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PL64
27th Jan 2010, 17:19
I was told that the correction you have to apply to the altimeter equals:
4% of the height (so AGL) per 10°C ISA deviation.

When flying in colder than ISA the altimeter will overread (indicate a higher altitude than your actual altitude) so you would have less terrain clearance than you would expect...

OK, let 's imagine you would like to fly over a mountain of 16000 ft with a vertical separation of 2000 ft. When flying at FL180 (let us assume the QNH is 1013 hPa) what would be the actual terrain clearance knowing that the temperature is ISA-10°C?

It is obvious that, when flying at FL180 I will not have the 2000 ft vertical separation, but what will be my actual vertical separation with that mountain?

Do you take the 4% of the 18000 ft or take the 4 % of the 2000ft (height)???? I always thought you had to take the 4% of the 18000ft but then somebody argued that the first 16000ft isn't air but mountain and can therefore not be compressed due to the lower temperature...:ugh:

Does anyone know which I should take and why?

Thanks!

It

Skyjob
27th Jan 2010, 20:21
The altimeter deos not know anything about the mountain, so cannot take it into account.
Altimeter corrections should always be applied AGL where this is operationally taken as runway elevation.
The corrections are then applied to any value below MSA entires in the FMC which need upwards correcting.

If your landing runway would be at 5000', then the perceived QHN pressure reported for that runway would read accurately 5000' when on the ground. Once above it, cold weather corrections are applied.
However, if your landing runway would be at 0', then the perceived pressure reported for that runway would read accurately 0' when on the ground, and when cold weather corrections are then applied to the 5000' referred to earlier, altimeter readings would underread actual altitude by 200' (by your calculation of 4%).

So you see, it does not just depend on temperature, it also depends on where the accurate reference datum is from.

Thus when flying over the mountain with a correct pressure reading at the top, would only need altimeter adjustments for 2000'. However when referring to the pressure reading from the airport at its base, much mroe needs to be added.

safewing
28th Jan 2010, 10:48
There are actually 3 elements to altitude correction.

1. ISA deviation even below 15 degrees celsius at a sea level airfield where in this instance altitude = height there is a technical correction. However it is considered negligible down to 0 degrees celsius. To offset the small correction the altimetry correction table used in PANS OPS and I presume EU OPS is based on a 2000 feet reference airfield

2 This is at the discretion of an operator but there is an phenomenon known as altimetry error caused by the Bernoulli effect. It states that a strong wing blowing across rugged terrain especially on approach can create an error. It gives a table but if I remember correctly 70kts generate 400 foot error.

3 The last one is based on true altitude based on degrees Kelvin where it states that ISA - deviations aloft can reduce true altitude over indicated which would be particularly pertinent passing over mountainous terrain. I have often wondered in a OEI altitude capability if this is considered. Many driftdown scenarios are based on ISA + deviations to maintain a terrain clearing altitude but equally important and I far as I can see not considered is regardless of what the altimeter is indicating you clear terrain by what is your true altitude?

rudderrudderrat
28th Jan 2010, 11:35
Hi PL64

Do you take the 4% of the 18000 ft or take the 4 % of the 2000ft

Good question. Suppoose you are flying over Mont Blanc and you are using Geneva's QNH as your reference for Altitude. Your altimeter will read correctly on the ground at Geneva - but will be in error as you climb. So technically you'd apply the correction to a column of air (18000 - 1400) 16600 feet tall.

blackmail
28th Jan 2010, 16:20
hello all,
correction rule for cold wx is : 4 FEET per °C isa deviation per 1000ft OR 1% per 2.5°C isa deviation, both rules give same result.
take FL180 & isa - 10°C gives 4ft x 10 x 18 = 720ft or (1% x 10)/2.5=4%. 4%from 18000ft= 720ft. so if QNH=std then at fl180 your true altitude = 720ft lower. so from high to low(pressure) & from hot to cold(temperatures) : look below.
i agree, its more user friendly to use correction tables than above formulae.
bm.:)

PL64
1st Feb 2010, 13:14
Hi all,

Thanks a lot for taking your time to writing these replies...

But I am still a little bit confused :rolleyes:.

Blackmail states that I should take the 4% on the 18000ft which would imply that the altimeter overreads by 720ft (terrain clearance of 1280ft).

Rudderruderrat states that you should take the 4% of 16600ft which would be 664ft (terrain clearance of 1336 ft).

Skyjob states that I should take the 4% on the 2000ft (intended vertical separation with the terrain (approx. Mont Blanc) which would imply that the altimeter overreads by only 80ft (terrain clearance of 1920 ft).

:O Who 's right?

Thanks a lot,

PL64

kijangnim
1st Feb 2010, 14:32
Greetings
This not the exact formulea, however it gives prety close results :ok:
4ft * Delta ISA per thousand feet
18000 delta ISA -10 so we have 4*-10*18 = -720 feet
Indicated altitutude is 18000 +(-720) true altitude 17280 feet

RobinMaiden
1st Feb 2010, 15:16
Here is a video I put together on the topic. It shows graphically why the formula has to use the sea level equivalent ISA temperature not the temperature at the airport which might be at 5,000'.

Hope this helps.

YouTube - Cold Weather Altimetry, Cold Wx Altimetry, Cold Weather Altimeter Corrections, Cold Weather Flying (http://www.youtube.com/watch?v=bkna6LnEMkE)

Robin Maiden

rudderrudderrat
1st Feb 2010, 20:38
Hi PL64,

Blackmail is correct. Strictly speaking you should use the sea level QNH setting. But where do you get the regional QNH from easily? If you use the nearest airport's QNH and do the maths, you won't be in error by very much. (56 feet error at 18,000 feet using GVA's)

(Extra maths. The QNH that the airport passes (Geneva in this case) is "fudged" to make your Altimeter read correctly on the ground. If Geneva passed you the real QNH, your Altimeter would be in error by 4% of 1400 feet = 56 feet due to the temperature. Thus the subscale setting is "fudged" by 2 mbs (in this case). So to be entirely accurate, correct GVA's QNH by the fudge factor and then do the maths.

punkalouver
2nd Feb 2010, 04:37
I would plan on correcting the altimeter for the column of air that you are flying in. that is what is affected by the cold.

So if you are descending into a coastal airport in the arctic with nearby 6,000 foot mountains and you are descending to an MSA of 8,000', correct for an 8,000 foot column of air, not 2,000 feet just because you are descending to a published altitude 2,000 feet above the mountains.

bookworm
2nd Feb 2010, 06:36
Skyjob states that I should take the 4% on the 2000ft

No, that's not what Skyjob said, and the problem is that you haven't given enough information to differentiate between the answers.

A QNH is calibrated to give the correct elevation at a reference point, usually an airport. What matters is your height above that reference point, and you don't tell us what that reference point is.

So if the QNH that you have "assumed" refers to a sea-level airport, you make the correction based on 18,000 ft. If the QNH refers to an airport at 1400 ft, you make the correction based on 16600 ft. If the QNH refers to an altiport at 16000 ft, you make the correction based on 2000 ft.

PappyJ
2nd Feb 2010, 06:58
The Pressure altimeter is calibrated to indicate true altitude in the Standard Atmospheric conditions. Due to continually heating and cooling, the atmosphere at any given point is seldom at the temperature of Standard Air.

In fact, the only time you can be certain that the altimeter indicates true altitude is when the aircraft is on the ground at the airport for which the current altimeter setting is set on the sub-scale of the altimeter.

When an aircraft is in flight, it can be assumed that the altitude indicated on an altimeter is always in error as a result of temperature variations.

The amount of error depends on the degree to which the average temperature of the column of air between the aircraft and the ground (Altimeter setting Location) differs from the average temperature of the standard atmosphere for the same column of air.

If the actual temperature of the air column in which the aircraft is flying is COLDER than the standard, the true altitude of the airplane above sea level will be lower than the indicated altitude. Conversely, if the temperature is WARMER than standard, the true altitude will be higher.

Obviously, since all altimeters in the same area are equally affected by temperature error, we don't apply any corrections for traffic avoidance purposes.

However, for terrain purposes, it's a different story.

Do you take the 4% of the 18000 ft or take the 4 % of the 2000ft (height)???? I always thought you had to take the 4% of the 18000ft but then somebody argued that the first 16000ft isn't air but mountain and can therefore not be compressed due to the lower temperature. Does anyone know which I should take and why?

Remember, the only time you can be certain that the altimeter indicates true altitude is when the aircraft is on the ground at the airport for which the current altimeter setting is set on the sub-scale of the altimeter.

The minute you're above that airport, temperature corrections become an issue. Why?

If the temperature was standard, and the lapse rate was also standard, the temperature at all levels (not altitudes) above that airport would be reasonably close to an equal indicated Altitude.

However, with a non-standard (colder) parcel of air, a greater number of levels can be found in that same column of air, therefore the level corresponding an indication of 5,000 feet (for example), may actually find itself at an actual height of 3,500 feet.

Conversely, with a non-standard (warmer) parcel of air, a lesser number of levels can be found in that same column of air, therefore the level corresponding to an indication 3,500 feet (example), may actually find itself at an actual height of 5000 feet.

Since you already know how to do the corrections, the answer should now be easy.

Help any?

PL64
11th Feb 2010, 15:56
Dear All,

Thanks a lot for taking the time to reply on this subject and for your very clear exlplanations...

Best regards,

PL64

Hope we get some warmer weather soon...:rolleyes: