Im doing some cad for something Im about to 3d print. I have a sphere made from a circle of diam 19.4mm rotated about its center axis. I want to start on the edge of the circle, make a second circle, make a sphere from it by rotating about its center axis, and then subtract the second sphere from the first (the larger one) as you can see from these images:







(the 3rd image is a cross-section view so you can see what is going on better)

I want to determine the size of the diameter in the smaller circle in the first image (the one that is labeled as diam = 3 mm such that the volume of the section that is removed from the larger circle after the subtraction is finished is equal to 1.23 cubic centimeters.

How can I go about determining the diam such that that will happen? Or more simply, what is that diameter?

http://mathworld.wolfram.com/Sphere-...ersection.html

Equation 16 is the one that you want. Take R = d = 19.4 mm, and solve for r.

You can solve this by vector calculus of even basic calculus using shell or disk integration if you want to see why the above is the result. But you can also do it by a computer simulation if you think about it.

Also make sure you use radius not diameter in the above formula for r,R,d.


Also are you sure about your numbers? Is 19.4mm diameter or radius? And is it really 1.23 cm^3 or 1.23 mm^3?

Volume is 1.23 cm^2 im sure

Diameter is 19.4 mm for sure

You realize then that 19.4mm diameter means 9.7mm radius and then the volume of all of it is 4/3*Pi*9.7^3=3822mm^2 or 3.822 cm^3 which means your solution is not a small sphere on top of a big one but nearly two of similar size intersecting each other. Only then you have such big fraction of its volume represent the intersection of the 2 spheres.

Didnt realize that, no. I can settle for 1/2 of the smaller spheres volume, or even a quarter. As long as it is evenly divisible by my desired volume it will work.

Can you explain a little bit more what you just said or what the overall objective here is?

This is the disk method solution to your problem by the way

https://www.wolframalpha.com/input/?...D,%7Bx,y%7D%5D


https://www.wolframalpha.com/input/?...F2%2FR,0%7D%5D

https://www.wolframalpha.com/input/?...3D%3D1230,r%5D


Change the last equation 1230 to any volume you want in mm^3 and the radius R 9.7 to anything and repeat the solution link with pressing enter. 9.82mm

Thanks. And thanks for all the replies, I have exactly what I need to solve the problem now.