Sunday, October 3, 2010

Wow

It had been a while since I drank myself to that level. Thank God I had a trash can in my room.

Thank God.

8 comments:

Master of the Craw said...

I too may have brought my blood alcohol level to "flammable".
I have no idea how I managed to bike to Hellen's. Apparently I could barely walk in a straight line.

Well, hopefully I wasn't too much of a drunk asshole.

Napoleon Bonerpants said...

I seem to recall climbing the walls...successfully

Master of the Craw said...

Fucking Jager...

Dementor said...

I wasnt that drunk. Hey, I didnt even bring booze to that party !
AND I manage to have fun.
There's a real progress here. I wonder if Rosa had as much fun. Hum.

Master of the Craw said...

Who?

Also, I vaguely remember you asked me a math question. Hopefully I wasn't completely off base.

Dementor said...

Rosa, you know, that little italian chick that was robbing against everyone with a pair of testicules?

Unless the dot product of two vectors doesnt give out their normal vector, then you werent offbase. But I seem to remember that the result was rather : the dot product of a vector with its normal vector is 0. But I'm not sure about that. Anyways, I found out a neat little function already in the code that did exactly what I was asking you about. Unfortunately its in a DLL so I cant review the implementation.

Master of the Craw said...

cross product dude...
dot product is |v||u|cos theta, that wouldn't help us at all.
cross product of u v produces a vector perpendicular to both u and v (provided u and v are co-planar).

I know there's also a trick in 2D where if you have a vector with slope s then another vector with slope -1/s will be perpendicular to it ( s cannot be 0, obviously). There might be an extension to this in 3 Space, IDK.

the other trick, which is much more costly, is to set your vector at the origin, figure out the reverse rotation transformation for one axis to the vector and apply that rotation to the standard axis (x, y ,z)

Master of the Craw said...

and yes, since cos 90 = 0 then if u.v = 0 you know that either u or v are zero vectors or the angle is 90