@3443nautical miles
 

I sort of figured it out after i posted it, but appreciate the explanation. cheers bro.

@9FLYJET9
 

No, no news :ugh:

@upilotg: ok, lets wait till we hear anything.

anyone with any info on september dates, please share.

radio aids questions
 

Q1. An a/c receives a reply pulse from a DME 1200micro sec after transsmission of the interrogation pulse.The DME has a fixed delay of 50 micro sec. the range of the a/c from the DME station is:
a.75nm
b.63nm
c.45nm
d.93nm- ans

Q2. The time interval b/w the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the a/c is 2 milli sec. the slant range is:

ans162nm

Q3. The accuracy of DME at 100 nm slant range is within:
a.3nm-- ans
b.4nm
c.2nm
d.1nm
{ i think ans shud be 1 nm as accuracy of dme is 1.25% of range.
so (1.25/100)*100=1.25}

plz shed some light.

Q4.A/c is following the ILS glidepath of 3degrees at an airfeild where the outer mrker is 4.2nm from the ILS touchdown point. the a/c approach speed is 130kt. the height of the a/c at the outer marker shud be:
a.1150ft
b.1200ft
c.1310ft---ans
d.960ft

{my ans does`nt match.........i did like..........i used 1 in 60 rule...
i.e. {(3degrees *4.2)/60}*6080=1276.8.................}
i think i m doing it in wrong way:ugh: plz correct...thanks

Q5. At 5.25nm frm threshold an a/c on an ILS Approach has a display showing it to be 4dots low on a 3 degree glidepath, using an angle of 0.15degree per dot of glideslope deviation and the 1 in 60 rule calculate the height of the a/c from the touchdown.
a.1375ft
b.1325ft
c.1450ft
d.1280ft
{as a/c is low for 4dots on ILS ...... i.e. 0.15*4=0.6
therefore, 3-0.6=2.4degree}
(now, using 1 in 60 rule, ((2.4*5.26)/60)*6080=1279.2}

plz ILS hed ILS ome light on this

Radio waves travel at 300000km/sec
The delay is 50 microsec therefore time is 1150 microsec.
Distance traveled by Radio wave (To and fro) in 1150 * 10 ^ (-6) sec = 345 km = 186 nm
one way distance = 93 nm.

2 milli sec = 2 *10^ (-3) sec
Distance covered by radio wave in 2 ms = 600 km = 324.3 nm
one way distance = 162 nm

Now ROD = Glide Angle * Ground Speed * 100/60
=> ROD = 3 * 130* 100/60 = 650 ft/min

Time to cover 4.2nm @ 130kts = 2 min
=> A/C would decend 1300 feet ~ 1310.

Dude even I'm getting 1280ft maybe they mean the GS indication is down 4 dots => present GS is 3.6 but then the answer is some 1900 ft ..does anyone else know how to solve this.

After solving everythg and pasting it on a word doc ... DJ Flyboy outraced me
Nice job !!
Q5. At 5.25nm frm threshold an a/c on an ILS Approach has a display showing it to be 4dots low on a 3 degree glidepath, using an angle of 0.15degree per dot of glideslope deviation and the 1 in 60 rule calculate the height of the a/c from the touchdown.
- ((2.4*5.25 )/60)*6080=1276} + 50 = 1326 ( the ils brings you 50 ft above touchdown )

:ok: do put up your explainations too.. they might be better than mine..

The Pulse takes 12.36 microseconds to travel 1 nautical mile to and back ,
2 millisecond [ms] = 2000 microsecond
2000/ 12.36 = 161.8 nm


-ICAO recommends accuracy of 0.25 nm plus 1.25% of the distance measured.
which would be 0.25 + 1.25 = 1.5
OR
- DME accuracy should be within 3% or .5 nm which ever is greater
( You will find both these answers in various books , so for the indigo exam go with 3% as it fits )

@ DJ Flyboy ... just following the footsteps of all in this forum ...To share all that i can !! :ok::ok:

ILS glide path is 2.8deg. At what altitude above threshold level is the upper limit of the glide path coverage at a distance of 1.5nms from the touchdown point?

a) 610feet
b)865feet
c)425feet
d)745feet --Ans


Does anyone know how to get this? I'm getting c as the answer.

LAsT set of radio aids doubts
 

Q1.What is the minimum level tht an a/c,at a range of113nm,must fly in order to contact the tower on R/T for a VDF bearing frm an airport sited 169ft above MSL
a.FL50
b.FL60-----ANs

Q2.An a/c travelling at330 m/sec transmit a signal at 10GHZ to a stationary receiver.if the a/c is flying directly towards the receiver and they are approximately at the same height the received frequency will be
a. 11MHz
b. 10.000011GHz

Q3.At 1000Z an a/c is overhead NDB AB enroute to NDB BC,track 075{M},hdg 082{M} at1029Z NDB AB bears 176 relative and NDB BC bears 353 relative.the hdg to steer at 1029Z to reach NDB BC is:
079{M}---ans

Q4. An a/c is mantaining track outbound from an NDB with a constant relative bearing of 184degree. to return to the NDB the relative bearing to maintain is:
356degree-------ans
{shud`nt the ans be 004degree......plz correct my concept.......i think i m making some mistake in understanding the question somewhere...}

thanks

yes...i got the ans
 

@ SAAB CAPT

ILS glide path is 2.8deg. At what altitude above threshold level is the upper limit of the glide path coverage at a distance of 1.5nms from the touchdown point?

a) 610feet
b)865feet
c)425feet
d)745feet --Ans


see.........as in the question they have mentioned that they are asking for the upper limit of glidepath coverage........and the coverage of gidepath .....lowerlimit is0.45of theta and the upperlimit is 1.75of theta....................in this case we require upper limit of glidepath............

so use 1.75of theta=1.75*2.8=4.9degrees

now altitude of a/c=({4.9*1.5}/60)*6080 feet= 744.8feet =745feet{approx}

i hope this solves ur problem...:ok:

@ Aviatroz

Thanks so much.

@ Aviatroz

Q1.What is the minimum level tht an a/c,at a range of113nm,must fly in order to contact the tower on R/T for a VDF bearing frm an airport sited 169ft above MSL
a.FL50
b.FL60-----ANs


113 = 1.25( sqrt rx + sqrt 169 )
113/1.25 - sqrt 169 = sqrt rx
rx = 5990.7 feet ~ Fl60

Q4. An a/c is mantaining track outbound from an NDB with a constant relative bearing of 184degree. to return to the NDB the relative bearing to maintain is:
356degree-------ans
{shud`nt the ans be 004degree......plz correct my concept.......i think i m making some mistake in understanding the question somewhere...}



ac is compensating for 4deg drift

@ SaabCapt ...got it drawn out !!

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radio aids questions
 

Q2.An a/c travelling at330 m/sec transmit a signal at 10GHZ to a stationary receiver.if the a/c is flying directly towards the receiver and they are approximately at the same height the received frequency will be
a. 11MHz
b. 10.000011GHz--ans

Q3.At 1000Z an a/c is overhead NDB AB enroute to NDB BC,track 075{M},hdg 082{M} at1029Z NDB AB bears 176 relative and NDB BC bears 353 relative.the hdg to steer at 1029Z to reach NDB BC is:
079{M}---ans

Sept indigo recruitment
 

can Someone plz find out what`s going out with indigo regarding Sept recruitment..........any info would be really helpful

Indigo assessment
21st and 22nd September

Just received the letter from Radhika H S for theIndiGo exam on 21st, 22nd September.