Why I'm getting a complex degree value while using "acos"?

           % To reproduce the snippet, you can use below code
            XYZc = [2489011.31135707, 7440368.1011554, 17.6551714555564];
            X0 = 2489018.662

X0 = 2.4890e+06

            Y0 = 7440333.989

Y0 = 7.4403e+06

            Z0= 10.091

Z0 = 10.0910

            dir_vec1 = XYZc-[X0, Y0, Z0];
            dir_mag1= norm(dir_vec1);
            dir_vec2 = [0,XYZc(2),0]-[X0, Y0, Z0];
            dir_mag2= norm(dir_vec2);
            alpha = acosd(dir_mag2/dir_mag1)

alpha = 0.0000e+00 + 6.7868e+02i

[Matt J]

I have edited your post for you to make your code output visible. It is always advisable to do this so we can see what output you are talking about.

Answer 1

[Matt J]

Because abs(dir_mag2/dir_mag1) is greater than 1.

 abs(dir_mag2/dir_mag1)

ans = 6.9710e+04

[Abb]

@Matt J thanks, Is there any solution for that?

[Matt J]

There is no problem to be solved. It is the correct result.

[Abb]

@Matt J I think it is incorrect, because I was looking for a reseanoble angle. Do you think normalization would work?

[Matt J]

What I mean is, it is not Matlab's fault. It is doing what you asked, correctly. Matlab has no way of knowing, just as I do not, what you were really trying to code.

[Abb]

I wanted to calculate the angle based on acosine. Please see this pic Reference

[Torsten]

In a right-angled triangle, the hypotenuse is always longer than the adjacant. So if you think dir_mag2 is the length of the adjacant and dir_mag1 is the length of the hypothenuse in the triangle you consider, you must have done something wrong.

[Matt J]

If you want the angle between dir_vec1 and dir_vec2, it would be,

acosd(  dot(  dir_vec1/dir_mag1 ,  dir_vec2/dir_mag2 )  )

Otherwise, you haven't said enough for us to guess what angle your input data is supposed to define.

[Abb]

@Matt J @Torsten thanks both of you, to calify, I want to compute this angle(screenshot), based on this equation, it gets me 78 degree, but visually it looks like a smaller angle that this value?!. Maybe something wrong, do you have any idea?

% i ran this code

vec1 = XYZc-[X0, Y0, Z0];

vec2 = [0,XYZc(2),0]-[X0, Y0, Z0];

acosd(dot(vec1, vec2) / (norm(vec1)*norm(vec2)))

[Torsten]

Looks like 78° :-)

XYZc = [2489011.31135707, 7440368.1011554, 17.6551714555564];
X0 = 2489018.662

X0 = 2.4890e+06

Y0 = 7440333.989

Y0 = 7.4403e+06

Z0= 10.091

Z0 = 10.0910

dir_vec1 = XYZc-[X0, Y0, Z0];
dir_mag1= norm(dir_vec1);
dir_vec2 = [0,XYZc(2),0]-[X0, Y0, Z0];
dir_mag2= norm(dir_vec2);
plot3([0 dir_vec1(1)]/dir_mag1,[0,dir_vec1(2)]/dir_mag1,[ 0,dir_vec1(3)]/dir_mag1)
hold on
plot3([0 dir_vec2(1)]/dir_mag2,[0,dir_vec2(2)]/dir_mag2,[ 0,dir_vec2(3)]/dir_mag2)

[Abb]

@Matt J@Torsten Thanks for both of you.

        % This was something I was looking for, this angle is correct
        X0 = 2489018.662;
        Y0 = 7440333.989;
        XYZc = [2489011.311, 7440368.10, 17.65];
        vec1 = [X0, Y0]- XYZc(1:2);
        vec2 = [ 0, vec1(2)];
        alpha= acosd(dot(vec2, vec1) / (norm(vec2)*norm(vec1)));

[Matt J]

You're welcome, but please Accept-click the answer to indicate that it resolved your question.

[Abb]

@Matt J I do not see "Accepted Answer" in this comment! can you please enable it?

[Image Analyst]

It's not in the comments, neither his nor yours. It should be above, at the very first answer post of @Matt J