How do we allocate equidistant points on a sinusoidal curve?

The analytic equation is given below.

y=-25*sin((2*pi/50)*x);

Graph of above equation is given below.

What I exactly want is to allocate discrete equidistant points on this sinusoid curve. The step distance is 0.004. Basically, I have to solve the below equation starting from initial (x,y)=(0,0) point and use next point to calculate further point one by one.

(x-x_p).^2+(y-y_p).^2=(0.004)^2

=> (x-x_p).^2+(-25*sin(2*pi/50)*x)+25*sin(2*pi/50)*x_p)).^2=(0.004)^2

Here (x_p,y_p) is the previous point. But above equation is a trancendental equation. I am unable to solve it.

Thanks in advance!

Answer 1

[Matt J]

You can use fzero,

x_p=0;

y=@(x) -25*sin((2*pi/50)*x);

fun=@(d) d.^4+( y(x_p+d^2)-y(x_p)).^2 - 0.004^2;

x=x_p+fzero(fun,0).^2

x = 0.0012

Check:

norm([x-x_p, y(x)-y(x_p)])

ans = 0.0040

[Satyajit Ghosh]

Thanks.

Answer 2

[Matt J]

You should also have a look at interparc()

https://www.mathworks.com/matlabcentral/fileexchange/34874-interparc?s_tid=srchtitle

[Satyajit Ghosh]

Thanks.