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Priyank Goel
Priyank Goel 2021-2-27
编辑： Star Strider 2021-2-27
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!

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回答（2 个）

Star Strider
Star Strider 2021-2-27
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.
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Priyank Goel
Priyank Goel 2021-2-27
Thanks for the answer,
Is there any other way as writing the complete equation 4 times can be little tedious?
Star Strider
Star Strider 2021-2-27
编辑：Star Strider 2021-2-27
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 2021-2-27
Ran in:
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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