# Aerodynamics. Airspeed.

Givens:

Indicated Altitude                    = 15,500 ft.
Baro-altimeter Setting             = 30.42 in.  Hg
Temperature                           = Standard plus 5o F
Indicated airspeed                   = 250 KIAS
Atmosphere                            = “Flight Theory and Aerodynamics” Table 2.1
Position Error                          = Chart below
Compressibility Error              = “Flight Theory and Aerodynamics” Figure 2.6 1.   Find the Pressure Altitude, ft
2.  Find Pressure Ratio
3  Find the Temperature ratio
4  Find the Density ratio
5  Find the Density altitude (feet)
6 Find Calibrated Airspeed (CAS)
7.  Find Equivalent Airspeed (EAS)
8.  Find True Airspeed (TAS)
9.  Find Dynamic Pressure of aircraft q, (lb/ft2)
Answers

1. Baro altimeter setting $= 30.42″$ Hg
Difference in altitude = Sea Pressure altitude – Baro altimeter setting $= -30.42 +29.92 = -0.5”$ Hg

$1 Hg = 1000 ft$ it means $-0.5 Hg =-500 ft$

Pressure altitude = Indicated altitude – Difference in altitude $= 15500 -500 =15000$ ft

2. Pressure ratio at sea level

$Delta$ = Pactual/Pstandardsealevel $= 30.42/29.92 = 1.0167$

pressure ratio at altitude

$Delta 1 = 0.5643$ from table 2.1 corresponding to an altitude of 15000 ft

3. temperature ratio at altitude

(temperature is given in deg R  (deg R = 460 +F)

$theta$ = Actual temperature/Standard sea level Temperature $= (59+5+460)/(59+460) =524/519 =1.00963$

4. Density ratio

$sigma =$ pressure ratio/temperature ratio $= 0.5643/1.00963 =0.5589$

5. Density altitude

from table 2.1 we have

at 0 ft  we have sigma =1
at 20000 ft we have sigma = 0.5328

at 15000 ft we have sigma = 0.6292

for every ft higher than 15000 ft we have $(0.6292-0.5328)/5000 =1.928*10^{-5}$ variation of $sigma$

total variation of $sigma$ is $0.6292-0.5589 =7.03*10^{-2}$

which means a variation in altitude of $7.03*10^{-2}/1.928*10^{-5} =3646.3$ ft above $15000$ ft

the density altitude is $15000 +3646.3 =18646.3$ ft

6. calibrated air speed= indicated air speed  +(-) position error

CAS = IAS – position error

For IAS = 250 KIAS we have from chart the error -2.4 Knots

CAS =250 -2.4 = 247.6 Knots

7. equivalent air speed is

EAS = CAS – Compressibility correction

compressibility correction is 4.25 knots from figure 2.6

EAS = 247.6 – 4.25 = 243.35 Knots

8. True air speed

TAS $= EAS/sqrt{sigma} = 243.35/sqrt{0.5589} = 325.5$ knots

9. Dynamic pressure

$q= sigma*V^2/295$

where V is the TAS in knots $=325.5$

$sigma =0.5589$

$q =(325.52)^2 *(0.5589)/295 = 200.74 (lb/ft^2)$