"I have a hard time believing 2.5g weight is giving only 1.5g outside force, especially as it attaches 1/2" forward of the pivot."
Please consult the Wikipedia page on levers. The 2.5g weight is going to provide much less than 2.5g of side force precisely BECAUSE it attaches to the arm wand so close to the pivot. We are concerned with the force generated at the other end of the arm wand (the lever), at the stylus tip. Work is Force (F) acting over a distance (s). W = F*s. According to laws of conservation of Work, the Work done by the dropping 2.5g weight must equal the Work done on the stylus tip (with minor losses due to friction). So, notice that the length of the arc traversed by the arm wand at its attachment to the string, as the string and weight drop, is much, much shorter than the length of the arc traversed by the stylus tip as it goes across the LP surface. Call the two distances s for the AS device and S' for the stylus tip. Because Work or Energy is conserved, F*s = F'(the AS force on the stylus tip)*S'. Because s<<S', then F'<<F. The force at the stylus tip must be much lower than 2.5g.
Please consult the Wikipedia page on levers. The 2.5g weight is going to provide much less than 2.5g of side force precisely BECAUSE it attaches to the arm wand so close to the pivot. We are concerned with the force generated at the other end of the arm wand (the lever), at the stylus tip. Work is Force (F) acting over a distance (s). W = F*s. According to laws of conservation of Work, the Work done by the dropping 2.5g weight must equal the Work done on the stylus tip (with minor losses due to friction). So, notice that the length of the arc traversed by the arm wand at its attachment to the string, as the string and weight drop, is much, much shorter than the length of the arc traversed by the stylus tip as it goes across the LP surface. Call the two distances s for the AS device and S' for the stylus tip. Because Work or Energy is conserved, F*s = F'(the AS force on the stylus tip)*S'. Because s<<S', then F'<<F. The force at the stylus tip must be much lower than 2.5g.