Draw an arc between two points --> (x1,y1,z1) and (x2,y2,z2)

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Huseyin Eldem
Huseyin Eldem 2013 年 2 月 23 日
コメント済み: Esther 2024 年 4 月 8 日
Hi. I have several points on the sphere. I want to combine these points with arc. I can draw these points with lines but I can't draw arcs Please help

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Youssef  Khmou
Youssef Khmou 2013 年 2 月 23 日
編集済み: Youssef Khmou 2013 年 2 月 23 日
hi, you can try drawing a circle but restrict it into certain region,
I started a code, try to adjust/enhance it :
% Draw an arc between two points
p1=[-2 -2 -3];
p2=[10 15 20];
% Circle x²+y²+z²=Constant
%=> z=sqrt(Constant-x²-y²-);
Center=((p2+p1))./2;
xc=Center(1);
yc=Center(2);
zc=Center(3);
% Radius
R=norm((p2-p1))/2;
%Min=min(min(p1,p2));
%Max=max(max(p1,p2));
%x=linspace(Min,Max,20);
%y=linspace(Min,Max,20);
x=linspace(p1(1),p2(1),200);
y=linspace(p1(2),p2(2),200);
z=zc+sqrt((R^2)-((x-xc).^2)-((y-yc).^2));
figure, plot3(x,y,z), grid on, hold on
plot3(p1(1),p1(2),p1(3),'*','MarkerSize',10),
plot3(p2(1),p2(2),p2(3),'*','MarkerSize',10),
hold on, plot3(xc,yc,zc,'*','MarkerSize',10)
  1 件のコメント
Huseyin Eldem
Huseyin Eldem 2013 年 2 月 24 日
thanks so much. I can draw arc with above code. But I couldn't draw for several points (example 200 points). Please help me.

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その他の回答 (5 件)

Huseyin Eldem
Huseyin Eldem 2013 年 2 月 24 日
thanks so much. I can draw arc with above code. But I couldn't draw for several points (example 200 points). Please help me.
  2 件のコメント
Youssef  Khmou
Youssef Khmou 2013 年 2 月 24 日
編集済み: Youssef Khmou 2013 年 2 月 24 日
hi Huseyin, then you didnt have to accept the answer because the problem is not solved yet , in the code above we used circle's equation, so to draw arc between 200 points, theirs coordinates must satisfy :
1<=i,j,k <=200 xi²+yi²+zi²=R²
IF the points do not satisfy the equation they dont BELONG to the arc :
ok lets try to draw an arc between 200 points : we generalize the above code :
clear, clc
p=round(10*randn(200,3));
% Circle x²+y²+z²=Constant
%=> z=sqrt(Constant-x²-y²-);
Center=((sum(p)))./2;
xc=Center(1);
yc=Center(2);
zc=Center(3);
% Radius
p2=max(p);
p1=min(p);
R=norm((p2-p1))/2;
%Min=min(min(p1,p2));
%Max=max(max(p1,p2));
%x=linspace(Min,Max,20);
%y=linspace(Min,Max,20);
x=linspace(p1(1),p2(1),200);
y=linspace(p1(2),p2(2),200);
z=zc+sqrt((R^2)-((x-xc).^2)-((y-yc).^2));
figure, plot3(x,y,z), grid on, hold on
plot3(p(:,1),p(:,2),p(:,3),'.')
As points do not satisfy the equation the line cant contain them all .
Esther
Esther 2024 年 4 月 8 日
Hello there can you help me with this one please y=x/(x2+1)

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Huseyin Eldem
Huseyin Eldem 2013 年 2 月 24 日
Hi Youssef. Sorry I shouldn't accept the answers. I dont know. Coordinates of my points satisfy this equation. My points on a unit circle that radius r=1; so
xi²+yi²+zi²= 1 my points satisfy this equation exactly...
  1 件のコメント
Youssef  Khmou
Youssef Khmou 2013 年 2 月 24 日
ok good, did you try the last code? it must work very well then .

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Huseyin Eldem
Huseyin Eldem 2013 年 2 月 25 日
didn't work :(
  1 件のコメント
Youssef  Khmou
Youssef Khmou 2013 年 2 月 25 日
ok can you upload the 200 points ? or copy and paste them here ?

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ChristianW
ChristianW 2013 年 2 月 25 日
編集済み: ChristianW 2013 年 2 月 25 日
function test_sphere
n = 200;
THETA = rand(1,n)*2*pi;
PHI = rand(1,n)*2*pi;
R = 1.1*ones(1,n);
[x,y,z] = sph2cart(THETA,PHI,R);
acc = 10; % accuracy: ~ lines per arc
V = track_arc(x,y,z,acc);
plot3(x,y,z,'.r'); axis([-1 1 -1 1 -1 1]); axis square; hold on
plot3(V(1,:),V(2,:),V(3,:),'g')
sphere(31)
colormap([1 1 1]*0.1)
function V = track_arc(x,y,z,acc)
v1 = [x(1:end-1);y(1:end-1);z(1:end-1)]; % Vector from center to 1st point
v2 = [x(2:end);y(2:end);z(2:end)]; % Vector from center to 2nd point
r = sqrt(sum([x(1); y(1); z(1)].^2,1));
v3a = cross(cross(v1,v2),v1); % v3 lies in plane of v1 & v2 and is orthog. to v1
v3 = r*v3a./repmat(sqrt(sum(v3a.^2,1)),3,1); % Make v3 of length r
% Let t range through the inner angle between v1 and v2
tmax = atan2(sqrt(sum(cross(v1,v2).^2,1)),dot(v1,v2));
V = zeros(3,sum(round(tmax*acc))); % preallocate
k = 0; % index in v
for i = 1:length(tmax)
steps = round(tmax(i)*acc)+1; %Edited +1
k = (1:steps) + k(end);
t = linspace(0,tmax(i),steps);
V(:,k) = v1(:,i)*cos(t)+v3(:,i)*sin(t);
end
For the shortest great circle path, I modified code from Roger Stafford: http://www.mathworks.com/matlabcentral/newsreader/view_thread/277881
  1 件のコメント
Huseyin Eldem
Huseyin Eldem 2013 年 3 月 3 日
Thanks Christian. It works.

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Huseyin Eldem
Huseyin Eldem 2013 年 2 月 25 日
For Example.
X Y Z R
0,355590319 -0,825650574 -0,438014446 1
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0,389588366 0,399438725 -0,8298612 1
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0,894045248 -0,244541832 -0,375343025 1
0,939506811 -0,210695624 -0,270063522 1
0,333053848 0,507811942 0,794482326 1
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-0,65966674 0,060786309 0,749096 1
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  1 件のコメント
Youssef  Khmou
Youssef Khmou 2013 年 2 月 25 日
hi, they numbers need more processing to convert them into Mat format, anyway try this with numbers :
X=reshape(x,20,10);
Y=reshape(y,20,10);
Z=reshape(z,20,10);
figure, surf(X,Y,Z), shading interp
If they satisfy the equation, them they must give a partial Sphere
...

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