I am planning a build on a simple 4-wheel-drive cart for hauling equipment for my business up/down yards with grade. 4 wheel cart will have weight capacity 1500 lb with potential of crossing up to 35 degree slope (slow RPM/high torque/motor brake necesary.) This will implement a 110V AC motor/110V motor-brake/gear box to increase torque and slow down RPMs. 1500 lb weight load 35 degrees slope max 110V AC service 15 Amps max 15 RPM roughly final RPM I do not have an estimate on HP/torque etc to handle this application. I do not have any specific electrical education so layman's terms are appreciated. Looking for input/formulas/advice on how to achieve the proper drivetrain ratings mostly HP, gear ratio, torque. We happen to have a motor 1/2 HP, 1750 RPM, motor clutch brake, and 20:1 gear ratio, but final RPMs are too fast for the application. (I am assuming a speed control would greatly reduce the torque and capability of this machine.) Thanks in advance!
Newbie here but I'll give you my 2cents: I would assume do calcs without grade first for simplification: 1 Horsepower is equal to the force needed to move 550lbs 1 meter in one second. So you would need about 2.7 horsepower to move that load (no slope) 1 meter per second (assuming no friction... as if the load was sitting on ice). So your 1/2 HP motor would (theoretically) move that load 1 meter every 5.3 seconds. That's a 1 to 1 gear ratio and assuming the motor can get it moving. Not sure about torque issues... I think a variac would allow you to slow that motor down but I'm not certain about torque/current draw limitations which might result (I would assume conservation of energy would come into play in any case). Alternatively, changing the gearing would allow that high speed to give you the correct rpm's on the wheel side obviously. So I think you're looking at a first problem of whether or not you'll have the amps to draw to overcome the friction of standstill??? And the in-rush current potential of that motor may not have what it takes (likely not). That's where I become unclear. I just haven't studied torque yet myself. One mechanic or hydraulic horsepower is equal to 745 watts. So you'd need 2013 watts to move it one meter per second. One electrical horsepower is equal to 746 watts, so 1/2 hp more is about 355 watts. Watts are always per second so it would take 355 * x = 1500. That's about 5 seconds (again). Same power/work calculations using watts vs hp. Usually with a motor like that you need a start capacitor to get the motor going. Perhaps a larger cap there would increase the initial torque into the range you need. There may be a run cap as well you need on the motor. But you're definitely (from my understanding) limited in rpms by the ac frequency, so you'll either need a variac or a gearing. That's all Iv'e got...
The typical AC motor like what is in a pump or compressor will not work well for variable speed/load applications. Adding a clutch/brake will mostly just add efficiency losses and not achieve variable speed control either. Having a gear box will allow you to trade some speed for torque and 20:1 puts you at a fair speed with 12” tires (87.5 RPM with a 12” tire is ~275' per minute or ~3MPH which should be a comfortable walking speed). Half a horse is probably not enough for this application, even geared down. The motor is not likely to develop more than a couple ft-lb locked rotor torque. This might get the load going on a flat surface but will not go up a 35 degree slope. On a hard flat surface with no wheel compliance, the force required to get a load moving can be small. For instance, you can push a car that is several thousand pounds with only 40-50 pounds of horizontal force. Once you start getting an angle on it though, the weight of the vehicle really starts to push back. Grade resistance = gross vehicle weight * sin(grade angle). For your 1500 pound vehicle a 10 degree grade angle gives you 260lbs of resistance and 35 degrees gives you 860. You will be drawing about 9amps at the rated load of the motor and a bit more than that when starting so you'll need a hefty power cord to run this at any distance. A 12ga 100' extension cord will loose you about 3 volts at 10 amps. May not seem like much but it's 2.5% lost just in the cord. I'd suggest trying to find a large DC motor and driving you vehicle with batteries if you want to keep it electric.
Thank you two for your input. We currently have a similar small machine with dozer type tracks on it. It will move weight loads greater than 1500 lb at over 45 degrees up. It incorporates a 110V 15 Amp Milwuakee drill "motor" and a gear box with chain drives to one axle and tracks around four guide wheels basically. It is very, very, very slow. This is what inspired my idea for a similar drive system. Simply a high gear reduction with small motor. Variable speed is not necesary. I dont mind starting with all new drivetrain components. I was just stating the 1/2 HP motor with 20:1 gear was still little too fast I think and I dont know how this converts to pull strength. My largest need here is to determine: xHP xgear ratio to move my load. I dont know what formula there is for this. Thanks!!
Careful. Power is not a force. To maintain the motion of a speeding body on a level frictionless surface requires zero power (assuming negligible air resistance). Calculations which follow are based on an erroneous understanding so are meaningless.
Thanks for the constructive criticism. So I'm curious then since you seem to be in the know: how many watts would it take to pull that 1500lb load for 1 meter in 1 second with no slope?
Actually I would go with DC on this - not AC motors, or is this something that will be plugged in while running? ( would still probably go with DC. If this is a one-off, could probably build with one or two treadmill motors. KISS - also the DC motor has better torque at zero/low speed. --- http://www.haulzall.com/v/frontend/industrial.htm for some other info on similar items.
Once you have it up to speed, the only losses after that are those due to friction. So to maintain speed, each second you have to add energy to make up for those losses. If there are no loses, there is no power input required.