Randomly pick an angle theta
and multiply that by the magnitude of the distance d
you want. Something like:
double theta = rand.NextDouble() * 2.0 * Math.PI;
double x = d * Math.Cos(theta);
double y = d * Math.Sin(theta);
Frage
I'm currently developing a 2-player Ping-Pong game (in 2D - real simple) from scratch, and it's going good. However Theres a problem I just can't seem to solve - I'm not sure if this should be located here or on MathExchange - anyway here goes.
Initially the ball should be located in the center of the canvas. When pressing a button the ball should be fired off in a completely random direction - but always with the same velocity.
The Ball object has (simplified) 4 fields - The position in X and Y, and the velocity in X and Y. This makes it simple to bounce the ball off the walls when it hits, simple by inverting the velocities.
public void Move() { if (X - Radius < 0 || X + Radius > GameWidth) { XVelocity = -XVelocity; } if (Y - Radius < 0 || Y + Radius > GameHeight) { YVelocity = -YVelocity; } X+= XVelocity; Y+= YVelocity; }
I figured the velocity should be the same in each game, so I figures I would use Pythagoras - the square of the two velocities should always be the same.
SO for the question:
Is there a way to randomly select two numbers (doubles) such that the sum of their squares is always a specific number - more formally:
double x = RandomDouble(); double y = RandomDouble(); if (x^2 + y^2 = 16) {/* should always be true */ }
Any help appreciated :)
Lösung
Randomly pick an angle theta
and multiply that by the magnitude of the distance d
you want. Something like:
double theta = rand.NextDouble() * 2.0 * Math.PI;
double x = d * Math.Cos(theta);
double y = d * Math.Sin(theta);
Andere Tipps
If the constant is C
, pick a number x
between 0
and sqrt(C)
.
Solve for the other number y
using simple algebra.
why not try this:
double x = RandomDouble();
double y = square(16-x^2);
as your application allow double type.
does this solve your problem?
if not, please let me know