Drag

Drag is the force of a fluid (air or water) on a moving object, parallel to the direction of motion. (The perpendicular component is Lift.)

Why $v^3$: the object has to push the fluid in front of it out of the way = give it kinetic energy.

• Volume swept per second $= Av$ → mass per second $\dot m = \rho A v$
• Each kg gets $\frac12 v^2$ → $P = \frac12 \dot m v^2 = \frac12 \rho A v^3$
• $c_{drag}$ corrects for shape and flow (laminar/turbulent): $P_{drag} = \frac12 c_{drag} A \rho v^3 \qquad F_{drag} = E / d = \frac{P_{drag}}{v} = \frac12 c_{drag} A \rho v^2$ $v^3 = v \times v^2$: one $v$ from the mass pushed per second ($\dot m = \rho A v$), $v^2$ from the kinetic energy each kg gets ($\frac12 v^2$). Double speed → 2× mass, 4× energy per kg → 8× power.

$A$ = frontal cross-sectional area (perpendicular to the motion).

• Car / plane: frontal area $A_{front}$
• Ship: water part $XW$, air part $(H - X)W$ → see Buoyancy. Archimedes’ principle
• Ship air drag ≈ 1% → ignore: $\rho_{water}/\rho_{air} \approx 1000/1.3 \approx 770$

Energy for a trip: $v^2$ or $v^3$?

$E = P \cdot t$ → what is fixed decides.

Fixed	Compare	Scales with
distance (A → B, per km, per trip)	$E = \frac{P}{v} d = F_{drag}\, d$	$v^2$
time (per hour, per day)	$E = P\, t$	$v^3$

Slower over a fixed distance: $P$ drops as $v^3$, but the trip takes longer ($t = d/v$) → net $v^2$. e.g. 80% of $v$: $P$ × 0.512, $t$ × 1.25 → $E$ × 0.64 (36% saved, not 49%)

Related: Lift, Rolling resistance, Braking power