Deforming sheet into conical structure

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Priyanshu 2024-7-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2134556-deforming-sheet-into-conical-structure

评论： William Rose 2024-11-6，14:51

采纳的回答： William Rose

I am trying to deform a flat sheet into a conical structure using MATLAB. (Pls see the image below)

L = length of sheet; R= bigger radius; r = smaller radius

Here's my approach:

First i create an upward ramp (as shown in the image above), with a displacement in z-direction given by:

u_1 = r/L * (pi-1) * x

2. Then curve the edges of rectangular sheet into the conical form, given by displacements:

3. Hence the net displacement of any point becomes u_1 + u_2.

I have written this as a matlab code:

for i=1:length(A)

A(i,2) = A(i,2) + (sqrt(2*A(i,3)*r - A(i,3)*A(i,3))-A(i,2))*(A(i,1)/L);

A(i,3) = A(i,3) + ((r/L)*(pi-1)*(A(i,1)) + (R-sqrt(R*R - A(i,2)*A(i,2))*(1-(A(i,1)/L))) + ((R-r)+ (2/pi)*A(i,2))*(A(i,1)/L));

A here is the matrix with initial coordinates of the flat sheet. Then using a for loop, i apply the u_1 + u_2 displacement, changing the y and z coordinates of the sheet.

However, after running the code, i get one side to be very nicely fitting the curvature of the cone, whereas on the other side, all the points coincide at one single point rather than forming a circular face. Please see the images below.

Original flat sheet

Curved conical sheet with the issue

Can someone please help me with this?

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Answer 1

William Rose 2024-7-5

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在 MATLAB Online 中打开

@Priyanshu,

You want to map points from the x-y plane (flat sheet) to half of a conical frustrum.

The initial points are points (x1,y1) in sheet 1: x1=[0,L], y1=[-0.5,+0.5]. The final points are on half of a conical frustrum, with axis from (x2,y2,z2)=(0,0,R1) to (L,0,R1), and radius R1 at x2=0, radius R2 at x2=L. I have changed your notation somewhat, because I want to define r(x2) as a function.

The mapping from x1 to x2 is simple:

x2=x1 eq.0

The mapping from y1 to (y2,z2) is more complicated. Define a radius function:

r(x2)=R1+x2*(R2-R1)/L eq.1

Since x2=x1, we can write

r(x1)=R1+x1*(R2-R1)/L eq.2

Define angle theta as the angle measured about the cone axis, in a plane perpendicular to the axis, i.e. theta is the angle in the y-z plane. Define theta so that theta=0 when z2<0 and y2=0; theta=-pi/2 when y2<0 and z2=R; theta=pi/2 when y2>0 and z2=R.

It follows from the definition of r(x2) and theta that (r,theta) and (y2,z2) are related as follows:

y2=r(x1)*sin(theta) eq.3

z2=R1-r(x1)*cos(theta) eq.4

Now we choose to map y1 (domain=-0.5 to +0.5) to theta (range -pi/2 to +pi/2).

theta=pi*y1 eq.5

Substitute eq.2 and eq.5 into eq.3:

y2=(R1+x1*(R2-R1)/L)*sin(pi*y1) eq.6

Substitute eq.2 and eq.5 into eq.4:

z2=R1-(R1+x1*(R2-R1)/L)*cos(pi*y1) eq.7

Equations 0, 6, and 7 completely define the mapping from the sheet to the conical frustrum.

In Matlab:

L=1; R1=0.5; R2=0.2;

x1=L*(0:1/6:1); y1=-0.5:0.1:0.5;

[X1,Y1]=meshgrid(x1,y1);

X2=X1;

Y2=(R1+X1*(R2-R1)/L).*sin(pi*Y1);

Z2=R1-(R1+X1*(R2-R1)/L).*cos(pi*Y1);

% Plot results

figure;

colr=['k','r','g','b','c','m','y'];

subplot(211)

for i=1:7

plot3(X1(:,i),Y1(:,i),zeros(length(y1),1),'*','Color',colr(i)); hold on

end

axis equal; grid on; view(45,20); zlim([0 R1])

xlabel('X1'); ylabel('Y1'); zlabel('Z1')

subplot(212)

for i=1:7

plot3(X2(:,i),Y2(:,i),Z2(:,i),'*','Color',colr(i)); hold on

end

axis equal; grid on; view(45,20)

xlabel('X2'); ylabel('Y2'); zlabel('Z2')

It seems to work.

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William Rose 2024-7-8

在 MATLAB Online 中打开

@Priyanshu,

You want to map points from a flat sheet to part of a conical frustrum.

The initial points are points (x1,y1,z1) in sheet 1: x1=[0,L], y1=[-0.5,+0.5], z1=0. The final points are on half of a conical frustrum, with axis from (x2,y2,z2)=(0,0,R1) to (L,0,R1), and radius R1 at x2=0, radius R2 at x2=L. I have changed your notation somewhat, because I want to define r(x2) as a function.

The mapping from x1 to x2 is simple:

x2=x1 eq.0

The mapping from (y1,z1) to (y2,z2) is more complicated. Define a radius function:

r(x2)=R1+x2*(R2-R1)/L eq.1

Since x2=x1, we can write

r(x1)=R1+x1*(R2-R1)/L eq.2

Define theta as the angle measured about the cone axis, in a plane perpendicular to the axis, i.e. theta is the angle in the y-z plane. Define theta so that theta=0 when z2<0 and y2=0; theta=-pi/2 when y2<0 and z2=R; theta=pi/2 when y2>0 and z2=R.

It follows from the definition of r(x2) and theta that (r,theta) and (y2,z2) are related as follows:

y2=r(x1)*sin(theta) eq.3

z2=R1-r(x1)*cos(theta) eq.4

We want the flat sheet to wrap half-way around the cone at x1=0, and we want the flat sheet to wrap all the way around the cone at x1=L. In other words, thetaMax=pi/2 at x1=0, and thetaMax=pi at x`=L. Therefore we have

thetaMax(x1)=pi/2+x1*pi/(2*L) eq.5

and we map (x1,y1) to theta by

theta(x1,y1)=thetaMax(x1)*y1*2 eq.6

Substitute eq.2 and eq.6 into eq.3:

y2=(R1+x1*(R2-R1)/L)*sin(thetaMax*y1*2) eq.7

Substitute eq.5 into eq.7:

y2=(R1+x1*(R2-R1)/L)*sin((pi+pi*x1/L)*y1) eq.8

Substitute eq.2 and eq.6 into eq.4:

z2=R1-(R1+x1*(R2-R1)/L)*cos(thetaMax*y1*2) eq.9

Substitute eq.5 into eq.9:

z2=R1-(R1+x1*(R2-R1)/L)*cos((pi+pi*x1/L)*y1) eq.10

Equations 0, 8, and 10 completely define the mapping from the sheet to the conical frustrum.

In Matlab:

L=1; R1=0.5; R2=0.2; % length, radii

nx=11; ny=13; % number of x-values and y-values

x1=L*linspace(0,L,nx); % x values

y1=0.5*linspace(-1,1,ny); % y values

[X1,Y1]=meshgrid(x1,y1);

Z1=zeros(size(X1));

X2=X1;

Y2=(R1+X1*(R2-R1)/L).*sin((pi+pi*X1/L).*Y1); % eq.8

Z2=R1-(R1+X1*(R2-R1)/L).*cos((pi+pi*X1/L).*Y1); % eq.10

% Plot results

figure;

colr=hsv2rgb([linspace(0,1,nx)',ones(nx,2)]);

subplot(211)

for i=1:nx

plot3(X1(:,i),Y1(:,i),Z1(:,i),'*','Color',colr(i,:)); hold on

end

axis equal; grid on; view(45,20); zlim([0 R1])

xlabel('X1'); ylabel('Y1'); zlabel('Z1')

subplot(212)

for i=1:nx

plot3(X2(:,i),Y2(:,i),Z2(:,i),'*','Color',colr(i,:)); hold on

end

axis equal; grid on; view(45,20)

xlabel('X2'); ylabel('Y2'); zlabel('Z2')

OK

William Rose 2024-7-10

在 MATLAB Online 中打开

@Priyanshu,

In the most recent code, I did initialize Z as an array with the same dimensions as arrays X and Y.

Yes, you can use mesh() (not meshgrid()). In that case you would not use the for loop. I used a for loop above so that I could color each row separately.

L=1; R1=0.5; R2=0.2; % length, radii

nx=11; ny=13; % number of x-values and y-values

x1=L*linspace(0,L,nx); % x values

y1=0.5*linspace(-1,1,ny); % y values

[X1,Y1]=meshgrid(x1,y1);

Z1=zeros(size(X1));

X2=X1;

Y2=(R1+X1*(R2-R1)/L).*sin((pi+pi*X1/L).*Y1); % eq.8

Z2=R1-(R1+X1*(R2-R1)/L).*cos((pi+pi*X1/L).*Y1); % eq.10

% Plot results

figure;

colr=hsv2rgb([linspace(0,1,nx)',ones(nx,2)]);

subplot(211)

mesh(X1,Y1,Z1);

% for i=1:nx

% plot3(X1(:,i),Y1(:,i),Z1(:,i),'*','Color',colr(i,:)); hold on

% end

axis equal; grid on; view(45,20); zlim([0 R1])

xlabel('X1'); ylabel('Y1'); zlabel('Z1')

subplot(212)

mesh(X2,Y2,Z2)

% for i=1:nx

% plot3(X2(:,i),Y2(:,i),Z2(:,i),'*','Color',colr(i,:)); hold on

% end

axis equal; grid on; view(45,20)

xlabel('X2'); ylabel('Y2'); zlabel('Z2')

William Rose 2024-11-2，1:24

编辑：William Rose 2024-11-2，2:16

@Priyanshu,

[Edit: Correct the last equation, at the end of this post.]

I did not expect that you would still be working on this.

In the most recent version of the sheet-wrapping algorithm, the initial sheet is specified by the set of points [X1, Y1, Z1], and the wrapped points are specified by the set [X2,Y2, Z2]. The wrapped points lie on the surface of a frustrum of a cone. The centerline of the cone is a line parallel to the x-axis, with Y2=0 and Z2=R1. This means the cross sections of the cone, perpendicular to its axis, are circles parallel to the Y2-Z2 plane. The circle radius

when X2=0, and the circle radius

when X2=L; i.e.

.

You requested that the wrapped surface extend half way around the cone at X2=0, and that the wrapped surface wrap all the way around the cone at X2=L. Therefore I made the total wrap angle,

, be a linear function of X2:

at X2=0, and

at X2=L; i.e.

.

With these choices, the max height (max Z2 value) of the wrapped surface, for each value of X2, is a constant plus the product of a linearly increasing function and a sinusoid:

.

Now you want

to be a linear function of x2. Is that right? Do you still want the points of the sheet to lie on the frustrum of a cone?

Priyanshu 2024-11-3，17:24

在 MATLAB Online 中打开

@William Rose, apologies for the poorly worded comment. Here's the fullscript.

clc; clearvars; close all;

% Define constants

length = 1;

radius1 = length / (2);

radius2 = 1 * length / (2 * pi);

% Define grid dimensions

numX = 10;

numY = 10;

% Generate grid points

xValues = linspace(0, length, numX) * length;

yValues = linspace(-1, 1, numY) * 0.5;

% Create mesh grid

[X, Y] = meshgrid(xValues, yValues);

Z = zeros(size(X)); % Initial Z values are zero

% Creating matrix of intial points

X_init = X'; X_init = reshape(X_init, [numel(X),1]);

Y_init = Y'; Y_init = reshape(Y_init, [numel(Y),1]);

Z_init = Z'; Z_init = reshape(Z_init, [numel(Z),1]);

Init_points = [X_init, Y_init, Z_init];

% Writing the intial points into a text file for further calculations

writematrix(Init_points, 'init_test.txt','Delimiter', 'tab');

% Adjust Y and Z for deformation

Y_deformed = (radius1 + (radius2 - radius1) .* X / length) .* sin((pi + pi * X / length) .* Y);

Z_deformed = radius1 + (radius1 + (radius2 - radius1) .* X / length) .* sin (((2 * asin((radius2 .* X / length) / (radius1 + (radius2 - radius1) .* X / length)) + pi) - pi) / 2);

Warning: Matrix is singular to working precision.

% Calulating displacement vectos and final displacement matrix

X_disp = X - X; X_disp = X_disp'; X_disp = reshape(X_disp, [numel(X),1]);

Y_disp = Y_deformed - Y; Y_disp = Y_disp'; Y_disp = reshape(Y_disp, [numel(Y),1]);

Z_disp = Z_deformed - Z; Z_disp = Z_disp'; Z_disp = reshape(Z_disp, [numel(Z),1]);

Disp_points = [X_disp, Y_disp, Z_disp];

% Writing the displaced points into a text file for further calculations

writematrix(Disp_points, 'def_test.txt','Delimiter', 'tab');

% Calculating net displacement magnitudes

Displacement = sqrt((X - X).^2 + (Y_deformed - Y).^2 + (Z_deformed - Z).^2);

% Initialize figure

figure;

% Plot initial shape

subplot(2, 2, 1)

mesh(X, Y, Z);

axis equal; grid on; view(45, 20); zlim([0 radius1])

title('Initial Shape')

xlabel('X'); ylabel('Y'); zlabel('Z');

% Plot deformed shape colored by displacement

subplot(2, 2, 2)

mesh(X, Y_deformed, Z_deformed, Displacement, 'FaceAlpha','0.5', 'FaceColor','interp');% Add displacement as color data

colorbar % Show color bar to indicate scale of displacement

colormap jet % Choose a colormap that spans a wide range of visible colors

axis equal; grid on; view(45, 20)

xlabel('X'); ylabel('Y'); zlabel('Z');

title('Deformed Shape Colored by Displacement Magnitude')

Priyanshu 2024-11-6，3:08

Thank you @William Rose!

William Rose 2024-11-6，14:51

@Priyanshu, you're welcome.

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Deforming sheet into conical structure

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Deforming sheet into conical structure

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@Priyanshu, you're welcome.

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