My requirements would be a 26" plate (instead of 20") and I need to hold and rotate ~25 lbs. The weight is distributed along the edge of the plate so I’m not sure how that translates to stepper motor torque requirements. The torque for a single point load I think would be 13in * 400oz = 5200 oz.in. That seems very high in comparison so does anyone have advice on this?
I did see there is a Big Easy Driver now that can drive up to 125 oz.in stepper motor. However the only thing I’ve found about motor and driver sizing is to make sure the max current draw can be matched by the power supply.
I was also surprised to see a friction fit gear wheel pushed onto the motor shaft. I see [Actobotics which should solve that problem for me with set screws and D shafts but I only see relatively small gear wheels. Did I miss something?
Hopefully I didn’t miss a blog post or tutorial that explains all this but I’ve been reading for a a few hours now and haven’t found anything that ties all the pieces together for me. My Arduino experience is limited to a small project a few years ago so I have plenty to learn.
To estimate the required torque, you need to know both the mass distribution (which determines the moment of inertia I), and the required angular acceleration alpha in radians/second^2. Torque = I*alpha
Since that won’t take into account bearing friction, you might just as well build the turntable, load it up and measure the torque required to spin it up to speed.
dschneiderch:
My requirements would be a 26" plate (instead of 20") and I need to hold and rotate ~25 lbs. The weight is distributed along the edge of the plate so I’m not sure how that translates to stepper motor torque requirements. The torque for a single point load I think would be 13in * 400oz = 5200 oz.in. That seems very high in comparison so does anyone have advice on this?
That would be the minimum torque required if the motor was mounted horizontally and the load hung vertically, sort of like a winch. It also assumes that the load isn’t centered.
Your turntable is the opposite with the motor mounted vertically and the torque wheel is horizontal. This is good because you can let a lot of the load rest on a bushing or bearing or other friction-reducing device. It’s sort of like the difference between lifting a 200 pound boulder vs pushing it around in a cart.
The particular torque needed will be difficult to estimate and depends a lot on how fast you want it to rotate from a stopped position. You might be able get some mechanical help with gears or belts.
As a worst case estimate, the moment of inertia of a plate with diameter 26 inches (radius .4 m), and mass 25 pounds (11.4 kg) distributed evenly on the rim is 4.6 kg-m.
To rotationally accelerate that wheel from zero to 10 degrees of rotation per second (0.17 radians/second) in 10 seconds gives angular acceleration 1 deg/sec^2 or 0.017 radians/sec^2.
The torque required is 4.6 kg-m*0.017sec^-2 = 0.08 N-m torque = 11 oz-in torque
As a worst case estimate, the moment of inertia of a plate with diameter 26 inches (radius .4 m), and mass 25 pounds (11.4 kg) distributed evenly on the rim is (M*R^2) = I = 1.8 kg-m^2. That is, a loaded wheel with assumed frictionless bearings.
To rotationally accelerate that wheel from zero to 10 degrees of rotation per second (0.17 radians/second) in 10 seconds gives angular acceleration alpha = (10 deg/sec) / (10 sec) = 1 deg/sec^2 or 0.017 radians/sec^2.
The torque Ialpha required is 1.8 kg-m^20.017sec^-2 = 0.03 N-m torque = 4.2 oz-in torque
If the mass is not concentrated around the rim, the torque required will be less. If you want the wheel to come up to speed faster, the torque required will be more.