What is the largest radius for 8 styrofoam spheres in a box?

In summary, the problem is to find the largest radius that 8 styrofoam spheres, placed on the corners of a box with a sphere of radius R inside, can have. The solution involves using a series and treating the problem as a three-dimensional geometry problem. The final answer is r = (2 - sqrt(3))R.
  • #1
VividVoid
4
0
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! ^^
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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
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.
 
Last edited by a moderator:
  • #4
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
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
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:
  • #7
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 [tex]R (2-\sqrt{3})[/tex].
 
Last edited:
  • #8
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
StatusX said:
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 [tex]R (2-\sqrt{3})[/tex].

Of course :cry: !
 
  • #10
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 [itex]2\sqrt{3}R[/itex]. 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 [itex]\sqrt{3}q[/itex]. Now, you know that the diagonal of the box should be [itex]2\sqrt{3}q + 2R = 2\sqrt{3}R[/itex], so [itex]q = \left (1 - \frac{1}{\sqrt{3}} \right )R[/itex].

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
[itex]z = a \pm \sqrt{r^2 - 2(b - a)^2}[/itex]

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

[itex]z = a - \sqrt{r^2 - 2(b - a)^2}[/itex] **

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

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

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 :eek: )

[tex]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[/tex]

Since r = R is inadmissible, the answer is [itex]r = (2 - \sqrt{3})R[/itex].
 
Last edited:
  • #11
AKG said:
Now, in the area where the sphere "kisses" the x-y plane, we'll have:

[itex]z = a - \sqrt{r^2 - 2(b - a)^2}[/itex] **

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
StatusX said:
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.
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)
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.
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
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
ok, but the xy plane generally refers to the z=0 plane.
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.
 

What is the formula for finding the radius of a sphere?

The formula for finding the radius of a sphere is r = √(3V/4π), where V is the volume of the sphere.

What is the relationship between the radius and volume of a sphere?

The radius and volume of a sphere are directly proportional. This means that as the radius increases, the volume also increases proportionally. Conversely, if the radius decreases, the volume will also decrease proportionally.

How do you measure the radius of a sphere?

The radius of a sphere can be measured by using a ruler or caliper to measure the distance from the center of the sphere to its surface. Alternatively, if the volume of the sphere is known, the formula mentioned in the first question can be used to calculate the radius.

What units are used to measure the radius of a sphere?

The radius of a sphere is typically measured in units of length, such as meters, centimeters, or inches. The unit used will depend on the specific measurement system being used.

Can the radius of a sphere be negative?

No, the radius of a sphere cannot be negative. It is a physical quantity that represents a distance and therefore must be positive. If the calculated radius is negative, it is likely that an error was made in the calculation.

Similar threads

  • Sci-Fi Writing and World Building
Replies
22
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
4K
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Quantum Physics
Replies
3
Views
1K
Replies
4
Views
1K
  • General Discussion
Replies
9
Views
2K
Replies
9
Views
2K
  • Mechanical Engineering
Replies
3
Views
2K
Replies
30
Views
561
Back
Top