Quick Met Question
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Quick Met Question
An aircraft flying at the 950mb level has a corrected outside air temperature of -6C . If the temperature and pressure at MSL is +10C and 1025mb, what will the aircraft height be AMSL.
I have the answer noted as 2005ft but can't seem to get that figure!
What do you get? If you can get 2005ft please show me how!
I have the answer noted as 2005ft but can't seem to get that figure!
What do you get? If you can get 2005ft please show me how!
As far as i know temp has nothing to do with it.
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How about, 1025-950 = 75 hPa at 27 ft each = 2025 ft and then subract 1% of ht for each 2.5 deg C different to ISA in this case 5 deg thus less 2% giving us -40.5 ft as colder therefore lower leaving 2025-40.5=1984.5 ft amsl. Cannot get to my old notes at present.
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The nearest exact figure I have for ISA is 954.6mb at 0.5km, or 1640ft. If ISA temps applied you would have to correct for 4.6mb at about 30ft/mb, putting you at 138ft higher at 950mb - 1778ft above the 1013mb datum. Correcting to QNH gets you a correction of 12 x 27mb = 324ft. QNH is a higher pressure than standard so this correction brings you to 2102ft.
You have two figures for ISA temperature deviation. At msl it is minus 5deg and at height it is about minus 17deg. Both figures imply a temp correction that will reduce your true height. The convention we normally use is to assume one single ISA deviation, and I can't suggest how to solve this. However, taking minus 5deg the correction would be 0.5 x 4% of 2102ft, or 42ft. Answer 2060ft. Using an ISA deviation of 12deg would give an answer near to 2005ft, but I don't see how to justify that, it isn't the average, and it isn't the ISA deviation at 950mb.
Maybe I got the height at 950mb wrong or there is some other glitch. Anyone out there got any view
Dick
You have two figures for ISA temperature deviation. At msl it is minus 5deg and at height it is about minus 17deg. Both figures imply a temp correction that will reduce your true height. The convention we normally use is to assume one single ISA deviation, and I can't suggest how to solve this. However, taking minus 5deg the correction would be 0.5 x 4% of 2102ft, or 42ft. Answer 2060ft. Using an ISA deviation of 12deg would give an answer near to 2005ft, but I don't see how to justify that, it isn't the average, and it isn't the ISA deviation at 950mb.
Maybe I got the height at 950mb wrong or there is some other glitch. Anyone out there got any view
Dick
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Thanks Mr Bristol? JBA, I think we may be looking too deeply at this - perhaps it is best to refer back to your provider as it is likely to have caused confusion before by the sound of it. If Dick is Mr Whittingham Senior then I suggest his opinion trumps all of ours! Sometimes altimetry questions are related to a particular part of the syllabus and I know I often complicated things by looking at the wrong part. I will watch with interest...
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My initial thinking was as per SkyCamMK.
Though found the formula h2 = h1 + 221.1*T*(log P1 - Log P2) for instances where pressure difference is greater than 50mb, where
h1 = known height
h2 = required height
T = mean temp in K
P1 = pressure at h1
P2 = pressure at h2
filling these in get h2 = 0 + 221.1*275*(log 1025- log 950) = 2006.5ft
JBA, using 96T/P if use the average T of 275 and average P of 987.5mb get a height change per mb of 26.74 ft and the multiply this by 75 (1025-950) get 2005.1ft. Just wondering did you miscalculate?
Like SkyCamMK will watch with interest.
Though found the formula h2 = h1 + 221.1*T*(log P1 - Log P2) for instances where pressure difference is greater than 50mb, where
h1 = known height
h2 = required height
T = mean temp in K
P1 = pressure at h1
P2 = pressure at h2
filling these in get h2 = 0 + 221.1*275*(log 1025- log 950) = 2006.5ft
JBA, using 96T/P if use the average T of 275 and average P of 987.5mb get a height change per mb of 26.74 ft and the multiply this by 75 (1025-950) get 2005.1ft. Just wondering did you miscalculate?
Like SkyCamMK will watch with interest.
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Full marks, mcgrath!
As neither the 96T/P nor the "log p" formula are part of the JAA syllabus I tried to solve it by the rules of thumb that are in the syllabus. Given the very close correlation between your answer and the book answer I guess that this is a question from a more advanced met course than the JAA ATPL theory requires.
Any comment JBA?
Dick
As neither the 96T/P nor the "log p" formula are part of the JAA syllabus I tried to solve it by the rules of thumb that are in the syllabus. Given the very close correlation between your answer and the book answer I guess that this is a question from a more advanced met course than the JAA ATPL theory requires.
Any comment JBA?
Dick
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I made the mistake of using dT at 289K instead of calculating an average temperature. I've added the sample to my formulae book so I don't make this mistake again. It wasn't BristolGS, I'm glad you guys could help me out.
I found the formulae McGrath noted in the Jeppesen Met Manual so i guess they are included in the JAA syllabus?
Thanks again
I found the formulae McGrath noted in the Jeppesen Met Manual so i guess they are included in the JAA syllabus?
Thanks again
Last edited by JBA; 3rd Mar 2008 at 16:05.
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Well, I'm impressed with your solution but why would I ever want to know that or be able to calculate it in the way shown above??? I am now a Grumpy Grandpa too its official!!
