# Drilling 3 orthogonal holes through a cube

I have a solid cube with its dimensions 1 unit in length. I have drill-bit of 1 unit in diameter.
I drill a hole through the centre of each face through to the other side.
When I have finished I am left with 8 identicle corner shapes.

Can anyone tell me what the volume is of one of these corner bits?
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Commented:
That page says that the intersection is an octahedron and that the volume of the resulting solid (that is, of the three cylinders) is (16 - sqrt(128)) r³.  Then, the volume left is:
8r³*[sqrt(2) - 1] =
For r = 1/2 you get a residual volume of sqrt(2) - 1 = 0.4142...
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Commented:
The intersection of the three holes is a sphere of diameter 1, because it is the only solid which has circles as projections along the three axes.

The volume of the pieces left is: volume of the cube, less 3 x volume of a cylindrical hole + 2 x volume of the sphere (because we want to count it only once).
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Commented:
I was convinced of what I said in my previous comment, but the numbers don't work out: if the edge of the cube is 2*r, the volume calculated as I said is (2r)³ - 3*(PI*r²*2r) + 2*(4/3 *PI* r³) = 8r³ - 6*PI*r³ + 8/3*PI*r³ = 8r³*(1 - PI*5/12).

With r = 1/2, the volume left is 1 - PI*5/12 = approx -0.309, but volumes cannot be negative!

Sorry, I have to think about it a bit longer...
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Commented:
Actually, the intersection is not a sphere.  The page http://astronomy.swin.edu.au/~pbourke/polyhedra/cylinders/
contains the solution you are looking for
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Author Commented:
PointyEars,

Yes, it would be quite impressive to see a perfect sphere being cut with just 3 shots of a circular drill saw.

What I really want is to see how the formula arrived at.

I tried to work it out for myself but got into a bit of a mess with the 3 dimensional integration.

JR2003
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Commented:
Well, your question asked for the value, not how to derive the formula...   ;-)

My clock shows 02:04 AM.  I'll have to go to bed now...
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Author Commented:
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Commented:
Thank you JR2003.

I did try to calculate the volume of one of the 8 pieces left.  To visualise it, draw the three axes as follows:

|  y
|
|*
| *
|   *
|      *
+----------*---------  x
/
/
/
/
/  z

The z-axis should be at 45 degrees.  I have also drawn the hole intersection with the xy face.  Consider the radii to be == 1.  At the end, you only need to multiply the result by r³ to have the generic volume formula.

The three circles have equations:
(x - 1)² + (y - 1)² = 1
(x - 1)² + (z - 1)² = 1
(y - 1)² + (z - 1)² = 1

If you draw the other two holes and imagine the three cylinders, you will see that they meet in a point that, looking at the drawing, should be a bit higher and on the right of the origin.  That point, being the intersection of the three cylinders, has all three coordinates = 1 - sqrt(2)/2.  Let's call this point C (C for cusp) and indicate with "A" the value 1 - sqrt(2)/2.

The parallel to the Z axis passing through C intersects the circle on xy in the point Pxy = (A,A,0).  Similarly, the parallel to the Y axis through C intersect the circle on xz in the point Pxz = (A,0,A).  The parallel to the Y axis through Pxy and the parallel to the Z axis through Pxz meet on the X axis in the point Px = (A,0,0).

Now, the four points C, Pxy, Pxz, and Px form a square of side A.  The volume of the pyramid formed by the square itself and the origin is given by A³/3, because the height along the X axis is precisely A.

Now we have to calculate the other funny shape left for x comprised between A and 1.  This is defined by the X axis and by three curves: an arc of circle on the xy plane, an arc of circle on the xz plane, and the arc of intersection between the two cylinders with axes perpendiclar to the xy and xz planes.

Now it is time to resolve (x - 1)² + (y - 1)² = 1.  It gives y = sqrt(2x - x²) - 1.  Similarly,
z = sqrt(2x - x²) - 1.

The intersections of the funny shape with planes parallel to the yz plane are all squares delimited by points with coordinates
(x,0,0)
(x,sqrt(2x - x²) - 1,0)
(x,0,sqrt(2x - x²) - 1)
(x,sqrt(2x - x²) - 1,sqrt(2x - x²) - 1)

The area of these intersections is therefore given by (sqrt(2x - x²) - 1)².  To calculate the volume you only need to integrate (sqrt(2x - x²) - 1)²dx for x between A and 1.  You then add A³/3 and you have the volume of 1/3 of one of the 8 pieces left after drilling the holes.

The integral should be calculable (but I am a bit rusty with those).  Whether the final result is
8*(1 - PI*5/12)
I cannot really say.  sqrt(2x -x²) = sqrt(1 - t²) with t = x-1.  You can then possibly set t = sin(a) and get rid of the sqrt.  The PI in the result will probably come out of an arctg (or arcsin or arccos) function.
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Author Commented:
Part 2 of the question is now open, if you want the points you need to put answers in there:

http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_21095962.html
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