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Messages - stirling

1811
VB and the development of wizards / Re: About a delay routine with G4
« on: April 12, 2008, 04:32:41 AM »
Hi Gianni

There's a couple of issues here. I'm not entirely sure why there appears to be a limit of around 22000 but for the moment, I'm not sure that is of practical importance - bear with me - it may become irrelevant.

I'm assuming you don't need a delay of 22000 seconds! so you probably have the "G04 dwell in ms" option in Config/general config ticked. If so and you untick it, your delay will be in seconds - so I think your problem will dissappear.

That said, I'm still a little confused as to how you're going about this. If I understand you correctly I think you're intending having all your gcode posted via the "code" sub to Mach from VB. If that is your idea there may be a better way to do it.

If you use the M3 macro to fire your torch (as you would if you were starting a spindle) just add your "wait capture code" prior to DoSpinCW() only use SetVar(1000, wait) instead of your 'code "G4 P" & wait'. Then in your gcode files have G4 P#1000. That way you'll be using a more flexible approach.

So...

In M3.m1s you'll have:

SetVar(1000, GetUserDRO(1037)) 'set gcode var 1000 to delay in SECONDS (no need to go via your wait variable - just wack it straight in there)
DoSpinCW()

In your gcode text files you'll have:

M3
G4 P#1000
G03 X100 Y100 I25 J25
...
...

Cheers

Ian

1812
General Mach Discussion / MOVED: About a delay routine with G4
« on: April 12, 2008, 03:55:26 AM »
This topic has been moved to [VB and the development of wizards].

http://www.machsupport.com/forum/index.php?topic=6519.0

1813
General Mach Discussion / Re: How to wire steppers???
« on: April 12, 2008, 03:32:43 AM »
I will wire the Gecko up and try it wired to give the full five amps to a motor connected in parrallel and let you know the result.

Jim - if you do this you will be overloading your motor current wise which is a really bad idea. You run the risk of significantly shortening its life. The rated current of your motor in parallel is 3.5Amps - not 5Amps. Putting figures to it, the coils will heat up to just over twice their rated value. (This is not to be confused with acceptable motor heating which is the result of overdriving VOLTAGE wise. I know it's confusing but the two are entirely different.)
the rated parallel current of your motor which is 3.5 Amps.

1814
General Mach Discussion / Re: How to wire steppers???
« on: April 11, 2008, 03:22:10 PM »
Sorry Jim but there is waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay more to it than you've suggested. You're talking pure Ohms Law which is wholly inapropriate other than to work out the static ratings for a motor. Once that motor moves you're into a whole different ball game.

Can I suggest you re-read Ian and my posts because we've moved a lot further on since you were last here.

Thanks

Ian

1815
VB and the development of wizards / Re: About a delay routine with G4
« on: April 11, 2008, 09:58:38 AM »
Hi Giannis

A bit more info would help. When you say "routine" - do you mean just these two lines or is there more?

wait = GetUserDRO (1037) * 770
code ("G4 P" & wait)

The thing is, this routine wouldn't be entered by just entering a value in a DRO and I'm not sure what you mean by "never stops" - do you mean it hangs? or do you mean you don't get the expected delay at the dwell?

Also I'm not sure what you mean by "escapes the routine", there's nothing to escape from. These lines of code will simply pass the command "G4 P?" to the Mach thread and then exit the vb thread. i.e. The VB thread will not remain "alive" for the duration of the dwell unless you follow it with the while isMoving() construct - and you'd only do that if there were more code following.

DoButton(0) is the cycle start button, so I'm a tad lost on why pressing that should stop anything.

Sorry Giannis - can you expand a little?

Thanks

Ian

1816
General Mach Discussion / Re: How to wire steppers???
« on: April 11, 2008, 06:36:34 AM »
Hi Ian

I think we're on the same page more or less with this. I understand that the driver does not induce a square wave of current. As I understand it, that's exactly why we use over voltage in order to try to get the leading edge of the current waveform as steep as possible.

I wasn't my intention to suggest that the output mechanical power of the motor is VA, that's why I said "your motor will have the same available power of 60W", meaning from the PS - which is VA. Clearly some of this is not converted into mechanical power otherwise we'd have a 100% efficient motor and we'd all be rich ;D

I agree that the output power of the motor effectively doubles as the PS voltage doubles - Mariss told me that too when he said this (I'm sure he won't mind me including it here.)

"Look at it this way: Doubling the power supply voltage doubles the mechanical power output of a step motor. The efficiency of the motor doesn't change, so twice as much electrical power has to be supplied to the drive. Electrical power is Volts times Amps. Because you doubled the voltage, you doubled the electrical power at the same amperage.

Conclusion: Power supply current is independent of supply voltage for a given motor. It is the same value regardless of power supply voltage.

Mariss
"

What I still don't really understand though is if Jim does indeed get more oomph by wiring in parallel as opposed to single coil, where is that extra oomph coming from? Like I said, there is 60W of electrical power available to drive the motor in both cases. As Mariss says - the efficiency of the motor doesn't change at the same amperage, and as you've said - not all of that will be converted into mechanical power, - but whatever proportion it is, is the same.

The only thing I can think of at the moment is prbably absolute rubbish but I'll throw it in anyway. We know that the current is induced in pulses. We know that the quicker we can get the current to rise to its 2.5A the better. Inductance of parallel is less than single coil. This presumably allows the current's leading edge to be steeper. Therefore the 2.5A is "present" in the motor for longer. Therefore the power is available for longer on each pulse than it is for single coil. - just a thought.

Ian

EDIT: As you correctly say - we're all learning. A thought just crossed my mind. A chopper will apparantly only draw about 2/3 of the rated current of the motor. Is this (I wonder) because as we know it's applying that current to the motor in pulses, it's only actually applying it for 2/3 of the time?

1817
General Mach Discussion / Re: How to wire steppers???
« on: April 10, 2008, 01:30:05 PM »
As your driver can only source 2.5A it can not supply the rated parallel current of your motor which is 3.5 Amps. The data sheet describes it as a PWM driver which I think is synonymous with it being a chopper driver. If so, then I don't think it will be damaged by your proposed parallel wiring scheme - but don't take my word for it. That said, at 24V in both cases your motor will have the same available power of 60W. Had you been able to source 3.5A then clearly the power would have been 84Watts in parallel as opposed to the 60W in single coil.
Despite the fact that in the first case you're "overdriving" voltage wise by a ratio of 3.2:1 and in the second case by 4.52:1, I can't see it actually makes any difference - power is power is power after all. But from your reaction when you briefly tried it I get the feeling it did make a difference. So if you feel it's better in parallel go for it. The only thing I can think of is that if it is better in parallel then I'm guessing it's because reluctance is coming into the equation - back to the books!

If you do switch to geckos (other great chopper drives are available) and up your V - make sure your rig's well bolted down - that thing's going to go so fast it'll make your head spin. ;D

1819
For a definitive answer to this, ask the same question in the Yahoo Gecko group. You will probably get an reply by THE expert, Mariss Freimanis, he is very good at explaining such problems.

Agreed - What little I know about steppers and drivers has mainly come from reading Mariss's posts and papers on one site or another.

As only half an expert, I think you are probably right in theory, but other factors come into the equation. Say you have a stepper with 200 full steps and a driver set at 10 microsteps and this setup yields 1000 microsteps per mm i.e. each microstep is 0.001 mm and a full step is 0.01 mm. Then if you were running in full step mode on two axes, you would generate a staircase of 0.01 steps, as you suggested. However, this would only occur on a G0 move, as at cutting speeds, the stepper will probably be in microstep mode, so your staircase steps would be too small to make any practical difference. . Mechanical inertia, cutting tool deflection and any latency in the driver and software will smooth this out. If you need a perfectly smooth finish with a stepper, then you are going to need a setup with more steps per mm.

Agreed again - as you say, except at really slow speeds, due to the momentum of the rotor (in an unmounted motor) plus the system (for a mounted motor) a stepper is rotating smoothly and not actually "stepping" at all.

Ian (who is still climbing the learning curve. :))

Likewise - may your learning curve be a smooth one. ;D

Cheers

Ian

1820
Can't help you directly because I was unable to get the probing plugin to work at all. However it certainly seems on the face of it unnecessary to record the rectraction Z level.

It was because I was looking for a little more than the available probing routines offer that I started http://www.machsupport.com/forum/index.php/topic,4352.0.html

You may want to take a look at http://www.razordance.co.uk/probe25D.htm if you don't want to plough through the whole thread.

Cheers

Ian