PRO`s and CON`s, .... who dares?!
Hello good people,
I`m a student and designing my own little hf VMC (milling SMALL volume alu).
On http://www.sureservo.com/default.htm i saw some nice servo`s that will fit my machine, i think.
My machine has a gantry off 20 Kg and a working area off 500 by 350 mm.
The servo`s off 200 watt (0,8Nm continuous duty / 1,9 Nm peak ) are my objective.
There is only one thing, i want to use a ballscrew with a lead off 5mm/turn,
so 200rpm = 1000mm/min and 400rpm = 2000mm/min (this is my working area )
Maybe for rapids a bit more, since there is almost no weight and a
Massive / solid 150 Kg frame.
3000 rpm (rated rpm) =15000mm/min ......... i know this is impossible
But will the servo`s work on the rpm off 200 to 400?
i dont want to gear because off the backlash (out off the question)
gearboxes with no backlash i cant afford.
Servo`s have a constant torque, right
power is only torque * 2 pi n
It is overdone because off the range till 3000rpm, but no lost steps and a dynamic system is a pro. the servo`s have 10.000 ppr encoders.
Cann somebody give me advice and tell the PRO`s and CON`s ?!?!?!
Kind regards and keep up the good work guys,
Here are some extra parameters, with acceleration enz.
The forces and torque in the base are going to pretty big, i have basic drawn the design in Solidworks and did a stress analyse on it. i can keep the stress nice spread and the maximal displacement off material (bending off the base by forces) = 0,000037mm
For forces that come from milling, friction from the thk rails and F=ma and stuff. i have taken 500 N. Milling isn`t taking any force because the maximum depth/width is 0,5 by 0,5 mm.
You won't get the full power out of the motors without gearing. Have you looked at the toothed timing belt drive? They're relatively low cost and the backlash is not really significant.
In fact you will get more power without the gearing, just not the torque.
by adding gearing into it you will lose up to 10% because of the efficiancy loss in the gear.
Saying that it is a good idea to add a small ratio of about 2-3:1 with a belt and pulley.
Also is the screw rated to 3000 RPM? over 500mm, it may whip a bit?
Originally Posted by H500
The screw is rated at 2870 rpm but this will never be the case, even if i gear down 3: 1 then the screw will do max. 500rpm since the lead is 5mm thus 2500mm/min.
the thrust from the screw with a 5mm lead without gearing will be max.
1,9= F * (0,005/2pi) * ( 1/0,9 ) F = 2150 N
3:1 = 3* 2150 = 6450 N
On the calculation sheet i get acceleration 0ff 2000mm/sec^2
zero speed - start - till 100mm/sec - stop in 0,1sec this in a distance off 5mm.
I don`t need more thrust, why do i need more rpm at the servo?
The torque so the thrust will be the same from 10 till 3000 rpm on the servo motor
will the servo have the same pro`s at 200 rpm as on 3000 rpm ?
You are ultimately looking for force to move the table and that takes power.
You have a choice in how you get power:
High torque and low RPM motor
Low torque and high RPM motor.
High torque requires high current flow to electric motors.
High voltage usually begets higher RPM potential.
Thus, you make the compromise of using higher motor speeds to get the high velocity table travel desireds. To get the torque, you then start using gearing or pulleys (2:1 is typical) to multiply the torque.
Timing belt drives are very efficient and dependable. They also allow for a bit better packaging as you can tuck the motor out of the way.
THere are hundreds, thousands of machines running with 2:1 drives and timing belts in commercial and home use. IT is a proven and reliable drive system.
RE: power vs rpm, it varies from motor type to motor type.
DC brush motors w/perm mags make peak torque at stall and it drops pretty much linearly as speed reaches max.
Power wise, the 2:1 drove DC servos on my BPT can break off a 1/2" diameter cutter if you go home THRU a part so power isn't the problem once you have the gearing and motors matched to your speed/torque requirements.
You measure the M, plug in the A and then look for the F you need to do the job. The faster you want to do it only means you are going to have to do a the math for a torque/speed/gearing optimization.
Can't speak for other types.
Hello Nc Cams,
i think i understand you ?!?!?!?!
But i can calculate the force deliverd bij the ballscrew when a torque by the servo motor is applied on it
torque=F*(lead/2 pi) * (1/ efficiency)
if i select a servo motor on Torque then i know the force it cann deliver.
the torque is deliverd on the hole rpm range off the motor ( i see in the spec. sheet)
i know witch speed off the gantry i want, when i devide this by the lead i know the rpm off the ballscrew, so the rpm off the motor (if there is no gear)
in my case this is not at the max. rpm off the motor.
is this not good???
i don`t need more force thus gearing.
i read that servo`s have a sweet spot between 2500/ 3000 rpm.
the torque (so force on ballscrew) is the same over the hole rpm range.
why should i need to use the hole rpm range.
i dont need more force and the speed on 400rpm is for me good.
Sorry for the bad situation sketch, my english isn`t that good.
please bare with me
Power = K * torq * rpm.
Originally Posted by mcpltd
A servo motor has a fixed continuous torque value that does not change with speed. From the equation, it is clear that the motor will deliver 10 times more power at 3000 rpm as it does at 300 rpm.
Motors also have a peak torque value, but if you try to run there continuously, the motor will burn out.
Yes, but that is not taking into account gearing, you sugested gearing to gain power, and that is not the case.
Originally Posted by H500
gearing only adds loss in power, gians in torque and reduced speed.
adding a 2:1 ratio to a motor rated at 1Nm will not double the torque, it will double the torque minus the efficiency.
and as nc cams has said torque will drop as speed increases.
It is simply incorrect to suggest that a servomotor's torque drop with speed. Just take a look at any torque curve.
If your motor was rated at 3000 rpm and you need to run at 300 rpm, it does not matter how inefficient the gears are. Even at 50%, you will still get 5 times more mechanical power than a direct coupling.
I think we are misunderstanding each other, i know that it is better to use a redution ratio to get a lower speed, rather than lower the speed of the motor, that is obvious; but saying that by adding a ratio to the motor will give you more power because the torque is higher is wrong.
If a motor is 1Kw (3000 RPM), and adding a 10:1 ratio to multiply the torque by 10 will not give you any more machanical power.
Originally Posted by H500
The classic equation for HP is
Torque (ft-lb) x RPM / 5252
1 HP = 745 watts or 0.745kw
simple math can convert the units to any convenient/more familiar units base.
Thus, If power is to remain constact,
Torque has to go up as RPM drops
Torque has to go down if RPM increases.
Thus, if you have a motor htat will have double the torque at half the speed, the power will stay the same (of course negating frictional/inductive losses).
The trouble with this axiom is this:
motor torque curves are typically NON-linear. Thus, if you double or halve one or the other, power does NOT and CAN NOT remain constant providing the above mentioned HP equation is to remain true.
I work with a lot of people who try to violate this naturaly law, day in an day out.
Non have succeeded in doing so yet....