how to calculate angle from coordinate

how to calculate all angle for this image if i have coordinate

>> S1=[X1 Y1];
>> S2=[X2,Y2];
>> S3=[X3,Y3];
>> S4=[X4,Y4];
>> S5=[X5,Y5];
>> S6=[X6,Y6];
>> S7=[X7,Y7];
>> S8=[X8,Y8];

[Walter Roberson]

Is it fully connected? I see some hints it might be, but from the diagram we cannot tell if the bottom left is connected to the mid left and upper left separately or if the bottom left is connect to the mid left and the mid left connected to the upper left without there being a lower left to upper left.

[Walter Roberson]

See https://stackoverflow.com/questions/1211212/how-to-calculate-an-angle-from-three-points

Answer 1

[Roger Stafford]

If A, B, and C are three points in two-dimensional space defined by 1 x 2 coordinate row vectors, here’s the way I like to compute the angle ABC - that is, the angle between line segments AB and CB:

AB = A-B;
CB = C-B;
ang = atan2(abs(det([AB;CB])),dot(AB,CB)); % Angle in radians

(Corrected)

[Walter Roberson]

Roger, the det(AB;CB) appears to have a typo?

[Roger Stafford]

You are right, Walter. I mistakenly left out the square brackets around AB:CB. I've made the correction.

Answer 2

[Walter Roberson]

Vectorizing Roger's formula:

X = [X1, X2, X3, X4, X5, X6, X7, X8];
Y = [Y1, Y2, Y3, Y4, Y5, Y6, Y7, Y8];

[Aidx, Bidx, Cidx] = ndgrid(1:8, 1:8, 1:8);
Ax = X(Aidx);
Ay = Y(Aidx);
Bx = X(Bidx);
By = Y(Bidx);
Cx = X(Cidx);
Cy = Y(Cidx);

ABx = Ax - Bx;
ABy = Ay - By;
CBx = Cx - Bx;
CBy = Cy - By;
det_term = ABx .* CBy - ABy .* CBx;
dot_term = ABx .* CBx + ABy .* CBy;

ang = atan2(abs(det_term),dot_term); % Angle in radians

Now, ang will be 8 x 8 x 8, and ang(I,J,K) is the angle between [X(I),Y(I)] to [X(J),Y(J)] to [X(K), Y(K)] . These will not be useful for any duplicated coordinates, but I leave it to you to extract the 336 useful angles, which will probably depend upon what you intend to do with the information.

[noor faizah zohardin]

base on this formula total angle= no of station x(no of staion -1), so I must calculate 56 angle because I have 8 station(point)

[Walter Roberson]

No, that is not correct unless you are measuring the angles relative to a fixed axes, which is not what Roger's formula does.

Roger's formula asks: given line segment AB formed by the line between A and B, and line segment BC formed by the line between B and C, what is the angle AB to BC ? So like the angle for the line between stations (5,2) and (2,6), the 5-2-6 angle.

With 8 possible sources, 7 possible intermediate stations and 6 possible destinations (that is, avoiding lines that double back on themselves), that is 8 * 7 * 6 = 336 angles.

It is certainly possible to instead ask what the angle is relative to the y axis, but that is not what Roger's formula calculates.

To ask the angle relative to the y axis, you would use

atan2(ydest - ysource, xdest - xsource)

[noor faizah zohardin]

now i know how to extract the right angle from Vectorizing Roger's formula.TQ very much