I’ve been using the Nema-17 motors that I bought from the store, but I long ago changed my Dewalt 660 to a spindle, which is much heavier. Works fine, no missed steps - but when I power the machine down now I have to support the tool or the tool will crash into the bed or project, whatever it’s over at the moment. I figure I need a different motor, but not sure what. I can’t be the first person to need to do this, but I don’t see options when I search the forum. Any help would be appreciated.
You can replace the 4 start leadscrew with a 1 start. It will not fall as easily. The downside is your top speed will be lower, because the rpms on the motor will be 4x for the same Z speed.
Just set a block of wood under the spindle nose, unless you do as Jeff said and go to a 1 start lead screw it is going to fall no matter the stepper motor once powered off.
I added some springs to offset the weight, it also helps fight backlash on leadscrew.
You may have to experiment with different springs, length, and stiffness to get what you
want. Maybe not he best solution but it worked for me, my spindle weighed 11 lbs.
Good point, I guess there would be some lift on the spindle. Here’s what I assumed when I looked at preventing the spindle from crashing down when stepper is off (which it did, and hard). I’m sure I missed something in my analysis, but it’s what seemed logical to me.
My MPCNC has about 70mm of Z-travel. I chose the spring rate and length to roughly offset 1/2 the weight of the spindle (12 lbs.) over it’s travel. Uncompressed, the spring is 130mm long.
Taking the two springs together, at maximum Z travel up, the springs (near 40% compression) push up 8.0 lbs. against the spindle (12 lbs.), so the stepper sees a net 4 lbs. downward instead of the full 12 lb. spindle. And at minimum Z travel down, the springs (near 90% compression) push upward by 16.4 lbs. against the spindle, for a net force of -4.4 lbs. upward on the stepper instead of the full 12 lb. spindle.
So the weight that the stepper has to drive changes from +12 lbs. always down to +/- 4lbs.(down/up).
If I’m correct, there’d be about 4 lbs. of net downward push at Z-max, decreasing to zero near Z-mid-travel. And to your point there’d be about 4 lbs. of net upward push at Z-min, also decreasing to zero near Z-mid-travel. (At Z-mid-travel the spring force is 12lbs., canceling the spindle 12 lbs.)
Wouldn’t the stepper be able to push downward against that 4lb. lift? The stepper torque is 76oz-in (4.75lbs-in) and with leadscrew radius of 4mm (0.16 in.) would theoretically be capable of 4.75/.16 = 30lbs-in.
I suppose “backlash” of the leadscrew is opposed by the downward force from Z-max to Z-mid , and opposed by the upward force from the Z-min to the Z-mid. However, the net force would be zero around Z-mid and backlash would likely be happening there. I haven’t tried to evaluate that, do you know of a good easy way to measure backlash that I could try?
