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Re: mach 4 learning the machine drive ratios
« Reply #10 on: December 14, 2020, 04:10:36 AM »
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.

My wife left with my best friend...
     and I miss him!
Re: mach 4 learning the machine drive ratios
« Reply #11 on: December 14, 2020, 10:31:31 AM »
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
Re: mach 4 learning the machine drive ratios
« Reply #12 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.
Re: mach 4 learning the machine drive ratios
« Reply #13 on: December 14, 2020, 01:40:01 PM »
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:

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'.


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:
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:

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


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

My wife left with my best friend...
     and I miss him!
Re: mach 4 learning the machine drive ratios
« Reply #14 on: December 14, 2020, 04:59:22 PM »
many thanks ;D
Re: mach 4 learning the machine drive ratios
« Reply #15 on: December 15, 2020, 10:42:17 AM »
also, is there an opportunity to correct any undesirable calculations with a drive gear change?
Re: mach 4 learning the machine drive ratios
« Reply #16 on: December 16, 2020, 05:39:14 AM »
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)
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:
    = (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)

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.

« Last Edit: December 16, 2020, 05:44:51 AM by joeaverage »
My wife left with my best friend...
     and I miss him!
Re: mach 4 learning the machine drive ratios
« Reply #17 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.

Re: mach 4 learning the machine drive ratios
« Reply #18 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. ;)
Re: mach 4 learning the machine drive ratios
« Reply #19 on: December 17, 2020, 01:44:17 PM »
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??

My wife left with my best friend...
     and I miss him!