0.3 Optimizing for Gas Instead of Time

In 0.2 the same stop constraints

x(T) = D, \qquad v(T) = 0, \qquad a(T) = 0

were paired with a rush: finish as soon as possible. With a hard force limit, that meant bang-bang — slam the gas ( $+F$ ), then slam the brake ( $-F$ ).

The clock is not the only score. Suppose the shop bills the driver for gas, not overtime. Same door, same rest conditions — different objective.

A simple gas proxy

Keep the vehicle: no drag, $m\ddot{x} = F$ . Real engines burn fuel in messy ways. Here we use a toy meter that matches the story:

G = \int_0^{T} |F(t)|\,\mathrm{d}t.

Gas used is proportional to effort — how hard and how long you push or brake. Units are N·s (impulse). Coasting with $F = 0$ adds nothing to $G$ , even though the clock keeps ticking. At rest with $F = 0$ , gas stays flat too.

(Other proxies exist — $\int F^2\,\mathrm{d}t$ punishes large forces more sharply. $|F|$ is enough to feel the trade-off.)

Gentle force, longer trip

Fix an accel-then-brake plan with equal $|F|$ and a halfway switch (no coast). Larger $|F|$ shortens $T$ , as in 0.2. Under $G = \int |F|\,\mathrm{d}t$ , that same slam raises gas: you pay a high rate for every second of thrust and brake. A smaller $|F|$ stretches the schedule but keeps the meter lower. The door conditions do not care which you pick — both can stop cleanly at $D$ .

Coasting is free (in this model)

You can also accel briefly, coast ( $F = 0$ ) at constant speed, then brake so $v = 0$ at the door. Coast distance costs time and zero gas. The plan still has to leave enough room to scrub speed before $x = D$ . Open-loop again: marks chosen ahead of time, not from a sensor.

Try it

Compare Min time (max F), Sip gas (small F), and Coast-heavy. Read both $t$ and $G$ after each run. Open Data to see gas climb only while $|F| > 0$ .

Discussion

Time-optimal and gas-optimal are different problems with the same boundary conditions. For minimum time with $|F| \le F_{\max}$ , extremes win: never linger below the limit, and coasting wastes the clock. For minimum $G = \int |F|\,\mathrm{d}t$ in this drag-free toy, the incentives flip. Soft force and long coast both look good on the meter; they look terrible on a stopwatch.

That is the first taste of multi-objective control. Changing the cost changes the “best” open-loop plan even when the vehicle and the destination stay fixed. Later chapters will meet the same idea with matrices ( $Q$ vs $R$ in Section 4.1, LQR) and with hard constraints (Section 4.2, MPC). Here the pizza run is enough: pick what you care about — minutes or gas — and watch the widget disagree with itself.

When brakes are free

The meter $G = \int |F|\,\mathrm{d}t$ charges for both throttle and brake. That is a fair “effort” score, but it is an odd gas story: cars do not burn fuel by pressing the brake pedal. (They do wear pads — a different cost.)

A closer toy for “tank of gas” is to bill only when the force is forward:

G_{\mathrm{throttle}} = \int_0^{T} \max\bigl(F(t),\, 0\bigr)\,\mathrm{d}t.

While $F < 0$ (braking) or $F = 0$ (coast / rest), the integrand is zero. Gas climbs only during the accel phase. In the playground above, switch the meter to throttle only (brakes free) and re-run Min time vs Sip gas. Under $G_{\mathrm{throttle}}$ , the brake half of a bang-bang plan is free on the gas plot — you still pay for the slam of $+F$ , and a smaller throttle still looks cheaper if you are patient.

The lesson is the same as in the Discussion: the cost function is a modeling choice. Absolute value treats push and pull as equal effort; $\max(F,0)$ treats braking as free. Real fleets mix fuel, time, wear, and comfort. The widget just lets you feel how flipping that choice moves the numbers.