# Rotational Dynamics - Modeling brake caliper deceleration of a chassis

• TheCondor
In summary: I'm not sure about the exact value.)2. In MODEL 2, I have drawn a static friction force at the tire-pavement contact. I have assumed this static friction is necessary to maintain the no slip condition of the tire and when it is exceeded the tire will slide. Is it appropriate to include it in the FBD as shown? Is a static friction force really necessary to maintain the no slip condition?Yes, a static friction force is necessary to maintain the no slip condition.
TheCondor
TL;DR Summary
Uncertainty about validity of 2 different models
I am a junior engineer tasked with modeling the dynamics of a small research UAV after landing. The UAV has 3 tires, 1 on the nose landing gear and 2 on the rear landing gear. The rear tires are equipped with disc brake calipers.

My coworker has explained that the simplified model (MODEL 1 below) is sufficient to find the horizontal deceleration on the aircraft CG caused by the brake calipers. This model assumes the tire radii are negligible and the caliper force acts at approximately the same location as the tire-runway interface. The coefficient of kinetic friction (μ_k) between the caliper and disc is then approximated from a textbook which plots μ_k decreasing with velocity.

Uncomfortable with the small radius approximation, I have attempted to model the wheel as a solid disk to find the horizontal acceleration of the wheel center (MODEL 2 below). Assuming the chassis is a rigid body, the horizontal acceleration of the wheel axle should give the acceleration of the whole aircraft.

I have difficulty understanding if either approach is correct and particularly the role and location of the frictional forces.

In MODEL 1, the maximum brake force is found directly from only the normal force and μ_k, but I don't understand why. The no slip condition and small radius approximation acting simultaneously are preventing me from understanding this clearly.

I can accept the approximation that if the tire radius is negligible, the kinetic caliper friction can be viewed as acting at the same point as the static tire friction. I also recognize that the brake force exerted by the calipers decreases at higher speeds (ie. the brake force should be a function of the kinetic friction coefficient of the caliper-disc interface).

Any help and corrections that you could provide to better intuit the physics at play would be greatly appreciated. Thank you very much for reading.

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Welcome, TheCondor!

If I understand it correctly, you are dealing with two very different coefficients of dynamic friction and with two normal forces: caliper-brake pads and tire's contact patch-pavement.

The normal force on each of the contact patches varies with the remaining wing's lift while the aircraft is still moving fast, as well as with some rolling and yaw adjustments to keep a straight trajectory on the runway.

After flare and pitch forward rotation has been completed, you have forces and weight distribuited between two calipers and three tires.
The stronger the deceleration the more weight is shared by the front tire.

If your aircraft has engine's thrust reversal, you should consider its help to the mechanical brakes, as well as its influence on pitch moment.

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Lnewqban,

Indeed the points you made are accurate. The lift, thrust, rolling resistance, drag and brake force are vector summed to estimate the total acceleration of the aircraft. My misunderstanding is specific to the calculation of the brake force.

The following questions in particular continue to bother me:
1. In MODEL 1, the brake force is drawn as acting at the tire-runway contact because the wheel is assumed to be small (wheel is a point). If the tire size is non-negligible, shouldn't this frictional force be acting at the caliper-pad contact point? In that case, won't the moment arm h be shorter?

2. In MODEL 2, I have drawn a static friction force at the tire-runway contact. I have assumed this static friction is necessary to maintain the no slip condition of the tire and when it is exceeded the tire will slide. Is it appropriate to include it in the FBD as shown? Is a static friction force really necessary to maintain the no slip condition?

MODEL 2 seems to give unrealistic brake force values. I'm tempted to redraw the FBD with only the caliper force contributing to the total braking force. If MODEL 2's FBD is redrawn with the caliper force only then clearly it approaches MODEL 1's results. If a static friction force is not necessary to maintain no-slip, then unfortunately I also don't understand when it should or shouldn't be drawn.

TheCondor said:
1. In MODEL 1, the brake force is drawn as acting at the tire-runway contact because the wheel is assumed to be small (wheel is a point). If the tire size is non-negligible, shouldn't this frictional force be acting at the caliper-pad contact point? In that case, won't the moment arm h be shorter?
The tire radius is NOT negligible. The torque from the caliper is the same as the torque from the tire, but the forces are not ... unless you have this set up:

The deceleration is related to the tire-pavement friction force. There is a maximum friction coefficient that can be found in textbooks or with tests. It doesn't matter how this force is created to calculate the deceleration. (Even though it is always at some amount of slip. Usually at around 10-20% slip.)

To design the brake system, you need to find how much torque it has to produce to create that tire maximum friction force (and how much energy it has to dissipate as well). That's it.

TheCondor and Lnewqban
TheCondor said:
Lnewqban,

Indeed the points you made are accurate. The lift, thrust, rolling resistance, drag and brake force are vector summed to estimate the total acceleration of the aircraft. My misunderstanding is specific to the calculation of the brake force.

The following questions in particular continue to bother me:
1. In MODEL 1, the brake force is drawn as acting at the tire-runway contact because the wheel is assumed to be small (wheel is a point). If the tire size is non-negligible, shouldn't this frictional force be acting at the caliper-pad contact point? In that case, won't the moment arm h be shorter?
You are welcome.

What I tried to say previously (sorry, English is not my native language) is that the weakest link in the mechanism of brakes is the contact patches of the tires, which greatly depend on the normal force on each tire, which depends on the dynamics of landing, like remaining wing lift, oscillations due to adjustments on roll and yaw, distribution of weight, bounce backs into the air, as well as nose down-tail up effect of brake application for maximum deceleration.

It is not until the full weight of the airplane is pushing the tires against the pavement and the front tire has touched down that the pilot can fully apply brakes.
By then, a good portion of available runway has been wasted and the rubber is pushed to the limits: there is no much safety factor there, especially when rain or snow are present.

I am sorry, I have not had time to properly understand the calculations that you have presented, but it seems that the pitch moment induced by the deceleration of the aircraft has not been included in Mode 1.

I believe that your concern is correct: the tangential force at the caliper-pads must be greater than the tangential force at the tire-runway, since resistive moment is the same.

The second big limitation of these brakes is their capability to convert all the kinetic energy of the landing aircraft into heat and to transfer it into the surrounding atmosphere in a few seconds, imposed by the available effective length of the runway.

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TheCondor
Thank you both very much for your replies, they are extremely helpful. Thank you also for the informative videos.

jack action said:
The tire radius is NOT negligible. The torque from the caliper is the same as the torque from the tire

It looks like this was my error in my FBD of the disk. I have a non-zero angular acceleration but it should be zero. I thought the angular acceleration is necessary because the angular velocity is decreasing.

How can the disk RPM decrease if the net torque on the disk is zero? The no slip condition means a=rα, so I thought this angular acceleration acts proportionally with the horizontal acceleration. How can the net torque be zero, but the horizontal acceleration be non-zero under no slip?

Thank you very much for your help.

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Lnewqban and berkeman
My mistake, the rule is still ##\sum F = ma## or ##\sum T = I\alpha##.

Looking at the whole vehicle, ignoring other resistive forces, i will be ##F_{brake} = m_e a## where ##F_{brake}## is the friction force at the tire contact patch and ##a## the vehicle deceleration. ##m_e## is the vehicle's equivalent mass, i.e. the mass of the vehicle plus the equivalent inertia from the rotating masses (more info here on the concept). You could of course get to the same result by using the Newton's 2nd law equation for all the individual mechanical components linked to one another and solve for n unknowns from n equations.

TheCondor and Lnewqban
Understood. So considering the whole vehicle braking under no slip, ##F_{brake}## is a static friction force which depends only on the pavement-tire slip ratio as shown in the chart above. In the case of the whole vehicle, the caliper friction force is internal so it is not drawn.

One thing I do not understand with this simplified model is that the brake force is generated by the caliper's kinetic friction which depends on velocity, heat etc. but ##F_{brake}## does not directly depend on velocity, only the slip ratio of the tire. If the wheels are approximately not slipping (eg. slip ratio = 0.1) does that mean the brake force is constant at every speed? Is that really correct?

In trying to model only the wheel as a solid disk (like the train brake shoe configuration) I get what looks like unreasonable acceleration (below). Can you point out where I'm going wrong?

Thank you very much for your help.

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TheCondor said:
generated by the caliper's kinetic friction which depends on velocity
I'm assuming the Coulomb's law of friction:
https://en.wikipedia.org/wiki/Friction#Laws_of_dry_friction said:
Coulomb's Law of Friction: Kinetic friction is independent of the sliding velocity.
TheCondor said:
Also, in trying to model only the wheel as a solid disk (like the train brake shoe configuration) I get what looks like unreasonable acceleration (below). Can you point out where I'm going wrong?
##F_{shoe}## is an internal force, forget about it.
$$m_e a = F_{brake}$$
##m_e## is equal to the total mass of the vehicle ##m## PLUS the inertia of the wheel ##I_w## converted to an equivalent mass ##m_{Iw}##, acting at the wheel radius ##r##, at the vehicle acceleration ##a##:
$$m_{Iw}ar = I_w\alpha = \frac{m_w r^2}{2}\frac{a}{r}$$
$$m_{Iw} = \frac{m_w}{2}$$
Therefore:
$$m_e a = F_{brake}$$
$$(m + m_{Iw}) a = F_{brake}$$
$$\left(m + \frac{m_w}{2}\right) a = \mu mg$$
Or:
$$a = \frac{\mu g}{\left(1 + \frac{m_w}{2m}\right)}$$
And in the case you presented, the mass of the wheel ##m_w## is the same as the mass of the vehicle ##m##, so ##a = \frac{2}{3} \mu g##.

TheCondor
Jack action,

I see. I understand now the reasoning behind excluding the brake caliper force from the entire system FBD. It looks like this was the major gap in my understanding of my coworker’s model. Thank you very much for helping to clarify that point and for reminding me about the Coulomb friction model's assumption.

Unfortunately, I don't agree that the brake shoe can be excluded from the FBD of the disk. It acts directly on the body so ignoring it is very uncomfortable, plus we know there must be a clockwise angular acceleration to decelerate the disk. An FBD with only the pavement friction force can not provide this clockwise angular deceleration. After some thought and a couple days I think I've found my error.

In the FBD I drew, the direction of the brake caliper / shoe friction force I drew is incorrect. The direction pointing to the right is arbitrary. If I drew it in a different location the sense would change, always apposing the angular velocity. In fact, if the brake caliper was a donut shape, the friction force would affect the disk the same way but would have no linear component.

The brake shoe or caliper can only provide a deceleration Torque, no linear Force. It does not contribute to the linear acceleration, which is why it can be incorrectly ignored in a model of the disk and the correct linear acceleration will still be produced.

The horizontal acceleration of the center of the wheel in the x direction is related to the angular acceleration by no slip: $$a_{o} = rα$$ This is why is it tempting but incorrect to say the sum of torques is zero ie. $$F_{friction}r = τ_{caliper}$$ Impossible, since there would be no change in ω, but correct to say $$a_{o} = rα$$ to describe the acceleration of the center of the disk.

I'll redraw what I think are the correct FBD and KD when I have a little more time.

Thank you both for your help!
C

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Lnewqban
The reality is that your caliper friction force does not come alone. To fully do the analysis including the caliper force on the disk, you must also include the same caliper force acting on the body (Newton's third law). This is how your caliper torque will be canceled. Then the body will interact with ground as well and your system will be closed.

The problem with your FBD is that the vehicle's body is missing. A caliper alone on a wheel will just turn with the wheel if it is clamped. Something must restrain it from moving. Add the body to your FBD and consider all forces and moments acting on each component, but consider also the constraints also (ex.: the linear accelerations of the body and wheel are the same in direction and magnitude, even though one's acting on the body mass and the other on the wheel mass).

Doing it the way I did, I just bypassed the internal forces that cancel each other out, and ended up with the overall result for the whole vehicle. The important thing to understand is that no matter how it works internally, it cannot produce more than what the overall system can provide. Obviously, if you design a brake system, you will have to analyze the internal forces at some point.

Lnewqban

## 1. How does a brake caliper decelerate a chassis?

A brake caliper decelerates a chassis by applying friction to the rotating disc or drum, which in turn slows down the wheels and the entire vehicle. This friction is created by the brake pads, which are pressed against the disc or drum by the caliper.

## 2. What factors affect the deceleration of a chassis by a brake caliper?

The deceleration of a chassis by a brake caliper can be affected by several factors, including the material and condition of the brake pads, the size and design of the caliper, the weight and speed of the vehicle, and the condition of the road surface.

## 3. How is the deceleration of a chassis by a brake caliper measured?

The deceleration of a chassis by a brake caliper is typically measured in terms of G-forces, which is the force of gravity acting on a mass. This can be calculated by dividing the deceleration rate (in meters per second squared) by the acceleration due to gravity (9.8 meters per second squared).

## 4. What is the role of rotational dynamics in modeling brake caliper deceleration?

Rotational dynamics is the study of how objects rotate and the forces that act upon them. In the case of brake caliper deceleration, rotational dynamics is important because it helps us understand how the rotation of the wheels and the disc or drum affects the deceleration of the chassis and how the forces from the brake caliper are distributed.

## 5. How can brake caliper deceleration be improved?

Brake caliper deceleration can be improved by using high-quality brake pads, ensuring proper maintenance and alignment of the caliper, and driving at appropriate speeds for the road conditions. Additionally, advancements in technology have led to the development of more efficient brake systems, such as anti-lock braking systems (ABS) and regenerative braking systems, which can also improve deceleration.

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