Figuring required motor torque for rolling load

[binfordw]

I need help with this. I need to find out how much power/ torque is required to move a rolling load up a 25 deg. incline. I need a drive motor for a electric cart. The cart weighs approx 1200lbs, and needs to travel only around 20 to 30 fpm. Anybody know a way to figure out how much torque the motor I need should have to do this?

thanks

 

[haggis]

binfordw

First of all, what you want to determine is motor HP, not torque.

You've given the basic info on your requirements but a few other pieces have to be taken into account. Type of wheel for example to determine friction factors...Rubber tire on concrete, steel wheel on rails, ball bearings in hub?

Let's ignore friction factors and do the basics. The diameter of the wheel has to figure into the equation.

For this example, I'll say the diameter of the wheel is 12" so to travel at 30 fpm, your rpm would be 9.5 (30/pi*dia).

The force required to hold the cart on the incline would be Weight * sin 25 or (1200lbs * .422 = 507lbs.)

Now the wheel dia being 12", and if the cart was being held stationary on the incline, the torque at the hub would have to be 507 * .5 (radius) = 253.5 ft.lbs.

The Horsepower reqired to move it up at 30 fpm would be Torque*rpm/5252 or 253.5*30/5252 = 1.44 HP

You'll also need a speed reducer with a ratio of about 190:1 if your motor speed is 1800 rpm.

The theory behind this is if you had a 507 lb weight 6" from the hub and rotated the wheel 30 times, the wheel would have travelled 94.2 ft. so you would have done (94.2*507) 47759.4 ft.lbs. of work. Since 1 HP = 33000 ft.lbs, 47759.4/33000 = 1.44 HP.

Hope this helps

Haggis

 

[sacem1]

Hello Haggis:

Your explanation was very good I'll just adjust some of your numbers, it seems in writing you got a little mixed up:

The requiered speed is 30 fpm and you got the wheel rotational speed right (9.5 rpm) but used the 30 feet instead of the 9.5 rpm in the HP calc which should state:

HP = Torque*rpm/5252 -> HP = 253.5 * 9.5 / 5252 = 0.46 HP

And the following shoul d read:
"The theory behind this is if you had a 507 lb weight 6" from the hub and rotated the wheel 9.5 times, the wheel would have travelled 30 ft. so you would have done (30*507) 15,210 ft.lbs. of work. Since 1 HP = 33000 ft.lbs, 15,210/33000 = 0.46 HP."

I might add for practical purposes to include a 60% eficiency (gross estimate) for the rig and you would have Real HP = Teoretical HP / efficiency -> Real HP = 0.46/0.60 = 0.77 HP

I would recomend a 3.5" between shaft distance worm speed reducer with a 50:1 ratio plus a final gear on the driving wheel (wheels?) of 15 teeth to 60 teeth D.P. 10 gears like those used in crane runways or the same relationship in RC 50 (5/8" roller chain and sprockets). The added benefit would be that the worm speed reducer will give binfordw a self braking rig just stop the motor and the rig stops, if you add an electronic soft starter (should be less than 150$ for that size of motor)you would have gradual speed increase when you start and when you stop, and if instead of that you but and inverter you will add speed control plus easy reverse.

Hope to have helped a little.

SACEM1

 

[binfordw]

Thanks guys, that is the info I was looking for.

To clarify a bit, I am using 10" pneumatic rubber tires to roll these carts. I am using bearings on the drive axle.

I do have a question though, 1/2 hp sounds right, but 250 + ft lbs?? That seems really high to me.


Another question, where is a good online site to find such motors? I havent found many so far. Most are used motors or are way too big for what I need.


Once again thanks so much for answering my questions!

 

[haggis]

sacem1

Thanks for picking up the typo and setting it straight. Sorry if I confused you binfordw

Haggis

 

[haggis]

binfordw

Try

http://www.globalspec.com

for a worldwide list of products and manufacturers.

There is also a good document at

http://www.falkcorp.com/products/pdf/281100/281100082.pdf

Haggis

 

[sacem1]

You can try Grainger at

http://www.grainger.com

they have the motors and the gear reducers plus the sprockets and chains youll need

Cheers

SACEM1

 

[sreid]

Just for fun as a backcheck; a 30 ft ramp at a 25 degree angle has a rise of 12.68 ft. The 1200 lb load does the 30 ft in one minute so the potential energy gain is

12.68 x 1200 = 15,214 ft-lb/min

HP is 33,000 ft-lb/min so

15,214/33,000 = 0.46 HP