# Calculating Additional Braking Torque Due To Momentum

1. Apr 15, 2009

### Gazzoo

Calculating "Additional" Braking Torque Due To Momentum

Hi,

First off this is not a homework problem :-)

I'm trying to understand how one could calculate the "additional" torque that vehicle brakes have to apply due to it's momentum?

I have been told that when you consider the vehicle a "closed system" then you don't have to worry about momentum.

However, my case would be more like if I could attach an external torque sensor to the brakes and see the values due to different masses at a given velocity.

The same set of brakes have to work harder to stop a fully loaded semi-truck versus an empty one in the same time/distance, right?

I know that momentum is p=mv and t=fd and I will have to look at an instance/moment of time as the velocity of the vehicle will change as the brakes are applied which is fine.

I would like to know how to make this calculation to determine the torque/forces applied to say the front wheel where it touches the ground.

So for example what would be the instantaneous momentum torque for a 1000kg vehicle travelling at 100km/hr at the front wheel the moment the brakes are applied?

I hope this makes sense.

Last edited: Apr 15, 2009
2. Apr 15, 2009

### Staff: Mentor

Re: Calculating "Additional" Braking Torque Due To Momentum

The momentum of a vehicle is not needed here, only the mass. Maximum braking force is a function of the static friction between the wheels and the road - a function of the mass of the vehicle and the coefficient of friction.

3. Apr 15, 2009

### xxChrisxx

Re: Calculating "Additional" Braking Torque Due To Momentum

: / wut

Max braking force isnt between the wheel and the road. Its between the pad and brake disc.

If you are using the coefficient of friction between the wheel and the road, then its gone horribly wrong and you've locked up the wheels :P

Also to the OP, the idea doesnt really make sense. As you can simply use F=ma to work out the braking force required to stop something in a given distance and time. As you are saying the amount of force to give a set deceleration.

Now this force can be said to act at the contact patch, so you can work out a moment by using t=fd. but why would you want to?

Last edited: Apr 15, 2009
4. Apr 15, 2009

### Cantab Morgan

Re: Calculating "Additional" Braking Torque Due To Momentum

Gone horribly wrong or no... that would be the maximum wouldn't it? I'd hate to think a vehicle's braking system wasn't powerful enough to lock up the wheels, so I'm inclined to agree with russ.

I think Gazzoo is basically asking a statics question, where torques are equally balanced. One torque comes from the force of the road on the tire, and that's equal and opposite to the torque that comes from the force of the brake pad. Now, since the wheel has a larger radius than where the pad applies its force, the fricative force of the pad must be larger than the fricative force of the road.

5. Apr 15, 2009

### xxChrisxx

Re: Calculating "Additional" Braking Torque Due To Momentum

You slow down much, much faster when you dont lock up the brakes. As then the decelerative force come from the tyre friction in sliding.You slow down much faster when the wheels are turning and the bakes are doing the decelerating.

The maximum force that the brakes should apply is that just below whats necessary to lock the wheel, this is the optimal squeeze point. Once you go over that point the brakes are doing nothing to slow you down. So the max coefficient of friction between the pad and disc is the theoretical maximum, not that of the wheel and road.

The calipers can apply a higher pressure (therefore force) to the brake disc, than gravity can on contact patch one its locked up. By extenstion this means that a rotating brake disc will provide a larger decelerating force than the sliding will.

It is a statics problem, but its more simple to work in terms of force. What you say about the wheel radius and brake disc radius is correct.

Last edited: Apr 15, 2009
6. Apr 15, 2009

### Staff: Mentor

Re: Calculating "Additional" Braking Torque Due To Momentum

If the brakes are locked, the torque they are generating is lower than if they are not locked, is it not...?
Coefficient of static friction, ie, the wheels are rolling because there is no sliding between the wheels and road.
Right, and why do the brakes lock up? They lock up because the static friction between the wheels and road isn't enough to keep the wheels spinning. So if, for example, you double the weight of the car, you double that friction force and double the maximum braking force the brakes can apply before locking up the wheels.

And as a result, if you double the mass of the vehicle, your a=f/m has a doubling of the available braking force and a doubling of the available tire friction and therefore no effect on braking distance. In other words, longer stopping distance due to higher weight is a self-correcting problem due to the higher available braking force.

I think this may be just an issue with me bypassing the original question somewhat: the OP is asking about torque on the wheels as generated by the brakes. I'm saying (Captain Morgan said it explicitly) that the brakes are designed to provide as much torque as is needed to lock the brakes up under any possible vehicle weight. If a truck's full weight is 5x its empty weight (guess), then when full its breaks can generate 5x as much torque without causing the wheels to lock up. Because the brakes were designed with this in mind, it is the load of the truck pressing down on the tires that enables an increase in braking force.

There's an important caveat, though: tires deform, so their friction coefficient isn't constant, it actually increases with load. That means that a truck that is full may actually be able to stop in a shorter distance than one that is empty, unless the brakes overheat and lock up...

Last edited: Apr 15, 2009
7. Apr 15, 2009

### xxChrisxx

Re: Calculating "Additional" Braking Torque Due To Momentum

Case in point for not reading what you put throughly enough. Just thought you were talking about max decelerating force occuring when you are sliding along with locked up wheels. Apologies.

The op question makes much more sense now i've just read the above.

8. Apr 15, 2009

### Staff: Mentor

Re: Calculating "Additional" Braking Torque Due To Momentum

No prob - based the knowledge you've displayed in your past posts, I assumed a minor misread.

9. Apr 16, 2009

### Cantab Morgan

Re: Calculating "Additional" Braking Torque Due To Momentum

Very interesting discussion! So, it would seem desirable to have some kind of sensor that applied the magic amount of braking force to maximize deceleration without locking up the wheels. A kind of "drive by wire." My understanding is that so-called anti-lock brakes don't actually do that. Their mission is to allow steering while breaking hard (by applying the brakes in a staccato burst), but not necessarily to shorten your stopping distance?

Anyway, since I like to warp questions to stretch my mind... Suppose the masses of the wheels themselves were a significant fraction of the mass of the vehicle. I wonder how all that angular momentum alters the problem.

We've left the realm of a realistic truck for sure, but does that qualitatively change the problem? Is there some mass ratio of truck to wheels such that it actually pays to lock up the wheels.

10. Apr 16, 2009

### rcgldr

Re: Calculating "Additional" Braking Torque Due To Momentum

I thought it was the other way around, but maybe that's just cornering friction as opposed to braking friction. It's why the "stiffer" end of a car (front or back) drifts (or slips) more in a turn, so that sway bars can be used to adjust oversteer / understeer.

Last edited: Apr 16, 2009
11. Apr 16, 2009

### Gazzoo

Re: Calculating "Additional" Braking Torque Due To Momentum

Hi Guys,

Here's another way to look at things regarding my initial post:

So if I'm interested in the forces/torque that the brake and ground are producing on a moving vehicle and for example I'm looking at a 1000Kg vehicle travelling at 100km/hr and want to understand the "extra" torque produced at the wheel due to momentum.

Now take the engine/transmission/front wheel assembly out of the car and set it up in a lab where the entire assembly is stationary with the wheel resting atop some rollers to allow it to spin when the engine is accelerated.

If I want to mimic/simulate the real world brake/torque situation (ignoring all the other different weight/friction etc factors regarding this setup) by attaching say another motor to the wheel assembly which will spin it in the forward direction.

Now the main engine is accelerated to 100km/hr, then the brakes are applied and at that moment the attached motor (which is spinning in the same direction and speed as the engine) is started and applies an "additional" torque.

So at that moment how much torque would this attached motor need to generate on this stationary wheel setup to simulate the real world momentum torque for a given velocity/time/mass when the brakes are applied?

I hope this helps....

12. Apr 23, 2009

### Jobrag

Re: Calculating "Additional" Braking Torque Due To Momentum

Go back to basics F=MA.
Assuming that you always want to slow down at the same rate (A is constant) then every time you increase the mass of your truck you need to increase the braking force by the same ratio, double the mass double the force.

13. Apr 30, 2009

### Gazzoo

Re: Calculating "Additional" Braking Torque Due To Momentum

Hi Jobrag,

Yes that makes sense.

So what I'm looking for is how to actually calculate this value from a set of parameters if I wanted to simulate this "momentum force/torque" generated by a fixed external motor operating on a fixed engine/transmission/shaft as I described earlier.

What set of equations could I use to get this torque value given that I know the

mass of the vehicle (m)
velocity of the vehicle (v)
some time parameter (t) (although if it's possible to calculate instantaneous torque that
would be ideal or a suggestion on how to do that)
radius/distance/size of rotor and other components etc...

14. May 3, 2009

### sganesh88

Re: Calculating "Additional" Braking Torque Due To Momentum

I did some math regarding this and obtained the value of the friction force generated at the contact patch while braking. (For a single equivalent wheel assumed to support the entire vehicle. I don't think this will distort the actual conditions as we're concerned only about the translational motion)
Friction force = (µkrNRM)/(I+MR2)

(This i obtained from the relation between the linear deceleration of the center of wheel and the angular acceleration of the wheel. The former is dependent entirely on the external force of friction while the latter requires brake torque as well as frictional torque)
where
µk - Kinetic friction coefficient (between the brake shoe and the drum)
r - Effective distance between the center of the wheel and the point of application of the brake.
R - Effective radius of the wheel.
N - Net Normal force applied by the brake shoe on the wheel (actually the drum)
I - Moment of inertia of the wheel (This is where the mass of the wheel alone as said by you enters the picture) (It can be assumed to be mr2/2 where m is the mass of the wheel alone)
M - Total mass of the vehicle

So for the critical condition, when this force becomes equal to the max static friction force possible,we get..

krNRM)/(I+MR2) = µsMg
µs - static friction coefficient between the road and the wheel

N = (µs g) (I+MR2)/ (µs r R )

This N relates to the intensity of push to be exerted at the brake pedal.

Someone else cross check the results. So i guess the OP's query has been answered.

Last edited: May 3, 2009
15. May 6, 2009

### swraman

Re: Calculating "Additional" Braking Torque Due To Momentum

Just thought Id shed some light on this topic...It looks like it is confusing as to whether the maximum force is gotten out of the tires/brakes when they are locked.

First off, the ABS system DOES decrease the stopping distance of a car.

Its a little different than plain old physics that you apply to everything else because of the way a tire works. As it turns out, the amount of force a tire exerts on the ground (and thus the force that accelerates the car) is directly proportional to what is called the Slip Ratio of the car.

The slip ratio is the ratio $$\frac{r \omega - V}{V}$$ during braking and $$\frac{r \omega - V}{r \omega}$$ during acceleration.
Where r is the radius of the wheel, omega is the angular speed of the wheel, and V is the velocity of the car.
Ideally, $$V = r \omega$$ but that is not true in the real world. The tire actually slips as it outputs a force on the ground. An easy way to think of this is: say the slip ratio is .2 and the car is accelerating, then every time the car travels the distance of 10 circumferences, the wheel rotates 12 times (would be 8 times if the car were braking).

The maximum forces to be gotten out of the tire usually occur around a slip ratio of .1. So, for every distance of 10 circumfrences the car travels, the tires actually rotate 11 times (or 9 times if it is braking). The ABS system attempts to keep the slip ratio at -.1 during braking, to maximize braking forces. Plus, as an added benefit, if the tires are not locked during braking, you can still steer the car, so you still remain in control of the steering. Analogus to the ABS, the Traction Control system in a car tries to keep the slip ratio of a car at .1 during acceleration.

If the tires are locked up, then the slip ratio is -1, and the force output of the tire are not optimal. (the graph of slip ratio vs output tire forces maximizes at .1 and -.1, and then decays for slip ratios greater (or less) than .1 (or -.1)

16. May 9, 2009

### sganesh88

Re: Calculating "Additional" Braking Torque Due To Momentum

ABS doesn't reduce stopping distance on all surfaces. On some surfaces with good friction coefficients skidding is much more efficient from the braking point of view. The translational energy is directly dissipated as heat without going through the intermediate rotational form. Of course no steering control and you got to bear the curse of your tires.