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I came across the following question while studying for the FAA PPL written test:

Determine the density altitude for these conditions:

Altimeter setting: 30.35
Runway temperature: +25°F
Airport elevation: 3,894 ft. MSL

I performed the following steps:

Temperature (°C) $\color{red}{= 5/9 × 25} = 14 $

Pressure altitude (PA) $= 3,894 + (29.92-30.35) × 930 = 3494 $

Density altitude (DA) $= \text{PA} + 120(14 - 15 + .002×\text{PA}) = 4200 $

Yet that answer is not listed, and according to the DA/PA graph provided, DA is 2000ft.

My final formula seems to be correct according to Wikipedia, plus it also matches my intuition of how PA works: According to the question we have 14°C at 4000ft, as opposed to the expected 7°C at 4000ft; seeing as temperature is higher, DA should be higher than PA.

Where am I going wrong?

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  • $\begingroup$ Why are you using 930 instead of 1,000 in you calculation of Pressure altitude? en.wikipedia.org/wiki/Pressure_altitude. Also, where did the second PA term in the DA formula come from? Not something I use. You might want to check out flyingmag.com/technique/tip-week/… $\endgroup$ – JScarry Jul 23 '17 at 23:56
  • $\begingroup$ @JScarry: 930ft/inHg is the correct factor at about 29.92 — 28.92 inHg; for higher altitudes it increases; thank you for bringing that to my attention, I'll probably use 1000 from now on. The $.002×\text{PA}$ is to correct for lapse rate. $\endgroup$ – Zaz Jul 24 '17 at 1:37
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Check your temperature conversion: 25F is -3.9C...

Should be (25-32)*(5/9).

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  • $\begingroup$ Yep, I suspected I'd goofed up with something as stupid as a conversion error. I'd answered several questions about change of density altitude with change of temperature, where you can forget the $-32$, and am embarrassed to say I went and did the same thing here. $\endgroup$ – Zaz Jul 21 '17 at 2:24
  • $\begingroup$ I'm studying for the commercial and CFI exams - there are plenty of ways to mess up! Happy to help. $\endgroup$ – sjdunham Jul 21 '17 at 18:24

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