Sorry for not responding quicker.

None of these methods are exclusive. The method that Quadart prefers is similar to receptors method, and includes a method for finding the vanishing points on the horizon, but lacks a way of linking the true height line to the width (its not a big deal to put it in, but its still missing.) receptors method includes a method to do that. Neither method includes a way to draw a square in perspective without drawing it in plan view or reproducing it off a square that was in plan view. My method solves this problem, and it allows for curvilinear placement of the 3rd vanishing point. Which defines the cube.

Quadart.

Yes we are reinventing the wheel. My pet project has been driving perspective using geometry (lines and circles). With a focus on drawing a single cube. I have been able to reach where I am today by looking at tutorials and examples, then testing them to see if they actually work. Quite often some of the major features do not hold up to testing. Its hard to follow a method if the math is shaky.

The website you linked to is a fantastic if very dense reference. It explains some things very well, but it can be hard to find an explanation of a term without trying to read the entire thing again. I have tried to explain what I understand, and fall into the same trap.

receptor

The geometry of your two methods is identical. You have both chosen to draw different lines, which is why they appear to be different. My method is also geometrically compatible with you method, even if I prefer not to use plan views most of the time because of difficulties translating from the plan view to the perspective system.

The 3rd vanishing point will always move. Here is the geometry that proves it: http://reflective-sentinal.deviantart.com/gallery/33441339#/d4n7ujj Additional example, look up at a tall building, the sides converge, yet when you look at the base of the building (horizon) the sides are parallel. There is always a smooth curve between these two states. Its not usually much of a problem in an image, but it sometimes is.

I define my centre of view by the front most corner of my square, thus that point defines the relationship to the 3rd vanishing point.

The proof for a square and a cube:

Its a graphic geometric proof. The only instrument you need to check the angles is a 45-45-90 triangle.

The 2 vanishing points on the horizon are the intersections with the circle. Draw a line from one of the VP to the circle then to the other VP. Measure the angle at the circle, and measure the angle to the opposing nadir/zenith (noon or 6 o’clock position position). Then try a different point on the circle and see if your answers change. *Edit* the diagonals of all the squares will point back to where the 45 degree angle crosses the horizon.*end edit*

After locating the 3rd vanishing point, you then need to find the square on the remaining 2 sides of the cube.

Regarding the angular measurements of vision. Its a pretty simple diagram, but I am pretty confused about it. As I understand it, the only time I will ever see a truly 90 degree angle in perspective is when I am perpendicular to it. Thus the 90 degree angle at the top of the circle must be above me, and 90 degrees from the horizon. But I know that the distance between the vanishing points on the horizon is also 90 degrees. Thus I am confused: radius=90 degrees and 2 x radius=90 degrees. The math seems to be broken but the geometry works very well. I don’t worry about it too much.

Re 2 dice tumbling. I think its a bit more complicated than that, although that would look close enough.