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high QNH on high Altitude airports

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Old 28th Oct 2013, 17:17
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high QNH on high Altitude airports

Has anybody got an idea about why is it that high elevation airports have commonly high daily QNH´s ?

I am aware of the uncompensated temperature effect over the altimeter but, shouldn´t this make us set the altimeter to a LOWER QNH value to compensate for it instead of a higher one ?

If this is so, then what is the ral reason for the high one ?


Thx in advance

( really worried about this. can´t sleep..... just kiddin´ )
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Old 28th Oct 2013, 19:54
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I know you know QNH is Sea Level Pressure AMSL?
Could it be that the air is less dense for an airport QNH . . .? at altitude?

Can`t be, if regional QNH is 1,000mb say, then it is 1,000 all the way up to Transition - leaving that aside for a mom, maybe they (the . . . ATC)
are trying to compensate for the density altitude.
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Old 28th Oct 2013, 22:13
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Hi BESTGLIDESPEED,

The QNH which ATC pass is such that your altimeter will read the correct reference altitude at the airport. If the lapse rate is not standard then it will be a higher QNH on a warm day, and a lower QNH on a cold day.

Edit. Consider a very cold day where the Altimeter will over read (hence the low temp correction chart). In order to get the Altimeter to read the correct airfield elevation, you will have to "fudge" the Airfield QNH. (to reduce the indicated Altitude on the altimeter, you'd "wind off" the appropriate error by changing the setting to a lower QNH.)

The converse is true for your higher temperature consideration.

Last edited by rudderrudderrat; 29th Oct 2013 at 16:00. Reason: extra text
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Old 2nd Nov 2013, 19:56
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That makes sense ruddrrudderrat.

Have been giving it a few twists in my head and i think i have finally got it pretty much figured out now.

Thanks for throwing a light down here.


Best regards
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Old 6th Nov 2013, 03:55
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A late comment, correcting a common error:

I know you know QNH is Sea Level Pressure AMSL?
QNH is station pressure reduced to mean sea level in a standard atmosphere. Important addition, as non-standard temperature will affect the true altitude of an aircraft overflying, or at a high elevation airfield.

Natstrackalpha, you were thinking perhaps of QFF, station pressure reduced to mean sea level pressure in the true atmosphere. At a 1000' elevation airfield, if you dig a hole in the ground 1000' deep and put a barometer at the bottom, it'll give you QFF.
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Old 10th Nov 2013, 16:48
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Thank you OKTAS8

Been having second thoughts about all this...


If, let´s say today we have in UIO ( Quito, Ecuador Ca. 9.400´ ) 20ºC, and we know for a fact that the QFF at "local sea level" is 1013hpa. to start from an easy base, then,
according to all the reasoning above, and doing the reverse calculation just to demonstrate the facts,
we should have to substract 30´/hpa. from 1013 all the way up to 9.400´of elevation.
9.400´/ 30´/hpa = 313 hpa. to be substracted to the 1013 hpa at MSL as we climb up all the way to our airport´s GND at 9.400´.

The local pressure in a standard atmosphere should read 700hpa. which, translated to our altimeter´s standard language "is" 9.400´.

Here´s where the problem arises, when, with our scale set to a real 1013 our altimeter is actually reading 12.280´!!! due to the uncompensated temperature effect ( see below ).

( 15º-2ºx 9,4+20º=24º Delta ISA >>> and being DA=pl+120xDelta ISA >>> 9.400+120x24 = 12.280´)


And here´s where I can finally re-post my question : " wouldn´t I have to wind the scale DOWN that is, to a LOWER presure base ( 96 hpa less than 1013 here due to the 24º delta ISA. That is, some exorbitated 917 hpa ) in order to reduce the excessive 2.880´of the standard " pressure column" and thus obly my altimeter to show me 9.400´on the GND , INSTEAD OF USING a crazy 1109 hpa ??

By the way, the USUAL QNH at this airport is neither, but around 1020 daily.

thx again for the patience....



Regards
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Old 10th Nov 2013, 20:14
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Hi BESTGLIDESPEED,
wouldn't I have to wind the scale DOWN that is, to a LOWER presure base
If you flew at an indicated Altitude of 9,400 ft with true sea level QNH set, and the average temperature of the air column below you is warmer than ISA, then you will be higher than UIO aerodrome. After landing, your Altimeter would read less than 9,400 ft, therefore you would need to increase the sub scale setting to get the airfield reference Altitude indicated correctly.

Your maths assumes the average temperature of the air column below you is ISA + 24. However it would appear +24 only applies to the air just above the airport (and up to say 1,000 agl). If the airfield QNH only changes by 7 mbs (1020-1013) i.e. 210 ft, then the average temperature of the air column below you is only about ISA + 6 C.
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Old 12th Nov 2013, 17:24
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Now I see it Rudderrudderrat,

It´s not that you have to bring the "PLANE" DOWN reducing your QNH, but instead, you have to lift the INDICATION UP to where you REALLY are in the new condition !


Those little shiny clue words that trigger the answer ....



Really thankful for that.



Ok, now,
finally liberated of my own-made mental knot, there´s only left the question of, why does the basic ATPL MET-math is not working here ...

( 4% altitude misindication per 10º delta ISA ) : +24 vs. real +6


Any clue ?


( you see, once I find a good lead, can´t let go until the source is totally dry... )


Sorry to be such a drag, but now I got most of the nightmare solved, just need to round up the result.



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Old 13th Nov 2013, 08:34
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Hi BESTGLIDESPEED,
why does the basic ATPL MET-math is not working here ...(4% altitude misindication per 10º delta ISA ) : +24 vs. real +6
That approximation is working OK.
According to Charles's Law, the volume of gas is proportional to the temperature measured in degs Absolute (Kelvin).
ISA = (273 + 15) 288K at sea level.
Therefore 10 degree warmer expands the atmosphere by 10/288 = 3.472% (4% approximately)
I think you are assuming that the local temperature at UIO is representative of a massive amount of atmosphere - it's not.

Last edited by rudderrudderrat; 13th Nov 2013 at 11:58. Reason: Corrected Boyle's Law for Charles's Law
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Old 13th Nov 2013, 10:12
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I'm impressed

I'll just take it as you have stated it here


Just learned something new today and also cleared something old

More than satisfied



Thanks a lot Rudderrudderrat


Tread concluded on my part



Regards
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