Equation of ball screw force given torque

 

For a given torque $T$ and radius $R$, the perpendicular force $F_1$ equals $\frac{T}{R}$.

If we unroll the thread of the screw and look at the right triangle that it forms, in order to take into account its slope, we graphically see that the relation between $F_1$ and the longitudinal (axial) force $F_2$ is the same as $L$ and $2\pi R$. The angle of both right triangles is the same, and so are the proportions. Intuitively, the shallower the angle, the higher the force $F_2$ is for a given force $F_1 = \frac{T}{R}$. The wedge effect is amplified.

Therefore: $\displaystyle \frac{F_2}{F_1} = \frac{2\pi R}{L}$

$$F_2 = T \frac{2\pi}{L}$$

Taking into account an additional possible gearbox and an efficiency factor:

$$F_2 = T \cdot \frac{2\pi}{L} \cdot GR \cdot \eta$$

$$T = F_2 \cdot \frac{L}{2\pi \cdot GR \cdot \eta}$$