What is torque required at wheel

1. Jul 10, 2014

vasim

i want to move a wheel of vehical in new test rig... i just knw tyre dia. 1.2 m .. load for perticular wheel is 8000 kg... required rpm 354 .. (i.e 80 km/h)
then what is torque required for wheel to rotate at 354 rpm..

2. Jul 10, 2014

Simon Bridge

Welcome to PF;
To rotate at a constant speed, the net torque is zero.
Does not matter what the speed is.

Torque is a kind of force - it produces accelerations not speeds.
Rework your question in those terms and the answer should occur to you ;)

3. Jul 12, 2014

litup

So the real question would be 'in how many seconds or minutes do you want to go from zero to 80 Km/hr.. You can see, for instance, if it had to be in one second, would require an acceleration of about 2+ G's so you could work out the torque required from there.

4. Jul 12, 2014

Simon Bridge

Thanks litup - I was hoping vasim would do something like that ;)

There is actually quite a lot we don't know about the rig in question - like what "8000kg load" means.
But vasim is looking at the test rig and the tire so should be able to see further than us.

5. Jul 13, 2014

vasim

thanx simon and ltup

i did some sample calculations for test rig as i am development engineer @ arabian axles manufacturing company i want to test my axles in perticular speed (80 km/h means 22.22 m/s )and load .... as my axles load carrying capacity is 16T SO i COnsider the the load @ each spring flap is 8000 kg + body weight 500 kg (only for test rig body weight mean bogie weight equally devided on 4 wheels)

now Accleration i take (assumptions) 80 km/h speed in 2 mins
therefore a= 22.22/120 = 0.1851 m/s/s
F = m X a = 8500 (KG) X 0.1851 (m/s/s)
= 1573 (kg m/s2)
= 1573 N .... (as per newtons law 1 Kg m/S2 = 1 N)

Now torque = F X r (tyre radius)
= 1573 *0.6 (m)
=944 Nm
i am going to give this torque to rollar which will going to turn on the wheel
so to rotate this rollar at torque 944

P = (2 X Pi xN x T)/60000
= 43 KW
so i has to purchase the motor of 50 kw????
or less as excess 7 kw will be useful to overcome frictional lossess between tyre and roller..

is my calculations correct??????

6. Jul 13, 2014

vasim

i did some sample calculations for test rig as i am development engineer @ arabian axles manufacturing company i want to test my axles in perticular speed (80 km/h means 22.22 m/s )and load .... as my axles load carrying capacity is 16T SO i COnsider the the load @ each spring flap is 8000 kg + body weight 500 kg (only for test rig body weight mean bogie weight equally devided on 4 wheels)

now Accleration i take (assumptions) 80 km/h speed in 2 mins
therefore a= 22.22/120 = 0.1851 m/s/s
F = m X a = 8500 (KG) X 0.1851 (m/s/s)
= 1573 (kg m/s2)
= 1573 N .... (as per newtons law 1 Kg m/S2 = 1 N)

Now torque = F X r (tyre radius)
= 1573 *0.6 (m)
=944 Nm
i am going to give this torque to rollar which will going to turn on the wheel
so to rotate this rollar at torque 944

P = (2 X Pi xN x T)/60000
= 43 KW
so i has to purchase the motor of 50 kw????
or less as excess 7 kw will be useful to overcome frictional lossess between tyre and roller..

is my calculations correct??????

7. Jul 13, 2014

Simon Bridge

Well done! Where did you get your degree?

That would be the force through the center of mass of the wheel needed to accelerate the center of mass to 80kmph.

... only if the force is applied at the outside of the tire - is this the case?

Nope.

The roller also has a moment of inertia.

You want the outside of a tire to be moving at a linear speed of 80kmph = 22.22m/s and you want to spin that up in 2mins by applying a constant torque to a roller pressing against the outside of the tire. Since the tire has radius 0.6m, that makes the angular speed about ω=37rad/s.

You can work out the average power from conservation of energy:

$$\bar P = \frac{\Delta E}{\Delta t} = \frac{1}{\Delta t}\left(\frac{1}{2}(I_{tire}+I_{wheel}+I_{axel}+I_{roller})\omega^2 + E_{losses}\right)$$

If you put $E_{losses}=0$ the result is a minimum power needed to get the acceleration you want. In general, this is a bad approximation - losses are seldom insubstantial and you won't get a constant acceleration for a constant applied torque. Even with all that, you still have to work out the moments of inertia, which depend on the exact geometry of the moving parts.

8. Jul 13, 2014

vasim

THANKS SIMON
can you pls give me sample calulation of MOment of inertia.. as i found difficulties to calculate... my roller dia is o.3 m

9. Jul 13, 2014

Simon Bridge

You can look up "moment of inertia" online to find examples.
You'll probably have to approximate the geometries to get a ballpark figure.
It is still best practice to measure the torque needed... you can also measure the moments of inertia.
An engineering course will usually involve exercises in doing this.

I'm trying to figure out what education level to aim my replies at.
In many jurisdictions the word "engineer" is legally reserved for people with a postgraduate Engineering degree and trade or professional certification. When you identified yourself as an engineer, I just assumed that you had one of these. Is this not the case?

10. Jul 14, 2014

vasim

dear simon thanx

i get my degree @ UNIVERICITY OF PUNE in MECHANICAL ENGINEERING in 2011

11. Jul 14, 2014

vasim

i knw how to calculate moment of inertia.. just i am become some rusty.... n my books are in india and i am working in saudi arabia... sill thanks for your help

12. Jul 14, 2014

dean barry

Assuming you want to use an ac motor, then run it up to top speed, which coincides with the top speed of the rig, so you need some gearing inbetween ( ac motor = 1470 rpm, rig = 354 rpm )
Example : motor sprocket 20 teeth, rig sprocket 83 teeth

What defines the motor is the time you accept for the rig to reach top speed, faster time = more motor power required.

A good manual on commercial ac motors with torque curves would help you.

I suggest a time step programme on an excel sheet so you can change the input (motor power) data easily.

13. Jul 14, 2014

Simon Bridge

OK, I won't labour the point, we all get rusty: you need a recap:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
... rotational stuff, scroll to the general form of the moment of inertia calculation.

http://en.wikipedia.org/wiki/List_of_moments_of_inertia
... list of common moments of inertia

If you are determined to go the calculation route - you'll need to make some approximations for the mass distribution. You may get away with it if you just want back-of-envelope figures.
But dean barry is correct, you need a manual.

You certainly need a lot more detail than you have got so far.
Good luck.

14. Jul 15, 2014

dean barry

You will need to consider the moment of inertia of the motor rotor itself, as well as the rotating parts of the rig.

With a 800 kg load and a Coefficient of rolling resistance of 0.03 your looking at a drag of 800 * 9.8 *0.03 =
235 N at the wheel radius, multiply by the wheel radius for rolling resistance torque.

The more i look at this the more id be tempted to build the rig and start with a small motor (say a washing machine motor) and leave room in the design for bigger motor options.

Id put some hefty gaurds in place as well.