Calculate motor torque requide to drive a cart

AI Thread Summary
The discussion centers on calculating the motor torque required to drive a cart weighing 4.5 tons on steel rails, with a desired speed of 0.8 m/s. Participants emphasize the importance of understanding friction, acceleration, and the forces acting on the cart to determine the necessary motor power. The conversation highlights the need for a detailed free body diagram and considerations for motor placement and design efficiency. Additionally, the impact of acceleration on power requirements is noted, suggesting that more power may be needed than for constant speed. Overall, the participants recommend thorough planning and possibly consulting an engineer for a more effective design approach.
Udaya
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Hello,
I'm in a design project of a cart. i want to decide a geared motor power and RPM. but I'm in a trouble with this because I'm unable to calculate the motor with radial friction.the cart is with steel wheel and drive on steel rails. place help.

Total weight cart+carrying material - 4.5 Ton
Radius of wheel - 80mm
Required speed of cart -0.8m/s
Thank you.
Udaya.
 
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That is not enough information.

Draw a free body diagram showing all forces acting.

The motor power required at constant speed will depend entirely on the friction, which will vary depending on the condition of the bearings, payload, etc. Of course this all goes out the window when you hit an incline.

It's best to overspec the motor and control the motor speed.
 
Hello Billy,

Thank you very much Billy for the reply. I want to calculate considering only weight of cart, friction of wheels and wheel diameter of wheel.can you please guide me go through the calculations.

Br,
Udaya.
 
So this is homework? You have no intention of building this thing?
billy_joule said:
Draw a free body diagram showing all forces acting.

At constant speed all forces sum to zero.
Or in other words, the torque applied by the motor will be equal and opposite to the torque from friction/s.
 
Hello Billy,

This is not a home work. Actually i want design this and fabricate.Hear i have attache my rough 3D drawing views.could you please explain the way i want to follow. The cart start at 0 m/s and i want to move it 20m in 0.8m/s speed and stop.
Thank you.
Udaya.

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Udaya said:
Hello Billy,

This is not a home work. Actually i want design this and fabricate.Hear i have attache my rough 3D drawing views.could you please explain the way i want to follow. The cart start at 0 m/s and i want to move it 20m in 0.8m/s speed and stop.
Thank you.
Udaya.
That FBD is poor - where is Bearing friction? Normal force?

That design is also poor and brings a lot of questions to mind.

How do you intend to power the motor? Why is the motor on the cart at all? why not stationary motor with chain or cable drive? Are the lateral forces during loading? What thrust loading will your bearings experience? Why isn't the motor parallel to the axle? why 0.8m/s? Why are the driven wheels so poorly supported? Why does the cart have a massive plate section? Why does the frame have no members to stop skewing? Why is such large box section used for the frame? Why ca't you use an off-the-shelf gantry like you'll see in most engineering workshops rather than design and build a cart? Or buy a cheap second hand forklift? I could go on..

Overall I think there are probably many better ways to solve your problem, I suggest you have an engineer look over your problem and design/find a more appropriate solution.
If you can define your problem with as much detail as possible someone on this forum may be able to suggest solutions.
As it is we can't help much, it looks like you haven't thought this through much and jumped on the first idea that came to mind.
 
Thank you Mr.Billy and Mr.Jack to guiding me. I will follow number of instructions of Mr. Billy and contact you when I'm required your help. Thank you
again for the helps.:w
 
Hello,
I am working on vehicle, for which i need to calculate following things

1. Required torque (Nm) to move vehicle from stationary position?
2. RPM of the engine at the time of torque mentioned in point number 1?

Available Input:
1. Vechile Weight(kg) - 4300
2. Engine Displacement - 3455
3. Gross Vechile Weight(kg) - 12990
4. Tyre Raduis (m) - 0.464
5. Gear Ratio 1st - 7.600
6. Gear Ratio 2nd - 3.7
7. Final gear Ratio - 6.142
8. Engine Max. Te(Nm) - 315 @ 1625RPM
9. Engine Torque = 230 Nm @ 1000 rpm , 241 Nm @ 1200 rpm (Actual Measurement)

Regards,
 
  • #10
"241 Nm @ 1200 rpm"

That gives a power of under 30kW, a third of the power of my last motorbike. Good luck shifting a 12 tonne vehicle with that. It's possible but probably not wise.

As for your questions:

1. Depends on the friction of all moving parts.
2. Zero. If the vehicle is stationary, so is the motor (unless you use a clutch).
 
  • #11
Thanks for help

We are using dry friction clutch
Vehicle stationary and engine is idling, but when i put vehicle on 1st gear at what point of Torque@RPM vehicle start moving?
 
  • #12
Hi,
I am designing a motorised rail transfer cart for industrial purpose. The total weight of the cart with load is around 10000kg.The cart is supported by four wheels of dia 8". The wheels are to be made of steel or cast iron. The motor is coupled to a gear box which is connected to the front shaft. Please help me to calculate the required motor capacity to move the cart. Assuming that the velocity is 20m/min and the travel distance is 80m.
 
  • #13
Welcome to PF.
Assuming there are no hills, energy is only required during the acceleration to 20 metres per minute. You need to specify how long it should take to accelerate, and how it will decelerate at the end of travel.

Solve the problem by energy flow. Power is the rate of flow of energy; watts = joules per second.

When traveling at 20 m/min = 0.3333 m/s, the kinetic energy will be; KE = ½ · m · v2
∴ KE = ½ · 10k · 0.33332 = 556. joule.
If the motor could develop 556 W, that would be 556 joule / sec so it would take about 1 second to accelerate to 20 m/min.
Halve that to 280 W and it will take twice as long, 2 sec to accelerate.
 
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