Rear brake usage?

[Ocean1]

Elegant simplicity is all very well, but I should have thought that on a bike there'd be some mechanism to limit the rear brake effort as the weight is transferred to the front. Like on my Ford Telstar SW.

Bike brake systems are generally way less sophistocated than a cars, perhaps one of the reasons is that independent control really is a desirable feature. I can see a GOOD independent ABS option being attractive though.

Also, that Telstar SW rear brake mechanism... you sure it's function isn't to increase bias to the back when there's a load in the back? Common feature on commercials.

[Badjelly]

Bike brake systems are generally way less sophisticated than a cars, perhaps one of the reasons is that independent control really is a desirable feature.

Yes, that's why I find it hard to believe that a manufacturer would discard independent control without including some mechanism to reduce the rear-brake contribution under hard braking.

There was a MotoGuzzi in the 70s that had a pedal that activated front & rear, plus a lever that activated only the front. So you could use the pedal for most braking, but add a bit of lever if you wanted to stop hard. Reviews suggested it worked quite well.

Also, that Telstar SW rear brake mechanism... you sure it's function isn't to increase bias to the back when there's a load in the back?

Er, yes.

[Mikkel]

There seem to be many here that disagree with your contention.

Wouldn't be the first time...

Anyway, if your bike and yourself constitute a total mass, M. Then the maximum amount of braking force you can achieve is equal to:

Ffriction = M*g*f

Where g is the gravitational acceleration (~9.8 m/s^2) and f is your static coefficient of friction. Unless there's a big difference between f for your front and rear tyre the magnitude of your braking force should remain unchanged regardless of the way weight is distributed between the two tyres.

If you want me to I'm happy to expand upon the above.

The front/rear brake issue is more important in regards to staying in control, maintaining stability and having the ability to maneuver out of danger.
I've failed at this once recently. All other times I've used my brakes I've been successful...

[discotex]

I understand trail braking isn't just the rear brake; it refers to the gradual release of the brakes (front, rear, or both) up to the apex.

Your definition is then different to mine. There is a big difference between what happens in a car and what happens on a bike.

Lost count of the times I've read it means front then read it means back. Seems to depend on the context.

In a racing context I believe it's talking about the front (maybe even both).

I.e. loading the suspension under braking then rather than having the forks rebound as you tip in you gradually let the brake off as the cornering forces take over.

I never understood why you'd do this before riding round Taupo but makes a significant difference to how the bike handles. It was somewhat a eurika moment when I got it right the first time.

On the street/open road I try to get my braking done well clear of the corner though.

[All]

Wouldn't be the first time...

Anyway, if your bike and yourself constitute a total mass, M. Then the maximum amount of braking force you can achieve is equal to:

Ffriction = M*g*f

Where g is the gravitational acceleration (~9.8 m/s^2) and f is your static coefficient of friction. Unless there's a big difference between f for your front and rear tyre the magnitude of your braking force should remain unchanged regardless of the way weight is distributed between the two tyres.

If the rear tyre has lost contact with the ground (due to really yanking on the front), then, yeah, there's a big difference between f for the front and rear. That's what I'm getting at. Take these two situations--in which would you stop in the least amount of time?

1. Really haul on the front hard so that the rear is 5 cm above the ground during the whole maneuver.
2. Brake hard on the front but not hard enough for the rear to rise. Also use the rear brake as much as possible, but keep it short of locking.

The consensus seems to be that the latter would provide the most effective braking. I genuinely don't know.

[All]

Lost count of the times I've read it means front then read it means back. Seems to depend on the context.

In a racing context I believe it's talking about the front (maybe even both).

I.e. loading the suspension under braking then rather than having the forks rebound as you tip in you gradually let the brake off as the cornering forces take over.

I never understood why you'd do this before riding round Taupo but makes a significant difference to how the bike handles. It was somewhat a eurika moment when I got it right the first time.

On the street/open road I try to get my braking done well clear of the corner though.

I ran a search on Google and most sources that I read say that it's just any braking past the turn in point.

[Mikkel]

If the rear tyre has lost contact with the ground (due to really yanking on the front), then, yeah, there's a big difference between f for the front and rear. That's what I'm getting at. Take these two situations--in which would you stop in the least amount of time?

1. Really haul on the front hard so that the rear is 5 cm above the ground during the whole maneuver.
2. Brake hard on the front but not hard enough for the rear to rise. Also use the rear brake as much as possible, but keep it short of locking.

The consensus seems to be that the latter would provide the most effective braking. I genuinely don't know.

No mate - f doesn't change with the weight distribution. f is a physical constant that tells you how well the tyre in question sticks to the road.
I won't sit here and type in an exhaustive explanation - I'm sure wikipedia can do it for me.

I'd agree that 2. is better - not because it'll make you stop faster, but because you have more control, more stability and more maneuverability.
If 1. reduces your stopping distance then it is only because you get to a point where you can not brake as hard without doing a cartwheel... 1. and 2. should give the same stopping distances if the wheel only just leaves the ground.

[All]

No mate - f doesn't change with the weight distribution.

How can there be any friction if the wheel is not in contact with the road?

[All]

1. and 2. should give the same stopping distances if the wheel only just leaves the ground.

I just wonder if, in a real world situation, this is the case.

[Mikkel]

How can there be any friction if the wheel is not in contact with the road?

f is the coefficient of friction not the friction force. If there's no normal force there's no friction either (i.e. if the bike and rider are weightless there's no gravitational force and consequently no normal force).

[All]

f is the coefficient of friction not the friction force. If there's no normal force there's no friction either (i.e. if the bike and rider are weightless there's no gravitational force and consequently no normal force).

OK, I'm totally out of my depth with physics terms; I never studied it.

[Mikkel]

I just wonder if, in a real world situation, this is the case.

If you do two experiments there's good chance you'll see a difference. If you do 1000 thousand I think you'll find that the average stopping distance for 1. and 2. are within one standard deviation of each other.

If you're not used to lifting the rear, doing so may cause you to freak out a bit and do stupid stuff. (i.e. locking the front and doing a faceplant) That is one thing I know from experience.

[avrflr]

Anyway, if your bike and yourself constitute a total mass, M. Then the maximum amount of braking force you can achieve is equal to:

Ffriction = M*g*f

Where g is the gravitational acceleration (~9.8 m/s^2) and f is your static coefficient of friction. Unless there's a big difference between f for your front and rear tyre the magnitude of your braking force should remain unchanged regardless of the way weight is distributed between the two tyres.

If you want me to I'm happy to expand upon the above.

You can't achieve that maximum on a motorbike in the dry with warm sticky tyres because (since the braking force is applied at the ground and the c of g is above the ground) you will raise the rear wheel, at which point the braking force you can apply decreases. Using the rear brake can help the suspension squat, lowering the c of g, and therefore increasing the amount of braking force you can apply without raising the rear wheel.

[Skyryder]

There was a MotoGuzzi in the 70s that had a pedal that activated front & rear, plus a lever that activated only the front. So you could use the pedal for most braking, but add a bit of lever if you wanted to stop hard. Reviews suggested it worked quite well.
Er, yes.

They still link brakes on the cruisers. Front is independant. Once you hit the brake pedal both front and back link up. I only use the brake pedal on the twisties then only for minor adjustments. Most of my braking is engine braking with adjustments prior to leaning the bike over............to hot then it's both front and rear hardish.

[dipshit]

OK, I'm totally out of my depth with physics terms; I never studied it.

Don't bother, he's talking engineering bollocks. What he is missing out on is the stability using the rear brake as well gives.

Have a look at these...

http://www.sportrider.com/ride/146_9...ips/index.html

http://www.sportrider.com/ride/146_0...lls/index.html