Torque Conversion Question

[Verboten]

Hello,
I have a stepper motor that has its holding torque listed as 32.4 kg-cm min.
Can this be converted to in-oz? That may be a silly question but I don't understand how time relates to torque (if it does).

Any help or explanation will be appreciated.
Thanks,
Jason

[BigDaddyG]

Jason,
Just ignore the "min" and do a standard conversion. kg=oz and cm=inch will give you the oz/in conversion. Multiply 32.4 x 2.205 = 71.442 lbs. (1143.072 oz) now the cm-inch.
1143.072 x .03937 = 45.003 oz/in. Hope this helps.
Good Luck,
Glen

[BigDaddyG]

Opps, the oz to in conversion was supposed to be .3937 making it 450.027 oz/in. See engineers have to check themselves all the time. No diff from measure twice, cut once! Good Luck.

[ger21]

Download this program and convert anything you can think of.

http://www.joshmadison.com/software/convert/

[garagefela]

Hi,
Regarding the torque rating on the motor, could this have possibly been minimum instead of minute? (just guessing here)

Cheers

[BigDaddyG]

Yes, that could very well be a "min" since the holding force is calculated on a per amp basis! 1 amp= 450 oz/in, 2 amp = 900 oz/in and so on.
Regards,

[arvidb]

The person who marked this motor didn't fully understand the difference between mass and force, and the use of a '-' instead of a '*' makes things even worse. The unit kg-cm (mass minus distance???) doesn't make any sense.

What is meant is the weight of the mass 1 kg times the distance of 1 cm - i.e. torque. Since the force of weight of 1 kg is about 9.81 N, this becomes about 3.18 Nm or about 450 oz*in, in units of torque.

My guess is also that the 'min' means minimum.

Arvid

[Verboten]

Hello All,
Thanks for the replies - they have all been helpful.
Just for reference - The motors in question are from Applied Motion Products ( AMP ).
The holding torque was not listed on the motor itself, I had to email the company to get the data sheet ( which was actually far easier and faster than my experiences with most other companies - what company do you work for ... etc.) They sent me the data sheet and the only difference from what I posted is that it is all in caps. (I didn't think that it mattered).

Which brings me to another question that I have from the data sheet: What is detent torque? The detent torque is listed as 400 G-CM MIN.

My brain was off in a different place and I hadn't thought that MIN could mean minimum.

This might not be off topic but I believe that my projects have led me in a different direction (toward more torque with accuracy being less important) and I think that I may list these motors so that I can move in a different direction (servos). My last question in this long-winded essay would be: What do you think would be a fair opening price for 2 motors as described (AMP Nema 34 450 oz*in - 8 lbs. each) unused with 1 EAD Nema 34 300 oz*in motor used tested working. I want them to sell but don't want to just give them away.
Thanks for all of your help,
Jason

[BigDaddyG]

Jason,
Glad to help! In laymans terms, Detent torque is about 5-20% of Holding Torque! The motor by nature has to overcome such things as the bearings, couplings etc. to actually provide movement. So, the detent torque is that number (Holding torque= no friction and Detent=actually what is rendered to the work once the motor has to move). In some cases, you could have to overcome enough forces such as the cut your taking (to deep), or the item you want to move (to heavy) and the motor will not actually move because it can not overcome these forces. When making your estimates for motor size, it is best to use the Detent torque, since it figures in the loss of "Power or Torque" within the motor its self. Like I said, usually 5-20%. Again, if the motor is just holding a point in space, it does not have to overcome the friction of the bearing etc, so that number is higher (again by 5-20%).

As and Engineer, I make it a point to add a multiplyer of about 2 to 2-1/2 what I actually figure it will take to make a cut of move an object. My father used to tell me to "Double it and add 50%", which has never failed me (god rest his sole) in years. That equation also works when estimating jobs for costs.

I hope this helps.
Regards,
Glen

[arvidb]

You can look at the detent torque as the magnetic friction of the motor. You can feel it as "steps" when you turn the motor with its windings unconnected. Detent torque can be both positive and negative depending on where within one step the rotor is positioned.

In this motor they guarantee that the motor has a detent torque of at least 0.04 Nm or 5.6 oz*in (again converting nonsense units to units of torque). The strange thing is that one would expect this quantity to be a max one, not a min one. Again maybe they don't understand what they're writing.

Bearing friction etc should not be included in the detent torque figure. If I understand things correctly, detent torque adds to the holding torque (provided the motor is positioned in between two full steps), but subtracts from the "useful" torque when the motor is moving.

Arvid

[BigDaddyG]

Arvid,
Bearing friction etc, although not included in the Detent torque figure as you stated, was a simple way for me to illustrate the "concept" or cause/effect of Detent torque. But Bearing Friction has a direct "bearing" on Detent Torque non the less.

The consequence is that if the load torque plus motor friction and detent torque exceeds the incremental torque of a microstep, successive microsteps will have to be realized until the accumulated torque exceeds the load torque plus the motor friction and detent torque. Simply stated, taking a microstep does not mean the motor will actually move. And reversing direction can require a large number of microsteps to get the motor to move.

But what if the motor is not loaded, as in some type of pointing or inertial positioning? The motor still has friction torque from its bearings. It also has a detent torque, in addition to other harmonic distortions. You'll have to wind up enough incremental torque to overcome the bearing friction. Even more disruptive than the bearing friction is the detent torque, which is typically 5 to 20% of the holding torque. Sometimes the detent torque adds to the overall torque generation. Sometimes it subtracts from the power torque generation. In any case, it wreaks havoc with overall accuracy.

Some manufacturers fabricate microstepping versions of their motors. Their aim typically is to reduce the detent torque, usually at the expense of holding torque, so the torque-versus-rotor position is closer to a sine wave. They also hope to improve linearity of torque versus current. These efforts reduce but do not eliminate the compromises associated with microstepping in regards to accuracy.

Regards,

[arvidb]

Yes, and anyone interested in the whole text (written by George Beauchemin of MicroMo Electronics Inc.), can find it here: http://www.machinedesign.com/ASP/str...tedArticle.asp

The point of microstepping is not to increase resolution, but to reduce ringing (make the motor turn smoother).

Arvid