Difference between revisions of "Stepper Motors"

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(Created page with "== Holding Torque == In our example configuration, we use NEMA23 motors with a holding torque of 9000 gf·cm — that is, 0.88 N·m in SI units, or 125 oz&mid...")
 
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=== How much torque ''is'' that? ===
 
=== How much torque ''is'' that? ===
With the supplied 18-tooth MXL pulleys,
+
With 18-tooth MXL pulleys,
 
one motor can hold a carriage against a force of about 15.5 kgf applied to
 
one motor can hold a carriage against a force of about 15.5 kgf applied to
 
it (that's 152 N, or 34 lbf).  This is at standstill, with the motor supplied with its rated current.
 
it (that's 152 N, or 34 lbf).  This is at standstill, with the motor supplied with its rated current.
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as the speed increases, the torque decreases almost linearly:
 
as the speed increases, the torque decreases almost linearly:
 
at half the maximum speed, the available torque will be about half the holding torque.
 
at half the maximum speed, the available torque will be about half the holding torque.
 +
 +
With 20-tooth GT2 pulleys, the force is about 14.1 kgf (139 N, or 31 lbf).
  
 
=== How much torque do you need? ===
 
=== How much torque do you need? ===

Revision as of 15:42, 16 September 2014

Holding Torque

In our example configuration, we use NEMA23 motors with a holding torque of 9000 gf·cm — that is, 0.88 N·m in SI units, or 125 oz·in in customary (US) units.

How much torque is that?

With 18-tooth MXL pulleys, one motor can hold a carriage against a force of about 15.5 kgf applied to it (that's 152 N, or 34 lbf). This is at standstill, with the motor supplied with its rated current. The torque remains almost constant at low speed, but after a certain point, as the speed increases, the torque decreases almost linearly: at half the maximum speed, the available torque will be about half the holding torque.

With 20-tooth GT2 pulleys, the force is about 14.1 kgf (139 N, or 31 lbf).

How much torque do you need?

The force the motor applies to the carriage must be enough to counteract all friction, the inertia of the moving parts (when accelerating), and the cutting forces on the tool (when milling). Even though 15.5 kgf is more than enough for this type of machine (and more than MXL belt is normally rated for), having a motor that powerful is still useful at higher speed, when its torque decreases. It is the available torque that limits the traverse acceleration and speed, the maximum cutting force achievable, as well as the feed rate for a given cutting force. That said, there's no point in going for a much higher torque than that. The motors become too big and too heavy for this type of machine.

What determines motor performance?

  • The type of motor. Generally, all things being equal,
    • 1.8° motors are faster and more powerful than 0.9° motors;
    • smaller motors are faster but less powerful than bigger motors;
    • motors with lower inductance (higher rated current/lower rated voltage) tend to be faster, and often more powerful too.
  • The driver:
    • more current capability can move the motors faster;
    • some advanced drivers use more complex techniques that can improve motor performance.
  • The controller:
    • more advanced movement algorithms may allow the motors to go faster, or at least use their acceleration capability more effectively;
    • a slow processor may limit the maximum speed.
  • The power supply voltage. Using a higher supply voltage:
    • can allow the motors to move faster;
    • can increase torque at medium and high speed (but makes no difference at slow speeds);
    • may decrease the accuracy of microstepping.

Does microstepping reduce torque?

No, it doesn't, but it's a common misconception. TO DO: add link to the Shapeoko forum, where I explain this at length.

How did we calculate the force?

This is how we arrived at the 15.5 kgf figure:

Belt and pulley pitch:

MXL = 0.08 in/tooth = 2.032 mm/tooth

Pitch circumference of 18-tooth pulley:

2.032 mm/tooth × 18 tooth = 36.576 mm

Pitch radius of 18-tooth pulley, which is also the arm of the force the motor applies to the carriage:

36.576 mm / 2π ≅ 5.82 mm

Motor holding torque:

9000 gf·cm = 90 kgf·mm

Force exerted on carriage:

90 kgf·mm / 5.82 mm ≅ 15.46 kgf