Position in seat?

[R-Soul]

I tried answering this but it has got a long way from where the OP was asking about how weight affects handling (I think that was the question??) so I'm stopping now.

See if you can find some info on how Kenny Roberts teaches riding skills, that may be an easier way to explain it.

No worries - I did not mean our banter as an argument - more as a discussion of the principles- I am more than willing to learn from anyone. and I agree with you that having dirt bike skills adds a lot to your abilities as a rider.

[R-Soul]

Must be time for me to give up riding, You have to be a mathematician to know how much to lean into a corner now !

We're just trying to psyche out everyone before the next track day..

[p.dath]

Actually for a given speed when the bike is equilibrium in a corner (i.e. not falling over more or less), the COM is fixed and does not move (remember the compnent . BUT, where the relative COM of each of the rider and bike is does make a difference (since you want the bike more upright for better suspension operation and therefore better traction). The further inward teh rider is, the more upright the bike is. But the COm of both the bike and rider stays the same.

I guess I can't agree with this. I don't see any point in leaning the bike over if your not going to lean with it. If your going to lean the bike over but move your body outward then you might just as well lean the bike less.

Why fight the lean of the bike - that you have put the bike into?

It is not the contact patch size that makes the difference. Its the operation of the suspension to keep that patch on the ground. There can be no traction when the wheel is in the air.

Assuming the contact patch is the same in both cases and the same road surface, there is no reason that the suspension should not be smooth both leaned over or when straight up.
The bike is not suddenly more likely to loose traction just because you lean it over into a corner.

[sinfull]

We're just trying to psyche out everyone before the next track day..

Ya got me ! I'll be taping a calculator to my bars for tomorrows Motott !

[R-Soul]

I guess I can't agree with this. I don't see any point in leaning the bike over if your not going to lean with it. If your going to lean the bike over but move your body outward then you might just as well lean the bike less.

Why fight the lean of the bike - that you have put the bike into?

Your own equation proves this. Do you see a designation for mass in there? No! The mass is not relevant because it gets cancelled out in the derivation of that equation. The centrifugal accelration is acting on the same mass, and the mass acting inwardly about teh contact pact is creating the balancing force against the centrfiugal accelration. And the whole equation relates to "lean angle required". Lean angle aof WHAT do you think? Bike and rider of course.

By the principle of conservation of angular momentum (for a system in equilibrium): if one mass - the rider- moves clockwise about a point (the contact patch when looking from the front), the other mass - the bike- another must move counter clockwise to remain in equilibrium. (Nature always works to keep a system in equilibrium).

Assuming the contact patch is the same in both cases and the same road surface, there is no reason that the suspension should not be smooth both leaned over or when straight up.
The bike is not suddenly more likely to loose traction just because you lean it over into a corner.

Traction is not about the size of the contact patch- its about the ability of the suspension to hold the contact patch against the ground. The back spring pushes directly downwardly in the line of the bike when viewed from the front. All the dampening and spring settings are designed around the bike operating in an upright position, and upright is the position in which the suspension is most effective.

This is not really debatable- this really IS fact and straight applied mathematics. When the back spring and forks are at an angle, only a component of the reaction force by the ground against the wheel is transmitted to the shock, and only that component is handled by it. the other component has effectively no suspension, and this will cause reduce the effectiveness of the suspension overall.

One day when we are at NASS at the same time, I will draw it out for you.

[R-Soul]

Ya got me ! I'll be taping a calculator to my bars for tomorrows Motott !

Its working- you already have increased mass! Now how am I going to get you take another ton and half on board (so that I have a hope of beating you..)

[aprilia_RS250]

Nope, lean angle.
http://en.wikipedia.org/wiki/Counter...o_lean_to_turn

Sorry you're right. The function however doesn't take into consideration the fact that as you lean you have less grip/contact with the road, it assumes you're on ice skates or something. This is not the case for bikes.

As bikes have wide tyres the more you lean the less contact you have with the road. The reason you lean normally is because you're changing momentum (momentum is a vector, i.e. it's direction and speed), so you're offsetting the force that's pushing you out of the turn by leaning in.

Now the reason why a biker leans out from the bike towards the apex is so the bike is more upright while maintaining the overall distributed lean angle (this is where your function above comes in where it gives you the "perfect lean" angle). The rider is effectively leaning out from the bike in order to get more grip from the tyres.

http://en.wikipedia.org/wiki/Bicycle...cycle_dynamics

[Headbanger]

This thread is no longer insane, Its just disappeared up it own backside.

[p.dath]

Your own equation proves this. Do you see a designation for mass in there? No! The mass is not relevant because it gets cancelled out in teh derivation of that equation. The centrifugal accelration is acting on the same mass, and the mass acting inwardly about teh contact pact is creating the balancing force against the centrfiugal accelration. And the whole equation relates to "lean angle required". Lean angle aof WHAT do you think? Bike and rider of course.

By the principle of conservation of angular momentum (for a system in equilibrium): if one mass - the rider- moves clockwise about a point (the contact patch when looking from the front), the other mass - the bike- another must move counter clockwise to remain in equilibrium.

I don't agree. If the rider is in equilibrium with the bike, as you suggest, nothing needs to move. The rider should be moving with the bike - not against it. A rider may momentarily move them self out of equilibrium to affect a change (such as commencing cornering), but otherwise should be in equilibrium with the bike.

If you lean the bike to much into a corner you *could* lean your body out to offset the effect - but what it means is you setup for the corner wrong in the first place, have got the bike off balance, and are having to take corrective action (more throttle would also work), and it won't help you take a corner faster. You need to be leaning with the bike into the corner.

Traction is not about the size of the contact patch- its about the ability of the suspension to hold the contact patch against the ground. The back spring pushes directly downwardly in the line of the bike when viewed from the front. All the dampening and spring settings are designed around the bike operating in an upright position, and upright is the position in which the suspension is most effective. This is not really debatable- this really IS fact and straight applied mathematics. When the back spring and forks are at an angle, only a component of the rection force by the ground against the wheel is transmitted to the shock, and only that component is handled by it. the other component has effectively no suspension, and this will cause reduce the effectiveness of the suspension overall.

I still disagree. Traction is directly affected by how much tyre you get onto the ground. Its directly related to the size of the contact patch. I do agree its directly related to how well the suspension holds the tyre onto the ground.

Are you familiar with camber thrust?
http://en.wikipedia.org/wiki/Camber_thrust

[Katman]

Are you familiar with camber thrust?

I'm about to make myself familiar with a knife thrust.

[sinfull]

Do you two realise your arguing the same argument ? Rsole aint saying lean the opposite to the turn !
Tyres are shaped the way they are so you always have the same amount of contact patch on the tarmat upright or leaning (unless the tyre is flat) this is of course if you still have chicken strip left ! If your using up all the chicken strip, then your contact patch will be lessening !
That is the time you need to start moving body off the bike into the corner to change the COG so the bike is more upright and back to the usuall contact patch !

There now fuck off for a ride !

[R-Soul]

I don't agree. If the rider is in equilibrium with the bike, as you suggest, nothing needs to move. The rider should be moving with the bike - not against it. A rider may momentarily move them self out of equilibrium to affect a change (such as commencing cornering), but otherwise should be in equilibrium with the bike.

No, a component of the rider AND bike's weight (when leaning over) is in equilibrium with the centrifugal acceleration acting against their combined weight. For a given speed, the lean angle must be a certain angle (like your equation says) regardless of where the bker and bike are placed individually- as long as their combined COM is at a particular lean angle. If one is far in, the other can be far out, as long as their combined COM is at a particular lean.

If you lean the bike to much into a corner you *could* lean your body out to offset the effect - but what it means is you setup for the corner wrong in the first place, have got the bike off balance, and are having to take corrective action (more throttle would also work), and it won't help you take a corner faster. You need to be leaning with the bike into the corner.

I am saying MX riders do it on piurpose to reduce traction and help break the back wheel out to realign it and to make its rotational power create a force slightly inwardly that substitutes for the loss of sideways traction. Thats why they lean outwardly in a corner.
But you would not want t in a road type situation when you want to maximise traction.

I still disagree. Traction is directly affected by how much tyre you get onto the ground. Its directly related to the size of the contact patch. I do agree its directly related to how well the suspension holds the tyre onto the ground.

Strictly speaking when you have a perfect surface, friction = coefficient of friction X force on contact patch. It has nothing to do with the size of the patch. But in practice, small or nano sclae forces do affect friction according to teh size of the contact patch.

See http://en.wikipedia.org/wiki/Frictionscale

These small scale nano forces are, however, heavily outweighted by the main players in the equation above (i.e. weight or downward force on the contact patch).

The suspension system's main job is to keep the weight acting down on the ground.

Are you familiar with camber thrust?
http://en.wikipedia.org/wiki/Camber_thrust

No I was not, but this would only affect the bike when the road camber is at an angle (since it effectively pushes the bike downhill). I also understand that thsi force is relatively minor.

[R-Soul]

There, now I am going for a ride.

[bogan]

I am saying MX riders do it on piurpose to reduce traction and help break the back wheel out to realign it and to make its rotational power create a force slightly inwardly that substitutes for the loss of sideways traction. Thats why they lean outwardly in a corner.
But you would not want t in a road type situation when you want to maximise traction.

Wrong order, we lean outward cos that way its a lot easier to cope if/when the bike does break traction

edit, oh and if you or p.dath want to write a wiki on 'whips' I would be greatly interested! some crazy forces going on in those.

[crazyhorse]

ha ha ha ha yeah right