I’m trying to understand the effects of a bike’s trail on its steering stability. There, however, is a concept of slippage velocity of the tires during steering of which I cannot make heads or tails of.

Book- Motorcycle Dynamics

Author- Vittore Cossalter

Page no. 13-14

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1st Question: Why should the final resultant velocity of the front tire be in the same initial direction of motion ? Shouldn’t it be in the plane or rotation of the front tire, now that a steering input has been given?

2nd Question: What is this slippage velocity ?

I’ve tried Google as well. But there isn’t any information on this topic in simple words.

  • $\begingroup$ For 1) because that is the direction you want to go $\endgroup$
    – Solar Mike
    Commented Dec 2, 2017 at 11:01
  • $\begingroup$ So, I want to go in the direction in which I have steered the handlebar. But then why is the resultant velocity of the tire still in the initial straight direction. Shouldn’t it change as I steer it ? Or else what’s the use of a handlebar ? $\endgroup$
    – RedHelmet
    Commented Dec 2, 2017 at 13:17
  • $\begingroup$ Also, I thought that (wR=V). But if that happens then Vslide =0. What is happening here ? $\endgroup$
    – RedHelmet
    Commented Dec 3, 2017 at 2:55

1 Answer 1


When a wheel is rolling in a straight line, the velocity of the wheel at the point of contact with ground is zero. That is, the backwards spinning speed of the wheel cancels the forward speed of the bike. In the frame of the wheel, everything is moving in the x-direction. Here $\bar{v}$ is the speed of the wheel at the contact point:

$$ \bar{v}_{straight,x} = v_{bike} - \omega r = 0 $$

$$ \bar{v}_{straight,z} = 0 $$

From the perspective of the wheel, the relative speed of the bike is different after the wheel is turned:

$$ \bar{v}_{turned,x} = v_{bike}\cos\theta - \omega r = -(1-\cos\theta)v_{bike} $$

$$ \bar{v}_{turned,z} = -v_{bike}\sin\theta $$

The terms $\bar{v}_{turned}$ terms define the sliding velocity of the wheel.


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