# How do I find the ground plane of a floating object in 3 point perspective?

I am learning 3 point perspective shadows and am having some trouble with this particular case. See image below:

I already understand how to cast shadows as you can see I have done with the bottom cube shape. The problem I am having now is that I can't determine exactly where the ground would be for the points of the shape that are above the ground. As indicated by the green lines, I have tried to find this for the top corners of the bottom shape, but I'm not quite sure if I am precisely correct. If anyone could better explain how I would approach this case I would be very grateful.

Edit:

Here's the object without the lines used to construct it if that makes things easier.

• This is my first time using this site so I didn't know I had to accept an answer in the first place. My bad for that since you pretty much gave me the perfect answer to my last question. Mar 25 at 0:42
• How the shadow is constructed? It isn't right if you used the isometric drawing shadow method that you got in your previous case. The shadow is a geometric item which should be drawn to the perspective drawing with the same perspective as the actual material items are drawn. Mar 25 at 1:14
• That's the thing. I didn't use the isometric method to cast this shadow or at least I tried not to. i.stack.imgur.com/Y9rCZ.png in this image you can see the shadow vanishing point (magenta) and light source (red). I will admit though that I did make some assumptions when attempting this which is why it might even be wrong. Mar 25 at 1:24

A perspective drawing is not reversible. It doesn't define what's the actual 3D structure. You said the bottom item is a cube. It's only your word. The same drawing can present right infinite number of different 3D shapes.

You have drawn the shadow like the cube stands on the ground. To make the shadow you have placed the light source and decided which point on your drawing is on the ground just below the light source. The items above the bottom cube do not have specified places. They can hover at much closer to the observer than the cube or as well they can be much more distant. You cannot draw the shadows before you have decided where the items actually are. Imagine they all stand on the ground on invisibly thin vertical feet, one foot per a corner. Draw to the drawing the grounding points of those feet. Note that the selected vanishing points are valid also for those imaginable lines which connect the decided footprints on the ground.

To get everything right it's better to start the construction from top and side view engineering drawings where the objects, light source, observer, ground plane and the imaging plane are exactly defined.

As a job engineering style construction of a perspective drawing is not a pleasure. Artists prefer to use mathematically valid shortcuts like vanishing points. You find them from drawing textbooks. Eyeballing based on experience is another major method in art. But the construction from perpendicular engineering drawings by using sight lines (a.k.a projection lines or projectors) is the waterproof elementary way to get it right.

Have you heard about forced perspective in the film industry? It is when you have two objects but in different locations where they should... or could be. So you can have a giant wizard and some hobbits in the same room.

I mention it because the object could be anywhere from a giant spaceship in the background or a tiny structure next to the camera.

Let me offset just some pixels of your shapes but with a tiny change. Can you spot it?

So the answer is there is NO way to tell unless you define 1 reference point on the ground.

So you need to make 1 decision. Where is the vertical projection of 1 point on that shape?

You could think the red spot corresponds to the same XY coordinate of other known points or choose another as you want. There could be the case the two shapes have the same center. There is no way of knowing.

See if the original orthogonal planes give you this information.