Originally Posted by 95geo
yes, all of this makes sense but the reason I mentioned the root diameter is because in aluminum im *assuming* the bolt with the smaller root diameter is still stronger than the amount of aluminum that is being captured by the threads.
or is all of this thrown out because the bolt is only as strong as the minor diameter (where it would break) because the threads in the aluminum are stronger?
Yup, you pretty much have it. It would really take a CAD layout to show it & I don't have the time to do it now but think of this assuming mean dimensions & no tolerances for clarity. Do this for both fine and coarse threads of a bolt. Don't worry about the diameter, just one side or radius. Pick any size bolt, they are pretty proportional.
Draw the major diameter.
Draw the minor diameter.
Divide up equally by pitch
Draw the threads (a bunch of triangles) Don't really need to do this but it is easier to visualize
Determine the amount of material "per thread" or "pitch" in a cross section.
Determine the circumfrence of the minor diameter (minor dia x Pi)
Multiply it by the cross section.
Without allowing for the lead angle, this is the amount of material per pitch that you have. Now, multiply this by the number of threads per inch.
See which one has more surface area........ coarse or fine It is not really surface area but cross sectional area. The one with more area is the stronger thread. I'm curious if it is close to 10% difference.
If you want to get really fancy you can also calculate the length of circumfrence with the lead angle factored in but my bet is it will be close to a wash.
I did this whole exercise 15 years ago but don't remember the exact outcome.