Need Motivation To Lean More?

[Jiggles]

Try this



Next time someone down talks the 250 I'm going to go tag up their tires







THIS IS A DISCUSSION THREAD ON CONTACT PATCHES, EVERYONE ELSE GTFO

[choneofakind]

Bahaha that's funny. Maybe I'll tag my tires before my next track day and see if it helps.

[Jono]

Or get off the bike and try to keep it more upright so you can lean it over more in a turn if need be. I am still learning the basics but trying to lean the bike over as far as possible just to wear down the side of the tire seems stupid. The farther on the edge of the tire the smaller the contact patch. But I like I said, I know nothing. Just another noob.

[Jiggles]

Your strips look good to me as far as I can tell

[choneofakind]

Jono, that's what I do. but it's kinda a mark of pride if you can be going fast enough at the track to still get to the edge while hanging off.

After checking out my BP, I can do a lot more hanging. But I also still have tire left to go, which means, I can go even faster

[csmith12]

Tie a short piece of dental floss to your triple and then your neck. Break the floss you fail at bp. haahhahahah

[choneofakind]

bahaa I should so do that too. Straight from the triple to my chin bar. haha

[dirty nasty]

This is all well and fun, but for a street bike I'd rather see chicken strips. There really isn't many times that you'll have the bike leaned over that far when riding under normal conditions on the street. It's a nice feeling knowing there isn't complete psychos out there using the road as a track.

Unfortunately I find myself to be one of those psychos sometimes.

On the track I don't see how you couldn't get rid of those strips.

[Motofool]

Originally Posted by dirty nasty

This is all well and fun, but for a street bike I'd rather see chicken strips. There really isn't many times that you'll have the bike leaned over that far when riding under normal conditions on the street.......

In motorcycling riding, "what is essential is invisible to the eye." - Antoine de Saint-Exupery

[choneofakind]

Yeah, I haven't touched the edges of my tire since my track day. I don't go crazy on the street because it makes me nervous. So I have nice fatty fat fat chicken strips.

I have about 1/4" chicken strips from my track day. My knee never touched, but I wasn't worried about that either. I think with better BP and a little less nervousness, I can a) go faster and run better lines through turns, and b) touch the knee (just a perk, not the goal) without running out of tire (the goal). The ninja can still deliver more than I can tame through turns

[Jono]

Originally Posted by choneofakind

The ninja can still deliver more than I can tame through turns

For many many more years! That's why I don't get why people "upgrade" so fast. I guess they are just superhuman and can handle it.

[dfox]

Originally Posted by Jono

Or get off the bike and try to keep it more upright so you can lean it over more in a turn if need be. I am still learning the basics but trying to lean the bike over as far as possible just to wear down the side of the tire seems stupid. The farther on the edge of the tire the smaller the contact patch. But I like I said, I know nothing. Just another noob.

Everyone talks about contact patch like its the end all.

Simple physics shows that until you lose traction, or are talking about melting rubber (think drag racing), contact patch means nothing.

[Motofool]

Would you may explaining that a little more?

[Jono]

Yeah seriously. What you said makes no sense to me.

[choneofakind]

Originally Posted by dfox

contact patch means nothing.

please explain.

[dfox]

Force Of friction equals the coefficient of friction times mass times the normal force. Nowhere in that equation is a contact patch.

[dfox]

A very quick google search yielded this question.

http://www.physlink.com/education/askexperts/ae200.cfm

[Alex]

But the coefficient of friction between the rubber and the road varies as its size varies. A tire run at the correct pressures will apply the correct size/shape contact patch, which will help ensure the correct running temperatures, which directly affect that coefficient.

[choneofakind]

dfox, no.

Force of friction is the coefficient of static friction times the downward force (which is equal in magnitude to the normal force).

The downward force is the mass times the acceleration due to gravity times the sine of the lean angle in reference to the ground.

However, the size of the contact patch determines the area that is in contact with the ground. Keep in mind that you want that normal force to be exerted over a larger area, reducing the pressure between the tire and the ground, because if you have all that normal force (and centripetal force) on a small area of tire, the tire will wash out more easily because you will need more grip than can be supplied based on the coefficient of static friction of that particular tire. Then you start sliding and have to worry about the coefficient of kinetic friction, which is always less than the coefficient of static friction. Aka, once you start sliding, you have less grip than when you're not sliding.

Your ways of generating a larger contact patch are by lowering tire pressure, (which means your tires will get hot, slick, and wear faster) standing the bike up more, or both (which is what riders on the track do.

[choneofakind]

Originally Posted by dfox

Force Of friction equals the coefficient of friction times mass times the normal force.

This is blatantly wrong.

You're multiplying a coefficient times kg times kg.m/s^2 and expecting to get kg.m/s^2

F is (kg m)/s/s

coefficient of friction mass times normal force gives you (kg kg m)/s/s

(kg m)/s/s =/= (kg kg m)/s/s


Just sayin.

[alex.s]

Originally Posted by dfox

Force Of friction equals the coefficient of friction times mass times the normal force. Nowhere in that equation is a contact patch.

here is what you linked:

Friction is proportional to the normal force of the asphalt acting upon the car tires. This force is simply equal to the weight which is distributed to each tire when the car is on level ground. Force can be stated as Pressure X Area. For a wide tire, the area is large but the force per unit area is small and vice versa. The force of friction is therefore the same whether the tire is wide or not. However, asphalt is not a uniform surface. Even with steamrollers to flatten the asphalt, the surface is still somewhat irregular, especially over the with of a tire. Drag racers can therefore increase the probability or likelihood of making contact with the road by using a wider tire. In addition a secondary benefit is that the wider tire increased the support base and makes it hard to turn the car over in a turn or in a mishap.

what you dont understand is that when you are leaned over on a bike, greater pressures are applied vertically on the tire increasing the PRESSURE applied to the tire against the asphalt. this increased pressure also increases the contact patch. unlike a car which does not have the same kind of vertical compression on the tires except on the outside tires in a turn

when people talk about contact patch, they are talking about the amount of pressure on the tire. how much weight is on it. if you have weight on it, the patch will be larger. less weight, smaller.

[menikmati]

I have faith in chone in these physics warz.

[ShadowStalk3r]

Relax everyone, it's F=ma. Trust me I'm an engineer.

[dfox]

in saying mass times "normal force", i will admit i was incorrect, but only in semantics, it is the normal acceleration. f is always equal to ma. i initially wrote f=coefficient time mass time gravity, however that's not always true, especially if you're on a banked turn, the acceleration actually has two parts, and will be higher then the acceleration due to gravity.

in nothing else stated do I agree with, and chances are, I'll never change your mind.

F(friction)=Mu(coefficient of friction) x m (mass of object) a (acceleration of object normal (perpendicular) to the force of friction). You show me in that formula, where area is. The mass of the object cannot change. The acceleration of the object is typically the force of gravity, however as I mentioned above it can be slightly larger if you're on a banked road. I see no area in that calculation.

As was pointed out, it is very true that F(force) = p (pressure) times a (area). The problem comes when you assume that the force changes due to area. When the area goes down, the pressure goes up. What are the units of pressure? pounds per square inch is one unit. If you have a given force, a reduction in contact area proportionally increases the pressure. If you as a person, lay on a scale, is your weight any different than if you're standing? no. does the pressure you exert on each square inch of the scale change? yes.



does contact patch help deal with inconsistencies in the road better? yes. does it increase the frictional force and allow you to turn harder? NO!

my credentials? a professional engineer who has taken a significant amount of static and dynamic physics classes at the undergrad and graduate level. now, you want to start talking about the coefficient of kinetic friction, THAT gets complicated, because now you're starting to talk about contact area making a difference, but at the microscopic level.

[csmith12]

Lol, I like this thread but the current direction sucks.

Here is what I sometimes do. I pick of a goal for the ride, then tape it over my speedo or tach. For example my last track day, I taped over my speedo and used a sharpie and wrote "SISSY" in it's place. While I know I shouldn't be looking at my speedo on track, every time I did.... well.... you know.

Jiggles wrote on his tire, what have you done?

[Motofool]

Originally Posted by csmith12

.......Jiggles wrote on his tire, what have you done?

To turn faster, to complete a circuit in less time or just to lean the bike more?

[alex.s]

Originally Posted by dfox

in saying mass times "normal force", i will admit i was incorrect, but only in semantics, it is the normal acceleration. f is always equal to ma. i initially wrote f=coefficient time mass time gravity, however that's not always true, especially if you're on a banked turn, the acceleration actually has two parts, and will be higher then the acceleration due to gravity.

in nothing else stated do I agree with, and chances are, I'll never change your mind.

F(friction)=Mu(coefficient of friction) x m (mass of object) a (acceleration of object normal (perpendicular) to the force of friction). You show me in that formula, where area is. The mass of the object cannot change. The acceleration of the object is typically the force of gravity, however as I mentioned above it can be slightly larger if you're on a banked road. I see no area in that calculation.

As was pointed out, it is very true that F(force) = p (pressure) times a (area). The problem comes when you assume that the force changes due to area. When the area goes down, the pressure goes up. What are the units of pressure? pounds per square inch is one unit. If you have a given force, a reduction in contact area proportionally increases the pressure. If you as a person, lay on a scale, is your weight any different than if you're standing? no. does the pressure you exert on each square inch of the scale change? yes.



does contact patch help deal with inconsistencies in the road better? yes. does it increase the frictional force and allow you to turn harder? NO!

my credentials? a professional engineer who has taken a significant amount of static and dynamic physics classes at the undergrad and graduate level. now, you want to start talking about the coefficient of kinetic friction, THAT gets complicated, because now you're starting to talk about contact area making a difference, but at the microscopic level.

you are absolutely correct. but your application of that thought i believe is misguided. lets say you increase the air pressure of the tire. take a turn exactly the same as you have before. the amount of force applied to the tire from the bike is the same, the weight of the bike and rider- everything is the same. but because of the increased air pressure the contact patch is smaller. the amount of pressure applied to the contact patch is the same, and the coefficients of the rubber are still the same. the amount of grip -- as in the maximum pressure applied to the tire before it breaks loose (pretending that the road is an even consistency as far as grip goes, and its not raining) is going to be very similar. nobody's arguing that.

when people talk about contact patch, they are talking about the dynamics of the bike. theres 500+ pounds of metal and meat moving around. lets take a basic concept for example, braking; the suspension isn't instantaneous. if you slam on the brakes, the front wheel is carrying roughly 45-50% of the weight at the time when your tire needs as much pressure down on the surface as it can to achieve that high grip level. you said it yourself its pressure, not contact patch size. contact patch size is a side effect of pressure. so when you brake smoothely instead of slamming on the brakes, you bring that pressure up slowly... you dont overload the tires traction limits as you increase the pressure. as you increase the pressure, the traction limits go up. what "was" the traction limit in the tires static state is completely pointless as the tires are never in their static state. its being forced into the ground by braking, accelerating, turning.

[csmith12]

Originally Posted by Motofool

To turn faster, to complete a circuit in less time or just to lean the bike more?

My best guess is to get rid of his chicken strips on a brand new tire. Only Jiggles knows for sure why he wrote it. hahahahahah

[dfox]

Originally Posted by alex.s

you are absolutely correct. but your application of that thought i believe is misguided. lets say you increase the air pressure of the tire. take a turn exactly the same as you have before. the amount of force applied to the tire from the bike is the same, the weight of the bike and rider- everything is the same. but because of the increased air pressure the contact patch is smaller. the amount of pressure applied to the contact patch is the same, and the coefficients of the rubber are still the same. the amount of grip -- as in the maximum pressure applied to the tire before it breaks loose (pretending that the road is an even consistency as far as grip goes, and its not raining) is going to be very similar. nobody's arguing that.

when people talk about contact patch, they are talking about the dynamics of the bike. theres 500+ pounds of metal and meat moving around. lets take a basic concept for example, braking; the suspension isn't instantaneous. if you slam on the brakes, the front wheel is carrying roughly 45-50% of the weight at the time when your tire needs as much pressure down on the surface as it can to achieve that high grip level. you said it yourself its pressure, not contact patch size. contact patch size is a side effect of pressure. so when you brake smoothely instead of slamming on the brakes, you bring that pressure up slowly... you dont overload the tires traction limits as you increase the pressure. as you increase the pressure, the traction limits go up. what "was" the traction limit in the tires static state is completely pointless as the tires are never in their static state. its being forced into the ground by braking, accelerating, turning.

You're saying the same thing as I am, except you're putting the chicken before the egg, per se.

The braking force, as you seem to be describing as increased friction force in the front tire, is not due to increased surface area, it's due to increased mass. What gives the front tire greater grip in a braking situation, is the fact that there is a change in momentum, and the dynamic affect of the suspension moves the center of gravity towards the front wheel. The m in f=mu*m*a goes up. That's why the front tire has more traction, not because the contact patch is larger. The braking force would be larger even if you had a completely soid wheel that didn't deform under higher loading because of the increased mass. Again, contact patch doesnt matter.

And what I am referring to as static, is the tire and the ground. You do not enter kinetic friction until you lose traction. Tires are always considered being in a static state, and the motorcycle is seen as being an indpendent object acting in a dynamic matter on the tires. What changes in a braking or turning situation is the mass or acceleration in the static friction force calculation. It doesn't all of the sudden revert you into a kinetic friction situation, unless of course you would prefer to low side.

[choneofakind]

I interpreted your posts the same way alex did.

Why do you propose that bikes with more power have wider rear tires then? If contact patch didn't matter at all for grip when leaned over, all bikes would be just fine running on a 130, like a 250 does. Last time I checked, liter bikes were running 190 or 200 rears, and 600's were running 180 rears.

[Confuzshuz]

Damn, I knew I should have taken physics back in high school.....

[dfox]

Originally Posted by choneofakind

I interpreted your posts the same way alex did.

Why do you propose that bikes with more power have wider rear tires then? If contact patch didn't matter at all for grip when leaned over, all bikes would be just fine running on a 130, like a 250 does. Last time I checked, liter bikes were running 190 or 200 rears, and 600's were running 180 rears.

The same argument could be had for cars and wider tires.

Wider tires equal less radius equals greater contact patch, equals less care about inconsistencies in the roadway. It may also allow for softer compounds to be used.

[choneofakind]

Originally Posted by dfox

Wider tires equal less radius equals greater contact patch, equals less care about inconsistencies in the roadway.

Again, care to explain that a little further? You may be an engineer, but you suck at communicating what you're thinking and I'm having a hard time following you here.

[Motofool]

Originally Posted by csmith12

My best guess is to get rid of his chicken strips on a brand new tire. Only Jiggles knows for sure why he wrote it. hahahahahah

Well, Chris.........we couldn't bring the thread back to the original direction.

Now I feel really confused about pavement traction, contact strips and chicken patches.

[Jiggles]

[choneofakind]

Dude, your chicken patches are huge!! You need more normal acceleration lean.

[Jiggles]

The problem is your clockwise steering, try with reversing time

[csmith12]

Originally Posted by Motofool

Well, Chris.........we couldn't bring the thread back to the original direction.

Now I feel really confused about pavement traction, contact strips and chicken patches.

lol yea, all this physics stuff is way above my head. I don't know how it scientifically works but I am learning.

[choneofakind]

Originally Posted by csmith12

lol yea, all this physics stuff is way above my head. I don't know how it scientifically works but I am learning.

I've got the 8th edition of Physics For Scientists and Engineers by Serway and Jewett. Remind me to bring it for you to peruse next time we're meeting up.

I suggest staying out of the E+M section unless you have to be in there.

[dfox]

Originally Posted by choneofakind

Again, care to explain that a little further? You may be an engineer, but you suck at communicating what you're thinking and I'm having a hard time following you here.

I have a hard time explaining things verbally, I need a big giant white board and some markers.

A wide tire will have a larger radius to the profile of the tire. This larger radius will allow for a slightly larger contact patch, and more tire very close to the ground compared to a smaller tire. The more rubber closer to the ground, the better chance you have of of making good solid contact with the roadway, at the microscopic level. Think about grooves, gravel, etc. in that respect, a wider tire gives a rider a better chance of obtaining the available friction force. but on a good roadway, with little abnormalities, a thin tire and wide tire should have the same available friction force.