Vector Motion in Java

Question

So, I'm making a game where you are to outrun enemies. Instead of having the enemies move at only up, down, left, right, and at 45 degree angles, I want the enemy to take the shortest linear path towards the player. Here's my code:

public void moveEnemy() {
    if (player.pos.x > enemy.pos.x) {
    enemy.vel.x = 3;
    }
    if (player.pos.x < enemy.pos.x) {
    enemy.vel.x = -3;
    }
    if (player.pos.y > enemy.pos.y) {
    enemy.vel.y = 3;
    }
    if (player.pos.y < enemy.pos.y) {
    enemy.vel.y = -3;
    }
    if (player.pos.x == enemy.pos.x) {
    enemy.vel.x = 0;
    }
    if (player.pos.y == enemy.pos.y) {
    enemy.vel.y = 0;
    }
}

So, what this does is sets the velocity in cardinal directions. What could I do to make this more accurate?

it does not increase. it SETS. when you want to increase use the `+=` operator — Philipp Sander, Oct 15 '13 at 15:44
I want the enemy to move at a constant velocity, but want it to move directly to the player — Derpster13, Oct 15 '13 at 15:45
After you have your angle, you'll probably want `enemy.vel.x=3*sin(angle);` and `enemy.vel.y=3*cos(angle)`. Or maybe I mixed those up, I often do. — Cruncher, Oct 15 '13 at 15:46
Math.atan: http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#atan(double) — Cruncher, Oct 15 '13 at 15:52
With all of these, there is one key thing that you need. Since these are going to give you `double`s, you need to save the absolute position of your enemies, then when you draw them, round them to the nearest pixels. — Cruncher, Oct 15 '13 at 15:55

score 1 · Accepted Answer · answered Oct 15 '13 at 15:53

Assuming you have the position of the player and enemy, and you want the enemy to always have a velocity of 3, then pull out your trigonometry textbook and do the following:

float h = Math.sqrt(Math.pow(enemy.pos.y-player.pos.y,2) + Math.pow(enemy.pos.x-player.pos.x,2));
float a = player.pos.x - enemy.pos.x;
float o = player.pos.y - enemy.pos.y;
enemy.vel.x = 3f*(a/h);
enemy.vel.y = 3f*(o/h);

What is this code doing, you ask? It's forming a triangle between the enemy and the player. You want the enemy to travel at 3 units/sec in the direction of the hypotenuse, so what you need to do is break that down into components that are parallel to the X and Y axes.

http://www.mathwords.com/s/sohcahtoa.htm

The floats h, a and o represent the hypotenuse, adjacent, and opposite sides of the triangle.

a/h is the velocity component parallel to the X axis.

o/h is the velocity component parallel to the y axis.

I had to add Math.round in a few places, because I am using points for the sprites, but overall, this runs quite nicely. — Derpster13, Oct 15 '13 at 16:27

score 0 · Answer 2 · answered Oct 15 '13 at 15:50

double spd=3;
double vX=player.pos.x-enemy.pos.x;
double vY=player.pos.y-enemy.pos.y;
double distance=Math.sqrt(vX*vX+vY*vY);
enemy.vel.x=vX/distance*spd;
enemy.vel.y=vY/distance*spd;

Calculates a vector pointing towards the position of palyer with a length of spd

Vector Motion in Java

2 Answers2