Direction Connection of Stepper Motor to Load

Stalling due to overload. Loss of load control with loss of power. Unwrap of cable(?)due to speed of step.

The weight will drop when the motor is turned off.

Not really, if the current stays on the rated holding torque will stay the rotation. Lets see, that is around 28.6 lbs and you don't want a reduction gear? A NEMA 34 can do it, but it is a heavy load.

You don't say what size stepper, nor how rapidly you need to lift the weight!

13 kg has a weight of 140 Newtons. The largest standard stepper from my source (NEMA 34 HT) has just under 4 N-m of torque. 4/140=0.028m radius, so as long as the center of the cable lifting the weight is closer than 2.8 cm from the center of the drum winding the cable, it can theoretically do the job with NO margin of error.

However, as someone else already pointed out, the instant you have a drop in, or loss of, voltage, the weight is going to fall to the bottom.

If you use a worm gear with a much smaller motor and driver, not only will it be cheaper than the larger motor and larger driver with no gearing, but it will be intrinsically safe: the friction in the worm gear will at least cause the weight to fall slowly in case of loss of power, and very likely prevent it from falling at all.

You will also have much finer position control, if that is required.

single-start worms are non reversible

I presume you are referring to 'reversible' in the sense of power flow. The friction in the system makes it easy for the single-start worm to turn the worm gear (in either direction, so it is reversible), but torque on the worm gear cannot cause the worm to rotate. Thus the intrinsic safety in terms of the weight not falling upon loss of power.

single-start worms are non reversible - cannot be back-driven.

As you have written it, your comment is untrue.

I know what you wanted to say, so do you agree you should correct your comment?

If you think of it in terms of the maximum output power of the motor multiplied by the efficiency of the lifting mechanism equals the rate of change of potential energy of the load being lifted, then you will succeed, Mildred.