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2 Answers

Interpolating – Performance Charts and Temperature at Altitude

Asked by: 9785 views Flight Instructor, Private Pilot, Student Pilot


I'm struggling with interpolating between figures on the Cessna 172N performance charts. Adding two numbers and dividing by two works fine and dandy when the figure you are interpolating for falls exactly bang on between the two. 

Regarding FD's, the reports only provide temperatures for 6000' and onwards.

For cruising altitudes at say, 4500', how do you interpolate to find a temperature at that altitude? I've been looking at nearest weather station enroute (halfway) for temperature as outlined in the Flight Training Manual. Can someone be kind enough to tell me exactly how to accurately interpolate? 

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2 Answers

  1. Wes Beard on Dec 28, 2012

    Interpolation is difficult sometimes but I have found a way to get a close of an answer as you like. Being extremely exact is nice but I have found that flying the airplane not to be so exact. I typically take the takeoff or landing distance of the next higher one and use that.

    The interpolation process though is this: Say you are at a field elevation of 2643′ and the POH gives you numbers for 2000 and 3000. The takeoff distance at 2000 is 1545′ and at 3000 is 1678′.

    The process is (1545 * .643) + (1678 * (1-.643)) = 1592.481′. Take the lower figure and multiply it by the difference between your altitude and the book altitude. Add the higher figure multiplying it by the difference between the higher altitude and your altitude. I put the decimal in front so we would not have to divide by 1000 later.

    The forecasted winds and temperatures (FD) report does not include winds where the reported altitude is within 1500′ AGL and the temperature is not reported when within 2500′ AGL. The reason this is the case is due to coriolis effect for the wind and surface heating for the temperature. I would either interpolate the surface temperature with the first temp at altitude or subtract 2° c / 1000′ for a close approximation. It isn’t going to affect your fuel consumption all that much.

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  2. Mark Kolber on Dec 29, 2012

    Wes, I think you may have kicked in a wrong number along the way. The midpoint between 2000 and 3000 is 2500 and the midpoint between the two takeoff distances is 1611.5. Your result give a shorter takeoff distance than the midpoint at an altitude higher than the midpoint.

    It’s probably too complex a process to describe in a short post, but let’s take a step back from a formula to the concept of interpolation. The whole idea is to take two sets of known values, find the rate of change between them and add the result.

    Take Wes’ example:
    at 2000 MSL, takeoff distance is 1545′
    at 3000 MSL, it’s 1678
    we want the takeoff distance at 2643 MSL

    This is the long way to do it. There are shortcut formulas, but the long way may help with understanding what the shorter formulas are doing.

    1. subtract 1545 from 1678, giving you 103
    2. divide that result by 1000, giving you 0.103. That’s the change in takeoff distance for each foot increase in altitude
    3. The 2643 MSL airport is 643′ higher than the 2000′ airport, so we take the 0.103 and multiply it by 643. The result, 85.5, is the additional takeoff distance you will need at 2643 MSL over the takeoff distance at 2000 MSL.
    4. Add that to the takeoff distance at 2000 MSL and you get a total takeoff distance of 1630.5′

    JunJug, If you run the same series of calculations at 2500, you’ll see you get the midpoint value that you would get simply by dividing by 2.

    One can definitely combine all of the steps into one formula but for the algebra-challenged among us, it may be easier to take the long way.

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