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Solid Geometry Nerd Assistance Please
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ghporter
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Mar 17, 2019, 08:52 AM
 
I have a geometry problem that I can’t let go. And so far I can’t figure out how to research it, because I seem to lack the appropriate terms for what I’m looking for.

Let’s say I have a larger cylinder that has a cylindrical hole through it at right angles to its axis and through the full diameter of the larger cylinder. Let’s say the larger cylinder has a radius of 10 and a height of 20, while the cylindrical hole through it has a radius of 1.

I need to find the volume of the cylindrical hole. The actual, precise volume. Here’s a picture of a similar situation, but with a larger hole through the main cylinder:


I can estimate it as “less than the volume of a solid cylinder through the larger one with a height of 20”, which is pretty easy to calculate, but that’s not precisely the volume of the cylindrical hole.

It seems to me that what I lack is some means to find the volume of the small, cylindrical segment where the hole “emerges” from the large cylinder. I may also need to find the “complimentary” volume - the difference between a penetrating cylinder (as in my estimate above), and just subtract, but still, that funky cylindrical volume is a puzzle. I don’t even know what to call that funky volume (and maybe this is the crux of the problem).

Any hints would be greatly appreciated. And telling me that I should have taken more calculus won’t help - I already know I am woefully lacking in calculus.

Thanks!

Glenn -----OTR/L, MOT, Tx
     
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Mar 17, 2019, 06:16 PM
 
So what you do is integrate over the length of the hole. Find a way to figure out the formula for the cutout as a function of the position. At y=0, this is easy: x2+z2=1, the area of a circle pi*r*r, r=1, ergo the area of the circle is pi. For as long as the hole is a cylinder - as in, you have not yet reached the edge of the solid cylinder - it remains the same. Once you reach the edge at a point we call y=a, that area will decrease with the curvature of the solid cylinder, and the area is 0 at y=10.

Then you set up the integration. For y goes -10 to 10, integrate f(y) dy. f(y) is going to be the constant pi between -a to a, so this part of the cylinder is just 2a*pi. The other bit is the interesting part. If your function for the area is A(y), you integrate A(y) from a to 10, and finally remember to double it to include the bit from -10 to -a.

So to find the function A(y). I think that the cutout will be an ellipse with a long axis of 1, and a short axis that is the third side of a straight triangle with the other sides being y and 10. That makes the side sqrt(10^2 -y^2), so A(y)=pi*1*sqrt(10^2-y^2). Not entirely sure though, I’m doing this in my head because I have no paper nearby, but I think it works out.
The new Mac Pro has up to 30 MB of cache inside the processor itself. That's more than the HD in my first Mac. Somehow I'm still running out of space.
     
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Mar 17, 2019, 06:37 PM
 
Forgot one point. a, the point where you transition from the straight cylinder to the interesting shape, is going to be the point where y^2+1^2=10^2, or put differently, y=sqrt(99), or y=9.95. If the figures in your example are the actual figures, just skip the complex math and just assume a cylinder of length 20 and radius 1, because the difference to that is going to be tiny.
The new Mac Pro has up to 30 MB of cache inside the processor itself. That's more than the HD in my first Mac. Somehow I'm still running out of space.
     
reader50
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Mar 17, 2019, 08:14 PM
 
The height of the large cylinder (20 units) is noise, to distract you from solving the problem. So long as the height is greater than the hole diameter, you can ignore the height.



Diagram of main cylinder seen from above, to scale. I extended the hole as a 2nd cylinder for illustration purposes.

The hole has a cross-section area of pi*r^2, or pi (square units) in your problem. You can obtain an approximate volume by multiplying by the length of the hole. An exact solution will require more trig than I'm comfortable with, and/or some calculus. The curved intersect between the two cylinders is the hard part. Below is an approximate solution - which should be quite close. Maybe close enough if this is a practical problem. If it's part of your kids' calculus homework, it won't be good enough.

The red lines all extend from the intersecting axis. Draw a vertical line on right edge, at right angles to the center red line. That gives you a pair of right triangles in red. You only need one to solve the problem.

For minimum hole length, draw the vertical line so it intersects the upper & lower red lines, where they meet the surface of the cylinder:

Minimum length of the hole becomes 2x a triangle with hypotenuse=10, short side=1, long side=X. Apply Pythagorean theorem, X=sqrt(100-1) or sqrt(99). Minimum hole length is 2x sqrt(99) or ~19.899 units. Multiply by hole area, to get minimum hole volume = 62.5169 cubic units.

Maximum length is simpler - the diameter of the cylinder - 20 units. So maximum hole volume is pi * 20 = 62.8319 cubic units.

Split the difference, and you get an approximate hole volume = 62.67 cubic units.
     
ghporter  (op)
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Mar 17, 2019, 09:31 PM
 
P, I see what you're doing, but I can't work it out myself. Not nearly enough of the integral calculus that I sat through was taken in, let alone stuck. (First and only math class taught by a TA... I learned more about his masters thesis - something about filters of infinite porosity - than about integration.) Just how "tiny" is that difference? If you can give me some sort of probable error, I might be able to short cut to the final formula you give.

Reader, I absolutely agree that the height of the large cylinder is useless for the main problem. I just threw it in for completeness. I think your approach is pretty close to the useful answer, but I'm not sure how close is "close enough." See above.

Glenn -----OTR/L, MOT, Tx
     
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Mar 18, 2019, 05:38 AM
 
I figured that you’d give the integral to some sort of numeric solver, because doing that stuff on pen and paper is a drag. I think Wolfram Alpha can probably do it?

Anyway: reader’s analysis is spot on. The difference between the minimum and maximum solution is 0.5%, so the error you make with either of them can never be more than 0.5%. An average between them will thus be within 0.25% of the right answer. That’s pretty close.
The new Mac Pro has up to 30 MB of cache inside the processor itself. That's more than the HD in my first Mac. Somehow I'm still running out of space.
     
ghporter  (op)
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Mar 18, 2019, 08:48 PM
 
0.25% is definitely "close enough" for me. Thanks both of you.

Glenn -----OTR/L, MOT, Tx
     
   
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