INSTRUMENT QUESTIONS

[AVIATROZ]

Q1. The ISA pressure for 18000feet are:
anS. 506hpa
plz Someone guide how to cal pressure change with altitude.

Q2.An inc. of 0.15Mach result in an inc. of 93kt TAs of an a/c. the local speed of sound is:
ANs-620KTs
PLZ guide how to solve this problem

Q3.An a/c is flying at 400ft from a high temp area to a cold temp area where the temp diff. is 20degree. what will be actual height of a/c:
Ans-3680feet
{As we are going from low density to high density area........the true height should decrease right!....i get that point but how to calculate barometric error......because QNH is not given:ugh:....only temprature diff of 20degree is given.........plz explain}

Q4.An inc. of 0.15Mach number result in an inc. of93kts in TAs.if temp deviation from IsA Is +9DEGREE,the FL is:
ANs--FL220{plz explain}

Q5.The atmospheric pressure at FL70 in a "standard +10" atmosphere is:
ANs---781.85HPa{plz expalin}

Q6.Cruising at FL390,M0.84 is found to give a TAs of 499kt.the IsA deviation at this level will be:
A. +17---------ans
B.+!9
C.-17
{i m getting my ans as +22degree plz help}

[AVIATROZ]

Q1. An a/c is flying a rate 1 turn at 480kt TAS. What is the diameter of the turn?

Ans-5nm..............{i used formula ROT=Vsquare/g tan{theta}............but did nt able to get the ans................any suggesions}


Q2.Assume a perfectly frictionless DI is corrected to give a zero drift on ground at 30N.The DI is set to read 270degree ib ab a/c tracking along 60degree parallel at a grdspeed of 545kts.The Di reading after 80 mins:

Ans--283degree{plz explain

[elitepilot20]

Q1. An a/c is flying a rate 1 turn at 480kt TAS. What is the diameter of the turn?

Ans-5nm..............{i used formula ROT=Vsquare/g tan{theta}............but did nt able to get the ans................any suggesions}

you can not calculate radius or diameter using this formula

your TAS is 480Kts
in rate one turn you need 2 mins to complete your turn
therefore you will travel 480/60X2 = 16nm which is also your circumference

And circumference = 2 X pie X r
therefore 16 = 2 X 22/7 X r
r= 2.54nm
n diameter = 5nm:8

[elitepilot20]

Q2.Assume a perfectly frictionless DI is corrected to give a zero drift on ground at 30N.The DI is set to read 270degree ib ab a/c tracking along 60degree parallel at a grdspeed of 545kts.The Di reading after 80 mins:

Ans--283degree{plz explain

is there any option 285/286 because i am getting this but not sure at all how far i am correct on this.

n waiting for other explanations as well.

[AVIATROZ]

can u plz explain how you got 285as ans
thanks for the help

[magnetic-north]

(Mach 1) 620 ktas = 100%

6.2 = 1 %

+ mach 0.15 = 15x 6.2 = 93KTAS

[keith williams]

Total wander = Earth Rate + Latitude Nut + Transport Wander

Total wander = ER + LN + TW

In the Northern hemisphere

ER in degrees per hour = -15 degrees x Sin Latitude

LN in degrees per hour = + 15 degrees x sin latitude nut setting

TW in degrees per hour = (East to West ground speed in knots) x Tan latitude / 60

Latitude nut setting is 30 degrees
Latitude is 60 degrees.
East – West ground speed = 545 knots
Flight time = 80 minutes which is 80/60 hours

So

ER = (- 15 x sin 60) x 80/60 = - 17.32050808 degrees

LN = (+15 x sin 30) x 80/60 = + 10 degrees

TW = (545 x tan 60) / 60 x 80/60 = 20.97705978 degrees

Adding them all together gives +13.6565517 degrees

Adding this to the initial heading of 270 gives 283.6565517 degrees.

[SuperflyTNT]

Can anyone please share the formula to calculate RADIUS OF TURN

Thank you

[elitepilot20]

@superfly
for radius use the circle circumference formula
i.e. 2.pie.r

[keith williams]

Radius Of Turn = TAS squared / g Tan AOB

But be careful to ensure that you are consistent with your units.

To get the radius in meters you need to use g in m/sec squared and TAS in m/sec. 1 knot = approximatley 0.515 m/s

If you want to get the radius in nm by using TAS in knots then you need to convert g = 9.81 m/sec sqaured into nm/hour squared. (Most people choose not to do this)

But for the question posted in this thread it is easier to use the circumference method as described by elitepilot.


For rate of turn use the following equation.

Rate Of Turn in degrees per second = g Tan AOB / TAS

[AVIATROZ]

Q.An a/c initiate a rate one turn at a Tas of 270kts.how far will it have travelled after 25 sec.

Ans--1.96nm

[elitepilot20]

Q.An a/c initiate a rate one turn at a Tas of 270kts.how far will it have travelled after 25 sec.

Ans--1.96nm

@AVIATROZ
you asking or explaining?

[AVIATROZ]

@ elite pilot

distance travelled by a/c will be
TAs* 25sec= 270*(25/3600)=1.87 Which is approx to 1.9nm


please advice if u think its not right method

thanks

[elitepilot20]

it is the right method according to me.

hey have you got your roll no. for indigo yet??

[AVIATROZ]

no buddy........hav`nt got the hall ticket yet..........it seem i `ll only be able to give the attemptt in oct{or whenever the next attempt be}..........hoping for the things to turn positive as they are supposed to be...............what about you buddy

[elitepilot20]

but you did receive the confirmation from radhika didn't you???

and about me, i didn't even get the confirmation from her. she asked for my details n went dark, haven't heard anything yet.



Ques-(from indigo q. bank)
echoes are received 30degree to left A/c hdg is 120M, find QTE?

no variation given so is this question incomplete or am i not getting the trick?

[AVIATROZ]

bro........i m also in same situation bro.........after the detail.........did`nt got any reply from them........anyways.........working hard for the next attempt........

about the question.........
as the echoes are being received..........30degree to the left..........so it should 90degree...........now the QTE{true direction from the station}
so i think it should be 090 + 180degree=270degree{true........if variation given then apply and get answer}

i am not sure if this is right but lets see if someone else give his views on this

[elitepilot20]

A (45N 10W) B (48*30*N 15W)
find A-B RL track? :O

[SuperflyTNT]

Thank you for the explanation and formula. Much appreciated.

[Shabez]

Aircraft with pressurized cabin in flight:
When switching to the alternate static pressure source, the pointer of the Vertical Speed Indicator:
A) indicates a climb, then settles down and reads incorrectly
B) indicates a descent, then settles down and reads incorrectly
C) indicates a slight continuous descent
D) indicates correctly

shouldnt this indicate a decent....high pressure inside cockpit so goes to static source indicates decent and then settles downs and indicates incorrectly...but the ans is listed as A

[Shabez]

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. :ugh:

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]

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]

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.