Why is the measuring point the same length as the line from the vanishing point to the station point (the line we curve up)? Why can't we just recede the scales on the ground line to the vanishing point instead of the measuring point, and then establish the unit squares in perspective?
In Andrew Loomis's book, Successful Drawing, he did not curve up to find his measuring point, instead he placed them without it. He only said to place them evenly along the horizon.
Last edited by Vay; April 8th, 2011 at 03:30 AM. Reason: more questions
share the source when asking stuff like this
is it this one? http://www.sonjebasa.net/Instruction/2PtPerspective.pdf
Sometime as far as rules don't make things fall apart everything is ok.
Loomis was satisfied with MPs near verticals without thorough measuring. In the example above author finds Measuring Points at the place where isosceles triangles are created by Base Line, Ground Line and line from Measuring Point.
Because lines from Vanishing Points, unlike from Measuring Points, don't cut same length units on the Ground Line and Base Line (or Station Line and VP-SP line).Why can't we just recede the scales on the ground line to the vanishing point instead of the measuring point, and then establish the unit squares in perspective?
not sure on this one.Why is the measuring point the same length as the line from the vanishing point to the station point (the line we curve up)?
How I try to understand it - imagine isosceles triangle without distortion(not in perspective), upper line is horizontal, another is equal to it and one more line to close it. put into perspective, look at it from above and increase so far that upper line becomes horizon line and lower point is at the station point. Distance is distorted but angles remained the same, triangle became part of perspective so parallel lines converge to points on horizon line, while at the station point angles remained undistorted. And if all of it is true - MP to VP length= VP to SP
Last edited by c0ffee; April 25th, 2011 at 10:31 PM.
Last edited by Vay; April 7th, 2011 at 05:33 AM.
ABD is isosceles, and angles in it are undistorted you are right. Distances are distorted due to perspective.
AC is parallel to AD, and BC parallel to BD so ABC is isosceles triangle too, but both distances and angles are distorted so we don't perceive it as such.
Last edited by c0ffee; April 7th, 2011 at 06:41 AM.
How do we determine the distance from the station point to the picture plane from a plan view to a picture in two point perspective, in other words, where do we place the ground line so that the distance between the SP to the ground line in perspective is in accord with the distance of the SP to the picture plane in the planned view?
I found a way, but I don't know if this is correct:
I got this question after seeing a planned view projection in "step 2" of "Two point perspective from plan" section in this PDF file below, because from the PDF, it looks as if the ground line is randomly establish and neglects the correct distance from between the SP to the picture plane from the planned view:
Remake of the example from the PDF:
Last edited by Vay; April 8th, 2011 at 03:26 AM.