# Do ALL non-instantaneous deceleration events transfer weight to front wheel?

There's a great question and discussion on hub vs disc brakes here. That got me wondering this: do ALL non-instantaneous deceleration events transfer weight to front wheel?

I've got a follow up theoretical (not practical) question.

Regarding the non-instantaneous qualification to my question, let's assume we're not interested in instantaneous or near-instantaneous deceleration events like slamming into a brick wall or the side of a car. Let's call those the singularities of the bike deceleration universe in which the regular laws of physics break down.

The common and I believe correct physics wisdom is that braking transfers weight to the front wheel. And I was thinking just now that probably all deceleration does so even if not due to brake application. But I'm not sure.

Assume you're cruising along on the level, pedaling to maintain a constant speed. There will be some weight distribution between the front and rear wheels, which will be the reference for discussing weight transfer below.

At some point you stop pedaling but you don't apply your brakes. Your speed will start to decrease due to a mix of (mostly) air resistance and (to a lesser degree) what I'll call "mechanical friction" in the the hub and tire-road and spoke-air interfaces. In that case is there still some non-zero increase of weight on the front wheel?

I think yes, albeit minimal, due to the friction the those mechanical friction factors which I think act like very weakly applied wheel-based brakes, even though air resistance on the rider and frame account for most of the deceleration in this case.

If I'm correct, does the answer change if the deceleration is purely due to air resistance on the parts of the bike and rider moving linearly in the direction of travel? E.g. if the mechanical friction described above was zero, so the only thing slowing you down was drag on your body and bike frame, would there still be a transfer of weight to the front wheel relative to the constant speed weight distribution?

My first instinct on this variation was "yes because you're slowing down", but now I'm wondering if the frontward force from your body's deceleration which I'm picturing moving weight to the front wheel in this case is actually balanced by the air drag so that the weight distribution remains unchanged.

What if we get rid of air drag too (in which case I think deceleration would be zero), but put the cyclist on an uphill trajectory so that after ceasing pedaling, they decelerate purely due to gravity?

• just a note that even in the singularity case (hitting a brick wall) weight still transfers to the front and the rear wheel comes off the ground. Angular momentum is a hell of a thing. Commented Mar 22 at 18:22
• @PaulH as the wheel makes first contact, yes. I've hit a kerb head-on and been ejected forwards from the saddle when the bike stopped dead (chrome rims in the wet, I ran out of road) Commented Mar 22 at 19:40
• @PaulH It depends on where you're hit. You usually flip forward because your front wheel is usually the first point of collision, but there is no reason that must be the case. If someone opens a car door into your chest, it's an instantaneous deceleration that transfers weight to the back of the bike. Not sure what you're getting at with angular momentum, the person doesn't have any angular momentum about their center of mass before the collision, they only start rotating because of the torque applied by the deceleration (if there is any). Commented Mar 26 at 13:56
• @NuclearHoagie The wheels (and tires) have angular momentum. You can observe this and the transfer of momentum from the wheels to the frame by putting the bike in the stand, spinning up the rear wheel, and then locking it up with the brakes. The bike still pitches forward. An open car door is not a brick wall. Commented Mar 26 at 21:49
• @PaulH Even accounting for the angular momentum of the wheels, there is no reason an instantaneous deceleration must pitch the bike forwards. Applying a sufficient torque above the center of mass can overcome whatever angular momentum the bike has to begin with. You're only guaranteed to flip forward if the force is applied at or below the center of mass. The duration of force application or how "instantaneous" it is is rather irrelevant, where it's applied matters far more. Commented Mar 27 at 12:55

There is no need for all forms of deceleration to transfer weight to the front wheel, although it is in practice usually the case due to how forces are commonly applied to a bike. The tendency of the bike to rotate forward or backward depends on where the torque is applied. Backward forces applied below the center of mass will tend to cause the bike to pitch forward (which is the typical case when applying braking forces through the tires), but backward forces applied above the center of mass will cause the bike to pitch backwards.

As a simple example, imagine a cyclist who bikes into a parking gate arm and gets "clotheslined" - the gate hits them in the torso, and the torque above the center of mass causes the front wheel to lift off the ground and the cyclist to fall off the back of the bike. There's no need for this to be an instantaneous or near-instantaneous force, any force applied backwards above the center of mass will apply a torque that reduces weight on the front wheel. This is basically how one performs a wheelie - pull back on the handlebars, and the front wheel rises.

All the questions about air resistance and hills can be answered by examining where the net force is applied, and what is the net torque around the center of mass. If you wear baggy pants, air resistance may be applied below the CoM and pitch the bike forwards, but if you wear a baggy jacket instead, the air resistance may be applied above the CoM and pitch the bike backwards.

As an aside, this principle of torque also explains why it can be advantageous to lower your center of mass during hard braking. By reducing the length of the moment arm between where the force is applied (the ground) and your center of mass, you reduce the amount of torque pitching the bike forward. If you could align the braking force with the center of mass, there would be no torque pitching the bike either way.

• So deploying a parachute from a backpack would, of course, transfer weight to the back wheel, and slow you down (though it wouldn't be good for actual stopping). Grabbing a fence to stop would probably hurt, but again would shift weight back. This is noticeable when learning to ride a unicycle - the front/back balance is far more obvious when there's no other wheel to take up the weight Commented Mar 22 at 19:38
• @ChrisH I like the unicycle observation ! Commented Mar 22 at 19:50
• Perhaps a useful rule thought experiment: "In circumstance ____ on a unicycle, do I want the wheel to be in front of me or behind me?" And in that same circumstance on a bicycle, that's the wheel that is loaded. Commented Mar 22 at 20:21
• @Michaelcomelately apart from being great fun, even trying to learn is very instructive for anyone interested in cycling physics Commented Mar 22 at 20:36

If the brakes are applied, mostly two points matter: the mass center of the bicycle + the rider and the braking point, the contact surface of the braking wheel to the ground.

As the mass center is well above the braking point, we get the rotation force around this point that is trying to dig the front wheel into the ground. This is true for both front and rear wheel braking.

No matter how the deceleration happens, if the wheels are spinning, they carry momentum; when the wheels stop spinning, that momentum has to go somewhere, it cannot vanish.

So, if the wheels are spinning forward, the bike will be loaded more on the front because of the transfer of momentum; if instead the wheels are spinning backwards, the rear will get more load, for the same reason.

• Heavy wheels at higher speed may have a notable momentum. When brakes attempt to lock the wheel to the frame, now all bicycle wants to rotate same direction as the wheel did so front down, rear up. This should even work with brakes applied while wheels are in the air. Commented Mar 27 at 8:14