Quote by FeX32
To answer the original post then since you don't like trig.
Solving the equations i wrote in the other post. If you are interested I rearranged them for [itex] \frac{\dot{d}}{\omega_2} [/itex] and then used the equivalence of this equal to Torque/piston force. Pluggin in your angles gives 862.5 ftlb of torque at the crank.
Not that this is with 1000lbs at the piston. Also note that this is independent of the length of the coupler link (rod), only dependent on the angles. I also assumed that your angles were all positive and starting in the first quadrant.
Cheers,

FeX32,
Not that I don't like trig. Just don't know enough of it. Many Thanks!!!
ed.