how to plot a plane using equations of planes?

Question

Sierra 2022-6-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1744970-how-to-plot-a-plane-using-equations-of-planes

评论： Sai Charan Sampara 2022-6-23

采纳的回答： Karim

在 MATLAB Online 中打开

I want to plot a plane using normal vecotr which is perpendicular to a plane.

my data is lon,lat,alt data(3D).

I used this code to plot a plane to which normal vector is perpendicular.

A = mean_trajectory(1).Longitude - mean_trajectory(2).Longitude;
B = mean_trajectory(1).Latitude - mean_trajectory(2).Latitude;
C = mean_trajectory(1).Altitude - mean_trajectory(2).Altitude;
D = -(A*mean_trajectory(2).Longitude + B*mean_trajectory(2).Latitude + C*mean_trajectory(2).Altitude);
x1   = linspace(126.6042,126.5995,10); % I manually limited the x axis and y axis.
y1   = linspace(37.2995,37.3048,10);
[x y] = meshgrid(x1,y1);
z = -1./C * (A.*x + B.*y + D);
surf(x,y,z)

but the result seems to be something wrong especially the 'z' value(alt).

I think it should be slightly titled. but it's not.

I draw what i want to make for your understanding.

left is what i want to draw and right is what i plotted so far.

If you more clarficiation, just tell me

Thanks!

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Karim 2022-6-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1744970-how-to-plot-a-plane-using-equations-of-planes#answer_990730

编辑：Karim 2022-6-22

在 MATLAB Online 中打开

does this result in what you need?

Edit: update to plot a plane at each point and plot the normal using quiver

XYZ = [ 126.6041 37.2994 3.3584e+3;

126.5994 37.3047 3.2463e+3;

126.5946 37.3101 3.1241e+3;

126.5898 37.3155 2.9994e+3;

126.5850 37.3208 2.8738e+3;

126.5802 37.3262 2.7481e+3];

MyNormals = XYZ(2:end,:) - XYZ(1:end-1,:);

MyNormals = MyNormals ./ repmat(sqrt(sum(MyNormals.^2,2)),1,3);

% plot the normals

figure

hold on

scatter3(XYZ(:,1), XYZ(:,2), XYZ(:,3),'r','filled')

quiver3(XYZ(1:end-1,1),XYZ(1:end-1,2),XYZ(1:end-1,3),MyNormals(:,1),MyNormals(:,2),MyNormals(:,3),0.2,'MaxHeadSize',0.01)

% create local x and y

gridPoints = linspace(-1,1,10);

[xx,yy] = ndgrid(gridPoints,gridPoints);

for i = 1:(size(XYZ,1)-1)

% calculate corresponding z

xv = XYZ(i,1) + xx;

yv = XYZ(i,2) + yy;

zv = (dot(MyNormals(i,:),XYZ(i,:)) - MyNormals(i,1).*xv - MyNormals(i,2).*yv) ./ MyNormals(i,3);

% plot the plane

surf(xv,yv,zv)

end

hold off

view(3)

axis tight

grid on

xlabel('longitude')

ylabel('latitude')

zlabel('altitude')

view([-43 5])

5 个评论
显示 3更早的评论隐藏 3更早的评论

Sierra 2022-6-22

Thanks ! KASR

I finally plotted the plane made from mean trajectory using your code, and I plotted every trajectory(other aircraft).

but the plane is so big that I cannot see well.

do you know how to define the size of plane? or is it already in your code?

Thanks again.

Sierra 2022-6-22

在 MATLAB Online 中打开

and I don't understand this code. could you explan it to me?

Have a good day KASR

MyNormals = MyNormals ./ repmat(sqrt(sum(MyNormals.^2,2)),1,3);

请先登录，再进行评论。

Answer 2

Sai Charan Sampara 2022-6-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1744970-how-to-plot-a-plane-using-equations-of-planes#answer_990765

For a given set of points or vector , the plane that is passing through them is always unique. So if the points that are being plotted in the surface plot are correct then the plane that has been plotted is the one you are looking for. From what I can understand you have made a plane passing through the point mean_trajectory(2) and perpendicular to the vector mean_trajectory(1)-mean_trajectory(2).If you are expecting the plane to also pass through mean_trajectory(1), it will not happen. Hence the z value will not reach 3.3584e3. If you want the point to be in the middle of the plotted plane just change the values of x1,y1 such that the required point is inside that range.Try

ThemeCopy

x1 = linspace(120,130,10);

y1 = linspace(30,40,10);

If I misunderstood your question , clarify it a bit more regarding what exactly you are looking for.

3 个评论
显示 1更早的评论隐藏 1更早的评论

Sierra 2022-6-22

I added one more image for clarification.

Sai Charan Sampara 2022-6-23

After changing the limits on x1,y1 the surface seems to be perpendicular to the line joining the 2 trajectory points. I have attached a screen shot below.

请先登录，再进行评论。