object array positioning-LibGdx

Question

In my game,if I touch a particular object,coin objects will come out of them at random speeds and occupy random positions.

public void update(delta){
if(isTouched()&& getY()<Constants.WORLD_HEIGHT/2){
         setY(getY()+(randomSpeed * delta));
         setX(getX()-(randomSpeed/4 * delta));
         }
  }

Now I want to make this coins occupy positions in some patterns.Like if 3 coins come out,a triangle pattern or if 4 coins, rectangular pattern like that.

I tried to make it work,but coins are coming out and moved,but overlapping each other.Not able to create any patterns. patterns like:

This is what I tried

int a = Math.abs(rndNo.nextInt() % 3)+1;//no of coins
        int no =0;

        float coinxPos = player.getX()-coins[0].getWidth()/2;
        float coinyPos = player.getY();
        int minCoinGap=20;
        switch (a) {
        case 1:
            for (int i = 0; i < coins.length; i++) {
                if (!coins[i].isCoinVisible() && no < a) {
                    coins[i].setCoinVisible(true);

                    coinxPos = coinxPos+rndNo.nextInt()%70;
                    coinyPos = coinyPos+rndNo.nextInt()%70;
                    coins[i].setPosition(coinxPos, coinyPos);
                    no++;
                }
            }
            break;
        case 2:
            for (int i = 0; i < coins.length; i++) {
                if (!coins[i].isCoinVisible() && no < a) {
                    coins[i].setCoinVisible(true);
                    coinxPos = coinxPos+minCoinGap+rndNo.nextInt()%70;
                    coinyPos = coinyPos+rndNo.nextInt()%150;
                    coins[i].setPosition(coinxPos, coinyPos);
                    no++;
                }
            }
           break:
          ......
          ......
          default:
            break;

may be this is a simple logic to implement,but I wasted a lot of time on it and got confused of how to make it work. Any help would be appreciated.

DMGregory · Accepted Answer · 2022-11-07 13:38:46Z

You can use sin and cos to get the points each coin should be from the object. This example will create 3 coins around the exploding object.

    explodedItems = new Array<Vector2>(); // array to hold coin positions
    Vector2 explodingObject = new Vector2(10,10); // object location that has coins explode from it
    int amountOfItemsToExplode = 3; // amount of items to explode out
    int distance = 5;
    float angleIncrement = 2 * PI / amountOfItemsToExplode;
    // use sin+cos to get points coins explode from
    for(float i = 0; i < amountOfItemsToExplode; i++){
        float x = (float) Math.sin(i * angleIncrement) + explodingObject.x;
        float y = (float) Math.cos(i * angleIncrement) + explodingObject.y;
        explodedItems.add(new Vector2(x*distance,y*distance));
    }

Arian_ki · Accepted Answer · 2022-03-23 22:26:46Z

0

Now I want to make this coins occupy positions in some patterns.Like if 3 coins come out,a triangle pattern or if 4 coins, rectangular pattern like that.

For this, I'll use a circle:

Use a random number as the radius
Let's say you have x coins, the position of each one can be calculated like this:

● new Vector2(cos(i * (360 / NumberOfCoins) + Offset) * Radius, sin(i * (360 / NumberOfCoins) + Offset) * Radius);

whereas Offset means how much to rotate the whole thing (can be used for more randomization and is in degrees) and i counts from 0 to NumberOfCoins - 1

Final Note: I'm not familiar with libgdx so the code I've written may need to be changed but the logic behind it is the same

edited Mar 23, 2022 at 22:26

answered Mar 23, 2022 at 22:15

Arian_ki

6915 silver badges25 bronze badges

\$\begingroup\$ Wow, the question was asked in 2017?? didn't notice that😅 \$\endgroup\$

Arian_ki
– Arian_ki

2022-03-23 22:16:37 +00:00
Commented Mar 23, 2022 at 22:16

Add a comment |

Stack Exchange Network

object array positioning-LibGdx

2 Answers 2

You must log in to answer this question.

Hot Network Questions

object array positioning-LibGdx

2 Answers 2

You must log in to answer this question.

Related

Hot Network Questions