How to brake 'beautifully'?

Question

Sometimes when I'm driving my car, I play a "game" against myself in which I try to minimize the deceleration felt by passengers (including myself) while still braking in a reasonable short space. I do know that the deceleration felt is not only dependent on what distance you're travelling until you reach a full stop, but also a function of velocity over time, or the "easing" with which you're braking.

Now, I do know this is an effect that has been studied and successfully applied. Not so long ago, I visited a hotel in which the elevator took me up 43 stories in about 20 seconds, with just a slight feeling of accelerating when starting and decelerating when stopping.

So, which is the trick? Which is the correct way to accelerate/decelerate to minimize acceleration perception while still maximizing distance traveled and maximum speed?

(Disclaimer: Please don't kill yourself by playing with your brakes. Safety is first, physics are second.)

Answer 1

We notice sudden changes in anything. We don't notice gradual change, whether in time or spread over space. If the car is moving uniformly along straight flat road, its acceleration a=0. Its velocity v is constant.

When the brake pedal is pushed, friction causes the car to decelarate. a = some negative number. You can't avoid that. You want to stop? Your velocity must decrease. (Duh.)

What the passengers feel is the change from a=0 to a=-5.6948 m/s^2 (making up a number.) You want that change to be gradual. Before braking, when a=0, and assuming you're sufficiently skilled and anticipate needing to stop, you light apply brakes. Suppose you make a linearly decrease from 0 to -5.6948, and once there, keep it there. This linear decrease takes place over, let us say, 5 seconds. Likewise, a linear let-up to a=0 timed perfectly as the car comes to a stop.

Remember acceleration is the time derivative of velocity. Velocity, therefore, is the indefinite integral of acceleration, which is a piecewise linear function of time. Plotting velocity vs time, you'd see pieces of parabolas joined together.

Now, I wonder if passengers would feel the change in acceleration during that time it's changing? It's changing gradually, but it is going from non-changing (zero) to changing linearly, to finally not changing (steady strong braking), with no sudden jumps. Maybe the game is to soften the rate of change of acceleration, to not suddenly go from flat to linear. Look at the time derivative of acceleration - it's all step functions in our example!

So make the time derivative of acceleration be piecewise linear, no jumps. Then the acceleration is smooth, no sudden changes in slope where parabolas join. Velocity is very smooth, described by 3rd degree polynomials. A good driver/pilot approaches this kind of smoothness by intuitive feel and experience.

Answer 2

The existing answers are in the right direction, but they've missed an important force. A car is not just a box that can provide some range of acceleration (via brakes). It's a box attached essentially with stiff springs (the suspension) to wheels which can provide that acceleration. So there are two big forces when you're coming to a stop - the brakes, and the spring force - and we want to minimize acceleration and jerk (derivative of acceleration) for the people, or effectively, the box.

This means that substantial acceleration, by design, results in the car shifting on the suspension. If the acceleration drops off suddenly, the car is shifted back into equilibrium by the suspension, resulting in the jerk we're all used to feeling when a car finally comes to a complete rest. It's not a huge displacement, but we're pretty good at perceiving that sort of thing. And it is jerk in the physical (change in acceleration) sense as well, because the springs are pretty stiff, so the car pretty much gets pulled straight into equilibrium and stopped there.

Avoiding this jerk means making sure the suspension is approximately in equilibrium when you stop, meaning essentially that acceleration has to be as close to zero as possible for some small amount of time before you stop. Really, this is just the result you'd expect for minimizing jerk - let the acceleration smoothly approach zero - but the springs exaggerate the problem. In practice this means smoothly releasing the brake such that it's almost fully released when you stop. This means that by the time you stop, it's just the brakes versus the car idling forward.

The idling, of course, is a third force we haven't accounted for, but it actually makes things easier in practice because of the way we control the brakes. It's roughly a constant forward force, so zero acceleration corresponds to the brake pedal being slightly depressed. This means that by shifting your foot slightly on the pedal, you're able to home in on the right spot, since the resistive force provided by the brake pedal is roughly constant through the whole region. If instead you were aiming for nearly fully releasing the pedal, the resistive force would be large with it depressed and zero once you release it, making it more difficult to control.

Note: sorry, no pictures/plots for now; I'll try to add some when I have some time. DarenW's plots are pretty much the right idea, though, it's just that the zero/near-zero acceleration region is more important than he realized.

Answer 3

In vertical motion, minimize the rate of change of acceleration (jerk). In an elevator, with different accelerations we have different weights, we feel comfortable at a constant acceleration since at constant acceleration we have constant weight.

In horizontal motion, minimize the acceleration. In a car, we feel comfortable at a constant velocity because of inertia. According to Newton's first law, "If an object experiences no net force, then its velocity is constant: the object is either at rest (if its velocity is zero), or it moves in a straight line with constant speed (if its velocity is nonzero)." [Wikipedia].

Note that both vertical jerk and horizontal acceleration are time dependent. The most beautiful braking ever would, sadly, take infinite time to complete (zero jerk and acceleration). However, this only applies if start out stationarily (or moving in the opposite direction). So if you want to consider optimization problems (e.g. calculate the motion of a vertically moving elevator and horizontally moving vehicle that would be the most comfortable for its passengers) we must set time limits/constraints.

Note that the person in the seat (gravity pushes driver against belts, etc.) is comfortable because a zero acceleration is also a constant acceleration. Similarly, a passenger of a vehicle (moving horizontally) would feel comfortable when the car is stationary because a zero velocity is also a constant velocity.

Paradoxically, the most beautiful braking ever is the one that does not exist. If an elevator or car starts out and ends at the same constant velocity, it would have both a minimized (zero) acceleration and minimized (zero) jerk, but would travel through a distance. The example I see most fit is the escalator: it moves both horizontally and vertically (at the same time) and at a constant velocity.

Of course one might argue that it isn't that comfortable to get onto something moving with a constant velocity (it's more comfortable to get into something stationary etc.) but that argument is too complicated... Happy Physics!

References. Images from...

Elevator: http://www.askamathematician.com/wp-content/uploads/2010/08/Elevator.gif

Car: http://www.force-dynamics.com/include/img/simulation-explanation.png

Escalator: http://upload.wikimedia.org/wikipedia/commons/3/37/Escalator_in_Dream_Mall.JPG

Answer 4

To be fair lets do a simplification of the problem and attach a spring (with coefficient $k$) between the car and the passengers to represent the seat and buttocks elasticity. If the position of the car is $x_0(t)$ and that of the passengers $x_1(t)$ then equation of motion of the passengers is

$$ m\;\frac{{\rm d}^2 x_1}{{\rm d} t^2} = k ( x_1 - x_0 ) $$

Now we apply a constant deceleration of the car of $\frac{{\rm d}^2 x_0}{{\rm d} t^2} =-\frac{v_0}{\Delta t}$ where $v_0$ is the initial speed and $\Delta t$ the time interval we want to stop at. The solution for the motion of the passengers is:

$$ x_1(t) = \frac{v_0}{\Delta t\,\omega^2} \left(1-\cos(\omega t)+\omega^2 t\, \Delta t - \frac{1}{2} \omega^2 t^2 \right) $$ and more importantly their acceleration

$$ a_1(t) = \frac{v_0}{\Delta t} ( \cos(\omega t)-1 ) $$

Now your goal is when the car stops $\dot{x}_0(t)=0$ to have the velocity of the passengers also zero.

Answer 5

A couple of other things that can be considered: - As the car's speed approaches zero, static friction will take over from dynamic friction in the brakes, which will suddenly increase the deceleration if constant pressure is maintained on the brake pedal. This effect can be avoided by reducing the pressure at the last moment. - The effect of the springs making the car rock just after stopping can also be reduced in the same way. If you release the brake pedal for a moment just as you stop, the wheels, having less mass than the body of the car, should do most of the oscillating.