Question for the resident boffins.....

[MSTRS:Me]

When your bike is stationary, it's weight (plus rider) is all on the two tyres' contact patches. However, once you start moving, centrifugal force comes into play. This means that (theoretically), the force throwing outwards at the tyre tread is altering the contact patch. What I am interested in knowing is, does this make your total unit 'lighter' in relation to it's stationary weight?? I realise that the arc of tyre etc below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.

[scumdog]

When your bike is stationary, it's weight (plus rider) is all on the two tyres' contact patches. However, once you start moving, centrifugal force comes into play. This means that (theoretically), the force throwing outwards at the tyre tread is altering the contact patch. What I am interested in knowing is, does this make your total unit 'lighter' in relation to it's stationary weight?? I realise that the arc of tyre etc below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.

I doubt that centrifugal force re the tyre has much bearing, the forces cancel each other out.
Lateral forces are more what affects your tyres grip. (I think)

[MSTRS:Me]

Ahhh, but do they?? I am talking of when you are travelling in a straight line only.

[sAsLEX]

centrifugal force comes into play

Cant do it doesn't exist

[MSTRS:Me]

Try swinging a bucket full of water then

[sAsLEX]

Try swinging a bucket full of water then

and? I will provide a force associated with centripetal acceleration, but no cenrifugal force.

and

The fictitious centrifugal force appears when a rotating reference frame is used for analyzing the system. The centrifugal force is exerted on all objects, and directed away from the axis of rotation.

[scumdog]

Try swinging a bucket full of water then

As said before, they cancel out unless you have a really out-of-balance rim.

Try swinging 80 buckets of water in an arc - all equally spaced out.

After all, if you spin a push-bike wheel while holding it by its axle does it leap violently up and down??

[sAsLEX]

Consider a ball that swings around a stationary pivot to which it is tethered by a light, strong rope. There is tension in the rope, pulling inwards on the ball (the centripetal force) and simultaneously pulling outwards on the pivot (the reactive centrifugal force). The tension is real, so these two forces still exist if we move to a corotating frame. However, in the rotating frame there is also a fictitious centrifugal force that pulls outwards on the ball. It is distinct from the reactive centrifugal force that pulls outward on the pivot.

This is what I meant, no cetrifugal force on the tread.

[MSTRS:Me]

After all, if you spin a push-bike wheel while holding it by its axle does it leap violently up and down??

No, of course it doesn't. We are talking about a state of balance, but not sure that upper and lower quadrants cancel each other out

[sAsLEX]

below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.

Your argument falls apart here, as the "force" you mention is equal in all directions on the tyre it cancels out completely. It does introduce gyroscopic implications though, but you are talking about going in a straight line so they dont come in to it.

[Skyryder]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

[MSTRS:Me]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

Absolutely- that is the point I was trying to get clarified

[scumdog]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

The scales would require some sort of downwards motion to cause a reading, if there was insufficient time for that motion to occur fully then I suppose you would get a lighter reading.
If the scales had a weigh-bridge of say 50 metres long you would then get a true reading.

[scumdog]

No, of course it doesn't. We are talking about a state of balance, but not sure that upper and lower quadrants cancel each other out

Lets re-look at the push-bike wheel thing, if it was on it's side and spinning it would weigh X weight, stand it vertical and I can not see it suddenly being lighter, it stays the same weight, X, ergo there is no force caused by the spinning that will 'lift' it and make it lighter.

Quadrants means in four sections, I cannot see how anything would affect the wheel in any particlar quadrant at random other than gravity.

[sAsLEX]

The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

not too sure how you could get this. If an object is not accellerating *sp then all forces are balanced, wether at 0 or 900 kmhr. No matter how fast one goes the vertical forces are balanced or else at a certain speed you would fly without wings ( I am ignoring lift associated with the rider/bike here).

What force associated with an increasing speed would counter the downwards force of gravity and reduce the bikes weight!?