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Altimetry - me dim, a Q

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Old 7th April 2004 | 13:18
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Altimetry - me dim, a Q

I've posted this onto Tech Log but from experience Rotorheads is something of a centre of excellence with regards to technical stuff (I appreciate that the below is noddy) and so are hoping that mod will let is stay for a bit.

I am much confused but beginning to see the light with regards to altimetry. If someone could be kind enough to explain the solution to the below Q, it would go some way to completing my understanding of some fundamental concepts in altimetry.

I have worked on it and think I have the right answer - an explanation will confirm it and assist my general understanding. Thanks very much in advance.

Minimum safe altitude in an area is 12,000ft. QNH at a local airfield is 1030hPa.
Average temp of atmosphere is ISA+10. What will altimeter read at this safe
altitude?
a) 11250
b) 11520
c) 12210
d) 1730
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Old 7th April 2004 | 15:23
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From: Alles über die platz
Surely an altimeter subscale set on standard setting of 1013.2mb (29.92inches) would read a pressure altitude !

Without a subscale isn't this altimeter a barometer?

At that height I would think you would have standard pressure set, so therefore at 12,000ft your altimeter would read 12,000ft.

Rule of thumb
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Old 7th April 2004 | 15:41
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From: Alles über die platz
As cron is from the UK, I thought fair to assume this was a UK based question, oh well silly me...

Are we talking safety altitudes/heights, minimum flight altitudes or Minimum Sector Altitudes here?

Whatever, what is the elevation of the local airfield?

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Old 7th April 2004 | 16:39
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I think I may be sharing cron's confusion here now.

If I'm standing atop a pinnacle at 12000 feet holding an altimeter set to current QNH (1030, in this case) in ISA conditions, the altimeter will read 12000 feet.

If the temperature now rises by 10 degrees, the air is hotter, therefore less dense, therefore thinner, therefore exerts less pressure, so the altimeter reading will surely rise ? The question asks what the altimeter would read at a height of 12000ft, not what altitude would you be at if the altimeter reads 12000ft.
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Old 7th April 2004 | 16:53
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From: Alles über die platz
12000ft on 1030 QNH It is stated that temp deviation is +10 deg cel from ISA
To calculate take 3 key figures

4 :Allow 4ft per 1000ft above press datum
10 :Temp deviation from ISA (+10deg cel)
12 :Min safe alt 12000ft (or height in question)

4x10x12=480 ft

ISA deviation warmer; alt underreads
ISA deviation colder; alt overreads

Temp is warmer; alt underreading; 12000ft - 480ft = 11520ft

No rules of thumb and I hope easier to understand.
Sorted
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Old 7th April 2004 | 17:08
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Charlie s,

Thank you, all is now clear.
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Old 7th April 2004 | 19:42
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From: Alles über die platz
We all get there in the end in this case, but it was the use of the term, "rule of thumb" which isn't one i tend to use or certainly one I would advise others to use when there are proven ways of reaching a solution.

Ever wondered how we got the phrase? Well;

A rule of thumb is something that works 80% of the time. 1 out of 5 times it is wrong, the thumb is one of 5 fingers,
hence the term "rule of thumb"
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Old 7th April 2004 | 22:33
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Well, if you read the question, it asks:
"Minimum safe altitude in an area is 12,000ft. QNH at a local airfield is 1030hPa.
Average temp of atmosphere is ISA+10. What will altimeter read at this safe
altitude?"

It says nothing about elevations. QNH and temperature are irrelevant. If an aircraft is flying at the "safe altitude" of 12,000', the altimeter will read 12,000'. Doesn't matter what QNH you have set, only the aircraft's elevation will vary.

But this dopey question doesn't have 12,000 as an answer, so they obviously mean something else. But who knows what?
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Old 8th April 2004 | 06:46
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I'm with AC

In Oz...anything above 10,000 ft and lower than 11,000 is within the "Transition Altitude"....above 10,000 you are in the Flight Levels...hence this LSALT of 12,000 is really FL120, based on ISA 1013hpa/mb.

If the UK is the same as Oz...any aircraft at "12,000" would in fact have 1013 set in the Altimeter, hence that aircraft would be reading and at 12,000 AMSL.

There will be some errors within the aircraft and positioning etc....

Bet this question was drawn up by a Navigator !!!, who has lost her way.

Duck.

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Old 8th April 2004 | 10:58
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Thank you, all contributors.
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Old 8th April 2004 | 11:21
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csc - I'm not going to be popular now. I don't have the answer, but I do recall this being from a JAA Specimen question paper. But I am grateful chaps - lost of useful stuff here for me to ponder.
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Old 9th April 2004 | 08:35
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? Just to throw another spanner in the works:

Doesn't it depend on the height of the "nearby airfield" ? I mean, if the airfield is at sea level then you have indeed got 12,000 ft worth of less dense air to correct for. This appears to be what's being assumed in the calculations.

But suppose the "nearby airfield" is also at 12,000 ft ? Their local QNH is the subscale setting that correctly registers 12,000 ft when you are sitting at the highest point on the airfield. In which case your aircraft is at the same level as the airfield and the answer to the question would also be "12,000 ft "

Wouldn't it ?
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Old 9th April 2004 | 12:40
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Guys....time out!
First of all before you go into all the formulae and whizz wheels etc etc - see what they are actually trying to glean from you. The question is about altimeter error. "What will the altimeter read at this safe altitude?" But as usual, it is ambivalent and confusing. Are you to assume that you have your altimeter set to 1013, or following procedure and have it set to 1030?

When you are flying into an area where pressure is greater than ISA the altimeter will sense it as a lower altitude. Your absolute altitude hasn't changed - the question is "What will the altimeter read at this safe altitude?" It will read lower.

Rate of change is 30ft per millibar.1030-1013=17*30=510ft. So one of two things - if your altimeter is set to 1013, it will be reading 11,490ft at safe altitude, while in the area of higher pressure, or if you have reset it to 1030 it will read 12,000ft, and then you would merely calculate temperature difference.

Looking at the choices, I think they want a pressure/temp calculation, so adjusting for temperature using the 4% per 10 degrees, 510x4%=20.4 + 11,490= 11,510 - Good enough for me.
 

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