Messages

1321

Mach4 General Discussion / Re: macro failed to compile

« on: December 16, 2020, 01:33:47 PM »

Hi,
good to hear you've found the solution.

If I delete it I can't do a thing.

I never imported a profile from anywhere, I just made up my own. You do not need an imported profile,
it does make life easier to start with but also requires that you comply with the assumptions and hardware arrangement
as envisaged by the supplier of the profile.....whether it suits your purpose or not.

Craig

1322

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 16, 2020, 05:39:14 AM »

Hi,
sorry about the delay, I've had a busy couple of days.

Lets do some simple calculations:

1) The first moment of inertia of the X axis ballscrew:
diameter = 32mm or radius =0.016m
Length=686mm or 0.686m

J_ballscrew=mass. (radius/2)²
and mass = pi .Radius².length. density of steel
m=3.14 (0.016)².(0.686). 8000                                      (8000kg/m³=denstiy of steel)
m=4.41kg
so:
J_ballscrew=(4.41). (0.016/2)²
  =2.8 .10 ^-4   kg.m²

2) The first moment of inertia of the servo/stepper armature:

Lets assume that you have a 34 size servo or steppper direct coupled  (or 1:1 belt drive) and the armature has a first moment of inertia of
1 . 10^-4 kg.m². For instance the spec sheet of my 750W servo list the armature inertia as 1.13 . 10^-4 kg.m²
so you can see the number I've suggested is a practical number.

Should we decide that we need a gear or belt reduction we will revisit this assumption.
J_armature=1 . 10^-4  kg.m²

3) The effective first moment of inertia presented by the axis mass.

Pitch=6mm or 0.006m
Assume axis mass of 100kg
J_linear=(axis mass). (pitch/2.PI)²
         =100 .  (0.006/2 . 3.141)²
         =0.911 . 10^-4  kg.m²

The total effective first moment of inertia is:
J_total=J_ballscrew+J_armature+J_linear
    = (2.8 + 1 + 0.911) . 10^-4 kg.m²
    =4.71 . 10^-4 kg.m²

Take a look at the proportions of the components that make up the total moment, the ballscrew is 60%, the armature is 21%
and the axis is the remaining 19% Clearly the ballscrew dominates the equation. Even if we over (or under) estimate the axis mass,
or over (or under) estimate the pitch of the ballscrew would still introduce minimal error in the total. Note how even the moment of the
armature exceeds the moment induced by the axis mass. This result is the norm and surprises many people, me included when I first
did the calculation, but physics does not lie.

Note that I haven't made any allowance for gear reduction (as yet) nor have I made any allowance for other rotating components like
pulleys and couplers. Making allowances for these items is simple enough but I did not wish to obscure the main ideas in the equation
by small correction terms.

Not also that I did not make any allowance for ballscrew efficiency or friction. Ballscrew efficiency is an easy correction, friction however
is not.

So in trying to keep the calculation as simple and illuminating as possible without introducing a whole bunch of small corrections
and confusing everyone.......please accept that this calculation is close, say within 10-20%.

Lets now do the acceleration calculation with a 1Nm torque:

dw/dt=torque/J_total     (w= angular velocity in rad/sec, dw/dt= angular acceleration)
    =1/ 4.7 . 10^-4
    =2127 rad/sec²
Converting that to linear:
A_linear=dw/dt  . pitch/2.PI
    =2172 . (0.006/2 . 3.141)
    =2.03m/sec²

Which is 1/5 of a g.....which is not too shabby!

Note that a standard 400W servo with 3000rpm rated speed will have rated torque of 1.27Nm, so would exceed 0.2g, which
is pretty impressive.

My 750W servo has 2.4Nm rated torque and 7.2NM peak for accels of 0.5g (rated) and 1.4g  (peak).

If you want to consider gear reduction let me know and we can rework this calculation.

The Z axis on a knee mill is often very draggy and so leaving out friction is unrealistic when applied to the Z axis. You really
need to replace the trapazoidal screw with a ballscrew if you wish to drive it at any realistic acceleration. Otherwise you will
need to use a really outsize Z axis motor or major gear reduction. Gear reduction means that even having fast X and Y axes is moot because
to stay coordinated they must slow to the pace of the Z axis.

Craig

1323

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 14, 2020, 01:40:01 PM »

Hi,
OK, as you say those are substantial ballscrews and are going to take a lot of torque to accelerate quickly.

This is a post I made on CNCZone a few months back:

Hi,
I've done the calclations for the acceleration potential of my machine including rotational inertia. I've had my calculation checked
by peteeng, a professional mechanical engineer. Additionally I have Hiwins calculation formulas and can double check all the calculations
and they all point to the same conclusion.

My previous calculations are widly optimistic.

The full calculation is:
Torque= Ieff. dw/dt

dw/dt is the angular acceleration.
Ieff is the total effective first moment of inertia.

Ieff is the sum of the individual components.

1)The rotational inertia of the armature of the servo. For my 750W 34size Delta servo that is (from the spec sheet)=1.13 .10-4 kg.m2
2)The rotational inertia of the ballscrew. For my 32mm dia screw of 650mm length =5.252 .10-4 kg.m2
Note that I got this figure from the THK spec sheet but both I and peteeng calculated it from first principles and all agree.
3)The effective rotational inertia implied by the axis mass. This forumla I derived from first principles and agrees with the theoretical treatment
published by Hiwin and also agrees with peteengs analysis.
Ilinear= m.(p/2pi)2 where p=pitch in meters and m is the axis mass.
Using numbers for my machine, p=0.005m and m=110kg:
Ilinear=110. (.005/2 x 3.141)2
=0.69 kg.m2

Ieff is the sum of these components:
Ieff= (1.13 + 5.252 + 0.69) .10-4
=7.07 .10-4 kg.m2

dw/dt= 2.4 /7.07 .10-4
=3395 rad/s2

And to convert that to linear acceleration:
Alinear= dw/dt x p/2pi
=2.7 m/s2

This figure is less than 1/10th of what I had calculated previously, and is my error. Note how despite the heavy axis, its contribution
to the first moment of inertia is small, even the first moment of the servo armature exceeds it, but both are dwarfed by the first moment of the ballscrew.
I had not made allowance for that factor previously.

The take away feature is that with my machine at least 'the rotational mass of the ballscrew dominates the acceleration equation'.

Craig

What we have to do is calculate/guess the three components that make up the total first moment of inertia, then we will know the torque
including any gear reduction necessary to achieve a given acceleration.

This is a post in the same thread whih details how I came to the formulas used:

Hi,
I thought I'd post my derivation of the forumla that combines linear momentum with rotational momentum. Its been many years since
I opened a physics book let alone read one!

My hypothesis is that there is some effective first moment of inertia, Jeff such that the total kinetic energy, Etot of a linearly
accelerating axis AND rotating ballscrew/servo assembly is described by:

Etot = 1/2 * Jeff * w2...............................equation [1]

Note I use MKS units:
Etot in Joules
Jeff in kg*m2
w in radians per second, rad/s

But we know that the total kinetic energy has two components, first the translational energy of the axis mass and the rotational energy of the rotating components:

Etot = 1/2 m*v2 + 1/2 Jcomb* w2.............. where Jcomb is the first moment of inertia of the rotating parts

But the axis velocity v is related to the angular velocity of the ballscrew, after all thats why we use ballscrews to precisely translate rotoational position
to linear position:

v = w/2pi * l where l is the pitch of the ballscrew in meters. Substituting:

Etot = 1/2 *m*(w*l/2pi)2 + 1/2 Jcomb*w2
=1/2*w2 { m*(l/2pi)2 + Jcomb }...................................equation [2]

Now we have two equations describing Etot, equation [1] and equation [2], using the principle of equating coefficents:

1/2 *Jeff * w2 = 1/2*w2 { m*(l/2pi)2 + Jcomb }

Jeff=m*(l/2pi)2 + Jcomb

So the component of the total effective first moment of inertia that is due to the linear movement of the axis is:
m*(l/2pi)2

For example my axis is 110kg, and the ballscrew pitch is 5mm or 0.005m:

=110 * (0.005/2pi)2
=110 * (0.005/6.283)2
=110 * (0.0007958)2
=0.69 *10-4 kg*m2

Craig

I have to disappear to work at the moment but will start applying your data soon.

Craig

1324

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 14, 2020, 04:10:36 AM »

Hi,
we could carry on a private email conversation....but there maybe others whom would get something out of it if it were on the forum,
after all many come here to learn. That would be my preference.

Craig

1325

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 13, 2020, 03:13:06 PM »

Hi,
not sure why the other pics did not attach...

Craig

1326

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 13, 2020, 02:54:50 PM »

Hi,
for my current build mill I have selected 750W Delta B2 series servos.

If your budget permits servos are preferred, if not you need low inductance steppers.

Don't be taken in by the hype around closed loop steppers, the manufacturers claim they are like servos.....they are not......
they are still steppers and suffer from rapidly degrading torque capability as speed increases. If you want closed loop performance
get servos.....don't mess around with Mr In-Between, aka closed loop steppers.

Low inductance 34 size steppers are probably right for your application.

It may surprise you but the dominating component of the inertia of your machine is not in fact the axis beds but is in fact the ballscrew.
I'm using rather large ballscrews, 32mm diameter, and they represent  85% of the inertia, the servo armature another 10% with the axis bed, 115kg,
representing the last 5%.

Because the ballscrew spins so fast, and by comparison the axis moves so slowly, it is the ballscrew that dominates the inertia equation.

If you can provide details of the ballscrews we can do some calculations that will at least put you in the right ballpark.
For these calculation we need an accurate measurement of ballscrew diameter, mucho important!. Next we need the length and the pitch,
important but not nearly as critical as diameter. An estimate of axis mass.

Craig

1327

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 13, 2020, 01:06:02 PM »

Hi,

and I thing I would like nema 34 1800oz motors.

Watch out for those high torque motors, many have bought them but found they are useless. They stutter, stall and vibrate like hell
above a few hundred rpm.

With 34 size steppers look for 2mH to 4mH, 2mH preferred and reject anything over 4mH.

A 34 size motor of 3mH inductance but only 750 oz.in will still be better than 1800 oz.in at 8mH. You must get low inductance motors
or your machine will b as slow as a wet week.

Craig

1328

Mach4 General Discussion / Re: Finding full list of functions

« on: December 12, 2020, 05:04:58 PM »

Hi,
to record a point cloud look in Mach4Hobby/LuaExamples/ProbeToFile and there you will find two macros m400 and m401.
They can handily be used a basis for a point cloud data file.

When I started with Mach4 I had, and still do have quite a battle understanding the wx.Widgets dialogues. The file dialouge in
the above examples have been huge in my learning curve to the extent I can now (haltingly!) roll my own file dialogues.

Craig

1329

General Mach Discussion / Re: mach 4 learning the machine drive ratios

« on: December 12, 2020, 03:46:51 PM »

Hi,

1. Can mach 4 learn from the ratios between the motor drive and driven pulleys, as I have two pulleys that I cannot source with what tooth count I think I need. I believe these were originally a 2 to 1 drive system for the x and y. Can I compensate in the software for a close, but different ratio

Fit what pulleys you can find, then do the calculation and install those numbers into Mach4. Mach4 does not 'learn' what you have done, you tell it
what it needs to know. Mach can quite handily accommodate non-integer ratios etc, and there is plenty of help to do the calculation and even
a 'suck it and see' type tuning aid available.

I have to find out what stepper drivers, and breakout board mach 4 needs. Is there any resource for this?

While Mach4 has a parallel port version called Darwin ($25 license fee applies) it is limited to 32 bit Windows 7 and earlier just like Mach3's parallel port.
In short Mach4 really requires an external motion controller, of which there are half a dozen manufacturers , some more expensive than others,
some more capable than others. There are a number of threads on the Mach4 board that cover the options.

I use and recommend an Ethernet SmoothStepper ($180) by Warp9TD, and many users pair that with an MB3 breakout board from
CNCRoom ($180). Thereafter you fit whatever stepper or servo drives you like.

There are cheaper breakout boards but few match the quality and well thought out design and balance of IOs of an MB3. If you are handy at
electronics and can add a handful of components to a cheaper and simpler (eg C10 for $23, single parallel port equiv) board you can have all the
flexibility AND save some dollars.....but you'll have to design and build a few extra circuits probably. There is help with those circuits if you need
it, electronics is my thing and they are all simple enough.

A somewhat cheaper alternative, albeit limited, is a PMDX411 and a Gecko G540. The PMDX411 is a single (parallel port equivalent) port
USB connected external motion controller with Mach4 plugin and the G540 is a 50V four channel stepper driver/breakout board combination.

I suspect you will become rather frustrated that much of Mach4's abilities are beyond reach because your controller (PMDX411=UC100=57CNCdb25)
only has 17 IOs, ie one parallel port equivalent, whereas a PMDX424 has 34 IOs, an Ethernet SmoothStepper has 51 IOs, a 57CNC has 57 IOs and a UC300
has 85 IOs. I highly recommend an external controller with 30 or more IOs.....you'll benefit in the end.

The Gecko G540 is an age old gem, many people use it. Note that it is a one parallel port equivalent input breakout board and four 50V stepper drivers for
about $350. 50V is not bad.....but 80V is better. Having separate breakout board and drivers is more flexible......but usually more expensive too.
Pay your money and take your pick.

When it comes to steppers, do not go for the highest torque units, they look attractive but they have high inductance and lose torque badly at speed
and will stutter and stall at only modest speeds. Lower torque units with comensurately lower inductance will run much MUCH faster without stuttering or stalling.
Manufacturers make high torque units to attract first time buyers who don't know about inductance. Don't waste your money.

With 23/24 size steppers look for inductance of 1mH to 2mH, 1mH preferred and reject anything over 2mH.
With 34 size steppers look for 2mH to 4mH, 2mH preferred and reject anything over 4mH.

Craig

1330

Mach4 General Discussion / Re: My tool runs off

« on: December 12, 2020, 02:14:22 AM »

Hi,
servos have two parameters, depending on the manufacturer, called the Following Error Window and the Zero Error Window.

Taking the most important of the two, the Following Error Window.....it means that you specify/program an interval in so many encoder counts
and should the commanded position and the actual position deviate by more the the Following Error then the servo drive will fault Following Error.

As an example I have an older 1.8kW AC Allen Bradley servo, it has an 8000 count per rev encoder. The default Following Error window is 20 counts.
So provided the servo can keep to within 20 counts (54 arc min) then it is deemed that the servo is adequately following the desired toolpath.
You could program the Following Error window to be much smaller, say 4 counts. Now what happens if the servo lags a commanded move by as little as
10.8 arc min then it will fault. If you have a slowly varying toolpath you might get away with such a tight Following Error window but if your
toolpath has a lot of high acceleration moves then it will nuisance trip all the time.

Ideally you would set the Following Error to a very low value so that you can be assured that the servo is accurately following the commanded toolpath,
but there is a balance to be found.

When a servo comes from the factory it is common for the Following Error window to be set very wide, maybe up to half a servo turn or several thousand
encoder counts. This will allow you to tune the servo without it faulting Following Error all the time. It is anticipated that you will reduce the Following Error
window when your done. Many first time servos users don't even know that its there and so you could have the situation where the servo deviates from
the commanded toolpath by as much as half a turn and still not fault out!

The second parameter, Zero Error Window, is specified in a number of encoder counts. Under normal circumstances a servo drive will apply current to
the servo motor to reduce the error between the commanded position (from Mach/controller) and the actual position as measured by the encoder.
If the motor overshoots the drive will apply reverse current so the motor backs up a bit. When it gets very close to the commanded position the
motor will go one step forward, realize its overshot and so go one step backward, only now to realize its overshot in that direction ......so it goes one step
forward.....and so on. Its called 'servo dither' and can cause buzzing/vibration and keeps the servo from cooling.

As a consequence most servo manufacturers have a Zero Error window so that if the commanded position an the actual position are within a few counts the
servo stop trying to reduce the error anymore, its considered good enough to perfect. Usually the manufacturers default value is adequate, it is still advisable
to consider the value applied to see whether it suits your machine.

Following Error Window is a most important but often overlooked parameter, and it is so important it can't just be left to chance, you must set it.

Craig