INSTRUMENT QUESTIONS

[Shabez]

Horizon errors

thanks

Following 180° stabilized turn with a constant attitude and bank, the artificial horizon indicates:
A) too high pitch-up and correct banking.
B) too high pitch up and too high banking.
C) attitude and banking correct.
D) too high pitch-up and too low banking.

i think answer ought to be A but it lists it as D

i think this is the order in conventional artificial horizon
90 under read
180 correct
270 overread
360 correct

pitch always too high except 360

[keith williams]

Your comments are all correct in this one Shabez.

[Anders S]

Hey guys, see if you can solve this one.

A gyro with no real drift is set to read the correct magnetic heading when the aircraft is heading 090° (T), variation 10°E. After 45 minutes flight eastwards along the parallel of 62N at 545k groundspeed, the gyro reading is:

a. 100.5°
b. 064.8°
c. 075.0°
d. 057.3°

[keith williams]

With 10E variation the initial indication when on 090(T) will be 080(M)

Earth rate wander in degrees per hour = -15 x Sin Lat

45 minutes = 0.75 hours so total Earth Rate wander over this period
is -15 x Lat 62 x 0.75 = -9.933 degrees


Transport Wander rate in degrees per hour = E to W groundspeed x Tan latitude /60

When flying W to E the groundspeed is considered negative

So over a period of 45 minutes
transport wander = -545 x Tan 62 x 0.75 / 60 = -12.812 degrees.


Total wander = -9.933 ER + -12.812 TW = -22.745


Final heading indication = initial indication + total wander

Final heading indication = 080 - 22.812 = 57.255 degrees

The closest option is 57.3 degrees.

[Anders S]

Ah, but of course! It helps if you calculate it the right order. I did 545/60 x tan(62), and obviously got an entirely different answer. Cheers.

[AVIATROZ]

Heelo aviator mates, need some help with couple of questions i am stuck with,
Q1.what will be the TAS if cruising altitude is 39000ft,temperature is ISA+5 and CAS200kts:
a. 388kts
b. 383kts..ans
c. 364kts
d. 370kts

Q2.At FL350 with a JSA deviation of -12,true airspeed when flying at M.78 is:
a. 460kts
b. 436kts..ans
c. 447kts
d. 490kts

Q3.An a/c is flying AT fl350 with a JSA deviation of +8,The mach no. is.83 and the TAS485KTS.IF A/C descends to FL300 and maintains the same mac no. and TAS,the JSA deviation will be:
a. +8
b. -2
c. +2...ans
d. -18
{i am getting ans B option,may be i am making some mistake plz correct}

Q4.The radius of turn @rate1,and TAS360kts is
a. 10nm
b. 5nm
c. 7.5nm
d. 2nm...ans
thanks for the guidance.

[keith williams]

Q1.

Using the POOLEYS CRP5

39000 ft is above the ISA tropopause so the ISA temperature will be –56.5 degrees Celsius. Adding +5 ISA deviation gives a temperature of –51.5 degrees Celsius.

In the airspeed window set -51.5 degrees against 39000 feet.

Against 200 knots CAS on the inner scale read off 388 knots TAS on the outer scale.

This is greater than 300 knots so we must apply compressibility correction.

Compressibility correction = (TAS / 100) – 3
Inserting the 388 knots TAS gives (388/100) – 3 = -0.88.
In the compressibility correction window move the pointer 0.88 units to the left.

Against 200 knots CAS on the inner scale read off 383 knots TAS on the outer scale.


Q2.

ISA temperature at 35000 feet = 15 (35 x 1.98) = -54.3 degrees Celsius.

Adding –12 temperature deviation gives –66.3 degrees Celsius.

Adding 273 to this converts it into 206.7 degrees kelvin

LSS = 38.94 x Square root of absolute temperature

So LSS = 38.94 x square root of 206.7 = 559.84 knots

Mach Number = TAS / LSS so TAS = Mach x LSS

So at M0.78 the TAS = 0.78 x 559.84 = 436.68 knots


Q3.

Mach Number = TAS / LSS so LSS = TAS / Mach

So at M0.83 if the TAS is 485 knots we have

LSS = 485 / 0.83 = 584.3 knots


LSS = 38.94 x Square root of absolute temperature

So Absolute temperature = (LSS /38.94) squared


Inserting the figures from above gives

Absolute temperature = (584.3 / 38.94) squared = 225.2 degrees Kelvin

Subtracting 273 converts this into –47.8 degrees Celsius.



ISA temperature at 30000 feet = 15 (30 x 1.98) = -44.4 degrees Celsius.

Temperature deviation = Actual temperature – ISA temperature

Inserting the figures above gives

Temperature deviation = ( -47.8 ) – ( -44.4 ) = +3.4 degrees Celsius.

The closest option is +2.


Q4.

A rate 1 turn takes 2 minutes to turn through 360 degrees.

Dividing 360 knots by 60 converts into 6 nm per minute.

So in 2 minutes the aircraft will fly 12 nm, which is the circumference of the circle.

Circumference = 2Pi x Radius

So Radius = Circumference / 2Pi

Dividing 12 nm by 2Pi gives a radius of 1.9098 nm.

The closest option is 2 nm.

[AVIATROZ]

some radio aids and instrument question

Aviator mates need some help with couple of question i am stuck with,your help will be really appreciated.
Q1. The time from transmission of interrogation pulse to the receipt of the reply from DME ground is 2000microsecs{ignore delay in dme}The slant range is:
a. 330nm
b. 185nm
c. 165nm...ANS
d. 370nm

Q2.Flying an ILS approach with 3degree glideslope referenced to 50feet above threshold,an a/c at4.6nm should be at approximate height of:
a. 1400ft
b. 1380ft
c. 1500ft
d. 1450ft......ANS
{acc. to me the ans should be height=glide path angle*range*100feet=3*4.6*100feet=1380feet i.e. B option but that is not the answer plz explain.}
..................................
INSTRUMENT QUESTIONS
Q3.An a/c is fitted with 2 altimeters.One is corrected for position error,the other is not corrected for position error.
a.Provided that ADC is working normally,there will be no error to either altimeter
b.at high speed the non-compensated altimeter will show lower altitude
c.at high speed the non-compensated altimeter will show higher altitude
(ANS is {c}....but i think ans should be {b}...please explain.)

Q4.A low altitude Radio altimeter used in precision approaches,following characteristics:
1. 1540mhz to 1660mhz range
2. pulse transmission
3. frequency modulation
4. height range between 0-5000feet
5.accuracy of +/-2ft between 0-500ft.

a. 1,4,5
b 3,4.....ANS...
c 3,5
d 2,3,5

{The height range of radio altimeter is from 50-2450 feet}explain,.....acc. to me and should be {c})
thanks
Q3.

[bayblade]

Q1. The time from transmission of interrogation pulse to the receipt of the reply from DME ground is 2000microsecs{ignore delay in dme}The slant range is:
a. 330nm
b. 185nm
c. 165nm...ANS
d. 370nm

this is simple formula based.
distance= speed * time.
= (3*10^8)*(2000*10^-6)
=600,000m
=(600,000*3.28)/6080 Nm
=323.68Nm
but note time of transmission and reception is given i.e twice the distance. hence actual distance is 323.68/2 = 161.8Nm.

[bayblade]

Q2.Flying an ILS approach with 3degree glideslope referenced to 50feet above threshold,an a/c at4.6nm should be at approximate height of:

use the formula:
note glideslope is referenced 50 ft above the threshold. but 3 deg takes you down to touchdown. hence you need to add 50 ft to the height obtained.
height (in feet) = distance(nm) * glideslope angle *101.3
= 4.6 * 3 * 101.3
=1397.94 ft.
hence height of the aircraft is 1397.94 + 50 =1447.9 ft.

[bayblade]

An a/c is fitted with 2 altimeters.One is corrected for position error,the other is not corrected for position error.
a.Provided that ADC is working normally,there will be no error to either altimeter
b.at high speed the non-compensated altimeter will show lower altitude
c.at high speed the non-compensated altimeter will show higher altitude

position error comes into picture during any of the following events:
a) during side slip
b) during ground effect
c) low speeds
d) when aircraft is in a certain attitude.
its got nothing to do with high speed.

i guess the question means non-compensated for compressibility error, which comes into play during high speeds.
the answer should be b according to me.

[Naveen Kalra]

Help Needed!!!

Hi All, need your help for following questions of Radio Aids (GSP), Doppler Radar Chapter, Page no. 167-168. Many thanks in advance.

Q3. An aircraft using a single beam Doppler system to measure its groundspeed is travelling at 300 m/s. Depression angle is 60 degrees. Transmission frequency is 10 GHz. What will be Doppler shift?
Ans. 10 KHz.

Q4. An aircraft uses a single beam Doppler system with depression angle 60 degrees to measure its ground speed. Transmission frequency is 10 GHz. What will the ground speed if the Doppler Shift is 5 KHz.
Ans. 300 kt.

Q7. An aircraft with a 4-beam Janus array has a depression angle of 60 degrees and an angle of 30 degrees horizontally between the beams and the aircraft longitudinal axis. What will be Doppler shift in the array if the transmission frequency is 11500 MHz and the ground speed is 600 kt.?
Ans. 20 KHz.

[bayblade]

doppler shift= 2(relative velocity/wavelength)*Cos(depression angle).

here,
relative velocity = 300m/s
wavelength = (3*10^8)/10G
Cos60=0.5
hence doppler shift= 10K Hz

the second sum is just the same type, you have to take care to convert the speed in kts to m/s though.

for Janus array modify the formula as such: (4*relative velocity*cos dep angle*cos horizontal angle)/wavelength.