Measuring g forces when stopping a car

[jack action]

Thank you, Jack.
Could you please explain the reasons for which the spin described in post #20 above happened?
"Then the rear wheels locked up, and the car spun 180 degrees before I could stop it. I was only going about 45 MPH".
I can't see one, since the front wheels were not braking at all.

It is the same instability but due to another reason: lack of rear lateral traction.

Let's say the tire friction coefficient is 1.0. This is true in any direction. Therefore, the combination of the lateral and longitudinal friction forces must respect that limit. It is often referred to as the traction circle.

In the case presented in post #20, the rear axle friction was completely used by braking (wheels were locked), leaving nothing to respond to lateral acceleration: for the car, in the lateral direction, its rear axle feels like it is on ice.

But the front axle isn't. Lateral friction force is available since none is used for braking. So it is the end that will react to the small lateral acceleration. Both forces - the other one being at the CG - will create an unbalanced yaw moment.

In the equations found in post #27, it is the equivalent of setting $C_{\alpha r} = 0$ which leads to $K_{us}=-\infty$ and $V_{crit} = 0$.

Use only the front brakes in the same manner, and you get a stable understeer condition.

This is also why we tend to spin out of control with a RWD when accelerating out of curve, and not with an FWD.

This is the kind of knowledge that is crucial to understand when comes time to balance the lateral weight transfer between the front and rear axle by selecting the proper springs and anti-roll bars:

The other function of anti-roll bars is to tune the handling balance of a car. Understeer or oversteer can be reduced by changing the proportion of the total roll stiffness that comes from the front and rear axles. Increasing it at the front increases the proportion of the total load transfer that the front axle reacts to—and decreases it in the rear. In general, this makes the outer front wheel run at a comparatively higher slip angle, and the outer rear wheel to run at a comparatively lower slip angle, increasing understeer. Increasing the proportion of roll stiffness at the rear axle has the opposite effect, decreasing understeer.

By playing with rear and front roll stiffnesses, we essentially play with $W_r$ and $W_f$ in the equations of post #27.

 

[jack action]

There are so many fearures of disc brakes that makes them attractive, I'm amazed they didn't catch on much sooner.

As surprising as it may be, designing a functioning caliper is very difficult. The main problem is the design of the seal and groove, which must both seal the bore AND retract the piston correctly. Nowadays, there are no more secrets as every avenue has been studied, but there are years of development behind that simple-looking piece of equipment.

It is well known that the design of the seal groove assembly in the brake caliper greatly influences the braking performance. The rubber seal performs the dual function of sealing the piston bore and retracting the caliper piston after a brake apply. However, the seal function is affected by the configuration of the seal groove, as well as the friction at the piston/seal and groove/seal interfaces. The material properties of the rubber seal are also important design parameters. Issues such as fluid displacement, piston retraction, piston sliding force, and brake drag are some of the critical brake performance parameters that must be considered in every caliper seal-groove design. Presently, the brake caliper seal groove design is still based on empirical rules established mainly from past experience and its performance is achieved through prototype testing.​

 

[sophiecentaur]

As surprising as it may be, designing a functioning caliper is very difficult.

I remember being impressed about how the profile of the seal does the two functions you describe but I assumes that a bit of trial and error would arrive at a reasonable soltion. Now you tell me it's actually a clever bit of design. Fools rush in . . . .again.

 

[Lnewqban]

In the case presented in post #20, the rear axle friction was completely used by braking (wheels were locked), leaving nothing to respond to lateral acceleration: for the car, in the lateral direction, its rear axle feels like it is on ice.

But the front axle isn't. Lateral friction force is available since none is used for braking. So it is the end that will react to the small lateral acceleration. Both forces - the other one being at the CG - will create an unbalanced yaw moment.

Appreciated detailed explanation, Jack.

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.

In motorcycles losing only the rear grip on asphalt, a fish-tail friction balance is reached, which prevents the rear from swinging all the way around.

 

[berkeman]

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.

There is at least rolling resistance.

 

[Lnewqban]

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.

 

[sophiecentaur]

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!

 

[sophiecentaur]

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.)

 

[jack action]

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.
$$G_{yaw} = \frac{\Omega_z}{\delta} =\frac{V}{l+K_{us}\frac{V^2}{g}}$$
$$G_{acc} = \frac{\frac{a_y}{g}}{\delta} = \frac{\frac{V^2}{gR}}{\delta} =\frac{V^2}{gl+K_{us}V^2}$$
$$\frac{\frac{1}{R}}{\delta} =\frac{1}{l+K_{us}\frac{V^2}{g}}$$

(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.

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.

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":

​

 

[Ranger Mike]

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.

 

[Lnewqban]

Thank you again, Jack.
Sorry, I don't understand this part:

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.

 

[Nik_2213]

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....,

 

[jack action]

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".)