Thrust Block Volume Much Greater than Shape?

MeyerFire has a great thrust block calculator that appears to match NFPA 13’s annex/appendix exactly.
https://www.meyerfire.com/blog/a-new-thrust-block-calculator-part-i
 
How can the thrust block volume fit within the dimensions of the thrust block?
 
The default entries, for example, give a thrust block volume of 57.8 ft³ (NFPA 13 Equation A.6.6.1c), a base height of 2.2 ft, and a base width of 4.27 ft.  This meets the bearing area required by the equations.
 
NFPA 13, Figure A.6.6.1(b) gives the angle between the base of the thrust block and the pipe as 45°. This makes the thrust block a pyramid shape but ends with a pipe instead of coming to a tip. The Figure shows both vertical and horizontal angles as 45°, but that wouldn’t work unless the height and width are equal.  If 45° is the minimum angle, then the width of 4.27 ft would control the distance from the base of the pyramid to the pipe. If each side has a 45° angle, this distance would be half the width or 2.135 ft.  These dimensions can be used with the pyramid volume formula to give another method to check the thrust block volume.
 
The pyramid formula is V = L*W*H/3, which gives a volume of 6.7 ft³.  This is very different from the NFPA equation result.   Even if the thrust block was poured as a cube, its volume would be V = L*W*H, which is 20 ft³ -- still much smaller than the NFPA value. – I am counting the tip of the pyramid, which is part of the pipe, not the thrust block. It would be more accurate to exclude the tip, lowering the required volume slightly, but it would be harder to calculate.
 
If the pyramid were 57.8 ft³ with a bearing area of 2.2 ft by 4.27 ft, then it would need a distance from the pipe of 18 feet.
 
I think there is something wrong with Equation A.6.6.1c in NFPA 13.  It doesn’t make sense that it cancels the thrust force with the weight of the fill material. If you could use some incredibly dense fill material (pretend fantasy fill that is 8670 lb/ft³), then the thrust block volume would be 1 ft³ by that formula, but how is that small thrust block volume going to transfer the thrust forces onto a large enough soil surface? 

The weight of the thrust block material should not be related to calculating the volume of fill material required.
 
Another issue I see with the equation is that it is suddenly using T = P*A*sin(theta) while all previous formulas were using T = P*A*sin(theta/2). 

Why did the angle change for this formula?

Thanks in advance for any insight you could shed on the difference in volume versus shape with the NFPA formulas.

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