# Calculating complex volume

Discussion in 'Math' started by pebe, Mar 16, 2006.

1. ### pebe Thread Starter AAC Fanatic!

Oct 11, 2004
628
3
Now here's a poser.
Take a cube of material with length of each side = 30mm, ie. volume=27mL

Drill a 10mm dia hole from the cenre of one face right through the material. Call that the X axis.

Now drill two similar holes in the Y and Z axes - again in the centre of the faces so all three holes intersect.

Question: What volume of material is left?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Starting with a volume of 27000 mm^3, I obtained an answer for the final volume once you drill the three cylinders of 20978.6142 mm^3.

My answer is based on my thinking that the intersection of three cylinders of equal length and diameter is a sphere. That would mean that having subtracted the volume of three such cylinders from the cube, I would need to add back the equivalent of two of these spheres to prevent double dipping.

I sure hope you have the answer.

Even if I am wrong it was fun to give it a go.

hgmjr

Apr 14, 2005
7,050
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4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Good find Ronh. That at least confirms what I already suspected about the likelihood of an error in my simplification.

hgmjr

5. ### pebe Thread Starter AAC Fanatic!

Oct 11, 2004
628
3
Good find. I was given this problem many years ago. I established the shape of the common part of the intersection of two holes. Its projection on two axes was a circle and on the third it was a square. I likened if to a football made up of only four panels each shaped like a Zulu's shield.

When it came to the third intersection I realised the shape has more facets on it, but my limited spacial capacity prevented me from imagining its true shape.

Now I'll try to follow the math.