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I'm studying for my AIRAT (Instructor Written) in Canada using Nizus, but I've always had trouble answering the questions about takeoff distances at 27 ºC when given a chart with 20 ºC and 30 ºC, even since the start of PPL. I want to understand it properly so I can effectively teach my students who have troubles with it.

The question is a two-parter, first calculating a takeoff distance with the following values:

Parameter Value
Airport Temp 30 ºC
Airport Elevation 3000' AMSL
Altimeter 30.92 inHg
Wind 10 kt tailwind
Flaps 10º
Runway Dry grass

With a pressure altitude of 2089', the nearest values are the 2000' and 3000'. I used the 2000' line of the chart.

At 2000' at 20 ºC the ground roll is 1080' and the total to clear a 50' obstacle is 1895'.
At 30 ºC it's 1155' and 2030'.

The chart says it's configured as follows: 2300 lbs, flaps 10, full power prior to brake release on a paved level dry runway with no wind. The notes state:
Headwind subtract 10% per 9kts
Tailwind add 10% every 2kts up to 10kts
Dry, Grass Runway or Gravel add 15% to ground roll.

At 30 ºC I calculate my takeoff distance being 3305'.

The second part asks by how much the takeoff distance will decrease if the temperature drops to 27 ºC.

I'm doing something wrong here as I calculate it to be 3250', with a distance change of 55', but that isn't an available answer.

How do I get the proper numbers to use at 27 ºC?

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  • $\begingroup$ this is the chart here i.imgur.com/9HnYWR3.png, the first question i.imgur.com/IGZSMxP.png, and the second i.imgur.com/kkKLH1p.png $\endgroup$
    – zeeb
    Commented May 28, 2022 at 16:39
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    $\begingroup$ Is your question really just about how to properly interpolate or is it more than that? $\endgroup$
    – Jim
    Commented May 28, 2022 at 17:21
  • $\begingroup$ If the reference question was designed to see how well you can interpolate then I’d think you’d need to interpolate for both temperature and pressure. $\endgroup$
    – Jim
    Commented May 28, 2022 at 17:28
  • $\begingroup$ In order to get the exact answer I'm sure you'd need to interpolate for 2089' too I guess, but I originally figured it'd be such a small difference at that scale that there'd be no need. For 27degC I originally found the numbers for 25 by adding em both together, divide by 2, then add the number for 30degC again and divide by 2. I figured that'd get me to 27.5degC, but maybe that's not how that works at all? And I appreciate you guys taking the time to help me work this out. $\endgroup$
    – zeeb
    Commented May 28, 2022 at 17:56
  • $\begingroup$ I have an ASA CX-3 Flight Computer and that's what it gives, though I've never understood how this works out pressure alt as when I do it for personal flight calcs I've always just done 29.92 - 30.92 = -1 * 1000 = -1000. $\endgroup$
    – zeeb
    Commented May 28, 2022 at 18:42

2 Answers 2

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I am getting a takeoff distance of 3,502' for a takeoff over 50 FT on a grass strip with a 10-kt tailwind.

2,030' * 1.5 * 1.15 = 3,502'

To interpolate you will need to figure out the percentage between the two temperatures.

(27-20) / (30-20) = 70%

We take the difference between the chart values for 20 and 30 degrees

3,502' - 3,267' = 235'

Multiply this value by the percentage

235 * .7 = 164.5'

Since we subtracted the 27 from the lower value we will add in this number ot the lower chart value

3,267 + 164.5 = 3,432

This is a difference of 70 feet.

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Pressure altitude with Kollsman at 30.92 would be 3000 feet - 1000 feet = 2000 feet. Dry grass only affects ground roll.


Distance to clear 50' obstacle at 30 C : 2030 feet
Ground roll distance: 1155 feet


30 C with 10 knot tailwind: 2030 × 1.50 = 3045 feet
Ground roll with tailwind: 1155 × 1.5 = 1732 feet
Ground roll with tailwind and grass: 1732 × 1.15 = 1992 feet
difference: 1992 - 1732 = 260 feet more

Total distance to clear 50' at 30 C: 3045 + 260 = 3305 feet


Distance to clear 50' obstacle at 20C: 1895 feet
Ground roll distance: 1080 feet


20C with 10 knot tailwind: 1895 × 1.5 = 2842 feet
Ground roll with tailwind: 1080 × 1.5 = 1621 feet
Ground roll with tailwind and grass: 1621 × 1.15 = 1864 feet
difference: 1864 - 1621 = 243 feet more

Total distance to clear 50' at 20 C: 2842 + 243 = 3085 feet


3305 - 3085 = 220 feet
220 feet÷10C = 22 feet per degree

22 feet per degree × 3 C = around 66 feet less at 27 C

Answers of 3305 feet and 66 feet difference seem ok!

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