1

I'm trying to understand linear perspective better.

Let's say I have a plane and a cube and a defined intersection line with the front face of the cube. The colored lines are 3 examples of the intersection with the front plane of the cube; red line (a), violet line (b), brown line (c).

How do I find the intersection with the top(1) and left(2) plane of the cube?

Also, when the cube is in 2-point perspective, how would I find the direction of the intersection lines (a, b or c)?

enter image description here

2
  • please see this answer on shadow generation, its a different case but the strategy is the same. Notice that i also have elevation of light marked. For a plane you need 3 elevations but the projection works the same way. I would draw this for you but i have no computer and drawing this on a phone is a bit awkward.
    – joojaa
    Jul 29 at 16:56
  • thank you for your shared resource. I will check that out.
    – james
    Jul 29 at 20:13
2

First of all. It will be very hard for you to find the right perspective if you are not using a ruler. Your convergent lines are all over the place, and that is an important aspect of using these types of grids in the first place.


To answer your question, just follow the intersections to wherever they take you. Imagine they are a pinball ball, bouncing whenever it intersects a viable line.

The first example:

The only way I can draw a line on the vertex of the red line with the cube is by following the line to the vanishing point of the plane(1), the Cyan lines until it collides to the background face of the cube, then, the only place to follow is upwards.

(1)I could "pinball" it upwards, but stick to the floor plane whenever you can.

enter image description here

Just keep "pinballing" all the intersections.

Using a 2 point perspective on the cube is exactly the same, just, instead of having a paralel X axis, follow the vanishing lines.


I must say. This plane only works because I am assuming it is a rectangular segment with internal 90° angles. This does not work on a trapezoidal plane. We would need to superimpose a rectangular plane and draw the other shape inside this plane.


I am preparing an aditional explenation. A long one...


Part 1

Allow me to drop the perspective and use ortographic projections. I'll add it later.

I have two random orange dots in space. They are A and B, but they are superimposed on the drawing (1)

enter image description here

The truth is that I have no real idea where they are. We need to define something. Let me define an X cordinate for A: Ax.

Now, the other two positions fall in place.

enter image description here

I can do the same with B. Let's assign a Z coordinate. Bz.

As you can see, I can define where in space the dot is only asigning one coordinate and projecting a cube, to find the other two.

enter image description here

Part 2

To intersect a plane, I need to define a line, not only a dot. To define a Line I need two dots.

enter image description here

I can choose another dot to define my line. In these both examples The planes are paralell to the X axis.

enter image description here

But in reality, a plane is defined by 3 dots or two lines. In this example I moved a line, where on the previous example was on the X axis. It was implied but it was there.

enter image description here

You can easily see which lines to use on an ortographic projection.

Part 3

A perspective projection is EXACTLY the same. The only change is that my vanishing point is closer to my drawing area.

enter image description here

At the end, it is only a matter of finding the coordinates I need, on the axis I can.

Pinball the lines and conect the dots.

4
  • Thank you very much for your time and detailed answer. Not every thing is answered but a lot. the first example makes sense. thank you (I just imagined the intersection wrong) the second example looks right but at the moment I cannot imagine why tracing with line c helps to find the intersection. It would be great to give me a more detailed so that I can visually understand whats going on and why this works.
    – james
    Jul 29 at 10:08
  • wow thank you very much for your very detailed answer. sometimes I feel a little bit stupid. of course a plane needs 3 points to be defined (in most cases a point and two vectors(en.wikipedia.org/wiki/Plane_(geometry)). if I understand you correct: my questions was underdefined, because I only defined 2 points (for a equation with 3 variables) so there are an infinite number of intersection possibilities on the left face of the cube. is that correct?
    – james
    Jul 29 at 20:12
  • To define a plane you need, either 3 nonlinear points in space, or one 3-Dimensional vector and 1 point. Imagine that you poke a piece of paper with a pencil. The pencil is the vector, and you only need to poke it in one place. Now, you can move the pencil... the vector all over the place.
    – Rafael
    Jul 29 at 23:40
  • Oh, and yes. Your lines need to be more defined. I will probably delete or correct the first part of the answer. I am realizing it is wrong.
    – Rafael
    Jul 29 at 23:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.