Expert: Paul Klarreich - 6/23/2006

Question
Hello Paul

Paul

Yes, your rewording of the problem is correct. I use the term "spherical cap" because the "observer" (man, or a machine capable of looking both ways, either sequentially, or simultaneously)can only see a part of each surface, which would seem to appear as two circles of different radii (R1 and R2, constant, or given) even though the original question was couched in terms of "visible surface area". Since L is > R1+R2, x, somewhere on L (also constant, or given, between the two spheres, could be viewed as sliding back abd forth until it settled at the point where its view of both surfaces resulted in the maximum surface area.

If the "surface area qualifier" is unnecessary-and it seems to me that it is, since x, on the line connecting the two centers, could only see two circles anyway, then the problem  appears to come down to:
1. We have two circles, whose planes are perpendicular to L and whose radii, say r1 and r2, are respectively functions of R1 and R2, and of the distances their  planes, say c1 and c2, are from your C1 and C2. (Its whether or not this view of the problem is correct that prompted my original request for advice.)
2. Determine r1 and r2
3. The distances between c1 and c2 are now L-(c1+c2), call it D,-also a constant(i.e., we are not adjusting the positions of these circles relative to some point on L)
4. Now it gets slightly more interesting from this point (at least for a near novice); we have some point on D, distant x from c1 and D-x from c2. Do we compute the length of the tangent drawn from x to the circumference (here x and the circle are not in tha same plane: x is on a line perpendicular to the plane of the circle and passing through its center) of each circle, add these lengths, and minimize that sum? If so, we would have a function f(x)= sqrt(x**2+ r1**2) + sqrt( (D-x)**2 + r2**2), where r1 and r2 are functions of R1, R2, and c1, c2. D is a function of c1 and c2.  C1 and c2 are functions of R1 and R2
5. I hope I have the above both described and typed correctly. If the above is not the way to proceed then I am lost!

I look forward to your next input, assuming you can stand this a little longer.

Regards

Ken

Followup To

Question -

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Followup To

Question -
Ken Rowe-Mathematical "Hobbiest"

I believe this to be an old problem from one of the Cambridge Mathematical Tripos Exams. Taks two spherical caps of radi R1 and R2, seperated by a distance L > R1+R2. Find the point on the line joining their centers where the viewable surface area is a maximum.

This does not appear that it should be that difficult; however, I am not making much headway with it.

Request advice on best way to proceed.

Thanks

Answer -
Hi, Kenneth,

Subject:  Spherical caps, maximum view
Question:  Ken Rowe-Mathematical "Hobbiest"

I believe this to be an old problem from one of the Cambridge Mathematical Tripos Exams. Taks two spherical caps of radi R1 and R2, seperated by a distance L > R1+R2. Find the point on the line joining their centers where the viewable surface area is a maximum.

This does not appear that it should be that difficult; however, I am not making much headway with it.

Request advice on best way to proceed.

Thanks
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Looks interesting, but I'm not sure exactly how things are laid out.  Where are these caps?  Are they actual spheres?  Hemispheres?  Your phrasing suggests a figure in which:

A. Two hemispheres lie on the x-axis.
B. The CENTERS are sepArated by a distance L > R1 + R2.

Or perhaps they are spheres, instead of hemispheres.  (Might not make a difference.)

So please specify.

The spherical caps are facing each other and an observer (a point) is somewhere between them on the line L, joining their centers. L exceeds the sum of the two spherical radii, R1 and R2, with one, say R1 >R2. The problem is seeking the position on L, between the two caps, where this observer can view the maximum surface area (including both spherical cap surfaces).

Sorry for the earlier confusion in my question.



Answer -
Hi, Kenneth,

We're getting closer, but one thing still worries me -- the term 'spherical caps' is not a well-known mathematical terms. (That means I don't know it.)  Why don't you just say 'spheres', if that's what they are?

Basically, I'm asking if the following is a correct wording of the problem:

Two spheres, which we can call Earth and Moon, are centered at C1 and C2 on the x-axis, with radii R1 and R2.  The distance between the centers, L = |C1 - C2| is greater than R1 + R2.  An observer stands on the line between Earth and Moon.  Where should he stand so that the sum of the surface areas he views is a maximum?

Of course, he can never see more than half of either sphere, just as we never see more than half of the moon.


Answer
Hi, Kenneth,

On the spherical caps problem.  (Really the difficulty lies in the fact that you British stopped speaking English about 100 years ago, and I have to translate everything.)

Since we want to determine surface areas, I thought it proper to first determine the area on a sphere cut out by a conical angle.  By this, I mean to take the following surface of revolution.  (Remember your second-semester calculus?)

Draw the circle x^2 + y^2 = a^2.  Draw the angle theta with one side along the x-axis, cutting off an arc.  Now rotate this around the x-axis.  The volume will be a cone-like piece and we want the surface area that is on the sphere.  

To do the integration, take a small piece of the circumference of the circle, call it ds, rotate that, and integrate from theta = 0 to theta0.

Assume that you have a piece of a circle, with radius a, rotated around the x-axis, to make a ring.  This is your area element, dA and the area of this ring is  

dA = 2pi y ds

Now ds is an intercepted arc, given by  ds = a dt  (dt means d-theta)

So the area swept out is:

{t=t0
|     2pi y a dt
}t=0

But we have  y = a sin t, so

A =
{t=t0
| 2 pi a^2 sin t dt
}t=0

- 2pi a^2 cos t [t = 0 to t = t0]

-2pi a^2[cos t0 - 1]

As a check: suppose that we let t0 go to pi.  That gives us the entire sphere:

-2pi a^2[cos pi - 1] = -2pi a^2[-1 -1] = -2pi a^2[-2] = 4 pi a^2, which is the correct formula for the area of a sphere.

So our surface area formula is:

A = 2pi a^2[1 - cos theta]

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That takes care of the background stuff, and we can turn our attention to the maximizing problem.

Let a person be at a point outside the sphere.  Assume the radius of the sphere is a1, and the person is at the point P, whose horizontal coordinate is x, where the sphere is centered at the origin.  Can we find an expression for theta in terms of x?  

Just look at the x-y plane.  The observer looks at the 'limb' of the sphere.  (That's the edge of his 'disk'.)

Draw a line from the observer, P, to the limb, at M on the circle, and thence to O, the center.  Is it clear that PM is perpendicular to OM?  Old geometry theorem -- a radius is perpendicular to a tangent at the point of tangency.

OP = a.
a/OP = cos theta

Combining that with our surface area formula gives:

A1 = 2pi a^2[1 - a/OP]

I think we are in business now.  

1. Let the distance between the centers C1 and C2 of the spheres = L.
2. Let the radii be  a1 and a2.
3. Let the observer be  'x' units from C1, therefore L-x units from C2.

The total area visible is:

A  = A1 + A2, where

A1 = 2pi a1^2[1 - a1/x]

A2 = 2pi a2^2[1 - a2/(L-x)]

Disregarding the 2pi part,

A = a1^2 - a1^3/x + a2^2 - a2^3/(L-x)
    + a1^3     a2^3
A' = ------ -  -------
      x^2     (L-x)^2

Set that equal to zero:

a1^3     a2^3
------ = -------
x^2     (L-x)^2
(L-x)^2     a2^3
-------- = -------
 x^2       a1^3

Now assign R = sqrt(a2^3/a1^3), i.e. the ratio of the 3/2 powers of the radii of the spheres, and we have:

L - x = Rx

L = Rx + x
     L
x = -----
   R + 1

I think that may be it.  Note that if the two 'planets' are the same size, R = 1 and  x = 1/2, meaning you should stand in the direct center between them, which seems reasonable.

How does this look?