What size stepper motor do I-need.

[Robin Hewitt]

How long is the cable on the parallel port.

A TTL signal might get you a reliable 10 feet without handshaking at that speed, if you were lucky

[magno_grail]

I have a couple of questions about your motor calculation spread sheet. The motor inertia is input in gm-cm^2 (B32). The rotor inertia given in I27 must be Kg-m^2 since the total inertia is the sum of the screw inertia, load inertia and motor inertia, the others are in Kg-m^2. I27 = G35 which is a lookup to convert units of the motor inertia.
The problem is 1000 gm = 1 Kg and 100 cm = 1 m so gm-cm^2 should be divided by 1000*100^2 = 1*E7 not 1*E8 as in G35. The motor inertia in the sum is 10 times too small.
I cannot find the time used for the acceleration torque (I34). The equation is G32*I28. G32 is labeled as V but it is really the number of pulses/second of the stepper based upon a 200 pulse/revolution motor and the screw speed (which is the Max. linear speed divided by the screw pitch). From looking at the equations, the running torque is small compared to the acceleration torque unless there is a lot of friction in the system. I have not found any references as to what range of acceleration is needed for a CNC. Obviously, when a stepper system is tuned it will limit the acceleration so positioning steps are not dropped due to the loads. Too small motor torque will result in a very slow CNC due to acceleration limitations. At some point a larger motor torque results in small performance gains for the cost increases. Are there any guidelines for this? Thanks.

[masinecc]

How can i calculate what motor size i need for my gantry if i use rack and gear on both sides. This example is for ballscrews. Looking forward for your answer.

[Boyan Silyavski]

How can i calculate what motor size i need for my gantry if i use rack and gear on both sides. This example is for ballscrews. Looking forward for your answer.

Backwards. Choose the RP size, gearing so that you have as a result normal rapid speeds with acceptable resolution. Then figure how many RPM you need in reality so you have mostly that in mind. Then you go back to the gearing ratio and gantry weight and from there you decide Nema 23 or 34.

Obviously as the RPM is the important part here cause most possible it will be well geared, then you finish with low impedance motors driven by high voltage.

And knowing that nema 23 motor driven at 70VDC moves a gantry max 50-60kg with gear 1:1 without problem at 10m/min via ballscrew, you can figure the motor you need for your case.

[Robin Hewitt]

Nobody works it out, they just assume that the biggest motor they can find has to be the best option. Then they buy a set of 3 because that saves even more thinking and they end up with a Z axis that could raise the Titanic and is so heavy the gantry has to be made from RSJ

[Boyan Silyavski]

Nobody works it out, they just assume that the biggest motor they can find has to be the best option. Then they buy a set of 3 because that saves even more thinking and they end up with a Z axis that could raise the Titanic and is so heavy the gantry has to be made from RSJ

[irving2008]

I think its a fact that microstepping, where you are actually stopping at the microsteps (i.e. using them for resolution) definitely is less torque, simply because the motor cannot hold that position accurately against load. In a dynamic situation however, lots of factors come into play. For example, if the axis is in constant motion the only torque required is to overcome friction and cutting loads, there is little torque required to provide acceleration. Unless the motor is being operated close to its torque limit (at that speed/volt/current combo) then microstepping should have little impact. The general rule I have used is in the spreadsheet is that the motor should provide 3x required dynamic torque at the maximum speed.

Large motors have high inductance so the torque drops off very fast with speed - the corner speed of those motors is 240rpm. I don't know how big your axis are, but I'm guessing its going to be around 1 - 1.2m? With your 10mm pitch screws 2.5m/min = 250rpm, so that is close to optimal (and 1/8 stepping = 6664steps/sec) and it looks OK at cutting speeds, but its very marginal at 10m/min rapids and that is where you may have lost steps (=27000steps/sec). You need to reduce rapids to 7m/min but it should be OK at 1/4 or 1/8 stepping.

[massimomb]

Thanks again, i will reduce the rapid to 5m/min and the working speed up to 2,2 m/min as top speed, setting the microsteps to 1/8.

M.

[DRITECH]

Hi,

I am building a gantry robot and for the Y-axis I am using a 1610 ballscrew (16mm diameter, 10 pitch, 1300mm long). I estimated that the mass is 36Kg and for the speed I planned to have 1800mm/min.

Using these calculations I concluded that the total torque is 0.23Nm. From the Excel the estimated torque resulted to 0.67Nm.

Now I have the following questions:

1) Can I use there calculations for a ball-screw system?

2) What is the calculation to determine the estimated torque (0.67Nm) from the total torque (0.23Nm)?

3) Is 0.67Nm enough for such mass? I did some research for similar projects and they use a much higher torque stepper motor.

All kind of help will be appreciated and sorry for my bad English.

Thanks in advance.

[Neale]

I'm trying to use the spreadsheet to check some parameters of my planned new router. This has thrown up a few questions:
1. Acceleration seems to be the largest contribution to torque required, but I can't find where load acceleration is built in. It must be in one of the formulae somewhere but I can't manage to unpick the formulae in the hidden columns to find it.
2. The spreadsheet recommends a motor torque based on an assumed safety margin of 3; if the safety margin isn't high enough with my intended motor, I presume that I could restrict the machine to a lower acceleration? But given that I don't know what acceleration is assumed, it's difficult to see what's happening. I do see that the acceleration torque number is linked to cutting speed.
3. For a twin ballscrew X axis, I assume that I can use the spreadsheet using actual ballscrew dimensions (single ballscrew) to give critical speed, but then use a double-length ballscrew in the spreadsheet to do the torque, etc, calculation.
4. If I go to a twin-motor setup, can I simply double the nominal motor torque? Or is it easier to halve the gantry mass and do the sums for a single motor/ballscrew?
To put some reality into this exercise, what would be sensible numbers to use for cutting speeds and accelerations? Bit of a "how long's a piece of string?" question, but my current router is so far out of the norm I don't feel that I can extrapolate from what I'm doing at the moment, and I don't know what kinds of numbers are reasonable. For starters, I would assume profile cutting, say, 9mm ply with a 6mm cutter in a couple of passes. Currently, with speed limited by machine rigidity, I would cut that at maybe 600mm/min.
Many thanks for any help available!