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8:01 PM
Is anyone here?
I'm here
but I can't answer your question, most likely. :)
do you knoe linear algebra?
I do know some yes
Oh :(
Oh good :)
I was expecting c# or some weird web technology
8:02 PM
blog.wolfire.com/2010/07/… <----I was reading that
But then......
It got to the part about rotating the x and y axis
looks like a pretty decent explanation. What part are you stuck on?
It said " The default axes are (1,0) for the x-axis and (0,1) for the y-axis, so we get the position 3(1,0) + 2(0,1). But the axes don't have to be (1,0) and (0,1)."
The part I'm stuck on: "To get the rotated x and y axes we just use the trigonometric function above. For example, if we are rotating by 49 degrees, then we get the new x-axis by rotating (1,0) by 49 degrees and we get the y-axis by rotating (0,1) by 49 degrees. Our new x-axis is (0.66, 0.75), and our new y-axis is (-0.75, 0.66)."
(1,0) is the unit X point
How can an axis have 2 axis?
if you rotate by an angle, it rotates into (cos(a), sin(a))
because it's a different coordinate system
8:06 PM
okay. the left coordinate system we'll call (X,Y)
the axes are (x,y)
the right coordinate system, we can call (X', Y')
X' is the slanted axis arrow
however, all those numbers on the right
are given in XY-space
if you represented them in X'Y'-space, the axes would still be (0,1) and (1,0)
and the triangle would still be (0,2), (-1,-1), and (1,-1)
Jimmy has it
the pink grid denotes X'Y' space
so your entire goal is to figure out howto convert between two different coordinate systems, one of which is rotated
8:10 PM
Ok so my problem is how do I convert X'Y' to actual X Y
if you figure out that "if I rotate 49 degrees, the X' unit vector is at (0.66, 0.75) and the Y' unit vector is at (-0.75, 0.66)"
then (0,2) in X'Y' is just 0*(unitX') + 2*(unitY')
and so on
the X' unit vector will be at (cos(a),sin(a)) (if you don't understand this, you might want to review your trig book)
A trig book? I should buy one
knowledge of basic trig is essential for game programming.
usually actually having the book is unnecessary
I do know the basics
Like sin and cos
8:17 PM
good :)
okay, the basics is probably enough
But I'm just having problems with this one thing
like, (1,0) rotates to (cos,sin) is basically the definition of sin/cos
As I said, how can one axis have two axes?
think of it as a unit vector
not an axis
so you can represent the "X-axis" by the unit vector (1,0)
and the "Y-axis" by the vector (0,1)
8:20 PM
A unit vector has a length (hypotenuse) of 1, right?
so therefore, the X'-axis is simply a vector (0.66,0.75) in the same way
of course, these coordinates are in XY-space
in X'Y' space, the X'-axis is just (1,0) again
and it's the old X and Y axes that have funny values.
strictly speaking, (1, 0) is a unit vector that points in the X' direction. The X' axis itself is a line.
hmmm yes.
if you go back to the last illustration
take a red grid point and try to visualize where it falls on the blue grid
and vice versa, taking a blue grid point and visualize where it falls on the red grid
That's all you're doing when converting from one space to another
x' = point.x * cos(angle) - y * sin(angle);
y' = point.x * sin(angle) + y * cos(angle);
Is this how you convert?
Ignore the "point."
at least for this example
8:33 PM
Wait....that's wrong because in the picture, you have -0.75 and 0.75
usually we'd represent this in matrix notation
hold up
your notation is a bit backwards
what we're really saying is "(0,2) in X'Y' space" and we need the XY-space coordinates
be right back
Diggin' the art.
8:47 PM
Though I'm getting a little sea-sick from the rocking...
Tetrad: could you help me with this linear algebra?
You know what, I'll figure this out some other time thanks to everyone who helped me so far.
@AidanMueller What exactly are you trying to do?
9:15 PM
@Tetrad Very nice! My only (minor) complaint is that the Splodes are cute, and I don't like popping them. :(
gotta shut down for software installation. back later
2 hours later…
11:27 PM
Q: Why doesn't this midpoint line algorithm draw vertical lines (or diagonals that approach the vertical)?

The RationalistI implemented this algorithm I found online, and it draws horizontal lines and half of all diagonals just fine. But as the diagonal line passes the "halfway" mark to becoming vertical, it doesn't update y and only draws the line with a smaller slope. It does draw it all the way to x2, but draws...

Using Bresenham to draw lines in OpenGL. What? :)
"Let's do this in the most complicated way possible." xD
11:43 PM
There are legitimate reasons to do it that way
otherwise: ouch
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