hello,
i don't know where to post this so can someone please move it for me if its in the wrong place

thankyou

i'm designing a CNC slantbed lathe and i wish to replace my steppers with servos for accuracy and noise issues

i am currently using these stepper motors
http://uk.rs-online.com/web/search/searchBrowseAction.html?method=searchProducts&searchTerm=5350439

these are 1.26nm force and are nice and strong and perfect for cnc machining

however i know nothing about servo motors and have this problem

is this servo more powerful than the stepper shown above as i want either equal or better for my lathe
http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=3237019

this one says its 0.16nm but is 3000rpm which is extremely fast for an x,z control
if i was to gear this 10:1 ratio so that its now running at 300rpm at the ballscrew would this mean the torque (nm) now equals 1.6nm???

im sorry if im completely wrong on all of this but as there expensive compared to steppers i want to make sure there 100% compatible and accurate before i buy

also the servo has the encoder included with it is this the right servo to be used in mach 3? and can i power the servo from this
http://geckodrive.net/g320x-p-28.html

sorry for all the questions but if anybody can help me on whats needed for a full servo setup or any links they would be greatly appreciated

thankyou

max

This is what I found from this website:    http://machinedesign.com/article/new-rules-for-sizing-servos-0712

Gearboxes offer a significant benefit in that they affect the inertia ratio by a factor of the gearbox ratio squared. This means, for instance, if a load has an arbitrary value of 100 and the gearbox ratio is 5:1, then a motor with an inertia value of 1 would result in an inertia ratio of 4:1. (100 ⁄ 5 2 = 4, or 4:1.) Without the gearbox, the motor would have to be extremely large to accelerate the load at the same rate.



Also from this site :  http://www.faulhaber-group.com/sprache2/n129872/n.html

Q: How can I calculate the final operating conditions (current, speed, etc.) for a motor+gearhead combination given the torque load at the output shaft of the gearhead?
A: To give you a generalized example, assume that the motor+gearhead combination 1724T012S+16/7, 43:1 is being used with 12 Volts applied to the motor terminals, and that a torque of 10 oz-in is desired at the output shaft of the gearhead.
 
Gearhead 16/7 with a 43:1 gear ratio has a data sheet efficiency value of 70%. This means that 30% of the power developed by the motor will be lost in the gearhead. The simplest method of accounting for gearhead losses is to increase the torque requirement by the appropriate amount and make the calculations as if the gearhead were 100% efficient. In this case, we increase the torque requirement at the gearhead output by 30% resulting in a torque (for calculation purposes) of 13 oz-in.
 
Total torque = 10 oz-in x 1.3 = 13 oz-in.
 
The torque reflected back to the motor is then simply the total torque divided by the gear ratio:
 
Motor torque = 13 oz-in / 43 = .302 oz-in.
 
The motor torque constant is the proportionality constant which defines the relationship between the torque at the motor shaft and the current in the motor windings. In this case, the torque constant for the motor 1724T012S is 2.01 oz-in / Amp. That is, for every 1 Amp in the motor windings, the motor will produce 2.01 oz-in of torque. The reciprocal of the motor constant in this case is .498 Amp / oz-in. Since we have already calculated the torque at the motor shaft to be .302 oz-in, we can use the reciprocal of the torque constant to calculate the motor current due to the external load:
 
Current = (.498 Amp / oz-in) x  .302 oz-in = .150 A = 150 mA
 
The motor has a small amount of internal friction which requires a proportionate amount of current to drive it. This current is defined as the motor no-load current. In this case, the value is 8 mA (taken from the data sheet). Since the motor requires 150 mA to drive the external load and 8 mA to drive its own internal friction, the total current required for this application would be 158 mA.
 
The speed of a DC motor is a linear function of the load which it is driving. The proportionality constant relating motor speed to the motor torque load is the slope of the torque versus speed curve. This slope is calculated by dividing the listed no-load speed (actually a negative value) of the motor (maximum speed and zero external load) by the stall torque (zero speed and maximum torque). In the case of motor 1724T12S, the slope of the torque versus speed curve is given by the following:
 
Slope = DY/DX = -8000 rpm  / 1.49 oz-in = -5369 / oz-in.
 
Note that the slope of the line is a negative value, indicating that the speed losses will be greater with increasing motor load. In this case, we calculated a motor load of .302 oz-in. Therefore, the motor speed loss due to this external torque load will be:
 
Speed loss = (-5369 rpm / oz-in) x .302 oz-in = -1621 rpm
 
With no load on the motor shaft, the motor speed will be 8,000 rpm. With a load of .302 oz-in, the motor will lose 1621 rpm from the no-load value. Therefore, in this application the motor speed is rendered by:
 
Motor speed = 8000 rpm - 1621 rpm = 6378 rpm
 
The speed of the motor at the output shaft of the gearhead under load is simply the motor speed divided by the gear ratio. In this case:
 
Output speed = 6378  / 43 =  148 rpm
 
We accounted for the power losses in the gearhead at the beginning of this exercise, so we need not be concerned about this factor again.
 
If you want help working out your particular application, please call one of our Applications Engineers, toll-free from the USA or Canada, at 1-800-807-9166.

   

Torque calculator  :   http://www.mnbigbirds.com/Servo%20Torque%20Caculator.htm

Jeff