Cycling UK Forum

I wonder if SB has confused the effect of a rotational mass rolling down an inclined plane where some of the potential energy is converted into rotational kinetic energy. At some angle the friction will be insufficient to cause rotation and the mass will slide. Due to the potential energy no longer being converted to rotational energy the mass would accelerate.
If there were no friction between a locked rear wheel and the road surface there would be a longer stopping distance (quite how the front brake would be effective under these circumstances I cannot imagine.)
I'm inclined to agree with Mick F that two brakes are more effective than one.

mark a. wrote:1) The front brake only applies force F1 so that the rear wheel is just lifting i.e. its best braking ability
2) Both brakes are applied, with front applying force F2 and the rear applying force R2, to maximise their combined braking ability
Mick's experiments suggests that F2 + R2 can be greater than F1 i.e. the bike can brake faster with both brakes applied. Sheldon implies that F2 + R2 cannot be greater than F1*.

Someone with more inclination could do some calculations (coefficients of friction, turning moments lifting the rear wheel etc) to see which scenario is best.

F2 + R2 cannot be greater than F1
As F2 gets closer to F1, the maximum possible R2 decreases, to zero in the limit, where F2 = F1

It's better to consider things graphically than to do complicated calculations. At F1, the combined vector force from the weight and the braking force passes through the front wheel contact patch. At this point the bike is effectively balancing on the front wheel with zero weight on the rear wheel. A wheel with no weight on it cannot contribute any braking, by skidding or otherwise.
I'll do some illustrations later (photobucket is blocked at work, to say nothing about actually working)

At the point of stopping, I agree.

But what about at 20mph?

I don't know about you, but I can stop PDQ, but at 20mph it takes one heck of a lot of effort and a greater stopping distance than you might think. In order for the rear to be lifting at 20mph, it would take quite a bit of a balancing act!

When I brake at speed, my back brake contributes one heck of a lot and I don't even get near to a head stand until I'm just about stopped.

andrew_s wrote:F2 + R2 cannot be greater than F1

I'm pretty certain this is correct, but I've left it deliberately vague above because I haven't done the maths to prove that it's the case. There might be some combination of braking forces, tyre choices or weight distribution where it's possible that F2 + R2 > F1 before the point of rear wheel lift.

Mick F wrote:At the point of stopping, I agree.

But what about at 20mph?

I don't know about you, but I can stop PDQ, but at 20mph it takes one heck of a lot of effort and a greater stopping distance than you might think. In order for the rear to be lifting at 20mph, it would take quite a bit of a balancing act!

When I brake at speed, my back brake contributes one heck of a lot and I don't even get near to a head stand until I'm just about stopped.

In theory if you could instantly go from no braking to applying the precise force in your front brake lever to reach F1, then the rear brake might not be necessary.

In practice though that's a pretty tricky proposition both in terms of knowing the correct force on your brake lever without tipping over and in the strength required. So in this real-world case the rear is useful.

Don't forget your feet for added stopping distance.

I could use my L foot to slow me down, hence...

F1 + R1 + L-Sole would be greater than F1 alone.

Anyone use their R-Sole?

As it will take a little time to hit F1max after you hit the brakes, It will depend on how grabby your front brake is as to how much braking you can do with the back brake.
The more grabby the front brake is, the less the back will cotribute to your total braking.

In an ideal world, you'd just grab the front brake and hit F1max straight away.
In that case the back brake won't do much, if any, braking.
In the real world, you'd grab both brakes.
It will take a little time to hit F1max due to cable streach, needing to heat the brake pad up to hit maximum braking, it takes a little time to unload the back wheel, etc, etc.
In this time the back brake is usefull in helping to stop the bike.
Hence Mick's results of a shorter stopping distance by useing both brakes.

I ride a recumbent trike.
The only time I've used the back brake to stop from any speed was because the front brake cables where frozen ........
Because of the lack of weight on the back wheel, any moderate stopping power from the back disk just caused the back tyre to skid.
So it was taking over 10x further to stop vs the twin front disks.
Plus being as low down as you can get, there is no way I can lift the back end with the front brakes, I skid the front tyres first.

Luck ...........

If we use the feet? do we have L1 left and R2. How can we use R2 to slow the bike down to get FO

Fatou wrote:Don't forget your feet for added stopping distance.

reduced, surely? now who's the r-sole

Fatou wrote:Don't forget your feet for added stopping distance.

I could use my L foot to slow me down, hence...

F1 + R1 + L-Sole would be greater than F1 alone.

Anyone use their R-Sole?

Fatou wrote:Anyone use their R-Sole?

Considering how far off the back of the saddle some mountain bikers get so they can brake on steep downhills, probably

The top cyclist is just braking normally, the bottom cyclist is braking as fast as is possible.
The yellow dot is the cyclist's centre of gravity.


Top cyclist
Weight W and braking force F2 add together to give a resultant force vector R.
This vector is in the direction of "down" as far as the cyclist is concerned, and meets the ground between the wheels.

The blue line is perpendicular to the resulting force vector. The wheel contact points are projected onto the blue line parallel to the force vector, and meet it at distances d1 and d2 from the force vector
The weight on each wheel is such that wt_rear * d2 = wt_front * d1. Distance d2 is 5 times d1, so 83% (5/6) of the weight is on the front wheel, and 17% (1/6) of the weight is on the back wheel.

Bottom cyclist
The cyclist is braking with a force F1 such that the force vector resulting from the braking force and the cyclist's weight passes through the front wheel contact patch.
In this case, distance d1 would, if drawn, be zero, and therefore the weight on the rear wheel is zero.

The maximum possible rate of deceleration is 1/arctan(A)g, where A is the angle the resultant force vector makes with the ground (W/F1=tan(A)). If we take A as the 60 degrees it looks like, the cyclist can decelerate at a maximum rate of 0.58g. This seems a bit low - I dare say the C of G should be further back.

Mick F wrote:(Pedantic head on)No, it's not pointless.

it is pointless to try and prove or disprove it. you can't do what sheldon's theory says, because you can't override years of learnt experience and risk smashing your face into the road. as andrew's excellent diagrams (nice work andrew ) show above, sheldon is right, in theory - but none of us have the recklessness to be able to get there, as mark & i have said.

mark a. wrote:In practice though that's a pretty tricky proposition both in terms of knowing the correct force on your brake lever without tipping over and in the strength required.

eileithyia wrote:yes I use mine, it was how I was taught to control the bike with the back brake adding the front brake as required.
Maybe he had gone over the handlebars himself but my dad taught me not to pull on front brake first due to this potential.

I was taught this too, but I don't agree. I have long since favoured the front brake for most purposes, pretty much as Sheldon argues - so much so that my rear cable has rusted up on my fixed (which doesn't need a rear brake to comply with the law). Must sort it out...

This is something that produces heated opinions, however, with everyone apparently telling everyone else who disagrees that they are wrong and riding dangerously. Same applies to which lever to connect to the front brake.

Actually it's a balance of risk, and people come up with different answers therefore.

Front = more effective, more controllable braking, lower risk of skidding (because more traction), higher risk if you do skid

Back = much greater tendency to skid, more chance of recovering if you do

Both = balance of risk, best and worst of both worlds

Most of my braking is gentle, as it should be if you think ahead, so it's not normally an issue. I'd expect to straighten up before braking if at all possible, or conversely brake before cornering, thus minimising risk of skids.

Sheldon is obviously right that front brake only is, in theory, the fastest way to stop. However, I for one am not that good, so if it is an emergency you'll see me on both brakes.

The key thing is to be able to "feel" the road, so that you know when you are at the limit of traction. Richard's Bicycle Book advocated pumping the brakes for this reason - braking a bit harder to find out how much you have in reserve, then letting off again. I asked my driving instructor if you did this in cars (before ABS), and he said it took years of practice. I'd had a decade of cycling by then, so I proceeded to do it without thinking, in my first practice emergency stop

Excellent post, Andrew. Thanks!