Determing the appropriate Motor.

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To determine the appropriate motor for a 500 lb cooler designed to travel at 4 MPH, calculations indicate a required wheel RPM of 150. For an electric motor operating at 1800 RPM, a gear ratio of 12:1 is necessary, while a 3000 RPM internal combustion engine requires a 20:1 ratio. The power needed at the wheels is calculated to be 1194 W (1.6 hp), with a recommended motor power of 1492 W (2 hp) to account for transmission losses and maintain performance on a 15% grade incline.

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Jmartuzzo
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Ok I figure you all may be able to help me with this one. I have a "Cooler", as in box that you put soft drinks, and other refreshments in), it’s in quotes due to the fact that it is huge and heavy (24x24x48). It is currently built on a set of 4 pneumatic wheels (4 inch wide x 9 in diamiter). At full weight it is 500lbs. trying to figure out what size motor I would need at what RPM, with what gear ratio, would produce the effect of 4 MPH. Also i was hoping for enough torque that a 15% Grade would only minimally reduce speed.

please let me know if i am leaving anything out, or if this should have been in the mechanical heading.

thank you for everyones/anyones assistance!
 
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I work better with SI unit, so let me convert your data:

m = 500 lb = 227 kg
v = 4 mph = 1.79 m/s
d = 9 in = 0.2286 m

The performance criteria you want to look at, is the maximum acceleration you will be able to achieve. 0.3g is the minimum acceleration you should consider. At 0.5g, that's when you begin "sinking" in your seat in your car. 0.15g corresponds roughly to the equivalent acceleration you will feel going up that 15% grade incline (So it is an absolute minimum). 1g = 9.81 m/s².

Let's go with the math:

RPM of the wheels needed to achieve 4 mph:

rpm = v / r = v / (d/2) = 1.79 / (0.2286/2) =15.64 rad/s = 150 rpm

I don't know what type of motor you will use, but typical electric motor works at 1800 rpm and typical IC engine works at 3000 rpm, so:

Gear Ratio = rpmmotor / rpmwheel = 1800 / 150 = 12:1 for electric motor
Gear Ratio = rpmmotor / rpmwheel = 3000 / 150 = 20:1 for IC engine

Power needed at the wheels:

Pw = Fv = mav = 227 * (0.3 * 9.81) * 1.79 = 1194 W = 1.6 hp

With this power, you will be able to climb that 15% grade hill and still have power to give your cart an acceleration of 0.15g.

Since there will be some kind of transmission (unless you can find a motor that produces that power at 150 rpm), you will suffer some losses from the engine to the wheels, which will depend on the type of transmission (gears, belt & pulleys, chain & sprockets, etc.). But with the kind of ratio we calculated, I would assume a max. efficiency of 80%, nothing less than 90% for sure. So the power of the motor needed is:

Pm = 1194 / 0.8 = 1492 W = 2 hp

That would be my estimation of what you need.
 

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