I have rigidbody issues. My character falling too slow when I moving to right or left. But if I don’t move anywhere It’s falling normally.
Here is the code
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
public class movement : MonoBehaviour
{
//I changed script name. Original form is karakterhareket.
public Rigidbody2D rb;
public float speed, jumpforce;
public bool IsJump;
void Start()
{
}
void Update()
{
if(Input.GetKey(KeyCode.D))
{
rb.velocity = new Vector2(speed, 0);
}
if(Input.GetKey(KeyCode.A))
{
rb.velocity = new Vector2(-speed, 0);
}
if(IsJump==false)
{
if(Input.GetKeyDown(KeyCode.W)||Input.GetKeyDown(KeyCode.Space))
{
rb.velocity = new Vector2(0, jumpforce);
IsJump = true;
}
}
}
public void OnCollisionEnter2D(Collision2D other)
{
if(other.gameObject.name==“Square”)
{
IsJump = false;
}
}
}
Make sure the rotations which make it fall over are locked on the rigidbody. A rigidbody is just a physics object
1 Like
Assume that gravity accelerates you at 1 meter per second per frame, and that you move at 1 meter per second, to make the numbers nice.
First, if we are not moving
- Frame 0: Velocity is (0,0)
- Frame 1: Velocity is (0,-1)
- Frame 2: Velocity is (0,-2)
This sounds reasonable. We go faster and faster as time goes on.
Now, what if we’re holding right?
- Frame 0: Velocity is (1,0)
- Frame 1: Velocity is (1,-1)
- Frame 2: Velocity is (1,-1)
But why?
rb.velocity = new Vector2(speed, 0);
If you’re moving, your velocity’s x and y components are both set every frame. You keep resetting your vertical speed!
I would suggest reading your current velocity, setting just the x component, and writing it back to the rigidbody. This way, you aren’t constantly setting your y-velocity to zero every frame.
1 Like
Reply for DevDunk Thank you so much. 
But I already locked the Z rotation. I am working on 2D project. This i
Reply for chemicalcrux: Thank you but I didn’t understand. Can you tell shorly?
You’re doing this:
Vector2 velocity = new Vector2(1, 2);
velocity = new Vector2(3, 0);
The original y value is gone.
You want to do something more like this:
Vector2 velocity = new Vector2(1, 2);
velocity.x = 3;
One important thing: you can’t just do this*:
rb.velocity.x = 3;
So, you’ll need to do something like this instead:
Vector2 vel = rb.velocity;
vel.x = 3;
rb.velocity = vel;
Does that make sense?
*this happens because Vector2 is a value, like a float or an int. So, when you say rb.velocity, you wind up with a copy of its value; it’s not a reference to the original that you can modify.
1 Like
To chemicalcrux: I saw your reply after I started to fix the problem. Thank you for your all replies. But I did something different. I made what I understood. Thanks anyway.
