Searching for math expert! angle of vectors in a loop using law of cosines

Hello,

I created a loop to get each angle in between two vektors as you can see in the picture. My aim is to get the angle on each sixth point as shwon in the diagramm.

Using the follwing scipt, my resulting angle is always 90°! I can not see the mistake! Probably you do!?

alphaans=int16.empty(1,length(x1),0);
for i = 1:(length(x1)-2)
      ax(i)=x1(1,i+1)-x1(1,i);
      ay(i)=y1(1,i+1)-y1(1,i);
      bx(i)=x1(1,i+2)-x1(1,i+1);      %second (b) vector
      by(i)=y1(1,i+2)-y1(1,i+1);
      cx(i)=x1(1,i)+ax(i)+bx(i);               %resulting, leading to third vector
      cy(i)=y1(1,i)+ay(i)+by(i);
      lcx(i)=x1(1,i+2)-x1(1,i); %third (c) vector 
      lcy(i)=x1(1,i+2)-x1(1,i);
      Ba(i)=sqrt((ax(1,i)^2+ay(1,i)^2)); % norm (a)
      Bb(i)=sqrt((bx(1,i)^2+by(1,i)^2)); % norm (b)
      Bc(i)=sqrt((lcx(1,i)^2+lcy(1,i)^2)); % norm (c)
      alpha(i)=acosd(((Bb(1,i))^2+(Ba(1,i))^2-(Bc(1,i))^2)/(2.*Ba(1,i).*Bb(1,i))); %law of cosines
      alphaans(1,i,1)=alpha(i); %put each angle in the array alphaans 
end

Thanks!

[Mostafa]

what are x1 and y1 ?

[J.P.]

x1 and y1 is a dataset where i put a linear regression on every sixth datapoint as you can see in the picture. the x and the y coordiantes

[Mostafa]

I'm sorry, I meant please put the data values of x1 and y1.

In any case, let's get started:

 %These two lines are redundant, delete them
 cx(i)=x1(1,i)+ax(i)+bx(i); %resulting, leading to third vector
 cy(i)=y1(1,i)+ay(i)+by(i);
 lcx(i)=x1(1,i+2)-x1(1,i); %third (c) vector 
 %This line is wrong
 lcy(i)=x1(1,i+2)-x1(1,i);
 %%%Corrected: lcy(i)=y1(i,i+2)-y1(1,i);

Other than that, you are not calculating for each sixth point, you are calculating for each two consecutive points (which will give you false results when the difference between them is very small).

Instead of i+1 and i+2, write i+6 and i+12 . You can modify the loop index increment also if you want with i = 1:6:(length(x1)-12)

So, finally:

 alphaans=int16.empty(1,length(x1),0);
 for i = 1:(length(x1)-12)
    ax(i)=x1(1,i+6)-x1(1,i); %vector (a)
    ay(i)=y1(1,i+6)-y1(1,i);
    bx(i)=x1(1,i+12)-x1(1,i+6); %vector (b)
    by(i)=y1(1,i+12)-y1(1,i+6);
    cx(i)=x1(1,i+12)-x1(1,i); %vector (c)
    cy(i)=y1(1,i+12)-y1(1,i);
    Ba(i)=sqrt((ax(1,i)^2+ay(1,i)^2)); % norm (a)
    Bb(i)=sqrt((bx(1,i)^2+by(1,i)^2)); % norm (b)
    Bc(i)=sqrt((cx(1,i)^2+cy(1,i)^2)); % norm (c)
    alpha(i)=acosd(((Bb(i))^2+(Ba(i))^2-(Bc(i))^2)/(2*Ba(i)*Bb(i))); %law of cosines
    alphaans(1,i,1)=alpha(i); %put each angle in the array alphaans
 end

[J.P.]

I am sorry, but I preprocessed the data, so that every sixth point is calculated already as you can see in the picture. But thank you so much. I should take a closer look, after copy-pasting the rows. Thanks a lot again!

[Mostafa]

I think you only need to replace each instance of i+6 with i+1 and i+12 with i+2

Cheers.

Answer 1

[Thorsten]

See also http://www.wikihow.com/Find-the-Angle-Between-Two-Vectors

% define sample values 
x = [1234.77 936.40  681.39  516.59  355.26  82.90];
y = [241.90  155.16  118.73  193.32  408.43  458.74];

% plot values
y = 500 - y; % subtract to make y axis pointing upwards, just for display
             % purposes
plot(x, y, 'ko-')
axis equal
grid on
box off

% get angle between successive line elements
vec = [diff(x)' diff(y)'];

for i = 1:size(vec, 1) - 1
  u = vec(i,:);
  v = vec(i+1,:);
  theta(i) = acos( (u * v')/(norm(u)*norm(v)) );
  % u * v' is the dot product between u and v
end

rad2deg(theta)