Two-Beam Formula
You have a weight (Gross Weight of Aircraft) suspended on 2 beams (the mains and nose wheel).
1. A is the nose wheel weight, and B is the weight on the mains, GW is gross weight.
2. a (little a) is the distance from where the gross weight is centered (aka the CG calculated using normal methods) to position A.
3. b (little b) is the distance from CG to position B.
Then the weight on A is the ratio b/(a+b) of the weight of GW. And the weight on B is the ratio a/(a+b) of the weight of GW.
So using their sample from the graph on 14-10, and distances from the datum on 14-8, I get this:
Let a = (CG - Nosewheel location) = (80.48 - 39.11) = 41.37
Let b = (Mains location - CG) = (93.96 - 80.48) = 13.48
A = GW * (b / (a+b)) = 1463 * ((13.48) / (41.37 + 13.48)) = 359.54
Which is pretty close to their graph of 362# on the nose wheel (and they fudge by saying CG was estimated).
If you do another sample, and choose places on the graph where lines intersect to try and be more accurate, I get:
GW = 1350
CG = 78.7
A (aka weight on NW) = 1350 * (15.26/54.85) .... = 375.58# on NW, corresponding exactly to the limit on the chart at that loading/GW.
The theory is correct, but I might have mistyped something in this post, so check my work