统一。 与法线的角度。 “打孔能力”的线性变化

Question

How to find the angle to the normal, under which the bullet hit the surface? And a bigger question, how to make a linear change of the "punching capacity" of a bullet depending on this angle? That is, so that at an angle of 45 degrees, the bullet lost half of the "punching ability"? Multiplying by standard trigonometric functions makes no sense, for they are all non-linear. Help please, because I do not understand trigonometry at all ...

The only thing, which I can find now - is the normal to the surface of the hit. enter image description here

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class Bullet : MonoBehaviour
{
public float BulletImpulse { get; private set; }
public Rigidbody BulletRigidbody { get; private set; }

void Start()
{
    BulletImpulse = 3f;
    BulletRigidbody = GetComponent<Rigidbody>();
    BulletRigidbody.AddForce(transform.up * BulletImpulse, ForceMode.Impulse);
    BulletRigidbody.AddTorque(-transform.up * BulletImpulse, ForceMode.Impulse);
}

void OnCollisionEnter(Collision collision)
{
    foreach (var item in collision.contacts)
    {
        Debug.DrawLine(item.point, item.point + item.normal, Color.green, 100, false);
        break;
    }
}}

Answer 1

To get the angle, you will need the direction of your bullet, which would be the normalized direction of the velocity at impact, and the direction of the normal at the point of impact. You can then use Vector3.angle() to get the angle between the two, and multiply that by the impact factor to get your result.

void OnCollisionEnter(Collision collision)
{
   Vector3 bulletDir = collision.getComponent<Rigidbody>().velocity.normalized;
   Vector3 collNormal = collision.contacts[0].normal;
   float angle = Vector3.angle(collNormal, bulletDir);

   float impactFactor = 1 - angle/90f; // impact on a scale from 0-1;
}

(You may need to use the inverse of the bulletDir)