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Value between two curves 2

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lupin77

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Jun 2, 2009
6
Hi there,
I have a graph with multiple curves of which I know their respective polynomials. The graph shows on the x-axis the angle of attack of a wing whereas the y-axis gives the lift coefficient for several aspect ratios (hence multiple curves)
How can I find the value in between two "adjacent curves" for a given value of x? I thought of linear interpolation but cannot apply it here because x does not change.

Many thanks

 
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Couldn't you just take each value of 'y' at your 'x' and find their mean?
[peace]

Fe
 
um, linear interpolation.

say you have y1(x) and y2(x), which are values from each of the shown curves. You want to find y`(x), which is a value at an intermediate point, between the curves y1 and y2, at the given value of x.

Assuming, as an example, that the curves y1(x) and y2(x) represent lift for angles-of-attack A1 and A2 respectively, and you want lift at A`, use linear interpolation as follows:

A` = y2(x) + (y1(x)-y2(x))*A`/(A1-A2)
alternatively,
(A'-A1)/(A2-A1) = (y`(x)-y1(x))/(y2(x)-y1(x))
 
umm, you say you know the data's respective polynominals ... substitute in the AoA into the two curves ???

i guess we're missing something ...
 
Thanks to all for the replies.
The graph I'm referring to is just one of a series of graphs which I have digitised and obtained polynomial expressions for the curves that fit the data. I had originally implemented a "mean" as suggested by FeX32. My only worry is that some curves cross each other
example for a certain value of x (AoA) the curve representing the lift coeff. for aspect-ratio 1.5 is below the one for aspect-ratio 3.5. As x increases the situation is reversed.
Hope this clarifies a bit.
 
it is minutely more difficult to find the AoA values on two curves corresponding to the same CL ... (and this is expressly what the OP doesn't ask for ... for a given AoA, what is the difference between two CL curves)

i'd've thought linear interpolation on each of the curves would get you there, or "goal seek" in excel (y=f(x) ... find "x" for a given "y").
 
my apologies, i should read more carefully ... the mean is exactly the answer you want ... the mean is always between the two curves, doesn't matter which is above the other.

if you've got polynominal expressions for teh two curves, you can average the coefficients to obtain the mean curve, so long as you're using the same powers for both curves.
 
I misread the original post - substitute "aspect ratios" for "angle of attack" in the description, and use linear interpolation as described. It doesn't matter which curve is on top.
 
The mean is a special case of linear interpolation where the interpolated aspect ratio is exactly halfway between the two aspect ratio's you have. You haven't said you are looking for an aspect ratio exactly halfway between so it seems appropriate to consider the more general case, and the more general solution is linear interpolation as suggested above.

Lift = f(AoA, AR)
Pick fixed a value of AoA. Then we have
Lift = f(AR) and apply simple linear interpolation based on aspect ratio AR

You want to interpolate to find lift for some AR between your fixed values AR1 and AR2
Lift = Lift1 + (AR - AR1) * slope where
Slope = (Lift2-Lift1)/(AR2-AR1)
Again all the above would be computed at the same value of AoA.

(it doesn't matter whether Lift1 or Lift2 is higher, the algebra will give the correct result)

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Thanks everybody for the valuable suggestions.
I'll implement the solutions posted and take it from there

Cheers
 
Why would one assume linearity of the aspect ratio? It's probably not true
It would seem to me that given several curves with Xp,Yp and aspect ratio ratio, Z, one could write the polynomial

Y= AmnX^nZ^m, where n and m range from 0 to N and M
from which you get the best fit
Standard techniques are available to get the best RMS fit. to wit
Error squared =sum(Y-Yp)^2 summed over p
To get the coefficients, take the partial derivative with respect o Amn to get a set of MN+ N +M linear equations in Amn which solution (resulting in minimizing the error of the fit)
yields the coefficients, Amn
 
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