interpolation

[cape]

Could someone post the correct method to interpolate from some of the performance charts or really any chart we would use in training.i think there is a standard method please feel free to comment thanks Cape.

 

[brokeflyer]

Could someone post the correct method to interpolate from some of the performance charts or really any chart we would use in training.i think there is a standard method please feel free to comment thanks Cape.

its just like finding an average number. take the 2 numbers and add them together, then divide by 2.

if you have 4 numbers, then add them together and divide by 4.

eg....landing distance at +10c is 4567ft and at +15c it 4968ft and your temp is +13c....add those and you get 9535 then divide by 2 = 4767ft.

 

[cape]

Clear enough thanks Brokeflyer.

 

[Mickey]

Not quite right. If you were looking for performance for 12.5 degrees that would have been right? You should take the difference from the 15c and the 10c and then multiply by 3/5. Add that to the 10c distance and there is your performance for 13c. (3/5 because 13 is 3/5 the way between 10 and 15. If you were looking for 14c you would use 4/5).

 

[Flying Ninja]

Mickey's correct.

Think of interpolation as a percentage.

4968 - 4567 = 401
15 - 10 = 5
13 - 10 = 3
3 / 5 = 0.6
401 * 0.6 = 240.6
4567 + 240.6 = 4807.6 --> 4808 ft required

 

[ackattacker]

Well, if one wants to get into the nitty gritty math of it... the above answers assume *linear* interpolation. There's all kinds, and linear is not going to produce the most accurate result.

The most accurate way would be to take the tabulated data, input it into a spreadsheet and make a x-y graph. Then curve fit a polynomial equation to the data, and use that equation to produce the interpolated point.

Yes, I'm a nerd. No, you shouldn't do this.

 

[Flying Ninja]

I just did a data analysis for the C172 landing distance table. Straight as an arrow. For you math freaks, that means linear.

 

[brokeflyer]

Not quite right. If you were looking for performance for 12.5 degrees that would have been right? You should take the difference from the 15c and the 10c and then multiply by 3/5. Add that to the 10c distance and there is your performance for 13c. (3/5 because 13 is 3/5 the way between 10 and 15. If you were looking for 14c you would use 4/5).

it safer to round up to the conservative side. His question wasnt that detailed so i answered properly with what info i was given.

Your apology will be accepted though.

 

[flyburg]

it safer to round up to the conservative side. His question wasnt that detailed so i answered properly with what info i was given.

Your apology will be accepted though.

Yet, your answer was LESS conservative than the correct way of interpolation!!!

The question was detailed enough to come up with the exact number!!!

Your answer to just average the values....not even close. if you want to be conservative, as you mention, then just round off the temp to higher figure and take the corresponding distance, but then again that wouldn't answer the OP's question.

I wouldn't count on those apologies ?

 

[flyburg]

Well, if one wants to get into the nitty gritty math of it... the above answers assume *linear* interpolation. There's all kinds, and linear is not going to produce the most accurate result.

The most accurate way would be to take the tabulated data, input it into a spreadsheet and make a x-y graph. Then curve fit a polynomial equation to the data, and use that equation to produce the interpolated point.

Yes, I'm a nerd. No, you shouldn't do this.

Completely correct from a math point of view. However, if the chart says, interpolation allowed without further explanation, then one may assume linear interpolation!!

Not a nerd, but have you ever seen an aviation related performance chart with a note that interpolation was allowed where they didn't mean linear?

So, ....your entire contribution was.......nerdy.