Altimetry calculations

Which approximations does an EASA altimetry question expect?

State the examination model before calculating. Keep it separate from an aircraft's operational correction procedure.

The specified calculation conventions

For the subject-050 learning-objective calculations, pressure values are rounded down to the next lower whole hectopascal before calculation: 1013.25 becomes 1013, and 1000.9 becomes 1000. The specified barometric conversion between MSL and 700 hPa is 30 ft per hPa. This is an examination approximation over the stated range; it is not a claim that the real barometric lapse rate is constant.

The 4% rule changes the height of the relevant layer by 4% for each 10 °C ISA deviation. If ΔT is the deviation in degrees Celsius, the layer-height factor is f = 1 + 0.004ΔT. For ISA−20, f = 0.92; for ISA+15, f = 1.06. Unless more information is supplied, the given ISA deviation is treated as constant through the layer. This is a rule of thumb, not a detailed integration of a measured sounding.

Account for aerodrome elevation. The temperature correction applies to the layer between the ground/setting datum and the aircraft, not blindly to its full altitude above MSL. First resolve the pressure-setting indication, then isolate that layer, apply the temperature factor and restore the datum elevation. Actual cold-temperature correction uses the applicable approved tables, systems and operational procedures, with any ATC coordination required.

How much does the indication change without moving the aircraft?

Use a direction check before applying 30 ft per hectopascal.

Higher setting, higher indication

At unchanged static pressure, indicated altitude changes by approximately 30 × (new setting − old setting) feet under the examination convention. Increasing the setting raises the indication. Reducing it lowers the indication. Write the signed setting difference instead of trying to remember a pressure-error slogan for every situation.

Suppose an aircraft indicates 6,000 ft on QNH 995 hPa. Changing to standard 1013 hPa increases the setting by 18 hPa, so the indication rises by 540 ft to a pressure altitude of approximately 6,540 ft. Changing back from 1013 to 995 lowers the indication by 540 ft. The pressure surface and the true position have not changed during the adjustment.

To compare QFE and QNH, use their pressure difference if supplied. In the approximate model, an aerodrome elevation of 1,200 ft corresponds to a QNH–QFE difference of about 40 hPa. QNH gives the elevation at the datum; QFE gives zero. Real supplied settings may not fit an exact 30 ft/hPa relation because the true atmosphere and the published reduction use more detailed relationships.

What should the instrument show before and after the reference change?

Use one pressure-setting equation for a parked aircraft, a climb and a descent.

Readings on the aerodrome

Consider an aerodrome elevation of 1,200 ft, with QNH 1003 hPa and QFE 963 hPa supplied in the exercise. On the ground the approximate indications are 1,200 ft on QNH, zero on QFE, and 1,500 ft on standard: 1,200 + (1013 − 1003) × 30. The pressure altitude of the aerodrome is therefore 1,500 ft in this calculation, although its elevation remains 1,200 ft.

Use the supplied aerodrome datum consistently. If QFE refers to a threshold or another defined datum, zero refers to that datum rather than every point on a sloping aerodrome. The on-ground check tests setting consistency; it does not remove temperature error in the layer above the aerodrome once the aircraft climbs.

A climb and a descent

Let the transition altitude be 5,000 ft and QNH 993 hPa. At the moment an aircraft climbing on QNH reaches 5,000 ft, changing to 1013 increases the indication by 600 ft to approximately 5,600 ft pressure altitude. Continue to the cleared available flight level; the aircraft has not suddenly climbed 600 true feet just because the reference changed.

On descent, suppose the assigned transition level is FL70 with the same QNH. At the moment the standard indication is 7,000 ft, selecting QNH 993 gives approximately 6,400 ft. This is a numerical illustration, not a method for inventing a transition level from an arbitrary minimum layer. The promulgated procedure and ATC information provide that level.

Why does a cold day put you lower than the indication suggests?

Apply the temperature factor to the vertical distance above the setting datum.

Cold air contracts the pressure layers

Between fixed pressure surfaces, cold air has a smaller geometric thickness than warm air. With the correct local QNH, an altimeter can be correct at the aerodrome datum but over-read true altitude above it in colder-than-standard air. At the same indicated altitude, the aircraft is then closer to that datum and potentially to terrain. Warm air produces the opposite thickness tendency.

For the examination 4% rule, true height above the datum = indicated height above the datum × f, where f = 1 + 0.004ΔT. True altitude = datum elevation + corrected true height. If QNH indicated altitude is 6,000 ft above MSL, the aerodrome elevation is 1,000 ft and the column is ISA−20, apply 0.92 to 5,000 ft: true height = 4,600 ft and true altitude = 5,600 ft. Applying 0.92 directly to 6,000 would wrongly correct the fixed 1,000 ft datum elevation.

The same temperature structure affects the geometric interval between two flight levels. Their pressure interval is fixed by the altimeter reference, but the vertical thickness of that interval is smaller in cold air and larger in warm air. A common pressure datum supports vertical organisation; it does not make pressure surfaces coincide with fixed geometric heights in all weather.

How much clearance remains beneath a flight level?

Resolve the pressure reference first, then the cold-air thickness, then the terrain.

A complete forward calculation

An aircraft flies FL80. QNH is 993 hPa, aerodrome datum elevation is 1,000 ft, the layer is ISA−20 and terrain under the aircraft is 4,000 ft MSL. Use the EASA examination pressure rounding and 30 ft/hPa convention. Standard becomes 1013 hPa, so changing the reference to QNH lowers the indication by (1013 − 993) × 30 = 600 ft. The QNH indicated altitude is 7,400 ft.

Subtract the datum elevation to obtain the layer to be temperature-corrected: 7,400 − 1,000 = 6,400 ft. At ISA−20 the factor is 0.92, giving true height above the datum of 5,888 ft. Restore the datum elevation: true altitude = 1,000 + 5,888 = 6,888 ft. Finally subtract the terrain elevation: clearance = 6,888 − 4,000 = 2,888 ft.

Two separate effects lowered the true altitude relative to the FL80 number: low QNH and a cold column. When flying towards lower pressure without updating a local pressure setting, the old higher setting can cause an over-read. This is the meaning behind the high-to-low warning, but the signed setting equation is more reliable than a slogan when the question asks for a particular instrument reading. Temperature must be handled separately.

How do you reverse the temperature correction?

Work backwards from the required true altitude and then select a permitted flight level.

Division reverses multiplication

Keep QNH 993 hPa, datum elevation 1,000 ft and ISA−20. Suppose the required true altitude is 7,000 ft. The required true height above the datum is 6,000 ft. Because true height = indicated height × 0.92, the required indicated height is 6,000/0.92 = 6,521.7 ft. Add the 1,000 ft datum to get approximately 7,521.7 ft QNH indication. Do not multiply by 1.08 as an exact inverse: 1/0.92 is about 1.087.

Convert the QNH indication to standard pressure altitude by adding (1013 − 993) × 30 = 600 ft. The minimum pressure altitude is therefore about 8,121.7 ft. If the explicitly available levels are FL80 and FL100, FL80 is insufficient and FL100 satisfies the numerical requirement. If other levels are offered, check them against the applicable route, direction, transition and clearance constraints rather than assuming FL90 is always available.

For a question that states a terrain elevation and a required clearance, add them first to establish the minimum true altitude. Then reverse temperature about the stated datum, convert the pressure reference and select a permitted level. Keep intermediate precision until the final choice; rounding a required value down can falsely make a marginal level appear sufficient. The examination rule's pressure rounding is a separate explicit convention.

Can you apply the ideas?

Four original questions. Decide from the facts, then review the explanation.