Force of brake pads on a wheel

[jbriggs444]

I don't want to solve a differential equation, I want to simulate it with programming :
I will set $v_0$ and the others parameters and will calculate $v$ and $x$ at each point by increments of time small enough so that $\text{d}v$ doesn't change a lot, this will give me a plot of distance traveled over time.

Fair enough. But my mind boggles at the length of the equations you are producing when your main loop should be something simple like:

Vanilla Euler:
10 x = x + v * delta_t
20 v = v + a * delta_t
30 t = t + delta_t
40 print t, v, x
50 go to 10

More accurate:
10 x = x + v * delta_t + 1/2 * a * delta_t^2
20 v = v + a * delta_t
30 t = t + delta_t
40 print t, v, x
50 go to 10

If $a$ is the key input that you need for your main loop then $a$ is the constant that you should be using your equations to solve for. It is dead easy to solve for. $a=\frac{F_e}{M_e}$. It falls straight out of Newton's second law.

 

[ROOT0X57B]

Yep, but as sadly it's mandatory for me to explain and detail everything, also I could be questioned about anything I might talk about

 

[ROOT0X57B]

Ok, I worked a bit but I'm stuck with a little problem now

I have $$\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)v_0 \text{d}v + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$
Let $M_{eq} = \displaystyle\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)$ (equivalent mass)
Let $E_B = 2F_BR_B$ (energy equivalent for braking)

So I have $$M_{eq} v_0 \text{d}v = E_B \theta$$

My code loop is like
10 t = t+dt
20 theta = theta + d_theta ?
30 dv = some_calculus_here
40 v = v + dv
50 x = x + v * dt

How can I manage to express $\theta$ with $\text{d}v$ so that I no longer have to handle it?

I know $\displaystyle v = {R_\text{ext}} \frac{\text{d}\theta}{\text{d}t}$
$v = v_0 + \text{d}v$

But I'm stuck here because I don't know how to calculate $\text{d}\theta$ in function of $\text{d}v$ as $\text{d}v$ depends of $\text{d}\theta$ too

 

[erobz]

Ok, I worked a bit but I'm stuck with a little problem now

I have $$\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)v_0 \text{d}v + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$
Let $M_{eq} = \displaystyle\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)$ (equivalent mass)
Let $E_B = 2F_BR_B$ (energy equivalent for braking)

So I have $$M_{eq} v_0 \text{d}v = E_B \theta$$

My code loop is like
10 t = t+dt
20 theta = theta + d_theta ?
30 dv = some_calculus_here
40 v = v + dv
50 x = x + v * dt

How can I manage to express $\theta$ with $\text{d}v$ so that I no longer have to handle it?

I know $\displaystyle v = {R_\text{ext}} \frac{\text{d}\theta}{\text{d}t}$
$v = v_0 + \text{d}v$

But I'm stuck here because I don't know how to calculate $\text{d}\theta$ in function of $\text{d}v$ as $\text{d}v$ depends of $\text{d}\theta$ too

It appears like you didn't let the angel change in your linearization. You introduced a small change in $v$ but you did not introduce a small change in $\theta$

Okay, as I want to be able to simulate the evolution of the distance over time, I will try to figure out $\text{d}v$

I worked by myself using all your stuff and a bit with my teacher who validated the main part

I use french physics notations, but it may not be a big deal ($J_\Delta = I$ for you and I think that's it)

I stopped here last time :
$$\displaystyle \frac{1}{2}M({v_0}^2-v^2) + 4 * \frac{1}{2}J_\Delta({\omega_0}^2-\omega^2) = 2F_BR_B \theta$$

As $v = v_0 + \text{d}v$ (a small step of speed):
$$\displaystyle \frac{1}{2}M((v+v_0)(v-v_0)) + 4 * \frac{1}{2}J_\Delta({\omega_0}^2-\omega^2) = 2F_BR_B \theta$$
$$\displaystyle \frac{1}{2}M((2v_0+\text{d}v)(\text{d}v)) + 4 * \frac{1}{2}J_\Delta({\omega_0}^2-\omega^2) = 2F_BR_B \theta$$
$$\displaystyle Mv_0\text{d}v + \text O(\text{d}v^2) + 4 * \frac{1}{2}J_\Delta({\omega_0}^2-\omega^2) = 2F_BR_B \theta$$

Because $\displaystyle\omega = \frac{v}{R_\text{ext}}$ :
$$\displaystyle Mv_0\text{d}v + 4 * \frac{1}{2}J_\Delta\bigg(\frac{{v_0}^2}{{R_\text{ext}}^2}-\frac{{v}^2}{{R_\text{ext}}^2}\bigg) + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$
$$\displaystyle Mv_0\text{d}v + 4 * \frac{1}{2}J_\Delta\bigg(\frac{1}{{R_\text{ext}}^2}\bigg({v_0}^2-v^2 \bigg)\bigg) + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$
$$\displaystyle Mv_0\text{d}v + 4 J_\Delta\bigg(\frac{v_0 \text{d}v}{{R_\text{ext}}^2}\bigg) + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$
$$\displaystyle\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)v_0 \text{d}v + \text{O}(\text{d}v^2) = 2F_BR_B \theta$$

---

From here, I have not shown my physics teacher

$\displaystyle v = {R_\text{ext}} \frac{\text{d}\theta}{\text{d}t}$ so $\displaystyle \text{d}v = {R_\text{ext}}\frac{\text{d}\theta}{\text{d}t}$ so $\displaystyle\text{d}\theta = \frac{\text{d}v\text{d}t}{R_\text{ext}}$

As $\theta = \theta_0 + \text{d}\theta$ (a small step of rotation) :
$$\displaystyle\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)v_0 \text{d}v + \text{O}(\text{d}v^2) = 2F_BR_B\bigg(\theta_0 + \frac{\text{d}v \text{d}t}{R}\bigg)$$
$$\displaystyle\bigg(M + 4\frac{J_\Delta}{{R_\text{ext}}^2}\bigg)v_0 \text{d}v + \text{O}(\text{d}v^2) = 2F_BR_B\frac{\text{d}v}{R}\text{d}t + 2F_BR_B \theta_0$$

So now with a little bit of python programming, I can simulate the distance traveled over time by little incrementations of time $\text{d}t$

 

[ROOT0X57B]

Yep, I can't figure out how to get things to work together

 

[erobz]

Yep, I can't figure out how to get things to work together

Well, the RHS linearizes as:

$$ 2 F_b R_b d \theta $$

and

$$ v = R_{ext} \frac{d \theta}{dt} = R_{ext} \frac{d \theta}{dx} \frac{dx}{dt} = R_{ext} \frac{d \theta}{dx} v $$

This implies:

$$ d \theta = \frac{1}{R_{ext}} dx $$

 

[ROOT0X57B]

Thanks !

Now I have
$$\displaystyle M_{eq}v_0 \text{d}v = 2F_B \frac{R_B}{R_{ext}}dx$$
But I need $\text{d}v$ to calculate $\text{d}x$ and $\text{d}x$ to calculate $\text{d}v$
I am probably missing something...

 

[erobz]

Thanks !

Now I have
$$\displaystyle M_{eq}v_0 \text{d}v = 2F_B \frac{R_B}{R_{ext}}dx$$
But I need $\text{d}v$ to calculate $\text{d}x$ and $\text{d}x$ to calculate $\text{d}v$
I am probably missing something...

$$ dv = a \, dt $$

 

[ROOT0X57B]

Newton's Second Law ?

 

[erobz]

Newton's Second Law ?

I think @jbriggs444 mentioned:

$$ a = \frac{F_e}{M_e} $$

And you were already intending to use that. I'm getting mixed up on what approach you want to take.

As it stands you have something like:

$$ \frac{dv}{dx} = K $$

What are you after? $x(t)$ ?

 

[ROOT0X57B]

I want to calculate $\text{d}v$ for each $\text{d}t$ so that I can follow to evolution of $v$ and $x$ over time (I don't care about $\theta$)

My goal is to use a code such as below
(For your understanding, "frein" = "brake", "plaquette" = "brake pad", "roue" = "wheel", "pneu" = "tire")

Do not take care of the expression in dv = ..., it's messed up

When I run the code, I get this kind of graph (this one is clearly wrong) :

 

[jbriggs444]

I am not going to try to explode that screen image of code to be able to read it. Try pasting it in a [code] block here. For example:

t = t + dt

 

[ROOT0X57B]

okay, did not really know how to do it...

 

[jbriggs444]

okay, did not really know how to do it...

Sorry, I was just about to edit the instructions into my previous post.

Start by turning off BB code rendering. You can click on the "[ ]" icon above your editing window to do so. That will let you see the "BB codes" in your post. Try it with this post

t = t + dt

You should see...
[CODE]
t + dt
[/CODE]

You can also go to Info => Help => BB Codes (https://www.physicsforums.com/help/bb-codes/) for more information than you may care for.

 

[ROOT0X57B]

I want to calculate $\text{d}v$ for each $\text{d}t$ so that I can follow to evolution of $v$ and $x$ over time (I don't care about $\theta$)

My goal is to use a code such as below
(For your understanding, "frein" = "brake", "plaquette" = "brake pad", "roue" = "wheel", "pneu" = "tire")

Do not take care of the expression in dv = ..., it's messed up
View attachment 302283

When I run the code, I get this kind of graph (this one is clearly wrong) :
View attachment 302284

EDIT: it is maybe not that wrong!

I compared the result (stop from 50km/h in 0.5s in 6 meters) with the official numbers : 14 meters
BUUUT, I did not consider the reaction time (which they do)
If I add to the result about 0.8 seconds of driving at 50 km/h (reaction delay is 1s, I take 0.8 to get the time there is little braking), I get 15 meters.
Not that far...

Anyway, here is the code in the image :

# Simulation de la distance de freinage d'une voiture en ville
# Aubin SIONVILLE
# TIPE 2023

# Cette simulation utilise les données d'une Citroën C1 de 2008
# Données disponibles à
# https://www.lacentrale.fr/fiche-technique-voiture-citroen-c1-(2)+1.0+68+confort+3p-2008.html

from utils import *
from matplotlib import pyplot as plt# Définition des constantes
masse  = 830 # Masse de la voiture
v0 = kmh_vers_mps(50) # Vitesse initiale : 50 km/h
rayon_pneu, rayon_interieur, largeur_bande = ref_roue_vers_dimensions('155-65R14')
rayon_frein = rayon_interieur * 0.8 # Approximation de R_B

pression_freins = bar_to_pascal(100) # Pression de freinage
surface_plaquette = 0.0057390 # Surface d'une plaquette de frein : 5739 mm² = 0.005739 m²

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

x = 0 # Position initiale
v = v0 # Vitesse initiale
t = 0 # Temps initial
theta = 0 # Angle initial

# Il faut un pas de temps pour que dv << v
dt = 0.0001

# On utilise la formule
# (masse + 4 * moment_inertie_roue / rayon_pneu**2) * v0dv = 2(pressions_freins * 2 *surface_plaquette)rayon_frein * theta

points_vitesse = [v0]
points_distance = [0]
points_temps = [0]

while v > limite_zero:
    t = t + dt
    theta = theta + x / rayon_pneu * dt
    dv = dt * (pression_freins * 2 * surface_plaquette * rayon_frein * theta) / (masse + 4 * moment_inertie_roue / rayon_pneu**2)
    v = v - dv
    x = x + v * dt

    points_distance.append(x)
    points_vitesse.append(v)
    points_temps.append(t)

points_vitesse = [mps_vers_kmh(v) for v in points_vitesse]

fig, ax = plt.subplots()
ax.plot(points_temps, points_distance, color='b')
ax.set_xlabel('Temps (s)')
ax.set_ylabel('Distance (m)', color='b')
ax.tick_params('y', colors='b')

ax2 = ax.twinx()
ax2.plot(points_temps, points_vitesse, 'r')
ax2.set_ylabel('Vitesse (km/h)', color='r')
ax2.tick_params('y', colors='r')

plt.title('Simulation de la distance de freinage d\'une voiture en ville')
plt.show()

 

[jbriggs444]

What I see that is relevant in that code segment is:

while v > limite_zero:
    t = t + dt
    theta = theta + x / rayon_pneu * dt
    dv = dt * (pression_freins * 2 * surface_plaquette * rayon_frein * theta) / (masse + 4 * moment_inertie_roue / rayon_pneu**2)
    v = v - dv
    x = x + v * dt

So you are counting up $t$ at an even rate. That seems fine.
But you are counting up theta at an accelerating rate. That seems broken. You are multiplying incremental time by current position instead of multiplying incremental time by current velocity.

You have also not exported the constant sub-expression outside of the main loop. But that is just a performance thing.

Wait a minute -- you have acceleration depending on theta. That's flat wrong.

 

[ROOT0X57B]

Yes, I told you everything in this part is messed up
I am currently sick, and everything in those two lines in actually messed upNow I just want to get an expression of $\text{d}v$ in which I get rid of $\theta$ so that I can use it in this piece of code

 

[jbriggs444]

Now I just want to get an expression of $\text{d}v$ in which I get rid of $\theta$ so that I can use it in this piece of code

dv = a * dt

 

[ROOT0X57B]

dv = a * dt

Yes...but I don't have any acceleration expression here
Do you mean I should use Newton's second law ?

 

[jbriggs444]

Yes...but I don't have any acceleration expression here
Do you mean I should use Newton's second law ?

I'd already given you $a$ previously$$a = \frac{F_e}{M_e}$$I'd already given you $F_e$ and $M_e$ previously. Both are constants that can be calculated from the givens of the problem.

 

[ROOT0X57B]

Thanks again!

Yes you have given me
$\displaystyle F_e = n_\text{wheels}F_B \frac{R_B}{R_\text{ext}}$
$\displaystyle M_e = n_\text{wheels}m_\text{wheels}/2$

If I understand well $n_\text{wheels}$ is the number of wheels involved

If so, is your $M_e$ equal to my $M_eq$ ?
$\displaystyle M_e = n_\text{wheels}m_\text{wheels}/2$
$\displaystyle M_{eq} = M + 4 \frac{I}{{R_{ext}}^2}$

At least your $F_e$ is equal to my $E_B$ (same expression)

Your post feels like you're a little bit upset, please excuse my dumbness (probably due to my illness thought)

 

[jbriggs444]

Thanks again!

Yes you have given me
$\displaystyle F_e = n_\text{wheels}F_B \frac{R_B}{R_\text{ext}}$
$\displaystyle M_e = n_\text{wheels}m_\text{wheels}/2$

One moment while I scroll back to check. The formula for $M_e$ above would count just the wheels and not the car. Surely I wouldn't have said that. [Scrolls back to check...]

As I'd expected, what I'd written was:

If we use $M_e$ (equivalent mass) as shorthand for $M+n_\text{wheels}m_\text{wheel}/2$ then...

If I understand well $n_\text{wheels}$ is the number of wheels involved

If so, is your $M_e$ equal to my $M_eq$ ?
$\displaystyle M_{eq} = M + 4 \frac{I}{{R_{ext}}^2}$

Yes indeed, it should work out to be the same formula. There is a factor of $\frac{1}{2}$ in the formula for moment of inertia that drops your 4 into a 2 and then the $R_\text{ext}$ cancels out entirely.

Your post feels like you're a little bit upset, please excuse my dumbness (probably due to my illness thought)

I apologize for any frustration that has leaked through into my language.

 

[ROOT0X57B]

One moment while I scroll back to check. The formula for $M_e$ above would count just the wheels and not the car. Surely I wouldn't have said that. [Scrolls back to check...]

Yep, copy mistake

 

[ROOT0X57B]

It works !

Here is the result :

If we take into account the delay of reaction (1s) :

Looks pretty accurate to me, (considering constant deceleration)

If anyone wants, here is the code (if french but well commented) :

from utils import *
from matplotlib import pyplot as plt
from parameters import *

points_vitesse = [v0]
points_distance = [0]
points_temps = [0]

compter_temps_reaction = True # Si on compte le temps de réaction

if compter_temps_reaction:
    ## Temps de réaction
    for i in range(int(t_reaction / dt)):
        t += dt
        x += v0 * dt
        points_distance.append(x)
        points_vitesse.append(v0)
        points_temps.append(t)

while v > limite_zero:
    t = t + dt
    M_eq = masse + n_roues_actives * moment_inertie_roue / rayon_pneu**2
    F_eq = 2 * (pression_freins * 2 * surface_plaquette) * rayon_frein / rayon_pneu
    a = F_eq / M_eq
    v = v - a * dt
    x = x + v * dt

    points_distance.append(x)
    points_vitesse.append(v)
    points_temps.append(t)

# On convertit les données en km/h
points_vitesse = [mps_vers_kmh(v) for v in points_vitesse]

# On affiche les données sur le même graphique
fig, ax = plt.subplots()
ax.plot(points_temps, points_distance, color='b')
ax.set_xlabel('Temps (s)')
ax.set_ylabel('Distance (m)', color='b')
ax.tick_params('y', colors='b')

ax2 = ax.twinx()
ax2.plot(points_temps, points_vitesse, 'r')
ax2.set_ylabel('Vitesse (km/h)', color='r')
ax2.tick_params('y', colors='r')

plt.title('Simulation de la distance de freinage d\'une voiture en ville')
plt.show()

and here is are the parameters used :

from utils import *

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

pression_freins = bar_to_pascal(100) # 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 = 0.0001

 

[ROOT0X57B]

Hum wait...

Something feels wrong
Where does $\mu$ (friction coefficient between road and tire) gets in ??
I think $F_e = n_\text{wheels}\mu F_BR_B$ instead of $n_\text{wheels}F_BR_B$

If I do that I have

And

Official data says a stop at 50km/h is done in about 28 meters
As I have no transition between "no brakes" and "full brakes" and I have a powerful force (emergency braking) it seems quite realistic

 

[jbriggs444]

Where does $\mu$ (friction coefficient between road and tire) gets in ??

It doesn't.

As long as you are in a controlled stop (not skidding), it is the force of the drivers foot on the pedal that controls the force of the brake pads on the rotors. The coefficient of friction of pad on rotor is relevant. The coefficient of friction of tire on road is not immediately relevant.

The coefficient of friction of tire on road is relevant as a limit. Brake harder than that and you start skidding. Sliding rather than rolling.

If you are sliding, the force of the driver's foot on the pedal, and the force of the pad on the rotor become irrelevant. No matter how hard the driver mashes the pedal, the car still slides.

The deceleration during braking is the minimum of the two -- the minimum of the braking commanded by the driver and the braking allowed by the tires on the road.

Edit: In some cases -- the braking system may be inadequate to cause the tires to skid. In those cases, the effectiveness of the braking system may impose a different constraint to minimize with. For instance, with drum brakes and "fade".

 

[ROOT0X57B]

For the next chapter, I will need the car to slide on the road so it may be logical to consider it here too.
How can I do that?
I suppose there will be another force that will get into account for "sliding friction" which will depend on $\mu$

 

[jbriggs444]

For the next chapter, I will need the car to slide on the road so it may be logical to consider it here too.
How can I do that?
I suppose there will be another force that will get into account for "sliding friction" which will depend on $\mu$

You have encountered the notion of the coefficient of static friction and of kinetic friction and the fact that the two can differ?

 

[ROOT0X57B]

No we have not seen the difference, we did simply call it friction for now

 

[jack action]

Hum wait...

Something feels wrong
Where does $\mu$ (friction coefficient between road and tire) gets in ??
I think $F_e = n_\text{wheels}\mu F_BR_B$ instead of $n_\text{wheels}F_BR_B$

The stop in 0.06s feels suspect too

Your equation doesn't assume there is a friction force anywhere.

Your equation doesn't assume that $F_B$ is a friction force; it is just a force. It could be a friction force if you simply define $F_B = \mu_{pad} F_N$ where $F_N$ would be the normal force applied to a brake pad.

But the true definition of that friction force depends on the brake system design and can be simple (disc) or very complex (duo-servo drum).

Furthermore, your model assumes that whatever the braking system will produce, the tire will accommodate. This is true until you reach the maximum friction force your tire can handle. To incorporate this you will have to constantly monitor if your braking force exceeds that limit. Read post ##99 again.

 

[ROOT0X57B]

Your equation doesn't assume there is a friction force anywhere.

Your equation doesn't assume that $F_B$ is a friction force; it is just a force. It could be a friction force if you simply define $F_B = \mu_{pad} F_N$ where $F_N$ would be the normal force applied to a brake pad.

But the true definition of that friction force depends on the brake system design and can be simple (disc) or very complex (duo-servo drum).

Furthermore, your model assumes that whatever the braking system will produce, the tire will accommodate. This is true until you reach the maximum friction force your tire can handle. To incorporate this you will have to constantly monitor if your braking force exceeds that limit. Read post ##99 again.

Yes I acknowledged this, that's why I'm asking actually
My brake system is simple (drum braking)

If my force exceeds the slip limit what do I do ? I make it slip by not decreasing the speed ? If so for how long ?

 

[jack action]

My brake system is simple (drum braking)

Drum brakes are not simple.

If my force exceeds the slip limit what do I do ? I make it slip by not decreasing the speed ? If so for how long ?

You replace the torque produced by your braking system with the tire friction force. Again, from post #99, $F_t = MIN(\frac{T_B}{r_t}, \mu mg)$.

 

[ROOT0X57B]

Drum brakes are not simple.

I meant there is no more complexity than what was there before.

 

[jbriggs444]

If my force exceeds the slip limit what do I do ? I make it slip by not decreasing the speed ? If so for how long ?

Are you contemplating a fixed constant braking force and a fixed constant limit on tire friction?

If so, you just check whether the one exceeds the other and have your model only consider the one that is limiting.Are you contemplating a variable braking force and variable road friction? For instance, are you braking to a stop and hit a patch of ice?

Then you have to use a model that incorporates both constraints. For instance, while the car is on the icy patch, the rotation rate of the tires will be decreasing rapidly. They may even come to a stop. When you regain contact with the road surface, the tires will spin back up to speed. By my count, you will have four regimes to think about:

1. Braking to a stop with no-slip rolling, limited by brake pressure.
2. Sliding on ice with the wheels slowing rapidly, limited by sliding friction on ice.
3. Sliding on ice with the wheels not turning, still limited by sliding friction on ice.
4. Sliding on road with the wheels speeding up, limited by sliding friction on pavement.

 

[ROOT0X57B]

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.