Getting an array of points along an arc

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MJ
MJ 2021-7-27
编辑： DGM 2021-7-27
How can I replicate the following code, which plots an arc given 2 endpoints and the radius, to instead output an array of points? (Which I can then plot using plot(x,y))
i.e. I want an array of points that will give me an arc, given two endpoints a and b, and a radius r.
figure;
a = [124.0102 62.2260];
b = [33.3602 77.9434];
r=250;
syms x y
[x,y]=solve((x-a(1))^2+(y-a(2))^2==r^2,(x-b(1))^2+(y-b(2))^2==r^2,x,y);
%plot arc
syms X Y
fig = ezplot((X-x(2))^2+(Y-y(2))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
min(a(2),b(2)),max(a(2),b(2))]);
set(fig,'color','black','LineStyle', '--','LineWidth',2)

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采纳的回答

KSSV
KSSV 2021-7-27
Ran in:
clc; clear all ;
figure;
a = [124.0102 62.2260];
b = [33.3602 77.9434];
r=250;
syms x y
[x,y]=solve((x-a(1))^2+(y-a(2))^2==r^2,(x-b(1))^2+(y-b(2))^2==r^2,x,y);
x = double(x) ;
y = double(y) ;
xi = linspace(b(1),a(1)) ;
yi = linspace(a(2),b(2)) ;
[X1,Y1] = meshgrid(xi,yi) ;
R = (X1-x(2)).^2+(Y1-y(2)).^2 ;
idx = abs(R-r^2)<15 ;
Xa = X1(idx) ; Ya = Y1(idx) ;
[Xa,idx] = sort(Xa) ;
Ya = Ya(idx) ;
plot(Xa,Ya,'-*b')

更多回答（1 个）

DGM
DGM 2021-7-27
编辑：DGM 2021-7-27
Ran in:
This is basically the same as this answer, but maybe the explanation helps a bit:
https://www.mathworks.com/matlabcentral/answers/327895-how-do-i-plot-a-circular-arc-with-given-two-endpoints-and-radius#answer_272201
a = [124.0102 62.2260]; % endpoints [x y]
b = [33.3602 77.9434];
r = 250; % radius
npoints = 100; % number of points to draw
% Finding the coordinates of the centers of circles
p = [a(:) b(:)];
c = sym('c',[2,1],'real');
eqs = [1,1]*(p - repmat(c(:),1,2)).^2 - r^2;
[cx cy] = vpasolve(eqs,c);
C = double([cx cy]); % solutions for centers [cx(:) cy(:)]
% each set of points with distance < 2*r
% will yield two solutions for the center coordinates
% each solution describes a circle, for which there are two possible arcs
% the arc angles are conjugate and inequal; we usually want the shorter one
% draw both short arcs and their corresponding centers
for ka = 1:size(C,1)
% find the shortest arc for this solution
v = C(ka,:);
p1 = p - v(:);
alp = atand(p1(2,:)./p1(1,:));
alp = alp + 180*(p1(1,:) < 0 & p1(2,:) > 0) - 180*(p1(1,:) < 0 & p1(2,:) < 0);
%alp = sort(alp); % only matters if you care about parameter direction
% generate points
phi = linspace(alp(1),alp(2),npoints)';
x = r*cosd(phi) + v(1);
y = r*sind(phi) + v(2);
% plot the short arc and center for this solution
plot(x,y); hold on
plot(v(1),v(2),'ok');
end
grid on
axis equal

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