How to calculate angle between two direction vectors that form a closed/open shape?

Question

I am trying to figure out the correct trig. eq./function to determine the following: The Angle-change (in DEGREES) between two DIRECTION VECTORS(already determined), that represent two line-segment. This is used in the context of SHAPE RECOGTNITION (hand drawn by user on screen).

SO basically,

a) if the user draws a (rough) shape, such as a circle, or oval, or rectangle etc - the lines that makes up that shape are broken down in to say.. 20 points(x-y pairs).

b) I have the DirectionVector for each of these LINE SEGMENTS.

c) So the BEGINNING of a LINE SEGMENT(x0,y0), will the END points of the previous line(so as to form a closed shape like a rectangle, let's say).

SO, my question is , given the context(i.e. determinign the type of a polygon), how does one find the angle-change between two DIRECTION VECTORS(available as two floating point values for x and y) ???

I have seen so many different trig. equations and I'm seeking clarity on this.

Thanks so much in advance folks!

Answer 1

If (x1,y1) is the first direction vector and (x2,y2) is the second one, it holds:

cos( alpha ) = (x1 * x2 + y1 * y2) / ( sqrt(x1*x1 + y1*y1) * sqrt(x2*x2 + y2*y2) )

sqrt means the square root.

Look up http://en.wikipedia.org/wiki/Dot_product

Especially the section "Geometric Representation".

Answer 2

You could try atan2:

float angle = atan2(previousY-currentY, previousX-currentY);

but also, as the previous answers mentioned, the

angle between two verctors = acos(first.dotProduct(second))

Answer 3

I guess you have the vector as three points (x_1, y_1), (x_2, y_2) and (x_3, y_3).

Then you can move the points so that (x_1, y_1) == (0, 0) by

(x_1, y_1) = (x_2, y_2) - (x_1, y_1) (x_2, y_2) = (x_3, y_3) - (x_1, y_1)

Now you have this situation:

Think of this triangle as two right-angled triangles. The first one has the angle alpha and a part of beta, the second right-angled triangle has the other part of beta.

Then you can apply:

You can calculate alpha like this: