304 reputation
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bio website ipeet.org
location Canada
age 25
visits member for 2 years, 2 months
seen May 20 '12 at 21:53

Feb
16
awarded  Yearling
Mar
24
answered How does a collision engine work?
Mar
22
revised AABB - AABB Collision, which face do I hit?
deleted 15 characters in body
Mar
22
answered AABB - AABB Collision, which face do I hit?
Mar
21
comment How do multipass shaders work in OpenGL?
As an aside, the behaviour referred to by @DavidC.Bishop is hardly specific to shaders and the GPU. Modern CPUs executing normal programs will speculatively execute at least one of the branches if the tested value is not yet available. If they turn out to be wrong, they roll back and redo. This ends up yielding a big speedup on average.
Mar
16
answered A* and Space partitioning
Mar
8
comment Server-side physic simulations with hundreds of players
As long as different clients are independent, you should have no problem scaling horizontally. Use something like EC2, bring capacity online as-needed.
Mar
8
comment Server-side physic simulations with hundreds of players
If the physics engine is deterministic (it should be) and the server simulates using the same start state and inputs as the client (which should be feasible), then the results should be within reasonable floating point error bounds (insignificant). Even if there are random effects, and it isn't feasible to pass over the RNG state, you can pass over the random numbers and use them for checking (and even check their distribution, in case fudging random rolls makes a big gameplay difference).
Mar
6
awarded  Editor
Mar
6
revised Rotate a particle system
added 66 characters in body
Mar
6
answered Rotate a particle system
Mar
4
answered What's the best head-bob formula?
Feb
28
answered Using Python what is the best way to perform heavy tasks in the background?
Feb
27
answered Recommended Polling Rate
Feb
20
awarded  Teacher
Feb
20
answered How to generate random points on the surface of a quadrilateral
Feb
16
comment Relating a point after an image is panned and zoomed
I haven't double checked, but it looks like Sx = zoomX, and dx = 0.5*mapWidth - panX*imageWidth*zoomX.
Feb
16
comment Relating a point after an image is panned and zoomed
A possibly clarifying thought that just occurred to me: the mappings from (displayLeft, displayTop) to (imageLeft, imageTop), from (displayRight, displayBottom) to (imageRight, imageBottom), and from (displayX, displayY) to (imageX, imageY) are all accomplished by an identical formula. Source lines 24-27 are all derived by substituting values into that formula. As far as I can tell, your formula is a re-arranged equivalent to mine.
Feb
16
comment Relating a point after an image is panned and zoomed
In the horizontal direction, you are applying the following transform: xt = (xo*Sx)+dx ; where xo is the initial x co-ordinate of the point (i.e pixel in image), xt is the transformed (i.e. pixel in display) x co-ordinate, Sx is the horizontal scaling (zoom), and dx is the horizontal translation (pan). Going from a display co-ordinate to image co-ordinate is a matter of solving for xo: xo = (xt - dx) / Sx. Swap 'y' for 'x' and you get the solution for vertical co-ordinates.
Feb
16
comment Relating a point after an image is panned and zoomed
That is not a disproof. Points are still being mapped to points. Even if it's trivial, the identity transform (which doesn't change any co-ordinate values) is still a co-ordinate transform. You clearly do not understand basic geometry. You are presenting a textbook linear algebra problem, and you are rejecting textbook answers. I find myself wondering: Since you seem to believe that you understand this better than anyone else trying to help you, why are you bothering to ask the question at all?