A question about pulleys, torque and a DC motor

Hello guys. I have roller blinds at my home that look like the one in the picture below, which I intend to automate using Arduino, a dc motor and 3d printed parts.
I want to add at the bottom of the chain a DC motor driven pulley which I will 3d print, which will move the chain instead of human hands.
The existing pulley of the blinds, like the one in the image, is 6cm diameter.
Currently it takes 7kg force to roll the blinds up when they're at the bottom.
I'm having a hard time understanding the pulley law's that are applicable for the second pulley I want to add, so I can't really plan it currently.
If I will make the second pulley 3cm diameter, (half the large pulley), will it only take 3.5kg (half of the original weight) to lift it now? I'm not sure if it's relevant in this case. I know it also cuts the RPM in half for top pulley.

Another question:
Assuming my motor's torque is X.cm, does the radius of the pulley it runs have any affect of the torque they will have together not in comparison to the other pulley? like if the pulley would be very large for instance, would it reduce the torque?
I hope it's understandable. Thanks

That looks like an IKEA product.
The smaller drum the less motor torque is needed, and the move goes slower.
Tie some bucket to the chain. Poor water in the bucket until the blind is moving. Convert the weight into force. Force times pulley radious equals the torque needed.

Torque of a motor is in Nm (Newton - meters) Yea I know metric :slight_smile:

So this means a torque of:

1 Nm = 1 Newton at a radius of 1 meter.
1 Nm = 2 Newtons at a rad of 1/2 meter ( here 2 * 1/2 = 1Nm)

So all thing being equal a pully of 1/2 the current diameter will have 2 x the torque of a pully of the current diameter.
Note the force in the chain will remain the same.

For electric motors torque varies with motor speed.

To complete your design you need a speed to raise and lower target.

So just making sure if I add a second pulley with half the radius of the top pulley, it will indeed need half the current torque it takes to lift the blinds?

I couldn't understand what you meant to say.

So just making sure if I add a second pulley with half the radius of the top pulley, it will indeed need half the current torque it takes to lift the blinds?

No. The top pulley is irreverent. Your measurement of 7kg** of force is all the info you need, well and the speed of the raising and lowering.

Lets work in different units than Nm. Lets use gram force - cm or gf-cm.
Your measured 7 kg force to move the shade. That = 7000 gram force.

We will look at two cases:

  1. motor pulley radius = 2 cm = 0.02m
    Motor torque will need to be: 0.02 m * 7 kg = 0.14 kgf - m

  2. motor pulley radius = 3 cm = 0.03m
    Motor torque will need to be: 0.03 m * 7 kg = 0.21 kgf - m

Keep in mind the minimum diameter will be that which the chain will wrap around. Some bead chains have a minimum bend radius.

** BTW 7kg force seems like a lot of force to move your shade. I doubt my grandmother could open this shade. You might want to verify this number.

No, a pulley of half the diameter requires half the torque for a given linear force.

Torque = Force x radius.

Torque is proportional to radius.

That's a secondary effect of the winding resistance - motor torque is proportional to motor current, and is independent of speed (unless that leads to current-limiting).

The unit is the newton, Newton is the guy its named after.

Oh please no.... SI makes everything easy, straightforward and almost impossible to make conversion mistakes, its always the best approach!

The answer for 6cm pulley is simply 7 kg x 9.8m/s/s x 0.03m = 2.06Nm That's a lot of torque BTW, a geared motor is indicated.

Pardon my misstatement:

The unit is the newton, Newton is the guy its named after.

Are you by any chance in academia?

SI makes everything easy, straightforward and almost impossible to make conversion mistakes, its always the best approach!

Until you have to imagine 0.03m (without a conversion you seem to be against).

That's a lot of torque BTW, a geared motor is indicated.

I think a re-measurement of the required force is indicated. Why would you sent the OP looking for a 2Nm gear motor for a shade? Clearly some common sense needs to be applied here.

Could you explain why? Why isn't the law of half the radius of second pulley = half the torque & half the speed apply here?

That means I'll need a motor torque of 21kgf.cm..? That's a lot.

Well it's quite big. About 3 meter wide and 2 meter high. I tried pulling it with 4kg weights, one weight didn't pull it, two did quite easily so I went for the worse case scenario.
It's all quite inaccurate, including the test I made with the weights but I believe it's an OK evaluation, and it's for the worse, so I won't be needing any more force than any result we'll reach.

Edit:
Another question please; How does the pulley's radius affect the motor's torque? Let's assume that at given load and current it has a 6kgf.m . how does the pulley's radius, that sits on the motor's shaft, affect the torque?

Slightly over one inch, in those ridiculously antiquated U.S. units.

Imagine what it costs the U.S., the only advanced country in the world refusing to convert to metric, where every mechanic has to have two complete sets of tools, and unit conversion errors lead to the complete destruction of $125 million research projects: Mars Probe Lost Due to Simple Math Error

It's already factored in - your weights were hung on the chain attached to that pulley.

But you had to convert it to something else.... which was my point.

Regarding US converting to Metric. Our idiots in congress cannot get even the simplest thing done.
I'm afraid the conversion will be forced slowly by Global industry. I worked in the automotive industry and they have gone all metric. But I still see new designs having dimensions like 2.54mm.

10 years ago I was working with a GM division in New York. They told me one 3 day weekend they went through the plant, removed every imperial ruler and left a metric ruler.

I was just trying to help you imagine that distance. I work in metric every day.

The reason its not part of the motor/pulley design is it is:

  1. not changeable
  2. your required force was determined with the upper pulley installed (as part of the shade).

So all you need to do is create a force (tension) on the chain of 7kg to move the shade.

I don't know your size limitation but perhaps a stepper motor would work. You could start movement slowly then pick up speed.

Have you thought about end stops?

OK I understand now, thanks.
I'm going for a DC motor with the circuit in the diagram.

Actually the only part I'm not yet sure about (regarding the circuit...) is the limit switches. My intention for the beginning is to set the limits by timing with Arduino. Then when I'll see the project actually works I might make it smarter, I was thinking IR/ magnet sensors.
Maybe you actually meant to ask about limit switches like for the whole circuit to be shorted in case things got out of boundaries and control. For this matter I will probably go for microswitches as seen in the diagram, I'm just not sure how to place them as one of them will have to be far from the setup so that's a long wire exposed. I will probably start without those at the beginning as well.

I didn't see the circuit, was there a link I missed?

You also have to decide what speed will be good.

If you are doing timing for end stops be sure if the motor stalls it doesn't damage your shade. It would also be good if it doesn't damage the motor.

Perhaps you could add a current limit to the motor circuit.

Sorry I now added the circuit below.

How is timing end stops and stalling related?

I'm powering it with only 1A PSU, according to the motor's performance graph (added below) it would run well under that current being able to give a torque and RPM that are enough for my needs in theory.
The motor's stall torque is 5.5A according to the data sheet.
But actually if it wouldn't be able to turn under the 1A it would still stall right? so how would a current limit help with this problem?
I intend to use a 1A fuse though.

It is much worse, but getting better.
For many years, all the hardware stores sold tools and other things with only Metric threaded fasteners, but refused to carry any replacement Metric fasteners. They also sold stuff with metric plumbing, but refused to stock replacements.
I had to use my German made lathe to make brass plumbing conversion pieces to mate the device to American standard threads.
Some of the stores are better, now.
Paul

I agree, but it hasn't been a result of our congress.
My local hardware store has a good assortment of metric nuts and bolts.

Funny thing though. In Germany if you purchase a metric socket set, the drive is American 1/4 inch. I'll guess this is common to most of the metric world.

Consider
Your driver has no current limit, I don't know your power supply has any current limit (likely is does but at an unknown current). Certainly is will not limit current at 1A.

If you accidently enable the drivers in a wrong sequence you will short out the power supply, just a caution.

Electromechanical systems like this nearly always have some failure mode where the cat gets caught in the roll up or whatever. It is wise to design for these events.