CL = (0.331±0.008) + (0.15±0.002)*ALPHA
I have to code this equation in MATLAB but don't know, how can I code this?
Thanks!

Réponses (2)

Star Strider
Star Strider le 27 Fév 2021

0 votes

One option:
CL = [(0.331+0.008) + (0.15+0.002)*ALPHA
(0.331-0.008) + (0.15-0.002)*ALPHA];
Experiment to get different results, depending on how you want to define the equations.

4 commentaires

Les Beckham
Les Beckham le 27 Fév 2021
Shouldn't there be four elements (for the four combinations (++, +-, -+, --)?
Star Strider
Star Strider le 27 Fév 2021
There can be. It depends on the desired result.
I kept it simple here to demonstrate the correct approach.
Priyank Goel
Priyank Goel le 27 Fév 2021
Thanks for the answer,
Is there any other way as writing the complete equation 4 times can be little tedious?
Star Strider
Star Strider le 27 Fév 2021
Modifié(e) : Star Strider le 27 Fév 2021
It’s quite possible to make this ridiculously complicated.
The easiest option is likely:
CL = [0.331 0.15] + [0.008 0.002].*[1 1; 1 -1; -1 1; -1 -1];
and just let it go at that.
EDIT — (27 Feb 2021 at 22:04)
Those are obviously only the limits, according to whatever criteria you wish to apply to them. Use them appropriately in your code.

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Walter Roberson
Walter Roberson le 27 Fév 2021

0 votes

syms ALPHA
syms delta_1 delta_2
assume(-0.008 <= delta_1 & delta_1 <= 0.008)
assume(-0.002 <= delta_2 & delta_2 <= 0.002)
CL = 0.331 + delta_1 + (0.15+delta_2)*ALPHA
CL = 
expand(CL)
ans = 
The matrix approach is not correct, as it gives specific values, whereas you need a range of values. For example you might have 1/(CL-0.15*ALPHA-0.331) as part of an expression, and if you only work with the edge cases you miss the fact that this passes through 1/0 .

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