While Dave’s “Optimum Handling” line may or may not be helpful, his explanation of forces in a turn could be down right dangerous, if someone where to depend on it:
Because the bike and rider are leaning, the rider’s weight is being pushed outwards and downwards by centrifugal force, thereby actually pushing the tires down onto the road, which increases traction.
Yikes! If only that were true.
The sad fact, however, is that the centrifugal force is perfectly parallel to the pavement, and so can contribute nothing to the force of the bike onto the road, and the lean angle of the bike and rider has nothing to do with it.
Instead, the centrifugal force must be resisted by the available friction between the tires and the pavement, which remains the same whether the bike is leaning or not.
I’ve posted a comment for Dave on his blog, and perhaps he can correct this mistake.
Bad Bicycle Science 
Dave is correct.
Both the Vertical & Centrifugal Force combine to give a new Resultant Force that’s greater than either one. For 100Kg rider & 50 Kg C force, R= 112Kg. I.e.the rider maintains his equilibrium by leaning into the curve until the Resultant passes through the tire contact patch again.
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Manfred,
Thank you for your comment.
Your two assertions, however, do not support your initial claim.
Yes, two forces can be combined into a resultant, and yes, one way to think of the equilibrium of a bike in a turn is that the resultant force at the contact patch is colinear with the plane of symmetry.
Neither of these points support Moulton’s claim that traction is somehow increased in a turn. By the Coulomb friction model, the friction force depends only on the vertical component of the ground reaction force, and the vertical component is not changed by being in a turn. The larger resultant force is irrelevant.
Cheers,
Andrew
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Your statement âfriction force depends only on the vertical component of the ground reaction forceâ should be restated more generally as âfriction always acts at right angle to any applied force. Ergo âRotorâ Amusement rides.
Rubber tires not only experience friction with the road surface, but more importantly also experience high resistance to slippage through the mechanical interlocking engagement with the road surface, analogous to a toothed rubber belt engaging a toothed pulley. The level of grip such mechanical engagement provides being entirely dependent on the force which acts upon it. I.e. Not restricted to just the vertical component of the force.
Why else would they mould âKnobsâ and âGroovesâ all around the running surface of a tire?
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Manfred,
Sorry for the slow replies, I’m travelling and have limited connectivity.
A more-general way to phrase my description of the friction between a tire and the ground would be “friction force depends only on the normal component of the ground reaction force.” To say that “friction always acts at right angles to any applied force” could not possibly be correct. If the applied force where parallel to the plane between the two surfaces, we certainly would not expect a friction force to be at right angles to either of them, right? That certainly would make a fascinating amusement park attraction.
I have mentioned already that I am using the Coulomb model of friction. Perhaps you could identify the friction model you are using, or provide a link to a paper or article describing it. I would be interested to read about a model of friction in which the applied force that friction resists is also responsible for the friction force that resists it. R. van der Steen has a nice survey of the tire literature as of 2007, but I can find no mention of the friction model to which you allude.
As for the knobs you mention, I believe it is well understood that they are designed to penetrate soft surfaces, such as sand, mud, or snow, and recruit the shear resistance of the penetrated material to help prevent slipping. On dry pavement, however, they provide poor handling characteristics, and that is why road racing tires usually have no tread pattern at all. The grooves you mention are either to allow water to escape and prevent hydroplaning or merely for marketing purposes. As Sheldon Brown explains “bicycle tires for on-road use have no need of any sort of tread features; in fact, the best road tires are perfectly smooth, with no tread at all!”
Cheers,
Andrew
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