tyre->road coefficient of friction

The message from Roger contains these words:

re "Mike"

Except that my killfile refused to work on such a bland address.

Oh well it did at least give me the chance to see that Mike had started a new thread to grovel to Dave, seemingly oblivious to the fact that it was Dave who originally called his random noise gibberish.

Reply to
Roger
Loading thread data ...

That's the average IQ of those reading this lowered back down again then.

If you won't accept references to Dan Gurney or Colin Chapman on what downforce is then there is little point.

Wonder if my killfile will accept you ?

Hardly. I checked his facts, compared them with mine and accepted that modern road tyres do have a far sharp fall-off in performance on lockup than I expected. If I could find some modern data on racing tyres I expect the data would be different but that is rather beside the point.

Reply to
Mike

I'm trying a free newsserver now and this is my first post so hopefully it will work.

The position of the CoG relative to the wheel centre has absolutely nothing to do with anything so far mentioned. The pivot point for any vehicle that overturns is the contact patch between the tyre and ground. The height of the CoG above this point and the wheebase length determine the lateral force required to overturn the vehicle. Think about it, draw out the force diagram and then maybe we can drop this one of many red herrings in this thread.

Reply to
Dave Baker

The message from "Mike" contains these words:

Arrogant little arse aren't you.

You haven't made any as yet but if you post an easily accessed reliable source I will look at it. But accept hearsay from you of all people, no chance.

Why not try it. On the other hand we could call it a draw at this stage and give both of us the chance to get on with what is left of our lives.

Seems like a grovel to me. Particularly that bit - "What I tried to say, obviously not at all well..." but elementary tactics I suppose not wanting to fight on 2 fronts but letting Dave get away with calling your bit of nonsense gibberish smacks of lack of gumption.

Reply to
Roger

Well you obviously didn't pay as much attention to your brake we did :-)

The vector sum of the deceleration and the bike weight exceeds the bike weight. At some point that can be expressed as the >weight all on the front wheel and a bit of upthrust on the back one. Mutatis mutandis. Its up to you whether you care to express that as > 100% weight transfer or a rotating couple round the front wheel.

Reply to
The Natural Philosopher

Think it through. A vehicle with CG at ground level auffers NO weight trasfer under braking, and ideal situation that every racing car designer tries to emulate.

Th higher the CG the greater the notinal weight transfer under a given decelarative force.

The argumanet sems to be anbout terminology only. What 'weight transfer' means, and whether upthrust or down thrust or > 100% transfer is meaningful.

In a sense, all are.

Reply to
The Natural Philosopher

HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.