Re: Cone in a groove

matt271829-news@xxxxxxxxxxx wrote:
> 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?
> >
> > 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.

That was the way I understood it too.

> 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.

I had wondered the same thing.

> After making that adjustment your answer seems to agree with
> mine (posted earlier).

Well, yes, it "seems to", but in fact it doesn't. I haven't looked
carefully enough to see which of the two is the correct answer. Curiously,
the wrong answer appears to give a fair approximation of the correct one.

David
.

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