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What effect does weight have on grip?


Tpoppa
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Outliers such as van der walls force, and mechanical interlocking increase with contact patch, and are more pronounced with static friction.

I'm pretty sure your coefficient of friction will include van der waals forces as well as any generalized contact factor due to surface roughness.  So, sure, we agree that static friction coefficient is larger than the kinetic friction coefficient.  But, I still don't see how contact area matters... equations and articles would be very helpful here.  Everything that I have learned is that when you have smaller contact area, your pressure increases (force over area), and the change in pressure compensates for the change in friction almost one for one, which is why you generally don't need to take into account your surface area... again, with exception to the already discussed reasons.

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Wouldn't it be obvious that a larger contact patch has more grip potential than a smaller one?

Smaller area would lose traction since the force is applied over a smaller area.assuming enough to break traction.

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Wouldn't it be obvious that a larger contact patch has more grip potential than a smaller one?

Smaller area would lose traction since the force is applied over a smaller area.assuming enough to break traction.

Not in the case of translational motion. If your normal force remains constant, the coefficient of friction is independent of contact patch. Not intuitive, but accurate.

A smaller contact patch will have more pressure.

Edited by magley64
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Wouldn't it be obvious that a larger contact patch has more grip potential than a smaller one?

Smaller area would lose traction since the force is applied over a smaller area.assuming enough to break traction.

Yes.  And the technology behind the profile of a motorcycle tire bears that out.

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Not in the case of translational motion. If your normal force remains constant, the coefficient of friction is independent of contact patch. Not intuitive, but accurate.

A smaller contact patch will have more pressure.

Please explain why tires get wider as bikes get heavier. If it weren't true, I could run 125 slicks on my big bike and it would all work out. But that's not how it works in real life. Bigger, heavier and more powerful needs more tire width and thus larger contact area to deal with the extra forces.

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Yes.  And the technology behind the profile of a motorcycle tire bears that out.

 

 

Ok, so I think we have two separate ideas here: 1. The idea of frictional forces that hold you to the road (independent of area), 2. The idea of optimal tire design (not independent of area).

 

We've established several valid reasons to have higher contact area:

1. heat dissipation

2. material limitations with shear stresses

3. road surface imperfections such as bumps or slickness from oil etc.

4. introducing lower tire pressure to increase your coefficient of friction as a result of material property change

others?

I'm only reiterating what I think I remember learning in classes, but if you don't believe me, I encourage you to google laws of tribology, or amontons 2nd law.  Perhaps there are people who will do a better job of explaining F=uN is the governing equation regarding friction and slip.  I'm not saying there aren't other effects coming into play, but that I think this is the most important in determining if slip will occur.

 

Edited by OsuMj
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I am not going to dive too deeply into the physics...But in real world application the tire compound, tread, and temperature (both tire & suface) have a huge impact on grip.

 

I was just curious on how weight impacted grip in the OP.  It doesn't affect it very much as far as I can tell.

 

carry on....

Edited by Tpoppa
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As mentioned above, the best answer is that far more factors play into cornering speed/grip.

 

Regarding effects of mass, the mass of the bike plays competing roles in cornering. As mentioned above, the force of friction

Ff = uN, where N = mg . So we have an equation for the maximum amount of friction force that can be developed which depends on the coefficient of friction u, the mass of the bike m, and the acceleration due to gravity g. This suggests that for a stationery motorcycle (if you locked both wheels so they couldn't spin), increased mass would in fact provide a stronger force of friction.

 

However, lets consider the cases of straight line and circular acceleration.

 

For a straight line, the fastest you could accelerate would be if the force of acceleration were equal to the maximum force provided by the tire. F=ma for force of acceleration, Ff=umg from above, setting these forces to be equal

 

                                                                Ff=F

 

substituting for our other equations we get

 

                                                                ma=umg

                                                               

                                 

And we can immediately see that mass can be canceled out of both sides and we get

 

                                                                  a=ug.  The maximum acceleration we can get is only dependent on the effect of gravity and the coefficient of friction, which is a property of the rubber in the tire.

 

 

Similarly, the centripetal force required to drive in a circle is given by F=mv^2 / r       where m is mass, v is the velocity or speed of the vehicle and r is the radius of the corner. Maximum corner speed will be when this force is equal to the maximum amount of friction available, so once again

 

                                                              Ff=F

 

Substitute in our equations and we get

 

                                                              umg = mv^2 / r

 

Once again, mass cancels out on both sides 

 

                                                              ug = v^2 / r

 

rearranging for velocity we have

 

                                                              v = (ugr)^.5  , or the square root of the product of u, g, and r.

 

This suggests that maximum corner speed is a product of friction coefficient, gravity, and corner radius alone.

 

 

The big picture here is that while increased mass does increase available friction force, it also decreases the amount you can accelerate (either in a straight line or around a circle), and these effects are both of the the same magnitude (strength) but counter each other.

 

That being said, there are a great deal many more factors in the real world that come into play such as lean angle, corner banking, suspension dynamics, and several others that complicate the issue, as well as the physical constraints of the materials in the tire.

 

Not sure this was helpful, but that's what I've got. And as mentioned above, surface area of contact (contact patch) does not affect the force of friction.

 

 

I am not going to dive too deeply into the physics...But in real world application the tire compound, tread, and temperature (both tire & suface) have a huge impact on grip.

 

I was just curious on how weight impacted grip in the OP.  It doesn't affect it very much as far as I can tell.

 

carry on....

I think that's essentially your answer as far as lateral tire slip - cornering, I guess it depends mainly on your coefficient of friction and your turning radius.

I can't think of any canceling effects for getting mass out of the longitudinal tire slip direction, so if you were simply to increase the force (from accelerating) in a straight line, you could cause tire slip easier with a lighter bike I would think?

Edited by OsuMj
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Lighter bike accelerates the same rate as a heavier bike with less energy therefore less grip needed to accelerate the lighter bike. Lighter bikes are preferred in drag racing too I would imagine.

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We are definitely saying different things.  I'm saying that I don't think that contact area [really] matters.  A random non-rotating object will be subject to both a static and kinetic friction coefficient as well. 

You are saying that contact area does matter because wheels are more complex situations...

 

and I don't understand what 'complexities' are making it so that the contact area matters, with exception to the reasons already discussed so far (such as heat dissipation, imperfections in the road, change in friction coefficient with varying tire pressure).

 

I agree here, I don't understand what complexities are making contact surface area suddenly matter. Magley, could you please provide some sources to enlighten us? Not being confrontational, if this is true I would like to know the details. My degree is in physics, so I kindof like these sorts of discussions. And I don't see why if we look at a rolling tire at an instant in time (t -> 1/ infinity) the rules of static friction wouldn't apply. (though I admit there may be some non-trivial contributions due to the elastic properties of our real world materials not matching the ideal incompressible inelastic assumptions for the aforementioned equations. 

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If surface area can be completely ignored in static friction applications, why is the interlaced phone book trick so impressive?

This is apples and oranges with the issue of grip for a motorcycle tire on the road. 

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Ok, so I think we have two separate ideas here: 1. The idea of frictional forces that hold you to the road (independent of area), 2. The idea of optimal tire design (not independent of area).

 

We've established several valid reasons to have higher contact area:

1. heat dissipation

2. material limitations with shear stresses

3. road surface imperfections such as bumps or slickness from oil etc.

4. introducing lower tire pressure to increase your coefficient of friction as a result of material property change

others?

I'm only reiterating what I think I remember learning in classes, but if you don't believe me, I encourage you to google laws of tribology, or amontons 2nd law.  Perhaps there are people who will do a better job of explaining F=uN is the governing equation regarding friction and slip.  I'm not saying there aren't other effects coming into play, but that I think this is the most important in determining if slip will occur.

 

 

Bingo!  We have a winner.  When it comes to the issue of friction vs contact patch,  #2 is the biggest reason we have larger tires/larger contact area on heavier bikes and high-HP bikes, and the /\-shaped tires on race bikes.  The capabilities of modern bikes have advanced to the point that weight, acceleration and centripetal force would result in the rubber at the tire/road interface deforming from both heat degradation and shear stress such that the safety--and of course, tire life--would be unacceptable.

 

As was posted above, Ammonton's Law states that for a given pressure, the surface area is directly proportional to force, so tire size isn't what 'creates' the larger contact patch; rather, it's the ability to run lower pressure while still accommodating a larger force (weight/acceleration/etc) without encountering other serious issues such as sidewall deformation and heat build-up.  Kind'a gives you pause next time you throw your bike into a hard, downhill corner at triple digit speeds….

Edited by Bubba
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This is apples and oranges with the issue of grip for a motorcycle tire on the road.

Why? Surely if it applies in one case that surface area has an effect on static friction in real world scenarios, it would apply in all cases to varying degrees.

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wow, congrats to everyone for discovering it's complicated and working through it. Been there and done that for aircraft on down to solar cars with bicycle wheels.

 

D.O.T. (Department of Transportation) data keeps it all straight. Based on only static distribution of weight on each type of tire and the load rating of the tire. Without a set of D.O.T. rules we would have serious problems with tires in practice. Load rating is based on design of tire, basically diameter, width and profile. All other considerations of motion and variables are covered by a very generous safety factor that basically overkills the requirements with a large safety factor.

 

It does not overcome the fact that a cheap tire might not do the job all that well, nor does it assume things like jumping cars in a monster truck or playing Dukes of Hazard with your vehicle.

 

Simply put, an increase in weight or downward force from any source requires an increase in load rating of tire. Contact patch is designed to be reasonable for the application and tire compound. And then tested to see if it's true.

 

Additional weight or less weight to increase cornering ability of a tire could be quite variable in results. Only testing it would determine if it's useful. As a guess, I'd say small changes could bring good results, if lucky. Large changes would upset the previous conditions.

 

Example of real world: You could change a rear tire to one size wider, and one size lower profile, and have approximately the same load rating. You could gain a larger contact patch, and have more traction or cornering friction. Keep in mind that the stock front tire could be less capable, and might break loose sooner than the rear.

 

edit: doah! One more thing. The actual tire pressure is a requirement of the D.O.T. standards also. A specific cold pressure is to be used for a specific static weight on a specific size of tire. There is no upper limit on pressure, but no tire is designed to operate below 26psi for extended lengths of time.

Edited by ReconRat
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Lighter is almost always better.

But Downforce is also a big factor.

If it is too light and there isn't something increasing the down-force, then there becomes a time when too light de-creases friction.

But this will not be a factor on anything we can afford. :)

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Why? Surely if it applies in one case that surface area has an effect on static friction in real world scenarios, it would apply in all cases to varying degrees.

Draw a free body diagram of 1 page of the book, then you can tell me why so much force is required

I say it's apples to oranges bc a wheel is slipping on a surface, not being dragged between two surfaces in compression

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Why? Surely if it applies in one case that surface area has an effect on static friction in real world scenarios, it would apply in all cases to varying degrees.

 

Let's be clear, you haven't proven that it applies in even one case.

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Let's be clear, you haven't proven that it applies in even one case.

Right.

When I said to draw the FBD I intended for you (magz) to see that it's still independent of area. If you have the phone book interlaced with half width of the pages, or a quarter, it should still require the same force to separate.

*edit* In theory.  In practice, the weight distribution of the phone book will change the separation force, but it still has nothing to do with surface area.

Edited by OsuMj
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