Measuring g forces when stopping a car

• Car guy
Car guy
Homework Statement: Not homework but I hope easy question
Relevant Equations: 32fps2

Please forgive me if this post is in the wrong area. Here is what I want to do. I want to rather crudely measure the g forces when stopping a car. I am able to work front and rear brakes independently and wish to demonstrate the difference in stopping power.

So I was thinking of having a camera aimed at a board that has radius lines on it. And a pendent would hang and while stopping I would record how much the pendant swung, indicating the G force generated.

My question is, what angle would be .5 Gs, 1 G, and 1.5 Gs?
The reason I am doing this is I have a rear engine car and everyone keeps saying that the rear brakes do most of the stopping because that is where the weight is. But the weight transfers forward while stopping taking weight off the rear and putting on the front.

So any help offered here would be appreciated.
Thank you!

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if you have an iphone, search the app store for 'accelerometer.'

Nik_2213, russ_watters and berkeman
Dullard said:
if you have an iphone, search the app store for 'accelerometer.'
I have an android, and thanks that's a good idea! (Darn and I thought I was smart!)
Thank you

Nik_2213
You can use the cell phone to take a video of the speedometer during a stop. Then manually record the time at appropriate speed increments, and plot the results. The image below is a coastdown test on my truck performed back in 2012. The plot shows the result of two runs in opposite directions with a light wind in the direction of the road. Test details are in this thread: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html.

Nik_2213 and berkeman
First. I agree - if you own a phone, you own an accelerometer. Use it.

There is also an app called "Torque" which can read your engine's parameters via OBD-II and record them, allowing you to cross-correlate with things like acceleration.

Nik_2213 and russ_watters
Suppose the steady angle (not "swinging"!) is ##\theta## degrees forward from straight down. Then the backward component of the acceleration is ##g \tan_d(( \theta)##.

Where ##\tan_d## inputs are in degrees.

Car guy said:
So I was thinking of having a camera aimed at a board that has radius lines on it. And a pendent would hang and while stopping I would record how much the pendant swung, indicating the G force generated.
FactChecker said:
Suppose the steady angle (not "swinging"!) is ##\theta## degrees forward from straight down. Then the backward component of the acceleration is ##g \tan_d(( \theta)##.

Where ##\tan_d## inputs are in degrees.

There is a problem with using a hanging weight to measure acceleration and deceleration in a vehicle. Since the vehicle has suspension, it tilts backward during acceleration and forward during deceleration. So you really can't use a hanging weight method inside a vehicle, unless you do something like take a video of it along with the horizon outside the vehicle to do a more complete calculation.

I realized this limitation when I got a simple inclinometer for my mountain bike, so I could get better calibrated on how steep the hills were that I was climbing in my workouts. I realized pretty quickly that my weight shift when climbing caused the suspension to add to that inclinometer reading, making it pretty much useless. It is no longer on my MTB (saves a few grams of weight...).

https://suncompany.com/products/lev...qBHSDgmT0_L2Ck8BCUuVvswKnOsLgXzhoCnqcQAvD_BwE

Nik_2213, Lnewqban and FactChecker
berkeman said:
There is a problem with using a hanging weight to measure acceleration and deceleration in a vehicle. Since the vehicle has suspension, it tilts backward during acceleration and forward during deceleration. So you really can't use a hanging weight method inside a vehicle, unless you do something like take a video of it along with the horizon outside the vehicle to do a more complete calculation.
Good point. I thought about the suspension and made the judgement that, unless he stopped really hard the suspension tilt would be relatively small. It's not clear how much accuracy is desired. I had not thought about the slope of the ground. I guess that might give a larger error. The pendulum could be leveled even if the car is not.
I think that a larger source of error might be that a pendulum might swing well beyond the angle of the force. This experiment might be more of a learning experience than the OP implies. :-)

Lnewqban and berkeman
Stick with the App. Let someone else do all the clever design.

russ_watters
Car guy said:
My question is, what angle would be .5 Gs, 1 G, and 1.5 Gs?
Welcome, @Car guy !

Those angles would be 26.6°, 45.0° and 56.3° respectively.

Car guy said:
The reason I am doing this is I have a rear engine car and everyone keeps saying that the rear brakes do most of the stopping because that is where the weight is. But the weight transfers forward while stopping taking weight off the rear and putting on the front.
Let's say that we tell you that roughly a 45° angle is caused by 1 g of deceleration.

How are you going to relate that g-force that is acting parallel to the road, the position of the engine, the pitch (nose down) moment of the chassis, and the forces acting on each contact patch, during breaking?

Please, note that the "weight transfer" (from the weight on each tire measured in static conditions) has more to do with the CG height to wheelbase length ratio than with the longitudinal location of the engine (that is only important for oversteering and understeering problems).

You can ask "everyone (who) keeps saying that the rear brakes do most of the stopping" to go check and compare the diameters of the rear and front brake discs or drums.
The manufacturer should know better than that person which tire can handle more brake power before skidding.

https://en.wikipedia.org/wiki/G-force#Unit_and_measurement

sophiecentaur
Reading the values from the accelerometer in your phone as others have said seems ideal as long as you can do it and the accelerometer can be trusted.

Recording the speedometer while driving feels bad to me. Maybe it can be done safely but I think it's the kind of thing I'd try and crash while doing it.

As an alternative, isn't the acceleration during braking constant? Or at least, almost constant. ABS should help with that because you can brake hard without worrying about how much pressure you apply on the brakes. You'll be braking close to the limit you can get from friction with the ground which is a constant force.
If that's the case, it'd be simple to start braking at a given velocity once you reach a landmark and then measure the braking distance.
As long as your velocity isn't too extreme and you have a straight area without people around it shouldn't be too hard or dangerous to perform a test like that.

NOTE: On second thought, the location of the center of mass and suspension affects how big is the normal force on each of the tires which affects the maximum braking force you can get from each wheel. Since there are brakes on the front (forces increase during braking) and the back (forces diminish during braking), maybe the total braking force available is still the same and remains constant (##\mu mg##) but I'm not sure now. I'd need to calculate it to confirm it.

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FactChecker and Lnewqban
Car guy said:
Please forgive me if this post is in the wrong area. Here is what I want to do. I want to rather crudely measure the g forces when stopping a car. I am able to work front and rear brakes independently and wish to demonstrate the difference in stopping power.

By the time I reached the end of the thread and answered, I had forgotten this is what you were trying to accomplish. What I proposed about measuring the braking distance is valid. Disregard the note I wrote.
Juanda said:
NOTE: On second thought, the location of the center of mass and suspension affects how big is the normal force on each of the tires which affects the maximum braking force you can get from each wheel. Since there are brakes on the front (forces increase during braking) and the back (forces diminish during braking), maybe the total braking force available is still the same and remains constant (##\mu mg##) but I'm not sure now. I'd need to calculate it to confirm it.
That note would only be relevant if you wanted to compare the actual braking you're achieving with the theoretical maximum you'd get. Since you don't care about the maximum achievable, as long as your ABS or your driving skills allow you to brake hard you'll be able to compare the braking distance using the front or rear brakes which is what interests you.
The expected result from the experiment is that the braking distance when using the front brakes will be significantly smaller due to the weight transfer shown in post #10 which increases their braking capacity.
Lnewqban said:

Car guy said:
I am able to work front and rear brakes independently and wish to demonstrate the difference in stopping power.
Car guy said:
The reason I am doing this is I have a rear engine car and everyone keeps saying that the rear brakes do most of the stopping because that is where the weight is. But the weight transfers forward while stopping taking weight off the rear and putting on the front.
Just measuring the brake distance - or time - will accomplish the same proof.

For any method you choose, you need front and rear tires with comparable friction characteristics - identical would be ideal.

russ_watters, Lnewqban and berkeman
Car guy said:
I am able to work front and rear brakes independently and wish to demonstrate the difference in stopping power.
How are you doing that?

sophiecentaur and Lnewqban
see post in THIS forum written in 2018
Type in

Front brakes provide 70% of stopping power. But why?​

I copied and pasted the header below

Front brakes provide 70% of stopping power. But why? (1 Viewer)​

Additionally, there is no such thing as "weight transfer". To the tires, it may seem there is additional weight under braking. In fact you can only transfer weight by physically placing lead weight bars to that corner of the car.
What you are dealing with is MOMENTUM and INERTIA. Get the problem identified and you can solve it.

Regarding post # 10 and note the wrong answer on the equation. It assumes the Center of Gravity is at the same height in before braking and during braking. This is obviously NOT the case as the front suspension will dive , compressing the front springs during braking and thus LOWERING the CG height and this throws off this equation. Using corrected CG height of 22 inches yields total value of the equation to 370 " pounds" if you MUST use weight transfer. You will need (simplistic answer not using motion rate spring rate and wheel rates) stiffer anti roll bar and front springs having 123 # more "spring rate " to counter the momentum. This is why ALL vehicles, ft engine , rear engine, mid engine need larger front brakes. It is not the weight, it is the momentum and inertia.

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berkeman
Ranger Mike said:
This is obviously NOT the case as the front suspension will dive , compressing the front springs during braking and thus LOWERING the CG height and this throws off this equation. Using corrected CG height of 22 inches yields total value of the equation to 370 " pounds" if you MUST use weight transfer.
It is not obvious how much the front end will dive or that it will dive at all. Even so, It is not obvious that the CG will drop as well, as the rear end may go up. The suspension geometry will dictate that (anti-dive / anti-lift).

Ranger Mike said:
This is why ALL vehicles, ft engine , rear engine, mid engine need larger front brakes.
With rear-engine vehicles, having the same brakes front & rear is a possibility because they can achieve a 50/50 weight distribution when braking; a very desirable condition. This is one of the reasons why the Porsche 911 can be so fast: braking quickly is as important as accelerating rapidly.

https://flat6motorsports.com/pages/911-brake-guide said:
Porsche 911 OEM Specifications
 Trim​ Front Brakes​ Rear Brakes​ 997.2 Carrera​ Four-Piston Aluminum Monobloc Calipers 330 mm x 28 mm rotors Four-Piston Aluminum Monobloc Calipers 330 mm x 28 mm rotors

Good one Jack. I should have said all cars have larger front brake force, not larger brakes.
True as you have posted regarding same front and rear Porsche brakes, regarding thickness, diameter and number of pistons in the caliper, But...
The Master cylinder has bigger piston for the front brakes vs rear brakes.
at least on the 944 car.

I find that the non-ABS cars use the 19mm/19mm Master Cylinder while the ABS cars use the 23mm/23mm. But, I have found multiple sources that say all NA 944's and 924S's use the 23mm/19mm master cylinder

Some 944S,& Turbo used brake fluid pressure regulator for the rear brakes, so there were variations in bias among the models. 944 Spec car will be best served by an actual OE new replacement or a rebuilt OE unit. These cars already have to much rear brake bias and anything other than the 23/19 split will only increase this problem.
So good info all around to consider.
Regarding the front end dive and I speak in the majority of cases. The exception does not make the rule. And we are governed by the majority in this society. Other wise every building would have a wheelchair accessible entry. It isa good reminder to frame answers correctly and not use incorrect phrases like ALL. But I maintain that if you have a suspension, your CG, RC and nose will dive on braking. If it does not, you are driving a go kart. Even F1 and Indy cars have suspension travel!

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berkeman
There is also an app called "Torque" which can read your engine's parameters via OBD-II and record them, allowing you to cross-correlate with things like acceleration.
Phone sensors too: it does acceleration graphing. No OBD adapter needed...or car, for that matter:

I have phyphox on my Android phone. Someone posted here on PF about it recently. By which I mean in the last 2 or 3 years, lol.

OmCheeto and berkeman
Car guy said:
I am able to work front and rear brakes independently and wish to demonstrate the difference in stopping power.
Easy to do on a bicycle. Using the front brake only will do a stoppie if the brake is in good shape. Rear brake alone will not do much at all, other than leaving rubber on the road. The center of gravity (CG) is high relative the the wheelbase, so there is a lot of weight transfer under braking.

Easy to do on a motorcycle. I never did a stoppie, but I did have a bike with soft springs in the front forks where the front suspension would bottom out under hard braking. That was interesting because the front tire would lock up instantly when the suspension bottomed out. The front brake stops the bike, the rear brake by itself will eventually get it to stop.

I once did an experiment to find if I could stop my car using the hand brake alone. The hand brake activated only the rear brakes. As I slowly applied the hand brake, the car slowly decelerated. Then the rear wheels locked up, and the car spun 180 degrees before I could stop it. I was only going about 45 MPH, it decelerated very rapidly during the short time it was sideways, so quickly came to a stop in a cloud of dust (gravel road).

If you plan to test rear brakes only, do it down the center of a wide road with no traffic at a low speed. Modern cars with antilock on the rear brakes will not spin out IF the antilock is working properly. It not, be prepared for an interesting time.

russ_watters and Lnewqban
Juanda said:
Recording the speedometer while driving feels bad to me.
There is room for two people in most cars. One to drive and one to measure?
jack action said:
With rear-engine vehicles, having the same brakes front & rear is a possibility because they can achieve a 50/50 weight distribution when braking; a very desirable condition.
Most times, when you apply the brakes you are not using anything like 100% pedal pressure. Large brake discs / drums can cope better with the heating effect (during hard driving) so the mean ratio of dissipation from front to rear will be more like the weight distribution than the max figures that this thread is discussing. Also, the suspension will allow some pitch, during which time, the front brake load will increase 'slowly' from the static situation to max braking. So a rear engine car will actually put more stress on the rear brakes than the "weight transfer" situation initially. Only a crazy driver will cause the car to dip forwards every time they apply the brakes. The diagrams higher up show the forces when braking is at the maximum.

The manufacturers know what they're doing.

Lnewqban
sophiecentaur said:
So a rear engine car will actually put more stress on the rear brakes than the "weight transfer" situation initially.
If I understood you correctly, I would disagree.

Say you have a car with a 60/40 F/R weight distribution and a possible 70/30 F/R maximum brake force distribution. (That would be a typical front-engine car but the same principle applies for any other weight distribution.)

If you want half the maximum brake force, you would ideally apply something below 35/15 (maximum weight transfer) and above 30/20 (no weight transfer) brake force ratio (compared to the maximum braking force) to use equally the front and rear brakes/tires.

But you could also use 50/0 (no rear brakes) and the front brakes would still be able to decelerate the car as they are designed to withstand 70% of the maximum braking force. But you could never use the rear brakes only as they cannot take more than 40% of the maximum brake force - and that is assuming no weight transfer.

Furthermore, a vehicle with the front wheels locked is considered safer (understeer condition) compared to one with the rear wheels locked (oversteer condition).

Most brake systems are made with a fixed ratio and that is the one for responding to the maximum weight transfer, in our case, 35/15 at half the maximum brake force. So you are always putting more stress than needed on the front tires than on the rear tires in every case except when at maximum braking force where you are using them to respect the maximum braking force that can be applied to each tire.

The good news is that the front brakes are designed to accommodate 70% of the maximum brake force and the rear ones 30%. So at a 35/15 ratio, you are still using half the braking power that each axle braking components - i.e. except the tires - can handle.

sophiecentaur said:
The manufacturers know what they're doing.
Yes, they do.

Porsche 911s are still rear engine because the weight distribution is more even under braking. Ultimately, the weak link in breaking is tire traction. Any manufacturer can easily make brakes that will lock the wheels, so that is not critical. But then tires skid or activating the ABS. Therefore, the key to the best braking is to get the best tire traction. That leads to the conclusion that the static weight distribution should be rear-loaded.

Lnewqban
jack action said:
If I understood you correctly, I would disagree.
Initially, the weight distribution will be the same as when static (the upwards added force on the rear suspension will be zero and the downwards added force at the front will be zero. Whether or not that is actually relevant to the whole braking effect, it has to be true. I'm just suggesting that the forces during the transition would change. Softer suspension would increase any forward tilt and, thus, the moment.

The action of dampers would affect the upwards and downwards forces differently:front springs would compress easier than the back springs would reduce. I haven't figured the oucome yet or the magnitude of the effect.

jack action said:
Furthermore, a vehicle with the front wheels locked is considered safer (understeer condition)
Except when on a tight curve with no guardrails, I assume.

jack action said:
... compared to one with the rear wheels locked (oversteer condition).
Isn't that situation comparable to purposely drifting a car only for show?
If I understand the maneuver properly, the driver still steers, controlling the trajectory of the drifting.

Copied from:
https://en.wikipedia.org/wiki/Drifting_(motorsport)

"Drifting is a driving technique where the driver intentionally oversteers, with loss of traction, while maintaining control and driving the car through the entirety of a corner or a turn. The technique causes the rear slip angle to exceed the front slip angle to such an extent that often the front wheels are pointing in the opposite direction to the turn (e.g. car is turning left, wheels are pointed right or vice versa, also known as opposite lock or counter-steering)."

FactChecker said:
Porsche 911s are still rear engine because the weight distribution is more even under braking. Ultimately, the weak link in breaking is tire traction. Any manufacturer can easily make brakes that will lock the wheels, so that is not critical. But then tires skid or activating the ABS. Therefore, the key to the best braking is to get the best tire traction. That leads to the conclusion that the static weight distribution should be rear-loaded.
The reason why we prefer 25/25/25/25 loading on tires is that the friction force is not linear with the normal load:
http://technicalf1explained.blogspot.com/2012/10/f1-tirespart-2.html said:
A tire’s coefficient of friction decreases as load increases. However, this does not mean that traction decreases. Traction will increase with an increasing vertical load.

This effect is illustrated in the graph below:

The end effect is that the increase in friction force with a lower vertical load never compensates for the decrease in the more loaded tire(s). Therefore, anything varying from 25/25/25/25 will reduce the overall maximum friction force available.

Lnewqban
Lnewqban said:
Isn't that situation comparable to purposely drifting a car only for show?
If you imagine the bicycle model:

The steering angle ##\delta## can be related to the slip angle of the tires:
$$\delta = \frac{l}{R} + \alpha_f - \alpha_r$$
The lateral tire forces ##F_y## can be expressed as a function of the slip angle:
$$F_y = C_{\alpha} \alpha$$
where ##C_{\alpha}## is the cornering stiffness of the tires. It can also be related to the mass and lateral acceleration:
$$F_y = \frac{W}{g}\frac{V^2}{R}$$
$$\alpha_f = \frac{W_f}{C_{\alpha_f}}\frac{V^2}{gR}$$
$$\alpha_r = \frac{W_r}{C_{\alpha_r}}\frac{V^2}{gR}$$
Where ##W_f## and ##W_r## are the front and rear weights. We thus end up with:
$$\delta = \frac{l}{R} + \left(\frac{W_f}{C_{\alpha_f}} -\frac{W_r}{C_{\alpha_r}}\right) \frac{V^2}{gR}$$
$$\delta = \frac{l}{R} + K_{us} \frac{V^2}{gR}$$
Where ##K_{us}## is referred to as the understeer coefficient for the vehicle.

With a neutral steer, ##K_{us}= 0##, and thus ##\delta = \frac{l}{R}##

With an oversteer condition, ##K_{us} \lt 0##. With this condition, a critical speed can be identified. It is the speed at which the steering angle required to negotiate any turn is zero:
$$V_{crit} = \sqrt{\frac{gl}{-K_{us}}}$$
Above this speed the vehicle's response becomes unstable. You can look for "snap oversteer" to find examples of it:

With understeer, you just need to stop what you were doing (braking or accelerating) to regain control of the steering; with oversteer, you usually need a diaper.

sophiecentaur and Lnewqban
jack action said:
With understeer, you just need to stop what you were doing (braking or accelerating) to regain control of the steering; with oversteer, you usually need a diaper.
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.

Very hard to determine the snap spin you had without more details. I assume the car had drum brakes on the rear. By their very nature, there is no way to know how close they were to activating when the emergency brake was applied. One brake shoe could have been out of "ideal" adjustment a great amount simply because the self adjusters were off, thus making one brake shoe activate much before the other one did. Their is so much mechanical mechanisms to try to fine tune on these and the whole production cost was such that the automotive industry went to disc brakes all around, finally. These Emergency brakes are good for one thing only. Parking on a hill and not depending on one crappy small plastic part to hold the car in park on a steep hill...crazy!
Now I know Jack gets it because he knew immediately about the fallacy of "weight transfer" He has heard me rant for years about this. One thing no one else (sophie came real close) picked up is the result of braking. You are loading the front tires by adding down force, thus traction. More traction, more grip, faster stopping.

sophiecentaur
Ranger Mike said:
the automotive industry went to disc brakes all around, finally.
There are so many fearures of disc brakes that makes them attractive, I'm amazed they didn't catch on much sooner. Presumeably drum brakes were used in the early days because discs need more input pedal effort (friction area) and 'leading shoes' have their own servo action. Power servo braking wasn't available in the early days. Rear drum brakes, with their 'one leading and one trailing' shoes were pathetic.
Driving in the early days, without power steering or braking was very hard work, particularly on heafy vehicles. You had to be tough and I could almost excuse those lazy devils who would always block tight junctions as they took the easy way round the turn without using full lock. They are just as lazy these days though, with no excuse.

Lnewqban said:
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:

https://en.wikipedia.org/wiki/Anti-roll_bar#Main_functions said:
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.

Lnewqban
sophiecentaur said:
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.

https://www.sae.org/publications/technical-papers/content/2002-01-0927/ said:
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.​

Lnewqban, sophiecentaur and gmax137
jack action said:
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.

jack action said:
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.

Lnewqban said:
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

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