Drag — pressure (form) drag, skin friction drag: Worked examples

This page is a drill sheet. The parent note built the two drag sources; here we hit every kind of number the topic can throw at you — small speeds, huge speeds, zero viscosity, flat plates, spheres, terminal fall, creeping crawls, and a nasty exam twist.

First, let's list what "every kind" even means.

The scenario matrix

Drag lives on one master formula (from the parent):

$D = \frac{1}{2} ρ v^{2} A C_{D}$

Read the symbols like a shopping list:

$ρ$ (Greek letter "rho") = density of the fluid, how much mass sits in each cubic metre. Units $kg/m^{3}$ .
$v$ = the relative speed between body and fluid. Units $m/s$ .
$A$ = a reference area (usually the frontal area — the shadow the body casts into the flow). Units $m^{2}$ .
$C_{D}$ = the drag coefficient, a pure number (no units) that packages the shape's messiness.

Every question about drag changes one knob and asks what happens. Here is the full table of knobs and edge-cases we must cover:

Cell	What varies / degenerate case	Which drag dominates	Example
A	Speed doubled ( $v \to 2 v$ )	either	Ex 1
B	Flat plate, two orientations (form vs friction split)	both, compared	Ex 2
C	Wall shear from a velocity slope ( $τ_{w} = μ d u / d y$ )	friction only	Ex 3
D	Degenerate: $v = 0$ (body at rest)	neither → $D = 0$	Ex 4
E	Limiting: inviscid fluid ( $μ \to 0$ )	d'Alembert, $D_{form} \to 0$	Ex 4
F	Real word problem: skydiver at terminal velocity	form (parachute)	Ex 5
G	Density change (sea level vs altitude)	either	Ex 6
H	Non-linear velocity profile (curved $u (y)$ )	friction only	Ex 7
I	Exam twist: they give you drag, ask for $C_{D}$ (invert the formula)	either	Ex 8
J	Limiting: creeping (Stokes) flow, tiny slow body, $D \propto μv R$	friction-like, linear in $v$	Ex 9

We now walk every cell. Prerequisites you may want open: Viscosity & Newton's Law of Viscosity, Boundary Layer & No-Slip Condition, Terminal Velocity, Dimensional Analysis & Drag Coefficient.

Intuition What "high Reynolds number" means, once, for the whole page

The Reynolds number $R e$ is a pure number that compares how strongly the fluid's inertia (its tendency to keep barrelling forward, scaling with $ρ v^{2}$ ) beats its viscosity (its stickiness, scaling with $μv / L$ for a body of size $L$ ). See Reynolds Number. When $R e$ is large (fast flow, big body, thin fluid) inertia wins: the flow separates into a wake and the drag coefficient $C_{D}$ settles to a roughly constant value that no longer changes with speed. That is exactly the regime where our master formula $D = \frac{1}{2} ρ v^{2} A C_{D}$ behaves as " $D \propto v^{2}$ with fixed $C_{D}$ " — the assumption Examples 1, 2, 5, 6, 8 quietly lean on. When $R e$ is tiny (slow, small, gooey — Example 9) viscosity wins instead, and the whole picture changes to $D \propto v$ . Keep this split in your head; it decides which formula is even legal.

Example 1 — Cell A: doubling the speed

Step 1 — Isolate the only thing that changed. Since $ρ, A, C_{D}$ are constant, the whole formula collapses to $D \propto v^{2}$ (the symbol $\propto$ means "is proportional to" — grows in lockstep with). Why this step? We are at high Reynolds number (a cyclist in air is fast and large — see the callout above), so $C_{D}$ is essentially constant and does not change when the speed changes. That is precisely what lets us treat $\frac{1}{2} ρ A C_{D}$ as one frozen constant and compare drags by their speeds alone.

Step 2 — Take the ratio. $\frac{D _{2}}{D _{1}} = \frac{v _{2}^{2}}{v _{1}^{2}} = (\frac{16}{8})^{2} = 2^{2} = 4$ Why this step? The unknown constants $\frac{1}{2} ρ A C_{D}$ cancel top-and-bottom, so we never needed them.

Step 3 — Scale up. $D_{2} = 4 \times 30 = 120 N$ Why this step? The ratio told us the new drag is $4$ times the old one, so we simply multiply the known old drag ( $30$ N) by that factor to land on the actual number.

Example 2 — Cell B: same plate, two orientations

The figure below draws both cases side by side — study it before the numbers:

Figure — Drag — pressure (form) drag, skin friction drag

Step 1 — Compute the dynamic-pressure block once. $\frac{1}{2} ρ v^{2} A = \frac{1}{2} (1.2) (1 0^{2}) (0.5) = 30 (this is a force, in N, per unit C_{D})$ Why this step? Everything except $C_{D}$ is shared by both cases, so compute it a single time.

Step 2 — Multiply by each $C_{D}$ . $D_{a} = 30 \times 0.005 = 0.15 N D_{b} = 30 \times 1.2 = 36 N$ Why this step? $C_{D}$ is the only knob that carries the orientation information — it silently encodes "how big is the wake" drawn on the right of the figure.

Step 3 — Ratio. $\frac{D _{b}}{D _{a}} = \frac{1.2}{0.005} = 240$

Example 3 — Cell C: wall shear from a linear profile

The figure plots this profile and marks the one slope that matters:

Step 1 — Read Newton's law of viscosity. $τ_{w} = μ \frac{d u}{d y}_{y = 0}$ Here $τ_{w}$ is the wall shear stress (force per area the fluid rubs along the surface, in Pa). See Viscosity & Newton's Law of Viscosity. Why this step? Shear stress is defined as viscosity times the rate of shear. No slope, no rub.

Step 2 — Differentiate the profile. For $u = 200 y$ , the slope is constant: $\frac{d u}{d y} = 200 s^{- 1}$ . In the figure this is why the lavender line is straight — one slope everywhere. Why this step? A straight line has one slope everywhere, so the value at $y = 0$ is just $200$ (the mint dot on the figure marks the no-slip point where $u (0) = 0$ , but the drag uses the slope, not the value).

Step 3 — Assemble stress, then force. $τ_{w} = (1.8 \times 1 0^{- 5}) (200) = 3.6 \times 1 0^{- 3} Pa$ $D_{friction} = τ_{w} A = (3.6 \times 1 0^{- 3}) (2) = 7.2 \times 1 0^{- 3} N$ Why this step? Newton's law gives a stress (force per area), but the question asks for a force; multiplying the stress by the wetted area $A$ sums that rub over the whole surface — exactly $\oint τ_{w} d A$ for a uniform stress.

Example 4 — Cells D & E: the two "nothing happens" cases

Definition Pressure

$p$ , and the form-drag piece $D_{form}$ Pressure $p$ is the normal (straight-in) push the fluid exerts per unit area on the surface, in pascals ( $Pa$ ). As fluid flows around a body its pressure is not uniform: at the front stagnation point the flow is brought to rest and pressure is high ( $p_{front}$ ); around the sides and rear it can drop ( $p_{back}$ ). The pressure (form) drag component $D_{form}$ sums how much more the front pushes than the back: $D_{form} = \oint (p_{front} - p_{back}) d A_{⊥}$ where $d A_{⊥}$ is a tiny patch of the body's frontal (cross-section) area — the area seen face-on into the flow, perpendicular to $v$ . We use $d A_{⊥}$ (not the full surface area) because only the part of each pressure force pointing along the flow survives as drag; projecting onto the frontal plane does that bookkeeping automatically.

Step 1 — Case D, plug $v = 0$ . $D = \frac{1}{2} ρ (0)^{2} A C_{D} = 0 N$ Why this step? No relative motion → no momentum change in the fluid → nothing to push back. The $v^{2}$ makes drag vanish quadratically as the body slows.

Step 2 — Case E, the subtle one. With $μ = 0$ there is no boundary layer and no flow separation. The pressure wraps symmetrically front-to-back, so $p_{front}$ recovers to an equal $p_{back}$ and: $D_{form} = \oint (p_{front} - p_{back}) d A_{⊥} = 0$ and $D_{friction} = 0$ because there is no viscosity to rub. Total $D = 0 N$ . Why this step? This is d'Alembert's paradox: viscosity is the hidden cause of both drags. Kill $μ$ and both vanish — even though the body is moving.

Example 5 — Cell F: skydiver terminal velocity (word problem)

Step 1 — Force balance. $D = m g \Rightarrow \frac{1}{2} ρ v^{2} A C_{D} = m g$ Why this step? Constant velocity means net force $= 0$ ; the only two vertical forces are weight down and drag up.

Step 2 — Solve for $v$ . $v = \frac{2 m g}{ρ A C _{D}} = \frac{2 ( 90 ) ( 9.8 )}{( 1.2 ) ( 25 ) ( 1.4 )}$ Why this step? We invert the $v^{2}$ by taking a square root — the same $v^{2}$ that punished us in Example 1 now helps us: it keeps terminal speed small because a little speed buys a lot of drag.

Step 3 — Number. $v = \frac{1764}{42} = 42 \approx 6.48 m/s$ Why this step? We finish the arithmetic under the root ( $1764/42 = 42$ ) and take the square root to turn the symbolic answer into the actual landing speed a designer can check against reality.

Example 6 — Cell G: same car, sea level vs altitude

Step 1 — Ratio, holding $v, A, C_{D}$ fixed. $\frac{D _{1}}{D _{0}} = \frac{ρ _{1}}{ρ _{0}} = \frac{0.9}{1.2} = 0.75$ Why this step? Every factor except $ρ$ cancels, so this is a clean proportion — no need for the actual value of $C_{D}$ .

Step 2 — Scale. $D_{1} = 0.75 \times 400 = 300 N$ Why this step? The ratio $0.75$ tells us the new drag is three-quarters of the old; multiplying the known sea-level drag ( $400$ N) by that fraction converts the proportion into the actual altitude drag.

Example 7 — Cell H: a curved velocity profile

Step 1 — Differentiate the whole profile. $\frac{d u}{d y} = 6000 y + 150$ Why this step? We need the rate of shear, which is the derivative — Newton's law only reads slopes, never the value of $u$ itself.

Step 2 — Evaluate at the wall $y = 0$ . $\frac{d u}{d y}_{y = 0} = 6000 (0) + 150 = 150 s^{- 1}$ Why this step? Skin friction is generated at the surface; the curved $y^{2}$ term contributes nothing there because its slope starts at zero (the no-slip condition anchors the curve).

Step 3 — Stress and force. $τ_{w} = (1.8 \times 1 0^{- 5}) (150) = 2.7 \times 1 0^{- 3} Pa$ $D_{friction} = τ_{w} A = (2.7 \times 1 0^{- 3}) (4) = 1.08 \times 1 0^{- 2} N$

Example 8 — Cell I: the exam twist (invert the formula)

Step 1 — Rearrange the master formula. $D = \frac{1}{2} ρ v^{2} A C_{D} \Rightarrow C_{D} = \frac{D}{\frac{1}{2} ρ v ^{2} A}$ Why this step? $C_{D}$ is the only unknown, so isolate it by dividing both sides by the dynamic-pressure block. This is exactly how drag coefficients are measured experimentally.

Step 2 — Compute the denominator. $\frac{1}{2} (1.2) (2 0^{2}) (0.3) = \frac{1}{2} (1.2) (400) (0.3) = 72 N$ Why this step? The denominator is the dynamic-pressure block (a force in N); working it out first gives us a single clean number to divide by, and its unit (N) guarantees the final answer will be dimensionless.

Step 3 — Divide. $C_{D} = \frac{54}{72} = 0.75$ Why this step? Dividing the measured drag (N) by the block (N) cancels the units, leaving the pure shape number $C_{D}$ that the experiment was designed to extract.

Example 9 — Cell J: creeping (Stokes) flow, drag linear in speed

Step 1 — Recognise the regime and pick the right law. $D = 6 π μ R v$ Why this step? At low Reynolds number viscous forces dominate and inertia (the $ρ v^{2}$ momentum-flux argument) is negligible, so the dynamic-pressure formula no longer applies — Stokes derived this viscosity-only result instead. It is essentially all friction-type drag, linear in $v$ , $μ$ , and $R$ .

Step 2 — Plug in the numbers. $D = 6 π (0.1) (1 \times 1 0^{- 4}) (2 \times 1 0^{- 3})$ Why this step? Each factor enters to the first power, so unlike the master formula there is nothing to square — this is what "linear in velocity" looks like in practice.

Step 3 — Number. $D = 6 π \times 2 \times 1 0^{- 8} = 12 π \times 1 0^{- 8} \approx 3.77 \times 1 0^{- 7} N$ Why this step? Collecting the constants ( $0.1 \times 1 0^{- 4} \times 2 \times 1 0^{- 3} = 2 \times 1 0^{- 8}$ ) and multiplying by $6 π$ gives the actual settling drag — a fraction of a micronewton, as forecast.