741 reputation
1411
bio website facebook.com/shadowsinrain
location Siberia
age 27
visits member for 2 years, 2 months
seen Mar 20 at 11:47

I'm using Google translator, yarrrr!

C++, Python, Gamedev, PCG, GUI, AI.


Jan
27
awarded  Yearling
Jan
7
awarded  Populist
Dec
23
comment How can I draw a ray from the camera in world space?
I pretty sure you should only see 1-pixel dot on screen.
Dec
23
comment How to rotate an object so that it is aligned with a vector?
What exactly is wrong with result of LookAt? Try passing C's position as &eye and either A's or B's position as &look, then apply whole matrix to C. Alsho check if you are using correct 'up' axis (DirectX uses +Y).
Aug
19
comment Using Behavior Trees and Events together
May be helpful: valvesoftware.com/publications/2009/…
Aug
19
revised How do I find the closest points(thereby forming a polygon) enclosing a particular point?(see image)
Got answer wrong in first revisions
Aug
19
answered How do I find the closest points(thereby forming a polygon) enclosing a particular point?(see image)
Jul
30
comment How can I correctly use an unordered_multimap as entity and component storage?
@natebot13 make_unique is in C++14, but may be implemented trivially in C++11: stackoverflow.com/a/12580468/1125702
Jul
27
comment How can I correctly use an unordered_multimap as entity and component storage?
@natebot13 Because you are passing instance of derived class, but your multimap wants unique_ptr to Component. Try inserting std::make_unique<Movable>().
Jul
25
answered How can I correctly use an unordered_multimap as entity and component storage?
Jul
18
answered Really simple count down timer
Jul
7
comment How do I resolve a collision of a circle with two rectangle corners?
@PandaPajama Issue with order of evaluation does not matters here, because, given high enough physics framerate, it wont be noticeable. And on low physics framerate, you will have enough bumping and glitches to worry about. If you want scientific-grade accurate physics engine, consider using continious integration.
Jul
7
comment How do I resolve a collision of a circle with two rectangle corners?
@mortusnegati Adding will break physics when rectangles overlapping, by causing anomally high penetration compensation vector.
Jul
7
comment How do I resolve a collision of a circle with two rectangle corners?
@PandaPajama I think if we are modeling infinitely solid objects, double collision like this is impossible, because given edges are colinear and not intersecting, so it is not possible to touch both of them with circle. Thus it makes sense to solve only one collision, then check again.
Jul
7
comment How do I resolve a collision of a circle with two rectangle corners?
Now this is clear. As answer already suggested, you may try to handle first collision only, and then re-detect colliding pairs again. I would recommend picking collision with shortest penetration vector (closest object). And of course you need to limit number of restarts, otherwise stuck object will hang you collision routine.
Jul
7
comment How do I resolve a collision of a circle with two rectangle corners?
In your example image, why diagonal vector used to compensate penetration? It is directed towards non-colliding rectangle. This is rather confusing. Can you show example with 2 actual collisions? And perhaps with good zoom. And ideally also with mentioned artifact.
May
15
answered Smooth moving on tile-based map, how to fix simple collisions?
May
15
answered Can't get the logic behind my AABB collision
Mar
19
comment Using uniform grids for collision detection - Efficient way to keep track of what a cell contains
@ArthurWulfWhite Tree is grid? Makes no sense for me.
Mar
18
comment Using uniform grids for collision detection - Efficient way to keep track of what a cell contains
@ArthurWulfWhite Well, if you want to handle 1k of dynamic entities, I think you should be serious and use another BSP. It does not imposes awkward limits. With carefull design BSP-tree may be laid into contigious array. Also I believe it will use much less memory (1024 * 12 * 2 or less). Not sure about actual search speed, pure opinion.