Calculating rotational motor power, to drive linear output

I have to build a motor to turn the front wheel steering of a small ATV. I would like 150 pounds of force, as fast as possible, 10+ inches / second. My goal for 150 pounds of force is from me taking a fishing scale, attach to steering arm and pulling the wheel at a dead-stop in moderately heavy sand - as my control test, so that I know my device would be strong enough to turn in harsh terrain.

Here is a picture of a the ATV:

There is no passenger. I’m building an autonomous R/C rover - so it’s going to have payload of lead batteries. I believe the strongest and fastest way would to create a leadscrew mounted to a motor, drive the steering arms. I can also just bolt a motor with a chain or pully to the steering shaft, except I think the amount of gears and pully’s it would take would use more space, than a rotation/linear solution. To go buy a pre-made linear servo assembly can be expensive, I want to build it with all my spare parts I’ve been salvaging over the years. I am not concerned with the feedback positioning at this point, I just want to make the motor/actuator.

I know how to calculate the rpm -vs- the screw pitch, will give me the speed. But I am confused on how to pick out my motor, test it’s lifting strength to drive the screw. There is information online, except - that I am dumb and when they start with the math and go on for paragraphs to explain about physics that I never learned, the information is not useful.

I know I can mount a stick with a weight to the shaft of a motor and spin it. Motor will stalls to give me AN ESTIMATE of the strength of a motor. Where I get confused, is the length of the stick with the weight, since the longer the stick acts like a lever and changes the rotational force.

The stick and weight is measuring torque. Torque is measured in units like foot-pounds or newton-meters. It is (exactly as you suggested) a force times a distance. 1Nm can be achieved with 1N (a very small force roughly equivalent to the weight of 100grams or 4 ounces) at the end of a 1-meter long lever. Or you can use a smaller lever like 0.1m and 10N force.

If you're trying to stall out a motor, then measuring this the simple way can be difficult. If you have just a tiny bit less than the stall torque then it will rotate your lever around until it's going down quite quickly, flinging your carefully-balanced weights all over the room. The not-so-simple way is to load up the arm so that the motor cannot turn it all the way to level. Start with the arm hanging down and measure how far it gets from straight down. Now you could measure the angle and do the trigonometry but if your weight is hanging on a string, just measure the horizontal distance from the string to the centerline of the motor. There's still a lot of errors in this method, but it gives you a good way of comparing motors.

Ok, let's back up a little bit. How about I replace the dangerous stick on a motor idea. That's too dynamic, with the length / weight of stick, and its probably dangerous.

Plan B, for my motor test-platform: I will bolt shackle-pully on the roof 10' - 25' high. I lace some fishing line, light string, or thin lifting wire. I have a bobbin or a spool mounted to the motor, with the string. The string goes up to the pully on roof, and then down to the floor connected to a weight (1 pound). I can easily just tie different weights to lift off the ground.

Now, I activate the motor bolted to my bench, it will wind the string up and pull the weight off the floor. With plan b: I do not have 1) Dangerous swinging sticks & weights, 2) wobbly centrifugal force against earth gravity 3) Eliminate the trig / geometry for stick length, etc...

So - if my motor can lift 1 pound, then I add weights - and find that it starts stalling at 1.5 pounds; is THAT the TORQUE of this motor ? Or do I need to measure the distance it can raise a known weight value?

Why not simply get hold f a linear actuator to drive your steering arm, such as one of these : http://www.ebay.co.uk/itm/DC12V-Electric-Linear-Actuator-Motor-Heavy-Duty-225lb-Stroke-2-4-6-8-10-12-/152702962975?var=&hash=item238dce4d1f:m:mlSAHxDBxk9_JNuxcI6JdOQ You'll find that static turning effort on any stationary vehicle is many times greater than that when the vehicle is actually moving (static stiction versus rolling friction) so I think your 150lbs force is way over the top.

What’s the diameter of the bobbin / spool? What units of torque are you wanting, inch-pounds, foot-pounds, Newton-meters, kg-meters?

The actuator on the link is 12mm/second. Its way too slow, that is grampy speed. They make stronger and faster ones, that get expensive. I have a bunch of motors that have all different speeds and voltages. Trying to use what I have, and better my understanding of calculating torque.

I did not think about the SPOOL diameter at first, but now realize - the radius of the spool, would be the same math as the stick with a weight. It is easier and precise for me to actually measure a spool diameter. If a 1" radius can lift 1 pound of weight 25 feet in 1 second, I could process the data and solve the math.

You can estimate the stall current from the motor resistance and the nominal supply voltage.

The motor constant can be determined from the speed it runs, unloaded, at nominal voltage. Motor constant is measured in Nm/A or V/(rad/s), its the same value in either.

So lets say your motor is 12V, 1ohm, then the stall current is 12A. If it runs at 3000rpm at 12V, thats about 300rad/s, so the motor constant K = 12/300 = 0.04, so its 0.04Nm/A, ie the stall torque must be 12A x 0.04 Nm/A = 0.48Nm. You won't get more that that torque from this motor at 12V.

Motor torque is roughly proportional to the volume of the motor, note, so that if you want high torques you either need a large motor, or (often much better) you need reduction gearing which trades speed for torque (at the expense of some mechanical losses).

The limiting factor for the size of motor is the power - product of speed (in rad/s) and torque (in Nm), gives watts directly.

So the starting points are desired torque and desired speed. From this you can determine motor power, and the gear ratio needed.