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   Altitude Correction
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  Using the Altitude Correction for Pressure Altitude Calculator

Example 1:

If the altimeter setting were 28.92 InHg what would be the difference in altitude between the pressure altitude and the indicated (or calibrated) altitude?:

Begin by highlighting the field that corresponds to InHg.

Type in 28.92 and read the result for the altitude correction in ft (a) which is 939 ft or 286 m (b). This means that the pressure altitude is 939 ft above the indicated (or calibrated) altitude that you are flying. You also get the 28.92 InHg pressure converted to HPa or mbars (c), in this case it is 979.34 HPa or mbar.

To illustrate this example, we have two altimeters. The one on the left is set to 28.92 InHg and indicates 2000 ft. If the pilot were to remain at that altitude and set the altimeter to pressure altitude, that is 29.92 InHg, the altimeter would display a 939 ft higher altitude which is 2939 ft.

Example 2:

What is the altimeter setting for the pressure altitude to be 500 m below the indicated (or calibrated) altitude if the altimeter is set to the altimeter setting?:

Begin by highlighting the field that corresponds to m.

Type in -500 on the Altitude Correction in meters. It is negative since the pressure altitude will be below. You can now read 31.74 InHg (a) or 1074.78 HPa (b) for the altimeter setting. The calculator also displays the -500 m value in feet, which is -1640 ft.

Example 3:

A pilot is flying at FL 340 (Flight Level 340) which means her altimeter is set to 29.92 for pressure altitude and is showing 34000 ft. She hears on the radio that the altimeter setting is 30.58 InHg. How much below or above is she flying from the altitude if the altimeter were set to 30.58 InHg, instead of 29.92 InHg, and she were to fly at the inicated altitude of 34000 ft.?:

Begin by highlighting the field that corresponds to InHg.

Type in 30.58 and, on the Altitude Correction in feet (a), you read -604 ft which means the pressure altitude is 604 ft below the indicated altitude if the altimeter were set to 30.58 InHg or 1035.56 HPa (c). Therefore she would be flying 604 ft or 184 m (-b) above the pressure altitude.

If another aircraft were flying at FL 350 in the oposite direction she would be 604 ft above FL 340 and therefore have only 396 ft (35000 ft - 34604 ft) of vertical separation instead of 1000 ft.

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