I'm trying to find the formula for calculating the tensile stress area of a bolt. According to page 1528 of The Machinery's Handbook (29th Ed.), that formula is given as

$$ A_S=\frac{\pi}{4}\left(\frac{d_m+d_p}{2}\right)^2 $$

Where the pitch diameter is $d_p=d-0.649519\times P$, and the minor diameter is $d_m=d-1.299038\times P$.

I was trying to understand where the definitions of $d_p$ and $d_m$ come from. The Wikipedia page for the Unified Thread Standard (link) was very helpful for understanding that. However, it is not clear where the equation for $d_m$ comes from. The value 1.299038 is equivalent to $\frac{3\sqrt{3}}{4}$. Translating this to be in terms of $H$ instead of $P$ results in $d_m=d-2\cdot\frac{3}{4}\cdot H$. In other words, it would appear that the boxed dimension in the drawing below (versions of which appear on pages 1807, 1912, and 1973 of the handbook mentioned above) should be $.75H$ instead of $.625H$.

Basic Profile of Screw Thread

I'm assuming I must be missing something obvious here because I'm seeing variations of the formula for $A_S$ listed above everywhere. The best I can figure is that the equation for $A_S$ assumes that the outermost 1/8th of the V is included in the nominal diameter of the bolt. However, I can't find any justification for this. If someone could explain it to me, I would greatly appreciate it.


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