Originally Posted by
RollCNY
STA is less relevant, as seatposts have offset. I would look at it as c=√(a²+b²). C is a constant, your leg extension, which does not need to be solved, as it will be the same before and after the change. If a is vertical and b is horizontal from BB, you now have √(a²+b²)=√((a+∆v)²+(b-∆h)²). Square both sides, use foil, and solve for ∆v.
That looks completely correct. Is it not true that the trig approach I used inherently incorporates the formula you provided? You're right about the lack of need for using the STA, but I wonder if the use of it, just as an approximation doesn't simplify the whole process.