Question

A wind tunnel test conducted on an airfoil section yielded the following data between the lift coefficient (CL) and the angle of attack (?): 12 1.40 16 1.71 20 1.38 de CL 0.11 0.55 0.95 You are required to develop a suitable polynomial relationship between ? and CL and fit a curve to the data points by the least-squares method using (a) hand calculations and (b) Matlab programming Hint: A quadratic equation (parabola) y(x)-aa,x +a x' can be used in this case for curve fitting The procedure of solution should be as follows: Write a Matlab function to fit a quadratic equation to the data points. Create[A and [b] according to your hand calculation, and use backslash operator x - (a) au Alb to find a vector of unknowns a (b) Use Matlab functions polyfit) and polyval). Plot original data (from the table), function y(x)-a,+ax+a.x' with coefficients obtained in (a); and values obtained using polyval 0 at the same graph. (Don't forget to create a vector for the range [0 20]!). Explain the results. MATLAB FUNCTIONS yy-interpl(x, y, xr,'linear') - uses the data of x (values of the independent variable) and y (values of the dependent variable), fits a linear equation using least-squares technique, and gives the interpolated values of y(yy) at the specified values of x(xx). The word linear' can be replaced by 'cubic', 'spline, or nearest' to achieve interpolation based on cubic equation, cubic spline or nearest value, respectively. If nothing is specified, linear interpolation is implied p-polyfit(x, y, n) - fits a least-squares polynomial of degree n for the specified set of data x and y. The coefficients of the polynomial are returned through the vector p. The elements ofp denote the coefficients of the polynomial starting from that of the highest power ofx. y- polyval(p,x) returns the value of a polynomial of degree n evaluated at x. The input argument p is a vector of length n+1 whose elements are the coefficients in descending powers of the polynomial to be evaluated.

Answer 1

KJosmd equation 2. 2 149 / 16 . 8 | 201 . 611T28 |2073c . ts8 400 27. 6 552 [8000 Go000 14400 Solvina fivel

This is MATLAB CODE---------

fprintf('Matrix b is defined as\n');

b=[6 60 880;60 880 14400;880 14400 250624]

fprintf('\n Matrix A is\n');

A=[6.1; 81.56;1260.96]

coefficient_vector=inv(b)*A

a_o=coefficient_vector(1,1)

a_1=coefficient_vector(2,1)

a_3=coefficient_vector(3,1)

% Question b

fprintf('\nAnswer to question b\n');

X=[0 4 8 12 16 20]

Y=[0.11 0.55 0.95 1.4 1.71 1.38]

A_coeff=polyfit(X,Y,2);

a_3=A_coeff(1)

a_2=A_coeff(2)

a_1=A_coeff(3)

y=polyval(A_coeff,X);

plot(X,y,'*')

xlabel('angle')

ylabel('CL')

title('Plot as per polyval');

RESULT--------------------------------------------

Matrix b is defined as
b =

        6       60      880
       60      880    14400
      880    14400   250624


 Matrix A is
A =

      6.1000
     81.5600
   1260.9600

coefficient_vector =

   0.0317857
   0.1674018
  -0.0046987

a_o =  0.031786
a_1 =  0.16740
a_3 = -0.0046987

Answer to question b
X =

    0    4    8   12   16   20

Y =

   0.11000   0.55000   0.95000   1.40000   1.71000   1.38000

a_3 = -0.0046987
a_2 =  0.16740
a_1 =  0.031786