Thread: Quite an Unusual one

07122014 #131
On axial type bearings, the fatter ring faces are where the thrust is applied. Because there is a taper or similar in there.

08122014 #132
Yes, you want the larger side of the inner rings facing outwards so that when you do up the nuts you apply a preload to the bearings. Don't put the bearings in the wrong way round, as they'll be pushed apart by the nut and probably be broken. Don't do the nuts up too tight else the bearings will fail prematurely and don't leave it too loose else the shaft can float in the housing.
From the machining marks in the picture it looks like the bearing bores were milled? If this is the case then it's a bit concerning  if their milling machine doesn't have zero backlash then the bore will be less round, which would distort the outer ring. It's also generally less accurate than boring, which is why bearing housings are usually bored.
When I make these I bore them on the lathe and make the bearings a light press fit, to get better rigidity and ensure the outer rings don't creep. Also, if you get more made I'd be inclined to increase the axial spacing of the angular contact bearings, as it's currently insufficient to get the best stiffness. You'll probably be OK, but the critical speed should be higher if the bearing spacing is increased a little.
[QUOTE=silyavski;64503]i am still not sure. When all is ready and unleveled, we will see if i am right or wrong. If i had made a table and not all on the floor, the frame would be even stiffer, so that would not be an issue. But As you can see the garage is very small, lets say it will be the machine enclosure :).
On paper if the table hangs in the air one side and somehow is supported only by one side/cantilever/ the bend will be around 0.03mm or less. In normal conditions with 500 kg gantry in the middle less than 0.03, so for the side twist.
So basically if i have not made any mistake all must be well under 0.05, which seems quite an achievement to me for 3000x1800frame, having in mind it has only 500kg steel in it.
Regarding the pulley selection, you know the dimensions of the rotating ballnut shaft and the rest can be estimated well enough, so work out the inertia with aluminium and steel pulleys and see if aluminium is worthwhile? Or post the following numbers here and I'll show you:
 Mass of gantry (i.e. mass that both rotating nuts are moving).
 Desired rapid feedrate.
 Ballnut size (RM2510?).
 Motor ratings.
I think you'll find aluminium pulleys aren't worth it, due to the relatively high inertia of the ballnut  but no point speculating...

08122014 #133
I was so tired yesterday, cause i had a flue, that i missed that. I guess i was overexcited .They were told to bore it on the lathe and i was promised that even it would be done in one pass without turning the block. I hope they did it right. Now i am worried a bit.
RM2510 nut
20t:30t
400w Samsung AC servo each side,
~100120kg gantry
Decided to drivel all 20:30t, only the Z is 20:20t. This after exausting 3 day calculations, belt lengths,inertia, tooth engagement, speeds, etc. and at the end common sense.
This is a separate thread of its own. I believe this the wisest decision between speed, acceleration, precision.
The rotating ball nut assembly. 20:30t hence 1.5:1 driven. Compromise due to common sense. At the end precision and power at the same time is the aim.
2048 encoder on the motor, so more or less 20m/min. I see on the motor specs that it can go up to 4500rpm, just the curve is straight till 3000rpm. The resolution will be ~0.003mm on all axis , 0.001mm on the Z/ 1605 screw and 2500ppr encoder/. That without any artificial gearing or micro stepping which is not needed it seems. Thats good, cause the Galil board is not micro stepping board.

08122014 #134
We can calculate a decent approximation to the inertia of all the parts in the system using this formula for the moment of inertia of a cylinder:
Just break down the parts into cylinders and add them all up  e.g. for the ballnut we can consider it as three cylinders  the flange, the smaller diameter bit and the hole.
To find the equivalent inertia of the 120kg gantry, you need to consider the effect of the ballscrew, so this formula is required (where L is the pitch of the screw):
The last thing to consider is the effect of the drive ratio  in your case 20/30. This causes the inertia of the parts on the driven shaft (the nut etc) to be scaled by this factor squared. The drive shaft inertia (i.e. motor + 20T pulley) is left as is and we work out the equivalent inertia as if everything was placed on the drive shaft.The total inertia of the system is therefore:
Putting all that together (see calculations attached) I get the following values for the total inertia of the system. I've had to guess the dimensions a bit and I don't know the inertia of the motor, so feel free to double check and make it more accurate if you wish.
motor pulley:nut pulley = inertia:
steel:steel = 4.54e4 kgm^2
steel:alu = 4.31e4 kgm^2
alu:steel = 4.42e4 kgm^2
alu:alu = 4.19e4 kgm^2
(For interest, equivalent inertia of the 120kg gantry is 1.36e4 kgm^2  about the same as just the RM2510 ballnut!)
So as you can see, in your case the material you select for the pulleys doesn't make too much difference  even with both pulleys made from aluminium the gain is less than 10%. Lets check your not really close to the motor torque rating, else those few % might be valuable. The motor is rated for 400W at 3000rpm, so the torque rating is . Torque is the product of angular acceleration and inertia, so if we say you want 3m/s^2 acceleration (reasonably snappy for a machine this size) then you need 20/30*2*pi*a/L=1257 rad/s^2 acceleration so using the worst case (steel pulleys), 1257*4.54e4=0.57Nm. Add to that the frictional losses using this and you're still well below the continuous rating of the motor (let alone the peak), so I'd stick with steel pulleys as they will last longer. Just noticed I momentarily forgot you have two motors, so the required torque is actually even lower...
The other thing to check is the inertia ratio of the system (I.total/I.motor), to make sure it is within the ratings of the servo drive system otherwise the servos may not be able to control the machine stably. The motor is probably around 4e5 kgm^2, so your inertia ratio is around 10 at the moment. Do check what the servo system you have(?) can tolerate as that's a bit high. If it's outside the recommended range you can change the pulley sizes to get a better match.

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08122014 #135
Thanks Jonathan,
i remember i calculated all this at the time of choosing, again with your help there, but as i did not understand everything very well, it was like monkey playing with the calculator buttons.
Now i understand a little better
It seems the actual ratio would be 12, same as the max allowed by manufacturer.
I still have no idea about the acceleration you are talking about, what will that mean in real life, but i would like the machine to be more or less snappy as you say

08122014 #136
This is good example for me being similar to mine situation.
I am wonder too what is the normal acceleration for a machine like this namely how fast would you like to reach the feed speed or the rapid speed.
My unchecked source warn to use over sized servo 
it says if you want to spin up to 3000 rpm a 400W motor during 10ms it require temporary 2500 W
It did not mention the inertia ratio that is as I see very important
factor.
At first sight this 10 ms is to fast for me when I put together the mosaic now you only have to think of a corner you mill .The machine has to stop and change the direction and run up again and the surface quality depends on the acceleration hereLast edited by vargai; 08122014 at 10:34 PM.

08122014 #137
What Jonathan suggests "3m/s^2 acceleration (reasonably snappy for a machine this size)" should be quite snappy.
I could judge only from my lazy small machine and my first build which was for a friend, where i left the acceleration to 0.8m/s^2 cause after that it started to seem a bit dangerous.
Imagining that beast now to move the gantry from one end to other for 1 sec, thats would be quite a view. But as i understand it, the acceleration has more to do with corners, circles and so, than moving the gantry with tremendous speed from one side of the table to the other. Cause at the end i control the machine paths so generally i always strive to make them perfect, without wasted movement.

09122014 #138
Hi, I would like to apply this to my machine and wonder how it does come from ( see behind the formula)
That means this is not enough to count the reduced inertia on motor shaft with only the rotating parts?
120 kg*(0,01/2pi)e^2= 3,4 e4 differs from your number.
Where I am wrong Am I understanding everything well?
ThanksLast edited by vargai; 09122014 at 07:08 AM.

09122014 #139
So at 600mm/min machining speed means 0,01m/s
t=v/a=0,0033s it reaches the speed during 3ms is quite snappy but no experience just feel
Actually I cannot see any point to apply high acceleration rather high rapid speed if you want a fast maschine
BUT
My ignorance does not pull me back in making opinion

09122014 #140
Just spotted I forgot to attache the calculations to the previous post  I've attached them to this post instead.
RotatingNut_inertia.txt
I'll make some general points which I think should answer the questions  do let me know if I've missed anything.
 Acceleration, why have it?
A CNC machine rarely moves at a constant velocity, but does generally aim to move at a constant speed. The critical point here is the difference between velocity and speed  you can have the machine moving at a constant speed, whilst the velocity of each axis is varying greatly. Simple examples are cutting arcs or simply moving round a corner. This changing velocity means you have an acceleration.
The acceleration value you enter in the motor tuning is the limiting value  the controller should never command the drives to exceed it. You can therefore work out the minimum radius of a corner the machine can achieve for a given feedrate and feedrate. From the formulas for circular motion we can express the acceleration caused by the change in direction in terms of the feedrate and radius of the path:
For example, suppose you're cutting at 10m/min (e.g. cutting MDF) and you want to cut squares with 10mm radius corners, without the feedrate reducing at the corners, we can work out the required acceleration:
You can play with the numbers to see when this matters  but as a general point we can say that the required acceleration is proportional to the velocity squared, so if you're mainly cutting at low speeds (e.g. cutting aluminium) then the acceleration is not so important. This should also explain the differences in toolpath you see between using constant velocity (G64) and exact mode (G61).
 Inertia ratio
It seems their limit for the inertia ratio is set by the capacitance present in the motor driver dcbus. When the motors are decelerated, some of the energy stored in the total inertia of the system is transferred into the capacitors, since the drivers are supplied from a simple rectifier which only lets power flow one way. The problem with this is it causes the voltage on the capacitors to rise and if the energy transferred is to high (due to high inertia ratio), the capacitors will be damaged. It's actually quite easy to work out roughly how much the voltage rises, just equate the energy stored in the inertia, to the capacitor energy, and rearrange for voltage. The things we can do to alleviate this problem are add more capacitance, add a breaking resistor to dissipate the energy as heat or reduce the input energy by lowering the system inertia.
All of these seem reasonable options. Recall from my previous post that the equivalent inertia of the system depends on the square of the drive ratio, so by changing the drive ratio a small amount you can lower the inertia ratio. e.g. going to 18:30 instead of 20:30 would be sufficient. You should still be able to get 20m/min as the motor has sufficient torque at 3333rpm. A more interesting way round it could be to connect the DCbuses of all the servo drivers in parallel, as in general when one axis decelerates another is accelerating so will absorb the energy. You might still need a braking resistor to handle an estop event though.
 Equivalent inertia of gantry
In my previous post I stated the following formula, without deriving or referencing it, which I agree was not good practice so I'll explain where it comes from:
When I use the word inertia, for a rotating system I mean the moment of inertia. For a mass moving linearly, such as the gantry, the inertial mass is equal to it's mass, . The issue here is we need to incorporate this inertial mass into a rotating system, so we can sum the inertias seen by the motor shaft. The general formula to calculate the moment of inertia, for a mass () rotating about an axis at a distance from that axis is:
For every revolution of the ballscrew (or nut) the gantry moves a distance L (the screw pitch), so we can relate the angular speed of the ballscrew (or nut) to the linear speed of the axis as follows:
From the definition of angular speed we know that . We can combine these equations to find the equivalent inertia:
Substitute:
We're not quite there though, as the ballscrew (or nut) isn't directly coupled to the shaft. This causes the angular velocity to be scaled by the drive ratio, lets call it . If you include that in the above derivation you'll find the inertia is scaled by the square of this ratio, so:
Last edited by Jonathan; 09122014 at 04:19 PM.

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