# Maximum Deceleration of a Bicycle

1. Nov 18, 2013

### Kaevan807

1. The problem statement, all variables and given/known data

I'm only trying to answer the maximum deceleration possible question, not the rest.

2. Relevant equations
Not really sure which equations are applicable, I know the Maximum deceleration of a body in general will be a = μ*g.

Although I would think that moments need to be taken into account for this question.

3. The attempt at a solution
I've had two trains of though with this question, the first was just assuming that between using both breaks, the maximum deceleration possible could be achieved and so we get:

a = (9.81*0.8) = 7.848 m/s2

Not a very engineering solution though and I know it's probably not the correct solution.

I would try to do it with moments, calculating the friction using each and both of the breaks, and then take moments about the axle of the front wheel, but I'm missing measurements to allow me to get these moments.

I know the maximum acceleration with the front break only is given by a = g* (distance from COM to front wheel/ height of COM).

i.e. a = (9.81 * (0.635/1.143)) = 5.45 m/s2

2. Nov 18, 2013

### tiny-tim

Hi Kaevan807!

If we ignore tipping, the maximum deceleration comes from the maximum friction force, which is µ(Nfront + Nrear).

(and if both wheels are braking, then N = mg)

And tipping will occur when Nrear = 0.

(alternatively, use the non-inertial frame of reference in which the bicycle is stationary …

then there are two forces on the bicycle: mg vertically down, and ma horizontally forward, making an "effective weight" of the combination of them …

tipping will happen when that weight misses the wheelbase)​

3. Nov 18, 2013

### Kaevan807

So does that mean I was right and that the maximum deceleration will be μ*g?

Or did I just completely misread what you just said?

4. Nov 18, 2013

### tiny-tim

"If we ignore tipping", yes

5. Nov 18, 2013

### Kaevan807

By tipping I presume you mean over the handlebars rather than to the side?

How do I calculate the maximum deceleration without ignoring tipping then?

6. Nov 18, 2013

### tiny-tim

yes
Tipping will occur when Nrear = 0.

(which you can find from τ = Iα)

7. Nov 18, 2013

### Kaevan807

Just getting more and more confused :P I understand tipping occurs when Nrear = 0, but what is T in that equation?

I is the moment of inertia right? But how do I calculate the moment of inertia of a bicycle? (Moments of inertia really aren't my strong point)

And how do I calculate the angular acceleration? ( which is what alpha is right?)

Just so I get this right though, assuming that Nrear is just = 0, (i.e. That point where we're using maximum breaking force on the front wheel only without tipping) that would mean that we're breaking with the front wheel only, in which case, I know how to calculate the maximum deceleration with just that wheel.

Would I be right in assuming that any breaking force which equates to a front wheel only deceleration less than that value would not cause tipping? And therefore would allow for full breaking power on the read wheel as well?

(Not really sure how I'd calculate that, but just want to get my head around where I'm going with this)

8. Nov 18, 2013

### tiny-tim

τ (tau) is the total torque
When it starts to tip, α = 0

(and now i'm off to bed :zzz:)

9. Nov 19, 2013

### Kaevan807

But if α = 0, then surely τ = 0? Which tells me what? Is there some equation that all of that goes into? Really don't get where I'm supposed to be going with this question

10. Nov 19, 2013

### tiny-tim

sorry, my late-at-night oversimplification

τ = dLtotal/dt = d/dt (Ic.o.mω + Lc.o.m),

where Lc.o.m is the angular momentum of the whole mass m as if it were concentrated at the centre of mass

that τ equals the moment of the weight plus the moment of Nrear

now set Nrear = 0 and α = 0

11. Nov 19, 2013

### Kaevan807

Sorry, still very confused..

First off, why are we calculating Torque?
Torque is a measure of how much a force causes an object to rotate right? So it makes more sense to calculate Torque than moments?

Where are we taking the Torque about? The object will rotate about the axle of the front wheel presumably? So where does this come into the equation?

Is ω in the equation the same as $\alpha$?

Looking through the link you put in, most of the formulae contain a velocity v, which we don't have, a mass m, which we don't have and a position vector r, which, relative to the axle of the front wheel, we only have an x-component for? I can see how you could leave some of these as constants, but I don't understand how you can go from Torque to Deceleration in the first place?

Thanks for taking the time to reply, and sorry that it's taking me so long to come to grips with this, some parts of mechanics I find really easy but this stuff trips me up a bit.

12. Nov 19, 2013

### tiny-tim

torque and moment (of force) are the same thing
no, it will tip about the bottom of the front wheel (the instantaneous centre of rotation)

we can take moments (torques), and use τ = dL/dt, about any stationary point, or about a moving point provided
i] it is the centre of mass
ii] it is the instantaneous centre of rotation, and the the instantaneous centre of rotation is moving in a straight line parallel to the centre of mass
sorry, no, i'm using the standard notation α = dω/dt = d2θ/dt2
m will cancel

v you don't need, because you only need dv/dt (a), which you can find from µN = ma

(and you do have the y-component, since it's the bottom, not the axle)

13. Nov 19, 2013

### Kaevan807

Finally got a chance to meet up with my lecturer today, who said that Max deceleration occurs when the front brake only is applied and the back wheel just comes off the ground, i.e. The equation that I have at the start.

Didn't realise that it tipped about the bottom of the front wheel which made everything make more sense. Thanks for the help.