Generate inner values of y for each x based on n

1 次查看（过去 30 天）
Saim Ehtesham
Saim Ehtesham 2022-11-25
评论： Saim 2022-11-25
Ran in:
I have the following code:
clear all
clc
n = 4; % number of divisions on the boundary
x1 = linspace(0,2,n);
y1_x1 = sqrt(1-( (x1.^2)/4 ) );
y2_x1 = -sqrt(1-( (x1.^2)/4 ) );
x2 = linspace(-2,0,n);
y1_x2 = sqrt(1-( (x2.^2)/4 ) );
y2_x2 = -sqrt(1-( (x2.^2)/4 ) );
x = [x1 x1 x2 x2];
y = [y1_x1 y2_x1 y1_x2 y2_x2];
scatter(x,y)
Now, is there a way to generate the inner values of y with respect to x?
For example, as we can see in the plot, an ellipse is formed, but the points are on the boundaries, I want points within the ellipse as well, such that it uses the values of x and y already calculated and connects the y values to x like at x = 0, y = 1.0000,0.9428,0.7454,0 and at next x on both positive and negative side (x = +-0.6667), y = 0.9428,0.7454,0 so on and so forth till x = +-2
But most importantly, it all stays connected, so that all I change is the value of n and the inner points are generated.
Thanks

请先登录，再回答此问题。

采纳的回答

KSSV
KSSV 2022-11-25
Ran in:
clc; clear all ;
clear all
clc
n = 4; % number of divisions on the boundary
x1 = linspace(0,2,n);
y1_x1 = sqrt(1-( (x1.^2)/4 ) );
y2_x1 = -sqrt(1-( (x1.^2)/4 ) );
x2 = linspace(-2,0,n);
y1_x2 = sqrt(1-( (x2.^2)/4 ) );
y2_x2 = -sqrt(1-( (x2.^2)/4 ) );
x = [x1 x1 x2 x2];
y = [y1_x1 y2_x1 y1_x2 y2_x2];
scatter(x,y)
% Arrange ellipe points properly
xCenter = mean(x);
yCenter = mean(y);
angles = atan2d(y - yCenter, x - xCenter);
[sortedAngles, sortOrder] = sort(angles);
x = x(sortOrder);
y = y(sortOrder);
[X,Y] = meshgrid(linspace(min(x),max(x)),linspace(min(y),max(y))) ;
Z = zeros(size(X)) ;
idx = inpolygon(X,Y,x,y) ;
Z(idx) = 1 ;
pcolor(X,Y,Z)
  4 个评论
显示 2更早的评论
Saim
Saim 2022-11-25
Thanks a lot for the clarification
Is it possible to use the values of X and Y that you generated in a delaunay connection?
Saim
Saim 2022-11-25
Also, is it possible to reduce or increase the points within the ellipse?

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Spatial Search 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by