What is the maximum lateral G's a bike can do?

[noamkrief]

Originally Posted by markjenn

...but in the corners, a car will usually has a significant cornering advantage due to having four contact patches

I agree with everything you said expect this:
4 contact patches have nothing to do with it. If you can imagine a bike with 4 wheels inline - it still wouldn't be able to roll through the corners as fast as a car.

I think it's WHERE the contact patches are placed that makes the difference. A mike must lean to turn and I think the factor that maxes out a bike's cornering speed is lean angle. A bike + rider hanging off as a total unit may produce a total lean angle of lets say 60 degrees. The footpegs are scraping, and the rider is hanging off so much that he is even tucking in his knee so that he can hang off farther. At this point, the balance of gravity and centrifugal force is at equilibrium and to gain more speed without increasing radius you must lean further which in this case - impossible.

Nomatter what tires you put on the bike, I think speed, radius, and lean angle would produce a certain amount of lateral G nomatter what bike you are on.
A Harley or even a bicycle at 30deg lean, at 40mph, in a 50 foot radius turn may produce .8 G's as an example. To increase speed, lean angle MUST increase. This equation is independant on tires. Your bike can be on a rail - literally - like a train and it would be the same.
Sports bikes compared to cruisers can lean farther but there is always the theoretical limit of 90 degrees lean if the bike had 0 width.

A car also has a theoretical limit but tires are always the weakest link before that theoretical limit.
Imagine a car with tires that produce unlimited traction. The limit would be where the centrifugal force will lift the inside tires off the ground and roll the car over.
I have never seen this happen unless the car hits a bump. Usually, the tires give out WAY before this limit is reached - nomatter how good your tires are. (Maybe a bus or Van is a different story).

In conclusion, without being an expert, I think the limiting factor of a bike's maximum lateral G's is not tires but the physical dimensions of the bike + rider that limit the lean angle possible.

[noamkrief]

I just input all the formulas into excel.

Here are some graphs of lean angles, speeds, and lateral G's of a 2 wheeled vehicle.

The following formulas were used for getting lateral G's:

acceleration in m/s = velocity² / radius (all in m/s)
to convert to G's: accelleration in m/s² / 9.80m/s² = G's

The following formulas were used for lean angles:
tana = .067*mph²/radius in feet

Then you have to inverse tangent of tana and you get degrees of lean angle.

Hope the graphs makes sense. if you want the excel spreadsheet PM me with your email address.


things I learned:
nomatter what Radius turn I input, 1 lateral G occurs at exactly 45 degrees lean angle.

after you reach 45 deg of lean, a little bit of extra lean starts exponentially adding more and more lateral G's. That means that if between 20 - 30 degrees of lean angle you have the difference of .25 lateral G's but between 40 - 50 degrees it's not an extra .25 G's, it's an extra .36 lateral G's.
I think that's the important lesson. As soon as you can break the 45 lean angle, any little bit more lean, can make a huge difference in the speed you can carry through the corner.

[markjenn]

Originally Posted by noamkrief

I agree with everything you said expect this:
4 contact patches have nothing to do with it.

When I said "four contact patches" I wasn't saying that the car has superior grip because there are four vs. two - I was saying that four patches give a larger total contact patch area vs. the bike's two.

Normally, the absolute area of the contact patches isn't very important - a smaller contact patch gives the same grip because the weight bearing on the contact patch is proportionally increased. But tires interact with pavement in ways where the rubber actually interlocks with the pavement and develops grip that is not proportional to weight. This gives some grip that is simply proportional to total contact patch area without respect to weight and cars have a lot more.

It is this phenomena that mainly drives why we have bikes with huge 190 wide rear tires.

- Mark

[Technomancer]

Originally Posted by markjenn

When I said "four contact patches" I wasn't saying that the car has superior grip because there are four vs. two - I was saying that four patches give a larger total contact patch area vs. the bike's two.

At max turn rate, the car's weight is almost entirely on the outside two wheels. The contact patch total is still bigger for the car, but at they point the car and bike are essentially both two-wheel vehicles. The car also has an effective lean angle because the two wheels with weight are on the outside of the turn relative to the center of mass. On a sports car, the effective lean angle should be closer to the ground than 45* in my estimation.

Noam, you should note that your lean angles are for the bike-and-rider combination. The bike itself will be leaned less and the rider leaned more. This will also affect a G-meter if it is giving separate x and y values, such that the total G will *not* be straight down relative to the bike. The total G magnitude will be unaffected.

- John

[markjenn]

Originally Posted by Technomancer

On a sports car, the effective lean angle should be closer to the ground than 45* in my estimation.

Maybe. Odd way to think about it - I have no idea what an "effective lean angle" is on a car. The dynamics of cars are wildly different. The chassis leans in the opposite direction from a bike.

My point was simply that there is some small advantage to cars having more rubber on the road which gives them a small advantage in cornering. Everything else being equal, on any given skid pad of a fixed radius, they will be able to go somewhat faster and pull somewhat more G's.

- Mark

[Technomancer]

Very simple. The effective lean angle is the angle from the outside two tires to the center of mass, when viewed from the front. This is valid when the inside two tires are *just* lifting off the ground, or close enough as to make no difference. Yes, cars lean the wrong way in turns, but the center of mass is still on the inside. Once the inside tires start to leave the ground, it is just a funny-shaped bike.

I think your main point is valid, more rubber is better; after all, shear forces are what causes the tire to leave rubber on the road and skid, and shear forces are measured per unit area. The more area, the lower the shear force on any given bit of rubber for the same total turning force.

I just don't agree that it is the only effect, or even the most important one for cars being faster at tracks than bikes.

- John

[bidwell]

Originally Posted by markjenn

The last few years, I've taken a few trips where we have mixed groups of cars/bikes. I can say from personal experience that keeping with a well-driven sports car on anything but the tightest roads is very difficult if not impossible.

- Mark

My experience is exactly opposite but then again they may have been **** drivers.