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Spherical shelter 3 (Posted on 2012-05-27)

A unit sphere is on a level surface in the rain. What is the volume of the largest convex solid that be placed under the sphere and not get rained on?

No Solution Yet Submitted by Jer

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For the convex shape to be maximum, I’m supposing that it is bounded by a cylindrical surface (the outer edge of the dry space), the base plane and a tangent plane to the sphere, which cuts the base circle in a chord, with mid-point M say.
Let A be the acute angle which this plane makes with the horizontal, and let t be the perpendicular distance, MC, from the chord to the centre, C, of the base circle.

To find the volume of the shape, note that a typical horizontal section at height y above the base, will always be a segment of a circle, radius 1, with a chord at a distance x (say) from the vertical axis through the centre, O, of the sphere,
where dy/dx = tan(A).

Area of this segment      = area sector – area triangle

                                    = arccos(x) - (1/2)*x*sqrt(1 – x²)

Volume of shape = Integral(x=t..1) of [arccos(x) – (1/2)*x*sqrt(1 – x²)] dy

            = Integral(x=t..1) of [arccos(x) – (1/2)*x*sqrt(1 – x²)]tan(A) dx

            = tan(A)[x*arccos(x) – sqrt(1 – x²) + (1/6)(1 – x²)^3/2] between t & 1

       V   = (1/6)tan(A)[(t² + 5)sqrt(1 – t²) – 6t*arccos(t)]               (1)

If the plane touches the sphere at B, angle BOC = A, and OM bisects this angle

so that tan(A/2) = |CF|/|OC| = t and therefore tan(A) = 2t/(1 – t²)

(1) now becomes V = (t/3)*[(t² + 5)sqrt(1 – t²) – 6t*arccos(t)]/(1 – t²)

Differentiating and simplifying eventually gives:

dV/dt = (1/3)[(5 – 2t⁴ +9t²)sqrt(1 – t²) – 12t*arccos(t)]/(1 – t²)²

dV/dt = 0 then gives (5 – 2t⁴ +9t²)sqrt(1 – t²) = 12t*arccos(t)           (2)

Apart from the root t = 1 (V indeterminate), equation (2) can’t be solved
explicitly for t. But the single root between 0 and 1 is shown numerically
to be t = 0.46506..., which gives a maximum volume of 0.3134..

Posted by Harry on 2012-06-02 22:24:01

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