Making Smoother the Curve fit to Airfoil data using cscvn function

Question

atharva aalok 2022-8-28

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1788565-making-smoother-the-curve-fit-to-airfoil-data-using-cscvn-function

评论： atharva aalok 2022-8-30

采纳的回答： John D'Errico

GOE802_DiscretePoints.csv

在 MATLAB Online 中打开

Hello I am using the following code to fit a curve to the following data

% Get the coordinates of the airfoil - airfoiltools.com - http://airfoiltools.com/airfoil/details?airfoil=goe802-il

GOE802_DiscretePoints = readtable('GOE802_DiscretePoints.csv', 'NumHeaderLines', 1);

GOE802_DiscretePoints = GOE802_DiscretePoints{:, :};

% Get the x and y coordinates of the airfoil for the upper and lower surfaces

x_lower_original = GOE802_DiscretePoints(:, 1);

y_lower_original = GOE802_DiscretePoints(:, 2);

x_upper_original = GOE802_DiscretePoints(:, 3);

y_upper_original = GOE802_DiscretePoints(:, 4);

x_all = [x_lower_original; flip(x_upper_original)];

y_all = [y_lower_original; flip(y_upper_original)];

x = x_all';

y = y_all';

select = [1:17, 17, 19:34]; % Repeating trailing point to make a sharp corner there

x = x(select);

y = y(select);

curve = cscvn( [x; y] );

fnplt(curve)

axis equal

How do I make it smoother for example I would like to break this boundary into N points where N>>n, say N = 100,000.

This is especially needed at the front tip where it should be smooth but is currently sharp.

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atharva aalok 2022-8-28

编辑：atharva aalok 2022-8-29

在 MATLAB Online 中打开

If Mr. John D Errico you are seeing this:

I am also trying to use your Interparc file from file exchange:

% Get the coordinates of the airfoil - airfoiltools.com - http://airfoiltools.com/airfoil/details?airfoil=goe802-il

GOE802_DiscretePoints = readtable('GOE802_DiscretePoints.csv', 'NumHeaderLines', 1);

GOE802_DiscretePoints = GOE802_DiscretePoints{:, :};

% Get the x and y coordinates of the airfoil for the upper and lower surfaces

x_lower_original = GOE802_DiscretePoints(:, 1);

y_lower_original = GOE802_DiscretePoints(:, 2);

x_upper_original = GOE802_DiscretePoints(:, 3);

y_upper_original = GOE802_DiscretePoints(:, 4);

x_all = [x_lower_original; flip(x_upper_original)];

y_all = [y_lower_original; flip(y_upper_original)];

x = x_all';

y = y_all';

select = [1:17, 19:34]; % Repeating trailing point to make a sharp corner there

x = x(select);

y = y(select);

xyint = interparc(5000,x,y,'spline');

figure(10);

axis equal

grid on

hold on

plot(x,y,'bo',xyint(:,1),xyint(:,2),'r-');

How do I get a sharp corner at the end?

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Answer 1

John D'Errico 2022-8-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1788565-making-smoother-the-curve-fit-to-airfoil-data-using-cscvn-function#answer_1036105

在 MATLAB Online 中打开

GOE802_DiscretePoints.csv

I saw your question. You don't need to direct a question at me.

However, it appears you are asking how to get a sharp corner at the END? Where, at x==1? You explicitly put that point in the middle of your curve, then used a spline to interpolate it.

There might appear to be two corners in that airfoil. One at x==1, and another between x=0.0125 and 0.025. There clearly is not a corner at x==0.

But at x == 1, the curve is not even continuous, with two distinct y values at x==1.

GOE802_DiscretePoints = readtable('GOE802_DiscretePoints.csv', 'NumHeaderLines', 1);
GOE802_DiscretePoints = GOE802_DiscretePoints{:, :};
x_all = [flip(GOE802_DiscretePoints(:, 3));GOE802_DiscretePoints(:, 1)];
y_all = [flip(GOE802_DiscretePoints(:,4));GOE802_DiscretePoints(:, 2)];
inddrop = @(n,ind) setdiff(1:n,ind);
xy = interparc(1000,x_all(inddrop(34,18)),y_all(inddrop(34,18)),'spline');
plot(xy(:,1),xy(:,2))

So what sharp corner might you have thought should be there?

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atharva aalok 2022-8-29

编辑：atharva aalok 2022-8-29

在 MATLAB Online 中打开

Dear Mr. John,

Regarding "But at x == 1, the curve is not even continuous, with two distinct y values at x==1."

Indeed the data points are such that there are two distinct y values at x == 1. To get rid of this I do the following in my previous mentioned code. (Sorry I didn't mention this explicitly.)

% select = [1:17, 19:34];
% x = x(select);
% y = y(select);
% Commented to avoid error "x not defined"

I see the problem in my code as you have explained "You explicitly put that point in the middle of your curve, then used a spline to interpolate it."

I have now fixed this issue as follows (I have changed the data file as well):

GOE802_DiscretePoints = readtable('GOE802_DiscretePoints.csv', 'NumHeaderLines', 1);

GOE802_DiscretePoints = GOE802_DiscretePoints{:, :};

% Put coordinates in a single vector starting from trailing edge and ending there - going lower -> upper edge

x_all_original = GOE802_DiscretePoints(:, 1);

y_all_original = GOE802_DiscretePoints(:, 2);

% Increasing the fineness of the discretization

N_airfoil = 1000;

% Get the finer distribution and overwrite the original coordinates

% Using interparc function by John D Errico from File Exchange - https://in.mathworks.com/matlabcentral/fileexchange/34874-interparc

Airfoil_points_fine = interparc(N_airfoil, x_all_original, y_all_original, 'spline');

x_all_original = Airfoil_points_fine(:, 1);

y_all_original = Airfoil_points_fine(:, 2);

figure(10);

hold on; grid on; axis equal

plot(GOE802_DiscretePoints(:, 1), GOE802_DiscretePoints(:, 2), 'bo', x_all_original, y_all_original, 'r-');

Now my question is the following.

I need a fine discretization of this shape (arbitrarily fine) such that the x values of each segment of the discretization on the top and bottom are the same. (that is for each point on the top edge there should be a point right below it on the bottom edge and vice versa)

That is something like a function that approximates the shape and then I can give my x vector of equally spaced values and it will calculate the y values.

Can this be done using interparc (instead of the equally spaced points w.r.t arc length)

Also, if possible please answer my original question regarding cscvn function.

atharva aalok 2022-8-29

I think I could go with the following solution:

Use the interparc function to make a finer discretization.

Fit a smoothing spline function to it.

Calculate at equally spaced x vector values.

Can this be done directly using the cscvn function by any chance?

atharva aalok 2022-8-30

After thiking about my above comment, I think its a horrible idea because then near the Leading Edge (x == 0) there would be way lesser points discretizing the curve since the slop is higher at that point than at other locations on the airfoil. Whereas, it should be the other way there should be way more points at x == 0 because of the highly curved nature (Lower radius of curvature).

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