INSTRUMENT QUESTIONS
 

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}

INSTRUMENT QUESTIONS
 

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

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

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.

eite pilot
 

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

question 2
 

(Mach 1) 620 ktas = 100%

6.2 = 1 %

+ mach 0.15 = 15x 6.2 = 93KTAS

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.

Can anyone please share the formula to calculate RADIUS OF TURN

Thank you

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

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
you asking or explaining?

i was asking for the solution but i figured it out now
 

@ 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

@AVIATROZ
 

it is the right method according to me.

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

@elitepilot
 

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

@AVIATROZ
 

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?

elite pilot
 

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

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

@keith_williams & elitepilot20
 

Thank you for the explanation and formula. Much appreciated.

why climb?
 

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