Question Analysis, Book Reference, Important Concepts, Formulae and topic wise Solutions for the topic "Airspeeds" are time-stamped below. Access the study materials, presentation, links to previous and next lectures and further information in the description section.

1. 1
FLIGHT MECHANICS
AIRSPEEDS (IAS, CAS, TAS & EAS)
TOPIC WISE SOLVED QUESTIONS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
1

2. 2
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2009 Equivalent, Calibrated, Indicated Air speeds 1 Mark
Q1. The relation between an airplane's true airspeed VTAS and equivalent airspeed VEAS in terms of the density
ratio (σ =
ρ
ρ0
), where ρ0 is the air density at sea-level and ρ is the air density at the shade at which the
airplane is flying, is given by the formula:
(A) VEAS
VTAS
= σ
(B) VEAS
VTAS
= σ2
(C) VEAS
VTAS
= σ
(D) VEAS
VTAS
=
1
σ
Explanation:
Equivalent airspeed (EAS) is that speed with which the aircraft needs to fly at sea level to duplicate the actual
dynamic pressure at a given altitude.so therefore, the relation can be written as;
Answer:
(C) VEAS
VTAS
= σ
1
2
ρvTAS
2
=
1
2
ροvEAS
2
VEAS
VTAS
=
ρ
ρ0

3. 3
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2016 Equivalent, Calibrated, Indicated Air speeds 1 Mark
Q2. Indicated airspeed is used by a pilot during
(A) take-off (B) navigation
(C) setting the engine RPM (D) setting the elevator angle.
Explanation:
During take-off, the density variation isn’t significant therefore TAS & EAS can be neglected. Simplest airspeed
reading is preferred, thus IAS shown on ASI is directly referred without further calculation
Answer:
(A) Take-off
Airspeed Purpose Remark
Indicated Airspeed (IAS) Take-off Reading shown by the Airspeed Indicator (ASI)
Calibrated Airspeed (CAS) Navigation GPS is used alternatively
True Airspeed (TAS) Navigation It is corrected for density variation
Equivalent Airspeed (EAS) Structural Analysis Direct function of incompressible dynamic pressure

4. 4
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2019 Equivalent, Calibrated, Indicated Air speeds 1 Mark
Q3. An airplane is in steady level flight with a true air speed of 50 m/s. The ambient air density and ambient
pressure at the flight altitude are 0.91 kg/m3 and 7x104 N/m2, respectively. At sea level, air density is 1.225
kg/m3 and ambient pressure is 1.01×105 N/m2. The equivalent or indicated air speed of the airplane is
__________ m/s (round off to 2 decimal places).
Answer: 43.05 to 43.15
Solution:
Given
VTAS = 50 m/s; ρ = 0.91 kg/m3; p = 7x104 N/m2; ρ0 = 1.225 kg/m3; p0 = 1.01×105 N/m2;
𝐹𝑜𝑟𝑚𝑢𝑙𝑎: VEAS = VTAS x
ρ
ρ0
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡: VEAS = 50 x
0.91
1.225
= 50 x 0.86189 = 43.09
m
s
.

5. 5
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2008 Equivalent, Calibrated, Indicated Air speeds 2 Mark
Q4. An aircraft is cruising at a true air speed (TAS) of 100 m/s under ISA conditions, at an altitude at which the
density of free stream is 0.526 kg/m3. What will be the equivalent air speed (EAS)?
(A) 65.5 m/s (B) 72.5 m/s
(C) 110.5 m/s (D) 152.7 m/s
Solution:
Given
VTAS = 100 m/s; ρ = 0.526 kg/m3;
ISA Conditions: ρ0 = 1.225 kg/m3; p0 = 1atm = 1.01325×105 N/m2 ; T0 = 288.16K
Answer:
(A) 65.5 m/s.
𝐹𝑜𝑟𝑚𝑢𝑙𝑎: VEAS = VTAS x
ρ
ρ0
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡: VEAS = 100 x
0.526
1.225
= 100 x 0.6553 = 65.53
m
s
.

6. 6
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2019 Equivalent, Calibrated, Indicated Air speeds 2 Mark
Q5. An aircraft with a gross weight of 2000 kg, has a speed of 130 m/s at sea level, where the conditions are: 1
atmosphere (pressure), 288 K (temperature), and 1.23 kg/m3 (density). The speed (in m/s) required by the
aircraft at an altitude of 9000 m, where the conditions are: 0.31 atmosphere, 230 K, and 0.47 kg/m3, to
maintain a steady, level flight is ________ (accurate to two decimal places).
Answer: 209.00 to 211.00
Solution:
Given
W = 2000 kg; VEAS = 130 m/s; p0 = 1atm = 1.01325×105 N/m2 ; T0 = 288.16K; ρ0 = 1.23 kg/m3
h = 9000m; ρ = 0.47 kg/m3; p = 0.31 atm; T = 230K
VTAS=
𝑉𝐸𝐴𝑆
ρ
ρ0
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡: VTAS =
130
0.47
1.225
=
130
0.6182
= 210
m
s
.
𝐹𝑜𝑟𝑚𝑢𝑙𝑎:
VEAS
VTAS
=
ρ
ρ0

7. 7
FLIGHT MECHANICS - BASICS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
GATE 2020 Equivalent, Calibrated, Indicated Air speeds 2 Mark
Q6. A Pitot tube mounted on the wing tip of an airplane flying at an altitude of 3 km measures a pressure of
0.72 bar, and the outside air temperature is 268.66 K. Take the sea level conditions as, pressure 1.01 bar,
temperature = 288.16 K, and density = 1.225 kg/m3. The acceleration due to gravity is 9.8 m/s2 and the
gas constant is 287 J/(kg K). Assuming standard atmosphere, the equivalent airspeed for this airplane is
_____________ m/s (round off to two decimal place).
Solution: Given
h = 3000m; ρ = 0.47 kg/m3; p0 = 0.72 bar; T = 268.66 K
Sea-Level Condt: pSL = 1.01 bar = 1.01×105 N/m2 ; TSL = 288.16K; ρ 𝑆𝐿 = 1.225 kg/m3 ; g = 9.8 m/s2; R = 287 J/(kg K).
𝐹𝑜𝑟𝑚𝑢𝑙𝑎:
P
pSL
=
𝑇
𝑇𝑆𝐿
−𝑔
𝑎𝑅 P
1.01
=
268.66
288.16
−9.8
−0.0065∗287
= 0.699 or 0.70 bar; Where a = Lapse Rate = -0.0065 K/m
VTAS =
2 (𝑝0 − 𝑝)
ρ ρ =
𝑃
𝑅𝑇
=
0.72 𝑥105
287 𝑥 268.66
= 0.9338 kg/𝑚3; VTAS =
2 (0.72 𝑥 105 − 0.70 𝑥 105)
0.9338
= 65.45 m/s
VEAS = VTAS x
ρ
ρ0
VEAS = 65.45 x
0.9338
1.225
= 57.14 m/s Answer: 57.10 to 60.00

8. 8