Interpolate value between arc

Note that now I have this function to create my arc

a=[1 1]; %P1

b=[9 9]; %P2

r=10;     %radius

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);

syms X Y

ezplot((X-x(1))^2+(Y-y(1))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
    min(a(2),b(2)),max(a(2),b(2))])

axis equal

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))])

axis equal

After plotting the arc, I need to know the points lie between the arc with random interpolate points n

How to do this?

[Roger Stafford]

Question: What do you mean exactly by "the points lie between the arc". What points are you talking about and what does it mean "between" - between what?

[TS Low]

ok, i think u comment on my previous post too

the points lie between point A and B

In this case, which is (1,1) and (9,9)

Now i am going to give matlab interpolate number of 7 (example)

Then, the result i want should be

(1.788,2.566)

(2.566,2.888)

(...)

(...)

(...)

(...)

(8.777,8.888)

*[Just example, not true value]

[Walter Roberson]

Those ezplot need to be changed to fimplicit

Answer 1

[Walter Roberson]

The equation you are using is a circle centered at x(1), y(1) with radius r. You know the endpoints; you can convert them into polar coordinates relative to the center. Now use linspace(first_theta, second_theta, 10) as the angle and r as the radius and put that through pol2cart, add x(1), y(1) to get the cartesian coordinates of the points of interest.

[TS Low]

So i need to change ezplot to fimplicit and also change my point and radius equation?

[Walter Roberson]

You need to change ezplot to fimplicit. You do not need to change your point and radius equation: it just is not of benefit to you to use them directly to calculate the intermediate locations.

[TH, R] = cart2pol([endpoint1_x - x(1), endpoint2_x - x(1)], [endpoint1_y - y(1), endpoint2_y - y(1)]);
TH10 = linspace(TH(1), TH(2), 10);
[X10, Y10] = pol2cart(TH10, r);
X10 = X10 + x(1);
Y10 = Y10 + y(1);

[TS Low]

ok thx, i will try on the suggestion u give me

[Walter Roberson]

%first input
% a=[1 1];
%P1
% b=[9 9];
%P2
A(1) = input('Please type your x1: ');
B(1) = input('Please type your y1: ');
A(2) = input('Please type your x2: ');
B(2) = input('Please type your y2: ');
a = [A(1) B(1)];
b = [A(2) B(2)];
r = input('Please type your radius: ');
%next solution
syms x y
[xsol, ysol]=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
figure(1)
fimplicit((X-xsol(1))^2+(Y-ysol(1))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
    min(a(2),b(2)),max(a(2),b(2))])

axis equal
figure(2)
fimplicit((X-xsol(2))^2+(Y-ysol(2))^2==r^2,[min(a(1),b(1)),max(a(1),b(1)), ...
    min(a(2),b(2)),max(a(2),b(2))])
axis equal

[TH, R] = cart2pol([a(1) - xsol(1), b(1) - xsol(1)], [a(2) - ysol(1), b(2) - ysol(1)]);
TH10 = linspace(TH(1), TH(2), 10);
[X10, Y10] = pol2cart(TH10, r);
X10 = X10 + xsol(1);
Y10 = Y10 + ysol(1);
figure(1)
hold on
plot(X10, Y10, 'rx');
hold off

[TS Low]

I try to solve it Now the figure one no longer shown and the fimplicit can't be use which i try before

[Walter Roberson]

Which MATLAB release are you using? That code needs R2016b or later in order to use fimplicit()

[TS Low]

i m using 2016a? So now the problem is the version? Any other alternative way to solve this?

[Walter Roberson]

Ah... I just tried again and this time ezplot worked, at least in R2017b. You could try changing to ezplot()

[TS Low]

Answer 2

[Roger Stafford]

I contend the right way to do that task is to calculate the center of the circular arc you have defined, and then generate the plotted arc using a varying angle that swings from the first point to the second point. You can carry out the desired interpolations in terms of values of this arc angle using 'interp1'.

You might be interested in the following "Answers" contribution:

https://www.mathworks.com/matlabcentral/answers/367126

(which I think was asked by you, TS Low.)

[Roger Stafford]

You should not expect the contributors who answer questions to do all your work for you. It should be sufficient to indicate the solution to difficulties you are facing. You should fill in the rest of the details yourself. Otherwise, how are you going to learn Matlab programming?

Answer 3

[Image Analyst]

Try spline(). See my attached demo.

[TS Low]

yes sir

i no need spline

I given user start point, end point, and radius

thus i need to create a curve and knowing the points that pass thru

[Image Analyst]

Of course you could use spline, but actually I think Walter's linspace idea is much simpler.

[TS Low]

thx bro