Need someone to verify that my math is correct

I've been sitting with a pencil for the better part of the evening trying to recall how to implement a scalable viewport that can navigate a 2D area. It's been a while since I've first heard of it, but I think I've figured it out, I just need to verify.

We have a 2D world with a "classic" cartesian coordinate system, x-axis points to the right, y-axis points to the top.

In the world area we have a rectangular viewport defined by 2 points Pmin and Pmax, where : Pmin(xmin, ymin), Pmax(xmax, ymax). Those points define viewport's size, location and scale

In the world area we have a point P, where Pmin < P(x, y) < Pmax. (P is in the viewport rect)

To display the whole damn thing, we've got a canvas (for example) that has an "altered" coordinate system, x-axis points right, y-axis points down. The canvas's size is MaxX and MaxY. The canvas's size is fixed.

Now, in order to display point P'(x', y') in the canvas I need to calculate it's position like this :

x' = (x - xmin) * Sx, where Sx = MaxX / (xmax - xmin)

y' = MaxY - (y - ymin) * Sy, where Sy = MaxY / (ymax - ymin)

*please note that y' coord is inverted due to canvas's coordinate system

In other words : the above math should take care of displaying a point while taking scale and 开发者_Python百科vieport's position into account. Am I correct ? If not, please prove me wrong.

Yes, that is correct. All points in the viewport will appear on the canvas -- and only those points -- and everything will appear right-side-up with distances preserved.

You may find it useful to create a class for the viewport which manages the scale and ranges. It can have methods such as

Point2 vp = viewport.transformFromWorld(Point2 pw);

and the inverse:

Point2 pw = viewport.transformToWorld(Point2 vp);

This is useful if you pick a point in the viewpoint with viewport coordinates and wish to transform to world.