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Bully For You (Posted on 2004-12-02) Difficulty: 4 of 5


Take a regular torus (doughnut) shaped object and cut a vertical slice through it at line A-A1(Fig1).

Now look at the cross section formed(Fig2). Is it possible to calculate the volume of the original Torus? Prove your results.

No Solution Yet Submitted by Juggler    
Rating: 3.6667 (3 votes)

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Solution What if the generating conic is not a circle? Comment 5 of 5 |
(In reply to solution by Charlie)

What if the generating conic section were a more eccentric ellipse than a circle? What if it's horizontal radius were twice its vertical radius?

Then, as the height of the torus is 12 mm, the smaller radius of the generating ellipse is 6 mm, while the larger is 12.  The cutting plane A-A1 leaves a band 2 mm wide, 1 mm either side of the horizontal center plane. How far is this cutting plane from the center of the generating ellipse? If it were a circle, then by the Pythagorean theorem, it would be sqrt(6^2-1^2)=sqrt(35). However this circle is "stretched out", so it's 2*sqrt(35).

Then, looking down on the torus from above, we have a line that's the projection of A-A1 onto the center plane.  The horizontal thickness of the tube is 24 mm, so from the outer circle to the A-A1 plane is 12+2*sqrt(35). Use this as one leg of a right triangle.  The other leg is half the projection of A-A1, or 25 mm.  The hypotenuse, extending from one point on the outer circle to another, forming a chord, is then sqrt(25^2 + (12+2*sqrt(35))^2) = 34.5394242799266.  The angle at the base of this triangle is then arccos((12+2*sqrt(35))/sqrt(25^2 + (12+2*sqrt(35))^2)), but more importantly, that argument to the arccos is the sine of the angle at the center of the torus that subtends half that chord that had been the hypotenuse, and becomes the base of a right triangle with the radius of the outer circle as the new hypotenuse. The radius is then (sqrt(25^2 + (12+2*sqrt(35))^2) / 2) / ((12+2*sqrt(35))/sqrt(25^2 + (12+2*sqrt(35))^2)) = 25.0286136737847.  Then R, the radius of the torus used in computing the area, is 12 less than this, or 13.0286136737847. We've already determined r1 and r2, the radii of the generating ellipse.

The volume formula is 2 * pi^2 * R * r1*r2. In this case, this works out to 18516.5658511024. For the record, that's ((sqrt(25^2+(12+2*sqrt(35))^2)/2)/((12+2*sqrt(35))/sqrt(25^2+(12+2*sqrt(35))^2))-12)*pi^2*6*12*2.

By the way, we did have to make sure R > r2, otherwise there'd be no hole.


  Posted by Charlie on 2004-12-02 19:11:04
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