How to generate a random vector3 with the same z distance relative to a map with rotation?

Question

How can I randomize the spawn location(x) of my gameobject while maintaining it's z distance from the map? Doing a transform.position.z won't do since the spawner and the map have a rotation.

 Vector3 spawnLoc = new Vector3(Random.Range(minX, maxX), transform.position.y, ?????);

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Answer 1

Best Answer

Answer by n3wb13 · Jul 06, 2018 at 10:18 AM

Hi guys solved it myself this is what I did.

     public float minX = -7.45f;
     public float maxX = -1.45f;
     float prevX;
     float prevZ;
 
     void Start () {
         prevX = transform.position.x;
         prevZ = transform.position.z;        
     }
 
  void Generate() {
 
         float newX = Random.Range(minX, maxX);
         float newZ = ((prevX - newX) - prevZ) * -1;
 
         prevX = newX;
         prevZ = newZ;
 
         Vector3 spawnLoc = new Vector3(newX, transform.position.y, newZ);
     }

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Answer 2

Answer by madks13 · Jul 06, 2018 at 08:28 AM

Does that mean your map is moving and rotating? Why not use relative positioning? If your spawned object is set as a child of the map, you won't need to know the rotation or position of the map. Just the relative position of your object map wise.

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n3wb13 · Jul 06, 2018 at 08:39 AM 0

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@madks13 sorry but I don't get it. BTW the map is not moving or rotating.

Answer 3

Answer by Zodiarc · Jul 06, 2018 at 08:42 AM

Why don't you keep the map and spawner rotation at 0 and just rotate the camera? Same visual effect, easier maths.

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n3wb13 · Jul 06, 2018 at 08:50 AM 0

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I already consider that but as a last resort. Since I would need to change a handful of scripts and existing map design.

Zodiarc n3wb13 · Jul 06, 2018 at 08:55 AM 0

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I think what you need is using maths specific for Polar coordinate systems

How to generate a random vector3 with the same z distance relative to a map with rotation?

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