2 - (r*sqrt(2)/4) = (2r-(r*sqrt(2)))/4 = r(2-sqrt(2))/4

Now, back to our original formula:

Area(ellipse) = (pi)*a*b = (pi) * r((1+sqrt(2))/2 * r(2-sqrt(2))/4 = (pi) * (r^2)*sqrt(2)/8

I could be wrong (and please for the love of God someone help me with the LaTEX, I think that's what it's called), but my answer is A=pi*(r^2)*sqrt(2)/8

Posted August 17th by Black Yoshi
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I got the same answer that Xhin did. When I have more time, I'll look more carefully at your answer Black Yoshi to see what happened.

Posted August 17th by EN
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@EN: @Xhin: @Black Yoshi:

The radius r of a circle with the same area as that of an ellipse with semimajor axis a and semiminor axis b, is the geometric mean of that semimajor axis and that semiminor axis.
r = squrt(a*b) .

You can easily get that from pi*r*r = pi*a*b.

———

Working out a compass-and-straightedge construction of such a circle is a different matter; maybe a more difficult and/or more complicated one, certainly a less obvious one.

————

If you use every part of the cut-up circle to fill the constructed figure, leaving no gaps and not covering anything twice, then they have to have the same area exactly.

That’s clearly not what you did.

Black Yoshi is on the right track.
And so is Xhin.
I didn’t check the details of either, though.

Edited August 17th by chiarizio
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Constructing this ellipse with a compass and straight should be relatively straightforward -- all you have to be able to do is construct both of these circles:

Then the other two ends of your ellipse are just circles with the centerpoint of each corner of the "eye" that trace one radius out.

The centerpoints of the small circles appear to be exactly one radius horizontally and one radius vertically away from each other, so that's pretty convenient.

It would probably take a lot of trace lines, but it would definitely work.

Posted August 17th by Xhin
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Xhin
Ground's what's around

@Xhin:
No, that’s an ovalish shape that’s not an ellipse.
No part of an ellipse is any part of any circle, unless the ellipse is that circle.
OTOH the figure EN asked about isn’t an ellipse. It’s a figure made up out of four pieces of a circle.
It’s not a smooth curve.

OTOH we’ve all been talking about how to calculate its area.
Its volume is zero.

Posted August 17th by chiarizio
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Yeah, it's an oval not a true ellipse. I meant area in the picture.

Posted August 17th by EN
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Whatever the case, you should be able to construct it.

I recently lucked into a really cheap compass set, so I might do just that.

Posted August 17th by Xhin
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Xhin
Ground's what's around

Yeah, it's an oval not a true ellipse. I meant area in the picture.

I believe the pieces have the same tangent at the joins, but the curvature of the ends is double that of the sides.
So it’s a smooth join to the first degree, but not the second.

Unless I’m wrong again!

Posted August 18th by chiarizio
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I spent the better part of the other night going back over my calculations...and only now do I realize that I was assuming the shape was a true ellipse, when in fact it's not. I ran the numbers by Xhin's method and got Xhin's answer, I ran them by my own method and got something that wasn't Xhin's answer and wasn't my original answer either. So Xhin's right in this one, I believe. Sometimes in mathematics you can make assumptions, but this is one of those cases where making assumptions can lead you into trouble.

Posted August 18th by Black Yoshi
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