# How to understand Z-Fighting?

I searched for Z Buffer on Google and I discovered that's a 2 dimensional array used by the graphical card and whose aim is to enable the handling of two pixels' depth in order to display one of these pixels in front of the other.

To illustrate my words, let an image which has only 1 pixel.

# What is a Z-Buffer? What is its utility?

## Introduction

Let $P_1$ be the pixel we want to draw in front of $P_2$ (thus, $P_1$ will hide $P_2$ at drawing). So $P_1$ and $P_2$ have the same $X$ and $Y$ coordinates.

We decide that the $Z$ coordinate of $P_1$ (ie. its depth) is less than $P_2$'s one (it means $P_1$ is closer to the computer's user than $P_2$): $Z(P_1) = 1$, and $Z(P_2) = 2$.

Our Z-Buffer has, by definition, the same size as our image: 1 case. Each case of the Z-Buffer contains one pixel's depth and each case is associated with one image's pixel.

First, we initialize the (only one) case of our Z Buffer at the maximum value we can store in its cases (depending on the Z Buffer's type, on the OS, etc.: double/float/integer).

## $P_1$

Now, we want to first draw $P_1$. The Z Buffer is used: the depth of the $(X(P_1); Y(P_1))$ case's value is compared to the $P_1$'s depth. Of course, $Z(P_1)$ is lesser than $Z( (X(P_1) ; Y(P_1)) )$. The results of this comparison are: first, we replace the old case value with the new $Z$ value (which is: $Z(P_1)$ ); then, $P_1$ is drawn.

# $P_2$

Finally, we want to draw $P_2$. Again the Z Buffer is used. $(X(P_2); Y(P_2))$ case's value is compared to $Z(P_2)$ (note that: $X(P_2) = X(P_1)$ and $Y(P_2) = Y(P_1)$ - cf. "So $P_1$ and $P_2$ have the same $X$ and $Y$ coordinates."). The results of this comparison are: $(Z(P_2) = 2)$ is superior to $( Z( (X(P_2); Y(P_2)) ) = 1)$; first, we don't replace any case; then, $P_2$ is not drawn.

That's how a Z Buffer is used to display a pixel in front of another.

## Example of application

http://raphaello.univ-fcomte.fr/IG/ZBuffer/Images/ZBuffer-03.gif

# Z-Fighting

## What is Z-Fighting?

If two polygons are very closed AND (note the "AND") if the camera is moving, Z-Fighting phenomenon appears: https://icculus.org/neverball/mapping/zfighting.png

## How to reproduce Z-Fighting (ie. how does it work? why does it appear?)

I know Z Fighting is caused by the resolution in depth of the Z Buffer, due to some conversion/casts mathematical and precision errors. But I don't understand it...

In particular, I don't understand why the camera must move.

My question is: could you explain this please?

• What is with all that mathematics that nobody understands
– user11352
Oct 22 '19 at 21:19