# Finding the radius of spheres

1. Nov 10, 2004

### VividVoid

Heya, this is my first post, I hope it isn't too hard :rofl:

Alright, the problem goes like this. I put a sphere into a box to be shipped somewhere, it fits perfectly into the box and the sphere has a radius R. Now I have 8 styrofoam spheres that are placed on the corners of the box, find the largest radius that these 8 styrofoam spheres can be.

I drew a little diagram, it's only a front view of what the inside would look like because I can't draw all that well in 3D...

http://douglas.flooda.us/Images/problem.jpg

I really have no clue how to approach this problem, so any help will be nice! ^^

2. Nov 10, 2004

### Moose352

You know that the sphere of radius R fits perfectly. With that you know that the size of the box and the length of the diaganol. Do you see that twice the radius of the big sphere plus the twice the radius of the smaller balls plus twice the distance from the center of the small balls to the corner must be equal to the length diagonal?

3. Nov 10, 2004

### VividVoid

Yeah, I've thought of that, but then I realized the radius of the small circle doesn't touch the corner. I think I might have made the drawing properly, but if you look closer than you'll see that there is a gap between the small circle and the corner.

http://douglas.flooda.us/Images/problemz.jpg

so 2R + 4r won't equal the diagonal of the box.

4. Nov 10, 2004

### jcsd

Thsi is what I reckon, thoguh I haven't put much thoguht into it:

You've got a series, whose sum to infinity is the diagonal of the box, whose first and second terms are 2R and 4r respectively and each term is related to each other term by: an + 1 = 2r/R an.

5. Nov 10, 2004

### jcsd

Sorry taht's not quite right the sequence should be

a1 = R, a2 = 2 an+1 = r/R * an and the sum to infinity is equal to L - (L - R)/2 where L is the length of the diagonal (which is of course sqrt(2)*R)

6. Nov 10, 2004

### VividVoid

hmm... I don't think it's suppose to be that hard... ^^;;

I looked in the back of my calc book, it says the answer is r = (2 - sqrt(3))R

I have the calculus: a new horizon 6th edition by anton book.

Oopies

Last edited: Nov 10, 2004
7. Nov 10, 2004

### StatusX

if the radii of the small spheres are r, then you know the center of the small spheres will be r from each side the spheres touch and r+R from the center of the larger sphere.

EDIT: I think the answer you gave is wrong. For one thing, it's bigger than R. I think it's supposed to be $$R (2-\sqrt{3})$$.

Last edited: Nov 10, 2004
8. Nov 10, 2004

### jcsd

I've made a mistake, that's not the right series. I'm too tired to solve it properly right now, but it's not thta diifcult to treat it as a series.

9. Nov 10, 2004

### jcsd

Of course !

10. Nov 10, 2004

### AKG

This is what I got:

Treat a corner of the box as the origin, and the edges from that corner as the x-, y-, and z-axes. The height of the box is 2R, so it's diagonal is $2\sqrt{3}R$. Now, imagine the plane x + y + z = L. You want this plane such that it just touches the big sphere. By symmetry, it will do so at a point (q, q, q). The distance from this point to the origin is $\sqrt{3}q$. Now, you know that the diagonal of the box should be $2\sqrt{3}q + 2R = 2\sqrt{3}R$, so $q = \left (1 - \frac{1}{\sqrt{3}} \right )R$.

The smaller spheres will have equation:

(x - a)² + (y - a)² + (z - a)² = r², where (a,a,a) is the center (by symmetry) and r is the radius. You know that it touches at the point (q,q,q), so:

3(q - a)² = r² *

If you imagine the sphere "growing" from the point (q,q,q), then it should touch the x-y plane at some point (b, b, 0) (by symmetry). So:

(x - a)² + (y - a)² + (z - a)² = r²
2(b - a)² + (z - a)² = r²
$z = a \pm \sqrt{r^2 - 2(b - a)^2}$

Now, in the area where the sphere "kisses" the x-y plane, we'll have:

$z = a - \sqrt{r^2 - 2(b - a)^2}$ **

and the derivative of z w.r.t. b should be zero here:

$dz/db = \frac{-4(b - a)}{2\sqrt{r^2 - 2(b - a)^2}}$

It's zero where b = a, and we see from this and ** that a = r. Returing to *, we have:

3(q - r)² = r²
3q² - 6qr + 2r² = 0
r² + (-3q)r + (1.5q²) = 0

EDIT: The stuff below was computed wrong (4 * 1.5 = 6, not 9 )

$$r = \frac{3q \pm \sqrt{9q^2 - 6q^2}}{2} = \frac{3q \pm \sqrt{3}q}{2} = \frac{3 \pm \sqrt{3}}{2}\left (1 - \frac{1}{\sqrt{3}} \right )R = R\ or\ (2 - \sqrt{3})R$$

Since r = R is inadmissible, the answer is $r = (2 - \sqrt{3})R$.

Last edited: Nov 10, 2004
11. Nov 10, 2004

### StatusX

I couldn't follow what you were trying to do with all those variables. Keep it simple, all you need is r and R. This step in particular seemed wrong, since z=0 in the xy plane.

All you have to do is put the small sphere at (r,r,r), the big one is at (R,R,R) and use the fact that the distance between them is r+R to solve for r in terms of R.

12. Nov 10, 2004

### AKG

No, it's not wrong, just poorly worded. Think of the sphere's intersection with the plane y = x. You'll get a circle with equation:

(x - a)² + (y - a)² + (z - a)² = r²
2(x - a)² + (z - a)² = r².

We are trying find a and r such that this circle just touches the x-y plane, so if you think of the equation above as a re-arrangement of z written as a function of x, then there should be some point b where z'(b) = 0, and z = 0. So, first we have the equation you quoted, but with x instead of b. Then, find the derivative of it. Then, find the point b where the derivative is zero, so substitute "dz/dx" with "0" and "x" with "b", so we solve and find b (the x value where the derivative is zero) is a. At this point, z should also equal 0, so we'll have:

2(x - a)² + (z - a)² = r².
2(b - a)² + (0 - a)² = r².
2(a - a)² + (0 - a)² = r².
and we get a = r, so the center is (r,r,r)
How do you know that the sphere is centered ar (r,r,r)? Of course, that's what I got, but how do you know that? For some reason, I don't see that immediately.

13. Nov 10, 2004

### StatusX

ok, but the xy plane generally refers to the z=0 plane. And I know the small sphere is at (r,r,r) because its tangent to the x=0, y=0, and z=0 planes. so its center is a distance r from each, measured perpendicular to the plane.

14. Nov 10, 2004

### AKG

What's your point? All I ever said was that the sphere needs to touch the xy plane, I never said that I was considering the intersection of the sphere and the xy plane, and looking at the set of points there. Look at the following:

2(x - a)² + (z - a)² = r².
2(b - a)² + (0 - a)² = r².

b is the point where the derivative is zero, so I substituted x = b. Why did I substitue z = 0 also though? Because for the biggest possible sphere, the derivative should be zero where z = 0, where it touches the xy plane. That's the only relation my solution(which I've corrected; it just had a stupid computational error) had to the xy plane.