How can I find the angle (from the vertical) between two points?

Question

Jeremy Salerno 2022-7-20

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回答： Star Strider 2022-7-20

I have two points, a blue circle and red star, and I want to find the angle between the two if a vertical line was drawn at the blue circle (see plot).

My code to make the points is below:

x1=2;
x2=5;
y1=2;
y2=4;
plot(x1,y1,'bo',x2,y2,'r*','MarkerSize',14)
xlim([0, 8])
ylim([0, 6])

I tried to compute the angle with atand like so:

atand((x1-x2)-(y1-y2))

and the output is -45. I am unsure if this is the correct method, and unsure why the value is negative.

Thank you!

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Answer 1

Star Strider 2022-7-20

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The atan2d function is likely more accurate —

x1=2;

x2=5;

y1=2;

y2=4;

plot(x1,y1,'bo',x2,y2,'r*','MarkerSize',14)

xlim([0, 8])

ylim([0, 6])

axis('equal') % More Accurate Plot View

r2b = atan2d((y1-y2),(x1-x2))

r2b = -146.3099

b2r = atan2d((y2-y1),(x2-x1))

b2r = 33.6901

hold on

plot([x1 x2], [1 1]*y1,'--k')

plot([x1 x2],[y1 y2],'--g')

plot([1 1]*x1,[y1 y2],'--k')

hold off

text(3.5,2.5,sprintf('\\angle %.1f°',b2r))

text(2.5,3.5,sprintf('\\angle %.1f°',90-b2r))

.

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Answer 2

Mathieu NOE 2022-7-20

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hello jeremy

small typo in your code

the angle is given by :

atand((x1-x2)/(y1-y2))

and not

atand((x1-x2)-(y1-y2))

so the result here is : atand((x1-x2)/(y1-y2)) = 56.3099

It was quite obvious from the x and y unequal length of your vector that the result cannot be 45 deg.

Personnaly I also prefer to think in terms of vector so the end coodinates comes first - minus the start point coordinates

the result is the same but I feel better this way :

atand((x2-x1)/(y2-y1)) = 56.3099

yet another possibilty, when you have two vectors is to use the following code - same numerical result

x1=2;
x2=5;
y1=2;
y2=4;
plot(x1,y1,'bo',x2,y2,'r*','MarkerSize',14)
xlim([0, 8])
ylim([0, 6])
v1 = [0; 1]; % vertical line = reference 
v2 = [x2-x1;y2-y1]
a = angle_btw(v1, v2) % in radians
ad = a *180/pi % in degrees
%% function to compute angle between 2 vectors 
function a = angle_btw(v1, v2)
% The traditional approach to obtaining an angle between two vectors (i.e. arccos(dot(u, v) / (norm(u) * norm(v))),
% as presented in some of the other answers) suffers from numerical instability in several corner cases.
% The following code works for n-dimensions and in all corner cases (it doesn't check for zero length vectors, but that's easy to add).
% This code is adapted from a Julia implementation by Jeffrey Sarnoff (MIT license),
% in turn based on these notes by Prof. W. Kahan (page 15).
    % Returns true if the value of the sign of x is negative, otherwise false.
    signbit = @(x) x < 0;
    u1 = v1 / norm(v1);
    u2 = v2 / norm(v2);
    y = u1 - u2;
    x = u1 + u2;
    a0 = 2 * atan(norm(y) / norm(x));
    if not(signbit(a0) || signbit(pi - a0))
        a = a0;
    elseif signbit(a0)
        a = 0.0;
    else
        a = pi;
    end;
end