0 个投票

Dear everybody,
I have a problem, I want to draw 3 line.
First line: In xy plan has to be sin(x) ;
Second line: In xz plan has to be cos(x) ;
Third line: In xyz plan has to be sin(x)+cos(x);
I have no idea for it. What is the solution?
I will appreciate your help.
Thank you.
Istvan

2 个评论
显示 无 隐藏 无

Sam Chak
Sam Chak 2025-2-12
Hi @Istvan, The first curve is , and the second curve is . But I cannot visualize the third curve on the so-called "x-y-z" plane. Can you sketch it?
Istvan
Istvan 2025-2-12
编辑：Istvan 2025-2-12
Hi @Sam Chak,
It would be an elliptically polar wave. in this picture the cos and sin function are replaced, but the point is the same.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Star Strider
Star Strider 2025-2-12
Ran in:

0 个投票

Ran in:
Peerhaps something like this —
t = linspace(0, 1).';
x = sin(2*pi*t);
y = cos(2*pi*t);
z = x+y;
figure
plot3(x, y, z)
grid on
xlabel('X')
ylabel('Y')
zlabel('Z')
axis('equal')
figure
stem3(x, y, z, '.')
hold on
patch(x, y, zeros(size(z)), 'g', FaceAlpha=0.5)
hold off
grid on
xlabel('X')
ylabel('Y')
zlabel('Z')
axis('equal')
figure
patch(x, y, z, 'r', FaceAlpha=0.5)
hold on
patch(x, y, zeros(size(z)), 'g', FaceAlpha=0.5)
hold off
grid on
axis('equal')
xlabel('X')
ylabel('Y')
zlabel('Z')
view(-27, 30)
.

6 个评论
显示 4更早的评论 隐藏 4更早的评论

Istvan
Istvan 2025-2-12
Hi @Star Strider
Dear Sam,
It would be an elliptically polar wave. in this picture the cos and sin function are replaced, but the point is the same.
Walter Roberson
Walter Roberson 2025-2-12
sin(x)+cos(x) does not get you an elliptic polar wave.
Something like sin(x)+cos(y) would be closer.
Sam Chak
Sam Chak 2025-2-12
编辑：Sam Chak 2025-2-12

But sin(x) + cos(y) would give a surface. Thus, I believe Star's parameterized equations is probably the best approach to generate the helical spiral trajectory (bolded black curve) that maps out the elliptical projection on the normal plane.

Star Strider
Star Strider 2025-2-13
Ran in:
Ran in:
I didn’t realise what you wanted.
Try this —
t = linspace(0, 4, 500).';
x = sin(2*pi*t);
y = cos(2*pi*t);
figure
patch(t, x, zeros(numel(t), 1), 'r', FaceAlpha=0.5, EdgeColor='r')
hold on
patch(t, zeros(numel(t), 1), [0; y(2:end-1); 0], 'b', FaceAlpha=0.5, EdgeColor='b')
plot3(t, x, y, 'g', LineWidth=2)
plot3(xlim, [0 0], [0 0], '-k', LineWidth=2)
plot3([0 0], ylim, [0 0], '-k', LineWidth=2)
plot3([0 0], [0 0], zlim, '-k', LineWidth=2)
hold off
grid on
xlabel('X')
ylabel('Y')
zlabel('Z')
text(0, 0, max(zlim), 'Y', Vert='bottom', Horiz='center')
text(0, min(ylim)-0.3, 0, 'Z', Horiz='left')
text(4.1, 0, 0, 'X', Horiz='left')
view(25,20)
Ax = gca;
Ax.Visible = 'off';
axis('equal')
●
Make appropriate changes to get the result you want.
.
Istvan
Istvan 2025-2-13
@Star Strider Perfect, that is exatly what i thought.
I am grateful to you.
Thank you.
Best regards.
Star Strider
Star Strider 2025-2-13
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

产品

版本

R2024b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by