One for the physics experts - Stopping distances at different speeds

[sal]

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A little rusty on the kinematic equations but here goes:

V2^2 = V1^2 + 2ad

V1 = 60mph = 26.6m/s
V2 = 0
d = 104ft = 31.7m

Solving for a, you get -11.2m/s2. Being that 1g = 9.81m/s2, 11.2m/s2 = 1.14g <- This number looks reasonable...

Someone be able to calculate the G force from one of the stops on a bike from the 60-0 chart?
The shortest stopped in 104.8 feet and the longest 148.6 feet.
or do you need to know the stopping time for the calculation?

 

[koorosh]

To answer the G-force question, the moment of your deceleration force must be greater than the moment of your weight both about the front tire patch area. In formula: m*a*h>m*g*d. where a is deceleration, d is horizontal distance of bike+rider's center of gravity to front tire patch and h is vertical distance of center of gravity to front tire patch. Cancelling "m" from both sides you get : a>g*d/h. Two assumptions are made here: 1 - You are riding straight only. 2 - Friction is high enough to avoid any slippage before stoppie, i.e. Friction force > m*a. 3- Front shocks have already reached their lowest positions during braking.

 

[sal]

Beat me to it! You can assume that's average acceleration, probably close enough to get a feel. 0.8-1.1g is well within the capibility of an aggressive street tire.

Assuming Constant acceleration (which is doubtful), 1.1 g for 105 feet, 0.8 g for 149 feet. Obviously some of my many assumptions were off in the first calculation.

 

[Brian P]

Sport bike max deceleration will normally be in the 1 g range, limited by the rear wheel having to remain in contact with the ground. If the center of gravity height is somewhere near half the wheelbase and centered between the wheels, braking will be limited to 1 g before a stoppie starts happening and acceleration will be limited to 1 g before it starts to wheelie.

MotoGP bikes may decelerate faster at 200 mph simply because the aerodynamic drag alone at that speed will be a significant portion of 1 g. The aerodynamic center of pressure is very much above ground level and possibly above the center of gravity, so the drag won't be anywhere near as significant a contributor to the overturning moment as normal braking forces are (applied at ground level, at the tire contact patch).

Sport bikes have a significant trade-off between cornering nimbleness (short wheelbase and high CG are favorable) and acceleration / braking capability (opposite up to a point - depending on how much friction is available between tire and ground).

 

[fastar1]

In general at what kind of G-Force of deceleration would the rear of a sport bike come off the ground and et to the edge of flipping? Assume the rider isn't trying to do a stoppie but rather trying to maximize braking.


..Tom

In general, around 1G. It will be exactly 1G if the C of G is the same distance above the front tire contact patch as it is behind the contact patch. If the C of G is higher or further forward, the max G will be less than 1. If it is lower or further back, it will be more than 1G.
For a bike to brake at 2G without lifting the rear, the C of G would have to be twice as far back as it is high above the contact patch.
In the example given in the original post, braking G averaged 1.05, so the C of G was about 1.05 x further from the contact patch than it was above it (probably a bit further still, since it probably wasn't a sustained 1.05G). Also that is with the front suspension compressed. When at rest the C of G obviously rises.

 

[Baggsy]

Sport bike max deceleration will normally be in the 1 g range, limited by the rear wheel having to remain in contact with the ground. If the center of gravity height is somewhere near half the wheelbase and centered between the wheels, braking will be limited to 1 g before a stoppie starts happening and acceleration will be limited to 1 g before it starts to wheelie.

MotoGP bikes may decelerate faster at 200 mph simply because the aerodynamic drag alone at that speed will be a significant portion of 1 g. The aerodynamic center of pressure is very much above ground level and possibly above the center of gravity, so the drag won't be anywhere near as significant a contributor to the overturning moment as normal braking forces are (applied at ground level, at the tire contact patch).

Sport bikes have a significant trade-off between cornering nimbleness (short wheelbase and high CG are favorable) and acceleration / braking capability (opposite up to a point - depending on how much friction is available between tire and ground).

Here's where I got the info from, so it's third or fourth hand:
http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=14283

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