Perfect cube two vanishing points

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  1. #1
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    Perfect cube two vanishing points

    Hi guys, I am needing some help to create a perfect cube using two vanishing points and I have got some issues on it, because I am not sure how to calculate the perfect symmetry of a cube in perspective.

    I have seen to many cases like the next picture to draw a straight line (red line) to calculate the other segment projection.
    Is this a rule of a cube in perspective?


    I have found this attached picture in google
    Name:  two-point-perspective-cubes.jpg
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Size:  370.9 KB


    and other question, is there some way to calculate the perfect blue long segment according the main black line with 4 cm?

    Name:  001.jpg
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    I am not sure what you mean by the "red line". The lines to VPs or the dashed red horizontals?

    The horizontal diagonals of that cube are purely incidental, entirely due to cube's position. If you rotate the cube, they will no longer be parallel to the horizon. (By the way, that first image is in no way representing a perfect cube.)

    But the lines to VPs in general can be used to construct lines parallel to any other line in the perspective plot. You just find a line's VP, and use it to add another line parallel to it, passing through any point. Cube or not, it is irrelevant; the rule is about sets of parallel lines. Since the cube has three sets of parallel sides, the rule can be used to plot a cube with three VPs (in two-point perspective, the third VP is degenerate, and the vertical parallels are plotted as really parallel.)

    The easiest way to make accurate measurements in perspective (including the measurement required to make all sides of the cube equal) is probably the "architect's method". You start with a top view of the scene and use the picture plane as the origin for the measurements. Look it up, it's been described several times on these forums.

    One thing that a lot of beginners make is placing the VPs way too close, and ending up with a very high distortion. Both your attached samples have the VPs placed too close.

    Actually, if you don't have "Perspective Made Easy" by Norling, it would be a good idea to get a copy. It has most of the stuff you'll need in practice, with good explanations.

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    Thanks. I have found the solution. It was so easy and I forgot to apply these old studies.

    Name:  Wide-angle-cube.jpg
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    Quote Originally Posted by gustavoahlen View Post
    Thanks. I have found the solution. It was so easy and I forgot to apply these old studies.
    Yep, that's the "architect's method."

    Just remember to move the VPs more apart to avoid distortion.

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    Yep...what you have there, in red, is not a perfect cube. Not sure where your construction is off...I question why your plan view width lines converge down to measuring point 1 on line AB? Which you seem to have labeled SP for Station Point? The Station Point is the location of the observer so...

    In general, for most "art", you don't need such precise, methodical construction. You need to understand the principles of course but the best way to get the hang of the cube is to estimate the receding sides. If you can't do that with some feeling of accuracy anyway...you're kinda screwed.

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    Quote Originally Posted by JeffX99 View Post
    Not sure where your construction is off
    It's not within the cone-of-vision, I believe.

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    Quote Originally Posted by JeffX99 View Post
    Yep...what you have there, in red, is not a perfect cube. Not sure where your construction is off...


    The construction is off because the corner of the base nearest to the viewer is nearly 90 degrees. The only time a square base has exactly a 90 degree angled corner is when you are looking down directly over it, that is when your line of sight is perpendicular to the ground plane, or when you are drawing a ground plan of a square. From an angle with the line of sight not perpendicular to the ground plane such as in the picture above, the corner base of the cube nearest to viewer should be greater than 90 degrees.

    You can test this by taking a piece of paper and imagining yourself as the sun as it appears in the sky on Earth, with the paper representing the Earth(in this geocentric case, you the sun will be moving). Start from one corner of the 90 degrees angle, and arch slowly over the paper and you will notice that, in your picture plane, the angles change. You will also notice that in the oblique view, the angle closest and farthest from you will never be less than or equal to 90 degrees, whereas the angle on the side will always be less than or equal 90 degrees. They will be equal to 90 degrees when your line of sight is perpendicular to the ground plane of the square or rectangle.

    A picture plane is different from the actual properties of an object. We all know a square has 90 degree angles, but if you were to take a photograph of a square's corner closest to you or farthest from you, from an angle, say of 45 degrees from the base of the surface the square is resting on, then you measure that photographed corner with a protractor, you will find that the angle is always greater than or equal to 90 degrees. In this case, the photograph is the picture plane.

    It might sound complicated, but its actually very intuitive when you understand it.

    Last edited by Vay; December 8th, 2012 at 04:41 AM.
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    Your red cube is not a cube. It’s too tall and is distorted at the bottom. As said, it is extending outside of the circle of view (the area of least distortion). The cube that would be right above it is too tall as well. The Station Point is too close to the Picture Plane, making the cube too narrow. The other thing making things difficult is you are representing a wide angle view (the Vps are either too close together or the cube is too large), compounded by being 2-pt perspective instead of 3-pt.

    Here’s a more cubical cube, based on your setup. The space is too wide-angle to properly represent any additional cubes, especially in 2-pt perspective.

    Name:  2p.jpg
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