FORCES ACTING ON AN AIRPLANE IN STEADY STATE FLIGHT
I. Wings Level Constant Altitude Flight
Newton’s First Law. A body at rest remains at rest, and a body in motion remains in motion at the same
speed and in the same direction, unless acted upon by an outside force.
For an aircraft in steady state straight and level flight, this means that thrust (T) = drag (D), and lift (L) =
weight (W), assuming that the thrust acts along the flight path (direction of flight).
Formally, Fx = Fy = 0, where the x-axis (thrust-drag axis) is by convention along the direction of flight.
An airplane in this situation will maintain a constant airspeed and altitude.
L
D
T
flight path
W
Scalar vs. Vector. A scalar quantity has only magnitude, while a vector quantity has both magnitude and
direction.
Thrust, drag, lift and weight are all vectors, and are measured in pounds. Airspeed is a scalar, since it has
magnitude but not direction. (Note: velocity—airspeed plus direction—is a vector; we will discuss this
point later in more detail) Other examples of scalars are temperature, pressure, density, heartbeat, GPA,
salary, and height.
Pilots usually measure airspeed in knots (nautical miles per hour). Since there are 6076 feet/nm, and 3600
seconds/hour, we can convert knots to feet/sec as follows:
(x nm/hour) (6076 ft/nm) / (3600 sec/hour) = x’ ft/sec.
That is, to convert knots to ft/sec (fps), multiply by 6076/3600. To convert to feet/minute (fpm), multiply
by 6076/60.
Example. 300 kts = 300  6076 / 3600 = 506.3 ft/sec = 300  6076 / 60 = 30,380 ft/min.
Exercise. Convert 500 kts to fps and fpm.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 1
Relationships in a Right Triangle: A right triangle has one right (90o) angle and two acute angles. The
hypotenuse c of the triangle is the side opposite the right angle. Let  be one of the acute angles. The other
two sides are a and b, with a adjacent to , as shown below. That is, b is the side opposite , and a is the
side adjacent to .
c
b

a
By the Theorem of Pythagoras, c2 = a2 + b2. Also,
sin  = opposite/hypotenuse = b/c;
cos  = adjacent/hypotenuse = a/c;
tan  = opposite/adjacent = b/a.
Note the following limits on the values of these functions, which derive from the fact that 0 <  < 90o.
As  increases, 0 < sin  < 1; 1 < cos  < 0; 0 < tan  < . This will help you avoid errors when using
your calculators. Recall also that
tan a = opposite/adjacent = (opposite/hypotenuse) / (adjacent/hypotenuse) = sin a / cos a.
Example.

20
10

17.32
Note that 102 + 17.322 = 100 + 299.9 = 399.9  202 = 400. Sin  = 10/20 = 0.5, so  =sin-10.5 = 60o. Cos 
= 10/20 = 0.5, so  =cos-10.5 = 30o. The values of the acute angles can also be calculated using the tangent
function. This triangle is the well known “30-60-90” or “one-two-square-root-of-three” triangle.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 2
Key Point: If c is the hypotenuse of a right triangle with acute angle , then the side opposite  is c sin ,
and the side adjacent to  is c cos .
c
c sin 

c cos 
Suppose the thrust vector of an airplane is not parallel the flight path. This is usually the case, since
ordinarily an aircraft in straight and level flight has a small nose up pitch angle p. The wing develops a
force called the aerodynamic force (AF), which we will study in detail later. The component of the AF
perpendicular to the flight path is called lift; the component parallel the flight path is called drag. (Of
course, drag created by the AF is not the only kind of drag.)
Key Point. Lift is always perpendicular the flight path; drag is parallel the flight path. We assume
thrust developed by an airplane is parallel the longitudinal axis. Weight always acts parallel to the earth’s
gravitational field, i.e., vertically downward.
L
T
T sin p
p
D
T cos p
flight path
W
Example, Suppose W = 3000#, T = 1000#, and p = 7 o. Since Fx = 0, F+x = F-x, and D = T cos p. Since
Fy = 0, F+y = F-y, and L + T sin p = W. Then,
F+x = F-x
D = T cos p
= 1000 cos 7o
= 992.5#
F+y = F-y
L = W – T sin p
= 3000 – 1000 sin 7o
= 3000 – 121.87 = 2878#
That is, 122 pounds of weight are being supported by the vertical component of thrust. This shows that lift
is not necessarily equal to thrust in straight and level flight. However, the relative values of T, W, and p are
not necessarily those which apply to light civil aircraft.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 3
Exercise: Suppose W = 5000, L = 4800, p = 5 o. Find D and T.
II. Wings Level Climbing Flight
Now consider the case of an aircraft in a wings-level steady state climb. The angle parameters are c, the
climb angle (flight path); p, the pitch angle; and r, the pitch angle relative to the flight path. Again, L is
perpendicular to the flight path, and drag parallel to the flight path.
T sin r
L
T
r
c
D
T cos r
W cos c
W
c
p
flight path
(x-axis)
p = pitch angle
c = climb angle
r = p – c = relative pitch angle
W sin c
Recall that the x-axis by convention lies along the flight path. Then since F+x = F-x, T cos r = D + W sin c;
and since F+y = F-y, L + T sin r = W cos c.
Example. Suppose p = 32.5o, c = 25o, T = 2000#, and W = 3000#. Then r = p – c = 32.5 – 25 = 7.5o, and
F+x = F-x
T cos r = D + W sin c
D = T cos r – W sin c
= 2000 cos 7.5 – 3000 sin 25
= 1982.89 – 1267.85 = 715.0#.
F+y = F-y
L + T sin r = W cos c
L = W cos c – T sin r
= 3000 cos 25 – 2000 sin 7.5
= 2718.92 – 261.05 = 2458#
Note that thrust is supporting 3000 – 2458 = 542# of the weight of the aircraft.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 4
To sum up, in a steady state climb:




Part of the weight of the aircraft is supported by thrust
Lift is less than weight
Part of the weight of the aircraft acts in the same direction as drag
If Tx and Ty are respectively the components of thrust parallel to and perpendicular to the flight
path, and Wx and Wy respectively the components of weight parallel to and perpendicular to the
flight path. Then
1. Lift = Wy - Ty; and
2. Drag = Tx - Wx
Exercise. Suppose weight is 115,000#, climb angle 20 o, pitch angle 30o, and thrust 52,500#. Find lift and
drag.
Exercise. Suppose weight is 100,000#, climb angle 10 o, pitch angle15o, and thrust 40,000#. Find lift and
drag.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 5
III Wings Level Power Off Descending Flight
The situation for steady state descending flight can be analyzed using an approach similar to that for
climbing flight. We consider the special case of power off gliding descents, i.e., no thrust descents.
a (glide angle)
L
D
W cos a a
W
W sin a
glide path (x-axis)
a
Since F+x = F-x, D = W sin a; and since F+y = F-y, L = W cos a.
Example. Suppose the glide angle is 10o and weight is 3000#. Find lift and drag.
F+x = F-x
W sin a = D
= 3000 sin 10o = 520.945#
F+y = F-y
L = W cos a
= 3000 cos 10o = 2,954.42#
Example. Suppose lift is 8000# and drag is 1200#. What is the angle of descent? The weight?
D / L = W sin a / W cos a = sin a / cos a = tan a
1200 /8000 = tan a
a = tan-1 (1200/8000)
= tan-1 0.15 = 8.530765610o
D = W sin a
W = D / sin a
= 1200 / sin 8.530765610
= 8089.499# = 8089#
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 6
How can we determine the glide angle for maximum glide distance?
L
D
L = W cos a
a
W
glide path (x-axis)
D = W sin a
a
Tan a = D / L. Since the tangent function of an acute angle increases as the angle increases, minimizing D /
L will minimize a. That is, the minimum glide angle (and hence maximum glide distance) is achieved
when the lift-drag ratio L / D is maximum, i.e., at (L / D)max.
(L / D)max corresponds to a unique angle of attack (AOA) for a given airplane. We will discuss AOA in
greater detail later in the course. As weight increases, the airspeed for maximum range power off glide
increases, but AOA remains constant. Note that maximum power off glide distance is independent of
weight; e.g., an airplane full of fuel glides just as far as the same airplane empty of fuel.
If the glide angle and altitude are known (or can be calculated), the glide distance can be determined as
follows.
Tan a = altitude / glide distance, so glide distance = altitude / tan a.
Example. Suppose altitude is 15,000 feet, and a = 10 o. Then glide distance (in feet and nm) can be
calculated as follows:
glide distance = altitude / tan a
= 15,000 / tan 10 = 85,069.2273 ft
= 85,069 ft / 6076 ft/nm = 14.0008 nm.
altitude AGL
a
glide distance
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 7
Exercise. Suppose weight is 20,000#, and drag is 2000#. Find power off glide angle, lift, glide distance in
nm from 5000 feet altitude AGL, and glide distance in nm per thousand feet of altitude.
III Constant Altitude Turning Flight
Here we discuss flight where speed, altitude, bank angle, turn radius, and G force are constant.
The circumference of a c circle of radius r and diameter d = 2 r is c = 2 r =  d.
Example. What is the rpm of a 7’ radius propeller with a tip speed of 975 ft/sec?
Diameter of prop d is 14’, so in one revolution a blade tip moves 14  = 43.98229617 ft/revolution.
Then (975 ft/sec) (60 sec/min) / (14  ft/revolution) = 1330.08 revolutions/min (rpm).
In general, RPM (rev/min) = tip speed (fps) (60 sec/min)/ (2  r ft/rev)
Exercise. Suppose RPM = 2500 and propeller radius is 4’. What is the tip speed?
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 8
Newton’s Second Law. A body of mass m acted upon by a force F experiences an acceleration a in the
direction of the force directly proportional to F and inversely proportional to m.
That is, F (#) = m (slugs)  a (ft/sec2). Acceleration a is the rate of change of velocity.
Time(sec)
0
5
10
15
20
Velocity (ft/sec)
0
10
20
30
40
If acceleration of object reflected in above table is constant, a = 40 ft/sec / 20 sec = 2 ft/ sec 2. That is, each
second the object increases velocity by 2 ft/sec.
Mass m is measured in m slugs = (F #) / (a ft/sec2) = (F/a) #-sec2 / ft. That is, the slug (#-sec2 / ft) is the
unit of mass accelerated at the rate of 1 ft/sec2 when acted upon by a force of one pound.
Mass is independent of gravity. Applying Newton’s second law, F = ma or W = mg, where g is the
acceleration due to earth’s gravity, i.e., 32.2 ft/sec 2. Thus, m = W / g, and F = W a / g.
Important fact. g = 32.2 ft/sec2. Memorize this.
Example. Suppose an airplane weights 10,000# and initial thrust on takeoff is 5000#. What is the
acceleration?
F = Wa/g, so a = Fg/W = (5000#) (32.2 ft/sec2) / 10,000# = 16.1 ft/sec2.
Exercise. Suppose 10,000# of thrust and initial acceleration of 20 ft/sec 2. Find weight of aircraft.
Let  (cap phi) be the bank angle of an airplane in steady state constant altitude turning flight. The forces
acting on the airplane are as follows.
L cos 
L

L sin 
W
For steady state flight, W = L cos . L sin  is the force causing the radial acceleration according to
Newton’s second law. This centripetal force is what causes the airplane to turn.
Note 1 : The diagram on p.188 of the text is misleading, since the forces are balanced and the airplane
would not turn. Centrifugal force is just the force opposite and equal to centripetal force according to
Newton’s third law. It causes the “G force” which you feel during turning flight.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 9
Newton’s Third Law. For every action there is an opposite and equal reaction.
Note 2: L cos  is the effective lift which must support the weight of the airplane. Since L cos 90 o = 0,
there is no effective lift in a 90o bank. (Discuss implications.)
Effective L
Total L

W
Cos  = effective lift / total lift = W / L. Since G force is defined to be G = L / W, cos  = 1 / (L/W) = 1 /
G. so G = 1 / cos .
As L increases, L sin  increases, and the G force on the a/c increases, while radius of turn decreases in
accordance with Newton’s Second law.
L
0o
L
L30
30o L30 = 1.1547 L
G
1
2
3
4
5
6
L
L45
45o L45 = 1.414 L
Cos  = 1/G
1
0.5
1/3
0.25
0.2
1/6
L
L60
60o L60 = 2 L
 (in deg)
0
60
70.53
75.52
78.46
80.41
Important fact. G force is independent of weight or speed in a steady state level turn. It depends solely
on the bank angle. No 90o steady state turn is possible.
Example. Suppose bank angle is 72o; what is the G force? Suppose G force is 3.75; what is bank angle.
G = 1 / cos 72 = 3.24. Cos  = 1 / 3.75, so  = cos-1 (1 / 3.75) = cos-1 0.26666667 = 74.53o.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 10
Radius of Turn as a Function of Bank Angle and Airspeed.
V
It can be shown that a body circling a point at a distance of radius r with a velocity
V experiences a radial acceleration of V2/r. Applying F = m a, we have
CF = m V2 / r =
r
CF
W # V 2 ft 2 / sec 2 W V 2
W V2
, or r 

ft .
g r
g ft / sec 2 CF #
g CF
Since W = L cos , and CF = L sin , r =
L cos  V 2
W V2
V2


ft .
g CF
g L sin 
g tan 
Note: V must be expressed in ft/sec. (Recall that multiplying by 6076/3600 converts knots to fps.)
L cos 
Example. Suppose airspeed is 300 kts at 45o angle of bank.
Then the radius of turn is
L
r = V2 / g tan 

2
L sin =CF
 6076

300 

506 .33333 2  7,961 .91 ft
 3600


=
32 .2 tan 45
32 .2 tan 45
W
Exercise. Suppose 380 KTAS at 50o angle of bank. Find radius of turn in feet and in nautical miles.
Rate of Turn as a Function of Bank Angle and Airspeed.
 r / 180 feet
1o
r
not to scale
Since 2  r is the circumference of a circle of radius r, an a/c
circling a fixed point at a distance of radius r feet travels
(2  r feet/revolution) / (360 degrees/revolution) =
( r / 180) feet/degree
Then, if the a/c has a speed of V ft/sec, rate of turn RT is
TR =
180 g tan  deg
V ft/sec
180 V
V2
180 V

.
, TR 

deg/sec. Since r 
2
r
V
sec
r
g tan 
 V 
ft/degree


180
 g tan  
As with radius of turn problems, V must be converted to fps if it is given in knots.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 11
Example. What is the turn rate in the previous problem?
TR =
180 g tan  (180 )(32 .2) tan 50

 3.43 deg/sec.
V
6076 

 380

3600 

Example. Suppose a/s is 380 knots. What angle of bank is required for a SRT (3 o/sec)? What is the turn
radius and G force developed?
TR =
180 g tan 
TR V deg/sec.
 tan  
V
180 g
  6076 
3380 

TRV
 3600 
tan  
 
 1.042897465 .
180 g
180 32 .2
 = 46.20293974o.
G = 1 / cos  = 1/ (cos 46.20293974) = 1.444865088.
2
2

 6076   ft


380


2


V2
 3600   sec

r

 12,248 .99 ft (or 12,248.99/6076 = 2.01596 nm).
ft
g tan 
32 .2
tan 46 .20293974
sec2
Exercise. Suppose 90 KTAS. What angle of bank is required for a SRT? What is the G force? What is
the turn radius in feet and nautical miles?
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 12
PROPERTIES OF THE ATMOSPHERE
I. Pressure.
Static Pressure P is the weight per unit area of a column of air. At sea level on a standard day (temperature
= 15o C or 59o F),
P0 = 2116 #/ft2 = (2116 #/fts) / (144 in2/ft2) = 14.69444 #/in2.
(Alternative measure of standard pressure is 29.92 inches of mercury; the millibar is a unit of pressure used
by meteorologists.)
Static pressure decreases as altitude increases. About ½ of the atmosphere by weight is below 18,000 feet,
so the standard pressure at that altitude is approximately ½ the standard pressure at sea level.
The pressure ratio  (small delta) at a given altitude with ambient pressure P is defined to be
 = P / P0
where P0 is as defined above. For example, 18000’ = 0.4992, reflecting the fact that about half the
atmosphere lies below 18,000 feet. Note that in general, 0    1;  > 1 only when ambient pressure
exceeds standard pressure at sea level.
II. Temperature.
To convert centigrade to Fahrenheit, C = 5/9(F-32). To convert Fahrenheit to centigrade, F = 9/5C + 32.
Example. 40oC = 9/5(40)+32 = 104oF. 70oF = 5/9(70 – 32) = 21.11oC.
Exercise. Convert 30 and 50 degrees centigrade to Fahrenheit. Convert 80 degrees Fahrenheit to
centigrade.
Must use absolute temperature (absolute zero = -273oC) in temperature calculations! -273o C = 9/5(-273)
+ 32 = -459.4  -460o F (absolute zero in Fahrenheit).
Add –273 (-460) to centigrade (Fahrenheit) to get Kelvin (Rankine). K is the symbol for Kelvin, R for
Rankine.
Sea level standard temperature:
T0 = 59o F = 59+460 = 519o R.
= 15o C = 15 + 273 = 288o K.
Temperature decreases with altitude until the tropopause (36, 089’), then stays constant at –69.7o F until
about 85,000’.
The Temperature ratio  (cap theta) at a given altitude where the temperature is T is defined as
 = T / To,
where T0 is as above.
Example. At the tropopause,  = (-69.7+460) / (59 + 460) = ( (5/9) (-69.7 – 32) + 273) / (15+273) = 0.752
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 13
III. Density.
Density  (small rho) is mass per unit volume. (Most important of the three properties in aerodynamics.)
 = mass / unit volume = slugs/ft3 = (#-sec/ft) / ft3 = (#-sec)/ft4
The density at sea level on a standard day is 0 = 0.002377 slugs/ft3.
The density ratio  (small sigma) at an altitude where the density is  is defined to be
= /0
where  0 is as above.
Example. 22000’ = 0.001183 slugs/ft3 in a standard atmosphere (to be defined shortly), so
22000’ = 22000’ /  0 = 0.001183 /0.002377 = 0.497686159  0.5.
From this we may conclude that about ½ the atmosphere by density lies below 22,000’. (Compare this to
18,000’ for pressure.)
Density of a gas is proportional to P/T; i.e.  = k P / T for some k > 0, where the value of k depends on the
properties of the gas. Thus,
 =  /  0 = (kP/T) / (kP0/T0) = (P/P0) / (T/T0) =  / .
In English, this equation says: “Density ratio is equal to pressure ratio divided by temperature ratio.”
These and related ideas are summed up in the Standard Atmosphere Table (p. 18 text). The ICAO
(International Civil Aviation Organization) adopted values in the table (based on observations and
averages) to promote uniformity in measuring and comparing aircraft performance worldwide.
IV. Pressure and Density Altitude
Pressure altitude is altitude corresponding to a given static pressure in a standard atmosphere; i.e., is
observed altitude corrected for non-standard pressure, assuming no altimeter error. (Set 29.92 plus or
minus altitude correction in altimeter pressure window.)
If pressure is 972 #/ft2, then  = 972 /2116 = 0.4594, which corresponds to approximately 20,000’ pressure
altitude in a standard atmosphere.
Density altitude is pressure altitude corrected for non-standard temperature. (Recall that  = k P / T, so that
density is directly proportional to pressure and inversely proportional to temperature.)
Density altitude is used to construct aircraft performance charts for various altitudes, since air density most
directly relates to performance parameters such as thrust, lift and drag.
We will return to these subjects later in the course.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 14
ICAO Standard Atmosphere Table
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 15
PITOT-STATIC AIRSPEED INDICATORS
I. Air Behavior in a Bernoulli Tube
Consider a circular tube with a uniform gradual construction in it. Suppose air flows through this tube
without being compressed (called incompressible flow, or subsonic flow in aerodynamics). Then the air
must “speed up” in the “narrower” part of the tube, and the speed is inversely proportional to the crosssectional area of the tube.
More precisely, let A1 and A2 be the cross-sectional areas (in ft2) of the tube at point 1 and point 2
respectively, and let V1 and V2 be the corresponding airflow velocities (in ft/sec) at these two points.
Bernoulli discovered that
A1 V1 = A2 V2.
If A1 < A2, then V2 > V1; i.e., the air speeds up at point 2. Suppose A1 = 1 ft/sq2 and A2 = 0.2 ft/sq2, with
V1 = 100 ft/sec, as shown in the Bernoulli tube depicted below. Then
A1 V1 = A2 V2, or V2 = (A1 V1) / V2 = (1 ft2) (100 ft/sec) / (0.2 ft2) = 500 ft/sec.
Point 1
A1 = 1 ft2
V1 = 100 ft/sec
q1  12 #/ft2
P1  2104 #/ft2
Point 2
A2 = 0.2 ft2
V2 = 500 ft/sec
q2  297 #/ft2
P2  1819 #/ft2
Now consider the dynamic air pressure q at points 1 and 2. As opposed to static air pressure, dynamic air
pressure is pressure due to the force exerted by moving air. It can be shown that dynamic air pressure is
proportional to the square of the air’s velocity. Specifically, dynamic pressure is given by
q =  V2 / 2 (recall that  is air density)
The unit for q is
# sec2 ft 2
4
2
# / ft 2 , which is the same as for static pressure P. Let H stand for total
ft
sec
pressure. Bernoulli also discovered that as dynamic pressure increases in a Bernoulli tube, static pressure
decreases, and vice versa. That is, total pressure remains constant:
H = P + q.
This principle may be stated as follows:
Bernoulli’s Principle: Static pressure is inversely proportional to air velocity as velocity changes, and
total pressure—static pressure plus dynamic pressure—remains unchanged.
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 16
Assume the Bernoulli tube depicted above is situated at sea level in a standard atmosphere. Then  =
0.002377 slugs/ft3, and H = P = 2116 #/ft2. (Total pressure H and static pressure P are the same when air
velocity V = 0, since then q = 0)
Then the dynamic pressures q1 and q2 at points 1 and 2 respectively is as follows:
q1=  V12 / 2 = 0.002377 (100)2 / 2  12 #/ft2; and
q2=  V22 / 2 = 0.002377 (500)2 / 2  297 #/ft2.
Then, applying Bernoulli’s Principle,
P1 = H1 – q1 = 2117 – 12 = 2104 #/ft2 at point 1, and
P2 = H2 – q2 = 2117 – 297 = 1819 #/ft2 at point 2.
This decrease in static pressure due to increased air velocity explains why an airfoil develops lift (as
discussed in detail later in the course), and is also the basis for the pitot-static airspeed system.
II. Pitot-Static Airspeed Systems
These systems measure airspeed in terms of the difference between total and static pressure acting on an
aircraft. As airspeed increases, dynamic pressure increases and static pressure decreases, while total
pressure remains constant according to Bernoulli’s principle. Thus low static pressure is related to high
airspeed, and higher static pressure to low airspeed.
Total pressure is measured at the tip of a pitot tube, where a stagnation point exists and the air velocity is
zero. At this stagnation point, the static pressure is equal to the total pressure H, since H = P + q.
Stagnation point of airflow at tip of pitot tube
(thus total pressure H is measured inside pitot tube)
static port (dynamic
pressure here is P)
Pitot Tube
pressure here is P,
static pressure, which
remains constant
as q increases
pressure here is H = q + P,
with q increasing as airspeed
increases
degree of diaphragm displacement
measures H – P = q, the dynamic pressure
due to airflow velocity (airspeed)
III. Types of Airspeed
IAS Indicated airspeed is the airspeed shown on the airspeed indicator.
CAS Calibrated airspeed is indicated corrected for pitot/static system errors. These are of two types:
Rogers’s AS309 Notes Part I (revised 30 August 2004: Page 17
a) errors in the a/s indicator itself
b) position error due to interference with airflow at the static port. Figure 2.6 (text p. 23) shows that
typical errors are plus or minus 5-7 knots, with the larger errors occuring at very high airspeeds.
(Good design minimizes the problem.)
EAS Equivalent airspeed is CAS corrected for compressibility effects. Such effects occur when ram air
effects on the pitot tube cause the air entering it to be compressed. This error always results in an a/s
reading that is too high. Fig 2.7 (text p. 23) shows this problem causes a/s indicators to read as much as 28
kts fast. Problems are worst at altitudes 20-40 M and airspeeds in the range 300-500 kts.
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TAS True airspeed is EAS corrected for non-standard density (i.e., for altitudes where density differs from
standard density at sea level). Useful primarily for navigation; has little or nothing to do with the way the
aircraft responds to control input from the pilot’s point of view.
The following derivation shows an interesting and important relationship between EAS and TAS:
q =  V2 / 2 and  =  / 0, so  =  0, and q =  0 V2 / 2 = (0/2) ( V2).
That is, dynamic pressure is directly proportional to ( V2), since (0/2) is a constant. It follows that
(TAS)2  = (EAS)2 0 = (EAS)2, or (TAS)2 = (EAS)2 / , and TAS = EAS
1 EAS
.



This helps explain why the ICAO Standard Atmosphere Table has a  column. Note: 1  is a value
that you will meet again in AS310 Performance. It is called SMOE, which is an acronym for standard
means of evaluation.
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