Choosing the right stepper motor

Hello,

I am working on a timelapse dolly project and time has come for me to choose a motor.
The rail is about 1.5 m long. the motor is planned to be placed next to the camera (hence - moving motor - not stable).
It will be connected to a pulley which will move the dolly using a timing belt.
The camera + motor + dolly + controller shouldn't weigh more than 5 kg.
Since this is for timelapsing, I need the motor to be able to move the dolly very slowly and ideally be able to pull at 60-70 deg angle.
I have read a little about stepper motors and still, not really sure which motor to pick.
What does the oz.in property mean? What torque do I need?
I am very sorry for bombarding you with all the questions and very thankful for any help.

Thanks a lot!
Amir

This is just a wild guess but I think oz. in. is a measure of the amount of torque necessary to move a mass of 1 oz a distance of 1 inch but I am uncertain if this means vertically against gravity or horizontally accross a surface having a friction coefficient. I will try
to research it further.

Does this help ?

Thanks a lot for the quick reply.
In that case then, do I just need a 180 oz.in stepper motor?
What other characteristics should I look at?
If anyone could recommend a stepping motor that might be useful for my project it'll be great !

Thanks!

raschemmel:
This is just a wild guess but I think oz. in. is a measure of the amount of torque necessary to move a mass of 1 oz a distance of 1 inch but I am uncertain if this means vertically against gravity or horizontally accross a surface having a friction coefficient. I will try
to research it further.

Does this help ?
Measuring stepper motors

You are right to be uncertain, that's not what torque is.

Torque is energy per unit angle, or equivalently tangential force times distance from axis

"oz-in" is actually erroneous as an ounce is not a unit of force, it should more
properly be "ounce-force-inch", but lets drop tedious imperial units, in SI this is
newton-metres, or perhaps more clearly joules/radian.

The calculation of torque is usually multiplying the radius of your pulley or gear by
the force it needs to exert on the belt.

In your setup the dominant forces are frictional so you really will need to measure
them once the mechanics are assembled, then choose an appropriate motor.

Alternatively estimate the worst case friction expected and hope you aren't
too far out. An ignorance factor of 2 or 3 is always worth throwing into the mix.

Note that with a timing belt setup the larger the pulley the coarser your movement
steps, so a small pulley is a good plan. The more flexible rubber/kevlar belts are better
for a light load like this, avoid polyurethane/steel belts.

If by "pull at 60-70 degree angle" you mean on a slope, then the gravitational
forces may be dominant - calculate them and add to your frictional forces.

Do you want to use this to make stop-motion video using a still camera, or do you need to be able to dolly a video camera while it is running? The former is a pretty easy task because you can let the whole rig settle for a second between moving the dolly and tripping the shutter; the latter is much more difficult in terms of vibration and you need very smooth motion.

Anyway, to expand on MarkT's answer, torque defines how much force is available from the motor.
Torque = Tangential_Force * Radius

Say your whole moving platform weighs 5kg: to support it vertically requires 59.8 = 49N of force. Or to support it on a 60-degree slope requires 49cos(60) = 25N of force. Now say you need to accelerate the trolley at ~0.25G = 2.5m/s^2, take F=ma and that gives you a required acceleration (and deceleration!) force of ~32N. So if you want to operate vertically, your motor system needs to be capable of supplying 49+32 = 81N of force.

Now, you choose a combination of pulley size (gearing factor) and motor. Say you use a 15mm pulley, directly attached to the motor, that has a radius of 0.0075m. Use the above equation:
T = F * R = 81 * 0.0075 = 0.6 N-m.

Therefore your motor needs at least 0.6 N-m of torque at all times, which is a bit less than 100 oz-in. Here's where it gets non-obvious: the specification on a stepper motor is holding torque, i.e. torque while stationary. The actual torque available while moving is less, and how much less will depend on how fast the motor is going. Read this introduction to get you started. And remember that torque is mostly proportional to current, so if you're running the motor at half of its max-rated current, you're getting only half the torque.

So being conservative, you could run your system with a stepper motor of about 2N-m holding torque with 15mm pulley and it would probably be OK. Or you could pick a smaller pulley size: going from 15mm to 10mm means you have 1/3 less radius, therefore 1/3 less torque is required to achieve the same force, but you will get 1/3 less speed too.

By "probably" above, I mean you still need to pay (a lot of) attention to your motion planning. You need to plan your motion so that the mechanical load (e.g. due to gravity and attempted acceleration) is not greater than what the motor can provide, otherwise it will skip steps and you won't get the position you expected or it might just halt & catch fire.

Have a google for "grbl", which is some firmware you can load into an Arduino that does stepper control in 3 axes, with text input over the USB port. You command it a position, and it goes there, obeying specified limits (that you choose beforehand) on speed and acceleration. Using grbl with an external PC (or even another microcontroller would be fine) to handle the motion planning (when to move, how far and how fast) would be IMHO an excellent way to implement a camera dolly. Even if you don't use it, I recommend you go and read up on things like grbl because it'll tell you a lot of the things about CNC motion that you will need to know to implement your own system.

Wow! Very helpful information. Thanks a lot.

The dolly will mainly function at Shoot-Move-Shoot move therefore a smooth movement is not very important.
What is, in your opinion, the best way to calculate the friction?

Again, many thanks. You guys helped a lot.

There seem to have been an awful lot of words to explain a simple thing like torque.

Think of a rotating shaft with an arm sticking out from it. A torque of 1 oz-inch is the force caused by a 1 oz weight connected to the arm 1 inch from the centre of the shaft. For the same force it would only be necessary to have a 0.5 oz weight if it was connected 2 inches from the shaft.

And, yes, SI units (Newton metres) make life much easier.

...R

Torque.png

An ounce is indeed a unit of force.

Thanks guys. I do know what torque is (I study EE), I was just confused since we don't use ounces in Israel.

Friction is not easily predictable, make the mechanism and measure it. Hopefully
your inertial and gravitation loads are dominant though, so you should have a reasonable
guide from them.

BTW the force to accelerate 5kg at 2.5 m/s^2 is 12.5N, not 32N.

A 60 degree slope means using sin(60), not cos(60), so its 42N for the gravitational
load. You might as well go for lifting the thing vertically anyway!