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

Abb
2023 6 10
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更新时间：2023 6 11
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% 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

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Matt J
Matt J 2023-6-10
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.

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 采纳的回答

Matt J
Matt J 2023-6-10
编辑：Matt J 2023-6-10
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Because abs(dir_mag2/dir_mag1) is greater than 1.
abs(dir_mag2/dir_mag1)
ans = 6.9710e+04

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Abb
Abb 2023-6-10
编辑：Abb 2023-6-10
@Matt J thanks, Is there any solution for that?
Matt J
Matt J 2023-6-10
There is no problem to be solved. It is the correct result.
Abb
Abb 2023-6-10
编辑：Abb 2023-6-10
@Matt J I think it is incorrect, because I was looking for a reseanoble angle. Do you think normalization would work?
Matt J
Matt J 2023-6-10
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
Abb 2023-6-10
I wanted to calculate the angle based on acosine. Please see this pic Reference
Torsten
Torsten 2023-6-10
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
Matt J 2023-6-10
编辑：Matt J 2023-6-10
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
Abb 2023-6-10
@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
Torsten 2023-6-10
Ran in:
Ran in:
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
Abb 2023-6-10
编辑：Abb 2023-6-11
@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
Matt J 2023-6-10
You're welcome, but please Accept-click the answer to indicate that it resolved your question.
Abb
Abb 2023-6-11
编辑：Abb 2023-6-11
@Matt J I do not see "Accepted Answer" in this comment! can you please enable it?
Image Analyst
Image Analyst 2023-6-11
It's not in the comments, neither his nor yours. It should be above, at the very first answer post of @Matt J

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