Torque and efficiency of a power screw

Notation: $W$ = axial load, $d_m$ = mean diameter, $p$ = pitch (lead $L$ for single start), $\alpha$ = helix angle $= \tan^{-1}\dfrac{L}{\pi d_m}$, $\mu = \tan\varphi$ = thread friction coefficient, $\varphi$ = friction angle.

One turn of the thread unwrapped is an inclined plane of angle $\alpha$ carrying the load $W$; the nut is pushed by a horizontal force $P$ at the mean radius.

                    /|
       P ------>   / |  W (load)
          ______  /  |
                 /   |
        reaction R at friction angle

Raising the load

Friction opposes the motion and acts down the plane. Resolving forces on the load:

$$\begin{aligned} P &= R\sin(\alpha+\varphi) \\ W &= R\cos(\alpha+\varphi) \end{aligned}$$ $$P = W\tan(\alpha+\varphi) = W\,\frac{\tan\alpha + \tan\varphi}{1-\tan\alpha\tan\varphi}$$

Torque to raise:

$$T_r = P\,\frac{d_m}{2} = W\,\frac{d_m}{2}\tan(\alpha+\varphi)$$

Lowering the load

Now friction acts up the plane:

$$P = W\tan(\varphi - \alpha), \qquad T_l = W\,\frac{d_m}{2}\tan(\varphi-\alpha)$$

If $\varphi > \alpha$, $P$ is positive: an effort is needed to lower the load. If $\varphi < \alpha$, $P$ is negative: the load falls by itself (overhauls).

Self-locking

A screw is self-locking (holds the load without a brake) if

$$\varphi \ge \alpha \quad (\mu \ge \tan\alpha)$$

Efficiency

Without friction, $\varphi = 0$ and the torque is $T_0 = W\dfrac{d_m}{2}\tan\alpha$. Therefore

$$\eta = \frac{T_0}{T_r} = \frac{\tan\alpha}{\tan(\alpha+\varphi)}$$

Efficiency is $\dfrac{\text{useful work}}{\text{input}} = \dfrac{W L}{2\pi T_r}$. For a self-locking screw, $\alpha \le \varphi$ gives $\eta \le \dfrac{\tan\varphi}{\tan2\varphi} = \dfrac{1-\tan^2\varphi}{2} < 50\%$.

Collar friction

If a thrust collar of mean radius $R_c$ and friction coefficient $\mu_c$ is used, the additional torque is $T_c = \mu_c W R_c$, and total torque $T = T_r + T_c$.