@peterdrake your left vanishing point is off page to the left?

@mrcompletely Yes.

@peterdrake what you have is correct, I think. Very few 2 point pieces have much content outside those external lines specifically bc it usually looks strange - it's kind of implying another vanishing point that doesn't exist. The front face of your third object looks strange, I agree, but it's lined up according to the rules of composition as I know them

@peterdrake what would happen to the image if you moved the off-page point up, such that it's between the middle and right boxes vertically? I can't picture if it would be more or less intuitive.

@educoder Aren't two-point vanishing points always on the horizon?

I'm not sure how to draw what you describe.

@peterdrake shooting from the hip— if you take all the guiding lines and vanishing points and reflect them about the vertical line through your central vanishing point, you get a third point on the horizon (off-page) and a new set of guiding lines. Objects to the right of the central vanishing would seem to need to be aligned with the guides that go towards the right hand off page point.

Does that make any sense?

@bfjvii Yes. I think that would be approaching 5-point perspective, but with the top and bottom VPs infinitely far away.

https://www.creativebloq.com/how-to/draw-5-point-perspective

@peterdrake

This is on the right track.

The problem is that your vertical lines (that is, representing vertical edges in 3D space) are all parallel on your 2D image plane.

This means that your image plane is itself vertical. But your viewpoint is above the horizon, and the horizon is above the midpoint of the page. This implies that your image plane is oblique relative to the direction of observation (red line on my diagram) instead of normal to it (green line). That's the difference between a two- and three-point perspective, and the cause of the distortion you see.

The solution is to also draw the vertical dimension with perspective. Because your direction of observation is slightly downward, the third vanishing point should be directly below the midpoint of the image, and probably at quite a distance (it would be infinitely far away only if you were looking perfectly horizontally, which is when your verticals should come out parallel).

I think that with some trig you could work out the proper distance as a function of the coordinates, relative to the centre of the page, of the two existing vanishing points.

@peterdrake hm, trying again with the image

@khird What if the horizon was at the (vertical) midpoint of the page?

@peterdrake I expect it'd look better - at that point everything ought to be lined up. Maybe try tracing that drawing onto another sheet, but offset vertically so it's like you describe? I'm trying to imagine it, and I think my brain is happier with the shape of the box when I picture it down near the bottom edge.

@peterdrake Here's an animation I just put together in GIMP from a decent quality photo, not a fisheye lens or anything. Note that each line in 3D space is parallel to the others of the same colour, but orthogonal to those of different colours. But in 2D space, all the lines of a particular colour converge to a single point - even the vertical ones.

@khird Yes, that looks like standard 3-point perspective.

My understanding is:

- Looking down? 3-point with the 3rd VP well below the horizon, as in your photo.

- Looking up? 3-point with the 3rd VP well above the horizon.

- Neither? 2-point (or possibly 1-point, if one set of lines is parallel to a line from the viewer straight toward the center of the page).

@peterdrake Yes, that's correct. It looks like uploading the image broke the animation, but imagine that it overlays another set of lines highlighting the dark square building along the right-hand edge, near the bottom (which is past the right-hand vanishing point, similar to the box in your drawing). The green lines corresponding to the 3D-vertical edges are nowhere near 2D-vertical, which is why the building doesn't look too distorted.

More broadly, imagine you're in the centre of an octahedron with its vertices at the cardinal directions (or whatever alignment corresponds to your drawing, if the boxes aren't oriented with their faces normal to NESW/UD). If the ray from your viewpoint through the centre of your image plane would intersect the octahedron at:

- a face, it's three-point perspective with the vanishing points at the three corners of the triangular face.
- an edge, it's two-point perspective with the vanishing points at the endpoints of the line segment defining the edge.
- a vertex, it's one-point perspective with the vanishing point at the vertex.

@peterdrake I tried working through the trig to figure out where to place the third point. I assumed you could measure the distances ℓ₁ and ℓ₂ from the image centre to the existing vanishing points, and the perpendicular distance 𝓱 from the image centre to the horizon passing through both points.

The method I came up with requires you to first calculate 𝓭², which is the square of the distance from the observer to the image plane:

𝓭² = √[(ℓ₁² - 𝓱²)(ℓ₂² - 𝓱²)] - 𝓱²

Then you simply divide by the distance to the horizon to calculate the distance from the image centre to the third vanishing point:

ℓ₃ = 𝓭²/𝓱

I don't know how I could test this rigourously, but it gives sensible results at a couple key cases:
- In the correct two-point perspective case, the horizon passes through the image centre, so 𝓱 = 0. The first formula gives the formula 𝓭 = √[ℓ₁ℓ₂] for altitude of a triangle, and the second one blows up as the third vanishing point goes to infinity.
- In the correct one-point perspective case, the line connecting the vanishing point to the centre of the image is perpendicular to the horizon, so ℓ₁ = 𝓱 and ℓ₂ = ∞; consequently, the distance is indeterminate. This corresponds to an extra degree of freedom, as the observer can move directly toward or away from the vanishing point without changing its position on the image plane.
- If ℓ₁ = ℓ₂ = 2𝓱, then ℓ₁ = ℓ₂ = ℓ₃. This corresponds to a pseudo-isometric orientation; all the vanishing points lie at 120° intervals on a circle surrounding the image centre.

@khird Here's a 3-point attempt. I always have trouble pointing rays at an imaginary vanishing point well beyond the page.

@peterdrake Nice! I extended the lines out further to see how you did. The blue lines, of course, are very good, the red ones have some variation, and the green ones are similar to the red but have worse outliers. If the extremes were improved a little bit I think it'd be quite good. The lower right corner is the only one I could tell looked wonky without the help of the extended lines.

@khird Wouldn't the lower right corner look even weirder if it was aimed properly?

Also, any advice on getting those lines aimed precisely at the vanishing point a meter away from the paper?

@peterdrake I think it'd look less weird - to me, the angle between the two rightmost "verticals" is too large for my brain to accept that they're parallel in 3D-space.

I don't know what an artist's advice would be, but from a geometry point of view:
- Fix a string at the vanishing point, and use it as a compass to make an arc near the centre of the page.
- Taking a compass, draw two overlapping circles, each with its centre on the arc.
- Lay your straightedge across the two points where these circles intersect, and draw a line along it.

The advantage to doing it this way is that once you've constructed the first arc, you can move the paper around as much as you'd like without worrying about keeping track of the vanishing point, because all the future work is relative to marks already on the page.

Of course you could come up with another way of constructing that first arc. If you have an angle θ cut out of something rigid or marked on something you can trace, then:
- place its vertex at the centre of the page with arms facing downward
- mark the points where its arms cross the edges of the page
- align the cardboard in any other orientation such that its arms still pass over the edge-marks
- mark the new position of its vertex.
The angle you'd need is related to the page width 𝓦 by θ = π - sin⁻¹[𝓦/(2ℓ₃)].

@khird I'll try a few variations in coming days.