Unit normal vector on an arbitrary curve

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MINAS KOUROUMPLAKIS
MINAS KOUROUMPLAKIS 2020-12-11
编辑： jessupj 2020-12-11
I need the unit normal vectors on a rounded triangle curve. I use an algorithm which was written by an idea in this forum and a Khan academy videoand is the following. I think though that it has flaws which at certain cases leads to false results e.g. boundary conditions. Does anyone know a very accurate method of finding these uniti normal vectors on a curve?
The code:
for j=1:1:N
f(j)=2*pi*(j-1)/N;
end
x=6.25*g*(cos(f)+a*cos(2*f));
y=6.25*g*(sin(f)-a*sin(2*f));
dy = gradient(y);
dx = gradient(x);
norm=sqrt(dx.^2+dy.^2);
nx=dy./norm;
ny=-dx./norm;
  1 个评论
jessupj
jessupj 2020-12-11
编辑：jessupj 2020-12-11
using 'norm' as a variable name is somewhat bad practice since you might want to use the norm() function at some point when dealilng with coordinate geometry.
isn't the unit normal you want here simply the normalized negative reciprocal of dy/dx computed from the function, like -dx./dy / sqrt(dx.^2./dy/dy) ?

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