Question for the resident boffins.....

[scumdog]

Sorry, you misunderstand. That was not the original question. The braking scenario is one - whereby impetus denied translates to a greatly increased downward pressure on the front contact patch, thereby making the attached bike 'weigh more'. Also when accelerating, the same force applies to the rear contact patch, for as long as acceleration continues.

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.

[Highlander]

there would be the same (more or less)

How does that work? if it is more or less how can it be the same?

[MSTRS:Me]

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

[sAsLEX]

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.

The coefficeint of friction would change once locked up, thereby reducing the friction force acting on the c.o.g therefor there would be less pressure.

[scumdog]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

It was more of a reference to the fact wherether the wheels were spinning or not the downwards force is (relatively) the same.

A wheelspin take-off should according to your theory provide more rearwards downpressure - yet in reality it would not. i.e. try it on a wet road with wheelspin vs a dry road with no wheelspin.

[sAsLEX]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

Its *baby* weight increased not in the vertical plane but in the horizontal actually. It didnt get heavier in that it would squish poor mummy holding it.

[MSTRS:Me]

Now you are just getting silly, I wasn't talking about traction. It is a good point however, since drag or friction has a bearing on how much pressure (weight) is exerted at the point of contact.

[MSTRS:Me]

Its *baby* weight increased not in the vertical plane but in the horizontal actually. It didnt get heavier in that it would squish poor mummy holding it.

yep, but thought it would help illustrate what I was getting at

[sAsLEX]

yep, but thought it would help illustrate what I was getting at

nope just illustrates you are getting confused really, completely different case that one. Can you provide any sort of theory or formula for this weight increase as something increases in speed?

[scumdog]

How does that work? if it is more or less how can it be the same?

O.K. pedant, I was allowing for 'minor variations'!

[Highlander]

O.K. pedant, I was allowing for 'minor variations'!

I knew what you meant, just had to do something to bring this back to Earth. Can anyone else hear that whooshing sound?

[MSTRS:Me]

nope just illustrates you are getting confused really, completely different case that one. Can you provide any sort of theory or formula for this weight increase as something increases in speed?

No I can't - and the theory was that a bike's 'weight' on the contact patches would decrease ever so slightly at speed (assuming acceleration has stopped) because of what I see as the lifting action of the tyre's weight being thrown outwards from the centre (axle).
Perhaps there is no lift because being circular the forces around the circumference keep everything equal as others have said.

[scumdog]

No I can't - and the theory was that a bike's 'weight' on the contact patches would decrease ever so slightly at speed (assuming acceleration has stopped) because of what I see as the lifting action of the tyre's weight being thrown outwards from the centre (axle).
Perhaps there is no lift because being circular the forces around the circumference keep everything equal as others have said.

That is what the rest of us are trying to tell you : the lifting as you call it is 360 degrees around the wheel, ergo it is pulling downwards, forwards and backwards just as much as it is upwards, therefore the forces cancel each other out.

As I said earlier, spin a bike wheel while hanging onto it by its axle - it certainly won't take off upwards into orbit.

[Virago]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

You're still confusing the downwards "weight" with the forwards momentum.

The "baby in the arms" scenario explains that beautifully. If the baby weighs 5Kg, it will still have a downwards force of 5Kg after impact. But attempts by the mother to stop the forward momentum make the baby seem many times that weight, but in a horizontal direction.

A bike accelerating or deceleratiing is exactly the same. A bike weighting 200Kg with a 50/50 balance between the wheels, will exert 100Kg on each point of contact. Under braking, the centre of balance shifts forward, putting a weight of say 150Kg downwards force on the front wheel. But there is now only 50Kg downwards force on the back wheel.

The additonal force on the front wheel comes from absorbing the forward momentum (deceleration), but this force is horizontal, not vertical.

An object simply cannot become "lighter" by travelling at speed. It's mass is fixed and unalterable. It's weight (mass x gravity) therefore cannot change either, unless you can magically change gravity. Only other external forces acting on the object can change the perceived weight.

[MSTRS:Me]

That is what the rest of us are trying to tell you : the lifting as you call it is 360 degrees around the wheel, ergo it is pulling downwards, forwards and backwards just as much as it is upwards, therefore the forces cancel each other out.

As I said earlier, spin a bike wheel while hanging onto it by its axle - it certainly won't take off upwards into orbit.

I know that now. As said was just a theory I was asking about. But as usual *some* members went so far off on a tangent (think about that one) that I couldn't resist seeing if digging in my heels would increase the pressure exerted on them