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Can someone please explain in simple terms how to work out your QNH from QFE?

For example, you take off at 0' but leave the circuit, what do you set the altimeter to? How do you know the QNH?

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    $\begingroup$ If you have it set to QFE, turn the knob to add the field elevation to the current altitude. Or set it to the field elevation on the ground. (Or get the QNH from a nearby ATIS.) :) $\endgroup$
    – falstro
    Commented Dec 30, 2016 at 21:21
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    $\begingroup$ I found this googling helpful, especially this rule of thumb: "Divide airport altitude (in feet) by 30 feet. Add resultant number of millibars onto the QFE that you are given. Then you have QNH.". But please (!) wait for more profound answers, this was just what I found at Google. $\endgroup$
    – PerlDuck
    Commented Dec 30, 2016 at 21:27
  • $\begingroup$ Appreciate your help guys, i've googled and confused myself silly. $\endgroup$
    – user13555
    Commented Dec 30, 2016 at 21:39
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    $\begingroup$ @user6035379 I think it's pretty much anywhere outside north america. Though QFE is very rarely used (I think the UK uses it more than others, it's also fairly common with gliders) $\endgroup$
    – falstro
    Commented Dec 31, 2016 at 11:55
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    $\begingroup$ @user6035379 Very common in the UK when operating within the airfield zone. It makes flying circuit height etc very easy. 1000 on the altimeter, job done. We switch to QNH once we transition to en-route or otherwise leave the zone. $\endgroup$
    – Simon
    Commented Dec 31, 2016 at 13:06

3 Answers 3

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In metric:

1 milibar is approximately equal to 30 feet. This calculation is therefore also approximate but is good for airfield elevations to several hundred feet since we round to the nearest millibar.

Divide the airfield altitude in feet by 30 to get the number of millibars above MSL. Add this to the QFE to get QNH or subtract it from QNH to get QFE.

For example, the airfield elevation is 200 feet. Dividing by 30 gives us 6.66r. The QFE is 1023. Add 6.66 to get 1029.66 and round up to 1030 millibars, which is the QNH.

In imperial:

1 inch mercury is approximately equal to 900 feet. This calculation is therefore also approximate but is good for airfield elevations to several hundred feet since we round to the nearest hundredth inches.

Divide the airfield altitude in feet by 900 to get the number of inches above MSL. Add this to the QFE to get QNH or subtract it from QNH to get QFE.

For example, the airfield elevation is 300 feet. Diving by 900 gives us 0.33r. The QFE is 30.12. Add 0.33 to get 30.45 which is the QNH.

If the airfield elevation is below sea level, subtract rather than add and vice versa.

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QNH and QFE are related by this equation:

$$\mathrm{QNH} = \mathrm{QFE}\left[1+\frac{hL}{T_0}\left(\frac{P_0}{\mathrm{QFE}}\right)^\frac{R_sL}{g}\right]^\frac{g}{R_sL}\tag{1}$$

where:

  • $h$: Elevation of the airport in meters
  • $L$: Temperature lapse $=0.0065 \mathrm{~K/m}$
  • $T_0$: Standard temperature $=288.15 \mathrm{~K}$
  • $P_0$: Standard pressure $=101325 \mathrm{~Pa}$
  • $g$: Gravitational acceleration $\approx 9.81 \mathrm{~m/s}^2$
  • $R_s$: specific gas constant for dry air $\approx 287.058 \mathrm{~J \cdot kg^{−1}K^{−1}}$

Check this post if you're interested in how equation 1 is derived (remember that QFE is the pressure measured at the airport).

It's easy to show how equation 1 is simplified into what Simon talks about in his answer. Using the Binomial approximation:

$$\mathrm{QNH} \approx \mathrm{QFE} \left[ 1+\frac{gh}{R_sT_0} \left(\frac{P_0}{\mathrm{QFE}}\right)^\frac{R_sL}{g} \right] = \mathrm{QFE} +\frac{gh}{R_sT_0} \mathrm{QFE} \left(\frac{P_0}{\mathrm{QFE}}\right)^\frac{R_sL}{g} \tag{2}$$

The above equation can be further simplified if we assume $\mathrm{QFE}$ never differs too much from $P_0$ (that is, $\mathrm{QFE} \approx P_0$, thus $\frac{P_0}{\mathrm{QFE}} \approx 1$). Hence:

$$\mathrm {QNH} \approx \mathrm{QFE} + \frac{gP_0}{R_sT_0}*h \tag{3}$$

And the constant $\frac{gP_0}{R_sT_0}$ is:

$$ \frac{ 9.81 \mathrm{~\frac{m}{s^2}} \cdot 101325 \mathrm{~Pa} }{ 287.058 \mathrm{~\frac{J}{kg \cdot K}} \cdot 288.15 \mathrm{~K} } = \frac{993998.25}{82715.7627} \mathrm{ \frac{m \cdot Pa}{s^2 \left( \frac{1}{kg} \right) \left( \frac{kg \cdot m^2}{s^2} \right)} } = 12.017 \mathrm{ \frac{Pa}{m} } $$

or in other words

  • roughly 0.12 millibars for every meter of elevation
  • 0.97 inHg for every 900ft of elevation
  • 1.0988 mb for every 30ft of elevation

which are the values that Simon talks about in his reply.


How accurate is this approximation?

Remember that we are applying some correction factor to QFE (a measured value) in order to estimate QNH. That is, we set the altimeter to equal 0 AGL; then we depart the local airport circuit and we want to properly set QNH on the altimeter to find our AMSL altitude without having to constantly add the displayed altitude to the known field elevation. We are performing some function, which has the format

$$ \mathrm{QNH} \approx \mathrm{QFE} + Ch \tag{4} $$

where $h$ is the AMSL elevation of the field where we set QFE and $C$ is a multiplier we apply to that field elevation. We have found three candidates for $C$ (based on the various approximation we applied to the original equation), namely:

$$ \begin{align} C_1 &= \frac{gP_0}{R_sT_0} \tag{5.1} \\ C_2 &= \frac{g}{R_sT_0} \mathrm{QFE} \left( \frac{P_0}{\mathrm{QFE}} \right) ^ { \frac{R_sL}{g} } \tag{5.2} \\ C_3 &= \frac{g}{R_sT_0} \mathrm{QFE} \tag{5.3} \\ \end{align} $$

In order to assess which ones is the most accurate we need to see how much equation 4 differs from equation 1 for the 3 values of $C_1$, $C_2$ and $C_3$.

Equation 1 has 2 independent variables (QFE and h) but in practice, once the elevation is given, QFE cannot have any arbitrary values (and viceversa). It's reasonable to consider only values within the range $P_{\mathrm{ISA}} -4\% < \mathrm{QFE} < P_{\mathrm{ISA}} +4\%$, where $P_{\mathrm{ISA}}$ is the pressure one would find at the corresponding elevation in the standard atmosphere.

With this in mind, let's see how well the 3 multipliers $C_1$, $C_2$ and $C_3$ approximate equation 1. The following table reports the exact QNH for a set of altitudes and QFEs and the errors in percentage for the three multipliers ($C_1$, $C_2$ and $C_3$ respectively).

AMSL(m) \ QFE P_ISA -4.0% P_ISA -2.0% P_ISA @AMSL P_ISA +2.0% P_ISA +4.0%
0 QNH 972.72 (+0.0%) (+0.0%) (+0.0%) QNH 992.99 (+0.0%) (+0.0%) (+0.0%) QNH 1013.25 (+0.0%) (+0.0%) (+0.0%) QNH 1033.52 (+0.0%) (+0.0%) (+0.0%) QNH 1053.78 (+0.0%) (+0.0%) (+0.0%)
500 QNH 973.17 (+0.34%) (-0.14%) (-0.25%) QNH 993.21 (+0.24%) (-0.14%) (-0.23%) QNH 1013.25 (+0.14%) (-0.14%) (-0.2%) QNH 1033.28 (+0.05%) (-0.14%) (-0.18%) QNH 1053.32 (-0.04%) (-0.14%) (-0.16%)
1000 QNH 973.62 (+0.96%) (-0.55%) (-0.88%) QNH 993.44 (+0.75%) (-0.55%) (-0.83%) QNH 1013.25 (+0.56%) (-0.54%) (-0.78%) QNH 1033.05 (+0.37%) (-0.54%) (-0.74%) QNH 1052.85 (+0.19%) (-0.53%) (-0.7%)
1500 QNH 974.07 (+1.84%) (-1.21%) (-1.85%) QNH 993.67 (+1.53%) (-1.2%) (-1.78%) QNH 1013.25 (+1.24%) (-1.19%) (-1.71%) QNH 1032.82 (+0.95%) (-1.18%) (-1.64%) QNH 1052.39 (+0.68%) (-1.17%) (-1.58%)
2000 QNH 974.52 (+2.97%) (-2.09%) (-3.12%) QNH 993.89 (+2.56%) (-2.08%) (-3.03%) QNH 1013.25 (+2.17%) (-2.06%) (-2.94%) QNH 1032.59 (+1.79%) (-2.05%) (-2.86%) QNH 1051.92 (+1.44%) (-2.04%) (-2.77%)
2500 QNH 974.97 (+4.34%) (-3.19%) (-4.67%) QNH 994.12 (+3.83%) (-3.17%) (-4.56%) QNH 1013.25 (+3.35%) (-3.14%) (-4.45%) QNH 1032.36 (+2.88%) (-3.12%) (-4.34%) QNH 1051.46 (+2.43%) (-3.1%) (-4.24%)
3000 QNH 975.42 (+5.95%) (-4.47%) (-6.46%) QNH 994.35 (+5.34%) (-4.44%) (-6.33%) QNH 1013.25 (+4.76%) (-4.41%) (-6.2%) QNH 1032.13 (+4.2%) (-4.39%) (-6.08%) QNH 1051.0 (+3.67%) (-4.36%) (-5.95%)
3500 QNH 975.87 (+7.78%) (-5.94%) (-8.46%) QNH 994.57 (+7.08%) (-5.9%) (-8.31%) QNH 1013.25 (+6.4%) (-5.86%) (-8.17%) QNH 1031.9 (+5.76%) (-5.82%) (-8.02%) QNH 1050.53 (+5.13%) (-5.79%) (-7.88%)
4000 QNH 976.32 (+9.83%) (-7.55%) (-10.65%) QNH 994.8 (+9.03%) (-7.51%) (-10.48%) QNH 1013.25 (+8.26%) (-7.46%) (-10.32%) QNH 1031.67 (+7.53%) (-7.41%) (-10.16%) QNH 1050.07 (+6.82%) (-7.37%) (-10.0%)
4500 QNH 976.77 (+12.09%) (-9.32%) (-13.0%) QNH 995.03 (+11.19%) (-9.26%) (-12.82%) QNH 1013.25 (+10.33%) (-9.2%) (-12.64%) QNH 1031.44 (+9.51%) (-9.15%) (-12.46%) QNH 1049.61 (+8.71%) (-9.09%) (-12.29%)
5000 QNH 977.22 (+14.54%) (-11.2%) (-15.48%) QNH 995.25 (+13.55%) (-11.14%) (-15.28%) QNH 1013.25 (+12.6%) (-11.07%) (-15.09%) QNH 1031.21 (+11.69%) (-11.01%) (-14.9%) QNH 1049.14 (+10.81%) (-10.94%) (-14.71%)

Observations:

  • $C_3$ performs the worst in almost all the cases, so it can be ignored
  • $C_1$ and $C_2$ perform pretty much the same and reasonably well (<= 2%) until 2000m
  • $C_1$ approximates QNH in excess, $C_2$ in defect (so $C_1$ is actually safer than $C_2$)
  • $C_1$ performs better with high pressure, $C_2$ is more accurate with low pressure

Conclusions:

  • The usual approximation of 1inHg/900ft or 1mb/30ft (i.e. $C_1$) is acceptable for practical purposes if the elevation is 2000m or lower.
  • Since it approximates the QNH in excess, $C_1$ is safer than the other 2 approximations evaluated (i.e. the pilot can fly to an altitude of 0 AGL without crashing on the ground)
  • In general, when possible, QNH should be taken from the METAR report (which likely uses equation 1 to compute it)
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  • $\begingroup$ I don't quite follow how we got from eq 2 to eq 3. I would have said: (P_0/QFE)≈1 which eliminates (P_0/QFE)^(exp) and now we are left with QFE plus QFE times (gh/R_sT_0). But it seems you canceled out QFE/QFE (ignoring the exponent!!) and kept P_0. Is that how the math works? $\endgroup$
    – randomhead
    Commented Feb 1, 2022 at 2:37
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    $\begingroup$ ...oh, of course, we are left with QFE times (const). But instead we write P_0 times (const) because we were assuming QFE≈P_0. Right. $\endgroup$
    – randomhead
    Commented Feb 1, 2022 at 2:48
  • $\begingroup$ @randomhead: Yes you got it right. And thanks for cleaning up the latex! $\endgroup$
    – fab
    Commented Feb 1, 2022 at 4:49
  • $\begingroup$ See my addition, especially the part hidden with an HTML comment block where I tried to quantify the error in calculating QNH with the different approximations (i.sstatic.net/HJVTp.gif). I got errors of greater than 2.5 inHg at high-but-not-crazy elevations (1100m/3600ft) so I feel like I must be doing something wrong. Or is the approximation just that terrible? Here's what I'm doing: i.sstatic.net/PTgq1.png $\endgroup$
    – randomhead
    Commented Feb 2, 2022 at 8:31
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – randomhead
    Commented Feb 2, 2022 at 17:57
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I just read about QNH and QFE so I thought I’d throw this out there to start the discussion. We don’t use QFE in the US but I think my logic is correct.

From the FAA Instrument Procedures Handbook

Barometric Pressure for Local Altimeter Setting (QNH)

A local altimeter setting equivalent to the barometric pressure measured at an airport altimeter datum and corrected to sea level pressure. At the airport altimeter datum, an altimeter set to QNH indicates airport elevation above mean sea level (MSL). Altimeters are set to QNH while operating at and below the transition altitude and below the transition level.

For flights in the vicinity of airports, express the vertical position of aircraft in terms of QNH or QFE at or below the transition altitude and in terms of QNE at or above the transition level. While passing through the transition layer, express vertical position in terms of FLs when ascending and in terms of altitudes when descending.

When an aircraft that receives a clearance as number one to land completes its approach using QFE, express the vertical position of the aircraft in terms of height above the airport elevation during that portion of its flight for which you may use QFE.

note that transition level is when you set the altimeter to a standard value. It varies by country. In the US it is 18,000' and 29.92 inches. Not relevant to this discussion.

In the US we don’t use QFE, but as a comment indicates, you would probably get that from the ATIS. If no ATIS is available, to convert QNH to QFE you would move the altimeter so that you decrease your altitude by the field elevation. Here’s an example:

My field elevation is 212'. If I set the altimeter to QNH it will show 212' as the altitude. To get QFE I need to change the altitude to 0'. In other words whatever altitude is showing on the altimeter, move the knob to make it show 212' less.

Note that when you turn the altimeter knob the altitude goes in the same direction as the pressure setting in the Kollsman window. In the US we use inches of mercury but the movement is the same: As an example, right now the altimeter setting at KSBP is 30.08. The setting at KSBA (62 nm away) is 30.01. If you fly to KSBA you would notice a change in altitude of -70 feet when you get the new altimeter setting from approach control.

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    $\begingroup$ "start the discussion" and "I think" are not good phrases to use in a Q&A site. Otherwise, good answer. You might consider an edit? $\endgroup$
    – Simon
    Commented Dec 31, 2016 at 11:33

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