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

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.开发者_如何转开发. 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!

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".

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))

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:

If I understand you correctly, you may just evaluate the dot product between two vectors and take the appropriate arccos to retrieve the angle between these vectors.