A bolt is a screw whose job is to stay put. A power screw is one whose job is to move: a jack lifting, a vise clamping, a press ram advancing. Unwrap one turn of its thread and you are looking at a ramp, as long as the circumference $\pi d_m$ and as high as the lead $l$. Turning the screw slides the load up or down that ramp, against friction.

Raising: $T_R = \dfrac{F d_m}{2} \left( \dfrac{l + \pi \mu d_m}{\pi d_m - \mu l} \right)$, read aloud T-R equals F d-m over two, times l plus pi mu d-m, over pi d-m minus mu l. $T_R$ is the raising torque in newton-metres, subscript R for raising. $F$ is the axial load on the screw, in kilonewtons here. $d_m$ is the mean thread diameter in mm, subscript m for mean: the major diameter less half the pitch, which is where the thread force acts. $l$ is the lead in mm, the distance the nut travels in one full turn. On a single-start thread the lead equals the pitch; on a two-start thread it is twice the pitch, and the relation always wants the lead. $\mu$, the Greek letter mu, is the thread friction coefficient, a bare number near 0.10 to 0.15 for lubricated steel. With $F$ in kilonewtons and $d_m$ in millimetres the torque lands straight in newton-metres. Use $\pi$ as 3.14159, never as 3.

Lowering: $T_L = \dfrac{F d_m}{2} \left( \dfrac{\pi \mu d_m - l}{\pi d_m + \mu l} \right)$, T-L equals F d-m over two, times pi mu d-m minus l, over pi d-m plus mu l, with subscript L for lowering. Same letters, two signs reversed. Going up, the load and the friction both fight you, so they add on top. Coming down, the load helps and only friction resists, so the lead is SUBTRACTED from the friction term.

That subtraction is the most useful line in the lesson. If $\pi \mu d_m$ is bigger than $l$, the lowering torque is positive: you have to push to bring the load down, and if you let go it stays. The screw is self-locking, which is what a jack is bought for. If the lead wins, $T_L$ comes out NEGATIVE, and that is an answer, not an error: the load drives the screw backwards on its own, and it needs a brake. The boundary is $\mu = \dfrac{l}{\pi d_m}$, friction against the slope of the ramp.

Two honest caveats. These are the SQUARE-thread forms, and an Acme or trapezoidal flank raises the friction term by a few per cent. And this is thread torque only: a screw turning against a plain thrust collar pays a second friction bill there, often as large as this one. A self-locking screw is never better than 50 per cent efficient, and that is the price of holding.