How to Calculate the Braking Momentum on a Wheel?

[jbriggs444]

I will have a homogeneously wet pavement (along with a water friction force on the bottom of the wheel) so $\mu$ might be changed between simulations but will be a time constant.

So you can determine outside the main loop whether road friction or brake friction will be the limiting factor.

Then use a main loop that models deceleration under that constraint alone.

 

[ROOT0X57B]

Ok now I want ABS to get into the equation, because if I don't, my results won't be realistic at all

I will use Pacejka's Magic Formula to get $\mu$ as a function of relative slip.
Now, how can I calculate the relative slip, I think of
$$\frac{R_\text{ext}\Omega - v}{|v|}$$
Where $\Omega$ is the longitudinal component of rotational speed $\omega$

To get $\omega$ I would have used $\displaystyle v = R_\text{ext}\frac{\text{d}\theta}{\text{d}t}$ because $\omega = \frac{\text{d}\theta}{\text{d}t}$
So $\omega = \frac{v}{R_\text{ext}}$

But this is assuming there is no slip...

What can I do then? Should I monitor $\theta$ thus $\omega$ at each step ?

 

[jbriggs444]

Ok now I want ABS to get into the equation, because if I don't, my results won't be realistic at all

I will use Pacejka's Magic Formula to get $\mu$ as a function of relative slip.

Can you please explain in your own words what you think Pacejka's Magic Formula is, how it relates to ABS and how you think that it is in any way a useful tool to determine the coefficient of static friction $\mu$ without knowing anything about the composition of the tire or of the pavement.

$\mu$ is normally something that you measure, not something that you predict.

 

[ROOT0X57B]

Yes, sorry my mind is so focused on it that I forgot to explain that...

Earlier (February - April), I was doing some research about ABS, relative slip, lack of friction...
I found Pacejka's "Magic formula" to be a good and quite simple model of $\mu$ when the tire slips.The formula is
$$\mu = A * (B * (1-e^{-C\text{slip_x}})-D*\text{slip_x}$$
Where A, B, C and D depend on the road conditions
(A more complex version using Arctan exists too)
Applied to data found online (verified through multiple websites), I have

Dashed area is when $\mu$ decreases

 

[jbriggs444]

So for any given set of road conditions, there is a maximum coefficient of friction that can be achieved if the brakes are modulated optimally.

If you want to simulate the motion of a car that is being braked optimally, model the coefficient of friction as having that value and call it done.

 

[ROOT0X57B]

If you want to simulate the motion of a car that is being braked optimally, model the coefficient of friction as having that value and call it done.

Okay, I will do that for now, but I'm curious about this :

What can I do then? Should I monitor $\theta$ thus $\omega$ at each step ?

 

[ROOT0X57B]

I think I will have to make $\mu$ to NOT be a constant (or it will feel too simple), do you have an idea on how to get $\Omega$?

 

[jbriggs444]

I think I will have to make $\mu$ to NOT be a constant (or it will feel too simple), do you have an idea on how to get $\Omega$?

$\Omega$ being the rotation rate that gives you the maximum coefficient of friction of tire on road?

Assuming yes then... If you modulate the brakes to allow the tires to rotate at $\Omega$ then the resulting coefficient of friction will be a constant.

How do you get $\Omega$? Easy. You read the desired relative slip off of the graph. If the relative slip is 20% then you set $\Omega$ to be the nominal rotation rate ($\frac{v}{R_\text{ext}}$) minus 20 %.

 

[jack action]

do you have an idea on how to get $\Omega$?

It's the angular velocity of the wheel, which is another input you have to feed to your program. Normally, in a vehicle, there is a sensor that measures the wheel rpm and there is another one that measures the velocity by evaluating how the vehicle changes position (similar to what your smartphone does).

I think I will have to make to NOT be a constant (or it will feel too simple),

I must say that I really have difficulty understanding what your trying to achieve. People usually try to simplify stuff, not complicate it.

You should have a function that solve this equation from $v_0$ to $v_f$:
$$\Delta x = \frac{m_e v_{avg}}{F_t (v_{avg}) + \frac{1}{2}\rho C_D A v_{avg}^2 + F_R} \Delta v$$
You just have to decide on a value for $\Delta v$ when solved by numerical analysis. One small enough in relationship with $v_0 - v_f$.

If you really want to know the time (which I think is unnecessary) you have to solve also:
$$\Delta t = \frac{\Delta x}{v_{avg}}$$

Note that I used the braking force $F_t$ as a function of velocity. That should be another function in your program. At first, to help you familiarize yourself with the equation, this should return a constant (independent of $v$) based on the maximum performance of the tire, i.e. $\mu mg$.

Afterward, you can add more precision by setting it to return $MIN(\frac{T_B}{r_t}, \mu mg)$, which is again a constant.

Then you can make $\mu$ a function of the normal force on each axle where - instead of $\mu mg$ - you would get:
$$\mu(N_f)N_f + \mu(N_r)N_r$$
Where $N_f$ and $N_r$ are the front and rear normal force on the respective axles and $\mu$ varies according to this normal force (another function to make). You will have to add inputs to the function $F_t$, such as vehicle dimensions and acceleration.

Then you can complexify the function $\mu$ even further if you wish with slip or anything else, but you will need even more inputs. You can even input the position $x$ of the vehicle to indicate when the quality of the road will change, thus affecting $\mu$ as the vehicle moves. But that is even more inputs for $F_t$.

But don't forget the first component of our braking force, i.e. $\frac{T_B}{r_t}$. The torque provided by the braking system can also vary, thus becoming another function in your program. This is where your ABS system would have an effect. Of course, many inputs could be needed, thus added to the function $F_t$ as well.

There are no limits on how detailed you want your model to be. But you have to start at the beginning. Make your functions simple but expandable.

 

[jbriggs444]

I must say that I really have difficulty understanding what your trying to achieve. People usually try to simplify stuff, not complicate it.

Thank you for saying what I was thinking.

 

[ROOT0X57B]

Excuse me, everyone, I messed up

First of all, at the contrary of what I've said before I'm working with disc brakes not drum brakes

After a deeper look into my previous research and results, this seems far too complex for the moment
Sorry for having wasted your time...

I have remake the code using all of your advice and I get this :

Seems fair with France's official numbers: 28 meters

But this is considering there is slipping all the time : $F_B > \mu Mg$
In fact after the calculations :

Brakes pressure : 5,000,000 Pa = 50 bar
Brakes torque : 45,744 Nm
Slip limit : 5,699 ($\mu Mg$)Why is the limit so low compared to the brake force ?? There has to be a maximum of 5 bar of pressure for the car not to slip, but according to what I find online, disc brakes tend to work around 30-50 bar, and a lot more if the brake pedal is hit hard, so I would expect emergency braking pressures to deal with 100 bar or so

 

[jack action]

With disc brakes, the braking torque of a single disc is:
$$T_B = 2\mu_p pA r_d$$
Where:

• $2$ is because there are two friction forces, one on each side of the disc;
• $\mu_p$ is the friction coefficient of the brake pads on the disc;
• $p$ is the brake line pressure;
• $A$ is the area of all the caliper pistons, on one side of the caliper only;
• $r_d$ is the radius where the friction forces are applied (middle of the pads).

From your program, it doesn't seem you are applying this right.

 

[ROOT0X57B]

With disc brakes, the braking torque of a single disc is:
$$T_B = 2\mu_p pA r_d$$

I thought it was
$$2\mu_B P_BS_BR_B$$
with
$mu_B$ the friction coefficient brake pads/disk
$P_B$ the pressure applied to the pads
$S_B$ the surface of one brake pad
$R_B$ the mean radius where the forces are applied

Why is it the surface of the pistons and not the one of the pads?

 

[jbriggs444]

I thought it was
$$2\mu_B P_BS_BR_B$$
with
[...]
$P_B$ the pressure applied to the pads

Why is it the surface of the pistons and not the one of the pads?

Because, in the formulation given by @jack action $p$ is the "brake line pressure" rather than the pressure between pad and disc.

The two formulas will deliver the same result because the force of the pad against the disc will match the force of the brake fluid against the piston.

 

[ROOT0X57B]

The two formulas will deliver the same result because the force of the pad against the disc will match the force of the brake fluid against the piston.

Yes that's what I thought

Why is so low then? Why only 5 bar?

 

[jbriggs444]

Yes that's what I thought

Why is so low then? Why only 5 bar?

Before we try to ask "why", let us try to confirm that brake fluid pressure is actually "only 5 bar". We might also ask "why does it matter"?

Most of the metal brake lines burst around 15,000 psi. The typical full-lock operating pressures are 900–1,000psi (69 bar) with manual brakes and 1,400-pluspsi (96 bar) with power-assisted brakes.

Now you might ask "why 69 bar or 96 bar". I would imagine that this is a convenient pressure that gives you a factor of 10 safety margin for the bursting strength of the brake lines, keeps the brake lines and pistons to a conveniently low diameter and does not blow out the seals around the pistons. As with all engineering solutions, you want a decent mix of "cheap", "effective" and "reliable"

More generic hydraulic cylinders tend to run at higher pressures than this.

Therefore, the weight that can be pushed or lifted by a hydraulic cylinder is equal to the pressure provided by the pump multiplied by the size of cylinder rod.

Most pumps have a standard range; usually from 3000 psi to 10,000 psi (210 to 690 Bar). The average tends to be at the lower end of the range; around 210 Bar. A pump can lift an infinite amount of weight, depending on how big the cylinder is. Increasing the PSI may require adjusting the seals, valves and cylinder design to cope with increased pressure.

I am not an automotive engineer. Not even close. But I would speculate that the reason for using lower pressures has to do with keeping the automobile braking system cheap and reliable. You do not need high pressures to make it effective.

 

[ROOT0X57B]

Before we try to ask "why", let us try to confirm that brake fluid pressure is actually "only 5 bar".
Brakes pressure : 5,000,000 Pa = 50 bar
Brakes torque : 45,744 Nm
Slip limit : 5,699 ($\mu Mg$)

There is the confirmation (it's linear about the pressure)

So there is only 0 - 5 bar of pressure in the brakes line?
It makes no sense with what you said :
Most of the metal brake lines burst around 15,000 psi. The typical full-lock operating pressures are 900–1,000psi (69 bar) with manual brakes and 1,400-plus psi (96 bar) with power-assisted brakes.

 

[jbriggs444]

There is the confirmation (it's linear about the pressure)

So there is only 0 - 5 bar of pressure in the brakes line?
It makes no sense with what you said :
Most of the metal brake lines burst around 15,000 psi. The typical full-lock operating pressures are 900–1,000psi (69 bar) with manual brakes and 1,400-plus psi (96 bar) with power-assisted brakes.

What is the source of the figure you are quoting for 0 - 5 bar of pressure in the brake line?

And exactly what do you think confirms what?

The pressure in the brake line will be related to the pressure of the pad on the disc by the fact that the total forces will match. So look at the surface area of the piston and compare it to the surface area of the pad. The area ratio is inverse to the pressure ratio.

 

[ROOT0X57B]

I calculate
$T_B = -2 * (P_BS_BR_B)/R_\text{ext}$
$\text{limit} = - \mu Mg$

The values I have given earlier are $||T_B||$ and $||\text{limit}||$ for $P_B = 50$bar

 

[jbriggs444]

View attachment 302387
I calculate
$T_B = -2 * (P_BS_BR_B)/R_\text{ext}$
$\text{limit} = - \mu Mg$

The values I have given earlier are $||T_B||$ and $||\text{limit}||$ for $P_B = 50$bar

What is the point of having posted this?

You seem to think that you have explained something here. Possibly something to do with pressure in the brake line. However, I see no mention of pressure in the brake line.

 

[ROOT0X57B]

There must be something I do not understand

I see no mention of pressure in the brake line.

?

 

[jbriggs444]

There must be something I do not understand

?

Definitions...

Brake line: The tubing that leads from the master cylinder to the brake cylinders.
Pressure in the brake line: [Gauge] pressure of the brake fluid in the brake line.

In the thread to this point, $P_B$ is the pressure between pad and disc, not the pressure between fluid and piston.

 

[ROOT0X57B]

Okay,
So the pressure in the brake line is not the one of the pad on the disc
the
There must be some "conversion" unit that makes the pressure on the pads so low

 

[jbriggs444]

Okay,
So the pressure in the brake line is not the one of the pad on the disc
the
There must be some "conversion" unit that makes the pressure on the pads so low

Please, please, post a complete calculation of the pressure on the pads that you think is "so low".

Complete in the sense that you explain where you obtained all of your inputs.
Complete in the sense that you provide all of the calculations using those inputs.
Complete in the sense that you provide the result of that calculation.

And then explain why you think that result is "so low".

Then we can proceed to find the error in the inputs, in the calculations or in the expectation.

 

[ROOT0X57B]

Please, please, post a complete calculation of the pressure on the pads that you think is "so low".

Complete in the sense that you explain where you obtained all of your inputs.
Complete in the sense that you provide all of the calculations using those inputs.
Complete in the sense that you provide the result of that calculation.

And then explain why you think that result is "so low".

Then we can proceed to find the error in the inputs, in the calculations or in the expectation.

I haven't done any calculation for the pressure on the pads as until now, I was mistaken and assumed the pressure in brake lines is the same as the pressure of the pads on the disc.

Now, I will have to calculate the pressure of the pads on the disc from the one in the brake line...

Sorry for the inconvenience again...

 

[ROOT0X57B]

Found online :

3) Clamping force: The clamping force of a caliper is the force exerted on the disc by the caliper pistons. Measured in pounds clamping force, it is the product of brake line pressure, in psi, multiplied by the total piston area of the caliper in square inches. This is true whether the caliper is of fixed or floating design. Increasing the pad area will not increase the clamping force.

I will call $P_{bl}$ the pressure in the brake line and $S_{bp}$ the surface area of the piston
Question 1 : isn't the area of the piston = the area of one pad ?

Then, I have $F_B = 2 * P_{bl} * S_{bp}$ I have to mention here that it's exactly what is was doing before but mistakenly
Question 2 : Have I made a mistake here ?

 

[jbriggs444]

Now, I will have to calculate the pressure of the pads on the disc from the one in the brake line...

Doing so will not bring you closer to a useful calculation.

Suppose that you take the pressure in the brake line as the input. What determines the pressure in the brake line?

The pressure in the brake line is determined by the force applied to the master cylinder and the area of the piston in the master cylinder.

Suppose that you take the force on the master cylinder as the input. What determines the force on the master cylinder?

The force on the master cylinder is determined by the force of the driver's foot on the brake pedal and the mechanical advantage provided by the pedal linkage (we can set power-assisted brakes to one side).

Suppose that you take the force of the driver's foot on the brake pedal as the input. What determines the force of the driver's foot on the pedal?

The force of the driver's foot on the pedal is determined by the driver's assessment of the vehicle's current speed, current deceleration rate and desired end state. He is part of a control loop.

The deceleration rate of a car is controlled by the driver within limits imposed by physics and engineering.

If we are trying to analyze a control loop then we should probably proceed by abstracting away all of the intermediate steps in the linkage (foot, pedal, master cylinder, brake cylinders, brake pads, wheels, tires) and think in terms of measurement precision and latency, controller latency, actuator precision and latency and physical limits.

[Welcome to the world of real-time programming where it is not just a matter of getting the right answer. It is a matter of getting an answer that is good enough in a time frame that is fast enough. A correct answer delivered too late is of no use].

I've never delved into the theory or practice of control systems beyond the basic ideas of positive and negative feedback, overcontrolling and under-controlling. I did see a presentation once where an astronaut decided to go for precise control of a landing of the Space Shuttle, just to see how close he could come to a defined mark. The interaction between the astronaut's over-aggressive control and the fly-by-wire electronics of the craft made for a nasty interaction that almost resulted in disaster.

Obviously, this sort of thing is well trodden ground in order to have self-driving cars.

 

[jbriggs444]

Question 1 : isn't the area of the piston = the area of one pad ?

No. The area of the piston is not equal to the area of the pad.

 

[ROOT0X57B]

For this force, I only want to take an approximation of reality, it's not the main point of my study.

I would be happy with a bare approximation (if online data says 60 to 100 bar I would be quite happy with 80)

It's just that 5 bars is not quite the same as 50 bars

 

[jbriggs444]

For this force, I only want to take an approximation of reality, it's not the main point of my study.

I would be happy with a bare approximation (if online data says 60 to 100 bar I would be quite happy with 80)

It's just that 5 bars is not quite the same as 50 bars

The key is that this is enough pressure to lock up the wheels.

Which means that the actual pressure is irrelevant. It is the tires on the road that are the limit to deceleration rate.

More generally, braking acceleration is the minimum of the amount commanded by the driver and the amount allowed by the tires. All of the details in between (pedal force, brake line pressure, brake pad pressure, braking torque, etc) are essentially irrelevant. Only those two factors come into it.

Edit: absent a source for the "5 bars" figure, there is nothing to explain.

 

[jack action]

Brakes pressure : 5,000,000 Pa = 50 bar
Brakes torque : 45,744 Nm
Slip limit : 5,699 ($\mu Mg$)

Let's start at the beginning. How did you get these numbers? Show us your calculations. Show us your sources for your input numbers. What is "Slip limit"? If it is $\mu Mg$, then there should be a unit of force coming with it. And I'm not even sure if you are always using the comma in your numbers as a decimal or a thousands separator.

I never heard about someone needing the "brake pad pressure". Nobody cares about those values, except brake pad manufacturers. Even brake caliper manufacturers will care more about brake pad pressure distribution than the pressure itself.

But you seem to want to determine the resulting brake torque at the wheels resulting from external forces acting on the vehicle. So why are you using the brake system as a starting point? You only need to find the brake torque needed by the vehicle (i.e. the tire friction force resulting from all the external forces times the tire radius). The torque coming from the braking system will necessarily be equal to this value, no matter how it is produced.

So are you looking at the effect of different forces acting on a vehicle as a whole, or are looking to design a braking system that will provide a given torque, no matter what it is used for?

 

[ROOT0X57B]

Let's start at the beginning. How did you get these numbers? Show us your calculations. Show us your sources for your input numbers.

These numbers are found by that (in loop):

couple_freins = -2 * (pression_freins * surface_plaquette) * (rayon_frein / rayon_pneu)

limite_glissement = -mu_sol * masse * g

if abs(couple_freins) > abs(limite_glissement):
    F_eq = limite_glissement
else:
    F_eq = couple_freins
 
M_eq = masse + n_roues_actives * moment_inertie_roue / rayon_pneu**2
 
a = F_eq / M_eq
v = v + a * dt
x = x + v * dt

I was just printing the absolute values of pression_freins, couple_freins, and limite_glissement.

The parameters used are

masse  = 830 # Masse de la voiture
v0 = kmh_vers_mps(50) # Vitesse initiale
rayon_pneu, rayon_interieur, largeur_bande = ref_roue_vers_dimensions('155-65R14')
rayon_frein = rayon_interieur * 0.8 # Approximation de la distance par rapport à l'axe où la pression est appliquée
n_roues_actives = 2 # Nombre de roues actives

mu_sol = 0.7 # Coefficient de friction entre la roue et le sol

pression_freins = bar_to_pascal(50) # Pression de freinage
surface_plaquette = 0.0057390 # Surface d'une plaquette de frein

masse_roue = 6 # Masse de la roue
moment_inertie_roue = (masse_roue * (rayon_pneu**2 - rayon_interieur**2)) / 2 # Moment d'inertie de la roue

t_reaction = 1 # Temps de réaction du conducteur

x = 0 # Position initiale
v = v0 # Vitesse initiale
a = 0 # Acceleration initiale
t = 0 # Temps initial

# Il faut un pas de temps pour que dv << v
dt = 0.001

For the most important ones, here are my sources :
$\mu$ : Wikibooks :

If I understand well, kinetic friction = sliding

$M, S_B, R_B, R_\text{ext}...$ : car/pieces manufacturer websites, the car is a citroen C1 from 2008, the wheel reference is
155-65R14
https://www.lacentrale.fr/fiche-technique-voiture-citroen-c1-(2)+1.0+68+attraction+3p-2008.html
Here for the brakes

$P_B$ pressure : see my previous post

I don't see any error from me in here
Maybe one but this is also my main question :
If the brakes have no impact on the braking distance because they make the tire slip, then what's the point on having brakes at all ?

 

[ROOT0X57B]

So are you looking at the effect of different forces acting on a vehicle as a whole, or are looking to design a braking system that will provide a given torque, no matter what it is used for?

I don't want to design any braking system, I only want to quickly model how the brakes make the car running decelerate, given as few inputs as possible.

EDIT : I am realizing just now that, because I'm in an emergency braking case, the brakes are supposed to be at their maximum capacities, so the car must slip

 

[ROOT0X57B]

I think I will stop this topic after your answers about that, try to refocus, and refactorize my project a bit.
I may come back in a new thread later for other questions, but I think we have already been too far from the initial subject of this thread.

 

[jbriggs444]

Maybe one but this is also my main question :
If the brakes have no impact on the braking distance because they make the tire slip, then what's the point on having brakes at all ?

You claim to have calculated a figure of 5 bar. But you refuse to show a calculation that yields 5 bar.

One uses brakes to bring a car to a stop in a controlled fashion with as little or as much deceleration as the driver desires.