I found this quote somewhere around here.

Not only was it unchallenged but the next poster gave it a +1. Other comments I've seen have pointed at similar misconceptions and it seems worth pointing out here in mechanics. A related misconception is probably that disk brakes transmit stopping force through the spokes and rim brakes do not.

Both notions are completely false of course or even somewhat upside down. In short momentum conservation dictates that ALL force to decelerate the forward motion of a bike MUST come through the ground (except for a little through the air), transferring momentum to the ground. For top mounted rim brakes (like all front rim brakes) this force is essentially transmitted through the spokes to the frame. (a bike hurling through a vacuum can never stop itself.. you have to push from the ground)

As the brakes make the wheel try to roll slower it is resisted by the ground since the weight of the frame and rider still want to continue over the ground at the same initial speed. Ignoring air resistance, the force at the ground is the full rearward acceleration (foward deceleration) force, m*a.

That force to stop the bike is then transmitted from the wheel to the hub through the rearward spokes which pull backwards and through the forward spokes which push backwards. This is true on radial spoked wheels essentially just as on cross wheels. What disc brakes do that rim brakes don't is disk brakes put TORQUE on the rim through the spokes, but only to slow down the wheel itself. Both put stopping force through the spokes radially because again, stopping force MUST come from the ground, not the brakes.

There is a force on the brake boss, a big one, but if the boss is on top (as usual for front wheels), the wheel is moving forward there and pulling the brakes, the boss, and the bike FORWARD! As it turns out the force on the brake boss IS equal in size to the force on the ground (for weightless wheels) in order to balance toque on the wheel (if the wheels have weight, SOME net torque is required to decelerate the rim, and then this force is even bigger). But if the brakes are top dead center, that force pulls the bike forward with m*a. That means to place net m*a on the frame, the force transmitted through the spokes is actually 2m*a, backwards!

It seems that saying the braking force goes through the bosses is at least misleading here. For chain stay mounted rear brakes the situation is a little different, and the partially rearward force on the bosses does contribute to frame deceleration, reducing the load on the spokes instead of increasing it.

I'm no engineer but I do know that the rim/tire is a continuous hoop running from the ground to the top (where rim brakes are usually found) and back to the ground again. I would suggest that braking forces are transmitted through the rim, around the rim, and to the caliper/fork. The spokes will see extra stress from the weight transfer but not from braking forces being carried through the spokes.


Consider that the braking forces are far greater then any ability for the rider to power the bike forwards. Yet under the lesser (compared to braking) forces of pedaling will and do wind up a fully radial rear wheel. Were the same spoke path of changing rotational forces be the case for braking (as for pedaling) then we would see many radial front wheels suffering from brake induced wind up. That we don't suggests to me that the spokes don't carry the braking forces when a rim brake is used.


It's really neat how this stuff has already been figured out in the lab of real world use. Regardless of what we think we need to look at real life models and apply their results into our theories. Andy.

You had a right idea but lost it (next time read the whole post). Yes forces transmit around the rim, but the force that the rim transmits to the brake is FORWARD. Spin the front wheel forward and then grab it with your hand just as the brake does. Does the wheel tug forward or backward on your hand? Forward of course. How can a forward pull on the brake/boss/frame slow the bike? Real world enough for you?

Where your idea was right is the brakes can and MUST resist the TORQUE on the wheel from the ground, and just a bit more to slow the wheel itself, and this can get passed around the rim (and/or through non radial spokes, but mostly the rim). The net force though (not torque) from the wheel to the frame WILL be m*a until you loose traction, and whatever the difference is between the force at the boss and m*a, MUST be transferred through the spokes to achieve that net. Or put the other way, the deceleration a, times m, WILL be whatever that NET force is. Either way you say it, the total force on the frame F adds up as F=m*a.

In the case of top mounted brakes, you start with approximately m*a in the wrong direction at the bosses and the spokes must actually take twice the force to resist the bike's forward motion AND counter the forward force at the brakes.

Makes me want to move my rear brakes to the seat tube to force the frame down on braking!

i have no clue what i just read, but braking force comes from my hands

Assuming all of this is factual, what's your point?

[tl,dr] (physics) It's friction turned into heat.

Of course it would force the frame down at the bosses, and this would be countered by an upward force at the hub. That's the whole point.

If you hold a wheel in your hand with a stick through the hub. If you spin the wheel so it spins away on top and grab it with your free hand on bottom to stop it, the wheel will try to roll away from you off that bottom hand, but you will counter with an inward resistance (force) at the hub. The two forces are transmitted through the spokes. And all this even happens without any net force on you at all (do it on roller skates, you won't go anywhere).

You might have been joking that my argument is liking picking yourself up by your feet, but if so, you were very wrong (well my argument was like that, but with full awareness that the very real force involved in that effort sums to zero, and yet does stress your arms).

It's magic. I just squeeze the brake levers. Or jam my foot behind the front fork.

I suppose you could say:

a) LARGE braking forces go through the spokes for both disk brakes and rim brakes.
b) braking force as in slowing down the mass of the bike, does not go through the boss, except for some of it in the case of chain-stay mounted brakes. In practice it's hard to assign responsibility to forces since they act in sum, but the force at the fork bosses does the opposite of this.

Great example. Try this barefoot. It will become very clear that your toes are squeezing very hard FORWARD into the fork and are not directly transmitting the force responsible for pushing back on the bike's weight. You're just stopping the wheel. The wheel stops the bike (through the hub and thus spokes, with an crucial support from the ground).

I had to think about this a lot. The braking force is transferred from the brakepad/disk interaction to the hub and from there through the spokes to the rim and tire and to the ground, then back from the rim and tire through the spokes and hub to the bearings and axle, and thence to the drops and frame. At the same time, the frame is stressed by the brake bosses and drops being pulled in different directions.

Ultimately it is the ground that stops your bike. If you applied the brakes while flying like ET they would have no effect.

Ah, no I was serious. Forcing the frame down would delay lifting of the rear wheel thus improving braking power.

It's the conversion of kinetic energy into heat by the brake pads that slows the bike. The tire/road interface powers the brake by turning the wheel and the brake is kept in place by its connection to the forks.


If you were flying ET-like and your wheels were connected to a propeller, applying a brake to the rim would slow you down.

Converse all stars, high tops only. I tend not to think that much about my brakes, ever.

Ouch!

And, why?

No! The downward force on the seattube (of a seattube mounted brake) would be opposed exactly and completely by an upward force by the hub.

One way to think about this is that we have two separate systems of forces acting here. The overall forces. Really simple. Gravity acting on the center of mass of the rider and bike. Upward forces from the road opposing that gravity force. Change of inertia seen as a horizontal force through the rider/bike center of mass. Horizontal friction forces from the road onto the tires exactly opposing that force. That's it. All other forces are internal and never seen by the road, the tires or the center of mass of the rider/bike.

So those internal forces must all cancel themselves out inside the "box" of the bike, rider and wheels. The road does not care where the brakes are located, how calipers are oriented or which tire is braked (until the bike changes inertia enough to shift the weight balance onto the front tire patch and off the rear tire patch, an effect of the overall picture and the location of the rider/bike center of mass relative to the tire patches).

To take the misconception of the location of the calipers mattering further, let's look at your conventional front disc brake. The caliper is behind the hub. The force that is applied to the caliper is up! If a seattube mounted caliper puts more weight on the rear end of the bike, then hard stops with a front disc brake causes wheelies, right? And as Andy points out, we can look at the real world and see that happening, right? (Obviously I need better eyewear. I have yet to see that hard stop wheelie.)


Again, no! Internal forces are not what slow you down. On a bicycle, the only things slowing you down are 1) friction between the tire and the road, 2) air resistance and 3) gravity if you are going uphill. All of the rest are internal forces (which may be reducing the power you can apply to the rear wheel and slowing you down when you don't want it ) On your ET plane, slowing the propeller causes drag and resisting lift from the air the plane is passing through. Similar to tire friction in the big picture. The fact that you may be heating a brake doesn't matter.

The heat you can measure coming off your rim brake will be (almost exactly) matched by the inertial energy lost by your rider/bike system as it slows, but that heat DID NOT slow the bike, the tires did! By mechanical friction, not the heat they generated also. (Double because it must match both the energy lost by the rider/bike slowing and the friction force applied to the tire at the road.)

Keep in mind that we are looking at two different systems. The really simple big picture and the more complex small picture. The only place they overlap is where forces are applied by one to the other. At the contact points of the two tires and by wind resistance on the overall rider/bike. That's it. Now the bike builder/designer has to think out how the internal mechanics are done and see to it that the sometimes very large internal forces are dealt with properly. But the road doesn't care!

Ben

Of course the generated heat is an issue. It's conservation of energy: to reduce the kinetic energy of your forward motion you have to convert it into something else such as heat. The tire patch isn't much (if at all) hotter during acceleration or deceleration unless the tire is slipping in which case the brake is no longer working.


If the heat generated during braking didn't matter we'd have brakes that didn't heat up.

+1
I let someone trained in physics design and manufacture a bike that pedals and stops efficiently. There is no need for me to reconstruct all of the math behind my bikes.

I know, I was trying to lead a horse to water. Never mind ...

I thought it came from a disturbance in the force?

No no, I'm wrong, it comes from turbulence in the phlogiston caused by the brakes activating the phlogistonator.

If you lift your bike in the air, spin the rear wheel and then brake, do you feel it push in any direction? Personally I expect it to lift a bit, no matter where the brake is placed, as the forward rotation of the wheel is transferred to the bike, causing it to also rotate forwards

So are you saying that the brakes wheel, with rim braking, is not seeing a rotational force (therefore no wind up) but a radial force? That I can agree with. Andy.

So you dig up a quote for 8 years ago according to google to complain about. And the full quote says force is transferred through the spokes radially.

What exactly are you complaining about?

cooker and 79pmooney have it all about right. As for the conundrum about heat, that's not wrong either. Yes the brake pads dissipate heat and this is an important function because the collision between the bike and the ground is an inelastic one where both mechanical energy and momentum cannot be conserved at the same time. While all momentum goes to the ground, all kinetic energy is dissipated. In an elastic collision you would bounce off a wall attached to the earth and would end up going the opposite direction.

You could instead slow the bike by running into a big rubber band. Right as you stop you could imagine clipping the rubber band so it doesn't spring back. In that case no heat was dissipated. That energy was stored in the rubber band and could be dissipated in some other braking mechanism later. Would you call that mechanism the stopping force? It does the same job as the brake pads. You might argue the band and the transfer of energy to it replaced the brake pads. That's another way to see it, but the band applied the same external force as the ground does during normal braking. In the end the point is the external force from the band is still the same as the force would have been from the ground for the same deceleration curve. That force requirement remains the same regardless where the energy goes.

These are somewhat separable issues and is related to why I said it's hard to exactly assign blame to forces. My force balance was however correct and the rearward force that decelerates the frame and rider requires large involvement of the spokes.