Ratchet handle force calculation

I assume by "open" you mean to release the strap. The force required to tighten the strap is determined just by the tension force in the strap and the ratio between the radius of the "spool" and the length of the handle.

There are two pawls: one in the moving handle, and one in the stationary part (you can see them in your diagram as the doohickies in the slots). Each is held in place by a spring. In order to release the strap, you pull a trigger that disconnects the handle pawl from the ratchet it normally engages, and rotate the handle until it disengages the other pawl. This, it does by means of a camming action: as you twist the handle around, the pawl is forced up a "ramp" (increasing radius). If there were no friction, the amount of force required to do this would be zero. So, to estimate the amount of force required, you need at least four things: the force between the ratchet and pawl, the coefficient of friction, the slope of the "ramp", and the ratio between the length of the handle and the radius of the ramp.

These are total guesses, but let's say the force between ratchet and pawl is 300 lb (it's less than 500 because the radius of the spooled webbing is less than that of the ratchet). Let's further say that the coefficient of friction is $0.25$. Then the outward force required to shift the pawl (neglecting the small amount of force needed to overcome the spring) is 150 lb (there will be a frictional force of 75 lb resisting the motion on each of the two faces of the pawl). Let's further say that the slope of the ramp is 1 (distance pawl is pushed out is equal to the tangential distance moved by the ramp), and that the ratio of handle length to ratchet radius is 5. Without friction, you'd need to apply a 30-lb force to the handle. If we make the assumption that the frictional force opposing the motion of the handle is dominated by the contribution from the pawl, (and that the coefficient is still 0.25) the amount of force required would end up at 40 lb.