How to Calculate the Braking Momentum on a Wheel?

[ROOT0X57B]

Is determining the stopping distance your objective?

The moment of the brake force is not correct. Why do you have ( presumably ) the mass of the vehicle in there? Also, the moment arm should be $r_b$, not $r_{ext}$ for that one.

I have the mass because of @erobz formula :
$F_B = \mu_B * \frac{M}{4}\frac{R_\text{ext}}{R_B}g$
$m = M/4$
Also, $\mathcal{M}_\Delta(F_B) = F_B R_B$ so $\mathcal{M}_\Delta(F_B) = \mu_B * mgR_\text{ext}$

 

[erobz]

Yes, copy mistake, my bad (otherwise I wouldn't even have mentioned it)

I have the mass because of @erobz formula :
$F_B = \mu_B * \frac{M}{4}\frac{R_\text{ext}}{R_B}g$

Well, maybe slow down a bit on that. That relationship would be used just to put an upper bound on $\mu_b$ such that the tire won't slip on the road. Also, there was no drag in that model.

I would recommend starting with "no drag" just so you get an ideal of what you are up against.

 

[ROOT0X57B]

What do you call drag exactly ? The friction force with the ground ?

 

[erobz]

What do you call drag exactly ? The friction force with the ground ?

The drag force from the water. Ignore it for the time being and figure out the stopping distance of your vehicle on dry road for a parameter $\mu_b$ to get a sense for what you expect and how to tackle it. Its likely that the drag on the vehicle will dominate the drag on the tire...unless this is a reasonably deep stream of water, and the vehicle is not moving very fast. That being said, if it is reasonably deep, some buoyant forces are going to develop, and the static friction will significantly fall. There are many complications here.

 

[ROOT0X57B]

The point of my calculations is actually to find out the impact of water fluid friction on the stopping distance.
I only consider the water to apply a force on the tire : $-\frac{1}{2}\rho C_xhlv^2$

The water layer is really thin (1mm to 2cm)

 

[erobz]

The point of my calculations is actually to find out the impact of water fluid friction on the stopping distance.
I only consider the water to apply a force on the tire : $-\frac{1}{2}\rho C_xhlv^2$

The water layer is really thin (1mm to 2cm)

Well, I have a feeling the "wet road" is going to have a much more significant impact acting through $\mu_s$ (the static coefficient of friction) than the torque of the drag force from that tiny bit of water. Maybe I'm wrong.

Another thing, the minus sign in front of the drag is not exactly needed here. The force is negative, but the torque ( whether its positive or negative ) from it is going to be established by your convention when you apply Newtons Second Law.

I understand, that was your original motivation. but it is still a wise idea to solve the less complex model first of "dry road stopping" for some reference.

 

[ROOT0X57B]

Well, I didn't think about $\mu_S$ being modified because the road is wet
How could I model that ?

For the Second Law, I copied the moment expression without $R_\text{ext}$

 

[erobz]

Well, I didn't think about $\mu_S$ being modified because the road is wet
How could I model that ?

You might be trying to find some experimental results for that one.

 

[ROOT0X57B]

Like finding values of $\mu_S$ for wet and dry road?

 

[jbriggs444]

Surely if there is enough water to slow the vehicle down significantly when the brakes are already being applied, there is enough water so that hydroplaning is a factor that needs to be considered.

 

[ROOT0X57B]

Ok, finally I will have 3 models two make
1 - Dry road, no water, temperature not taken into account
2 - Wet road, water, temperature not taken into account
3 - Wet road, water, temperature taken into account

 

[ROOT0X57B]

Surely if there is enough water to slow the vehicle down significantly when the brakes are already being applied, there is enough water so that hydroplaning is a factor that needs to be considered.

I don't go fast enough I think (50kmh)

 

[jbriggs444]

I don't go fast enough I think (50kmh)

Think again. It is not just the possibility that the tires will be lifted out of contact with the pavement but the possibility that the normal force of tire on pavement will be significantly reduced.

 

[ROOT0X57B]

I will consider there is no aquaplaning then, to simplify equations
This is my project for engineering school entry exam (not supposed to be too complex though)

 

[erobz]

Well, something to consider. Depending on how much physics you know (or are expected to know), just figuring out the stopping distance on dry road is not quite as simple as you first might expect.

 

[ROOT0X57B]

Well, something to consider. Depending on how much physics you know (or are expected to know), just figuring out the stopping distance on dry road is not quite as simple as you first might expect.

Well, stopping distance on dry road would be a little bit of level I think, but could actually be given in some high ranked school entry test.
I want to keep it simple somehow, but I am expected to work on my project 2 hours a week for a year and half.

I also have to create something myself, this is why I am so obsessed with modeling the general situation.

 

[erobz]

Well, stopping distance on dry road would be a little bit of level I think, but could actually be given in some high ranked school entry test.
I want to keep it simple somehow, but I am expected to work on my project 2 hours a week for a year and half.

I also have to create something myself, this is why I am so obsessed with modeling the general situation.

This problem isn't 156 man hrs.

 

[hutchphd]

How does the height of the temperature layer and the heat generated by the friction with the ground affect the stopping distance?

With respect, this statement of the problem makes no sense to me. What is the "height of the temperature layer" ?

I know all of this may feel a bit confusing

We will do simpler
I have a disc of radius R spinning around its center axis
I now press an undeformable brake pad onto the disc
Thus, the pad is fixed while the disc keeps rotating, but there is friction between the disc and the pad
This friction obviously acts against the rotation of the disc
This is what torque I want to figure out.

So first solve the simple dry case (with help as necessary) forthe torque. That will be useful regardless.
What is the exact description from the engineering school for this exercise?? I personally think your choice is at best ill-formed and perhaps ill-conceived. Do you have other ideas and interests?

 

[ROOT0X57B]

With respect, this statement of the problem makes no sense to me. What is the "height of the temperature layer" ?

It's the height of the water layer, not temperature, my bad

Actually, one of the exams to be admitted to engineering school is what we call here "TIPE", it's a project of research within a given thematic (this year thematic is "the city"). We have to ask ourselves a question that could not be answered immediately, but not too complex either because we have to answer it in the end.
The research must be led in one of our fields of study (maths, physics, and computer science). Most importantly we have to produce something new (a physics model for me).

My question is "What is the impact of humidity and temperature of braking on the stopping distance of a car?"

So as said before, I planned to create three different models :
One with dry friction when not considering the temperature from braking (base case)
One with a water layer, no temperature, to get the impact of humidity
The last one with water and temperature considered, to get the impact of temperature

 

[erobz]

So as said before, I planned to create three different models :
One with dry friction when not considering the temperature from braking (base case)
One with a water layer, no temperature, to get the impact of humidity
The last one with water and temperature considered, to get the impact of temperature

Better check your experimental design. You're not going to get the impact of a parameter unless you hold the other parameters constant.

 

[ROOT0X57B]

Hmm, okay I guess?
I whished my final result could be a 3D graph :
For a chosen car (so mass, brake pressure...), to plot the stopping distance as a function of the height of water layer and braking temperature?

 

[erobz]

How are you going to get the braking temperature? You mean you are trying to model the temperature of the brakes?

 

[ROOT0X57B]

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

 

[erobz]

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

Thats a different type of friction all together. Rolling resistance is not static friction. I think you are being a bit ambitious for your experience level ( just my opinion ) it this topic.

 

[ROOT0X57B]

I did nothing for now about that so I am actually free to change up things
I thought only considering water could be a little bit less than expected so I had the idea of temperature.
Would you recommend moving to brakes temperature?

 

[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 $\theta_o = 0$