reverse engineering of a gear/motor

[thepieman]

First of all nice forum.

Currently having a few issues and just want someone to check my method.

Im trying to spec a new motor and gearbox for a conveyor system inside a machine. The machine is quite old and the manuals nor the manufacture still exist. It currently uses a system of open gears (as in not in a gear box) and a motor but we are planning on changing to a gearbox and motor.

Basically to spec a new motor and gearbox we need a torque figure. My question is by reverse engineering what is there currently can we find a torque figure or am i missing something?

Say the gears are 100% efficient, that means that the total motor power (2.2kw) is at the output shaft. We know the output rpm (.1818 rpm) so using these figure we can work out the torque which the system is currently using as a maximum (at the output shaft). Then factoring in efficiency we can get the required torque of the system. Is this the way to go or is there a basic step im missing, just seems to simple

 

[DesignerGuy16]

T = HP x 5252/rpm

T = torque (lbf*in)
HP = horsepower
5252 = constant
RPM = revolutions per minute

HP = 2.2kw = ~3HP
RPM = .1818

T = 3 x 5252/.1818
T = 86666 lbf*in
T = 7222 lbf*ft

Something like that.

James Spisich
Design Engineer, CSWP

 

[drawoh]

thepieman,

Is there any chance you can disconnect the existing motor and gears, and test the conveyor for torque? You should be able to do this with a torque wrench, or with a standard wrench and a spring scale.

You do not know what the motor was originally intended to do. It may be dirty and corroded.

Look for badges on the motor. Typing part numbers into Google works most of the time for me.

JHG

 

[DesignerGuy16]

If he has a 3HP input that's presumably 1750rpm, I don't think he'd have a torque wrench large enough to see anything.

We have 1/2hp gearboxes with a 3500 lbf*in output torque at those kind of speeds. From the sense of scale what I posted sounds reasonable. Your output torque is directly related to the input power vs rotational speed. Yes you're going to get losses, but good luck figuring that out on an old/ancient machine without a standard gearbox. I'd assume more like 30-50% efficiency for that kind of reduction on a custom setup. You might see 50% up to even 70-80% efficiency from a standard box depending on the type out get, but expect to pay the higher you go.

James Spisich
Design Engineer, CSWP

 

[thepieman]

Thanks for your help. This conveyor is moving a load of about 11 tonnes so i don't think the torque wrench idea will work, unless you know of a huge wrench!

I just needed someone to confirm that you could use the input power and the output speed to get an output torque and that i wasn't losing my mind!

 

[thepieman]

Just one other thought. Say you calculate the require torque at the output to be 115,000 N.m with the gear at 100% efficiency.

If the gears are actually 50% efficient does this mean the torque required is going to be 172,500 N.m (150%) of required or 57,500 N.m (50%) or another figure all together?

Thanks

 

[DesignerGuy16]

If you want my honest opinion on your situation:

Talk directly with a gearbox manufacturer (Lenze, boston, sumotomo, winsmith, grove, rossi, etc etc). There's way too many to name. Give them your belt loading, RPM, input HP and discuss the situation. They'll help you size the application and correct style of gearbox.

Most gearboxes are rated by input HP/output torque. The output torque shown is a maximum loading (most gearboxes have a safety factor on these numbers). Basically you want to get a gearbox sized appropriately for a 3HP motor, maybe upsized slightly for higher duty and longer life. If you undersize you risk overheating the oil, and ultimately eating the gears. With that type of loading you're dealing with I'd be working through the numbers with a vendor and get their guarantee in writing.



James Spisich
Design Engineer, CSWP

 

[thepieman]

Thanks for that. I have spoken to a few suppliers about our requirements.

Just coming back to the efficiency bit, if efficiency decreases what happens to the output torque of a gearbox, does it increase or decrease?

 

[dvd]

Jspisich: you've got a bull in your formula, it should be
T=3*63025/.1818= 1040017 lb-in of torque

thepieman:
If you tell a gearbox rep that you want a gearbox that has a 9600:1 ratio and 3 hp input he will get that happy dollar sign look in his eyes. Your system may not have ever been engineered, so maybe instead of reverse engineering it you should just engineer it.

the load is 11 tonnes (2204 lb)
what is the conveyed velocity?
does it go up/down an incline? horizontal?
what is the load conveyed on belt? rollers? slats? chains?
you can calculate chain/belt pull force based on some friction factor x 11 tonnes.

Horsepower will be roughly = chain/belt pull x velocity / 33000

Torque is roughly = chain/belt pull x pulley dia./2

but, since I don't know specifics....

You can get some nice references on conveyor design at:

http://www.cemanet.org/publications/index.html

but post more details here and you will get some help also

You can ask about shaft sizing in another thread.

If you are looking for someone to do some engineering and refurbishment, that can be arranged also.

 

[DesignerGuy16]

That's what I get for using the internet instead of pulling out the references.

It all boils down to talk to a vendor and come up with a realistic change-out.

James Spisich
Design Engineer, CSWP

 

[diskullman]

I'd advise you to look in an old Rexnord book and reverse your engineering. Rather than work backwards from what is there currently, you should determine your load and service factors based on the conveyors specifications, 11 tonnes is a start, feet per minute, and so on. Complete directions from Rexnord as they manufacture many conveyor components as well as gearmotors.

Russell Giuliano