<p>Let's ignore <span class="math-container">$K_d$</span>'s dependence on wavelength for a moment (which @lightxbulb mentioned in the comments). Consider a bright white surface, and a grey one. The difference between them is precisely that the latter reflects less light; this, has lower value of <span class="math-container">$K_d$</span>.</p>
<p>In general, one might loosely think of <span class="math-container">$K_d(\lambda)$</span> as the colour of the material (usually represented as a triple of values for <span class="math-container">$\lambda$</span> equal to reed, green, and blue lights wavelength).</p>

CC BY-SA 4.0
2022-05-23T06:56:59.870Z
My problem with the notion of "color of the material" is that it then get's multiplied by light's color and color multiplication makes little sense.
Essam
CC BY-SA 4.0
2022-05-23T09:26:34.453Z
@Essam Color multiplication is reflection. Take a material that reflects $k_{d,r}=80\%$ of incoming red light (in a trichromatic model). And let the arriving red light be $L_r$, then the reflected light (ignoring Lambert's cosine law) is $k_{d,r} L_r = 0.8L_r$. This simply means that $20\%$ was absorbed.
lightxbulb
CC BY-SA 4.0
2022-05-22T19:19:56.270Z
2022-05-22T19:19:56.270Z
lisyarus
<p>I'm a little bit confused with the transition from just setting solid colors to object to involving light and its different components. For instance, the following formula
<span class="math-container">$R = I K_d max(0, n.l)$</span> that computes the involvement of the diffuse component of light in the object's color. The part where we compute the dot product makes sense because indeed visually points in the object where the normal is orthogonal on the light vector should be dark. However, I can't get to understand what's the purpose of <span class="math-container">$K_d$</span>. Is there any example of an object in real life that would for instance have low <span class="math-container">$K_d$</span> (hardly react to the existence of a light source)? Why would changing <span class="math-container">$K_d$</span>'s value ever make sense?</p>
<p>Thank you.</p>

CC BY-SA 4.0
2022-05-21T22:46:39.350Z
You can think of $k_d$ as a descriptor for how much light the object reflects. Also $k_d$ is typically a 3 component vector in order to rrprrsent color (RGB).
lightxbulb
<lighting>
Understanding the diffuse component of light
Essam
# Understanding the diffuse component of light

(Asked by: username_0 on 2022-05-21)

I'm a little bit confused with the transition from just setting solid colors to object to involving light and its different components. For instance, the following formula
$R = I K\_d max(0, n.l)$ that computes the involvement of the diffuse component of light in the object's color. The part where we compute the dot product makes sense because indeed visually points in the object where the normal is orthogonal on the light vector should be dark. However, I can't get to understand what's the purpose of $K\_d$. Is there any example of an object in real life that would for instance have low $K\_d$ (hardly react to the existence of a light source)? Why would changing $K\_d$'s value ever make sense?

Thank you.

**Tags:** lighting

> **Comments:**
> **username_1**: You can think of $k_d$ as a descriptor for how much light the object reflects. Also $k_d$ is typically a 3 component vector in order to rrprrsent color (RGB).  

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## 1 answers

### Answer by username_2 on 2022-05-22. Score: 1 | Accepted answer

Let's ignore $K\_d$'s dependence on wavelength for a moment (which @username_1 mentioned in the comments). Consider a bright white surface, and a grey one. The difference between them is precisely that the latter reflects less light; this, has lower value of $K\_d$.

In general, one might loosely think of $K\_d(\lambda)$ as the colour of the material (usually represented as a triple of values for $\lambda$ equal to reed, green, and blue lights wavelength).

> **Comments:**
> **username_0**: My problem with the notion of "color of the material" is that it then get's multiplied by light's color and color multiplication makes little sense.  
> **username_1**: @username_0 Color multiplication is reflection. Take a material that reflects $k_{d,r}=80\%$ of incoming red light (in a trichromatic model). And let the arriving red light be $L_r$, then the reflected light (ignoring Lambert's cosine law) is $k_{d,r} L_r = 0.8L_r$. This simply means that $20\%$ was absorbed.  

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