New answers tagged vector
Simply calculate UVs (0,0 point is left bottom, 1,1 point is right top) assuming you v got sprite texture dimensions sp in pixels just divide it by texture size: SpriteDimensions sp; //that you must know Vector2f UVList = new Vector2f; UVList = new Vector2f(sp.x / tex_width, sp.y / tex_height); // 0,0 UVList = new Vector2f(sp.x / tex_width, ...
Short answer: by smart compositing rotation and translation. In the image below you can see the process (radius r is distance of your planet from star). If you rotate the moon by rot_m degrees(updated in main loop), it will circle the origin point. If you first rotate and then translate by radius r it will circle in right distance but wont follow your ...
First, the behavior you're describing is that of an aerodynamic vehicle with a single vector of force. A spaceship, flying in space, would operate in an entirely different manner. If this is still the desired behavior: Second, to utilize SpriteKit's efficiency, it might be best to use actions on your nodes. This isn't a requirement, but they exist ...
to complete the answer: angle = atan2((C-A).y, (C-A).x) + PI/2;
Calculate a vector from B to A, normalize it (divide by the vector's length), then multiply by the circle size: vx = A.x - B.x vy = A.y - B.y length = sqrt(vx*vx + vy*vy) C.x = vx / length * size + A.x C.y = vy / length * size + A.y For the angle you can use the atan2 function, if your language has it.
You add a constant force by doing pretty much what you're doing already. The problem you're likely having is a debugging one, or your expecting more force to be added then you're actually adding. Keep in mind that if you want the force to affect the object, you'll probably want to apply a larger force when the touch has ended. Remember that how much the ...
All you need is the object to have a velocity. Then, every tick you add your force (gravity) to it's velocity, and add your velocity to it's position.
Multisampling, as the people over at Metanet call it. Basically, you do a racket/ball collision test at frequent enough time intervals between frames such that the ball can't be passed by the racket without a collision being detected. This gives a rough estimate as to when the collision occurred, after which you can start subdividing your time step to find ...
What you could do is give the racket a defined constant "range" that is +- half the distance the ball moves every update. This way the ball would never pass through the racket. This will of course result in minor inconsistencies with the ball's rebound trajectory, but you'd have to test this out to determine their actual effects. Yes the logical size of the ...
Top 50 recent answers are included