The manual that came with my Jeppensen E6B has the following sample...

If an aircraft is flying at 12,500 feet with an outside air temperature of -20C and the altimeter is set on 30.42 inches of mercury, what is the true altitude?

The explanation goes on to find a pressure altitude of 12,000 feet. After placing -20C over 12,000 we find 12,500 (12.5) on the B scale and read the true altitude of 12,000 (12.0) on the outer ring.

Looking closely at the outer ring, I see that the true altitude is actually just slightly below 12,000. Now for practical purposes I realize we don't worry about differences like this and round the result. But I'm working on a project in which I need to calculate the true altitude precisely. I'm having a difficult time finding the formula. The closest thing I have found says the correction is 4 feet per thousand feet indicated per degree off of ISA.

When I attempt to apply that formula to our sample problem I get 4 * 12.5 * -35 = -1750. Applying that correction gives a value far below expected so obviously I'm doing something wrong. Can someone straighten me out here? References to documentation with educational value are especially welcome.

  • $\begingroup$ Can you add more detail and context? In practice, it is impossible to calculate true altitude from pressure since the atmosphere does not act in a linear way. One day A, with ISA standard at sea level, the pressure altitude for say 900Hpa can be different for day B with ISA standard day. It will also change from minute to minute. So when you say you must calculate it exactly, we need to find a different approach. $\endgroup$ – Simon Mar 11 '16 at 6:46
  • $\begingroup$ If you need true altitude, you should not use pressure altitude, especially at higher altitudes. Use GPS instead. Out of curiosity what do you need true altitude for? There are corrections you can apply based on meteorological models that allow you to convert pressure altitude to true altitude, but they involve quite some math. $\endgroup$ – DeltaLima Mar 11 '16 at 8:27
  • $\begingroup$ @DeltaLima, I'm creating an E6B app as a programming excercise. I wanted it to be mathematically true to the paper versions. Determining the location of the individual tick marks on each scale sometimes requires a little reverse-engineering. Knowing the precise formula used by manual E6B's (even if they weren't the most accurate formula) would help me recreate the most consistent experience. $\endgroup$ – dazedandconfused Apr 12 '16 at 12:17
  • $\begingroup$ I don't know the approximation used by E6B, but I do know the official mathematical model used by ICAO. Would that help? $\endgroup$ – DeltaLima Apr 12 '16 at 15:56
  • $\begingroup$ Couldn't hurt, if nothing else it would be an educational opportunity for me. Thank you. $\endgroup$ – dazedandconfused Apr 12 '16 at 15:58

I would think the OAT in the example is what your on-board thermometer is showing, so at 12,500 ft, not at sea level. ISA is +15°C at sea level, but at 12,500 ft it is -9.8°C, so you are just -10 off ISA. Let's see. 4 ⋅ 12.5 ⋅ (−10) = −500, so the formulas match.

Note though, that this is still just an approximation. The temperature lapse rate might also differ and that would have to be taken into account as well. The more precise equation is also nonlinear.

  • $\begingroup$ This filled in a big missing piece... I wasn't accounting for the lapse rate and so I was thinking -35 off ISA rather than -10. But as you said, this is an approximation. Do you happen to know "the more precise equation?" I'm trying to replicate the result obtained from the E6B which is slightly lower than 12,000. I suspect we use "4" for convenience but the actual value is 4.xxx $\endgroup$ – dazedandconfused Mar 11 '16 at 16:02
  • 1
    $\begingroup$ @dazedandconfused, I've seen some calculations in simulators, but never took the time to truly understand them. In any case while the temperature lapse rate is considered constant in the troposphere (up to 36,000 ft), the pressure and density still decrease exponentially, so the calculation is rather complicated. $\endgroup$ – Jan Hudec Mar 11 '16 at 22:01

True Altitude= P.A. + ( (4/1000)* P.A.* Temp. Dev.)

Where P.A.= Pressure Altitude

Temp. dev.= Temperature Deviation from IsA temperature at that level

  • $\begingroup$ Q: Given the following information, what is the true altitude? (rounded to the nearest 50 ft) QNH: 983 hPa Altitude: FL 85 Outside Air Temperature: ISA - 10° $\endgroup$ – MBC870 May 27 '20 at 7:38

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