Torque Req'd to Start Movement

[bobert]

My apologies in advance...I may not supply all the necessary info. on the first try. I need to build a material cart (on casters of some 8" to 12" dia.) that will carry a #2,500 load (total weight: cart + load = approx. #3,500). Cart is approx. 4'-0" x 8'-0" x 3'-0" tall. The cart will be manuevered by a human. The environment is hazardous (fumes from process) so the device should be non-sparking, explosion-proof, etc. The idea, so far, is to look into using an air gear-motor to drive the cart. I don't know if this will work as one of the first things I need to know is "[red]How much torque will I need to overcome the "at rest" position, break friction, start to move, whatever the term is[/red]". Also, I will probably be interested in horsepower requirements and I don't know that formula either. Is HP simply the value by which the cart remains in motion? My need is probably for slow movement, low rpms i.e., "Granny Gear". So I'm thinking that the important factor is the torque. (?)
Another thought (if the air motor can't produce enough torque) is to have a sort of "power assist" thing where the cart does not move because of the air motor but with a (little) effort by the operator the cart can be manipulated and possibly stopped sooner / easier. So, is there any formulae for calculating "power-assist" forces? I would greatly appreciate any help, ideas or thoughts regarding this. In fact, there may be a better way and if someone has an idea, please let me know! Thank you.

By the way, does anyone know if this is even remotely possible without the need for power? Given the right size wheels, proper bearings, etc., is it possible that a human could move this much weight without unreasonable effort & stress?

 

[insideman]

Decently lubricated device should have near zero starting torque under ideal conditions. However, I doubt that your conditions are ideal!

There will be slopes, irregularities, dirt, other stuff on the rolling surface that will affect this.

Wild guess is that you will need a force of 5-10% of the weight to start motion. Convert this force to torque.

 

[butelja]

How about assuming a maximum "bump" size for irregularities in the floor? If you do this and look at the geometry of the wheel rolling over the bump, you can calculate the torque required to cause the wheel to climb over the bump. Compare this to INSIDEMAN's 5-10% recommendation and take the greater of the 2.

 

[GregLocock]

A human could easily move such a weight if the tyres were of sufficient diameter, and the floor sufficiently smooth, and you were able to spend enough money. Typical coefficient of rolling resistance for a good bicycle tyre once it is rolling on smooth bitumen is 0.006, and allow twice that for stiction. However, given the weight of your truck this could exceed Occupational Health and Safety limits for some countries, where workers are not allowed to exceed loads of about 30lbf.

Also there are many reasons why this would be unlikely to be achieved in practice, including the obvious point that warehouse floors have slight gradients, which might be very exciting.



Cheers

Greg Locock

 

[bobert]

Thanks guys for all the help; I really appreciate your taking the time to respond. However, I still don't think I have what I need. I can tell that you know what you are talking about, but I just don't get it. I have been able to find all sorts of examples, diagrams & formula tips for calculating the torque applied to an object. But my question is HOW MUCH force do I need to get the cart moving? I'm sure I will probably be humbled when I find out how obvious the answer is....and I apologize for my slowness. If anyone could help me understand this better I would greatly appreciate it.

 

[jfl]

As I recall, F=ma. How fast do you want the cart to accelerate? The torque at the wheels is needed to overcome the friction in the wheel bearings and the additional torqe will translate into a force at the floor divided by the rolling radius of the wheels. I hope this clue will get you thinking in the right direction (no pun intended).

 

[chameleon]

ok, so u need to calculate the braking (starting from rest) torque for your manually operated cart. firstly u would need a traction force (between the wheels and the ground surface to prevent slippage). Rubber is a good enough material. However, careful not to make the coeffient of friction to high as the braking torque would increase as such. Frictional force = coeff. of friction x weight of cart. Now, u need overcome this frictional force plus any bearing friction. Since i assume that u will design your cart so it can be properly lubricated. u cay say: braking torque > or = frictional resistance x radius of wheel. This would give u an idea of the required braking torque. now, u must re-iterate this value to account for undulating surfaces, which can complicate the design if its used in different environments.
power required = torque (N-m) x angular speed (in rad/s not rpm)
power will be in watts.746 watts = 1 HP
the formulas are all there, good luck!

 

[GregLocock]

It was encoded in my message. As a /minimum/ 2*.006*(the weight of the truck). This is on good, large, high pressure tires on a perfectly smooth floor.

For a 1.5 ton load this would be about 36 pounds.

In practice you would need more
than that, you'd probably want at least 0.05* the weight of the truck.


Cheers

Greg Locock

 

[Ralph2]

Probably not the formula that you want but..... As a kid I remember moving railway cars with a standard 6 foot crow bar. Get the point of the bar behind the wheel and push down. Amazing what you can move with a bar.....
Use the largest size of wheels possible. Use steel wheels and roller (needle) bearings. Rubber will flaten and require a lot more initial force. On a level floor a "good" man can move 4000 pounds without too much trouble. We do it all the time in our shop, (concrete floor with cracks) but on a heavy load only the steel wheeled carts are used.
Good luck
Ralph

 

[bobert]

thanks guys!....I appreciate all the help!