Using trigonometry to find projection of a point at $(x, y, z)$

Question

The following slide is part of a lecture on computer graphics, where the author is explaining the pinhole camera model:

I don't understand what "trigonometry" the author is referring to here. It seems like the author just defines the coordinates as to define the equations $\dfrac{x_p}{z_p} = \dfrac{x}{z}$, $\dfrac{y_p}{z_p} = \dfrac{y}{z}$, and $z_p = - d$ – no trigonometry involved.

I was wondering if people would please take the time to clarify what "trigonometry" the author is referring to, and then explain the derivation of these equations.

Answer 1

There’s no “trigonometry,” as the term is commonly understood, directly involved here. However, the root meaning of the word is “measuring triangles,” so in that very loose sense once could say that trigonometry is used.

The equations are derived via a straightforward application of similarity: the two triangles in the right-hand diagram with hypotenuses defined by the ray are obviously similar.

Answer 2

Strictly speaking, the author does not define the projections, but (as he writes) has deduced them from more general principles of trigonometry. If you look at the image there are alot of triangles involved and if you know some of the lengths or angles you can compute others by means of trigonometry like pythagoras or sine (which is the ratio of the opposite line to the hypothenuse) etc.

There are many good webpages, where you can find the derivation of the projection coordinates. I recall this site to be one of them.

Answer 3

There is no trigonometry. There is a line passing through the origin and $(x,y,z)$ and the goal is to see where it intersects the plane $z=-d$.

The line can be parameterised as $t \mapsto (tx,ty,tz)$ and assuming $z \neq 0$ this line intersects the plane at $t_p$ where $t_p z = -d $, or $t_p = - {d \over z}$. Hence the intersection point is $x_p = - {d \over z} x$, $y_p = - {d \over z} y$, $z_p = -d$. The formulae ${x_p \over z_p} = {x \over z}$, ${y_p \over z_p} = {y \over z}$, $z_p = -d$ follow from this.