I'm no expert, anyone know how 1.87Nm would go?
This one's probably been answered but for everyone else...
You can easily figure out the torque you require on the shaft, just need a very simple equation.
All you need to do is test out your manual hand crank first. If you can figure out how hard you need to push on it then you can easily find out the rest. You can do this by hanging a known weight off the handle when it's horizontal, or simply push on the handle with a scale and take the reading.
The required torque on the inputshaft in Nm is simply: T = P * L * 9.81 where
P = Equivalent hand force on crank, how hard you push in kilograms
L = Length of crank (Centre of shaft to centre of handle)
Say you find it takes 5kg to push down on a 300mm long hand crank, then the torque required at the input shaft to crush grain is:
T = 5kg * 0.3m * 9.81 = 14.7 Nm
This would be how much torque you would need at the input shaft after gear reduction.
If you want 100 RPM at this torque then you have Required Power in watts = (RPM * Torque) / 9.55.
Take required power and divided by motor rated RPM and multiply by 9.55 will give you required motor torque before the step down. Ie
P = (100 RPM * 14.7Nm) / 9.55 = 153 W (motor power)
If i have a motor that runs at 600 rpm then the required motor torque is:
Tmotor = (153 * 9.55) / 600 = 2.435 Nm
BTW in this case you would have needed a 6:1 reduction so
600 RPM / 6 = 100 RPM and 2.435 Nm * 6 = 14.7 Nm
Don't forget to leave a bit up your sleeve. Hope this makes sense as I'm rushing it during my lunch hour.
Cheers,
Rob
