What effect does weight have on grip?

[Tpoppa]

Does a heavier motorcycle have any less/more grip than a lighter one...assuming same tires, similar suspension setup. etc?

[ReconRat]

yes more, dry force of friction (parallel) is based on the qualities of the material contact (size and type) and the downward force (perpendicular).

 

and weight shift plays an important part in the downward force.

 

edit: useful friction is also a balance between weight vs material plus downward force when trying to stop.

total weight might be more than the effective material contact patch can handle.

Edited August 6, 2014 by ReconRat

[Bubba]

Does a heavier motorcycle have any less/more grip than a lighter one...assuming same tires, similar suspension setup. etc?

Yes.  This is why you see most of the top-level Moto-GP racers weighing >250 lbs--to increase traction.  OK….JK.

 

Pretty sure traction is a function of the dynamic forces between the contact patch and the road.  The contact patch is function of force vectors pushing the rubber to the pavement and tire pressure.  So in the simplest terms, the greater the force, the larger the contact patch and the lower the pressure, the larger the contact patch and thus, the greater the traction.  Problem with your question is that the actual coefficient of friction at the tire/pavement interface isn't affected that much by the weight of the bike/rider.  In a turn, the larger forces are created by centripedal force, which are greater for a large mass (heavier bike/rider) than a lighter one.  Just as a lighter car can out corner a heavier one--given that they have similar tires--so a lighter bike can out corner a heavier one.  You have to think of the force vs CoF vectors involved….and I'm far too lazy an SOB to try to draw something and post it here.  You'll have to go find it on the interwebz for yourself.  The same is true for straight-line acceleration, i.e., drag races--you get better traction with greater weight transfer to the rear wheels, but the extra mass you must accelerate by adding weight over the drive wheels more than diminishes the benefit you gain from the added traction.

[Helmutt]

Pretty sure traction is a function of the dynamic forces between the contact patch and the road.  The contact patch is function of force vectors pushing the rubber to the pavement and tire pressure.  So in the simplest terms, the greater the force, the larger the contact patch and the lower the pressure, the larger the contact patch and thus, the greater the traction.  Problem with your question is that the actual coefficient of friction at the tire/pavement interface isn't affected that much by the weight of the bike/rider.  In a turn, the larger forces are created by centripedal force, which are greater for a large mass (heavier bike/rider) than a lighter one.  Just as a lighter car can out corner a heavier one--given that they have similar tires--so a lighter bike can out corner a heavier one.  You have to think of the force vs CoF vectors involved….and I'm far too lazy an SOB to try to draw something and post it here.  You'll have to go find it on the interwebz for yourself.  The same is true for straight-line acceleration, i.e., drag races--you get better traction with greater weight transfer to the rear wheels, but the extra mass you must accelerate by adding weight over the drive wheels more than diminishes the benefit you gain from the added traction.

Well put. Although I do understand the physics at work here, my vocabulary to explain it was found wanting

[turnone]

I think the American car companies called it road hugging weight in the 70's. Lighter is better.

[magley64]

Short answer, yes, but mass has a detrimental effect because of inertia. You have more grip, but you need more grip because you have more inertia.

[Tpoppa]

Here is why I am asking:

 

More weight does give you more grip and traction when launching off the line (ask a drag racer).  More down force gives more grip in corners (ask an Indy car driver why they have wings front and rear). I'm not sure if the same principles apply to 2 wheeled vehicle.

 

CBR600RR 420ish lbs with Angel GTs

VFR1200 590ish lbs with Angel GTs.

 

On the CBR, I can get both tires to break loose and go into a small controlled slide.  I haven't reached that point yet on the VFR.  I get that 590 lbs is going to require more lateral grip, but the weight is also creating a larger contact patch and more downward friction due to gravity.  There are a few factors at play.  

 

I know my limits pretty well on the CBR, and still trying to find them on the Viffer.   

Edited August 7, 2014 by Tpoppa

[Tpoppa]

 The same is true for straight-line acceleration, i.e., drag races--you get better traction with greater weight transfer to the rear wheels, but the extra mass you must accelerate by adding weight over the drive wheels more than diminishes the benefit you gain from the added traction.

This.  I am asking specifically about grip.  Not the effect on acceleration, braking, or even suspension performance.   

[jschaf]

I think you're going to have to do more testing to determine the answer to your question. Please report back with results.

[Josh1234]

Here is why I am asking:

 

More weight does give you more grip and traction when launching off the line (ask a drag racer).  More down force gives more grip in corners (ask an Indy car driver why they have wings front and rear). I'm not sure if the same principles apply to 2 wheeled vehicle.

 

 

Whoah, stop right there.  Indy car drivers have a "wing," sure, but for a very different reason.  The "Wing" actually provides DOWNforce to press the car INTO the pavement.  This is so that air doesn't get under the car and cause a loss of control.  At high speeds, air underneath your car, lifting it off the ground, is a bad thing

 

Traction is based on friction, and friction is increased with weight.  As to the DEGREE of increase, I'll leave that to smarter minds.

[Tpoppa]

Whoah, stop right there.  Indy car drivers have a "wing," sure, but for a very different reason.  The "Wing" actually provides DOWNforce to press the car INTO the pavement.  This is so that air doesn't get under the car and cause a loss of control.  At high speeds, air underneath your car, lifting it off the ground, is a bad thing

 

Traction is based on friction, and friction is increased with weight.  As to the DEGREE of increase, I'll leave that to smarter minds.

http://en.wikipedia.org/wiki/Formula_One_car

 

 

"The aerodynamic designer has two primary concerns: the creation of downforce, to help push the car's tyres onto the track and improve cornering forces; and minimising the drag that gets caused by turbulence and acts to slow the car down."

[OsuMj]

Here is why I am asking:

 

More weight does give you more grip and traction when launching off the line (ask a drag racer).  More down force gives more grip in corners (ask an Indy car driver why they have wings front and rear). I'm not sure if the same principles apply to 2 wheeled vehicle.

 

CBR600RR 420ish lbs with Angel GTs

VFR1200 590ish lbs with Angel GTs.

 

On the CBR and get get both tires to break loose and go into a small controlled slide.  I haven't reached that point yet on the VFR.  I get that 590 lbs is going to require more lateral grip, but the weight is also creating a larger contact patch and more downward friction due to gravity.  There are a few factors at play.  

 

I know my limits pretty well on the CBR, and still trying to find them on the Viffer.   

I'm digging memories up from a class I took 8 years ago... someone chime in if I'm remembering incorrectly.

 

The frictional forces are dependent on two factors: mass and coefficient of friction.  The coefficient of friction is determined based on the interaction between your surface and tire - but is NOT dependent on the amount of surface area. The equations are Ff=uN where Ff is force of friction, u is coefficient of friction, and N is mass* gravity.  When the applied force exceeds Ff, that is when you will break loose.

So, there are two reasons that your heavier bike might not be breaking loose for you yet, 1. It weighs more, so the static frictional forces are higher, 2. Your coefficient of friction may be different due to different tires or surfaces that you're riding on.

[Tpoppa]

but is NOT dependent on the amount of surface area. 

Good information.  

 

Question on this line specifically...Are you saying the size of the contact patch doesn't effect the amount of grip?  Conventional wisdom says that dropping pressure a few PSI will increase the size of the contact patch and result in more grip.

 

[OsuMj]

The idea is that if you increase the surface area, you are decreasing the amount of pressure (Force divided by Area).

This is all by equations^

I'm sure there are some "real world" reasons for having larger surface area... heat dissipation? Spreading out the amount force to reduce damage in one very small location?  not sure.

[Josh1234]

The idea is that if you increase the surface area, you are decreasing the amount of pressure (Force divided by Area).

This is all by equations^

I'm sure there are some "real world" reasons for having larger surface area... heat dissipation? Spreading out the amount force to reduce damage in one very small location?  not sure.

 

The larger the contact patch, the easier it is for your tire to track the surface and not lose traction.  A very small contact patch only takes a very small imperfection to break that contact.  A larger contact patch can read the road and tolerate more interruptions without a total loss of contact/traction/grip.

[OsuMj]

The larger the contact patch, the easier it is for your tire to track the surface and not lose traction.  A very small contact patch only takes a very small imperfection to break that contact.  A larger contact patch can read the road and tolerate more interruptions without a total loss of contact/traction/grip.

Makes sense, things like a little oil slick or a hole in the road.

[fizzer]

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.

[fizzer]

The idea is that if you increase the surface area, you are decreasing the amount of pressure (Force divided by Area).

This is all by equations^

I'm sure there are some "real world" reasons for having larger surface area... heat dissipation? Spreading out the amount force to reduce damage in one very small location?  not sure.

 

 

My understanding is that the effects of tire surface area have more to do with the physical limitations of the materials and ways we build our wheels/tires such as heat dissipation, shear stresses, etc. These theoretical equations assume a perfect friction producing material which does not warp, break down, stress etc. under the effects of acceleration. Also likely to include effects mentioned above such as ability to handle imperfections. Both the road and the wheel/tire would have to be flawless, indestructible, incompressible materials to make the above equations apply.

[Tpoppa]

So...the VFR is the heaviest motorcycle I have ridden with sporting intentions.  It felt a little bit out of my comfort zone at first.  Side to side transitions take a little more effort.  Corner stability and grip have a lot better than I expected.  I won't be trying to ride it at the same pace as the CBR, but I always want to know what I'm dealing with

 

Lot's of good information.

[magley64]

I'm digging memories up from a class I took 8 years ago... someone chime in if I'm remembering incorrectly.

The frictional forces are dependent on two factors: mass and coefficient of friction. The coefficient of friction is determined based on the interaction between your surface and tire - but is NOT dependent on the amount of surface area. The equations are Ff=uN where Ff is force of friction, u is coefficient of friction, and N is mass* gravity. When the applied force exceeds Ff, that is when you will break loose.

So, there are two reasons that your heavier bike might not be breaking loose for you yet, 1. It weighs more, so the static frictional forces are higher, 2. Your coefficient of friction may be different due to different tires or surfaces that you're riding on.

This only applies to translational motion. Because a rolling mass has momentary static friction, the contact area does have an effect. If you're dragging an object, all of your equations are correct, rolling wheels are slightly more complex. (Also why abs stops you faster than locking up) Edited August 8, 2014 by magley64

[magley64]

And as mentioned above, surface area of contact (contact patch) does not affect the force of friction.

Not true in the case of rolling wheels, momentary static friction does take contact patch into consideration. Contact patch makes no difference when sliding, but does matter when rolling.

[Tpoppa]

Further research will be conducted this weekend

[OsuMj]

This only applies to translational motion. Because a rolling mass has momentary static friction, the contact area does have an effect. If you're dragging an object, all of your equations are correct, rolling wheels are slightly more complex. (Also why abs stops you faster than locking up)

I was pretty certain that abs lets you stop faster because the coefficient of static friction (wheels in direct controlled contact) is larger than the coefficient of kinetic/dynamic friction (wheels sliding), so when your tires can actually make controlled contact with the surface, they are able to create a larger "stopping" force....  Again, I'm not expert, but this was my understanding. 

I could see changing the tire pressure so that you can change the coefficient of friction, since rubber is such a complex material, I'm sure that adding air and making it stiffer or taking air out and making it less stiff will change its properties.

Just so I can understand what you're saying about the complexities of the rolling wheel, maybe you could direct me to some equations or an article? 

[magley64]

I was pretty certain that abs lets you stop faster because the coefficient of static friction (wheels in direct controlled contact) is larger than the coefficient of kinetic/dynamic friction (wheels sliding), so when your tires can actually make controlled contact with the surface, they are able to create a larger "stopping" force.... Again, I'm not expert, but this was my understanding.

I could see changing the tire pressure so that you can change the coefficient of friction, since rubber is such a complex material, I'm sure that adding air and making it stiffer or taking air out and making it less stiff will change its properties.

Just so I can understand what you're saying about the complexities of the rolling wheel, maybe you could direct me to some equations or an article?

exactly, momentary static friction. I was saying exactly what you were saying, maybe I wasn't as clear as you are on it.

[OsuMj]

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).