Calculating amount of torque needed to move a vehical.

[professor]

what torque(fource causing rotation)(NM) is needed at the pinion to move a 1000kg rig with a 10:1 diff ratio forward on a 45 deg climb.(tyres are 1000mm high 39"or so)

I need the formulars and the theory to get my head around this.

may need to factor the losses in the diff after i get my head around the theory!!! and im not interested in the motor and gear box powers and losses only how much it will take to rotate the pinion.


I'm trying to understand what stresses the axels and pinion are under.




Chad

[-Scott-]

Excluding losses, you need enough force to push 1000kg up the slope.

First, calculate the component of the vehicle's weight acting along the slope.

Weight = mass x gravity = 1000kg x 9.8m/s^2 = 9800N
Fraction = weight x sin(angle) = 9800N x sin(45) = 6930N

You need enough torque at the axle to generate this force at the tyre's radius.

Torque = force x radius = 6930N x 0.5m = 3465Nm (at the axle.)

Torque at the pinion = Torque at axle / diff ratio = 3465Nm/10 = 346.5Nm.

Going one step further, torque at the flywheel = 346.5/gear ratio.

Note that this is the minimum torque required to theoretically move the vehicle up the hill (you're actually accelerating it against gravity.) To make it go at any significant speed you actually need to accelerate faster than this, so you'll really need more than this.

And 45 degrees sucks, because sin(45) = cos(45), so my numerical answer is right even if that sin should be a cos - but I can't be stuffed drawing a diagram, and did it all in my head. So I don't guarantee that the sin shouldn't be cos. Try the calcs for 30 degrees; the correct one is whichever gives the lower number.

Cheers,

Scott

[DamTriton]

I'll have a go, more by logic and simple maths than by any single great all-encomassing formula

Diff ratio: 10:1
Tyre radius: 0.5m
Weight of vehicle: 1000kg
Angle of inclination: 45 degrees
Friction losses in diff ~10% (usually considered to be a total of 30% from engine to wheel)

Converting the weight to mass in Newtons of the vehicle gives 9800N
Therefore for a wheel radius of 0.5m wheel total torque would be 0.5 x 9800 = 4900Nm torque to act only against gravity, more needed to procede up slope, effectivly a 90 degree slope at this stage,and ideal grip of tyres at 90 degrees. We have 4 wheels so assuming equal load spit between them we would divide the 4900Nm by 4 giving 1225Nm per wheel. We are driving two wheels so we need 2450 Nm.
Using Pythagorases theorem a slope of 45 degrees would resolve to a 0.7071 fraction of the "dead lift" 90 degree slope against gravity (ie the ground is supporting some of the weight) so 2450 x 0.7071 = 1734Nm (think of it as a winching scenario, easier to winch on the flat than up a 90 degree step
10:1 ratio in your diff would reduce the 1734Nm to 173Nm (ie one tenth of the torque)
Factor in your friction losses by adding 10% to the 173Nm gives 190Nm at the pinion.

Remember that this 190Nm at the pinion would only be enough to oppose gravity, to actually procede up a hill would require more torque, the excess torque being applied to everyones favourite science formula, Force = mass x accelleration with appropriate scaling to cater for the 45 degree incline.

Happy to stand corrected, and I hope it all makes a bit of sense.

Edit: Scott you beat me too it, but you forgot about the fact there are 4 tyres contacting the ground and we are calculating the drive to the pinion connected to two of them (I made the assumption given the tyre sizes that we are talking about a comp rig driving all 4 wheels). The scaling factor for 30 degree incline is .5 and for 60 degrees it is .86 (ie Sin x)

[Bush65]

Calculate static force parallel to slope due to mass of vehicle and gravity acting on the incline: ie. force (Newton) = mass (kg) x 9.81 m/s^2 x sin 45

Calculate force required to accelerate vehicle: ie. force (Newton) = mass (kg) x accel (m/s^2)

Calculate rolling resistance of tyres on track surface - varies wildly due to tyre dia and inflation and the nature of the surface. Lets take a guess and say 25% of the normal reaction ie. force (Newton) = 0.25 x mass (kg) x 9.81 m/s^2 x cos 45

Sum these 2 forces ie. force1 + force2 + force3

This is total for front + rear axle, so split this to determine force for rear. This depends on location of centre of gravity from rear axle - looking at side view on slope. This may well be in the order of 90% at rear tyres and 10% at fronts.

Now we have the force at the contact point between the rear tyres and ground.

Estimate distribution between left and right tyre. Worst case is 100% to one with most traction and 0% at the one with no traction (if this were not the case ARB would not be able to sell air lockers).

Then toque at axle is: torque (Newton m) = force (Newton) x rolling radius of tyre (metres).

But this neglects friction in wheel bearings, brake drag, seal drag etc. So increase the torque by say, 5%.

Now there are some variables used here that you may not have realistic values for, without conducting a series of test.

The other approach is to use a worst case static situation.

Assume that the maximum coefficient friction between the tyres and ground that can be expected is 1 (some soft rubber compounds when hot can exceed this on a good surface, but use 1).

And that the maximum force that can be distributed to the rear wheels is the total weight of the vehicle (this is conservative, so justifies not using a friction coefficient greater than 1).

Then max torque is: Torque (Newton m) = coefficient of friction x mass (kg) x 9.81 (m/s^2) x rolling radius of tyre (metre)

Note: none of this takes into account shock loading. Shock loading does not so much change the torque but it greatly affects the strength of the axle.

If you want to do this you have to determine how much impact energy can be converted into strain energy in the axles. This depends upon the volume of material and resilience of the axle (how the volume is distributed along the length of the axle - ie. diam changes).

[professor]

wow

There is some smart people out there.


This is gold


Now to jam all that into a spread sheet so I can put some real figures into it


So to factor in a SWL (Safe working load) a maximum load with losses and some left over so I can run a rig with enough torque to move the rig but not enough to break it.
Is it around %15 of losses at the diff?(maybe factor it in with portals)

I may have to have another good read

Thanks for your input people I will keep you informed of my progress

Chad

[Bush65]

wow

There is some smart people out there.


This is gold


Now to jam all that into a spread sheet so I can put some real figures into it


So to factor in a SWL (Safe working load) a maximum load with losses and some left over so I can run a rig with enough torque to move the rig but not enough to break it.
Is it around %15 of losses at the diff?(maybe factor it in with portals)

I may have to have another good read

Thanks for your input people I will keep you informed of my progress

Chad

With hypoid gears (used in diffs) there is a lot of sliding contact, so much more loss compared to gears on parallel shafts (spur or helical).

I don't know what value to use, but 15% sounds ok.

Portals would have additional losses. 2 gear portals like hummers should have lower losses than 3 gear portals like volvo, which should be less than 4 gear portals like maxi-drive.

Safe working load is different concept to what you want. This is applied to lifting components, before being changed to working load limit (WLL).

Nearly all design codes are changing to "limit state design", where you evaluate both the "strength limit state" and "seviceability limit state", to which a capacity factor is applied.

The loading code applies a factor of 1.25 to dead loads (due to self weight) and 1.5 to live loads. There are other loads eg. wind loads and earth quake loads, but you do not need to be concerned with these.

To account for operator abuse we use a gorilla factor

[-Scott-]

So, if I understand some of the other posts here, the amount of drive force you can achieve from an axle depends on the number of tyres on the axle?

So, a motorbike, driving one wheel with a 1m diameter, with 100Nm at the axle, can generate 200N of drive force. But, if I add a SECOND wheel to that axle, I can double the drive force?

Wow!

So, if I put dual wheels on each corner of my 4by, I can double it's drive force! I'll be able to accelerate twice as fast!

Better still, I'll get triples. Holy sh!t. If I can get FOUR wheels per corner, my Paj will be the fastest drag car in the country.

Scott

[MR04WD]

[quote="professor"]wow

There is some smart people out there.


I never doubted anyones knowledge on here,but like you Chad I didn't realise how many smarts are on here until I read the thread 'RIDDLE ME THIS' in 'General Chit Chat'.........OMFG!!!!!!!! I had to refrain from reading it as it was giving me headaches.