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Thread: Volumes

20090224, 13:28 #1
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OK, so this isn't really a puzzle, but I'm sitting here at work stumped on it And I knew if anyone could answer it, the guys @ Woody's could do it
Say I had a right triangle and wanted to rotate it around a diameter where like so (attachment)... I'm looking for a formula to calculate the volume of this where the base, height, and diameter are all variable. So this is basically a chamfer that follows a circular path. Does anyone have any ideas? I'm having trouble finding anything on google that really spells it out for me since I've forgotten all of the calculus I've learned a long time ago.
Thanks in advance!!
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<big>John</big>

20090224, 19:45 #2
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I will let you work out the details and the simplification. It can be done without calculus.
If we are given the 3 dimensions: D, the diameter of the circle, x the base of the triangle, and y the height of the triangle, you will be working with 3 shapes to get the volume of the projected piece.
The first piece is the cylinder defined by the diameter of the circle and the height of the triangle. Its volume is pi*y* (d/2)²
The volume is reduced by the bottom part of an inverted cone whose base has a diameter of d whose height is y and ends at a circle whose diameter is D2x.
The volume of this bottom segment of thhis inverted cone is the volume of the entire cone minus the volume of the cone starting at the D2x portion.
So the top portion of the cone has a radius of (D2x)/2. Its height can be determined by the similar triangles:
Ht / [(D2x)/2] = y/x
Ht = y/x * [(D2x)/2]
Since the volume of a cone is 1/3 * pi * h* r², the volume of this cone is 1/3 * pi * (y/x * [(D2x)/2] )* [(D2x)/2]²
The whole cone has a radius of D/2 and has a height of y + ht of the upper cone = y + y/x * [(D2x)/2]
Thus its volume is: 1/3 * pi * (y + y/x * [(D2x)/2]) * (D/2)²
So the volume of the rotateted triangle is the volume of the Cylinder  the volume of the entire cone + the volume of the top cone:
[pi*y* (d/2)²]  [1/3 * pi * (y + y/x * [(D2x)/2]) * (D/2)²] + [1/3 * pi * (y/x * [(D2x)/2] )* [(D2x)/2]²]
I leave it to you to expand out the terms, collect like terms and simplify...
Steve

20090225, 11:27 #3
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[quote name='sdckapr' post='761252' date='25Feb09 01:45']I will let you work out the details and the simplification. It can be done without calculus.
If we are given the 3 dimensions: D, the diameter of the circle, x the base of the triangle, and y the height of the triangle, you will be working with 3 shapes to get the volume of the projected piece.
The first piece is the cylinder defined by the diameter of the circle and the height of the triangle. Its volume is pi*y* (d/2)²
The volume is reduced by the bottom part of an inverted cone whose base has a diameter of d whose height is y and ends at a circle whose diameter is D2x.
The volume of this bottom segment of thhis inverted cone is the volume of the entire cone minus the volume of the cone starting at the D2x portion.
So the top portion of the cone has a radius of (D2x)/2. Its height can be determined by the similar triangles:
Ht / [(D2x)/2] = y/x
Ht = y/x * [(D2x)/2]
Since the volume of a cone is 1/3 * pi * h* r², the volume of this cone is 1/3 * pi * (y/x * [(D2x)/2] )* [(D2x)/2]²
The whole cone has a radius of D/2 and has a height of y + ht of the upper cone = y + y/x * [(D2x)/2]
Thus its volume is: 1/3 * pi * (y + y/x * [(D2x)/2]) * (D/2)²
So the volume of the rotateted triangle is the volume of the Cylinder  the volume of the entire cone + the volume of the top cone:
[pi*y* (d/2)²]  [1/3 * pi * (y + y/x * [(D2x)/2]) * (D/2)²] + [1/3 * pi * (y/x * [(D2x)/2] )* [(D2x)/2]²]
I leave it to you to expand out the terms, collect like terms and simplify...
Steve[/quote]
apparently it's called a frustum cone.
V = (pi*h(r1^2+r1*r2+r2^2)/3)(pi*r^2*h)
Thanks!<img src=/w3timages/blueline.gif width=33% height=2>
<big>John</big>

20090225, 12:01 #4
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[quote name='officespacer' post='761478' date='25Feb09 11:27']apparently it's called a frustum cone.
V = (pi*h(r1^2+r1*r2+r2^2)/3)(pi*r^2*h)
Thanks![/quote]
I sat down and worked out the simplification of my formula over lunch today:
V =PI/6*y*x*(3*D2*x)
This is in the form of the variables you gave (X,Y, D) and not the radii as you have in your equation. You can substitute your x,y, d for your various r's and it should be the same... [I did work out the calcs and in excel using iterative calculations to sum the rings of various radii (the "calculus approach") and as the number of rings increased, my iterative sum approaches the value from the above calculation]
Steve