# Need help choosing a motor.

1. Apr 17, 2012

### beeohbee

Hi everybody,

I'm making a machine at home which will cut a predetermined length of ribbon from a reel. I've designed it all but I'm not sure what sort of spec motor to use.

I'll be using a microcontroller to take the users input and a sensor will sense the size of the wheel. The microcontroller can then calculate when to cut the ribbon so that it's the correct length.

Essentially, a motor will drive a reel of ribbon which will be held vertically on a horizontal shaft. The reel will weigh no more than 2kg and is about 0.3m in diameter. I'm happy for it to rotate at any speed between 1 and 5 rps so I've modelled it on 3 rps.

I've calculated the torque by modelling the wheel as 2kg in mass and assuming that that weight is located at a distance of 0.15m from the shaft. I used that to find the power needed, which is roughly 55W.

Could anybody advise me on the sort of motor which would be useful in this situation? I'm not sure if it would be best using a high torque DC motor at about 12V or if an AC motor would be better. I can use a relay to switch an AC circuit on but I don't know if that would be even more over the top. The types of motor are a little over my head at the moment so any help is appreciated.

I know it's a completely over-engineered project but I thought it would be a good exercise to educate myself on electrical systems and electronics.

Cheers,

Rich

2. Apr 17, 2012

### DragonPetter

I think a DC motor is easier to use for position control.

I'd look at stepper motors, since they are simple to control with a microcontroller. They also will make it easier to stop where you want it to stop, where overshoot from the inertia of the spinning reel could occur.

I'm not so sure about your torque and power calculation, it seems really high. Is your calculation equivalent to finding the torque/power to swing a 2kg ball on a .15m rope? That would be intuitively different from rotating 2kg distributed radially about an axis which has a moment of inertia that you would calculate. I think you'd need to use an integral to find the total torque as the mass is distributed radially from 0 distance to whatever the maximum radius of the wheel is, assuming a rigid and homogenous material. How did you get your power number from the torque calculation? Were you assuming a 3 rps angular velocity?

Edit: see the cylinder as an example of your reel http://en.wikipedia.org/wiki/List_of_moments_of_inertia to compute torque. If you are rotating horizontally (around the z axis), such that gravity is not playing a part, you will need very little torque to spin the reel at a constant rate (just enough to overcome friction on the motor's rotor), and only more torque as you accelerate or deccelerate, which you will use this angular acceleration for the torque calculation.

Last edited: Apr 17, 2012
3. Apr 17, 2012

### DragonPetter

I just thought of a second component you may need for your torque calculation, depending on how precise you want your spec to be.

Above, I'm only talking about the torque of the moment of inertia, its calculated like:
$T = \alpha I$ where alpha is the angular acceleration (hence, at a constant rpm you will ideally need 0 torque, but practical losses of the motor require torque to keep it moving).

The other torque component will be from the force the ribbon exerts on the reel. Assuming I have the right image in my head of what you are building, there will be a tangent force to the wheel that is based on the tension of the ribbon. This is calculated the easy way since it is tangential (90 degrees, so cross product turns into pure multiplication):
$T = F*d$

4. Apr 19, 2012

### beeohbee

Thanks for your response. I can see now that my calculations were very naive! I had based my calculations on the equivalent of a 2kg particle on the end of a string as you thought.

From your help, I have recalculated the moment of inertia, assuming the reel to be a cylinder with a height of 0.04m, a radius of 0.15m and weighing 2kg. This comes out as 0.00288kg.m^2

To find the torque now, I need to use T=aI and to find the angular acceleration (a) I have used w=w0 + at.
Assuming that the reel starts at rest, so that w0=0: the angular acceleration must equal the angular velocity I require divided by the time it takes to reach that speed.

If I model the system so that the motor turns the reel at 2rps, the angular velocity is equal to 12.566 rads/sec. I am assuming that it takes a second to reach this speed, so the angular acceleration can be assumed to be equal to the angular velocity.

Therefore, with an angular acceleration of 12.566 rads/sec^2: the torque is equal to 0.036Nm or 36mNm.

This seems like a reasonable torque to move that mass to me. Although, this doesn't take into account the component of force that the ribbon exerts on the reel. Can you clarify this for me?

To make it clear, what I'm attempting to build is a machine which holds a reel of ribbon in the same plane as a car tyre. It would be driven similarly, in that it is placed on a driven shaft.

5. Apr 19, 2012

### PajoTheDwarf

Don't base anything on the rotation of the reel, and for reasons of reduced power requirements without needing to have it centered precisely on the axis of rotation, put the reel on a vertical axis rotating in a horzontal plane.

Assuming that the ribbon is not made of tissue paper and can handle the strain, have the ribbon wind around a a feed roller and pull the it off of the reel.

The reel might be able to rotate freely or can be driven by PWM DC motor with a tension sensor (possibly two stage) as the control input.

I second the recommendation that you use a stepper motor for the "measurement" which is one reason that you want a precise diameter of feed roller delivering the ribbon to the cutter.

To allow using a smaller cheaper stepper I would use a large feed roller on a similarly large pulley with the stepper motor driving it from a small pulley. That will also give you unbelievable precision in your measurements since it will take a few thousand steps to turn the feed roller once, and multiply the torque.