Although it is not possible to tell by viewing this small reproduction of

A viewer's eye is located at the point E=(0,0,-d) in the (x,y,z) coordinate system located in 3-space (ie x=0, y=0, z=-d). Notice that just one eye is used. Out in the real world is an object, represented by a vase here. As light rays from points on the object (such as the point P(x,y,z) on the vase) travel in straight lines to the viewer's eye, they pierce the picture plane (the x-y plane where z=0), and we imagine them leaving behind appropriately colored dots, such as the point P'(x',y',0). The collection of all projection points P' comprise the perspective image (the perspective drawing) of the object. Take a look at the picture below and re-read this with this picture in mind.

x' = (d x) / (z+d)

y' = (d y) / (z+d)

where d is the distance from the viewer's eye at (0,0,-d) to the picture plane (z=0).

Hence, given a real-life 3-D object, the artist will draw x' and y' on their 2-D sheet.

I don't expect you to understand how these formulas were obtained, but I do expect you to know that these are formulas for the 2-D coordinates on the picture plane of the real-life object that will be drawn in correct perspective (see the picture above).

Suppose the viewer is 3 units from the picture plane. If P(2,4,5) is a point on an object we wish to paint, find the picture plane coordinates (x', y') of the perspective image of P.As a second example, we might want to make a perspective drawing of a real-life Christmas tree. We first put a dot at the image (x',y') of a point (x,y,z) where the coordinates of x' and y' are given by the perspective theorem as above. Then we continue to trace all possible such lines, accumulating all possible points P' associated with our original object. Once we have done this, we will end up with a perspective drawing of our Christmas tree.

SolutionWe have d=3, x=2, y=4, z=5. Thus

x'=(d x) / (z+d) = (3*2)/(5+3)=6/8=3/4 and

y'=(d y) / (z+d) =(3*4)/(5+3)=12/8=3/2.

You will see a chart that is partly filled in with real-life x, y and z coordinates of a house (in columns A, B and C, respectively). We will use the viewing distance of 15 (as in column D) to calculate x' and y', and create a perspective drawing of it in Excel. So, we want to mathematically project the three dimensional house onto the mathematically precise perspective image in the plane (where we can draw it).

So, we want to transform x, y and z to new coordinates x'=(d x)/(z+d) and y'=(d y) /(z+d).

=d2*a2/(c2+d2)

so type this formula (on the above line) into E2 and hit return. You should now see -1.875. Click on E2 again. At the bottom right corner of E2 scroll until you get a square with arrows which looks like . Then click, hold down, and fill down the Excel column by scrolling down and releasing in E18. The number you will see there is -2.7631579.

Your mouse should now be a thin cross when you take it to your picture.

We want to connect the dots to make the picture represented above.