Re: Cone in a groove


Gary Wooding wrote:
>
> Imagine a horizontal symmetrical V-shaped groove with an included angle
> of V, and right circular cone with an included angle of C lying in it.
> The apex of the cone touches the bottom of the groove and its sides
> touch the inside planes of the groove - in other words, the planes of
> the groove are tangents to the cone. What is the angle between the axis
> of the cone and the bottom of the V?

SPOILER





































My solution works with half angles, i.e. the axial
angles of the groove and cone are 2G and 2C resp,
and the angle to find is 2X.

Assuming the groove base is the z axis, and its
planes are symmetric about the x axis, these have
equations:

+- x.sin(G) - y.cos(G) = 0 [1]

For a cone with vertex at the origin and axis in the
xz plane, a unit normal in that axis at the origin is
(cos(X), 0, sin(X)).

So in one of the groove planes, say the +ve option,
the unit line[s] of intersection with this cone are
as follows, by definition of the scalar product:

x.cos(X) + z.sin(X) = cos(C) [2]

Replacing [1] with the +ve option, as y = x.tan(G),
in x^2 + y^2 + z^2 = 1 gives:

x^2 = (1 - z^2).cos^2(G)

In [2] after subtracting z.sin(x) from both sides
and squaring, we can replace the latter to obtain
a quadratic in z, which has solution[s]:

z = cos(C).sin(X) +- cos(G).cos(X).sqrt(D)

where:

D := sin^2(X) + cos^2(G).cos^2(X) - cos^2(C).

However, in this problem the line(s) of contact
of the cone and the groove are tangential, and
thus there is one (double) root, i.e. D = 0,
which gives:

sin^2(X).sin^2(G) = cos^2(C) - cos^2(G)

Sanity checks:

sin(X) not real for G < C - YUP

sin(X) = 0 for G == C - YUP

sin(X) undefined for G == PI - YUP



Cheers

John R Ramsden

.

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