Airfoil Theory

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

25 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Airfoil Theory within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what airfoil theory describes physically and when it applies.
State every one of the 25 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Airfoil Theory is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: airfoil theory.
Capstone Studio instructional photograph

Fluid Mechanics — Airfoil Theory: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 25 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

CL	Quantity produced by "CL = lift coefficient" — read its definition and unit from the handbook line directly above the equation.
ρ	Quantity produced by "ρ = fluid density" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = velocity (m/s) of the undisturbed fluid and" — read its definition and unit from the handbook line directly above the equation.
AP	Quantity produced by "AP = projected area of the airfoil as seen from above (plan area). This same area is used in defining the drag" — read its definition and unit from the handbook line directly above the equation.
k1	Quantity produced by "k1 = constant of proportionality" — read its definition and unit from the handbook line directly above the equation.
α	Quantity produced by "α = angle of attack (angle between chord of airfoil and direction of flow)" — read its definition and unit from the handbook line directly above the equation.
β	Quantity produced by "β = negative of angle of attack for zero lift" — read its definition and unit from the handbook line directly above the equation.
CD	Quantity produced by "CD = CD3 +" — read its definition and unit from the handbook line directly above the equation.
where CD∞	Quantity produced by "where CD∞ = infinite span drag coefficient" — read its definition and unit from the handbook line directly above the equation.
AR	Quantity produced by "AR = b = 2" — read its definition and unit from the handbook line directly above the equation.
b	Quantity produced by "b = span length" — read its definition and unit from the handbook line directly above the equation.
Ap	Quantity produced by "Ap = plan area" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The lift force on an airfoil FL is given by
CL tv 2AP
where
coefficient for an airfoil.
The lift coefficient CL can be approximated by the equation
where
The drag coefficient CD may be approximated by
The aspect ratio AR is defined
2 Ap
Ap c
where
The aerodynamic moment M is given by
CM tv 2Apc
where the moment is taken about the front quarter point of the airfoil.
AERODYNAMIC MOMENT CENTER
CHORD
Properes of Water (SI Metric Units)
Absolute Dynamic Kinematic
Temperature Specific Weight Density Viscosity Viscosity Vapor Pressure
(kN/m3) (kg/m3) (Pa•s) (m2/s) (kPa)
0 9.805 999.8 0.001781 0.000001785 0.61
5 9.807 1000.0 0.001518 0.000001518 0.87
10 9.804 999.7 0.001307 0.000001306 1.23
15 9.798 999.1 0.001139 0.000001139 1.70

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory

A wing of chord 1.7 m and span 11.0 m flies at 78 m/s through air of density 1.20 kg/m³ with C_L = 0.95 and C_D = 0.055. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.7 m × span 11.0 m
V = 78 m/s, ρ = 1.20 kg/m³
C_L = 0.95, C_D = 0.055

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.7 (11.0) = 18.70 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.20) (78)^{2} = 3, 650 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.95 (3, 650) (18.70) = 64.85 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.055 (3, 650) (18.70) = 3.754 k N$
Ratio and power
$L / D = 17.3; P = D V = 292.8 k W$

Answer: L = 64.85 kN, D = 3.754 kN, L/D = 17.3, P = 292.8 kW

Why the other options are there

L = 129.7 kN (½ omitted)
L/D = 0.058 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 2

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (2)

A wing of chord 1.1 m and span 4.5 m flies at 47 m/s through air of density 1.11 kg/m³ with C_L = 0.70 and C_D = 0.070. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.1 m × span 4.5 m
V = 47 m/s, ρ = 1.11 kg/m³
C_L = 0.70, C_D = 0.070

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.1 (4.5) = 4.95 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.11) (47)^{2} = 1, 226 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.70 (1, 226) (4.95) = 4.25 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.070 (1, 226) (4.95) = 0.425 k N$
Ratio and power
$L / D = 10.0; P = D V = 20.0 k W$

Answer: L = 4.25 kN, D = 0.425 kN, L/D = 10.0, P = 20.0 kW

Why the other options are there

L = 8.50 kN (½ omitted)
L/D = 0.100 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 3

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (3)

A wing of chord 1.7 m and span 12.5 m flies at 89 m/s through air of density 1.20 kg/m³ with C_L = 0.70 and C_D = 0.075. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.7 m × span 12.5 m
V = 89 m/s, ρ = 1.20 kg/m³
C_L = 0.70, C_D = 0.075

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.7 (12.5) = 21.25 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.20) (89)^{2} = 4, 753 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.70 (4, 753) (21.25) = 70.69 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.075 (4, 753) (21.25) = 7.574 k N$
Ratio and power
$L / D = 9.3; P = D V = 674.1 k W$

Answer: L = 70.69 kN, D = 7.574 kN, L/D = 9.3, P = 674.1 kW

Why the other options are there

L = 141.4 kN (½ omitted)
L/D = 0.107 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 4

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (4)

A wing of chord 2.3 m and span 6.0 m flies at 37 m/s through air of density 1.10 kg/m³ with C_L = 0.60 and C_D = 0.025. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 2.3 m × span 6.0 m
V = 37 m/s, ρ = 1.10 kg/m³
C_L = 0.60, C_D = 0.025

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 2.3 (6.0) = 13.80 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.10) (37)^{2} = 753.0 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.60 (753.0) (13.80) = 6.23 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.025 (753.0) (13.80) = 0.260 k N$
Ratio and power
$L / D = 24.0; P = D V = 9.6 k W$

Answer: L = 6.23 kN, D = 0.260 kN, L/D = 24.0, P = 9.6 kW

Why the other options are there

L = 12.47 kN (½ omitted)
L/D = 0.042 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 5

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (5)

A wing of chord 1.3 m and span 9.5 m flies at 81 m/s through air of density 1.21 kg/m³ with C_L = 0.40 and C_D = 0.035. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.3 m × span 9.5 m
V = 81 m/s, ρ = 1.21 kg/m³
C_L = 0.40, C_D = 0.035

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.3 (9.5) = 12.35 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.21) (81)^{2} = 3, 969 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.40 (3, 969) (12.35) = 19.61 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.035 (3, 969) (12.35) = 1.716 k N$
Ratio and power
$L / D = 11.4; P = D V = 139.0 k W$

Answer: L = 19.61 kN, D = 1.716 kN, L/D = 11.4, P = 139.0 kW

Why the other options are there

L = 39.22 kN (½ omitted)
L/D = 0.088 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 6

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (6)

A wing of chord 1.3 m and span 13.0 m flies at 78 m/s through air of density 0.99 kg/m³ with C_L = 0.35 and C_D = 0.040. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.3 m × span 13.0 m
V = 78 m/s, ρ = 0.99 kg/m³
C_L = 0.35, C_D = 0.040

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.3 (13.0) = 16.90 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (0.99) (78)^{2} = 3, 012 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.35 (3, 012) (16.90) = 17.81 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.040 (3, 012) (16.90) = 2.036 k N$
Ratio and power
$L / D = 8.8; P = D V = 158.8 k W$

Answer: L = 17.81 kN, D = 2.036 kN, L/D = 8.8, P = 158.8 kW

Why the other options are there

L = 35.63 kN (½ omitted)
L/D = 0.114 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 7

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (7)

A wing of chord 1.9 m and span 6.0 m flies at 68 m/s through air of density 0.91 kg/m³ with C_L = 1.20 and C_D = 0.085. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.9 m × span 6.0 m
V = 68 m/s, ρ = 0.91 kg/m³
C_L = 1.20, C_D = 0.085

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.9 (6.0) = 11.40 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (0.91) (68)^{2} = 2, 104 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 1.20 (2, 104) (11.40) = 28.78 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.085 (2, 104) (11.40) = 2.039 k N$
Ratio and power
$L / D = 14.1; P = D V = 138.6 k W$

Answer: L = 28.78 kN, D = 2.039 kN, L/D = 14.1, P = 138.6 kW

Why the other options are there

L = 57.56 kN (½ omitted)
L/D = 0.071 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 8

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (8)

A wing of chord 0.8 m and span 8.5 m flies at 76 m/s through air of density 1.09 kg/m³ with C_L = 0.95 and C_D = 0.050. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 0.8 m × span 8.5 m
V = 76 m/s, ρ = 1.09 kg/m³
C_L = 0.95, C_D = 0.050

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 0.8 (8.5) = 6.80 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (1.09) (76)^{2} = 3, 148 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.95 (3, 148) (6.80) = 20.34 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.050 (3, 148) (6.80) = 1.070 k N$
Ratio and power
$L / D = 19.0; P = D V = 81.3 k W$

Answer: L = 20.34 kN, D = 1.070 kN, L/D = 19.0, P = 81.3 kW

Why the other options are there

L = 40.67 kN (½ omitted)
L/D = 0.053 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 9

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (9)

A wing of chord 1.3 m and span 8.0 m flies at 41 m/s through air of density 0.97 kg/m³ with C_L = 0.55 and C_D = 0.060. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 1.3 m × span 8.0 m
V = 41 m/s, ρ = 0.97 kg/m³
C_L = 0.55, C_D = 0.060

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 1.3 (8.0) = 10.40 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (0.97) (41)^{2} = 815.3 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.55 (815.3) (10.40) = 4.66 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.060 (815.3) (10.40) = 0.509 k N$
Ratio and power
$L / D = 9.2; P = D V = 20.9 k W$

Answer: L = 4.66 kN, D = 0.509 kN, L/D = 9.2, P = 20.9 kW

Why the other options are there

L = 9.33 kN (½ omitted)
L/D = 0.109 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Example 10

Lift, drag and lift-to-drag ratio of a wing section — Airfoil Theory (10)

A wing of chord 2.1 m and span 10.0 m flies at 47 m/s through air of density 0.94 kg/m³ with C_L = 0.40 and C_D = 0.085. Compute the dynamic pressure, lift, drag, L/D ratio and the power needed to overcome drag.

Given

Chord 2.1 m × span 10.0 m
V = 47 m/s, ρ = 0.94 kg/m³
C_L = 0.40, C_D = 0.085

Find

q, L, D, L/D and the drag power

Start with the thinking

Both lift and drag use the same planform area and the same dynamic pressure — only the coefficient changes.
Velocity enters squared in the force and cubed in the power.

Step-by-step solution

Planform area
$S = 2.1 (10.0) = 21.00 m^{2}$
Formula
$q = \frac{1}{2} ρ V^{2}$
Substituting
$q = 0.5 (0.94) (47)^{2} = 1, 038 P a$
Formula
$L = C_{L} q S$
Substituting
$L = 0.40 (1, 038) (21.00) = 8.72 k N$
Formula
$D = C_{D} q S$
Substituting
$D = 0.085 (1, 038) (21.00) = 1.853 k N$
Ratio and power
$L / D = 4.7; P = D V = 87.1 k W$

Answer: L = 8.72 kN, D = 1.853 kN, L/D = 4.7, P = 87.1 kW

Why the other options are there

L = 17.44 kN (½ omitted)
L/D = 0.213 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Airfoil Theory

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Airfoil Theory contains 25 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Airfoil Theory

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

FL =Needs technical review

CL=liftcoefficient

ρ=fluiddensity

v=velocity(m/s)oftheundisturbedfluidand

AP=projectedareaoftheairfoilasseenfromabove(planarea).Thissameareaisusedindefiningthedrag

CL = 2πk1sin(α + β), which is valid for small values of α and βNeeds technical review

k1​=constantofproportionality

α=angleofattack(anglebetweenchordofairfoilanddirectionofflow)

β=negativeofangleofattackforzerolift

CD=CD3​+

whereCD∞=infinitespandragcoefficient

AR=b=2

b=spanlength

Ap=planarea

c=chordlength

M=Needs technical review

CM=momentcoefficient

ρ=fluiddensity

v=velocity

αCAMBERLINE

(∘C)γρµυPv

γViscosityViscosityPv

(×10lbf−sec/ft2​)

−5(×10−5ft2​/sec)

VALUEOFVDFORWATERAT60∘F(V=ft/sec,D=in.)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section