How to Calculate the Braking Momentum on a Wheel?

[ROOT0X57B]

Okay, so if I got it all good :

$$\displaystyle \frac{1}{2}M{v_0}^2 + m{R_\text{ext}}^2{\omega_0}^2 = \frac{1}{2}Mv^2 + m{R_\text{ext}}^2\omega^2 - 2\int F_Bdx$$

$dx = R_\text{ext}d \theta$ so

$$\displaystyle \frac{1}{2}M{v_0}^2 + m{R_\text{ext}}^2{\omega_0}^2 = \frac{1}{2}Mv^2 + m{R_\text{ext}}^2\omega^2 - 2\int \mu_BP_BS_B.R_\text{ext}d \theta$$

$\theta_0 = 0$, so

$$\displaystyle \frac{1}{2}M{v_0}^2 + m{R_\text{ext}}^2{\omega_0}^2 = \frac{1}{2}Mv^2 + m{R_\text{ext}}^2\omega^2 - 2 \mu_BP_BS_BR_\text{ext} \theta$$

$v = R_\text{ext}\omega$, so

$$\displaystyle \frac{1}{2}M{v_0}^2 + {mv_0}^2 = \frac{1}{2}Mv^2 + mv^2 - 2 \mu_BP_BS_BR_\text{ext} \theta$$

$$\displaystyle \bigg(\frac{1}{2}M+m\bigg){v_0}^2 = \bigg(\frac{1}{2}M+m\bigg)v^2- 2 \mu_BP_BS_BR_\text{ext} \theta$$

$$\displaystyle \bigg(\frac{1}{2}M+m\bigg){v}^2 = \bigg(\frac{1}{2}M+m\bigg){v_0}^2 + 2 \mu_BP_BS_BR_\text{ext} \theta$$

$$\displaystyle v^2 = v_0^2 + \frac{4 \mu_BP_BS_BR_\text{ext}}{M+2m}\theta$$

As $v = R_\text{ext}\omega$, $x = R_\text{ext}\theta$ ($R_\text{ext}$ is constant)

$$\displaystyle {\dot x}^2 = {{\dot x}_0}^2 + \frac{4 \mu_BP_BS_B}{M+2m}x$$

So I have to solve the linear differential equation :

$$\displaystyle {\dot x}^2 - \frac{4 \mu_BP_BS_B}{M+2m}x = {v_0}^2$$

 

[erobz]

Okay, so if I got it all good :

error in the integral defining the work done by the brake in what you wrote:

$$\displaystyle \frac{1}{2}M{v_0}^2 + m{R_\text{ext}}^2{\omega_0}^2 = \frac{1}{2}Mv^2 + m{R_\text{ext}}^2\omega^2 - 2\int \mu_BP_BS_B.R_\text{ext}d \theta$$

error below in consequences of Dot Product of $F_b$ acting with its displacement. Another thing, do yourself ( and me ) a favor for now and just call the moment of inertia of the wheel $I$ and rework the algebra for the general case.

$$\displaystyle \bigg(\frac{1}{2}M+m\bigg){v_0}^2 = \bigg(\frac{1}{2}M+m\bigg)v^2- 2 \mu_BP_BS_BR_\text{ext} \theta$$

There are errors that floated down stream. I was going to ask you to do this...later The question I pose to you is it necessary to solve the ODE for $x(t)$ to find the stopping distance?

$$\displaystyle {\dot x}^2 - \frac{4 \mu_BP_BS_B}{M+2m}x = {v_0}^2$$

 

[jbriggs444]

$$\frac{1}{2}Mv^2 + mv^2$$

Recognize that this is the same as $\frac{1}{2}(M+2m)v^2$

$$\displaystyle \frac{4 \mu_BP_BS_BR_\text{ext}}{M+2m}\theta$$

Recognize that this is the same as $k(x-x_0)$ for some $k$

Which means that you are looking at something very simple. The equivalent of a brick of mass $M+2m$ sliding on a surface with a particular coefficient of kinetic friction that you can calculate.

 

[ROOT0X57B]

Recognize that this is the same as $\frac{1}{2}(M+2m)v^2$

Recognize that this is the same as $k(x-x_0)$ for some $k$

Which means that you are looking at something very simple. The equivalent of a brick of mass $M+2m$ sliding on a surface with a particular coefficient of kinetic friction that you can calculate.

I was going like that because it looks like my school books equations, I had the same idea as you do for the mass, and $-k(x-x_0)$ looks like a damping force
But according to @erobz :

error below in consequences of Dot Product of $F_b$ acting with its displacement.

Didn't understand it was a dot product because nothing was written as a vector
I will suppose $F_B$ and $d\theta$ to be vectors

$F_b = \mu_b P A$

I used this to make the integral

The question I pose to you is it necessary to solve the ODE for $x(t)$ to find the stopping distance?

The ODE solution would give me $x(t)$ for every $t$ but I only need the stopping distance actually

So with a little rework :

$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 - 2\int F_B \cdot dx$$
Because $F_B$ acts in the opposite direction of $d\theta$, $F_B \cdot R_\text{ext}d\theta = -F_BR_\text{ext}$

So I have now
$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 - 2\int -F_BR_\text{ext}$$

Problem is, I don't have an "integrator" (don't know how to call it) in my integral because $d\theta$ is gone
If I integrate anyway by $\theta$ :
$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 + 2F_BR_\text{ext}\theta$$

I feel so wrong and so sorry for bothering you and taking your time because of my incompetence...

 

[erobz]

I was going like that because it looks like my school books equations, but according to @erobz :Didn't understand it was a dot product because nothing was written as a vector
I will suppose $F_B$ and $d\theta$ to be vectorsI used this to make the integralThe ODE solution would give me $x(t)$ for every $t$ but I only need the stopping distance actually

So with a little rework :

$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 - 2\int F_B \cdot dx$$
Because $F_B$ acts in the opposite direction of $d\theta$, $F_B \cdot R_\text{ext}d\theta = -F_BR_\text{ext}$

So I have now
$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 - 2\int -F_BR_\text{ext}$$

Problem is, I don't have an "integrator" (don't know how to call it) in my integral because $d\theta$ is gone
If I integrate anyway by $\theta$ :
$$\displaystyle \frac{1}{2}M{v_0}^2 + I{\omega_0}^2 = \frac{1}{2}Mv^2 + I\omega^2 + 2F_BR_\text{ext}\theta$$

I feel so wrong and so sorry for bothering you and taking your time because of my incompetence...

Slow down! You're doing fine.

 

[jbriggs444]

Didn't understand it was a dot product because nothing was written as a vector
I will suppose $F_B$ and $d\theta$ to be vectors

It is not that complicated. It is simple if attacked properly.

But you need to be clear on your notation.

What is $F_B$? What is $F_b$? Are they the same or different?
What is $S_B$?

I assume that $F_B$ is the frictional force of brake pad on rotor/drum. If so, why are you multiplying that force by the distance traversed by the tire treads rather than the distance traversed by the brake rotor/drum?

 

[erobz]

What displacement does $F_b$ act over?

 

[hutchphd]

feel so wrong and so sorry for bothering you and taking your time because of my incompetence...

My PhD advisor once told me I did physics calculations by successive approximation. (We actually got along very well!)
If you are not making mistakes you are not trying hard enough. You do need to develop the habit of rechecking everything all the time. So not to worry.

 

[ROOT0X57B]

It is not that complicated. It is simple if attacked properly.

But you need to be clear on your notation.

What is $F_B$? What is $F_b$? Are they the same or different?
What is $S_B$?

Okay I'll redefine things a little bit
$F_B$ is the frictional force of the brake pad on the rotor
I use no $F_b$
$S_B$ is the surface of a brake pad

why are you multiplying that force by the distance traversed by the tire treads rather than the distance traversed by the brake rotor/drum?

So the $R_\text{ext}$ in my integral is actually $R_B$ ?

 

[ROOT0X57B]

What displacement does $F_b$ act over?

The rotation of the wheel

 

[ROOT0X57B]

Slow down! You're doing fine.

I just saw a little mistake in my equations, $I$ is the inertia of one wheel so I have to multiply it by 4
Moreover, the actual equation you gave me before is
$$\displaystyle \frac{1}{2}M{v_0}^2 + 4*\frac{1}{2}I{\omega_0}^2 = \frac{1}{2}Mv^2 + 4*\frac{1}{2}I\omega^2 - 2*\int F_Bdx$$
So have actually have
$$\displaystyle \frac{1}{2}M{v_0}^2 + 2I{\omega_0}^2 = \frac{1}{2}Mv^2 + 2I\omega^2 +2F_BR_B\theta$$

 

[erobz]

The rotation of the wheel

Think about what part of the wheel the force is acting on. Specifically, what location. And think about how much distance it will cover vs the wheel given some angle of rotation.

Other than that you have fixed the error I was referring to with the dot product. If you are getting flustered, give yourself 15 min, take a walk. Sometimes it helps, sometimes it doesn't, but with the exception of an exam...it almost never hurts!

 

[erobz]

I just saw a little mistake in my equations, $I$ is the inertia of one wheel so I have to multiply it by 4
Moreover, the actual equation you gave me before is
$$\displaystyle \frac{1}{2}M{v_0}^2 + 4*\frac{1}{2}I{\omega_0}^2 = \frac{1}{2}Mv^2 + 4*\frac{1}{2}I\omega^2 - 2*\int F_Bdx$$
So have actually have
$$\displaystyle \frac{1}{2}M{v_0}^2 + 2I{\omega_0}^2 = \frac{1}{2}Mv^2 + 2I\omega^2 +2F_BR_B\theta$$

This is looking much better. Good job!

 

[ROOT0X57B]

Other than you have fixed the error I was referring to with the dot product. If you are getting flustered, give yourself 15 min, take a walk. Sometimes it helps, sometimes it doesn't, but with the exception of an exam...it almost never hurts!

I am a little bit stressed, this project is important but is for next year. Next week will be a full week of mock exams, this explains that...

 

[erobz]

When you get back from your break, just put in your force $F_B$ in terms of the other parameters. Change from $\theta$ ( the angel the wheel rotates) to $x$ ( the displacement of the vehicle ) and be sure you revisit what allows you to do that in the assumptions of this model.

 

[jbriggs444]

Okay I'll redefine things a little bit
$F_B$ is the frictional force of the brake pad on the rotor
I use no $F_b$
$S_B$ is the surface of a brake pad

Ahhh, I see. Personally, I'd never have used $P_B$ and $S_B$ since neither is particularly important. Only their product is relevant -- the normal force of brake pad on rotor/disk. But that force is already determined by the pressure in the hydraulic fluid, the area of the brake piston and the mechanical advantage (if any) of the arrangement that transmits that force to the brake pad.

So the $R_\text{ext}$ in my integral is actually $R_B$ ?

Right. But since you want to be able to express everything in terms of $x$ and get rid of $\theta$ and $\omega$, you should be thinking about $(x-x_0)\frac{R_b}{R_\text{ext}}$.

Maybe think about defining a variable name for the ratio of energy dissipated per distance travelled. And then make the intuitive leap: "That is another word for force".

 

[erobz]

Personally, I'd never have used PB and SB since neither is particularly important

I always like to "over" parameterize... so they are not alone! I would say "different strokes for different folks". Perhaps you might say "cleanliness is next to godliness"

 

[jbriggs444]

I always like to "over" parameterize... so they are not alone! I would say "different strokes for different folks". Perhaps you might say "cleanliness is next to godliness"

Maybe it is my mathematics background showing through. If I see an expression involving a bunch of parameters and constants, I want to sweep those under the table, call the expression $k$ and save myself the work of carrying the expression through all of the algebra.

If that let's me convert the problem to $\frac{1}{2}mv^2 = kx$ then I can recognize a SUVAT equation as well as the next guy.

 

[ROOT0X57B]

Guys, no need to fight that off, I actually do a mix of you both usually do : I keep all the parameters, knowing they are constants I can control but I didn't write the $x$ in the fraction so it looks like $kx$

 

[erobz]

Maybe it is my mathematics background showing through. If I see an expression involving a bunch of parameters and constants, I want to sweep those under the table, call the expression $k$ and save myself the work of carrying the expression through all of the algebra.

If that let's me convert the problem to $\frac{1}{2}mv^2 = kx$ then I can recognize a SUVAT equation as well as the next guy.

Yeah, I agree. I think you're conditioned through higher education! Mine... I would say is an engineering background (the only background I have) showing through. If I'm designing something, I want to have every possible thing I can adjust explicitly written out in front of me!

I do usually group parameters into some Greek letter to do the computation and algebra though.

 

[ROOT0X57B]

Right. But since you want to be able to express everything in terms of $x$ and get rid of $\theta$ and $\omega$, you should be thinking about $(x-x_0)\frac{R_b}{R_\text{ext}}$.

Maybe think about defining a variable name for the ratio of energy dissipated per distance travelled. And then make the intuitive leap: "That is another word for force".

Ok, I tried something but I'm kind of stuck now...
$\big[2F_BR_B\theta\big] = \big[E\big]$
So for "ratio of energy dissipated per distance traveled :
$\Big[\frac{2F_BR_B\theta}{(x-x_0)}\Big] = \big[F\big]$

But while the car travels to the positive $x$ values, $\theta$ gets negative do $\Delta x = \frac{-\theta}{2\pi}R_\text{ext}$
So $\frac{2F_BR_B\theta}{(x-x_0)} = \frac{-4\pi F_BR_Bx}{(x-x_0)R_\text{ext}}$

From here I see $\frac{R_B}{R_\text{ext}}$ but $(x-x_0)$ isn't at the right place...

 

[jbriggs444]

Ok, I tried something but I'm kind of stuck now...
$\big[2F_BR_B\theta\big] = \big[E\big]$
So for "ratio of energy dissipated per distance traveled :
$\Big[\frac{2F_BR_B\theta}{(x-x_0)}\Big] = \big[F\big]$

What I have in mind is something that you already almost have. Let's start over with the formula for kinetic energy.$$KE=\frac{1}{2}Mv^2 + n_\text{wheels} \frac{1}{2}I \omega^2$$Now we want to get that $\omega$ out of there and use $v$ instead. So we make the substitution of $\frac{v}{R_\text{ext}}$ for omega:$$KE=\frac{1}{2}Mv^2 + n_\text{wheels} \frac{I v^2}{2 {R_\text{ext}}^2}$$We should also take the opportunity to express that $I$ as $\frac{1}{2}m_\text{wheel}{R_\text{ext}}^2$. After a bit of cancelling, that results in:$$KE=\frac{1}{2}Mv^2 + n_\text{wheels} \frac{m_\text{wheel} v^2}{4}$$Now this is a good equation. We can put it in a more familiar-seeming form:$$KE=\frac{1}{2}(M + n_\text{wheels}m_\text{wheel}/2)v^2$$If we use $M_e$ (equivalent mass) as shorthand for $M + n_\text{wheels}m_\text{wheel}/2$ then we can rewrite that formula as:$$KE=\frac{1}{2}M_ev^2$$
Now let's flip over to the work side of the equation. We start with the equation for work extracted from the car+wheels assembly as a function of rotation angle $\theta$.$$W= -n_\text{wheels} F_B R_B \theta$$But we do not want this in terms of $\theta$. We want it in terms of $x$. Or, $x-x_0$ to be more precise. Fortunately, $\theta = \frac{x-x_0}{R_\text{ext}}$. So we have:$$W = -n_\text{wheels} F_B R_B \frac{x-x_0}{R_\text{ext}}$$We can make that more familiar-looking by rearranging this as:$$W = -(n_\text{wheels} F_B \frac{R_B}{R_\text{ext}}) (x - x_0)$$If we use $F_e$ (equivalent force) as shorthand for $n_\text{wheels} F_B \frac{R_b}{R_\text{ext}}$ then we have:$$W = -F_e (x-x_0)$$.
That is what I had in mind when I suggested energy per unit distance.

Now you are almost home. You invoke the work-energy theorem and almost immediately write:$$\frac{1}{2}M_ev^2 - \frac{1}{2}M_e{v_0}^2 = -F_e (x-x_0)$$

 

[jack action]

I'm sorry I missed that thread earlier.

If you are looking for a distance, you start with an equation that has a distance in it. You want to relate it to a force? We have a simple one that uses both, the definition of work:
$$\Delta E = F\Delta x$$
Where $F$ is an applied force, for a distance $\Delta x$, that will require (or dissipate) an amount of energy $\Delta E$. So:
$$ \Delta x = \frac{\Delta E}{F}$$
How can we relate this to a braking vehicle? Obviously, the force $F$ is the braking force. For the energy, we know the car will change speed during braking, thus changing its kinetic energy $\frac{1}{2}mv^2$.

But what is the braking force? Well if we assume we are braking with the friction force created by all the tires supporting the car and that all tires have equal characteristics, then the braking force is $\mu mg$. So:
$$\Delta x = \frac{\frac{1}{2}m(v_f^2 - v_0^2)}{\mu mg}$$
Or:
$$\Delta x = \frac{1}{2}\frac{v_f^2 - v_0^2}{\mu g}$$
That is the base in a very simplistic manner. The stopping distance required for the car speed to drop from $v_f$ to $v_0$ only depends on the tire-road coefficient of friction $\mu$. We don't even know how the torque is applied to wheels, we just know that it creates a friction force; a friction force that has an upper limit set by $\mu$.

But let's elaborate on the equation a little bit further.

First with the energy of the car. A car has also parts that rotate, like the wheels but also the braking system, driveshaft, transmission gears, etc. These parts will also reduce their RPM as the car slows down, which means more energy to dissipate. The good news is that it will be in proportion with the total mass of the car, thus introducing the concept of equivalent mass $m_e$ as shown in post #92. I also suggest this article for a deeper study on the subject. So, to resume, we get:
$$E = \frac{1}{2}m_ev^2$$
Where $m_e = \lambda m$, and $\lambda = 1 + \frac{\sum_x I_x g_{r\ x}}{mr_t}$.

$\lambda$ depends on the summation of the product of the rotational inertia $I_x$ and gear ratio $g_{r\ x}$ with respect to the wheel for all rotating component $x$, and the tire radius $r_t$. (see the article for the derivation).

Second, let's look at the braking force. Yes, there is the tire-road friction force $F_t$, but there is also the drag $F_D$ and the rolling resistance $F_R$. So now $F = F_t + F_D + F_R$.

Oh shoot! $F_D$ has the car speed in it ($\frac{1}{2}\rho C_D A v^2$). We now need to use the differential version and integrate:
$$F(v)dx = m_e vdv$$
Or:
$$\Delta x = \int_{v=v_0}^{v_f} \frac{m_e v}{F_t + \frac{1}{2}\rho C_D A v^2 + F_R} dv$$
(It's a fun one!)

Now, you can use the maximum tire-road friction force as $F_t$ to get the minimum braking distance required to slow down the car. But what if the braking system is unable to make enough torque to create that maximum friction force? Obviously, the distance will be greater. And now we're getting into the heart of your problem.

I'm going to let you digest this introduction and see where you want to go from here.

 

[erobz]

I'm sorry I missed that thread earlier.

If you are looking for a distance, you start with an equation that has a distance in it. You want to relate it to a force? We have a simple one that uses both, the definition of work:
$$\Delta E = F\Delta x$$
Where $F$ is an applied force, for a distance $\Delta x$, that will require (or dissipate) an amount of energy $\Delta E$. So:
$$ \Delta x = \frac{\Delta E}{F}$$
How can we relate this to a braking vehicle? Obviously, the force $F$ is the braking force. For the energy, we know the car will change speed during braking, thus changing its kinetic energy $\frac{1}{2}mv^2$.

But what is the braking force? Well if we assume we are braking with the friction force created by all the tires supporting the car and that all tires have equal characteristics, then the braking force is $\mu mg$. So:
$$\Delta x = \frac{\frac{1}{2}m(v_f^2 - v_0^2)}{\mu mg}$$
Or:
$$\Delta x = \frac{1}{2}\frac{v_f^2 - v_0^2}{\mu g}$$
That is the base in a very simplistic manner. The stopping distance required for the car speed to drop from $v_f$ to $v_0$ only depends on the tire-road coefficient of friction $\mu$. We don't even know how the torque is applied to wheels, we just know that it creates a friction force; a friction force that has an upper limit set by $\mu$.

But let's elaborate on the equation a little bit further.

First with the energy of the car. A car has also parts that rotate, like the wheels but also the braking system, driveshaft, transmission gears, etc. These parts will also reduce their RPM as the car slows down, which means more energy to dissipate. The good news is that it will be in proportion with the total mass of the car, thus introducing the concept of equivalent mass $m_e$ as shown in post #92. I also suggest this article for a deeper study on the subject. So, to resume, we get:
$$E = \frac{1}{2}m_ev^2$$
Where $m_e = \lambda m$, and $\lambda = 1 + \frac{\sum_x I_x g_{r\ x}}{mr_t}$.

$\lambda$ depends on the summation of the product of the rotational inertia $I_x$ and gear ratio $g_{r\ x}$ with respect to the wheel for all rotating component $x$, and the tire radius $r_t$. (see the article for the derivation).

Second, let's look at the braking force. Yes, there is the tire-road friction force $F_t$, but there is also the drag $F_D$ and the rolling resistance $F_R$. So now $F = F_t + F_D + F_R$.

Oh shoot! $F_D$ has the car speed in it ($\frac{1}{2}\rho C_D A v^2$). We now need to use the differential version and integrate:
$$F(v)dx = m_e vdv$$
Or:
$$\Delta x = \int_{v=v_0}^{v_f} \frac{m_e v}{F_t + \frac{1}{2}\rho C_D A v^2 + F_R} dv$$
(It's a fun one!)

Now, you can use the maximum tire-road friction force as $F_t$ to get the minimum braking distance required to slow down the car. But what if the braking system is unable to make enough torque to create that maximum friction force? Obviously, the distance will be greater. And now we're getting into the heart of your problem.

I'm going to let you digest this introduction and see where you want to go from here.

If anything I suspect this will be a bit of an eye opener to the actual complexity of this problem. Perhaps, stopping on dry road will be complex enough for their project. You kind of cut to the chase, ( I think ) we were "teaching them to fish". But the cat is out of the bag now, so to speak.

 

[jack action]

You kind of cut to the chase, ( I think ) we were leading them to water. But the cat is out of the bag now, so to speak.

No, I began at the starting point: I want to know the braking distance. Now we can detail and complicate this equation as much as we want such that we reach a 156 hours project.

 

[erobz]

No, I began at the starting point: I want to know the braking distance. Now we can detail and complicate this equation as much as we want such that we reach a 156 hours project.

If you are looking for a distance, you start with an equation that has a distance in it. You want to relate it to a force? We have a simple one that uses both, the definition of work:
ΔE=FΔx
Where F is an applied force, for a distance Δx, that will require (or dissipate) an amount of energy ΔE. So:
Δx=ΔEF
How can we relate this to a braking vehicle? Obviously, the force F is the braking force. For the energy, we know the car will change speed during braking, thus changing its kinetic energy 12mv2.

But what is the braking force? Well if we assume we are braking with the friction force created by all the tires supporting the car and that all tires have equal characteristics, then the braking force is μmg. So:
Δx=12m(vf2−v02)μmg
Or:
Δx=12vf2−v02μg
That is the base in a very simplistic manner. The stopping distance required for the car speed to drop from vf to v0 only depends on the tire-road coefficient of friction μ. We don't even know how the torque is applied to wheels, we just know that it creates a friction force; a friction force that has an upper limit set by μ.

Yeah... when I said that earlier, I was referring to the portion of your response above "will not take 156 hrs". I said if you can't produce this model $\uparrow$ your not going to produce the other models.

We were trying to explain to them that even stopping on dry road is significantly more complex than it first appears and there might not be a need to add extra complexities of "water drag" on the tires etc...

I was planning on asking them to think of extra possible discrete additions of complexity as the model progressed. The OP has not entered college yet.

 

[erobz]

This is just pure speculation, but I'd imagine if we actually considered the full complexity of the automobile and all its systems one might spend a few years modeling stopping on a dry road!

 

[ROOT0X57B]

I'm sorry I missed that thread earlier.

If you are looking for a distance, you start with an equation that has a distance in it. You want to relate it to a force? We have a simple one that uses both, the definition of work:
$$\Delta E = F\Delta x$$
Where $F$ is an applied force, for a distance $\Delta x$, that will require (or dissipate) an amount of energy $\Delta E$. So:
$$ \Delta x = \frac{\Delta E}{F}$$
How can we relate this to a braking vehicle? Obviously, the force $F$ is the braking force. For the energy, we know the car will change speed during braking, thus changing its kinetic energy $\frac{1}{2}mv^2$.

But what is the braking force? Well if we assume we are braking with the friction force created by all the tires supporting the car and that all tires have equal characteristics, then the braking force is $\mu mg$. So:
$$\Delta x = \frac{\frac{1}{2}m(v_f^2 - v_0^2)}{\mu mg}$$
Or:
$$\Delta x = \frac{1}{2}\frac{v_f^2 - v_0^2}{\mu g}$$
That is the base in a very simplistic manner. The stopping distance required for the car speed to drop from $v_f$ to $v_0$ only depends on the tire-road coefficient of friction $\mu$. We don't even know how the torque is applied to wheels, we just know that it creates a friction force; a friction force that has an upper limit set by $\mu$.

But let's elaborate on the equation a little bit further.

First with the energy of the car. A car has also parts that rotate, like the wheels but also the braking system, driveshaft, transmission gears, etc. These parts will also reduce their RPM as the car slows down, which means more energy to dissipate. The good news is that it will be in proportion with the total mass of the car, thus introducing the concept of equivalent mass $m_e$ as shown in post #92. I also suggest this article for a deeper study on the subject. So, to resume, we get:
$$E = \frac{1}{2}m_ev^2$$
Where $m_e = \lambda m$, and $\lambda = 1 + \frac{\sum_x I_x g_{r\ x}}{mr_t}$.

$\lambda$ depends on the summation of the product of the rotational inertia $I_x$ and gear ratio $g_{r\ x}$ with respect to the wheel for all rotating component $x$, and the tire radius $r_t$. (see the article for the derivation).

Second, let's look at the braking force. Yes, there is the tire-road friction force $F_t$, but there is also the drag $F_D$ and the rolling resistance $F_R$. So now $F = F_t + F_D + F_R$.

Oh shoot! $F_D$ has the car speed in it ($\frac{1}{2}\rho C_D A v^2$). We now need to use the differential version and integrate:
$$F(v)dx = m_e vdv$$
Or:
$$\Delta x = \int_{v=v_0}^{v_f} \frac{m_e v}{F_t + \frac{1}{2}\rho C_D A v^2 + F_R} dv$$
(It's a fun one!)

Now, you can use the maximum tire-road friction force as $F_t$ to get the minimum braking distance required to slow down the car. But what if the braking system is unable to make enough torque to create that maximum friction force? Obviously, the distance will be greater. And now we're getting into the heart of your problem.

I'm going to let you digest this introduction and see where you want to go from here.

I think you considered the force to be tire-road friction instead of brake pads-disk friction

 

[jack action]

I think you considered the force to be tire-road friction instead of brake pads-disk friction

Yes, because that's what is actually governing the maximum braking force $F_t$ you can get.

But you can get a smaller braking force $F_t$ by modulating the brake pads-disk friction. So you can still extend the equation further by setting $F_t = MIN(\frac{T_B}{r_t}, \mu mg)$, where $T_B$ is the braking torque from all wheels and $r_t$ the tire radius.

 

[ROOT0X57B]

Yes, because that's what is actually governing the maximum braking force $F_t$ you can get.

But you can get a smaller braking force $F_t$ by modulating the brake pads-disk friction. So you can still extend the equation further by setting $F_t = MIN(\frac{T_B}{r_t}, \mu mg)$, where $T_B$ is the braking torque from all wheels and $r_t$ the tire radius.

Okay, seems a lot clearer...
How can I have $T_B$ now?

EDIT : It seems like @jbriggs444 answered it earlier in the detailed post

 

[jack action]

Okay, seems a lot clearer...
How can I have $T_B$ now?

EDIT : It seems like @jbriggs444 answered it earlier in the detailed post

$T_B$ is the friction torque produced by your braking system (what you actually started with, in this thread). You might even add a braking force from the engine or electric motor (regenerative braking).

 

[ROOT0X57B]

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]

Second, let's look at the braking force. Yes, there is the tire-road friction force $F_t$, but there is also the drag $F_D$ and the rolling resistance $F_R$. So now $F = F_t + F_D + F_R$.

Actually, I do not count the drag as I deal with relatively slow speeds and my wheel is a perfect cylinder so there is no rolling resistance either

$T_B$ is the friction torque produced by your braking system (what you actually started with, in this thread). You might even add a braking force from the engine or electric motor (regenerative braking).

I consider the motor to be inactive as it's the braking time and regenerative braking is none of my topic

 

[jbriggs444]

So now with a little bit of python programming, I can simulate the distance traveled over time by little incrementations of time

This is a case of constant acceleration. That means that there is a SUVAT equation for this. No need for any differential equations:$$s=v_0t + \frac{1}{2}at^2$$.
Sure, one can do it with Euler integration and Python or your choice of language. But with an analytic solution so close at hand, there seems little point in resorting to numerical methods.

Having a canned solution against which to compare a numerical result might give you a little bit of experience seeing the behavior of different numerical approaches. Here is an article to give you a taste.

http://faculty.olin.edu/bstorey/Notes/DiffEq.pdf

Things get interesting if $F_B$ is a function of $v$ (first order differential equation) or of $x$ (second order).

 

[ROOT0X57B]

This is a case of constant acceleration. That means that there a SUVAT equation for this. No need for any differential equations:$$s=v_0t + \frac{1}{2}at^2$$.
Sure, one can do it with Euler integration and Python or your choice of language. But with an analytic solution so close at hand, there seems little point to doing so.

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.