I'm having what feels like a very dumb problem when it comes to scaling.

I'm trying to draw the interior of an apartment while also drawing a city view in the background. I'm struggling to properly scale the objects in the interior with the city exterior.

I've provided a visual demonstration of my issue here https://i.sstatic.net/jBFnV.jpg

enter image description here

I am even struggling when I attempt to do the same thing with an image: https://i.sstatic.net/ETt3C.jpg

enter image description here

2 Answers 2


In the first image, you have one wrong assumption:

That the guy is standing in the same lines of perspective as the buildings you are drawing with magenta lines. If that were the case, he could not see their facades of them.

He is standing in a building that does not share the same perspective lines. He is (using the vanishing point as a reference) standing in a building in the middle of that street.

He is like 50cm to the left of the camera that is on the orange line.

enter image description here

The guy that would be standing on the same plane as the facades of the building would be more or less where the purple lines are:

enter image description here

(To find the exact position we need to make a tiny person, or a person´s head, on the buildings and then project perspective lines to cross the green ones)

Here is a rough model of a person, and the black lines are just a zoom of the person.

enter image description here

And here is a close-up of where the person is according to the windows of the building.

enter image description here

Then, the lines are more or less, at what a head would be regarding that person. The scale is too imprecise but is just the idea.

The purple lines are not at the height of the tiny person, because your person's head at the first plane (green lines) is not at the same height as the other building's floor. (You can see that on the green lines to the right) Nothing indicates that the floor should be at the same absolute height. They could be just two Kryptonian flying around.

I am using the green lines on the left to indicate that our second subject, (red circle) is on, just by chance, the same plane where our initial subject is.

In the second image, you also have some confusion. In general, you are assuming that in real life we all share the same orthogonal vanishing points. We don't, there is an INFINITE number of vanishing points. Organic shapes could not even have one.

One street could be a bit narrower than the other, one could be a diagonal. A building could be on a descent.

Let me span you with some perspective-related answers.

Finding the possible vanishing points in a landscape

How to draw an object that's tilted in two perspective system

At what point does 1 point perspective become 2 point perspective?

My final idea is that you do not need to match everything into an orthogonal, 90°, square-based grid.

Your initial person is just standing in a random window, and it is OK to be standing on a random window.

  • Thank you so much for your answer, i am still a bit confused, how were you able to determine that the person in the purple lines you added is standing on the same plane as the facades of the building?
    – LonelyExo
    Feb 20, 2023 at 14:57
  • I am complementing the answer. 2 mins...
    – Rafael
    Feb 20, 2023 at 15:43
  • Done. I added a couple of diagrams and complemented the description.
    – Rafael
    Feb 20, 2023 at 15:58

Let's imagine there's a camera behind the man in the room. Then it's quite easy to compare the height of the head with the heights of those city items which are further from the camera, but approximately in the same direction as the head. The corner you used as your measure stick is not such item, it's totally in different direction.

Let's assume

  • your drawing is geometrically valid
  • the camera lens stands at distance Dm behind the man
  • the height of the head of the man is Hm
  • the distance from the camera lens to a city item (in the same direction as the head ) is Di
  • the real height of the city item is Hi
  • in the image the apparent height of the head is equal with the apparent height of the city item

If all those things are true then the next proportionality is also true:

Hi/Hm = Di/Dm

That's an elementary geometry fact. It's got by applying the proportionality of uniform triangles to lines of sight in projection imaging. If we know 3 of the 4 numbers we can calculate the 4th number. That's basic school math, for example Hi = Hm(Di/Dm) or as well Hm = Hi(Dm/Di), Di = Dm(Hi/Hm) and Dm = Di(Hm/Hi).

One calculation example:

In your image one corner of the buildings just in front of the head seems to be half as high as the head. If Dm is (say) 5 ft, Di is 1200 ft and Hm is 1 ft the equation gives Hi = 1200/5 ft = 240 ft. That's the height of an item which is apparently as high as the 1 ft high head. Thus the height of the mentioned corner is 120 ft, which is not at all impossible.

The city item you used as your measuring stick is in different direction than just in front of the camera. But you can move it's height to the front of the camera with a line which points to the left vanishing point of your boxes.

If you do not know 3 of the numbers in equation Hi/Hm = Di/Dm the fourth stays unknown. That's because the same scene can as well have a giant man much further from the camera than in the same room or the city is a miniature, not much further from the camera than the man.

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