Figuring out Gear ratio and motor torque on a flat conveyor 1

[abrown1965]

I am looking at using a MK conveyor that you can mount fixtures on to hold parts that will be driven by a servo motor and a gear box. The total weight of all of my fixtures and parts that will be on the conveyor is 100 pounds. The conveyor will have to move 360mm every 18 seconds and needs to move this distance in 1 second. The conveyor has a belt radius of 52mm.
I wanting to know how to figure the gear box ratio that I will need and the servo motor torque that I will need. Can anybody help me out? I do not want to know the answer, I want to know how you can get the answer so that I can figure this out on my own next time.

 

[BrianPetersen]

Figure out the acceleration and top speed needed to give the required motion profile at the conveying surface.
In the interest of "just making sure", add at least 0.1 g to the theoretical acceleration to account for friction, etc., and that is for clean conditions, if you are conveying rocks through a mine in a dirty environment, the friction could be much bigger than this.
Figure out the total mass of all the parts, the fixtures, and the belt.
F = m x A
T = F x R (at the conveyor pulley)
Figure out the rotation speed of the conveyor pulley at the conveyor's needed top speed
The ratio between that rotation speed and the rotation speed of commonly available motors will tell you the gearbox ratio.
Then figure out the theoretical torque without considering frictional losses.
Obtain the efficiency of the gearbox and obtain the required motor output torque accounting for gearbox losses.

It would be wise to add some percentage to the calculation results, both in terms of torque and speed, just to make sure.

 

[BrianPetersen]

I should add that in many cases, the inertia of the gearbox and motor can be significant, and the above does not account for that. Inertial effects will be on top of the actual torque needed to move the load.

 

[abrown1965]

This will be in a clean environment. With the conveyor having to travel 360 mm in a second then that would equal .36 meters.
With the conveyor having a radius of 52mm that would equal the circumference of 0.32656 meters. So with one revolution of the conveyor I almost travel the distance needed. I will be using a servo motor and the gear ratio will be determeind on the RPM of the motor I pick. Could you show me how you calculate the torque so that I understand it

 

[abrown1965]

This will be in a clean environment. With the conveyor having to travel 360 mm in a second then that would equal .36 meters.
With the conveyor having a radius of 52mm that would equal the circumference of 0.32656 meters. So with one revolution of the conveyor I almost travel the distance needed. I will be using a servo motor and the gear ratio will be determeind on the RPM of the motor I pick. Could you show me how you calculate the torque so that I understand it

 

[abrown1965]

This will be in a clean environment. With the conveyor having to travel 360 mm in a second then that would equal .36 meters.
With the conveyor having a radius of 52mm that would equal the circumference of 0.32656 meters. So with one revolution of the conveyor I almost travel the distance needed. I will be using a servo motor and the gear ratio will be determind on the RPM of the motor I pick. Could you show me how you calculate the torque so that I understand it

 

[abrown1965]

I also read that on servo motors that if I have to use a gear reducer (gear box) to achieve my speed of my conveyor that what ever ratio I use multiplies the torque of the motor also.....So if I end up using a motor with 5 torqu and I then use a 3:1 gear box then the real torque of the motor is 15 coming out of the gear box......Is this correct....

 

[chicopee]

The torque coming out will be 1/3rd of the torque in, as long as you neglect frictional forces, otherwise, it will be lowered. It can be easily proven, however, what you should be more concerned is how much horsepower is available from the output shaft and that answer will be the same horsepower delivered by the input shaft, ie, theoretically, otherwise again, it will be lower.

 

[BrianPetersen]

Pick a motion profile.

There is no single unique solution to moving 360mm in 1 second so you just have to pick something that works. For example, you could instantaneously accelerate to 360mm/s and then run at constant 360mm/s and instantaneously decelerate. This gives the lowest top speed but of course instantaneous acceleration requires infinite force and so this solution is not practical. An attempt to come close to this would be violent on your fixtures and workpieces.

Another choice is to accelerate uniformly over 1/2 second to some top speed that we don't yet know and then decelerate uniformly to zero over the next 1/2 second. This motion profile will minimize the acceleration.

From basic dynamics

D = 1/2 * a * T^2

We know D (= 180mm, half the distance that you need to travel) and we know T (1/2 second, half the time available) so solve for a:

a = 2 * D / T^2 = 2 * 0.18 / (0.5^2) = 1.44 m/s2 (For reference, gravity is 9.81 m/s2 so this is quite plausible and achievable)

We need that top speed ...

V = A * t = 1.44 * 0.5 = 0.72 m/s

Other motion profiles are possible. You could accelerate a bit harder to a slightly lower top speed, plateau there for some period of time, then decelerate slightly harder. It's an arbitrary choice at that point. But if your drive is capable of giving the speed above, and capable of delivering something greater than the calculated acceleration given above, you will have some room to make that decision later.

Can you take it from there? I know you said you didn't want numbers, but it's too hard to explain this concept without applying some numbers to it, might as well use yours.

 

[tygerdawg]

The key to your calculation should be sizing for 'peak torque' requirements. This is the sum of all the worst case torque needs. Includes torques needed to move translational & rotational masses, torque to accelerate your inertias, torque for friction/stiction, torque for phase of moon (wink), torque for gravity effects, etc and so on and so forth. Acceleration values come from deltaVelocity / deltaTime. These are not hard calculations, but can be a bit tedious. I also remember a significant factor of "number of starts". Starting the motor at a high enough starts-per-hour frequency would heat the motor, so the recommendation was to upsize the motor to accommodate extra heat. It has been a number of years since I was doing this type of work, but I remember that SEW Eurodrive had an online sizing calculator that was quite useful. Based on various application 'models' (like your conveyor), one inputs the values and the calculator spits out one or more recommended gearmotor part numbers specifying horsepower & gear ratio. I suspect that now many other gearmotor manufacturers have something useful also. If nothing else, the gearmotor catalogs usually have an extensive application engineering section to guide one through the sizing process. Go through the sizing calculations, develop required power & ratio, and select from the catalog lists. And the catalog lists only give (usually) the most popular products. An almost infinite variety of power/ratio can be fabricated. Then after all of that, you must select the shape style of the gearmotor that best suits your application needs.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[MintJulep]

BrianPeterson has laid things out nicely.

I'll just add that each of his suggested motion profiles have distinct discontinuities in the velocity vs. time profile - hence theoretically infinite jerk at those transitions.

A sinusoidal motion profile would avoid that.

 

[mikekilroy]

just use sizing program like this good one:

http://www.kollmorgen.com/en-us/service-and-support/technical/motioneering/

http://www.KilroyWasHere<dot>com