How to Calculate the Braking Momentum on a Wheel?

[erobz]

Would you recommend moving to brakes temperature?

No, heat transfer is complex.

 

[erobz]

I'd recommend you produce (as I and others have mentioned) the most simplistic model that captures the " gravel size " resolution of stopping on a dry road as an exploration into whether or not you wish to explore other topics instead.

 

[ROOT0X57B]

No, heat transfer is complex.

We have thermodynamics in our physics lessons, but maybe it goes too much further
We have been taught about enthalpy, heat flux, heat transfer...

I'd recommend you produce (as I and others have mentioned) the most simplistic model that captures the " gravel size " resolution of stopping on a dry road as an exploration into whether or not you wish to explore other topics instead.

Didn't understand this one, what do you call "gravel size resolution"?
If I have to consider only friction with the ground and brakes, that would be too easy or I would have to consider some mechanical engineering parameters (axle, suspensions...) which are actually off-topic (I asked teachers about that already)

 

[jbriggs444]

No, the temperature coming from the friction between the tire and the road

At the risk of repeating what @erobz has said, you only get temperature rise from wheels on the road if there is something other than pure rolling without slipping going on.

One way that this happens is with the flexing in the treads and sidewalls as the contact patch flattens out as it hits the road and then un-flattens as it is pulled back up. The tire material is fairly elastic -- it tends to return to its original configuration after being deflected. But the force that it absorbs during deflection is a bit larger than the force that it returns during the rebound. This means that the upward force of road on tire at the front edge of the contact patch is larger than the upward force of road on tire at the back edge. That means that the support force from road on tire is a little bit forward of the wheel's axle. That means that it results in a net torque. That net torque is rolling resistance.

If you work out the simultaneous equations for a coasting car, this rearward torque from rolling resistance must be (almost) matched by a forward torque from static friction of road on tires. Which means a rearward force of static friction of road on tires. (The push from the road helps keep the tires rotating forward while is is also slowing the car as a whole down).

In addition to rolling resistance, there may be some slip between tires and road. Rubber (like most things) does not behave exactly in the ideal fashion that the laws of static and kinetic friction call for. When near the maximum force called for by the coefficient of static friction, it will slip. This tendency is enhanced by the existence of tread "squirm". The contact patch does not always stay consistently motionless with respect to the road. Especially at the edges or with deep treads, pieces of it will move differently than other pieces -- squirming.

Both rolling resistance and squirming will result in mechanical energy being lost as heat.

Braking will, ideally, not result in any heat at the treads. It will instead result in heat at the rotors/pads/drums/disks because that is where the slipping is occurring.

 

[ROOT0X57B]

Okay, so you're right, calculations would be way too complex if the material can deform itself...

 

[erobz]

My point is that you say that "stopping distance on dry road is too simplistic". That may very well be the case, but that also implies that you should be able to produce that model for us, here...without much effort. However, if you can't produce that model you have a minuscule ( to nil) chance of accurately describing the complex phenomena you wish to describe ( with your current level of education - not necessarily smarts or determination), and you are going to waste much time and effort without achieving any reasonable results or greater understanding. Why waste the time?

To get to the model you desire, might very well require undergraduate course in Classical Mechanics. and to the next simpler model, undergraduate Physics 1. There is at least a full undergraduate physics course between the understanding of those two models. To put it this way. I didn't fully connect the dots and I have a Bachelors of Science in Mechanical Engineering and 10 years of work experience in the field ( sad maybe, but none the less true )!

Before an artist can make a wood carving, they have to have a very good idea of what size of log they need to make the piece they desire.

 

[erobz]

At the risk of repeating what @erobz has said, you only get temperature rise from wheels on the road if there is something other than pure rolling without slipping going on.

One way that this happens is with the flexing in the treads and sidewalls as the contact patch flattens out as it hits the road and then un-flattens as it is pulled back up. The tire material is fairly elastic -- it tends to return to its original configuration after being deflected. But the force that it absorbs during deflection is a bit larger than the force that it returns during the rebound. This means that the upward force of road on tire at the front edge of the contact patch is larger than the upward force of road on tire at the back edge. That means that the support force from road on tire is a little bit forward of the wheel's axle. That means that it results in a net torque. That net torque is rolling resistance.

If you work out the simultaneous equations for a coasting car, this rearward torque from rolling resistance must be (almost) matched by a forward torque from static friction of road on tires. Which means a rearward force of static friction of road on tires. (The push from the road helps keep the tires rotating forward while is is also slowing the car as a whole down).

In addition to rolling resistance, there may be some slip between tires and road. Rubber (like most things) does not behave exactly in the ideal fashion that the laws of static and kinetic friction call for. When near the maximum force called for by the coefficient of static friction, it will slip. This tendency is enhanced by the existence of tread "squirm". The contact patch does not always stay consistently motionless with respect to the road. Especially at the edges or with deep treads, pieces of it will move differently than other pieces -- squirming.

Both rolling resistance and squirming will result in mechanical energy being lost as heat.

Braking will, ideally, not result in any heat at the treads. It will instead result in heat at the rotors/pads/drums/disks because that is where the slipping is occurring.

Thanks for explaining that some. I have all my textbooks. I can find very little more than a gloss over treatment of rolling resistance anywhere in the ciurriculum ( at least the curriculum 10 years ago). Even my Classical Mechanics text (John R. Taylor) doesn’t really address it.

 

[ROOT0X57B]

Ok so I am now dealing with the first, basic model
No rolling resistance here, only "simple friction forces"

As you said I will get into energy considerations :
I call $M$ the mass of all the car and $m_\text{wheel}$ the mass of one wheel

$$\frac{1}{2}Mv_0^2 + 4*\frac{1}{2}I\omega_0^2 = \frac{1}{2}Mv^2 + 4*\frac{1}{2}I\omega^2 - 2*\int F_Bdx$$
I multiply the integral by 2 because there are two wheels with brakes and $F_B$ acts on one wheel

Because wheels are similar to filled cylinders $I = \frac{1}{2}m_\text{wheel}R_\text{ext}$
So I get

$$\displaystyle \frac{1}{2}Mv_0^2 + m_\text{wheel}R_\text{ext}\omega_0^2 = \frac{1}{2}Mv^2 + m_\text{wheel}R_\text{ext}\omega^2 - 2\int F_Bdx$$

The question is, how do I link that to $\theta$ (and then to the stopping distance $L$ will be easy) ?
I acknowledged the $dx$ of the integral is $R_Bd\theta$ but this means I have to integrate $F_B$ between $\theta_0$ and $\theta$ but I have no theta in my force expression :
$F_B = \mu_B \frac{1}{4}Mg\frac{R_\text{ext}}{R_B}$

I would automatically say $\displaystyle\int F_Bdx = F_BR_B(\theta_0 - \theta)$ as $F_B$ doesn't depend on $\theta$ but it looks kinda strange to me.

 

[jbriggs444]

The question is, how do I link that to $\theta$

$$v=R_\text{ext} \omega$$ $$dx = R_\text{ext} d \theta$$
If you use these identities, there is some simplification that you can do to your energy equation. Remove the $R_\text{ext} \omega$'s and replace them with $v$'s.

There is a rule of thumb that avid cyclists use: Rim weight counts for twice as much as frame weight. It's not quite the same rule for solid discs, but similar.

 

[erobz]

Ok so I am now dealing with the first, basic model
No rolling resistance here, only "simple friction forces"

As you said I will get into energy considerations :
I call $M$ the mass of all the car and $m_\text{wheel}$ the mass of one wheel

$$\frac{1}{2}Mv_0^2 + 4*\frac{1}{2}I\omega_0^2 = \frac{1}{2}Mv^2 + 4*\frac{1}{2}I\omega^2 - 2*\int F_Bdx$$
I multiply the integral by 2 because there are two wheels with brakes and $F_B$ acts on one wheel

Because wheels are similar to filled cylinders $I = \frac{1}{2}m_\text{wheel}R_\text{ext}$
So I get

$$\displaystyle \frac{1}{2}Mv_0^2 + m_\text{wheel}R_\text{ext}\omega_0^2 = \frac{1}{2}Mv^2 + m_\text{wheel}R_\text{ext}\omega^2 - 2\int F_Bdx$$

The question is, how do I link that to $\theta$ (and then to the stopping distance $L$ will be easy) ?
I acknowledged the $dx$ of the integral is $R_Bd\theta$ but this means I have to integrate $F_B$ between $\theta_0$ and $\theta$ but I have no theta in my force expression :
$F_B = \mu_B \frac{1}{4}Mg\frac{R_\text{ext}}{R_B}$

I would automatically say $\displaystyle\int F_Bdx = F_BR_B(\theta_0 - \theta)$ as $F_B$ doesn't depend on $\theta$ but it looks kinda strange to me.

$I = \frac{1}{2}m_\text{wheel}R_\text{ext}$ There is an error in this.

$F_B = \mu_B \frac{1}{4}Mg\frac{R_\text{ext}}{R_B}$ this also should be reworked. I thought you were trying to generalize based on your initial post?

$F_b = \mu_b P A$

There is a problem with letting $F_B = \mu_B \frac{1}{4}Mg\frac{R_\text{ext}}{R_B}$. Its best if you revisit the assumptions here. You are actually going to be solving an inequality for $F_b$ which will have consequences for $\mu_b$.

You can drop $\theta_o$ in $\int F_b r_b \cdot d \theta$ as the initial condition $\theta_o = 0$

 

[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.