Effects

Draw shadow of a specifc polygon

羽化

Creates a new Image containing the same image only rotated

Explaination of the calculations

The trig involved in calculating the new width and height is fairly simple; the hard part was remembering that when PI/2 <= theta <= PI and 3PI/2 <= theta < 2PI the width and height are switched.

When you rotate a rectangle, r, the bounding box surrounding r contains for right-triangles of empty space. Each of the triangles hypotenuse's are a known length, either the width or the height of r. Because we know the length of the hypotenuse and we have a known angle of rotation, we can use the trig function identities to find the length of the other two sides.

sine = opposite/hypotenuse cosine = adjacent/hypotenuse

solving for the unknown we get

opposite = sine hypotenuse adjacent = cosine hypotenuse

Another interesting point about these triangles is that there are only two different triangles. The proof for which is easy to see, but its been too long since I've written a proof that I can't explain it well enough to want to publish it.

Just trust me when I say the triangles formed by the lengths width are always the same (for a given theta) and the same goes for the height of r.

Rather than associate the opposite/adjacent sides with the width and height of the original bitmap, I'll associate them based on their position.

adjacent/oppositeTop will refer to the triangles making up the upper right and lower left corners

adjacent/oppositeBottom will refer to the triangles making up the upper left and lower right corners

The names are based on the right side corners, because thats where I did my work on paper (the right side).

Now if you draw this out, you will see that the width of the bounding box is calculated by adding together adjacentTop and oppositeBottom while the height is calculate by adding together adjacentBottom and oppositeTop.

A new Bitmap that is just large enough to contain the rotated image without cutting any corners off.