# Help with calculating motor requirements

## Main Question or Discussion Point

I have a problem I am hoping someone can help me with. I am designing a moving cart that will need to transport a 2000 lb load from point A to point B. The speed of the cart will need to be approximately 2.3 MPH. I have done some initial calculations. If the electric motor were to spin at 2500 rpm, I would need to reduce it by a ratio of 39.06:1. The gear reduction calculation was based on the spur gears driving a 12" tall tire. This gave the final drive speed of 2.3 MPH. Considering this ratio of reduction, what horsepower requirement would be needed to acheive this. Second, if the cart had to travel up a moderate slope of say 9 degrees, what power requirement would then be needed.

Can anyone help me with these numbers, as the math involved is just simply is over my head. Or if anyone knows if there is a spreadsheet or calculator out there that I can put my numbers in to get the appropriate results. The variables in this project may change slightly, so it would be helpful to be able to figure it out myself.

Thank you in advance for any help!

Last edited:

Related Other Physics Topics News on Phys.org
http://www.engr.colostate.edu/~allan/fluids/page8/page8.html
till P = Fd * V

But i see they haven't included internal resistances(which contributes around 5-7% of the total resistance) and the grade resistance (which equals mg*sin9(deg)). Including these factors and substituting in the above expression, you will get the value for power.

First, I would convert everything to the mks system. The rolling resistance coefficient (RRC) for rubber tires (automobile) is about 0.01, while the RRC for steel on steel is about 0.001, but steel tires have too small a coefficient of friction for a 9 degree slope. So for rubber tires, the power is RRC*mass*velocity = watts (in mks system), and watts/746 = HP. Your more significant power requirement is the slope requirement. As stated in previous post, Force = RRC*M *g (in Newtons)
Also, Force*dx =work , and Force*dx/dt = Force times velocity = power (watts)
If you need more than 1 HP (you will), you have a signifcant motor requirement, and you may have to consider another gear ratio to match an available motor to the required RPM. You also need to consider what happens if the motor loses power while on the slope. It may need either a mechanical or electrical (brake on power-off) brake. Because of the cart acceleration/deceleration requirements, you might want to consider a dc brush motor or a brushless dc motor with a Hall-Effect sensor for rotor angle. Motors in your HP range might be available (with speed controllers) from golf cart or small-electric car manufacturers/suppliers.

Could you possibly put those equations in a little bit easier to follow format. I don't know if you would be willing, but could you use your equation as an example so I can see how you use each values? I don't know what the MKS format you are reffering to is. And yes it will be on rubber pneumatic tires. Can anyone actually show me how to calculate it, I mean use my numbers and show me how you came up with it. I really would be greatfull.

As for the gearing, I was planning 4 gears. Drive gear (8 tooth) to a 50 tooth gear on a shaft with another 8 tooth gear driving a final 50 tooth gear. This should allow for proper strength. I haven't decided yet whether or not to be directly driven or by chain.

It seems overkill to use a golf cart motor. Those are designed to move a large load up to 20 miles an hour or more. Mine only has to go 2-3. We are hoping to use a motor in the 4-5 hp range, but the math needs to prove it, obviously we can't guess.

As for braking, I was thinking about regen braking. I have been reading that this system will allow for electric braking. This would prevent runaway situations, and a mechanical parking brake for actual long term braking. The system would require a momentary switch to be pressed to make it move, and if you let go, it would stop.

I look at my daughters hot-wheel jeep as an example. It has a TINY motor, but it will push around 200 lbs at 3 MPH. Also, when she lets off the pedal, the wheels STOP it. Kind of like a cordless drill as well. Would a system like this work for this application.

The power to go up a 9 degree slope is
P=M*g*v in the meter kilogram second units
M=2000 lbs/2/2 = 909 kilograms
g=9.81 meters/sec squared
v = 2.3 mph x 1609/3600 = 1.03 meters/sec

So power = 909 kg x 9.81 x sin(9 degrees) x 1.03 = 1436 watts = 1.93 HP

Thank you SO much. I think I understand it, and it at least will give me an idea what to do if I need to change the variables. Thanks again I greatly appreciate it.

If anyone has experience with motor control and may have a suggestion for regen braking, I would be open to any suggestions or advice you have.

The power to go up a 9 degree slope is
P=M*g*v in the meter kilogram second units
M=2000 lbs/2/2 = 909 kilograms
g=9.81 meters/sec squared
v = 2.3 mph x 1609/3600 = 1.03 meters/sec

So power = 909 kg x 9.81 x sin(9 degrees) x 1.03 = 1436 watts = 1.93 HP
Ok, I do have a couple questions about this formula. I'm sorry if I sound ignorant, but where do you get the 9.81 meters/sec squared? What is this derived from?

Also, if I want to calculate different slopes, can I just substitute the SIN(9) for SIN(some other degree)?

I noticed if use the same formula without the SIN function, the answer is obviously wrong as it is higher than if the slope os figured in.

I don't know if it is correct with other amounts of slope.

The 9.81 meters per sec-squared is the gravitational force on the surface of Earth. The equation comes from college freshman (or high school) physics. For different slopes, use formula with different sin(degree) values. This may be on your pocket calculator, or math tables. Also, you can get approximate values by multiplying the degree value by pi/180.

I do have another question for you. Will you be going only in a straight line path, or will you be turning? If you are driving more than one wheel, you will need a differential.

russ_watters
Mentor
The 9.81 meters per sec-squared is the gravitational force on the surface of Earth.
Well, gravitational acceleration on the surface of the earth - used to calculate the gravitational force (by multiplying by mass).

Dear Bob...thank you very much for the post.it helped me also solving a problem.i really appreciate the way you explained it.simple and direct to the point.
i was also facing a problem something like this. in my case it is a mass of 600kg with a motor and gear meshed with a fixed rack and pinion.the load is hanged on a pulley with a counter weight.this needs to travel up and down at a speed of 2m/min (0.02m/sec) approx. i worked out with your equation.as the load is counter weighed, i considered the slope value as zero.i reached in 118 Watts. please review and advise..

Anyone can help me to calculate the Torque for Drive Pulley of Rolling Conveyor?
The details below:
- Roller Conveyor : length = 18m (speed of the belt conveyor is 2.5m / sec)
- Conveyor consists of 25 roller, 1 drive pulley & 1 tail pulley
- Total load of the conveyor : 5,500 kg

To move the conveyor, we install Drive Pulley with diameter of 550mm, and connected to Gear Box (RPM = 95) + Motor (5,5 HP ; RPM 1,500)

How do we calculate how much is the Torque of the Drive Pulley?

CWatters
Homework Helper
Gold Member
Do you happen to know if a 5.5hp motor is powerful enough for this job? If you do then you can estimate the max torque assuming the motor is delivering 5.5HP at 1500rpm......

5.5 HP = 4125W

So the torque at the motor is 4125/157 = 26Nm

The gearbox reduces rpm and multiplies the torque so torque at the output of the gearbox/drive pulley should be something like..

26 * 1500/95 = 410Nm

But that could be greatly exceeded. I can think of a lot of ways the torque could be higher for periods of time. Some examples..

Depending on the type of motor and how it's controlled the starting torque might be much higher.

You say the load is 5500kg. The transit time is 18/2.5 = 7.2 seconds. So every second 760kg is being dumped on one end of the conveyor belt and has to be accelerated to 2.5m/s. If the load is lumpy the torque might spike higher?

What happens if the belt with 5500Kg on it has to be stopped in a hurry or there is a power cut?

I'm not an expert on these systems. Google found a guide to the basics. Looks complicated to me.

http://www.krk.com.br/html/produtos/phoenix/Design_Fundamentals.pdf [Broken]

Last edited by a moderator: