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Well actually what I need is a little more than simple math...
Is there a site with information or a sample calculation on how to obtain the torque and rpm needed to move a gantry with a x weight?
What I need is a a procedure to calculate what servos I'll be needing. having:
a. the weight of the gantry
b. the max cutting speed
c. the max repositioning speed
this is for a plasma table, so there will not be reaction forces acting against the gantry as in a router table.
Any help, is very appreciated...need to buy the servos right away
Thanks,
Hector
Look at the servo manufacturers web sites for free graphic sizing programs, most are allied to their particular product but you can get pretty close and get a feel for what you need. e.g. http://www.electromate.com/technicalsupport/?c=software , parker-hannefin, www.motiononline.com , Galil.com Pacific-Scientific, Reliance Electric, Allen-Bradley, Electro-Craft.
Al
CNC, Mechatronics Integration and Custom Machine Design (Skype Avail).
“Logic will get you from A to B. Imagination will take you everywhere.”
Albert E.
There are a few variables you need to know in order to figure this out:
1. The friction of your linear axis.
- are you using high quality linear slides, or something else?
2. The friction of your screw.
- an acme screw has an efficiency of around 36%
- a high quality ballscrew has an efficiency of around 91%
3. The weight of your gantry.
For example, a hypothetical 10 oz-in (.63 lb-in) motor with direct (1:1) gearing to an acme screw with pitch of 0.25" will be able to exert about 5.7 pounds of force on whatever it is moving:
thrust = gear mutiple * [1 / ( [(screw lead / 2*pi) * (1 / screw efficiency)] / torque)]
thrust = 1 * [1 / ( [(0.25 / 2 * pi) * (1 / 0.36)] / 0.63)]
thrust = 5.7 pounds
(Note: if we used a 1:2 gear off the motor, we'd get double the thrust)
I imagine that for a moving gantry with no other forces acting on it, the peak load on the motor is going to occur when you want to reverse the direction of the gantry very quickly (say within 50 milliseconds). If the gantry weighs 200 pounds (75Kg), and it is moving at a speed of 6 cm/sec (roughly 140 ips), then we really want this thing to change its speed by a total of 12 cm/sec
(remember, it is reversing direction and accelerating back up to 6 cm/sec) within 50 milliseconds.
Force = mass * acceleration
So for acceleration of 0.12 m/sec / 0.050 seconds, we'd have F = 75*0.12/0.050 -- or F = 180N
180N is about 40 pounds of force (1 pound-force = 4.448 N)
So, in this case, with a 200 pound gantry we'd like to run at 140 ips and reverse direction in 50 milliseconds or better, our little 10 in-oz motor will not cut it. We'd either need to go directly to about a 100 oz-in motor, or we'd need to do a combination of the following:
1. Use a higher efficiency screw
2. Use a stronger motor
3. Decrease the pitch of the drive screw (this will slow down our gantry, too)
4. Use a lighter gantry
5. Use a different gear ratio to drive our screw (this will also be slower)
6. Relax our requirement that the gantry change directions in 50 milliseconds.
We ignored friction in all of this, but if using a good linear slide, the variance shouldn't even be worth calculating.
For just the gantry moving speed, as a separate calculation, you'd do this:
linear speed = motor rpm / gearing * screw pitch
Example: 2500 rpm motor / 1:4 gearing * 0.250 pitch
= 2500/4*0.25
= 156 inches per second
Note: In our above example, if we did use a 1:4 gear, we'd get 32.5 pounds of force at the screw. We could then think about going to a 30 oz-in motor, and get 97.5 pounds of force, which should be plenty. We'd have 156 ips gantry jog speed, and we could easily push around a 200 pound gantry.
-Chris
Hi Chris,
Thanks for so helpfull information...regarding your comments,
1. I'm thinking on using either camfollowers over a machined flat or v bearings over either regular steel angle or v railings(expensive! for a 6'x12'or20' table). I don't have a remotless idea on the friction on this kind of linear axis.
2. Im planning to use rack and pinion in x,y,z axis, don't know which one is better 14.5 degree angle or 20 degree, and I don't know either what kind of friction they impose on the system.
3. The weigh of the gantry, well, Im planning to do it from steel so it will be pretty heavy. the maximum speed for cutting is 240ipm, I'm building this table for a Hypertherm HT2000 High Speed Plasma.
I'm planning to use servos with drive belt reduction...on x and y axis. for the z just a small dc motor, I'll be using the rutex 510 board.
What kind of calculations will be involved in this type of design...in conclusion
---big table
---heavy gantry
---fast speeds
---rack and pinion
---v bearings or cam followers
Regards,
Hector
In addition to the forces associated with friction, don't forget to add in the force associated with acceleration. Most of the forces associated with friction are independent of the speed (I assume that you are not going so fast that air resistance is a consideration).Originally Posted by kanankeban
Ken
Kenneth Lerman
55 Main Street
Newtown, CT 06470
The 20 deg. angle spur gear/rack takes a heavier load than the 14.5
Remember on a horizontally controlled axis, especially if there is minimum side thrust as in Plasma cutting, the maxim torque required is not at rest or at final travel speed , but at acceleration & deceleration (inertia calculation) for very good design information, check out this site and if you get their free library of catalogue, half of them is dedicated to Engineering topics.
http://www.sdp-si.com/D220/D220cat.htm#S1
BTW are you driving both sides of the gantry? At that weight you may need to.
Al
CNC, Mechatronics Integration and Custom Machine Design (Skype Avail).
“Logic will get you from A to B. Imagination will take you everywhere.”
Albert E.
Hector, the only calculation that has any importance is the force required to accelerate (and to stop) the gantry.
Here is how to work backwards to a motor size specification. I'm going to do this entirely in metric (SI units), then we can convert the answer to Imperial units when we get toward the end.
1. Determine the mass of your gantry, in Kg. You stated that the gantry is steel and pretty heavy. I couldn't even guess how heavy it is, so let me just pick a number of 400 pounds, or about 182Kg.
2. Determine how fast you need this gantry to move. You said 240 ipm, so we will convert that to metric and say it is 610 cm/min or 11 cm/sec (rounding up).
3. Determine how quickly the moving gantry has to be able to slow down to a complete stop, then reverse direction. In my other post, I used 50 milliseconds. I suppose we could also use 100 milliseconds (or maybe even 200 milliseconds). It depends on what you are milling. For me, doing woodworking, I don't want a router bit to linger any more than 1/10th of a second when cutting a hard wood like maple.
Okay, so now we have the mass of the gantry, its constant speed, and how quickly we need it to come to a stop, then reverse direction. Let's look at the formulas:
1. F=ma (Force equals mass times acceleration). Again, if we say that we want the gantry to completely reverse direction in 100 milliseconds, this means that we are doing this:
- traveling at 11 cm/sec
- coming to a dead stop
- traveling in the opposite direction at 11 cm/sec
- doing all of this in 100 milliseconds
So, the net change in acceleration is 22 cm/sec (since we are reversing direction), and we need to effect that change in acceleration in 100 milliseconds. Our gantry has a mass of 182Kg, so the equation is:
F = 182 * 22 cm/sec / .1 seconds
This reads as "Force equals 182 kilograms times 22 centimeters per second, per 100 milliseconds"
One minor touch-up (since SI deals with meters, not centimeters) is that we will convert 22 cm/sec into .22 m/sec:
F = 182 * .22 m/sec / .1 sec
Solving, F = 400N (Newtons, the SI unit for force).
So, we need a drive system that can impart 400N of force on our gantry. How do we build such a drive system? Read Arvid's (Arvidb) white paper for the details, but in summary you need to select a screw pitch and step-up pulley system that will give you the force you need at the screw. At this point, I'm going to convert 400N into Imperial units. 1 pound-force = 4.448 N, so our 400N turns into 90 pounds-force (rounding up).
So we need a drive system that can impart 90 pounds of force at the screw. Well...that cuts it too close, so let's pick a multiplier to give us enough headroom (for friction, or mabye the gantry isn't perfectly level, etc.). Let's use 2x as the multiplier, and agree that we need to generate 180 pounds of force (or 'thrust'):
The rest of this will be in Imperial units.
(gear multiple * motor torque) = thrust * (screw lead / 2*pi) * (1 / screw efficiency)
or
motor torque = (thrust * (screw lead / 2*pi) * (1 / screw efficiency)) / gear multiple
We need to pick a couple of values, so let's use:
gear multiple = 3 (i.e. a 1:3 pulley configuration)
screw lead = 0.250" (4 turns per inch screw movement)
screw efficiency = 0.90 (ball screw)
solving the equation:
motor torque = (180 * (0.250 / (2 * 3.14)) * (1/0.90)) / 3
motor torque = (180 * 0.0398 * 1.11) / 3
motor torque = 2.65 lb-in, or 43 oz-in
So, for this example, it says that a 43 oz-in continuous torque servo motor will work. You'd need to run that motor at around 3000 RPM to end up with 240-250 ips linear speed through that 1:3 gear and 4 TPI screw.
-Chris
Hey Chris,
Thanks so much for so usefull information, now let me get some time to fully understand the process, and I will come back with some more doubts for sure
Regards,
Hector
WoodSnarfer,
First, thank you for posting such complete (as far as I know it's cmplete) example. I have taken your equations and put them in an Excel spreadsheet. My goal is to get your help in confirming/authoring the spreadsheet to a final form, then convert it to a php web form for universal use.
Please have a look and let me know if this would be something your would want to work on.
Jay
Last edited by Jay C; 02-08-2005 at 04:12 PM.
Jay,
It's not 100% complete. That's why I throw in a 2x 'fudge' factor. To really do this the right way, the set of equations should take into account:
1. the mass, diameter, angular velocity and acceleration of the drive screw(s).
2. friction (for every contributor of friction in the system).
3. estimated peak cutting force required.
Even after that, we'd probably throw in the 2x 'fudge' factor, since it's never a good idea to spec anything at the hairy edge of your math results.
I learned (well, relearned) most of this stuff from reading Arvid's white paper, which you can find here (and I think you've already seen) - along with revisting physics notes on the internet, and checking out manufacturer's product specs and engineering equations.
http://www.cnczone.com/forums/attach...achmentid=4954
But, even so...it would be pretty cool to have a web page that does the calculations for a given set of specs for an axis. I think such a utility ought to accept input in any form (Imperial or Metric), for any units (oz-in, lb-in, Nm, etc.) and be smart enough to do the appropriate conversions and spit out the answer in the format(s) the user wants.
I had cobbled together a worksheet to help me evaluate motors...here's a screen shot if it helps you think about it at all.
I'll work out the equation for screw angular acceleration/force and see if it can easily be added in to the mix...unless someone reading this beats me to it, of course
-Chris
Chris, I updated the spreadsheet (twice now) after reading avidb's (while you were posting) whitepaper.
Anyone care to provide feedback on the equations ... or is this a useless tool?
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