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The motor power required relies on a lot of factors, and some of the most important ones it looks like you haven't decided on yet (although if you have, we'll need to know to help you).
The force required to move something, at a minimum, is (ideally) equal to the coefficient of friction multiplied by the normal force (gravity*mass) of the object you're trying to move. What is the weight of your gantry? Calculate this in pounds. Try to overestimate, not under. Then look at the specifications for the linear bearings you intend to use. I imagine they come with specifications on their friction, along with maximum load etc. which you'll want to check.
Saying your mu (friction coefficient) was 0.05, and your gantry was 100lb, the needed force would be 5lb-force. Now you need to look at what is supplying that force - a lead screw of sorts, let's say a ball screw (smaller diameter screws will not work at this size very well, FYI). Find the efficiency of this screw from its documentation. A good ball screw has an efficiency of around 90% at minimum, so let's say 0.90. Divide your needed force by this (5lb-force / 0.9 = ~5.6lb-force). Now you need to do calculations on the force supplied through the ball screw (before losses due to efficiency) by the stepper motor.
Imagine a ball screw of diameter 2". That means the point at which force is being supplied by the motor at a radius of 1". Motor torques are rated usually, in our case, in ounce-inches. This means it is the number of ounces the motor could move if attached an inch away from the center of the motor's axel (as you can imagine, if you put the weight on a lever 20" long, it can only hold 1/20th the weight. Holding something on the end of a long beam will be more difficult than on a short, as it works as a lever, with the fulcrum at the point you're holding it).
Anyways: Now you take the required force, and take into consideration the number of turns the screw must undergo to move the gantry a single inch. Say there was 2 turns/inch. That means each turn of the screw 360 degrees (2*pi radians) advances your gantry by half an inch.
You multiply the needed force by the distance the gantry progresses per turn (0.5") and divide it by the radius that the torque acts on (1"). Remember that one turn is 2*pi radians. All physics formulas work in terms of radians (it's a lot simpler), as when you multiply a radius by a distance in radians you get an arc length, and so on, without any need for conversion. Since the force is acting over the diameter of the circle that is the cross-section of the ball screw, you divide the number you have so far by 2*pi to get the equivalent torque needed for your linear force.
So finally, that's your 5.6lb-force * 0.5" (inches/turn) / 1" (radius of ballscrew) / (2*pi radians) = 0.44lb-inches or just over 7oz-in. (Remember, we started with pounds, not ounces, so we have to convert!)
This number is the amount of force required to _keep your gantry in constant velocity_. That means applying this amount of force won't move it anywhere unless it's already moving. So your motor needs to be rated a good bit above this. How much? That depends on the acceleration you want. Go back through all the calculations above, except this time leave out mu (friction coefficient). This gives us the torque required to accelerate the gantry at 1G (9.8m/s^2) after friction is taken care of. So that's 7oz-in / 0.05 = 140oz/in. So to accelerate (from rest) at 1G you'd need a stepper capable of ~148oz/in.
These are all best case numbers, so now you have to add a bit of fudge room. The faster a motor turns, the lower torque it can output (in general, for steppers). Steppers are usually rated for their _holding torque_, that is, the torque they can apply at stall (zero motion, simply holding a weight, fighting some force, etc. without any turning). As soon as your gantry starts to move, the motor is speeding up and losing torque. Very quickly the acceleration you started with goes down to 0.5G's or 0.25G's. You'll need a full data sheet on a motor to figure out exactly the rate at which its torque decreases in relation to its speed (If you know enough calculus: You can solve dV/dt = (torque - friction)/mass for V (conversion of units is very likely necessary) and watch its limit as t goes to infinity to find the ultimate top speed of your router).
I'd say at least double the calculated required torque for the acceleration you want, and remember, in reality you'll never reach that acceleration due to all manners of losses of energy. Even if you could, it'd only last for a small fraction of a second, as the rate of change of acceleration is decreasing over time.
(Hopefully someone can double check my math?)
