Question

i’m writing a program that click JPanel and then a circle will spawn then its x & y will add 10 pixel per frame. The following two lines are temporarily to determine the next step after the circle spawn.

    destinationX = -10 + (int)(Math.random() * ((10 - (-10)) + 1));
    destinationY = -10 + (int)(Math.random() * ((10 - (-10)) + 1));

The above code is not the best to determine the position of the second steps of the cirlce. I want apply a method which make possibilities much more even. The following way is i about to do…..

           x2,y2     
             /
            /
           /
          /
         /
      h /
       /
      / a
      ----------------------------
   x1,y1

x1,y2 are known, a is random 360 degrees, h is 10, how to cal x2, y2?

Answer 1

Are you familiar with trigonometry? You have drawn a right triangle with h as the hypotenuse, a as the angle in question, an imaginary leg opposite (opp) of the angle, and a leg adjacent (adj) to the angle. Here’s an updated diagram:

           x2,y2     
             /|
            / |
           /  |
          /   |
         /    | opp
      h /     |
       /      |
      / a     |
      ---------
   x1,y1  adj

Trigonometry defines a function
sine (sin) as opp / h and a similar function
cosine (cos) as adj / h.

In your case, you want to compute x2 = x1 + adj.
Since cos(a) = adj / h,
adj = h * cos(a); so by substitution we arrive at
x2 = x1 + h * cos(a).

We can do a similar derivation for y2 and get y2 = y1 + h * sin(a).

If you want to move a point by a fixed amount in a random direction, you need Math.sin and Math.cos:

double angle = Math.toRadians(Math.random() * 360);
double amount = 10;
int x2 = x1 + (int)(amount * Math.cos(angle));
int y2 = y1 + (int)(amount * Math.sin(angle));