Mathematical brains required - the space between 4 spheres

peteruk

peteruk

peteruk

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Hello Forumites,

I'm currently doing a modelling assignment regarding the use of radiation treatment on tumours, but for my own research to help I'm looking at the tessalation of spheres.

If I place 3 identical spheres together, touching and at 120 degrees, I know that the shape within the middle resembles a triangle but with 3 separate quadrants cut out. After some studying I can work out the area of this shape.

However, I concluded trying to calculate the volume bound between a plane at the top's of the spheres and another at the bottom was particularly hard, so to make it easier I changed the experiment to include a sphere placed on top of these 3, so we now have something that resembles a tetrahedron made up of 4 spheres.

The shape inside these 4 spheres (bound by a border drawn by connecting all the midpoints of the spheres) I now understand to be a tetrahedron-like shape, but each side has the deformed-triangle shape mentioned previously. I'm now unsure how to calculate this volume, although considering the curvature of it I will require some kind of integration.

Anyone with a vivid imagination and a black-belt in geometry, how would I go about doing this?

Many thanks to any help :)
Peter
 
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I'd try joining the centres of the circles to get a tetrahedron. Minus from the volume of that the volumes of the portions of sphere. I'm not sure quite how to get those, but perhaps investigate placing the tetrahedron inside a cube, working out the %volume and then placing the section of sphere inside the cube, doing the same and comparing.

Sorry that doesn't make much sense, it's late. :(
 
Im confused with the mixing up of 2d and 3d shapes in op, but here is my effort:
mathy.jpg

You need to first calculate the area of the triangle (diameter of circle squared /2 )
Then you need to take away the area of half of one circle. (pi*radius)
which should give you the area of the gap between the circles.
 
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OK, for the version with 4 spheres you need to work out the area of the pyramid you get if you join up the centre points (1/3 * area of base * height)
then you need to take away half the volume of a sphere (pi * radius squared)

edit: no, the second bit is not right. You might need to take away a quarter of the area of a sphere. Im having trouble visualising it. :(
 
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touch said:
Im confused with the mixing up of 2d and 3d shapes in op, but here is my effort:
mathy.jpg

You need to first calculate the area of the triangle (diameter of circle squared /2 )
Then you need to take away the area of half of one circle. (pi*radius)
which should give you the area of the gap between the circles.
Click to expand...

That was the starting of my thinking, and I can work out how to get this area. :)

But, if I then put a sphere on top of those 3 spheres and create a shape within the gap with 4 of these sides, how would I calculate the volume of this? :eek:

Thanks for posting!
 
touch said:
OK, for the version with 4 spheres you need to work out the area of the pyramid you get if you join up the centre points (1/3 * area of base * height)
then you need to take away half the volume of a sphere (pi * radius squared)
Click to expand...

This simple?

Beautiful. Thank you.
 
Wow, most "maths questions" on this forum are depressingly simple, but this is really quite difficult. I congratulate you on posing a question I can't simply answer, and I have a degree in maths.

Sphere packing is not something I have studied, so the best I can do is point you to this Wikipedia page. I hope that helps.

Edit: I thought you were trying to find the volume enclosed by the spheres? And it's just occurred to me that 4 spheres packed together wouldn't actually enclose a volume, so curse my brain-fart. Listen to touch.
 
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peteruk said:
This simple?
Click to expand...

No, sorry. I have edited my post.
I'm sure it's similar to the way you calculate the 2d version but its not half a sohere you take away. Im having trouble imagining it but i think it may be a quarter.
 
peteruk said:
However, I concluded trying to calculate the volume bound between the top's of the spheres and the bottom was particularly hard,
Click to expand...

If my quick working out is correct then this is actually very easy:

A(triangle)= (BxH)/2 B=diameter of sphere H=Bcos30=B(sqrt3/2)

A(triangle)=(sqrt3xB^2)/4

V(prism)=HxA(triangle)=(sqrt3xB^3)/4

V(sphere)=(4/3)pi.r^3

Three spheres with 60 degrees of their volume are taken out of the volume of the prism, or the equivalent of half a sphere

V(half sphere)=(4/6)pi.r^3

V(prism) - V(half sphere) = (sqrt3xB^3)/4 - (4/6)pi.r^3 where B=2r

(2r)^3=8r^3

so:

V(prism) - V(half sphere) = (sqrt3x8r^3)/4 - (4/6)pi.r^3


V(prism) - V(half sphere) = sqrt3x2r^3 - (2/3)pi.r^3

Which should give you the area of the shape i think you are looking for.
 
SlyReaper said:
Edit: I thought you were trying to find the volume enclosed by the spheres? And it's just occurred to me that 4 spheres packed together wouldn't actually enclose a volume, so curse my brain-fart. Listen to touch.
Click to expand...

Yes, you are right, I didn't explain what I meant too well: Imagine drawing a border by joining up all the midpoints of the spheres, so you have a tetrahedron shape, but of course only some of it is 'empty space'.

That link you've presented is extremely useful however, hexagonal - rather than the triagonal packing I am currently attempting - appears to be where my thoughts should head once I've put this to bed.
 
jak731 said:
If my quick working out is correct then this is actually very easy:

Wall of maths

Which should give you the area of the shape i think you are looking for.
Click to expand...

I'll follow this through in the morning, if you've managed to simplify it, I'll be most impressed. :)
 
Thanks for the advice so far everyone; I've edited my original post to make it a little clearer what I'm after. I'll get back on this in the morning *stretches*
 
jak731 said:
Three spheres with 60 degrees of their volume are taken out of the volume of the prism, or the equivalent of half a sphere
Click to expand...

This is where i got stuck. There are 4 spheres. I guess that it will be equivalent to a quarter of a sphere but i have no idea how to work it out correctly.
 
touch said:
This is where i got stuck. There are 4 spheres. I guess that it will be equivalent to a quarter of a sphere but i have no idea how to work it out correctly.
Click to expand...

The op said he thought the 3 sphere problem was too difficult so tried it with 4 instead, i think the 3 sphere is trivial but the 4 sphere is harder : /
 
Well using some crude calculations and a bit of modelling, I can estimate that the section of each sphere, created by the regular tetrahedron between the centers of the spheres is 1/24 of the volume of the sphere.

So, to find the empty space in that tetrahedron, you simply need to find the total volume of it, and remove 4/24, or 1/6 of the volume of one of the spheres.

Assuming your spheres have a radius of 5cm:

The volume formula for a regular tetrahedron with an edge length of A is sqrt(2)/12*A^3

So sqrt(2)/12*10^3 is 117.85cm^3, which is the volume of the entire main tetrahedron.

The volume of each sphere is 4/3pir^3. 4/3 * pi * 5^3 is 523.6cm^3

523.6 / 6 is 87.27cm^2, which is the volume of the 4 sections of each sphere added together.

117.85 - 87.27 = 30.58cm^3, which should, hopefully be the volume of the empty space inside that tetrahedron.

I'm afraid I can't offer anything concrete, this is mostly guesswork that appears to give a correct answer. Hope it helps though.
 
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