Physics Debate - cornering 101.

# Thread: Physics Debate - cornering 101.

1. ## Physics Debate - cornering 101.

This is not about handling of heavy vs light motorcycles, just about grip. I have been wondering what type of bike would have better grip in a given corner and same speeds with same tires.

Would a heavier bike hold the road better that a lighter bike - same turn, same tires, same speed, same wheel base, same rider.... etc.

Thoughts?

2.

3. Originally Posted by Fury
This is not about handling of heavy vs light motorcycles, just about grip. I have been wondering what type of bike would have better grip in a given corner and same speeds with same tires.

Would a heavier bike hold the road better that a lighter bike - same turn, same tires, same speed, same wheel base, same rider.... etc.

Thoughts?
I would hazard a guess that a lighter bike would have better grip.

Given the same turn, tires, speed, wheel base and rider, the only aspect that would change would be the angle of attack needed to balance the centrifugal force and gravity. An infinitely light bike could turn this hypothetical corner with zero angle of attack and an infinite heavy bike would need to attack the corner at 90 degrees (given infinite co-efficient of friction between the rubber and the road). Since we're working with real bikes, the lighter bike could attack the corner at a lesser angle.

However, most bike wheels are round enough that the contact patch is more-or-less the same regardless of angle (until one runs into the chicken strips), at which point the harsh mistress of gravity smacks you one and low side you to mother earth!

What does that mean? Nothing, until you think about the fact that the majority of the time, you're riding down the middle of the tire and it's going to be warmer than the sides. A warm contact patch sticks better to the road than a cold contact patch.

My 5c. Flame away, what do I know, I ride a cruiser and can't get near leaning before the pegs scrape.

Discuss.

4. I would venture to guess a lighter bike would hold better as well since less force is being applied to the coeffient of friction. Interesting question though. It begs the question of how much weight/force is neccesary to overcome the friction of the tire given equal road conditions, equal tires, equal speed and equal corner radius.

The contact patch is also a big variable since with increased weight = increased force transfered to the tire would mean a larger contact patch which may nullify the weight difference?

Tee Tee?

5. Thinking back on Physics 11/12...

Frictional force is the product of the coefficient of friction and the normal force exerted by the surface. The coefficient would be the constant if all that is different between the two bikes is their weight. Normal force is the force the ground is using to keep your bike from falling into the core of th earth. On a level surface, that force would just be a product of mass and 9.81 (acceleration due to gravity).

To summarize, frictional force is higher when the bike is heavier.

I hope I didn't make any mistakes, grade 11/12 was so long ago...

6. E=1/2 m v^2

also, a moving object has a tendancy to keep on moving in the straigh line (vectors)

heavy motorcycle has more kinetic energy than light one going the same velocity in a straght line (refer to formula)

when turning the bike you are fighting that straight energy as you are breaking the bike's tendancy to go in the straight line, so the ammount of energy pushing the bike to the outside will be greater for a heavier bike

yeey i am a nerdo

7. Originally Posted by Vili
E=1/2 m v^2

also, a moving object has a tendancy to keep on moving in the straigh line (vectors)

heavy motorcycle has more kinetic energy than light one going the same velocity in a straght line (refer to formula)

when turning the bike you are fighting that straight energy as you are breaking the bike's tendancy to go in the straight line, so the ammount of energy pushing the bike to the outside will be greater for a heavier bike

yeey i am a nerdo
Interesting!!!! It all makes sense too!!

8. Originally Posted by Vili
E=1/2 m v^2

also, a moving object has a tendancy to keep on moving in the straigh line (vectors)

heavy motorcycle has more kinetic energy than light one going the same velocity in a straght line (refer to formula)

when turning the bike you are fighting that straight energy as you are breaking the bike's tendancy to go in the straight line, so the ammount of energy pushing the bike to the outside will be greater for a heavier bike

yeey i am a nerdo
but would the fact that you lean a bike over relative to the speed and radius of the turn change anything? If you think of a car you are blatently changing the direction(vector) of the machine with only the friction of the tires, whereas a bike its using the friction of the tires as well as the lean angle to change its vector. Its not only the coefficient of friction changing the direction of the bike?

9. After more physics brainstorming, I realized that a bike making a turn is simply frictional force acting as the centripetal force. In order for the turn to be successful, the frictional force must exceed the centripetal force needed for the turn. Centripetal force is a product of mass and centripetal acceleration. And as I posted above, frictional force on a level surface is a product of the coefficient of friction, gravitational force, and mass. A turn will look something like:

F(f) >= F(c)
u*g*m >= m*a(c)

Where:
u = coefficient of friction
g = gravitational acceleration
m = mass
a(c) = centripetal acceleration

Cancelling m out and you'll get:
u*g >= a(c)

Which means mass does not matter, what matter are the coefficient of friction and the centripetal acceleration (which depends on velocity and radius of the turn).

10. Originally Posted by dmanrevived
After more physics brainstorming, I realized that a bike making a turn is simply frictional force acting as the centripetal force. In order for the turn to be successful, the frictional force must exceed the centripetal force needed for the turn. Centripetal force is a product of mass and centripetal acceleration. And as I posted above, frictional force on a level surface is a product of the coefficient of friction, gravitational force, and mass. A turn will look something like:

F(f) >= F(c)
u*g*m >= m*a(c)

Where:
u = coefficient of friction
g = gravitational acceleration
m = mass
a(c) = centripetal acceleration

Cancelling m out and you'll get:
u*g >= a(c)

Which means mass does not matter, what matter are the coefficient of friction and the centripetal acceleration (which depends on velocity and radius of the turn).
right, exactly. Here's my point though, would the increased mass of the bike translate into an increased contact patch given equal tire pressures, hence giving an increased coefficient of friction over the ligher bike. Or would the decreased centripetal force needed for the ligher bike nullify that?

oh hells yeah, out comes the wine!

but would the fact that you lean a bike over relative to the speed and radius of the turn change anything? If you think of a car you are blatently changing the direction(vector) of the machine with only the friction of the tires, whereas a bike its using the friction of the tires as well as the lean angle to change its vector. Its not only the coefficient of friction changing the direction of the bike?
we are comparing two bikes in the same situation except one is heavier ...

so no it would not matter, since other forces, like the lean angle and the rotation of the tire would be prety close

12. I'm not sure if coefficient of friction changes. I looked it up at wikipedia just now and it doesn't say anything about the size of the contact area, just that there are two types of coefficients of friction for a pair of surfaces, one kinetic and one static (which does not apply here).

13. Originally Posted by Vili
we are comparing two bikes in the same situation except one is heavier ...

so no it would not matter, since other forces, like the lean angle and the rotation of the tire would be prety close
so you are saying that both bikes will theoretically break traction at the exact same point? given the increased coefficient of friction and centripital force from the heavier bike. they would essentialy cancel each other out?

14. would the coefficient of friction change with mass? I dont recall ....

If it would then the heavier bike slides first

15. Originally Posted by dmanrevived
I'm not sure if coefficient of friction changes. I looked it up at wikipedia just now and it doesn't say anything about the size of the contact area, just that there are two types of coefficients of friction for a pair of surfaces, one kinetic and one static (which does not apply here).
But granted the heavier bike will have a larger contact patch since more force is being transfered to the tires? Given the same tire pressure on both bikes? This would give the heavier bike a larger coefficient of friction, no? But as well a higher centripetal force in the corner?