I have problem understanding what kind of answer is expected in meteorology question as follows:

How many feet must you climb through to reach FL 85?

Given: FL 85
Departure field elevation 1500 ft
QNH 1013 hPa
Temperature ISA -10deg C
1 hPa = 27ft

Is it asking about an indicated altitude change or about a true elevation difference?

  1. If this was about indicated, altimeter set to 1013 hPa at 1500ft aerodrome at ISA-10deg C would show as 1560ft (am I applying compensation of 1% per 2.5deg C correctly?). Therefore to change its indication to 8500ft, aircraft must make a climb of indicated 6940ft? That makes answer makes sense for me, since it would represent something that is visible on AI and VSI.

  2. On the other hand, if the question is about true elevation difference, FL85 would be indicated around 8160ft elevation at ISA -10deg C (again, is this correction in the right direction)? To climb from elevation 1500ft, it would require 6660ft elevation change. That answer feels pretty useless, since no instrument would show that (apart from a GPS or some radar).

  3. The answer accompanying the question suggest to calculate difference of 7000ft and compensate 1% per 2.5deg C that to get 6720ft. That does not feel to correct to me at all, since it completely ignores initial elevation/altitude.

If I am using incorrect terminology anywhere, I would be happy to correct that.

  • 6
    $\begingroup$ The post has both "FL75" and "FL85" - which is correct? Also, please understand that a vague question is vague for everyone, as nobody here has mind-reading insight into what an author intended about his question. $\endgroup$
    – Ralph J
    Jul 3 at 14:26
  • 1
    $\begingroup$ I agree with the previous comment. The test question appears to be designed to test your comprehension of the difference between true altitude of 1500' at the field, and flight level altitude at some height (7500' or 8500'...your own post is unclear on that point), as flight level altitudes are not the same as true altitudes. But a) your own post is presented in a hard-to-understand way, and b) asking a group of strangers what some third party meant when they wrote something is only going to get opinion-based answers. No one here can definitely state what the test question means. $\endgroup$ Jul 3 at 15:03
  • $\begingroup$ In the queue to be closed because it would be “opinion based”? I”m voting to keep this question open on a site that is supposed to be where experts answer questions, from the general public. $\endgroup$
    – Koyovis
    Jul 4 at 1:40
  • $\begingroup$ @RalphJ thank you, corrected to FL85 $\endgroup$
    – kpx1894
    Jul 4 at 16:07
  • $\begingroup$ @PeterDuniho b) I believe it is still worth asking this here, maybe this is better explained in other ground schools materials - I was only provided explanation 3). $\endgroup$
    – kpx1894
    Jul 4 at 16:08

Turns out the answer 3 is indeed correct, but a shortcut is being taken in the solution that muddies things up.

First, consider the more sensible question about climbing to $8,500\ \mathrm{ft}$ indicated at the airfield altimeter setting. Clearly you need to apply the $10°\mathrm C\times\frac{1\%}{2.5°\mathrm C}$ correction to the $7,000\ \mathrm{ft}$ of difference. That's how the correction works.

Now you have FL85 instead of $8,500\ \mathrm{ft}$. Alas, the altimeter setting is $1013\ \mathrm{hPa}$, which means the FL85 does, after all, coincide with $8,500\ \mathrm{ft}$ indicated.

The step to flight level seems to be there just to confuse the otherwise easy question, while also making it less realistic. If the airport was surrounded by $8,000\ \mathrm{ft}$ peaks, you would want to know whether flying $8,500\ \mathrm{ft}$ indicated still gives you enough terrain clearance, but you never fly flight levels unless you are sufficiently far from the terrain that you don't need to care.


Flight levels appear to be based on a standard sea level pressure of 1013.25 hPa or 29.92 inches mercury.

Above a Transition Altitude one switches from local "QNH" to standard pressure.

Standard temperature is 15 °C at sea level, decreasing 2 °C every 1000 feet.

Since (cooincidentally) your QNH is 1013 hPa, you would climb to 8500 feet Indicated. (You would not have to reset your pressure gauge). Easy. Everyone else above the TA is at the same pressure setting.

To calculate your actual distance climbed figure the standard temperature at FL 8500 should be -2 °C and ISA -10 °C. This means that your Outside Air Temperature or OAT will be -12 °C at 8500 feet. (This can be verified!).

The rule is 1% correction every 2.5 °C, so a 4% correction, or 8500 feet × .04 = 340 feet.

That works out to actually climbing 8500 - 1500 - 340 = 6660 feet to FL 85.

Note that the ISA deviation is given for 8500 feet. When your altimeter, set to 1013 hPa, reads 8500 feet, because of the temperature difference, your actual altitude is 8160 feet.

Therefor, the temperature correction must be applied to the entire 8500 feet, not just 7000 feet above the airfield.

  • $\begingroup$ The temperature is specified as ISA-10°C, not as -10°C@FL85, so you should be getting 4% correction. $\endgroup$
    – Jan Hudec
    Jul 5 at 19:00
  • $\begingroup$ @JanHudec Difficult to answer without clarity, but we try! QNH 1013 must read out as 1500 feet, or it might be a bad day at that field. Standard temperature at 8500 feet is - 2 C. Need clarification where - 10C is at, but as John K pointed out, colder than standard will show higher altimeter than one really is. So the climb to 8500 indicated will be somewhat less than 7000 feet. Let's hope there aren't any 8200 foot peaks there with trees on top. $\endgroup$ Jul 5 at 21:42
  • $\begingroup$ You are right, and I missed the QNH, which makes the third option actually correct, although the explanation is missing a rather crucial step then. $\endgroup$
    – Jan Hudec
    Jul 5 at 23:12

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