Met Help please!
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Met Help please!
Dear all,
Can anyone help me answer this question?
1) You intend to fly over a mountain range. The recommended minimum flight altitude is, according to the aviation chart, 15000' AMSL. The air mass that you will fly through is on average 15'c warmer than the standard atmosphere. The altimeter is set to QNH (1023hPa). At what altitimeter reading will you effectively be at the recommended minimum flight altitude?
A) 14100' <-------------- Correct answer
B) 13830'
C) 14370'
D) 15900'
Many thanks in advance.
WB
Can anyone help me answer this question?
1) You intend to fly over a mountain range. The recommended minimum flight altitude is, according to the aviation chart, 15000' AMSL. The air mass that you will fly through is on average 15'c warmer than the standard atmosphere. The altimeter is set to QNH (1023hPa). At what altitimeter reading will you effectively be at the recommended minimum flight altitude?
A) 14100' <-------------- Correct answer
B) 13830'
C) 14370'
D) 15900'
Many thanks in advance.
WB
Join Date: Feb 2002
Location: oxford
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The numerical value of the QNH has no significance in this question. As long as you are flying on QNH, you are measuring with mean sea level as a datum, whether it is 1023, 1013, 993, or any other value.
The only factor is going to be temperature. The column of air between the ground and you is 15ºC warmer than ISA, meaning that it is less dense. Therefore you will need to climb through more of it get a particular millibar change (indicated altitude) indicated on your altimeter. True altitude is going to be higher than indicated altitude.
Using the rule of 4 feet per thousand feet per degree of temperature deviation, you have 15 lots of 1000 feet. so that is 15 x 4, ie, 60 feet per degree centrigrade. The ISA deviation is +15, so multiply that 60 feet by 15, ie, 900 feet.
So, with a true altitude of 15,000 feet, the indicated altitude will be 14,100 feet.
Or, of course, you can use the 'True Altitude' correction function on your CRP5, which I find easier, and teach.
The only factor is going to be temperature. The column of air between the ground and you is 15ºC warmer than ISA, meaning that it is less dense. Therefore you will need to climb through more of it get a particular millibar change (indicated altitude) indicated on your altimeter. True altitude is going to be higher than indicated altitude.
Using the rule of 4 feet per thousand feet per degree of temperature deviation, you have 15 lots of 1000 feet. so that is 15 x 4, ie, 60 feet per degree centrigrade. The ISA deviation is +15, so multiply that 60 feet by 15, ie, 900 feet.
So, with a true altitude of 15,000 feet, the indicated altitude will be 14,100 feet.
Or, of course, you can use the 'True Altitude' correction function on your CRP5, which I find easier, and teach.
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Agree with OB about using the CRP-5, however to be pedantic the QNH is given as it technically gives slightly more accuracy as you can work out pressure altitude and a more precise temperature. You should notice in the calculation that this altitude precision cannot be used as the scales are too small, but it does affect the temperature by 1°.
Taking minimum flight altitude of 15,000 feet, the pressure altitude would be 14,730 feet (reduce by 270 feet for the 10 HPa drop to 1013 HPa). ISA temperature -14°C, so OAT is +1°C (you'll get -15°C and 0°C if you use 15,000 feet, and make very little difference to the final answer!).
In the "Altitude" window there is a temperature scale to match against the altitude scale. Match +1°C against 15,000 feet (see, I was being really pedantic, and am using 15,000 because the scale cannot be read to 270 feet). Read from true altitude of 15,000 on the outer scale to just over 14,000 on the inner, say about 14,050 feet. You get about 14,100 if you use 0°C.
That is how I would teach it too, although I make sure the students know the formula so they can choose.
Send Clowns
Gen Nav
BCFT
P.S. The reason for the pedantry is really for consistency. It should not affect this answer unless the question ask for more precision than is fair, but it keeps you thinking about what reference your altitudes are in!
Taking minimum flight altitude of 15,000 feet, the pressure altitude would be 14,730 feet (reduce by 270 feet for the 10 HPa drop to 1013 HPa). ISA temperature -14°C, so OAT is +1°C (you'll get -15°C and 0°C if you use 15,000 feet, and make very little difference to the final answer!).
In the "Altitude" window there is a temperature scale to match against the altitude scale. Match +1°C against 15,000 feet (see, I was being really pedantic, and am using 15,000 because the scale cannot be read to 270 feet). Read from true altitude of 15,000 on the outer scale to just over 14,000 on the inner, say about 14,050 feet. You get about 14,100 if you use 0°C.
That is how I would teach it too, although I make sure the students know the formula so they can choose.
Send Clowns
Gen Nav
BCFT
P.S. The reason for the pedantry is really for consistency. It should not affect this answer unless the question ask for more precision than is fair, but it keeps you thinking about what reference your altitudes are in!
