[DAAD]

Machine details: Avidcnc Pro 5x10
Weight X gantry double motor 220kg
Weight Y gantry single stepper 125kg
R&P 3.2/1 reduction, 1" pulley pitch diameter
rotor inertia steppers 220G/cm2

Adam

[joeavaerage]

Hi,
as the calculation shows the inertia of a machine is dominated by the ballscrews.

I need the length, the pitch and especially the diameter of the ballscrews.

Craig

[DAAD]

Craig, no ballscrews, its a rack and pinion setup.

[joeavaerage]

Hi,
OK, then no rotational inertia from ballscrews.

Still need the pitch....or equivalent to it. I assume when you say 'pulley' you mean pinion, that is the gear that engages with the rack?

If that's the case then the 'pitch' is 25.4mm x pi=79.8mm.

Does that sound right to you? Does the gantry move approx 80mm (about 3 inch) per revolution of the pinion?

Craig

[joeavaerage]

Hi,
I see your steppers have a first moment of inertia of 220g.cm²

Do yourself a favor and do ALL calculations in MKS units, its just so much less confusing.

In MKS that is 0.22 x 10^-4 kg.m².

Just as a point of comparison my 34 size 750W servos have a first moment of inertia of 1.13 x 10^-4 kg.m².
So approximately four to five times the inertia of your steppers. Does that sound right to you? I am just trying to compare to give us
an indication whether we are on the right track, its real easy to do a calculation but because you did not think about whether the numbers you are applying
are sensible or not, you can go WAY wrong.

Craig

[DAAD]

this is what i can give:

MACH Settings for R&P:
* The R&P system is based on a pinion with a 1" pitch circle. The total linear distance traveled per revolution of the pinion is thus 3.14159". With the 3.2:1 reduction, this means that the distance traveled per motor revolution is 3.14159 / 3.2, or 0.9817". If you have a stepper with 200 steps per revolution, this means you have 200 / 0.9817" = 203.718 steps per inch, or 0.004908" per step. With 10x microstepping , you would have 2037.18 steps per inch, or 0.0004908" per step.

80mm sounds about right
inertia is 2200 g/cm2 had a typo.

thanks for the effort.

Adam

[joeavaerage]

Hi,
the data sounds good.
I'm glad I asked the question about the rotational inertia of the stepper....we could have gone off into the boondocks otherwise.

They must be quite big steppers? That spec puts them at nearly double the inertia of my Delta servos......which certainly would be correct for
42 size steppers and even some of the larger 34 size steppers. Does that match what you have?

I'll look at the numbers this weekend.

The calculation I've done is for ballscrews and I was pleased, proud if the truth be known, that I could derive the formula that relates linear momentum
to angular momentum. I was even more pleased when I could compare my derivation with the formulas published by Hiwin and Yaskawa and find perfect agreement.
I need to do the same thing for your machine but with RP rather than ballscrews.

Craig

[peteeng]

Hi DAAD - Its good to have data to make a decision so I looked at your mechanical system and its capabilities:
1) In terms of cutting forces (no accels) you have heaps of grunt available up to 100kgf in the Y (2 motors) and 50kgf in X (one motor) at your stated 500ipm. 50kgf is big enough to cut aluminium and heavy timber easily
2) In terms of accelerating at the stated 0.83m/s/s you again have potentially all the torque you need.

So either tune up your mechanicals as you may have lots of drag or friction to overcome or:

The result is I think you lack voltage so go hard on the voltage first. If you look at the dielectric strength of typical steppers they are around 500VAC. I suggest you run the drivers at their max voltage will make a huge difference. The motors only put out what is required to overcome the load so as the V goes up the current drawn goes down but power stays the same. (electronic people please correct if this is wrong). I had a little router that I tried to make portable. I used a 12v car battery first and it was soooo disappointing, then I use 2 so 24V and the delta was remarkable so put on another 36V and WOW. So then bought a 48V switching PS. But will go higher next time round and go to max driver voltage... Leadshine say 72V input for your drivers so go there...Peter

Oh yes check my numbers I have been known to be incorrect also you may want to reduce the reduction to get more speed. Maybe 2:1 seems speeds what you want?

[peteeng]

Hi DAAD - Some clarification needed. You said "reduction" but if the pinion is 1" (pinion is connected to the rack) then the crown wheel is 3.2x bigger so its a step up drive? ie you have a 3.2" wheel on the motor? This would make some my calcs wrong.... Plus what is the material and width of the pulleys so I can include them in the inertia calc....Peter

Looked up AVID you have two pinions confusssed...

so the picture is your drive? and its 3.2x total reduction. What is the PCD of the rack pinion ? is it 1" ? Peter

[DAAD]

Hey Peter,

Thanks for the input.
The photo of the drive is correct.
If i look at the avid cnc specs on the site the pinion (on the rack) is 1" PCD. The pionion has 2OT to 64T reduction wich makes the 3.2 reduction.
The pulley on the motor is also 20T, PCD i'm not shure off. Attached is a pic of the pulley on the motor.

At work atm, will look at the calculations this evening.

Adam

this is what avid gives on there site:

MACH Settings for R&P:
* The R&P system is based on a pinion with a 1" pitch circle. The total linear distance traveled per revolution of the pinion is thus 3.14159". With the 3.2:1 reduction, this means that the distance traveled per motor revolution is 3.14159 / 3.2, or 0.9817". If you have a stepper with 200 steps per revolution, this means you have 200 / 0.9817" = 203.718 steps per inch, or 0.004908" per step. With 10x microstepping , you would have 2037.18 steps per inch, or 0.0004908" per step.

[DAAD]

Hi,
the data sounds good.
I'm glad I asked the question about the rotational inertia of the stepper....we could have gone off into the boondocks otherwise.

They must be quite big steppers? That spec puts them at nearly double the inertia of my Delta servos......which certainly would be correct for
42 size steppers and even some of the larger 34 size steppers. Does that match what you have?

I'll look at the numbers this weekend.

The calculation I've done is for ballscrews and I was pleased, proud if the truth be known, that I could derive the formula that relates linear momentum
to angular momentum. I was even more pleased when I could compare my derivation with the formulas published by Hiwin and Yaskawa and find perfect agreement.
I need to do the same thing for your machine but with RP rather than ballscrews.

Craig

Stepper size is nema 34 with 6.8NM of peak torque.
See first post for specs.

[DAAD]

The result is I think you lack voltage so go hard on the voltage first. If you look at the dielectric strength of typical steppers they are around 500VAC. I suggest you run the drivers at their max voltage will make a huge difference. The motors only put out what is required to overcome the load so as the V goes up the current drawn goes down but power stays the same. (electronic people please correct if this is wrong). I had a little router that I tried to make portable. I used a 12v car battery first and it was soooo disappointing, then I use 2 so 24V and the delta was remarkable so put on another 36V and WOW. So then bought a 48V switching PS. But will go higher next time round and go to max driver voltage... Leadshine say 72V input for your drivers so go there...Peter

Oh yes check my numbers I have been known to be incorrect also you may want to reduce the reduction to get more speed. Maybe 2:1 seems speeds what you want?

Thanks,

There is already a 24VPSU on the way to place in serial with the 48V i have.
Rodw Mentioned this also.
So first point i will attack.

I have data for the motors at 72V with an 6arms driver, but my drivers only deliver 5arms if i look at the specs.
Still would be +_15% i gain with higher voltage.

[joeavaerage]

Hi,
the derivation of the momentum equation:
E_total=E_rotational + E_linear

Or:
1/2. J_effective.w_motor²=1/2 .J_rotational.w_motor² + 1/2 .m.v².........[1]

The challenge is to write the linear kinetic energy term in terms of w_motor:

The angular velocity of the pinion is w_pinion, so the revolutions of the pinion per sec are:
Pinion (revs/sec) = w_pinion/ 2 .pi and so the velocity is:

v=w_pinion/ 2 .pi * 0.08 ......................................... given that the circumference of the pinion is 0.08m

But with the gear reduction:
w_motor= w_pinion * 3.2

So;
v = (w_motor /3.2) / 2 .pi *0.08
v= w_motor * 0.004

Or:
v²=w_motor² * 1.6x10^-5

substituting in [1]:

1/2 . J_effective . w_motor²= 1/2 . J_rotational .w_motor² + 1/2 . m * 1.6x10^-5. w_motor²

Equating coefficients:

J_effective= J_rotational + m * 1.6x10^-5

Lets for the moment neglect the rotational moments of the pulleys and pinion so the only rotational moment is the stepper itself, ie 2.2 x10^-4 kg.m²
The Y gantry is 125kg, therefore the first moment of inertia of the linear axis is:

J_linear= 125 * 1.6 x10^-5
=19.8 x 10^-4 .kgm²

And so the total effective first moment of inertia is:
J_effective= (2.2 + 19.8 ) x 10^-4 .kg.m²


So unlike my new build mill where the ballscrews dominate the momentum equation, with the servo momentum being much smaller and the linear component smaller again,
in your machine the linear component of the equation dominates. So much so that it would be permissable to ignore the rotational momentum of the stepper without introducing
more than a few percent error.

Craig

[peteeng]

Hi DAAD - I'm not a electronics person but there can be issues attaching switch mode power supplies in series. Are they SMPS's ?? especially mismatched ones?? Anyone else up on this? Peter

https://electronics.stackexchange.co...crease-voltage

[DAAD]

Hi DAAD - I'm not a electronics person but there can be issues attaching switch mode power supplies in series. Are they SMPS's ?? especially mismatched ones?? Anyone else up on this? Peter

https://electronics.stackexchange.co...crease-voltage

Checked the manual, and it's possible to place this kind of PSU in serial/parallel if needed. Just need to provide a correct diode to get it working.

[DAAD]

Hi DAAD - Its good to have data to make a decision so I looked at your mechanical system and its capabilities:
1) In terms of cutting forces (no accels) you have heaps of grunt available up to 100kgf in the Y (2 motors) and 50kgf in X (one motor) at your stated 500ipm. 50kgf is big enough to cut aluminium and heavy timber easily
2) In terms of accelerating at the stated 0.83m/s/s you again have potentially all the torque you need.

So either tune up your mechanicals as you may have lots of drag or friction to overcome or:

The result is I think you lack voltage so go hard on the voltage first. If you look at the dielectric strength of typical steppers they are around 500VAC. I suggest you run the drivers at their max voltage will make a huge difference. The motors only put out what is required to overcome the load so as the V goes up the current drawn goes down but power stays the same. (electronic people please correct if this is wrong). I had a little router that I tried to make portable. I used a 12v car battery first and it was soooo disappointing, then I use 2 so 24V and the delta was remarkable so put on another 36V and WOW. So then bought a 48V switching PS. But will go higher next time round and go to max driver voltage... Leadshine say 72V input for your drivers so go there...Peter

Oh yes check my numbers I have been known to be incorrect also you may want to reduce the reduction to get more speed. Maybe 2:1 seems speeds what you want?

Thanks for the effort,
Trying to understand what you have done here.

The cutting page is clear to me.
About the torque page, if i understand correct, i need 0,42Nm just to run my system without other forces? I see you did calculate this on 1000ipm (max rapid rate) Any reason for this and not calculate at max cutting speed?
Third page is a difficult one... how do you get the 0.105 and the t=.5 sec? Not clear to me.

Adam

[DAAD]

Hi,
the derivation of the momentum equation:
E_total=E_rotational + E_linear

Or:
1/2. J_effective.w_motor²=1/2 .J_rotational.w_motor² + 1/2 .m.v².........[1]

The challenge is to write the linear kinetic energy term in terms of w_motor:

The angular velocity of the pinion is w_pinion, so the revolutions of the pinion per sec are:
Pinion (revs/sec) = w_pinion/ 2 .pi and so the velocity is:

v=w_pinion/ 2 .pi * 0.08 ......................................... given that the circumference of the pinion is 0.08m

But with the gear reduction:
w_motor= w_pinion * 3.2

So;
v = (w_motor /3.2) / 2 .pi *0.08
v= w_motor * 0.004

Or:
v²=w_motor² * 1.6x10^-5

substituting in [1]:

1/2 . J_effective . w_motor²= 1/2 . J_rotational .w_motor² + 1/2 . m * 1.6x10^-5. w_motor²

Equating coefficients:

J_effective= J_rotational + m * 1.6x10^-5

Lets for the moment neglect the rotational moments of the pulleys and pinion so the only rotational moment is the stepper itself, ie 2.2 x10^-4 kg.m²
The Y gantry is 125kg, therefore the first moment of inertia of the linear axis is:

J_linear= 125 * 1.6 x10^-5
=19.8 x 10^-4 .kgm²

And so the total effective first moment of inertia is:
J_effective= (2.2 + 19.8 ) x 10^-4 .kg.m²


So unlike my new build mill where the ballscrews dominate the momentum equation, with the servo momentum being much smaller and the linear component smaller again,
in your machine the linear component of the equation dominates. So much so that it would be permissable to ignore the rotational momentum of the stepper without introducing
more than a few percent error.

Craig

Craig,

Thanks for the calc!
If i understand correct my inertia ratio would be 19.8/2.2 = 9
Reading about inertia mismatch the say "best but almost impossible" would be 1/1. Ok would be be max 1/10. Safe would be 5.

So i'f im getting it right a higher rotor inertia would benefit my setup? Say motor inertia 4.5 kg.m2 instead of the 2.2 kg.m2, wich would result in:
19.8/4.5 = 4.4?

I'm i correct with this reasoning?

Adam

[Buoyen]

Inertia matching says something about how much of the total energy has gone info accelerating the system vs the motor. In the design stage this can gice you an idea about how big motor you need.

In practice the only thing that matters to me is: can I achieve the acceleration and speed I want with a certain motor? To do this take the total inertia (motor+weight+ screw+load etc) into an equation and see what acceleration you can get with the available torque that you have at a certain speed. If you calculate at your desired max cutting speed and your desired rapid speed (no load) and those two check ok you should be safe with steppers.

I'm using the calculator tha oriental motor has on their site: https://www.orientalmotor.com/motor-...ew-sizing.html

It does not take rotor inertia into account so you have to manually do an adjustment to the calculated inertia.

[joeavaerage]

Hi,
the inertia ratio is a figure of merit, and yes 5:1 is the 'sweet spot', however 9:1 is not out of the way.
As I posted if you ignored the rotational inertia of the motor you introduce about 10% error, however I have provided
the total effective moment of inertia including both rotational and linear so you don't need any course approximations.

You have the effective first moment of inertia so you can now calculate the kinematics, ie accelerations, speed etc.

dw/dt (rad/s²=Torque / J_effective

So for a 1Nm motor the angular acceleration is:
dw/dt= 1 / 20x10^-4
=500 rad/s²

Converting to linear:
a (m/s²)= dw/dt /(3.2 * 2.pi) *0.08
=1.98m/s²................................So about 0.2g accel with a 1Nm motor, is not to shabby. A 5Nm motor could get you 1g accels!

Craig

[DAAD]

Inertia matching says something about how much of the total energy has gone info accelerating the system vs the motor. In the design stage this can gice you an idea about how big motor you need.

In practice the only thing that matters to me is: can I achieve the acceleration and speed I want with a certain motor? To do this take the total inertia (motor+weight+ screw+load etc) into an equation and see what acceleration you can get with the available torque that you have at a certain speed. If you calculate at your desired max cutting speed and your desired rapid speed (no load) and those two check ok you should be safe with steppers.

I'm using the calculator tha oriental motor has on their site: https://www.orientalmotor.com/motor-...ew-sizing.html

It does not take rotor inertia into account so you have to manually do an adjustment to the calculated inertia.

Thanks for the link!
Unfortunate but there is no rack and pinion option on there site.

- - - Updated - - -

Hi,
the inertia ratio is a figure of merit, and yes 5:1 is the 'sweet spot', however 9:1 is not out of the way.
As I posted if you ignored the rotational inertia of the motor you introduce about 10% error, however I have provided
the total effective moment of inertia including both rotational and linear so you don't need any course approximations.

You have the effective first moment of inertia so you can now calculate the kinematics, ie accelerations, speed etc.

dw/dt (rad/s²=Torque / J_effective

So for a 1Nm motor the angular acceleration is:
dw/dt= 1 / 20x10^-4
=500 rad/s²

Converting to linear:
a (m/s²)= dw/dt /(3.2 * 2.pi) *0.08
=1.98m/s²................................So about 0.2g accel with a 1Nm motor, is not to shabby. A 5Nm motor could get you 1g accels!

Craig

Thanks!