# Calculating required torque of stepper motor for manipulating blinds

I have a outdoors wooden window blinds and currently I am raising/lowering them manually - by pulling/releasing a rope that's hooked on them using pulley. It's the most basic setup but still I can't understand how the calculation of torque works.

I know the formula is Torque = Length x Force and Force is gravity pull (mass x 9.81 N*kg-1) (+ some friction). But I am not sure what "Length" is in this case. And where in this formula is considered the motor shaft diameter and step size. From what I understand the torque should be bigger when using microsteps.

So, what is the correct formula in this setup to estimate required torque for the step motor to be able to pull the blinds all the way up and back?

• Length is the pulley radius. Jan 8 '21 at 23:09
• You might want to change "weight" to "mass". (It will help clarify your own thoughts on the matter.) Watch your units. 'Kg' is kelvin-grams (nonsense) where you mean 'kg'. (Fixed.) SE supports superscript <sup>...</sub> and subscript (but not in the comments). Jan 8 '21 at 23:16
• The pully is fixed (not rotating) and its radius is 0.35cm. I am not sure about the mass of the blinds though. I have tried to stand on the scale while pulling the rope and it lowered my mass displayed on scale by ~10kg (when the blinds were completely down). So if we calculate that with 10kg for mass and 0.35cm for length, the result is 31.7835Ncm. Does that seem real? And can you clarify, why the length is the pulley radius? from the wiki image en.wikipedia.org/wiki/Pulley#/media/File:Polea-simple-fija.jpg it does not seem to be depended on that. Jan 9 '21 at 0:03
• if I think about it, even if the pulley was 1 meter in radius and I did one step on the motor , the rope would still get pulled by the same length (depending on motor shaft radius) Jan 9 '21 at 0:06
• but maybe I am thinking about different pulley. In the inside, before the rope enters window, there is a small 0.35cm radius fixed pulley. (maybe the pulley is a strong word - it's just a stationary cylinder) It "bends" the rope by 90 degrees and the rope continues to the inner window and outside, where there is probably some other pulley mechanism that does the blinds rolling. Jan 9 '21 at 0:15

Two things.

• What your 0.35cm pulley/wire is doing is just redirecting the tension in the rope. You need a separate pulley to wind the rope around.

• I see something is wrong in the dimension of your pulley.

let's say your blind is 180cm long and the rope touched the pulley 4cm away from its support at each end, and is lifting 5kg/ half of the weight of the blind. A 0.35 cm wire can support the following moment.

• I= moment of inertia

• s= section modulus

$$I = \frac{πr^4}{4}=0.00073cm^4, \quad and\ S=\frac{I}{C}=0.0042cm^3$$

$$M_{max}= Fy*S=2500*0.0042=10.5kgcm$$

$$5kg*4cm*2=40kgcm$$

It means your wire/pulley should have bent by now.

Ignoring that Your pulley here is the pulley attached to the motor, or to the wall and say it is a 5cm radius pulley, then you need a torque

$$\tau= 0.05*10kg*9.8*2_{safety-factor}=9.8Nm$$

• The 0.35cm pulley I was describing in the comments is not important at all - as I understand now, it's just used to change the direction and has no impact on torque. Sorry for misdirection. Apart from that, there is currently no pulley - just a 0.5cm diameter (probably nylon) rope and ~2m of outdoor wooden blinds that weights around ~10kg. Thanks for the formulas. Jan 12 '21 at 13:36
• So if understand, what matters most is the radius of motor shaft (or something that I am winding the rope around) - the smaller radius, the less torque I need. But as the rope is winding around itself (on the motor shaft), the radius is getting bigger and bigger the more I wind up. Which means that required torque is getting bigger too. Is that correct? And if that is true, I somehow need to ensure, that the rope would not wind around itself, but it needs to wind symmetrically along the motor shaft, that needs to be long enough. Jan 12 '21 at 13:36
• @Tomik, yes. but the safety factor of 2 we put there should cover that. Also, you can use a wider pulley. Jan 12 '21 at 14:44

To raise the blinds, you will need a motor and a drum/pulley of some sort to wind up the rope. The act of winding up the rope will exert the force required to raise the blinds. When you pulled on the rope to raise the blinds, it decreased your measured weight by 10kg (these must be some major blinds, mine are more like 10oz (US based blinds) that would be 0.34kg on your scale).

For the sake of calculation, I'll say we have a drum that is 40 cm in diameter (since these are monster blinds, you'll need a big drum). That would be 20 cm (or 0.2m) in radius. The torque required will be:

(0.2m) * 10kgf*9.81N/kgf = 19.6Nm

As you observed, the torque is the force applied times the "length" (the radius from the center of rotation from where the force is applied).

• Why the big drum? A small one would reduce the torque requirement. Jan 10 '21 at 8:51
• @Transistor The big drum because they have a big rope for 10kgf worth of blinds. Realistically, I just supported myself with some 3mm nylon line I got for my blinds (the record will show that I am not a small person). If we went with a 2cm drum, that would be (0.01m)*10kgf*9.81N/kgf=~ 1Nm but it will require a lot more winding. Meanwhile, I was mostly trying to show how to calculate the required torque. Jan 12 '21 at 5:09
• The radius of my rope is 0.25cm. I was hoping to get a small stepper motor with similar radius, but as you say, it would require much of the winding and the radius would get eventually bigger anyway by winded rope ... So probably it would be easier with some bigger drum as you recommend,maybe small stepper motor + some gears. And yes, pulling the blinds (~2m in height) really lowered my weight around ~10kg, but of course in these 10kg there is already a friction accounted, since I measured it but pulling rope, so I have no idea what is they real weight, but luckily that's not important now. Jan 12 '21 at 13:53