Perspective - how to find a perpendicular angle on an inclined plane?

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  1. #1
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    Perspective - how to find a perpendicular angle on an inclined plane?

    I really don't know if this is something obvious that I'm missing or it's something that most people just eyeball? Basically, if I have an inclined plane in perspective, such as a roof of a house - how can I draw a box on top of it, that is perpendicular, i.e. at right angle to the inclined surface that it's laying on top of?

    I know that if I could establish a perfect square that's laying on top of this inclined plane, I could put an ellipse in this square and the minor axis of the ellipse would be the perpendicular angle. But do I need to find a square? Please help, y'all!!!

    I guess this perpendicular angle would be called a "surface normal" in 3D modelling, hehe...

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    Learn to construct your vanishing points, and you can reconstruct your normal...

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    Well, unless you can elaborate or demonstrate anything, that is a meaningless "answer".

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    Study a decent manual on perspective, like http://handprint.com/HP/WCL/tech10.html or Raynes' Complete Guide to Perspective. If you expect answers on a silver plate, perspective is probably not going to work for you...

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    It's a very specific question. Your answer however is not. It's like me asking which tuberosity is a specific muscle in the human body connected to and you saying go study anatomy.

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    Quote Originally Posted by z01ks View Post
    It's a very specific question. Your answer however is not. It's like me asking which tuberosity is a specific muscle in the human body connected to and you saying go study anatomy.
    No. Your question is simple, but the answer is not. I am really not going to lecture about perspective for the rest of the day, unless I am getting paid. The construction of vanishing points has already been elaborated upon by various manuals, which is why I am redirecting you. Whichever route you take, it going to take you hours or days to understand this at a level where you will be able to benefit from it.

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    You can hardly call that redirecting. Can you redirect me to a chapter instead of a whole book for example? Can you redirect me to a page or a paragraph? If explaining how to find a perpendicular angle of any given inclined plane takes a whole day, then I guess I will just eyeball it. But you are not saying that. You are not saying much anything.

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    OK, I've actually found a great course by Erik Olson that covers this on the www.newmastersacademy.org page - you can sign up for a 7-day free trial on that page, which I did, so you can watch any videos you want. The selection that they have is pretty awesome...

    Looks like it's a matter of plotting the angle on a measuring line, that's parallel to the picture plane.

    Last edited by z01ks; June 5th, 2014 at 11:33 AM.
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    So, if anyone else is interested or unaware of this:

    Name:  perpendicular to slope.jpg
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    Very simple answer after all. The angle of corresponding measuring point and auxiliary vanishing point to the horizon line is the same as the angle of the given slope (31deg in the illustration). This can also be transferred to a measuring line in front of the object (similarly to the way measuring points are ordinarily used). To get the auxiliary vanishing point of the direction perpendicular to the slope (the lowest VP in the illustration), you have to draw a line, starting from the measuring point, that is at right angle to the line which goes from measuring point to auxiliary vanishing point of slope. This is probably too confusing to understand just in text, so I added the drawing.

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    Nice to see you are getting there! What puzzles me, however, is that you seem to confuse the left measuring point MPL and the Auxiliary viewpoint...

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    I think this is a different method than what you are referring to. It's not an auxiliary viewpoint, there is no drawing an arc through an auxiliary vanishing point on a vertical vanishing line - this is not the method here. In my eyes this one is a simpler method. The main point is the 90degree angle that's been marked at the measuring point (just an ordinary measuring point of 2-point perspective - same distance on the horizon line from the corresponding vanishing point as the station point is from that vanishing point) - that 90 degree angle is the key to finding the Auxiliary Viewpoint of the Perpendicular Angle to the Slope...

    If I'm understanding it incorrectly, then prove me wrong.

    On a side note - in the picture there are some lines that are irrelevant - a slope that goes to a lower Aux VP than the one drawn in blue, and the corresponding Aux VP of Perpendiculars, which went off page, so I made the slope slightly higher to fit it on the page.

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    Quote Originally Posted by z01ks View Post
    I think this is a different method than what you are referring to.
    That is very well possible, and I seriously doubt whether it gives the same results. Feel free to do whatever you want, but at least take a look at the standard method, as is given on the handprint site...

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    The empty posturing continues. "Seriously doubt"? Why seriously doubt instead of trying it for yourself? As far as I can tell it works. Hmm, maybe that's why its being taught by an Art Center professor. You can feel free to use the more complicated method.

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    Quote Originally Posted by z01ks View Post
    The empty posturing continues. "Seriously doubt"? Why seriously doubt instead of trying it for yourself?
    I did.

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    z01ks’ Illustrated method is correct for a subject in a 2-pt perspective scene, where the line of sight is parallel to the ground plane and where vertical lines (lines perpendicular to the ground plane) remain vertical. Note that the skewed boxes are in 3-pt perspective.

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    Last edited by bill618; June 7th, 2014 at 05:33 PM.
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  18. #16
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    Quote Originally Posted by bill618 View Post
    z01ks’ Illustrated method is correct for a subject in a 2-pt perspective scene, where the line of sight is parallel to the ground plane and where vertical lines (lines perpendicular to the ground plane) remain vertical. Note that the skewed boxes are in 3-pt perspective.
    In that case, you can do the art community a favour if you could write an erratum for the handprint site, to point out the construction posted there is wrong.

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  19. #17
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    I did a quick double check, and I owe you guys an apology! The handprint site and the Erik Olsen course boil down to the same construction, although the presentation is slightly different (Note to self: when scribbling on the back of an old envelope, use an envelope that is big enough). Sorry!

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