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What would be a projection of a finite right cone with base radius $R$ and height $H$ which is tilted about an angle $\theta$ with the vertical on a vertical.

I have an intuition that it may look like combination of a triangle and a semi ellipse, but I can't prove it. Also it is only applicable to some extent of tilt. I wish to know the detailed analysis of the shape.

Any help is appreciated.

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  • $\begingroup$ Is your cone finite (i.e., does it have a 'base', or extend infinitely downward)? Is it a right circular cone? $\endgroup$ Commented Sep 18 at 13:36
  • $\begingroup$ @JohnHughes Yes, the cone is finite, with base radius R and height H $\endgroup$ Commented Sep 18 at 13:37
  • $\begingroup$ @LuvGupta Pls refer an older post here for your interests. $\endgroup$ Commented Sep 18 at 19:35

2 Answers 2

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Whether the projection is parallel or central (which you should specify), the projection of the base is the intersection of an elliptic cylindre or an elliptic cone with the viewing plane, and is a conic. If the cone wholly lies behind the plane, that projection is an ellipse.

If the base is facing the observer, you see the whole ellipse. Otherwise only a part of it (a half for a parallel projection).

The apparent shape may also include two line segments from the apex, that tangent the ellipse. But not if the projection of the apex lies inside the ellipse.

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  • $\begingroup$ Any mathematical proof $\endgroup$ Commented Sep 18 at 15:38
  • $\begingroup$ @LuvGupta: " the intersection of an elliptic cylindre or an elliptic cone with the viewing plane is a conic" is a mathematical proof. $\endgroup$ Commented Sep 18 at 15:39
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You can see the projections when the original vertical axis of cone is tilted through angle (th) about y-axis as shown.

$(a=R)$ in parallel projection, but $b=R \cos(th)$ in projection onto original cone base plane or $(x,y)$ plane.

The cone generators are tangent to projected ellipse as per your intuition.

enter image description here

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