Morris Minor Owners Club

OK heres a question for the engineer types.
Which diff is kinder to halfshafts? the ultra low one I have in my van, or a high one from a saloon? van ones are easier to spin up so does it take less torque, than one thats not so easy to get going but is easier on the revs. Or is the torque the same because its a function of the vehicle weight being moved a set distance in a set time?
I hope this makes sense to everybody as I,m troubled by the number of broken halfshafts I read about I,ve bought two spares but I dont know which side of thecar they came off And I hear they get 'handed' over the years and snap almost immediatly if put on the wrong side.

A lower ratio diff will supply more torque to the wheels for a given amount of torque at the prop shaft.

Or is the torque the same because its a function of the vehicle weight being moved a set distance in a set time?

If you accelerate at the same rate, then the half shaft torque is the same.

A low ratio diff allows you to put more torque on the axle, but doesn't mean you HAVE to.
(low ratio diff has a higher number e.g. 4.55:1 is lower than 4.22:1)

The torque on the axle shaft is limited to the resistance of the wheel to rotate. More torque may be available but the shaft will only experience what is being used.
There once was a theory that wheel bearings contributed to axle failure. What's the opinion these days? I've only had two fail and were on the same car, same size while I have one sided sealed bearings in. put original style bearings back - no failure.

It's not foolproof but more wear on the spines one side can give a clue as to which side they had been running, difficult to explain in words but you can work it out, assuming forward motion is used more.

I would say the lower ratio diff (ie higher numerically) will put more torque into the half-shafts - because it allows more of the available power from the miserable A series engine to go through to the half shafts - simply by allowing ot to rev more for a given speed. If the torque was unlimited - ie big powerful V8 etc - then the ratio would make no odds - it would then be purely down to tyre grip!

Surely the torque available is limited purely by the engine? A diff change can't increase the torque available because its not capable of producing a turning force without another force (engine) acting upon it ? The more torque you put through the drivetrain (ie by fitting a V8) would either result in the diff/driveshafts/gearbox exploding or if the frictional load of the tyres was overcome first you would get masses of wheel spin........

Aye - but the (numerically) higher ratio diff allows more torque multiplication from the prop shaft into the half-shafts! And the A series normally doesn't have enough torque getting through to be a serious problem. Change the diff - and the torque is more (for given road speed, due to now higher engine speed) and the half shafts see more torque.

Surely the torque available is limited purely by the engine?

nope - as Andrew says ;-)

For a quick test - try accellerating from 0 to 20 in 1st gear and then again in 4th gear... If the torque was the same at the wheel (from the halfshaft) the acceleration would be the same. Yeah - ok you already know that and wouldn't need to try but it's an attempt to give it in non numeric terms.

I'm guessing you've got torque and power confused. Power will be the same regardless of gearing (power in this case = torque x angular velocity)

assuming no losses with an 80Nm engine

first gear 5:1 and diff 4.5:1 would mean torque is 80x5x4.5 = 1800Nm at the wheels
second gear 2.5:1 and diff 4.5:1 would mean torque is 80x2.5x4.5 = 900Nm at the wheels
third gear 1.4:1 and diff 4.5:1 would mean torque is 80x1x4.5 = 504Nm at the wheels
forth gear 1:1 and diff 4.5:1 would mean torque is 80x1x4.5 = 360Nm at the wheels

first gear 5:1 and diff 4.2:1 would mean torque is 80x5x4.5 = 1680Nm at the wheels
forth gear 1:1 and diff 4.2:1 would mean torque is 80x1x4.5 = 336Nm at the wheels

its why you get better take off in first than you do in forth gear, its why you sometimes have to change down a gear to get up a hill because more torque is required to get you up the hill

torque = force x radius so if your wheel radius is 0.45m then linear force is 4000Nm for first gear, 2000Nm for second gear, 1120Nm for third gear and 800Nm for forth gear (from the first set of figures)

going up a 1:6 hill (one metre up for 6 metres horizontal) would equate to a vertical acceleration of 1.64 ms. to propel a 800kg morry minor up this hill at any speed (external forces excluded for simplicity) then you would need 1312Nm of linear force meaning you could go up that hill in first and second gear as they have sufficient linear force allow sustain speed.

The torque on the axle shaft is limited to the resistance of the wheel to rotate. More torque may be available but the shaft will only experience what is being used.

So diff ratio is not the only factor. Wheel diameter and tyre width and compound has a bearing too.

jaekl wrote:The torque on the axle shaft is limited to the resistance of the wheel to rotate.

i disagree with that. you can put 1200Nm of torque through the axles to turn the wheels. if there is little resistance to change then the car will speed up. if there is heaps of resistance then the car wont speed up.

your statement would only be the case if the limiting factor was if hte engine couldnt speed up fast enough (like if you were revving up with the rear wheels off the ground)

The torque on the axle shaft is limited to the resistance of the wheel to rotate.


i disagree with that. you can put 1200Nm of torque through the axles to turn the wheels. if there is little resistance to change then the car will speed up. if there is heaps of resistance then the car wont speed up.

But you need a great deal of torque to start the wheel rotating and less to keep it in motion. Likewise you'd need more torque to start a wheel spinning (as in wheelspin not just motion) with a wider than standard tyre and less to keep that wheel going. Is it what they call 'stiction'?
I would imagine you could probably get away with using narrow tyres with a tuned 1275, but with 185's you'd be breaking standard halfshafts

Don't forget the resistance to rotate the wheels includes accelerating the entire vehicle or just spinning the wheels using the invention of the wheel. As stated the diff ratio provides the advantage to the engine as does the gearbox. The diameter of the wheels, the drag and weight of the car, and the rate of acceleration is what affects the torque transmitted through the axle. Where would the excess torque go?

but if you put your foot down at peak torque rpm then the engine is only capable of putting a certain amount of torque into the system.

you would only exceed that amount of torque if you dropped the clutch from high revs

of course the torque isnt always going to be peaked for that engine rpm, it depends entirely on how much you push your foot down

if you did a wheel spin (without dropping the clutch but relying on the raw power of a morry minor engine) then you wouldnt have any more torque at wheels than if you had the traction ie bigger tyres, because the engine can only develop a certain amount of torque

Who would have thought we would be having a power discussion - with a Morry?
A lower, higher numerical ratio (kinda like fuel consumption versus fuel mileage) allows more torque for a given engine at a given rpm but lower axle rpm hence the same power.
As typical with discussions there really is no disagreement, just a different point of view. Don't look at it from the engine side look at it from the load side. Remember the engine rating is established under unlimited available load conditions and that is the only time the engine produces that amount of power. If you could accelerate a vehicle the same while only changing engines and ratios, the torque peak on the axle shaft would be exactly the same. Therefore different engines or in this case ratios allow you to increase the acceleration and hence the torque on the axle but at the expense of top speed, fuel mileage etc.

torque = force x radius so if your wheel radius is 0.45m then linear force is 4000Nm for first gear, 2000Nm for second gear, 1120Nm for third gear and 800Nm for forth gear (from the first set of figures)

Impressive that someone got as far as doing numbers
Your 'Nm' should be N btw.


As for where the peak torque is 'limited' - if the grip on the tyre (estimate approx 0.3 coefficent between tyre and road, and ~35 to 40%vehicle weight on rear axle) is less than the linear force exerted then you will get wheelspin. That assumption is fine in a static condition but other conditions such as dropping the clutch from high RPM and or weight transfer during accelleration will change the torque at which the wheel slips.
Any 'spare' torque above that required to spin the wheel is used to speed up the wheel, so the entire torque is still applied to the halfshafts.

you're onto it! thanks for pointing out that mistake of mine.

My only comment to all this is that when I was a kid it seemed to take a lot more pressure on the pedal of my pushbike when starting off in third gear as opposed to first gear. That is toque isnt it force times the length of the pedal?

My only comment to all this is that when I was a kid it seemed to take a lot more pressure on the pedal of my pushbike when starting off in third gear as opposed to first gear. That is toque isnt it force times the length of the pedal?

- yes that's the ticket ;-)