<p>Light that is blocked will mean that $L(p*, -w_i)$ under the integral is 0 plus how much light the blocking object itself reflects. </p>

<p>In other words the shadowing is embedded in the incoming light function.</p>

CC BY-SA 3.0
2016-05-08T12:53:05.680Z
2016-05-08T15:40:27.513Z
2016-05-08T15:40:27.513Z
ratchet freak
<p>This is how the rendering equation is written in the textbook</p>

<p>$$L(p,\omega) = L_e(p,\omega) + \int f(p,\omega_i,\omega) \, L(p*,-\omega_i)\cos \theta \, d\omega_i$$</p>

<p>which component of this equation handles the shadowing?</p>

CC BY-SA 3.0
2016-05-08T18:32:01.207Z
BTW, what is the "*" notation in this formula? I've never seen that in the rendering equation before.
Nathan Reed
CC BY-SA 3.0
2016-05-08T18:53:54.177Z
@NathanReed p* is just a point from another surface, L(p*,-wi) is just the radiance from point p in direction wi
user2976568
<rendering><shadow>
How does the rendering equation incorporate shadowing
user2976568
# How does the rendering equation incorporate shadowing

(Asked by: username_0 on 2016-05-08)

This is how the rendering equation is written in the textbook

$$L(p,\omega) = L\_e(p,\omega) + \int f(p,\omega\_i,\omega) \, L(p\*,-\omega\_i)\cos \theta \, d\omega\_i$$

which component of this equation handles the shadowing?

**Tags:** rendering, shadow

> **Comments:**
> **username_1**: BTW, what is the "*" notation in this formula? I've never seen that in the rendering equation before.  
> **username_0**: @username_1 p* is just a point from another surface, L(p*,-wi) is just the radiance from point p in direction wi  

---

## 1 answers

### Answer by username_2 on 2016-05-08. Score: 10 | Accepted answer

Light that is blocked will mean that $L(p\*, -w\_i)$ under the integral is 0 plus how much light the blocking object itself reflects.

In other words the shadowing is embedded in the incoming light function.


---

