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berkeman

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Welcome to the PF. Is this question for schoolwork?

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Hello, no, it's a cuirosidad. I try to resolve, however, I can not remember how.

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berkeman

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Fair enough. Do you remember how to use integration to calculate the volume between planes and other surfaces?Hello, no, it's a cuirosidad. I try to resolve, however, I can not remember how.

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For example if my quick calculations are correct, for angle 0 degrees of plane parallel to ground a solution would be: ## h-\frac {h} { \sqrt[3] 2} ##.

- #6

mfb

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The angle will still depend on the height/radius ratio but that dependence is trivial.

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I see..

The angle will still depend on the height/radius ratio but that dependence is trivial.

Still it depends on the orientation of the plane then. Is the line of cut of the tilted plane with the z=0 plane the x=0 line, the y=x line? there are y=nx solutions...

For the simplest case, here's a reference: http://mathworld.wolfram.com/EllipticCone.html

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jbriggs444

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One assumes the intersection of the tilted plane with the z=0 plane forms a line which is expected to be tangent to the base of the cone. And one assumes a right circular cone.Still it depends on the orientation of the plane then. Is the line of cut of the tilted plane with the z=0 plane the x=0 line, the y=x line? there are y=nx solutions...

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Good luck with the calculations!

- #10

mfb

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- #11

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Huh?

Volume of elliptic cone is half the volume of original cone. So, ##

π.a.b.h/3 = π.h.r^2 /6 =>

a.b = r^2 /2

##

Where did you see an area?

- #12

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a.b = r^2 /2 ## is not the right formula to be used...

The definition of an elliptic cone given in http://mathworld.wolfram.com/EllipticCone.html is not applicable in this case. What we are having here is a right circular cone cut by an oblique plane.

So in ##

π.a.b.h'/3 = π.h.r^2 /6 ## I would be using h' instead where h' is the height of the perpendicular drawn from the apex of the cone to the oblique cutting plane.

Again I am not certain about this so you must wait until I verify this.

- #13

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π.a.b.h'/3 = π.h.r^2 /6 ## where h' is the height of the perpendicular drawn from the apex of the cone to the oblique cutting plane and h,r the height and radius of the original cone is the correct formula to be used!

- #14

mfb

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That is better, and h'<h is the reason your previous approach didn't work.

π.a.b.h'/3 = π.h.r^2 /6 ## where h' is the height of the perpendicular drawn from the apex of the cone to the oblique cutting plane and h,r the height and radius of the original cone is the correct formula to be used!

Define ##L=\sqrt{h^2 + r^2/4}##, the distance between apex and the circumference of the intersection between cone and horizontal plane. Define ##\alpha## to be the angle between cone and horizontal surface and ##\gamma## to be the angle between cut plane and horizontal surface. Define ##\beta= \pi - 2 \alpha## (angle at the apex). Then ##h' = L \sin(\alpha - \gamma)##.

From the law of sines we get ##\frac{L}{\sin(\alpha+\gamma)} = \frac{a}{\sin(\beta)}##.

Now we just have to find b.

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