stepper motor problem?

[songguako4さん]

A question oft asked on MYCNCUK is "how big a motor do I need?". There is no simple answer to this, and the options are usually:
a) follow someone elses build and copy theirs; 
b) take a guess and try again if you are wrong; or
c) work it out, which is the subject of this tutorial.

When choosing a motor you need to know:
a) what power and torque output is required at a given speed
b) what electrical characteristics are appropriate to acheive that

What I have tried to do here is the engineering approach, by showing the calculations needed to get some idea of power, which then dictates motor size. I am concerning myself with a stepper motor directly driving a leadscrew to move the gantry, table, etc. Similar calculations can, however, be done for belt drive or geared up/down with timing pulleys.

DISCLAIMER: This tutorial is to give you an insight into how to approach the selection of a motor. I take no responsibility for any consequences of following this tutorial and you alone are responsible for your choice and purchase of motors etc. 

The second aspect to accelerating the moving item is to overcome its inertia (the tendency of something to remain at rest) - this is true even if friction were zero.

The motor turns the leadscrew to convert rotational motion into linear motion. There is friction here too, expressed as the efficiency of the leadscrew. This is typically 80% for ballscrews and as low as 30% for trapezoidal screws (bronze or delrin nuts on steel) and inertia, as the screw itself has inertia which is dependent on its mass and its length. 

Now we have all the elements we need.

So, considering the frictional component of the torque, this is given by:

Torque = F * p/(2pi * e)

[1]

where F is the force to be overcome in Newtons, p is the screw pitch in metres and e is efficiency.

For this example I shall assume a TR12x3 trapzoidal screw 12mm dia, 3mm pitch. 

The force to be overcome is, as said above, either the stiction or the kinetic friction plus the cutting forces. For the purposes of simplicity 
assume the cutting forces range from 5N for wood to 20N for alloy using the sort of spindles/cutters found in hobby sized machines up to 75N for steel on a mill. 

The frictional forces are calculated from the mass of the load and the friction coefficient:

F = M * g * Fc

[2]

where g is gravity, which can be taken as 10

Typical static friction coefficients for common sliding mechanisms are:
0.003 for a ball slide, 
0.01 for low-end ball races on aluminum channel, 
0.05 for teflon on steel,
0.16 for bronze on steel
1.10 for cast iron on cast iron.

For most of these the kinetic frictional coefficient can be taken as the 
same, although it is around 0.2 for greased cast iron to cast iron.

Assuming a low cost router using ball races and our 20kg load the frictional force (from equation 2) is 20 * 10 * .01 = 2N. Add to this the cutting force for wood at 5N and the force to be overcome is 7N, therefore the torque (from equation 1) is:

T = 7 * .003/(2pi * .3) = 0.01Nm

This doesn't sound a lot when motors are rated at 1 - 3Nm, but we haven't finished yet.

The second calculation is the inertia of the moving item, expressed in terms of the inertia seen by the motor. The symbol we use for this is J(load) and it is calculated thus:

J(load) = mass(load) * pitch^2/(2 * pi)^2

[3]

where mass in Kg, pitch in metres gives inertia in kg m^2
[note: ^2 means raise to the 2nd power, e.g. square it]

In our example we will use a trapezoidal TR12x3 single start screw to move this 20Kg gantry, so from equation 3, J(load) = 20 * 0.003^2/40 = 4.5 x 10E-06 kgm^2 (the 40 is a good approximation to 2pi squared). To this we add the inertia of the screw, which is given by:

J(screw) = 1/2 Mass * radius^2 

[4]

where the mass is given by:

mass(screw) = pi * radius^2* length * density

[5]

In our worked example a 12mm screw 800mm long has a mass of 3.1416 * .006^2 * .8 * 7800 = 0.71kg and therefore an inertia of J(screw) = 1/2 * 0.71 * .006^2 = 1.28 x 10E-05, so the screw has a higher inertia than the load! 

The total inertia to be overcome is the sum of J(load) and J(screw) = 1.72x10E-05 kgm^2. (Note the spreadsheet also adds in the rotor inertia of the motor)

Next we have to decide how fast we want the gantry to move under load. Typically for a wood router anything from 500 to 1000mm/min would be suitable, for cutting aluminium you might want to look at 1800mm/min or better when using small cutting tools. The maximum traverse speed is given by:

Smax = max motor rpm * screw pitch. 

[6]

You can see that the torque required is very different to the 'torque rating' of the motor. It is important to note that the holding torque of a stepper motor online is to some extent of little relevance. This is the physical torque required to overcome the electromagnetic forces holding the rotor stationary and is the torque the motor tends towards as speed drops to zero. 
Vd = 32 * sqrt(L)

[9]

where Vd is operating voltage, and L is the motor inductance in mH. If your drivers are limited in voltage a low inductance motor is essential if you want any reasonable speeds.

So, lets look at the motors available. Pick any website, such as Zapp Automation's, and look at the list of NEMA17 and NEMA23 motors. Here are the options:

Motor V A mH Nm Inertia

23hs22-2804s-1684B 2.80 1.6 2.8 0.44 68

17hs16-2004s 9.24 0.7 32.8 1.00 275

SY57STH51-3008B 3.10 2.1 3.6 1.00 275
SY57STH56-2008B 5.04 1.4 10.0 1.24 300
SY57STH56-3008B 3.15 2.1 4.4 1.24 300
SY57STH76-3008B 4.00 2.1 6.4 1.85 480

Plugging any of these into the spreadsheet gives similar results, so which to choose? Next calculate the ideal voltage for each (the spreadsheet shows this as the 'ideal voltage')

Motor V A mH Vd
SY42STH47-1684B 2.80 1.6 2.8 54
SY57STH51-1008B 9.24 0.7 32.8 183
SY57STH51-3008B 3.10 2.1 3.6 60
SY57STH56-2008B 5.04 1.4 10.0 101
SY57STH56-3008B 3.15 2.1 4.4 67
SY57STH76-3008B 4.00 2.1 6.4 81

Lets assume we want to use a low cost driver board, such as the System3 from DIYCNC which is OK to 2.5A but limited to 30v max, or the TBA6560 boards available on eBay. None of those are going to manage 60v, indeed 24v is the likely voltage, but the motors that are the lower ideal voltage will perform better with those drivers. So on this basis the SY42STH47-1684B or the SY57STH51-3008B would be contenders. I'd probably opt for the 1Nm NEMA23 motor over the 0.44Nm NEMA17 to give a bit more leeway and scope for upgrades. Anything bigger would be a waste of money and would perform no better (and usually worse - there is such a thing as too big a motor).

Below shows similar calculations repeated for a number of examples

25kg gantry 4' Rockcliffe oilite bronze on steel, TR12x3 1.2m long. 1000mm/min. Torque = 0.1Nm, power = 3W so a 1Nm - 1.5Nm motor.

35kg gantry 2m ballrace on channel, 16mm ballscrew 5mm pitch, 1.8m long, 2000mm/min. dense hardwood capable. Torque = 0.4Nm, power = 12W (typical 2Nm NEMA23 motor)

50kg dovetail table + 5kg workpiece + 5kg vice, 20mm ballscrew 5mm pitch, 900mm long, 1200mm/min, light alloy/steel cutting. Torque = 0.9Nm , power = 32W (8Nm NEMA34 motor)

50kg dovetail table + 10kg workpiece + 5kg vice, 25mm ballscrew 5mm pitch, 900mm long, 1800mm/min (with slightly reduced acceleration), heavy alloy/steel cutting. Torque = 1.1Nm , power = 64W (possible with 12Nm NEMA34 motor, but this is starting to get into servo motor territory to meet that speed/accel requirement)