Re: Cone in a groove



john_ramsden@xxxxxxxxxxxxxx wrote:
> 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
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> 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
>
>

If C = G then your formula gives 2X = 0. The way I understood the
problem, the required angle in this case is pi/2. Since your "angle to
find" is 2X, you ought to have 2X = pi/2.

I wonder if your X (not 2X) is actually the complement of the required
angle. After making that adjustment your answer seems to agree with
mine (posted earlier).

.

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