ATPL NAV questions
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ATPL NAV questions
Hey guys just studying for ATPL NAV with AFT notes and got a couple of questions wrong in the revision tests so thought id post a couple to get an explanation.
Q1: You are cruising at A070 directly above a mountain of ELEV 6420ft. The QNH is 1001 HPA with an OAT of -14 dec C. The radio altimeter would indicate -
a) 188 ft
b) 972 ft
c) 390 ft
d) 580 ft
Q2: On the ALICE SPRINGS ILS, the glide path intersects the Outer Marker at an altitude of 3120 ft. An aircraft is making an ILS approach. The local QNH for AS is 1028 and is set corrctly. The surface temperature is +3 degrees C and the elevation of the airport is 1789 ft.
If this aircraft crosses the outer marker with its altimeter indicating the height shown on the instrument approach, its position in relation to the glidepath is -
a) on the glidepath
b) 25 ft above the glidepath
c) 50 ft below the glidepath
d) 25 ft below the glidepath
Thanks in advance
Q1: You are cruising at A070 directly above a mountain of ELEV 6420ft. The QNH is 1001 HPA with an OAT of -14 dec C. The radio altimeter would indicate -
a) 188 ft
b) 972 ft
c) 390 ft
d) 580 ft
Q2: On the ALICE SPRINGS ILS, the glide path intersects the Outer Marker at an altitude of 3120 ft. An aircraft is making an ILS approach. The local QNH for AS is 1028 and is set corrctly. The surface temperature is +3 degrees C and the elevation of the airport is 1789 ft.
If this aircraft crosses the outer marker with its altimeter indicating the height shown on the instrument approach, its position in relation to the glidepath is -
a) on the glidepath
b) 25 ft above the glidepath
c) 50 ft below the glidepath
d) 25 ft below the glidepath
Thanks in advance
Last edited by xxRED BARONxx; 18th Feb 2013 at 07:20.
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Baron.
I remember struggling at times with the ol' NAV!!
The first question the answer is 580ft. They chuck in the QNH and temp just to throw you off the scent. Im assuming you have yet to do ATPL systems because it covers they way the radio altimeter works. In short it has nothing to do with QNH or TEMP because it bounces radio waves off the groung to give its reading.
Im not sure but I think the second one is "on the glidepath" once again I think they're chucking numbers at you to throw you. If the height oin the plate reads the same as the altimeter when you cross it, you should be on the slope.
hope this helps
Rocket
I remember struggling at times with the ol' NAV!!
The first question the answer is 580ft. They chuck in the QNH and temp just to throw you off the scent. Im assuming you have yet to do ATPL systems because it covers they way the radio altimeter works. In short it has nothing to do with QNH or TEMP because it bounces radio waves off the groung to give its reading.
Im not sure but I think the second one is "on the glidepath" once again I think they're chucking numbers at you to throw you. If the height oin the plate reads the same as the altimeter when you cross it, you should be on the slope.
hope this helps
Rocket
I think they're chucking numbers at you to throw you
ISA-8 gives an error of 41 feet at the Outer Marker height!
If you were on the GS at the OM, the Altimeter would read 3161'.
If you were at 3120' at the OM, you would be 41' below the GS.
If it's cold watch out below!
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The answer to the first question is NOT 580', and the temp and QNH is NOT there to throw you off.
Study True Altitude, this is where your answer lies. I don't have time to work it out for you, but in this case the altimeter will be overreading (due cold OAT), therefore it's either A or C. I see you've checked that it's A.
Study True Altitude, this is where your answer lies. I don't have time to work it out for you, but in this case the altimeter will be overreading (due cold OAT), therefore it's either A or C. I see you've checked that it's A.
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AH thanks bleeding air, i figured it out now.. you have to convert the calibrated altitude indicated on the altimeter to the true altitude (calibrated corrected for temp) to be able to calculate the height AGL, which is what the radio altimeter indicates.
I used the whiz wheel way to get the true altitude and then minused the elevation to get an answer of 230'AGL which would make the answer A.
"You are cruising at A070 directly above a mountain of ELEV 6420ft. The QNH is 1001 HPA with an OAT of -14 dec C. The radio altimeter would indicate -
a) 188 ft
b) 972 ft
c) 390 ft
d) 580 ft"
I used the whiz wheel way to get the true altitude and then minused the elevation to get an answer of 230'AGL which would make the answer A.
"You are cruising at A070 directly above a mountain of ELEV 6420ft. The QNH is 1001 HPA with an OAT of -14 dec C. The radio altimeter would indicate -
a) 188 ft
b) 972 ft
c) 390 ft
d) 580 ft"
Last edited by xxRED BARONxx; 18th Feb 2013 at 14:55.
how did you come up with the number 41'
However there is a rough calculation you can use which is reasonably accurate for small ISA Deviations and Heights, to convert Indicated Altitude to True Altitude.
+/- 4ft per degree of ISA deviation * (height / 1000)
Height is the height above the temperature source (normally the airport elevation).
In your example, the ISA deviation is -8, and the height is (3120-1789)=1331'
(-8 * 4ft) * (1331/1000) = -42.6
There is a very good explanation here by Chimbu Chuckles!
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Getting ready for the Nav exam and I have been studying using the AFT notes and using Avery practice exams as well.
Avery prac exams have a lot of questions not currently covered in the AFT notes but can be found in JAA ATPL books. Anyway I came across a question that AFT and Avery both have but both have similar workings out but one omits a step and the other includes the step giving completely different answers.
From Avery prac exam
"aircraft at 36000ft and approaching a DME station sited at 6000ft. What is the max theoretical distance which you receive a signal"
Formula is 1.5 x height above station then square rooted (1.5 X 30000 square rooted = 212nm
Obviously the site at the station is subtracted from the aircraft height to work out the answer. Fair enough
In AFT notes
If transmitter was at 5000ft and aeroplane is at 10000ft what is the theoretical range.
AFT working out uses 1.5 X 10000ft square rooted + 1.5 X 5000ft square rooted = 122.5 + 86.6 = 209.1NM
So which is the correct working out for theoretical range, Avery or AFT
thanks
Avery prac exams have a lot of questions not currently covered in the AFT notes but can be found in JAA ATPL books. Anyway I came across a question that AFT and Avery both have but both have similar workings out but one omits a step and the other includes the step giving completely different answers.
From Avery prac exam
"aircraft at 36000ft and approaching a DME station sited at 6000ft. What is the max theoretical distance which you receive a signal"
Formula is 1.5 x height above station then square rooted (1.5 X 30000 square rooted = 212nm
Obviously the site at the station is subtracted from the aircraft height to work out the answer. Fair enough
In AFT notes
If transmitter was at 5000ft and aeroplane is at 10000ft what is the theoretical range.
AFT working out uses 1.5 X 10000ft square rooted + 1.5 X 5000ft square rooted = 122.5 + 86.6 = 209.1NM
So which is the correct working out for theoretical range, Avery or AFT
thanks
