Found a way to pick a random point around the viewport boundaries

Unity Engine
,
1

Hi there,

Been searching quite some time for a proper solution on how to get a random position, around the orthographic boundaries.
Since i wasn’t able to find any suitable solution, i managed to create one myself.

I’m just putting this snippet out there, for anyone sitting in my position and just needs a solution for now.
Im sure this could be prettier in some way but it does the job for now :slight_smile:

Have fun.

/**
     * Get a random position outside our orthographic boundaries
    */
    private Vector3 GetRandomPosOffScreen() {

        // Booleans for up, down, left and right
        bool isXAxis = Random.Range(1f, 10f) < 5f; // True = X axis | False = Y Axis
        bool isLeftSide = Random.Range(1f, 10f) < 5f; // True = Left side | False = Right side
        bool isUpSide = Random.Range(1f, 10f) < 5f; // True = Up | False = Down

        // Getting the actual height and width of the game scene (viewport)
        float screenHeight = Camera.main.orthographicSize * 2.0f;
        float screenWidth = screenHeight * Camera.main.aspect;

        // Some extra space along the boundaries
        float offset = 2f;

        // Defining the edges of the boundaries with added offset
        float minusWidth = -screenWidth / 2f - offset;
        float plusWidth = screenWidth / 2f + offset;
        float minusHeight = -screenHeight / 2f - offset;
        float plusHeight = screenHeight / 2f + offset;

        if (isXAxis) {
            // Returning a random x axis coordinate
            return new(isLeftSide ? minusWidth : plusWidth, Random.Range(minusHeight, plusHeight), 0);
        } else {
            // Returning a random y axis coordinate
            return new(Random.Range(minusWidth, plusWidth), isUpSide ? minusHeight : plusHeight, 0);
        }
    }
1 Like
2

Prettier way:

Vector3 randomPoint = new(Random.Range(0f, 1f), Random.Range(0f, 1f));
randomPoint.z = 10f; // set this to whatever you want the distance of the point from the camera to be. Default for a 2D game would be 10.
Vector3 worldPoint = Camera.main.ViewportToWorldPoint(randomPoint);
3

You’re right, i just don’t see this going outside the screen borders which is what i was searching for.

4

I believe it would work if you provide values less than 0 and greater than 1.
If you’re asking only for points OUTSIDE the screen, you should be able to do something like this:

float x = Random.Range(-0.2f, 0.2f);
float y = Random.Range(-0.2f, 0.2f);
if (x >= 0) x += 1;
if (y >= 0) y += 1;
Vector3 randomPoint = new(x, y);

Then continue as above with ViewportToWorldPoint.

2 Likes
5

@PraetorBlue If i had just found your example sooner, i could’ve saved a few hours :open_mouth:
Turns out, this works just fine like my example does. Nice work :slight_smile:

/**
     * Get a random position outside our orthographic boundaries
    */
    private Vector3 GetRandomPosOffScreen() {

        float x = Random.Range(-0.2f, 0.2f);
        float y = Random.Range(-0.2f, 0.2f);
        x += Mathf.Sign(x);
        y += Mathf.Sign(y);
        Vector3 randomPoint = new(x, y);

        randomPoint.z = 10f; // set this to whatever you want the distance of the point from the camera to be. Default for a 2D game would be 10.
        Vector3 worldPoint = Camera.main.ViewportToWorldPoint(randomPoint);

        return worldPoint;
    }
6

See my updated example code, as I believe using Mathf.Sign is a bug (since you’ll get -1.2 instead of -0.2 for example)

2 Likes