How can i get value of alpha using trail and error

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Priyank Goel 2021-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/810245-how-can-i-get-value-of-alpha-using-trail-and-error

编辑： DGM 2021-4-22

I have equation CL = 2*(sin(alpha))^2 * cos(alpha)

and value of CL = -0.525

I need value of alpha using trail and error running for loop in MATLAB but not getting, Please help me with possible soution

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Scott MacKenzie 2021-4-22

编辑：Scott MacKenzie 2021-4-22

在 MATLAB Online 中打开

I'm not going to put together a solution, but here's some code to get you thinking about how to do this.

>> alpha = linspace(0,2*pi);
>> cl = 2*sin(alpha).^2.*cos(alpha);
>> lower = cl < -0.525;
>> idx = find(lower, 1);
>> clBracket = [cl(idx-1) cl(idx)]
clBracket =
      -0.4951     -0.58416
      
>> alphaBracket = [alpha(idx-1) alpha(idx)]
alphaBracket =
       1.8405        1.904
       

You can see above that the CL value you specified is between the two CL "bracket" values. And the corresponding alphas are also shown. So, at this stage we know that alpha is between 1.8405 radians and 1.904 radians. The mean of these two values is a first-level approximation of alpha. You could follow this up with additional code that gets you closer to alpha. For example, you can repeat the statements above, except beginning with

alpha = linspace(alphaBracket(1), alphaBracket(2))

This will get you much tigher bracket values, and get you closer to an acceptable solution. Are you with me?

You can put all this in a loop and compute

abs(alphaBracket(1) - alphaBracket(2))

as the last step. When this value is below some acceptable value, usually called episilon, you're done.

Good luck.

Priyank Goel 2021-4-22

Thanks a lot for a well descriptive explaination!

I also got the exact solution for different alphas using "fzero" most close value to your solution being 1.8607.

Also, I would be grateful if you could tell how to get smooth curve for given points for x & y axis

x1 = [3.34, 3.93, 3.65, 2.50, 1]

x2 = [0.19, 0.36, 0.64, 0.84, 1]

y = [0.10, 0.20, 0.40, 0.60, 1]

DGM 2021-4-22

编辑：DGM 2021-4-22

在 MATLAB Online 中打开

If your goal is simply to get smooth curves through those points, you can just do spline interpolation:

x1 = [3.34, 3.93, 3.65, 2.50, 1];
x2 = [0.19, 0.36, 0.64, 0.84, 1];
y = [0.10, 0.20, 0.40, 0.60, 1];
yf=linspace(0.1,1,50);
x1f=interp1(y,x1,yf,'spline');
x2f=interp1(y,x2,yf,'spline');
% draw the original coarse curves
plot(y,x1,':'); hold on; grid on 
plot(y,x2,':');
% draw the smoothed curves
plot(yf,x1f);
plot(yf,x2f);