Measuring g forces when stopping a car

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  • #36
Perhaps I am trying to say that, in cases of recovery from a skid, there is a natural limit to the slide and angle of tail-fishing.
The sliding rear tire recovers some grip as the sweeping angle increases, due to the direction of the translation to increasingly diverging from the rolling plane.
It could be that the instinctive steering correction can be quicker in a motorcycle than it is in a car.

motorcycle-broom.gif


v7GP5-.gif

Fy2N50.gif
 
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  • #37
berkeman said:
There is at least rolling resistance.
Yes - particularly on a gravel road. It seems to me that the experiment had several unknowns. A gravel road couldn't be relied on to have a uniform surface so the net (unbalanced) rolling resistance could easily not act on the CM of the car. That could start it rotating.

Once the car starts to rotate, the front wheels still have lateral grip and the rear ones will just slide over the gravel. . . . . wheeeeee!
 
  • #38
Lnewqban said:
It could be that the instinctive steering correction can be quicker in a motorcycle than it is in a car.
As I mentioned before, there are many unstated variables in any of this. You can't really compare driving a domestic car with speedway riding. Rally drivers use a lot more tail slip, intentionally, and 'handbrake turns' used to be a favourite amongst those of my contemporaries who were into that sort of driving. I remember them saying that they'd change the direction of the pawl spring on their handbrake so they could let go quickly without needing to 'release' the brake with the button. (I have a feeling that the MOT test would spot that, these days.)
 
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  • #39
Lnewqban said:
I am still unable to see what makes the rear end, and the CM, move all the way forward and ahead of the front one, which free rolling seemly presents no resistance to the forward movement.
I guess you have to appreciate what it means to be at the critical speed in an oversteer condition in a steady-state response to steering input. It means infinite yaw velocity gain, infinite lateral acceleration gain, and infinite curvature response.
$$G_{yaw} = \frac{\Omega_z}{\delta} =\frac{V}{l+K_{us}\frac{V^2}{g}}$$
$$G_{acc} = \frac{\frac{a_y}{g}}{\delta} = \frac{\frac{V^2}{gR}}{\delta} =\frac{V^2}{gl+K_{us}V^2}$$
$$\frac{\frac{1}{R}}{\delta} =\frac{1}{l+K_{us}\frac{V^2}{g}}$$

yaw-velocity-gain.jpg

accel-gain-curvature-response.jpg

(source)​

Therefore, you can already be a goner as soon as the steering angle just slightly moves.

Plus, if the front wheels do not follow the vehicle's line of motion, it is not pure rolling anymore as there's a lateral force acting on them.

Lnewqban said:
motorcycle-broom.gif


v7GP5-.gif

Fy2N50.gif
In these cases, the throttle and steering responses - by a pro driver who knows his machine - definitively modify the bike's behavior. Especially in drifting where the bike is thrown out of balance before entering the curve.

drift-Guide-2.jpg

Lnewqban said:
The sliding rear tire recovers some grip as the sweeping angle increases,
You cannot recover the grip of an already sliding tire. The friction force is at its maximum value. Worst, the friction circle is more of a friction ellipse where the longitudinal maximum friction forces are greater than the lateral ones.

Finally, I think these examples show that there is not always a "natural limit to the slide and angle of tail-fishing":


 
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  • #40
Now lets be realistic here!
On a motorcycle , you can change the Center of Gravity real easy by shifting your body position. Not so with a vehicle. In racing, we always had two master cylinders on the race car and had a bias bar to tune front to rear brake engagement so the front hooked up before the rear. Otherwise you spin out. Pulling the parking brake ( and not knowing ifin the braking is equal for both rear tires will cause you to spin out. If it was the same the car would stop straight line.

Do not try to compare a two wheel motorcycle to a 4 wheel car. Totally different scenario.
prove me wrong! Jack up our car that spun out. Take a torque wrench and measure foot pounds it takes to rotate each rear wheel after you apply the parking brake (in increments ,do not lock up the brakes on maximum setting on the parking brake, try half way) I bet you find a true difference in torque required to turn the wheels and I bet the side with less torque is the side you spun toward.

 
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  • #41
Thank you again, Jack.
Sorry, I don't understand this part:
jack action said:
You cannot recover the grip of an already sliding tire. The friction force is at its maximum value. Worst, the friction circle is more of a friction ellipse where the longitudinal maximum friction forces are greater than the lateral ones.
The fundamental of a high-side fall is the recovering of the grip of the sliding rear tire.

829b6f590add84ac06e8584e56f0ff5a.gif
 
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  • #42
Back to car braking:
Does the vehicle have anti-lock braking / traction control ? They would make analysis difficult...
Unless they have a diagnostic port to supply such data ??

FWIW, I can only remember one (1) incident when, in extremis, my duly-paranoid, road-reading anticipation was almost foiled by a WTF-grade 'low flying ijit'. I had to 'totally slam-on'. Despite non-ideal road conditions, the 'system' provided maximum sustained braking, maintained steering control. I shed enough speed soon enough to go around the oblivious perp.
My beloved wife swore like a brace of fish-wives, demanded I fit Sparrow missiles on the roof-rails. a 'remote-turret' machine gun. Oh, and paired wind-horns. All at her control...
Yes, deployed, the 'Wrath of Kath' was awesome....,
 
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  • #43
Lnewqban said:
Thank you again, Jack.
Sorry, I don't understand this part:

The fundamental of a high-side fall is the recovering of the grip of the sliding rear tire.

View attachment 349617
If you are sliding under braking - i.e. the problem at hand, a locked wheel - the fact that you begin sliding sideways won't increase the friction braking force (still the wheel locked but partly sideways).

The images of motorcycles you are showing involve most likely variations in the wheel torque which, if decreased, will allow more lateral friction force on the tire, i.e. possibly no more sideways sliding. This is not steady-state, it doesn't happen "naturally": the driver must vary the input. (Or the road becomes somehow "stickier".)
 

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