# Torque required to move a wheeled weight

• Stephen123
In summary: Sorry misread acceleration for velocity. The acceleration is what i was trying to find with (Vf^2 - Vi^2)/2*D.The small wheels don't seem like an issue to me as the object is only 10 m high and the wheels are designed to take the load.
Stephen123
I have a problem where I have a large mass on wheels that I want to power with an electric motor. My hope is that I can find the torque i need to start and drive the motor and this is what I have so far.

Mass = 400t.
Initial velocity = 0
Final velocity = 2.5m/s
Distance required to travel = 101m

I used acceleration = (Vf^2 - Vi^2)/2*D for the acceleration but i don't think that is right because does that mean I am wanting it to reach full velocity at the 101 metre mare with that equation?

Once i find the acceleration, my plan is to use F=Ma to find what I think will be the force to overcome being stationary, and knowing I have two 600mm diameter wheels being driven, is the torque just half the force multiplied by the wheel radius?

Any input would be great thanks as I need some guidance.

Thank you.

400 tons? What are you moving, a skyscraper?

You need to provide more information. What is the desired / allowable acceleration, based on wheelbase, track width, height of CG, and desired travel time. Is there any gradient at all in the travel path? What is the rolling resistance? Is the load large? Does air resistance need to be considered - a headwind against a tall structure can create a lot of force. Or is this some sort of homework problem where most practical realities can be ignored?

russ_watters
Stephen123 said:
Mass = 400t.
I think the Space Shuttle and empty external fuel tank are about 200 tons (with a quick Google search). I guess you should study the design of "The Crawler"...

http://image.trucktrend.com/f/91045...00-hp+nasa-diesel-shuttle-crawler+shuttle.jpg

#### Attachments

• 0705dp-00-hp%2Bnasa-diesel-shuttle-crawler%2Bshuttle.jpg
34.3 KB · Views: 713
russ_watters
jrmichler said:
400 tons? What are you moving, a skyscraper?

You need to provide more information. What is the desired / allowable acceleration, based on wheelbase, track width, height of CG, and desired travel time. Is there any gradient at all in the travel path? What is the rolling resistance? Is the load large? Does air resistance need to be considered - a headwind against a tall structure can create a lot of force. Or is this some sort of homework problem where most practical realities can be ignored?

The desired / allowable acceleration is 2.5m/s. Travel time will be long since it is a 101m overall distance but it terms of time I was going to estimate a number for maybe 7 seconds for it to reach max acceleration and then travel at a constant speed, seem reasonable?

Wheelbase, track width, are still undetermined at this point. Height of CG is also still unconfirmed but somewhere around the 7m high mark.

No gradient in the travel path.

Rolling resistance will be rubber and concrete but I haven't found a reputable source for the rolling resistance just yet.

Yes the load is large, and wind resistance will be something to look at later.

Thanks.

Stephen123 said:
The desired / allowable acceleration is 2.5m/s.
That is a velocity, not an acceleration. Do you know the difference?
Stephen123 said:
Yes the load is large, and wind resistance will be something to look at later.
Did you see my post? Did you look into The Crawler to see what its acceleration and top velocity are? Why do you think you can do this with electric motors and small wheels?

And is this for a schoolwork assignment? For which class? What is your ME background so far?

berkeman said:
That is a velocity, not an acceleration. Do you know the difference?

Did you see my post? Did you look into The Crawler to see what its acceleration and top velocity are? Why do you think you can do this with electric motors and small wheels?

And is this for a schoolwork assignment? For which class? What is your ME background so far?

Sorry misread acceleration for velocity. The acceleration is what i was trying to find with (Vf^2 - Vi^2)/2*D.

Hydraulic motors are also an option I have thought about but whichever option I would still need to find the torque required initiate the movement so i figured that wasn't all too relevant compared to the calcs at this point.

The small wheels don't seem like an issue to me as the object is only 10 m high and the wheels are designed to take the load.

No not a school assignment, work for a drafting company and doing some prelim work for a new job.

When doing concept work on a project that is clearly far outside your organization's experience, the best way to start is to look at similar machines and extrapolate. A Caterpillar 797F mining truck is rated to carry 400 tons at speeds to 42 MPH. If you scale the power by the ratio of your speed to 42 MPH, you will need about 530 hp. That truck weighs 288 tons empty, your vehicle would weigh slightly less, but almost certainly over 200 tons.

Those tiny wheels may be designed to take the load, but somebody should calculate the strength of the concrete they are running on. A very rough guess is the concrete needs to be at least three feet thick if it has sufficient properly engineered steel in it.

Please do not be offended, but the questions that you are asking tell me that you not only do not know where to start, but have nobody in your organization who can guide you. A 200 ton vehicle is a ten million dollar project. A project that size needs a team of experienced engineers and a project engineer/manager with deep experience in large structures to have a chance of succeeding. Not to mention the purchasing, manufacturing, and the field support functions.

berkeman
jrmichler said:
When doing concept work on a project that is clearly far outside your organization's experience, the best way to start is to look at similar machines and extrapolate. A Caterpillar 797F mining truck is rated to carry 400 tons at speeds to 42 MPH. If you scale the power by the ratio of your speed to 42 MPH, you will need about 530 hp. That truck weighs 288 tons empty, your vehicle would weigh slightly less, but almost certainly over 200 tons.

Those tiny wheels may be designed to take the load, but somebody should calculate the strength of the concrete they are running on. A very rough guess is the concrete needs to be at least three feet thick if it has sufficient properly engineered steel in it.

Please do not be offended, but the questions that you are asking tell me that you not only do not know where to start, but have nobody in your organization who can guide you. A 200 ton vehicle is a ten million dollar project. A project that size needs a team of experienced engineers and a project engineer/manager with deep experience in large structures to have a chance of succeeding. Not to mention the purchasing, manufacturing, and the field support functions.

The 400t load includes both the load and the vehicle. It is the total mass of both components and also allowing a little extra.

The Concrete below the vehicle has been checked by the client already as they use it for similar applications.

As i said this is preliminary work. If the job goes ahead then a team may be required but for some simple checks early on to see weather the design is feasible is all i am after.

Calculating the torque needed to accelerate at the required rate is one thing, it's quite another to calculate the torque required to overcome other forces such as rolling resistance or static friction. You might need experimental data.

And how on Earth do you support 400 tons on 600 mm (24") diameter wheels? A 747 needs 18 wheels to support it.

I have headed teams that built and shipped large, high speed machines. Designing a driveline for a 400 ton vehicle is not the place to learn basic engineering.

billy_joule and berkeman
+1 regarding suggestions you get proper advice.

Here is a brief attempt to make sense of your proposed numbers...

You say you want it to start from rest and reach 2.5m/s in 7 seconds.

V=U+at

U=0 so...

a=V/t = 2.5/7 = 0.36m/s2

400 metric tons = 400,000kg

F=ma = 400,000 * 0.36 = 144,000 Newton's

Power = force * velocity = 144,000 * 2.5 = 360,000W = 360kW or 480HP

And that's just to accelerate the mass. No allowance for rolling resistance, friction, losses in drive train etc.

berkeman

## What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is typically measured in units of Newton-meters (Nm) or foot-pounds (ft-lb).

## How is torque related to moving a wheeled weight?

When a force is applied to a wheeled weight, it creates a torque that causes the wheel to rotate. The amount of torque required to move the weight depends on the weight of the object, the size and type of wheel, and the surface it is moving on.

## What factors affect the torque required to move a wheeled weight?

The torque required to move a wheeled weight is influenced by several factors, including the weight of the object, the size and type of wheel, the surface it is moving on, and any external forces acting on the object.

## How can torque be calculated?

To calculate the torque required to move a wheeled weight, you need to know the weight of the object, the radius of the wheel, and the coefficient of friction between the wheel and the surface it is moving on. The formula for torque is T=Fr, where T is torque, F is the force applied to the object, and r is the radius of the wheel.

## How can torque be reduced?

To reduce the torque required to move a wheeled weight, you can try to reduce the weight of the object, increase the size of the wheel, or decrease the coefficient of friction between the wheel and the surface it is moving on. Additionally, using lubricants or changing the shape of the wheel can also help reduce torque.

Replies
8
Views
5K
Replies
4
Views
1K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
7
Views
7K
Replies
6
Views
3K
Replies
3
Views
2K
Replies
6
Views
3K
Replies
3
Views
1K
Replies
4
Views
2K