On Thu, 2 Aug 2012, cogoman wrote:

> Date: Thu, 02 Aug 2012 23:57:14 -0400
> From: cogoman <cogo...@optimum.net>
> Reply-To: "Enhanced Machine Controller (EMC)"
>     <emc-users@lists.sourceforge.net>
> To: "Enhanced Machine Controller (EMC)" <emc-users@lists.sourceforge.net>
> Subject: [Emc-users] .9 degree vs. 1.8 degree stepper used in RepRap
>
>   I was looking up something and came across a thread from a gent in
> Australia who was wondering if he could buy .9 degree steppers for his
> RepRap and get them to work, and one guy commented on stepper torque
> with a half truth for the situation.  Apparently the norm for RepRap is
> 200 steps per rev at 1/16th stepping, and he was checking out going with
> 400 steps per rev at 1/8th stepping.
>
>   I was hoping you'all could check my work before I put in their
> forum.  So,... without further ado...
> ********************************************
>
> --------------------
>
> That is a common misconception. If you set the current on a
> micro-stepping driver to give the same power dissipation you get the
> same torque. I.e. set the single coil on current to 1.4 times the two
> coil on rating.
>
> --------------------
>
> This is somewhat true, but not for the case presented. With a bi-polar
> motor, when running both coils at the same current in full step mode you
> /are/ limited to .707X amps. When you go to 1/2 step with .707X amps
> driving both coils, the next step will have one coil at 1.0X amps, and
> your 1/2 step torque will be roughly the same.
>
> Now the trouble comes when you use 1/4 stepping. If you raise the single
> coil current to 1.414X, you will degauss the magnets, or burn up the
> windings or both. If you keep the current set to .707X amps, your next
> step will be (according to the table in the A4988 data sheet) either:
>
> from 70.71% to 92.39%
>
> OR
>
> from 70.71% to 38.27%
>
> this will be a change of .217 amps OR .324 amps. If you ran the motor
> full step at less than 1/3 the current it was designed for, would you
> not get less torque out of it?
>
> At 1/16 stepping the current changes in increments of:
>
> 100% to 99.52% to 98.08% to 95.69% to 92.39% to 88.19% to 83.15% to
> 77.3% to 70.71% ... to 9.80% to 0%
>
> I believe the biggest change in current here is from less than 10% to
> 0%. How much torque would you get running the motor full step at 10% of
> the rated current?
>
> You don't have to do the math if you Google up the answer. Search for
> "micro-step torque" and you can come across this link:
>
> http://machinedesign.com/article/microstepping-myths-1009
>
> This article lists the torque available from full step to 1/256 step torque.
>
> 100.00% 70.71% 38.27% 19.51% 9.80% 4.91% 2.45% 1.23% 0.61%
>
>   He doesn't take into account that the single coil driven current can
> be 1.414 times the both coils driven equally current, so you can
> multiply the lower torques by 1.414 to get:
>
> 100% 100% 54.11% 27.59% 13.85% 6.94% 3.46% 1.74% .86%
>
>   This still shows that your CURRENT-PER-STEP drops when you go beyond
> 1/2 step, and is significantly lower at 1/8 step. At 1/16 step, this
> loss of torque is very serious for a milling machine or a router. For a
> filament deposition 3d printer it's probably noticeable, but not very
> serious.
>
>    The problem presented here begs to have the difference between full
> step and micro-stepping put into perspective. Let's take an example of
> the same machine, the same stepper motor, and two different methods of
> getting the same resolution.
>
> Solution 1: run the motor full step, but use a 16 to 1 (backlash free)
> gear ratio to get 16 time the resolution.
>
> Solution 2: run the motor direct drive, and use a 1/16 step stepper driver.
>
> Solution 1 has full torque for every step, while for solution 2 there's
> less than 15% of the full torque available for each step. If there's a
> lot of static friction in solution 2, small movements may require 4 or 6
> steps before the head moves at all. In a typical 3d printer you will
> break out of static friction when movement starts, and the head will
> continue moving while laying down your plastic. While the loss of torque
> with solution 2 fighting against the dynamic friction may keep the head
> behind the commanded position by a few steps, the movement per step
> should be fairly close until an axis turns around. X and Y should be
> pretty close to what you'd expect. Z typically is the opposite. Z
> usually goes from one stable position to another stable position very
> close to it, and sits there for a while. Here solution 1 is the clear
> winner.
>
> Of course with solution 1 you'll have to eliminate any resonances in the
> system, and perhaps put up with a little noise when you're not moving 1
> or 2 steps at a time.
>
>
> Direct drive with a 400 step motor running 1/8 stepping compared to a
> 200 step motor running 1/16 stepping comes out marginally in favor of
> the 400 step motor for the Z axis, and only in favor of the 200 step
> motor if the X and Y axis would excite a resonance at 1/8 stepping which
> doesn't show up at 1/16 stepping. There are mechanical methods to
> eliminate the resonances too.
>

I think its important to understand that the torque per uStep decreases as the 
number of uSteps per step increases _BUT_ the stall torque does not, nor does 
the slope of the torque vs position displacement from null (Q=0), that is the 
stiffness does not change, just the settability of the null position.

So there is no disadvantage of high uStep ratios (we've found that ratios less 
than 32 are a lot noiser than need be) But you should not expect much 
accuracy improvement unless you have very low loading.

High uStep step drives are also usable in a servo mode with encoders where the 
ramp up torque curve (uSteps from null) can be driven by the PID output. This 
works fairly well dispite the large non-linearity in the torque curve



Peter Wallace
Mesa Electronics


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