Question by Major · Jun 25, 2014 at 07:50 PM · vector3 character gravity planet

Finding "UP" while using planetary gravity?

I am trying to create planets and I have the gravity figured out, but whenever my character walks beyond a certain point, it is no longer grounded and the character just falls. So how do I find the normal direction for a planet so that I can have my character walk on the planet? Thanks!

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Hanarchist · Jun 25, 2014 at 10:01 PM 0

I can't help you personally, since I too am having difficulty with gravity, but have you checked this video out? He seems to do what you're looking to do, and his tutorials are pretty reliable.

https://www.youtube.com/watch?v=gHeQ8Hr92P4

Major · Jun 25, 2014 at 11:11 PM 0

Yeah I have seen that tutorial, but the problem with it is that I want there to be the possibility of having say three players on one planet.

Also if you would like some help with gravity you can use my script. It's only a slightly modified version of a public script anyways.

AlucardJay · Jun 25, 2014 at 11:26 PM 0

wouldn't it just be :

 planetaryPull = ( player.position - planet.centerOfGravity.position ).normalized * -1.0 * gravity; // assuming gravity is a positive value

Major · Jun 25, 2014 at 11:52 PM 0

Besides this being off topic, here is the script that I am using:

 using UnityEngine;
 using System.Collections;
 
 public class Gravity : $$anonymous$$onoBehaviour {
 
     public float range;
     public Collider[] objects;
 
     void Start () 
     {
     
     }
 
     void Update () 
     {
         objects = Physics.OverlapSphere(transform.position, range);
         float force = rigidbody.mass / 100;
 
         foreach(Collider c in objects)
         {
             Rigidbody rb = c.attachedRigidbody;
 
             if(rb != null)
             {
                 if(c.gameObject.tag != "Planet")
                 {
                     Vector3 offset = transform.position - c.transform.position;
                     rb.AddForce(offset / offset.sqr$$anonymous$$agnitude * rigidbody.mass * force);
             
                 }
             }
         }
     }
 }