On DirectX coordination system, which is left hand coordinate, the center of the screen is ( x, y, z ) = ( 0, 0, [0~1] ); "[0-1]" means from zero to one. Top right corner of the screen is ( x, y , z ) = ( 1, 1, 0 ) and bottom left of the screen is ( x, y, z ) = ( -1, -1, 0 ).

When we show the 3D points or lines on the screen, we use perspective way rather than orthographical way. In other word, a far object appears to be smaller while a closer object appears to be bigger.

The way we do this is to divide x and y values by z value. When z value is bigger, x/z and y/z get smaller. In other words the points or lines get closer to the center of the screen when the z value is bigger, because the center is ( 0, 0, [0-1] ).

The matrix will be like this:

[ 1, 0, 0, 0 ] [ x ]

[ 0, 1, 0, 0 ] [ y ]

[ 0, 0, 1, 0 ] [ z ]

[ 0, 0, 1, 0 ] [ 1 ]

=

[ x ]

[ y ]

[ z ]

[ z ]

We always divide x, y, and z values by the last value, w, in order to make the w value to be 1.

[ x / z ]

[ y / z ]

[ z / z ]

[ 1 ]

Now we see there is a problem on Z value. Z will be always 1. In computer graphics, Z value 1 means deepest inside of the screen. In order words, the most far objects have z value one, while the closest objects have z value zero in DirectX.

In order to solve this problem we use this matrix instead:

[ 1, 0, 0, 0 ]

[ 0, 1, 0, 0 ]

[ 0, 0, 1, -1 ]

[ 0, 0, 1, 0 ]

The result is:

[ x / z ]

[ y / z ]

[ ( z - 1 ) / z ]

[ 1 ]

Note the z value is now ( z - 1 ) / z. It is equals to ( 1 - 1 / z ).Now far objects will have less than one but closer to one. Since the Z value is inversed the values are not evenly distributed. For far objects, the projected Z values would not vary much. For example, projected Z for Z value, 100, is 1 - 1/100 = 0.99 and projected Z of Z value, 200, is 0.995. The Z values are double times different but the projected Z value differs only 0.005. In other words, for far objects the projected Z value is less sensitive while for closer objects the projected Z value is more sensitive.

This is a preferable result on perspective view, because we can notice depth difference of closer objects more than far objects.

One problem I realized is that by this projection matrix the negative Z value may result in a very big positive value. For example, for Z value, -2, which means the object is behind of the camera, the projected Z value is 1 + 1 / 2 = 1.5, meaning that it is at infinitely far front side not on back side. To solve this problem, we need to make sure the Z value always ranges from zero or positive value.

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