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Front and rear wheel speed difference?

Started by h106frp, August 26, 2015, 10:22:17 PM

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Playing with my widget i can view front and rear wheel speed and the difference whilst riding.

With the default 125 i can observe that front braking with any useful but controlled degree of force (not locking the wheel) and the bike upright, straight and stable causes a difference of up to 3m/s difference in wheel speed with the front rotating slower than the rear. Not tried it with any of the more powerful bikes yet.

This seems to suggest either the front is always sliding to some degree or the rear is spinning up. Pretty sure i am not lifting the rear tyre and the throttle is shut, all braking assists are off.

Not sure it means anything, i am just wondering what causes this to happen, any ideas?


From my knowelage rear wheel just slides. I read about it in some motorcycle book. At high speeds for example rear wheel spins faster to the degree that speedometer from front wheel can show up to 20 Km/h lower speed than from the rear wheel.


Thanks, i was wondering how it could occur without the bike tying itself in a knot. Strangely the default 990 does not seem to exhibit it to the same degree and seems to be about 1m/s difference with similar braking force.


Just a thought, but are the wheels the same size? 17inch rear to 16 inch front maybe? or even rim width would have an effect I'd have thought? In that scenario the front wheel would surely spin faster? :-\

Just a thought.  ;)



Normal riding the wheel speeds are reported as identical - i guess they must be 'normalised' somehow to allow for the variations in contact diameter. Difference only shows when braking.


Essentially, any tyre must slide to generate a force. As G.Clooney used to say: no sliding, no party (not 100% sure of the quote).

So when you open the gas, the rear will slide. Even when you don't feel it, even on a Fiat Panda.
Same when you brake: the tyre will slide, in order to generate the necessary force (to slow down the entire bike).
If a tyre is not sliding, then it's neither accelerating (i.e. generating a force that pushes the bike forward) nor braking.

Same goes when cornering: the tyre slides (laterally) to generate a cornering force.

Side note: one does not need to lock a tyre to be sliding, but most people do not realise that.


For the extremely picky ones around here (just me ?): the above is not absolutely true, tyres can generate little amount of forces even without sliding (e.g. rolling resistance, camber thrust, ...).
But for significant amounts of force (be it braking or accelerating), the tyre will slide.


Thanks, makes sense now you have explained it.  :)


Do you have a source that explains it? Don't think I understood.


Quote from: RIDER on September 01, 2015, 04:57:37 PM
Do you have a source that explains it? Don't think I understood.
Just to grasp the idea, look at the black line on that graph:

It gives tha ratio Fx/Fz as function of the slip ratio.

Fx is the longitudinal force the tyre can generate (to accelerate or brake). Fz is the normal load (how much "weight" is on the tyre).
The slip ratio is equal to (wheel rotation speed times the wheel radius - V) / V, where V is the speed at which the bike (actually, the contact patch) is moving with respect to the world.

  • Slip ratio < 0 ==> wheel is rotating slower than the bike velocity, usually when breaking (-1 = wheel is locked)
  • Slip ratio = 0 ==> no slipping (i.e. pure rolling motion)
  • Slip ratio > 0 ==> some slipping, usually when accelerating

As you can see in the graph above, when the slip ratio is zero, the black line shows zero for Fx/Fz: in order to have the tyre generating a force, the slip ratio must be bigger (or smaller) than zero.



The idea seems simple but impossible, that's why I feel like im missing something. Would the same idea affect tracked vehicles?


Quote from: RIDER on September 01, 2015, 10:49:11 PM
The idea seems simple but impossible, that's why I feel like im missing something.

Impossible ?! What do you mean ?
It's not an idea, it's how a tyre behaves. I mean: you put a tyre on a complex machine capable of measuring these properties and the graph above is what comes out.

Quote from: RIDER on September 01, 2015, 10:49:11 PM
Would the same idea affect tracked vehicles?

Yes, it's a property of the tyres in general. It applies to cars, bicycles and all.
I think it even applies to train wheels (no rubber, metal-metal contact).



Maybe a simpler concept is that you must have friction to give the driving (or retarding) mechanism between the tyre and surface, to get friction you need two sliding surfaces in contact, therefore the tyre must be sliding to some degree.

Maybe too simple? 


A bit too simple (as there's also static friction for example), but yes that's more or less it.



September 02, 2015, 11:42:53 AM #13 Last Edit: September 02, 2015, 11:46:50 AM by SKD
You even have a little bit of slip when a tire is just rolling. I won't explain this but when you think of the tire (maybe in 2D) on the ground it's a circle with a flat spot. Since the hub of the wheel has a constant RPM and the velocity at the tire is proportional to the radius and this rpm the velocity at the surface of the tire must change at the flat spot. So the tire speed in the middle of this spot is less then the surface speed of the tire not on the ground. So the speed of the bike will be somewhere between this speed of the free spinning wheel and the speed the tire has at the point of coontact to the street.
So one parameter for this effect is how much tire pressure you're using. An ideal stiff tire wouldn't show this effect at all.


Quote from: SKD on September 02, 2015, 11:42:53 AM
You even have a little bit of slip when a tire is just rolling.
Right, but that depends on the exact definition of "just rolling". As soon as you take into account flexible tyres, all becomes a bit messier and one has to be very precise.

In general: pure rolling = no slipping (i.e. contact patch is not slipping with respect to the ground/track). But if there's tyre deformation, this does not mean that "tyre radius * omega = V" (as you explained, due to the deformation).