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Mach Discussion => General Mach Discussion => Topic started by: Retriever on December 11, 2020, 02:52:30 PM

Title: mach 4 learning the machine drive ratios
Post by: Retriever on December 11, 2020, 02:52:30 PM
Good day everyone. Welp, I have just purchased a bridgeport cnc series 1 from an auction. I have to upgrade all the motors, drivers, breakout board, and have to  find drive pulleys for the motors, as all this stuff has been removed before the auction. My questions are
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
2. I have to find out what stepper drivers, and breakout board mach 4 needs. Is there any resource for this?
The system will be windows 10, 64bit
Thanks in advance. 
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 12, 2020, 03:46:51 PM
Hi,

Quote
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.

Quote
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
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 13, 2020, 11:53:45 AM
Thanks Craig. I'm not that proficient with electronics. I'm a retired automatic transmission builder. I will follow your guidance. I am a firm believer that once right beats twice cheap every time. My needs will be for moving weight on the x/y axis, and I thing I would like nema 34 1800oz motors. Every time I try to save a buck, I end up sending 2$ after a dime.
This machine is in nearly new condition, and it is pristine. I bought it from a research university that did medical prototypes, and was quickly shoved in the corner when they upgraded to new machines. Again, I appreciate your knowledge, and willingness to share such knowledge.
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 13, 2020, 01:06:02 PM
Hi,

Quote
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.

Quote
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
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 13, 2020, 02:02:22 PM
Perhaps a nema 23?. Again, I appreciate your assistance
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 13, 2020, 02:24:50 PM
I guess if you had a 2hp machine, that you wanted to do heavy metal on, which motors would you use?
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage 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
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 13, 2020, 03:13:06 PM
Hi,
not sure why the other pics did not attach...

Craig
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 13, 2020, 07:49:50 PM
Super thanks, I will get all the data I can. My email is coloradochris20002gmail.com if this is more convenient to communicate. I'm really excited about this project. This looks like amazing fun.
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 13, 2020, 07:50:34 PM
correction, coloradochris2000@gmail.com
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage 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
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 14, 2020, 10:31:31 AM
perfect.
The major on the ball screws are 32mm
Pitch is approx 6mm as measured with a pair of vary nears
length of x axis is 686mm
length of y axis is 737
These are measured roughly with the machine assembled.
I measured rotating torque with a snap on torque meter.
7.5 inlbs break torque
5 in lbs rotating.
resolution is poor at best.
no motor or drive pulleys
Thanks again. Chris
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 14, 2020, 10:45:21 AM
btw, both driven pulleys are HTD 40 tooth, 8mm, steel
the z axis is 26 driven steel trapezoid with a 5 inch travel
I may have the same ball screws that you do.
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage 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:

Quote
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:
Quote
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
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 14, 2020, 04:59:22 PM
many thanks ;D
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 15, 2020, 10:42:17 AM
also, is there an opportunity to correct any undesirable calculations with a drive gear change?
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage 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

Jballscrew=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:
Jballscrew=(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.
Jarmature=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
Jlinear=(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:
Jtotal=Jballscrew+Jarmature+Jlinear
    = (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/Jtotal     (w= angular velocity in rad/sec, dw/dt= angular acceleration)
    =1/ 4.7 . 10-4
    =2127 rad/sec2
Converting that to linear:
Alinear=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
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 17, 2020, 11:11:33 AM
Craig, I'm blown away by your expertise. The z axis belt and pulleys are trapezoid, the spindle is ball screw. The spindle has an air operated variable pulley system for speed control, and an air operated transmission brake for the spindle. This is a pretty beefy machine. It weighs in at just under 3200 lbs. The z axis ball screw has a barely  perceptible "load", or drag when moved with hand motion on the belt. Its really free with no backlash. I think that's why Bridgeport used trapezoid belts and pulleys for the z axis.

I think that these drive systems on the x/y were 1:2. The nearest drive pulleys I can find in HTD 8mm is an 18 tooth. available in aluminum, or plastic.  I can go larger. I think this might be an opportunity to maximize motor capacity, or resolution. What do you think. I cant begin to express my gratitude for your assistance. This is a really fine machine. Thanks again. Chris.











Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 17, 2020, 11:16:29 AM
BTW. There is absolutely no hurry. I'm way beyond that. ;D. Being in a hurry is a young mans sport. Right once is always faster than twice fast. ;)
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 17, 2020, 01:44:17 PM
Hi,
I'm a little confused by your terminology.

You say 'Z axis and spindle'....what exactly do you mean?. My interpretation is the the Z axis is the Knee casting/screw/dovetail ways
and the spindle is the Quill, which also goes up and down?

Usually the Knee screw is very stiff, after all it is supporting the X axis , Y axis, the saddle and the Knee casting.
The quill on the other hand is much lighter but also usually only has 4 inches or so of travel.

With deep reduction (belt or gears) you can have a small motor do a big axis but at reduced speed.

My suggestion is specify both the acceleration and max speed you want of each axis and then do the calculation to see what sort
of motor and reduction is required to achieve it. If you choose really high accelerations and speeds (for production say) then don't
be surprised that you will require large and expensive motors.

To follow a toolpath high acceleration capacity is very desirable. Consider a toolpath that changes 90 degrees, like a corner.
In order for the machine to follow it must slow down and stop in one direction and the accelerate off in the new direction. If the machine
is to do this change quickly acceleration/deceleration must be high. Production industrial machines can be a s high as 10g.
My new build mill will be 0.25 g at rated torque and 0.75g at peak torque.

Maximum speed is another question to be pondered. There are two aspects, the first is the rapid or traverse speed. When traversing there is no cutting
forces and so all of the available motor torque is used to accelerate and maintain a given maximum axis speed. In industrial machines its common
to see traverse speeds of 30m/min and production machines up to 100m/min. My own new build mill will be 15m/min and with field weakening
(a  technique for exceeding rated servo speed) 25m/min.

The other speed aspect is maximum machining speed. Now the available motor torque has to accelerate and maintain axis speed AND supply cutting thrust.
Its not uncommon to see industrial machines with speeds of 40m/min traverse and 20m/min cutting say.

I would suggest for a hobbyist machine you aim for accelerations in the order 0.2g to 1 g and speeds of 10m/min rapid and 5m/min cutting.
Note that your machine has the mass and rigidity to go faster.....but do you have the budget to do it??

Craig
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 17, 2020, 09:32:28 PM
I was referring to the quill as the z axis. I am thinking that the knee will be "set" for a given operation. Once set, than the tool heights, fixturing and work can be programmed. My understanding of the machine is that all the dynamic functions of all the motors will be
x/y and the quill I'm referring to as z. I will focus on getting this machine functioning as well as I can get it with my limited skill set. I don't want my personal skill limits to compromise the build, but its not going into production either. Our discussions have already corrected a huge amount of mistakes not yet made. If I can get this machine to run well, maybe it will enable me to run well. That's the goal. I'm looking for the "goldilocks" outcome for my little situation.  8) That said, I would like to be conservative, but not cheap. I seem to want to overbuild everything, but I have in the past underbuilt, and have reaped the consequences.
Thanks again, Chris
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 18, 2020, 04:16:48 AM
Hi,
not sure what to make of the quill, I don't know how it works and the forces involved. The X and Y axes are a little more straight forward.

Lets assume that you wish to get close to the performance that I posted earlier, ie 10m/min rapids and 0.5g.

If the ballscrews have a pitch of 6mm, per your previous post, then to achieve 10m/min the ballscrews must rotate at
10,000/6 =1666 rpm.

There is your first hurdle, steppers lose torque badly with speed, and only very few of them can maintain any useful torque above 1000rpm.
You could use an increasing gear reduction but that would require even more torque from the steppers, I suspect its unlikely to happen.

Per my previously posted calculation you can get about 0.2g with 1Nm, so to get 0.5g will require 0.5/0.2= 2.5Nm. Note also that this makes no allowance
for friction nor ballscrew efficiency, so I would suggest that around 2.5Nm should be considered the minimum torque. Note that this torque is required
at top speed.

A very low inductance (1mH) stepper may retain 40-50% of its torque at 1000 rpm, which would suggest that the stepper needs at least 5-6Nm at low
speed. Steppers that have that sort of torque and low (sub 2mH) inductance are rear beasts. A more normal 34size stepper may well have adequate torque
at low speed but 34 size steppers have inductances of 4mH and more. Such a stepper is going to run out of 'puff' at 500-750rpm.

Side note: 1Nm= 130oz.in
                2.5Nm=325oz.in
                5Nm=650oz.in

All in all I think steppers may well provide enough torque at low speed but I suspect that they will fail to satisfy at speed. High speed operation
is always steppers Achilles Heel.
I think 34 size steppers of around 800oz.in and 4mh (or less is preferred) would work OK, but maybe fail to get to 10m/min, but more likely 5m/min.

The other alternative is servos. Servos come in a wide variety of specs, but the readily available brands likely to be affordable for hobbyists are:
200W 3000rpm =0.65Nm cont, 1.9Nm peak
400W 3000rpm = 1.27Nm cont, 3.81Nm peak
750W 3000rpm = 2.4Nm cont, 7.2Nm peak

If you used any of these three servos with a 2:1 reduction then you would get very close to your desired 10m/min rapids.

The 200W servo would, after the 2:1 reduction, have barely enough torque without dipping into overload all the time.

The 400W servo would after the 2:1 reduction would have plenty of torque, (1.27 x 2 =2.54Nm) without using overload.

The 750W servo has enough torque without reduction. Without reduction you could have rapids of 20m/min, alternately
with reduction you could still get near 10m/min but have a surfeit of torque for even higher accelerations.

So servos will certainly provide the best performance but they cost more than steppers, and steppers will do OK except
at high speed.

For my new build mill have bought three (one braked) 750W Delta B2 series servos. They cost (excluding the braked one) $460USD each, which
includes the drive and cables, plus shipping. I found that the 750w models were only $40 more than 400W ones, so I got the 750W's.

I notice that many US buyers favor Clearpath (US) over DMM (Canadian made in China) or Delta (Taiwanese made in China) but they are
more expensive, about $480USD for 350W. For various reasons, cost primary among them I went with Delta, almost twice the power for less money!

Ultimately your budget will determine what you do. I would recommend 750W servos if you can afford them.

Craig
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 18, 2020, 02:54:58 PM
Your counsel is super immensely appreciated. I think that long after I have forgotten how much they cost, I will be living with the results. Right once beats twice cheap every time. Therefore, I will be focusing on servos. Its tempting to go with steppers, but I want to continue to like/love my machine instead of the alternative. Success breads success.
 I already feel like you have steered me around big trouble. I think I will still do a 2:1 drive, and the 750w servos. I think I will be very happy with those results.
Again, much appreciated, and thanks, Chris.
BTW, what profile and size are your servo drive shafts?
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 18, 2020, 06:43:44 PM
Hi,
my 750W Delta B2 series are 80mm across the flange and 19mm shaft with keyway and seal.

I think you are right to go with servos despite the initial cost. If you get good quality and capable units they will give
twenty years untroubled service and even hold resale value whereas underperforming units will not.


Craig
Title: Re: mach 4 learning the machine drive ratios
Post by: Retriever on December 19, 2020, 03:26:31 PM
That makes perfect sense. The  ballscrews are the same. Also, as I understand, servo hold is determined by position. How does the "brake" function work on your braked servo?
Is there a possibility for a conflict between programmed stop, and applied brake stopping the servo short of its programmed satisfied position? In other words, could the brake apply causing the servo to continuously keep attempting to reach perfect position?
Thanks again for your generosity, and understanding dealing with someone with entry level , if that, skills. Chris
Title: Re: mach 4 learning the machine drive ratios
Post by: joeaverage on December 19, 2020, 04:08:07 PM
Hi,
a braked servo has an electromagnetic brake that comes on when its depowered.

With a normal servo when its depowered the armature may turn with only modest amounts of torque to overcome
friction and bearing drag. With such a servo if and when the Z axis is depowered gravity will act on the axis and potentially
the ballscrew will turn and the axis will sag.  With a brake however, when the servo is depowered the brake comes on to prevent
unintentional movement.

The brake could also be used to lock an axis say. If you drive a servo to a given position the servo drive will use its available power
and control to hold its position. There will always be a small amount of deflection about the controlled position however. It may be
that an electromagnetic brake is better in that circumstance.

I am not planning to use the brake it that manner....but who knows...things may change. My intention is to use the braked servo
on the Z axis so that on intentional, or otherwise, de-power the Z axis does not sag into the work zone. The spindles (two)
I have and use on my existing mini-mill will be used on my newbuild mill and they are light enough that I do not expect the Z axis
to sag under their weight.

I do, at a future time, plan to build a third spindle based on a largish AC servo of 2.7kW at torques of 14Nm. That servo and associated spindle,
which I envisage will be ATC, would be heavy enough that Z axis sag would be an issue. I decided therefore to bite the bullet and buy
a suitably equipped servo for that future plan.

I paid  $460 (excl shipping) for a 750W un-braked servo kit (servo, drive and cables) and the braked one cost $600....so you pay a premium for them.

In the case of my 750W Delta servo, and I imagine other brands would be similar, the brake is 24V at 250mA. So I need an auxiliary
power supply of 24V for the purpose. The brake is switched on or off by a electronic switch within the drive and that switch is in turn controlled
by the motion controller. There are a number of programmable parameters that govern how and when the brake operates within the drive,
but the typical brake-on and brake-off delays are 20ms.

Craig