Braking: Newsletter Challenge (03/28/06)

A constant rate slows evenly (drag) but the distances needed to eventually stop would be really-really long.(not realistic) A constant GRADUAL (evenly increasing) pressure provides the smoothest shortest stop.ABS not included. In ABS the car brakes pulse (pressure-no pressure-pressure etc.) The pulse rate is so fast you rarely notice the jerking it causes so it FEELS smooth.

By the usage, twerp, we must assume this scenario takes place at least two generations ago, before the advent of ABS. But, lonster's take seems quite succinct: she's right; and she's wrong. But the instructor is most definitely wrong in not sticking to driving, leaving physical science in the classroom.

I would have to wonder if you have actually experienced ABS action. On the rare occasions I have had to brake hard enough for it to engage I would hardly characterize the loud hammering as smooth or hardly noticeable. I have never had the ABS kick in under normal braking conditions nor would there be a reason for it to do so. I suppose on ice the effect would be negligible, along with your subsequent deceleration. When it kicks in on dry pavement, you will notice it.

If you want to generate the most efficient braking maneuver you have to stab the pedal with authority. This raises the brake pad compund and its mating surface to operating pressure and efficiency quickly. As the brakes begin working you have to modulate the pressure so that wheel lockup does not occur while speed continues to decline. If properly done the car will come to rest quickly, smoothly and efficiently. The trick is to be able to reduce the intial pressure quickly since many systems will simply lock the wheels and contribute nothing to the slowing of the car, leaving only the tire patch to dissipate the heat which it won't do very well. The reason ABS works so well for untrained drivers is that the stab and hold method is automatically modulated and the pads and rotors can come to operating temperature quickly. The secondary advantage is that you can steer, since the steering wheels are not locked up. In racing situations, however, you want to be able to lock the brakes, particulary in oval racing, because the car will begin to spin towards the center of the turn or away from the wall. Meanwhile you hold the wheel straight, tuck in your arms, brace your body in the seat and wait for all the noise to stop.

Green engineers ... Got to love 'em. AllState and the passenger drinking a cup of coffee won't like that instructor. The distance to stop at a constant rate will vary with the pressure applied, and you WILL have a sudden jerk at the end. (either because too much pressure at the end overcomes the energy left and the car jerks to a stop (coffee spilled) or too little pressure and the car in front of you removes the rest of the energy (Allstate unhappy). Bell curve it.

Are you guys forgetting that the driver feels the G's and adjusts the pedal force accordingly, without going thru equations? Ferget all this geek calculation, brakeing happens, and if your daughter does not pay attention and get with it, S... will happen too.

Since it's the KE of the car that you're trying to dissapate in order to reduce the velocity of the car, but the effects that you notice are the decrease in velocity and acceleration.

What the instructor wants is a smooth deceleration...i.e. a smooth rate of reduction of velocity.

But....your constant level of braking reduces the KE at a constant rate and velocity by a square relationship so you actually want to reduce the level of braking in a square root function as your velocity drops (in the limit, no braking necessary at zero velocity).

I think you are pretty close. I see the deceleration curve as being more S shaped. You want to ease into the pedal to avoid the initial head jerk, so squeezing down on the pedal is required. As you approach the end of your stop you ease back up on the pedal to avoid the head jerk when you come to rest. It is really a seat of the pants experience.

The final result you want is a smooth entry into the braking and a smooth exit to the minimum pedal force required to stop the car. The physics between the brake pedal and the stopping of the car is more complex than a simple linear function. Friction between the brake pad and the disk/drum changes as temperatures increase due to friction. Road surface and inclination change the equation as do vehicle speeds and loads. There are other factors, too. Long braking, such as going down long inclines should include time where you get off the brake to let them cool down (less of an issue with high performance cars designed to scrub off large amounts of heat) so as to avoid brake fade and rotor damage.

So, back to my seat of the pants measurement philosophy. You want to apply pressure to the brake pedal in a fashion that is not digital, but contains a short period where deceleration builds up to the desired deceleration level, maintain that level through most of the braking distance, and complete the stop with a slight flaring action to avoid the neck jerk at the end. If done correctly you should not notice you have stopped without looking to see that the car is no longer rolling.

Lastly, and most important, you should always leave enough reserve braking to allow you to shorten your intended stopping point if the situation changes and requires emergency action. Do not modulate (pump) the brake pedal if you have engaged the ABS system.

Nicely put. We are the control element in a servo loop. We learn (or some of us do) to do the comfort thing at the beginning and the end but the middle often is subject to adjustment based on an assessment of the progress of the deceleration. Some of the first automatic braking on a subway system had to be modified to "feather" the end of the deceleration curve to prevent the "jerk" that would occur at the end when the energy stored in the elasticity of the system is recovered to quickly.

S-shaped is exactly right, but with a longer tail on the "s". Next time you initiate normal braking in your car, notice the pedal pressure you are exerting. You gradually apply pressure initially, to avoid pitching forward, until the desired rate of deceleration is reached,at which point you begin to release that pressure, even more gradually, because there is increasingly less momentum for the brakes to overcome as the vehicle slows.

The real question being asked here, is how can we prevent our daughters from driving?

for the car to decelerate smoothly, it requires a constant deceleration. as the car slows, it has less KE and therefore needs less pressure on the brake to maintain the retardation.

The Kinetic energy of the car depends upon two variables: the mass (m) and the speed (v). The following equation represent the kinetic energy (KE) of the car. KE = ½ * m * v2 where m = mass of object v = speed of object This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold decrease in speed, the kinetic energy will decrease by a factor of four; for a threefold decrease in speed, the kinetic energy will decrease by a factor of nine. The kinetic energy is dependent upon the square of the speed. Therefore would decreasing the KE at a constant rate bring the car to the smoothest finish as the driving instructor states ??

You've got it. Constant rate of energy removal will bring you to a screeching halt, depending on what constant rate you choose... Instructor's wrong.

I second this statement. If the energy removal "rate" is based upon time (Joules/second,for example), it has absolutley nothing to do with ABS or cubic spline. The cubic spline will smooth the passengers' experience of the deceleration force, but at the price of violating the condition of maintaining a constant rate of energy removal. To maintain a constant rate of energy removal would require extreme deceleration as the velocity approaches zero: certainly beyond the capability of tires on pavement. Maybe you could acheive it with a very robust rocket-sled braking system (a multi-stage caliper system gripping a rail). Well, I have to concede that you could set a modest rate of energy removal intially and maintain that. You'd need a couple of miles to get stopped, depending upon initial speed. It would be interesting to put the numbers to that!

Forget all that nonsense about brake pressure, stopping distance, and antilock brakes. Can't you guys stay on topic? The question is, does removal of the car's kinetic energy at a constant rate give a smooth deceleration? Yes it does. With E for kinetic energy, v for velocity, m for mass, A for acceleration: We have F=mA and E=(mv^2)/2. From the latter, dE/dt=mdv/dt=mA=F, so if dE/dt is a constant, F is a constant and the deceleration is smooth. As another poster points out, sudden changes in force result in a shock so ideally the rate of change of energy removal (and deceleration force) should be increased from zero smoothly at the start of braking, and decreased smoothly to zero as the velocity reaches zero. A nice cubic spline function should do it.

I think you have most of the equations there but are forgetting that there is not much energy left in a car at slow speeds, requiring exponentially faster deceleration to achieve the same rate of energy transfer.
If we take power, the rate of energy transfer to be Power = Force * Distance/Time, and Distance = Velocity * Time, then Power = Force * Velocity
Now, if we take Force = Mass * Acceleration, then Power = Mass * Velocity * Acceleration
Now finally, if we assume, from the teacher comment that power is constant and mass is constant, then by dividing mass on each side, we get the equation K = Velocity * Acceleration, K being an arbitrary constant equal to Power/Mass. Then, divide by velocity, and you'll find that K/Velocity = Acceleration, so as velocity approaches zero, acceleration would have to approach infinity.
I think we can all agree that cars tires can only stop the car at close to 1g, so I hope we can also agree that 1g is not close to infinity and that it will cause the tires to skid, probably not stopping the car the fastest or smoothest.

Seriously, I would expect the instructor to be talking about smooth deceleration from a seat of the pants experience, not instrumenting a car with test equipment and giving a physics lesson. Remember, this is a High School driving education class. If this is a modern class in a typical school, most of the students are only reading at sixth grade level or worse and their idea of history was yesterday's lunch menu.

The bottom line when you drive is to be smooth. The same applies in spades when you are racing. Smooth braking is accomplished with a smooth application of the pedal to gently (not violently) apply pressure to the brake pedal, set and maintain a constant deceleration rate (as felt by the seat of the pants), and complete the stop (if coming to a complete stop) by flaring the amount of pressure on the brake pedal to reduce deceleration smoothly to zero.

The lead in and trailing flare to the braking action is a bit more advanced, but the concept of maintaining a constant rate of deceleration is a sound practice versus a PIO (pilot induced oscillation) effect by modulating the brake pedal. PIO is when you apply too little or too much force and then overcorrect and so the cycle continues ad nauseum.

The main reason you want to maintain a constant rate of deceleration is to allow the driver to estimate as accurately as possible where the car will stop. If your deceleration rate is constant it is easier to determine where you will come to a stop than if the deceleration rate is changing. The sooner in your braking cycle you can accurately estimate your final stop point the better you can modify and manage your stop should you need to make changes or if the situation itself changes.

Most people do not even consciously think about that estimation, but it is an inherent part of driving. One of the things we do when landing a sailplane is to maintain a constant sink rate and airspeed. Doing this makes it infinitely easier to determine exactly where your touch down point will be. You only get one landing for each takeoff in a sailplane, so it has to be right the first time.

I agree that the comments about pads, discs, ABS etc are beside the point, and the question is about the basic physics of the situation. But there's a mistake in the calculus here. dE/dt=mvdv/dt so to keep dE/dt constant, which is what the instructor seems to be asking for, requires dv/dt to increase as v falls (as pointed out by halcyon) so the braking force on the car has to increase, giving anything but a smooth stop. And when dv/dt reaches its limit of ~ 1g, dE/dt falls from then on. Instructor should have asked for constant deceleration. But if he meant constant rate of energy loss wrt distance travelled D, rather than wrt time, then dE/dD=mvdv/dD=mv(dv/dt)*(dt/dD)=mdv/dt (as dt/dD=1/v) so constant deceleration achieves this. That would me my defence if I were the instructor!

I always thought dE/dt was power not force (check you dimensions). If you differentiate E=mv^2/2 we should get something like dE/dt=2mv*(dv/dt)=2mva. So rate of change of energy is proportional to the product of instant velocity (v) and instant acceleration. Although this doesn't answer the core question, does constant rate of energy loss mean a smoother ride when it comes to braking :) I think we need to look beyond equations here!

Well when I think more about it, the instructor's arguement doesn't make sense from personal experience. I think I can say that constant loss of energy is not the answer because from the equation dE/dt proportional to product of velocity and acceleration. From personal experience I can tell that acceleration shouldn't increase at lower speed. Just wondering what is it (as in physics) that leads to a smoother ride. I think it is constant acceleration, except when you step on the brake and step off the brake. Both of these should be done slowly. So acceleration should increase from zero to a constant, then stay constant. What happens when the car comes to a complete stop seems a little more complicated though. Why do we get that jerk in the end?

Been thinking about the jerk at the complete stop.. I think this is what happens. We get the final jerk because the acceleration is no longer there. For a second I take myself into a non inertial reference frame (inside the decelerating car). Now when the car is decelarting there's a fictitious force pushing me backwards and to maintain my position I push myself forward. The situation stays like this through out the braking time. And when the car stops completely, the fictitious force is gone (because acceleration is gone), but I keep pushing myself forward for another moment and that's the jerk we experience... our own pushing to counteract the fictitious force. So I think the final acceleration needs to be as low as possible. So my final answer is slow increase in acceleration until we reach the constant level, maintain the constant level of acceleration.... and finally before the complete stop try to reduce the acceleration again (for the final jerk). Now all we have to do is buy our own private roads.

Interesting that several people have used the term "jerk" intuitively. It's actually a technical term, the derivative of acceleration. A ride that is perceived as "smooth" will have a continuous jerk curve over time. More explanation at http://en.wikipedia.org/wiki/Jerk.

I think part of why we get the jerk is that stopping the forward motion of the unsprung weight releases the compression of the springs between the sprung (body) and unsprung (wheels)weights. The body is, in effect, snapping rearward by the energy stored in the compressed springs. The harder you stop, the more those springs are compressed.

You're right, dE/dt has dimensions of power. Who said it was a force? But dE/dt=mvdv/dt as I said in earlier comment, not 2mvdv/dt (the 2 cancels) This doesn't affect the conclusion, that to keep dE/dt constant as v falls, deceleration has to increase.

So far evey one who answerd this question has gone to far into left feild, the simple answer is if you want the car to slow and loose energy at the smoothest possible rate than just remove youre foot from the gas pedal and the engine will loose feul therefore losing energy. Applying the brakes only retards the kinetic energy at the tires but the weight of the car is still in motion which causes backlash and skids.

The simple answer is the daughter is right. If you remove energy from the car at a constant rate, you will begin braking rather abruptly and come to a halt with a jerk. That is not a reference to the instructor by the way. This is not the smoothest way to stop a car.

The fact that dv/dt = c/v from the derivation results in a late response behaviour, the deceleration being rather clumsy at first. In fact, the time for the vehicle to reduce its velocity to v/2 this way happens to be 3/4 of the time to completely stop. The ideal velocity profile should be vice versa, quick to stop, like an exponential decay, if it has the claim to be safe.

I haven't read a response that gets everything right, and I may wind up in the same category. "perkisc" and "halcyon" are on the right track. To "remove energy from the car at a constant rate," as the Challenge question stated, will NOT result in the smoothest stopping of the car. Some "Anonymous Coward," "rcapper," and "Munky" are right about smoothness being related to an "S-curve" shape of velocity. Another "Anonymous Coward" states, contradictingly, that " . . . does removal of the car's kinetic energy at a constant rate give a smooth deceleration? Yes it does," and then states, "A nice cubic spline function should do it." A nice haversine function should do it too. But, the question, "Is she right," about her statement, "I think he's a twerp — that's not going to work very well," remains. Simply put, the car's energy is a function of velocity squared, and to remove energy at a constant rate requires that braking (deceleration) must increase at an increasing rate (braking squared?). There will definitely be a "jerk" at the point when velocity reaches zero, which is NOT a smooth stop. SHE'S RIGHT!

I guess I should have continued with my explanation, because I was only concerned about proving that a constant energy curve would be inverse, and hence not most efficient at lower speeds, not even efficient at higher speeds if the goal is stopping the car quickly. Regardless, she is right, the instructor should have used the term constant deceleration not constant energy. That would at least reduce the max acceleration on the car, but then comes the question of how do you approach and receed from the acceleration, and the answer to that is the feathering to reduce 'jerk'. I think we all know she's right, she usually is. I was never really concerned with what makes a good stop, I think we all pretty much know what that is. What I was trying to prove was that a constant energy curve would be a bad idea, leaving the rest for everyone else to infer.
Not a bad challenge though, I've seen worse.

well guys, there are a lotof interesting opinions on this one. Its correct that its the constant deccelaration and not constant energy which brings the car to a smoother stop. However, the one very important aspect is the inertia. The car moving with a velocity attains an inertia in motion and considering the car to be a two component system, the upper body being one and the transmisson and and tyres being the second one, and both of them joined by a flexible connection, the inertia is the criterion which will make the passenger feel the jerk or asmoother stop. If the car brakes suddenly or at a faster rate of braking, the lower portion of the car will have an abrupt reduction in inertia whereas the upper body in which the passenger is seating continues to be having a higher inertia. The difference in the inertia is what makes the human body feel the jerks. Again, if the difference between the inertias of the upper and lower bodies is reduced by constant decceleration, the braking will be smoother as the body gets adjusted to the difference and does not fel the jerks. Well, guess this makes sense....

The difference between the two components are called spring weight (the car's body and drive train) and unsprung weight (the wheels, tires, brake system, and portions of the suspension itself).

Rereading the question causes me to give pause to what is meant by removing energy at a constant rate (actually transferring the car's KE potential into heat). So I ran two simulations.

First, if constant means dissipating X Joules of energy per unit time, then the car's deceleration will increase geometrically until it has stopped rather violently.

If, on the other hand, constant means the same percentage of its instantaneous KE per unit time (i.e., 10% of its KE is dissipated every 1 second), the car will decelerate in a non-linear fashion so that the velocity will decay exponentially per unit time (1 second). You will approach a velocity asymptote, but never actually come to a complete stop!

So, in the first example the car's deceleration will geometrically increase. That is not a very smooth stop!

The second example is extremely smooth and gets smoother as time goes by, but you never really stop! That is not very useful.

The answer appears that the instructor's method is not right! The smoothest stop is where the deceleration is constant, at least during the majority of the braking.

Is there any way comments could be identified by which contribution they're replying to? i.e. identify "this" on reply heading when doing "Reply to this". I appreciate you can change the entry in the subject box but there are often a lot of entries under each subject title (for a particular challenge) and it's not easy to see which one is being replied to. If it can't be built in can I suggest commenters add it at start of reply?

The question is interestingly phrased. There are multiple types of energy involved with a moving vehicle: Potential Energy, if there is a change in elevation during the braking. Kinetic Energy strictly due to the motion itself. The only energy that the brakes of a car can actually remove is the kinetic energy (KE=0.5*mass*Velocity^2). By removing this engergy at a constant rate, the velocity of the car will not change in a linear fashion because it is squared in the equation. If the girl was to remove energy at a constant rate, the car's velocity will not drop at a constant rate, rather the deceleration will be small at first, but by the time the car is almost at rest, the deceleration will have increased exponentially. As the car finally stops, the girl is practically standing on the brake pedal. This is not a smooth stop by any means. The girl is correct if the insrtuctor is to be taken literally. She should understand that he is a driver-ed teacher, and he is probably not concerned with the laws of physics in detail

Velocity and deceleration are two different things. you would have to intergrate your velocity function to determine your rate of acceleration (or decel) Anyways, what matters here is inertia (the resistance to changes in momentum), that is what gives one the feeling they're in motion. And momentum is a linear product of inertial mass. LINEAR. Thus, removing velocity at a constant rate - as previously stated - results in the square drop-off of Energy, not constant.

HOLD ON EVERYBODY! Dispite the driving instructor's lack of accuracy in his description of the braking action of the car as it relates to the drivers input, THIS IS A TEENAGE GIRL! She made her comment because she couldn't get away with something, he dresses funny, she doesn't like his hair, he talks 'weird', not enough '...like this or like that...' To her, he's a twerp. So anything he said or will say she'll admonish. I'm not going to discount the fact that she might be able to prove she's right. BUT SHE DIDN'T. The next lesson, he may redeme himself. For me, I'll design a security lock keypad. If I'm not in the passenger seat, the car doesn't start.