Your argument seems valid to me. Dan goes deep in his calculations and does it right determining the barrels deflections and moments of Inertia.
Explaining all of that in the link will take some studying and a lot more than can fit in a post. No way around it.
A simpler way of looking at it, is through Moments (or shear Force). M = r x F where M = moment r = distance from static point (for us the upper receiver) F = Force (for us at the end of the barrel).
So we assume a 10lb Force for ease of calculations and a 20" barrel has a moment of 200 lb in and a 12.5" barrel has a moment of 125 lb in. If the contours are identical I believe this will be effective for demonstrating the stiffness of a barrel compared to another barrel.
Explaining all of that in the link will take some studying and a lot more than can fit in a post. No way around it.
A simpler way of looking at it, is through Moments (or shear Force). M = r x F where M = moment r = distance from static point (for us the upper receiver) F = Force (for us at the end of the barrel).
So we assume a 10lb Force for ease of calculations and a 20" barrel has a moment of 200 lb in and a 12.5" barrel has a moment of 125 lb in. If the contours are identical I believe this will be effective for demonstrating the stiffness of a barrel compared to another barrel.
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