This is the only video (series) on six point perspective on TH-cam, so thank you! Everything I draw will be in six point curvilinear perspective from now on.
Yessssssss thank you sooooooo much Sorry for it being a pain I remember the intro lol how you said it was a pain thank you so much I’ve always wondered if there was a level beyond the 1,2,3,4,5 points and cylindrical perspective lol thanks to you I appreciate it 👍👍👏👏
This was great - I'd like a little more info on how this relates to other perspectives, and how this has strengths & weaknesses. Maybe that's in the other parts? Thx again.
Lets see . . . 6-pt perspective is the curvilinear equivalent of 3-pt perspective, you use it when you're not looking square-on at anything. The viewpoint is looking a little up or a little down, and a little to either side. The strength is that it allows you to see a MUCH wider view than straight-line perspective does. It's weakness is that it's slow and complicated to do. It's strength/weakness is that it looks weird.
@@zy_load So, 5-point curvilinear perspective does the same job as 1-point perspective (looking straight ahead at the primary objects in the scene), 4-point curvilinear does the same job as 2-pt perspective, and 6-pt curvilinear does the same job as 3-pt perspective
(6) point perspective is a city wide perspective (non euclidean vr) (5) a Point of view (VR) (4) looking at 3d modeling software (screen) (3) a 2d representation of 3d models (graph paper) (2) a 1d representation of 2d models. (Drawing paper) (1) a 1d model (finger paints)
83 mm is the radius of the circle, and the radius of the circle is 90 degrees. So for this circle, 90 degrees is 83 mm (on a line that passes through the center). If your circle was 100mm, you'd use 100mm as your 90 degree measurement to place the distance from a VP to it's opposite horizon line.
@@Jasoncm I had the same question as the original person about how you arrived at this, and I must honestly tell you that I found this explanation incomprehensible. "and the radius of the circle is 90 degrees"??? Relative to what? I feel like you're using mm and radius interchangeably here, but one is a measurement of length (i.e. the radius is the length of a ray emanating from the centerpoint of a circle out to its perimeter, or half the diameter), and the other is a measurement of an angle (implying two lines orginating at a common point). There's a piece of important info missing. Not trying to be difficult here---I would really like to understand this and keep going with your videos. But you've lost me very early on with this one point that Luke brought up about where 83mm is coming from (in the video you said centimeters). Any further clarification would be most welcome. Thank you.
@@anahata2009 Hi! Thanks for explaining your question. It's tricky, and I'll try to do it in a comment, but if that doesn't work, maybe a little video answer would be best. It's about the "cone of vision," or "what width of scene can you see in the picture?" In this perspective system, you are seeing a 180 degree field of vision. So the center of the circle is looking straight ahead, the top of the circle is straight up (90degrees away from straight ahead), the rightmost edge of the circle is straight to the viewer's right (90 degrees away from straight ahead) etc. The whole circumference of the circle represents 90 degrees away from looking straight ahead. So in this system, the radius of the circle represents a 90 degree change in the direction you're looking (as long as it's on a straight line that passes through the center of the circle). Let me know if this doesn't clear things up!
@@Jasoncm Hi Jason. Thanks for taking the time to try to further clarify. I'm still bewildered, like some crucial piece of information to orient myself is missing . . .but maybe getting closer? I understand the cone of vision representing everything in front of the viewer from left to right (arc of 180 degrees), and up and down. So when you say "The whole circumference of the circle represents 90 degrees away from looking straight ahead" I guess what you're describing is essentially a flat plane (picture a round plate) that is oriented perpendicular to the viewer's straight-ahead line of vision? Maybe? Even if I've understood that correctly, what you said next is still confusing to me: "in this system, the radius of the circle represents a 90 degree change in the direction you're looking (as long as it's on a straight line that passes through the center of the circle)" There are an extraordinary number of angles such a 'straight line through the center of the circle" on that plane could pass through. Are you talking about a straight line on the aforementioned perpendicular plane at the viewer's eye level, paralell to the ground? And I still don't know where a measurement of 88mm comes into play . . . A little video answer would be great, but at this point I feel like I'm going to need an elaborate 3D animation to untwist my neurons. ;-)
@@anahata2009 exactly! And that round picture plate, your eye itself is right at the center of it, because that 90 degree cone (or 180, depending on if you're thinking radius or diameter) lets you look straight up and straight down and fully to the left and fully to the right, just not behind you.
Nice demonstration of how to construct a curvilinear perspective, and super cool that you're constructing a 6-point perspective; I haven't seen that before! However, a small detail (that is somewhat involved to explain)-it seems to me like you're almost using a stereographic projection but not quite. The fact that you're drawing circles makes it seem like you're using a stereographic projection, because all lines becomes circular arcs (or lines) in that projection. However, the fact that you equate a right angle with a distance on the paper, as you do at 5:52, makes it seem like you're using an equidistant projection, as in an equidistant projection, a right angle will always have the same length as long as it either intersects COV or is part of a line that does, while in a stereographic projection it will be differently long on the paper depending on where it is located (even if it intersects COV in both cases) since it is not a distance-preserving projection. In your case, specifically, the distance on the paper should be slightly shorter between VP1 and the point where the horizon intersects the line that intersects VP1 and COV, compared with the radius of the 90 degree cone of vision (because the half-way point between VP1 and the horizon is more centered, i.e. closer to COV, than a half-way point between COV and the 90-degree cone of vision, and objects become larger in the projection the farther away from COV they are). (Thus, if you make them equally long, that is an error that could potentially lead to inconsistencies later down the road.) Apart from that, it seems like you have worked out a very rigorous way to construct the perspective!
That's really interesting stuff. There's some expediency to this system to make it more practical for creating drawings and paintings. It's never been intended to be mathematically precise. I don't know that I'd give up the arcs -- having to plot sine curve sections instead of being able to use a compass I think would make this process so onerous I'd never use it. I wonder if there's a relatively elegant way to plot the distance between VPs other than equidistant. Thanks for writing, I enjoy talking about this stuff.
@@Jasoncm I completely get that, and what you do still seems to work very well for you which is the most important thing. I also admit that I don't know how to find out where the horizon arc should be located in a stereographic projection in any easy way (without using a calculator). The only way I have found to actually work in a stereographic projection when drawing is by using a Wulff net (a.k.a. stereonet), but from what I have seen those always have one vanishing point in COV. And of course, even if you could get a similar net where no vanishing point is in COV (which I think should be possible), having to use a computer to print a visual guide on a separate piece of paper would still be very cumbersome and make the process more complicated. I think you're process is elegant in its (relative) simplicity.
@@Jasoncm I created a proof-of-concept for a curvilinear perspective based on the stereographic perspective in GeoGebra here if you are interested, with the code buexkeus (I couldn't paste the URL without TH-cam removing my comment, but if you are familiar with GeoGebra, you may be able to figure it out). You can move the blue points to change the perspective, and you can move around and zoom in/out like in Google Maps. You can also show some of the supporting shapes I used to create the perspective by going to algebra mode (by pressing the calculator) and checking the shapes you want to show, but I have used some constructions that are not pure straightedge and compass constructions, which may make it slightly more cumbersome if you want to draw in on paper.
@@kristoferkrus I don't know geogebra I'm afraid. I went to geogebra and put buexkeus in the search and got nothing. Any hints to help me find your stuff?
Problem: find the arc that passes through three specified points. Solution: three non-colinear points describe a triangle. Construct the perpendicular bisectors of each side of the triangle (method demonstrated in video). The intersection is called the circumcenter of the triangle. The circumcenter is the same center as the circle that inscribes the triangle. Therefore, the circumcenter is the center of the desired arc, and the radius is the distance between that center and any of the specified points.
Follow-up... *why* isn't it exact? Do we need to take any considerations for the angles between the vanishing points? vp1, vp2 seem to be chosen somewhat arbitrarily. Next step... how do we fill in a grid? Equidistant spacing along the first two axes we drew, same as in three point perspective to find the intersections on the surface of the sphere, and then draw arcs of increasing curvature passing through the poles orthogonal to the corresponding vp?
@@innovationsanonymous8841 there are some great tricks for equal spacing in 5-point, and some of them extrapolate pretty well to six-point. VP 1 and VP two are chosen in one sense arbitrarily, but it's really to give the angle of view you want. If your up VP is close to the edge of the circle, you're looking a little up, if it's closer to the center, you're looking WAY up.
@@innovationsanonymous8841 why isn't it exact? I honestly don't know. Flocon and Barre said it wasn't exact, and I read through their explanation years ago. I remember being convinced at the time, but I don't remember the exact explanation. You can probably find a pdf of their book "curvilinear perspective" if you want to read it. I THINK if you were to make this space with sine curves instead of arcs, it might be more exact.
Learning from someone that says it's a pain and complicated can't be a good thing if rather hear it's fun and easy so I'm not stressed out by a teacher that sounds stressed out
7 points or more is definitely possible. First I should say that you could easily have 7 points or more in a standard 2-pt perspective drawing if there are multiple objects in the scene that don't line up with the main scene. If you have chairs around a circular table, each chair will use separate vanishing points. But that's not what I'm talking about here, I'm just talking about the primary scene's vanishing points. Each vp represents a direction. So you could have vps for up, down, left, right, forward, backward (that's six) and then another point for forward again, and up again, etc and your directions would show up multiple times in the same drawing (maybe with changes, maybe the same)
What is 90 degrees in terms of mm. I didnt understand... what do you exactly mean by that, what am i supposed to do? Will you please help. Where did that 83 mm come from? What's the measurement i shall take and from where. Please clarify it. You should be specific in terms of geometry, this is really not a great way to explain the geometrical measurments and the rays you are drawing... i hope you understand.
So, the main circle that circumscribes the drawing represents a 90 degree cone of vision. So the radius of that circle is 90 degrees. So whatever size you draw your circle, the radius is your 90 degree measurement (that can only be applied with a ruler if it passes through the center of the circle, otherwise the line of measurement would curve and the measurement would change.
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This is the only video (series) on six point perspective on TH-cam, so thank you! Everything I draw will be in six point curvilinear perspective from now on.
I'd love to see one of your 6 pt drawings!
Bro I saw a TH-cam short on perspective earlier and now I know what 1, 2, 3, 4, 5, and 6 point perspectives are 💯
Yessssssss thank you sooooooo much
Sorry for it being a pain I remember the intro lol how you said it was a pain thank you so much I’ve always wondered if there was a level beyond the 1,2,3,4,5 points and cylindrical perspective lol thanks to you I appreciate it 👍👍👏👏
This was great - I'd like a little more info on how this relates to other perspectives, and how this has strengths & weaknesses. Maybe that's in the other parts? Thx again.
Lets see . . . 6-pt perspective is the curvilinear equivalent of 3-pt perspective, you use it when you're not looking square-on at anything. The viewpoint is looking a little up or a little down, and a little to either side. The strength is that it allows you to see a MUCH wider view than straight-line perspective does. It's weakness is that it's slow and complicated to do. It's strength/weakness is that it looks weird.
What would the the curvlinear equivalent of 2 point perspective ?
@@zy_load So, 5-point curvilinear perspective does the same job as 1-point perspective (looking straight ahead at the primary objects in the scene), 4-point curvilinear does the same job as 2-pt perspective, and 6-pt curvilinear does the same job as 3-pt perspective
(6) point perspective is a city wide perspective (non euclidean vr)
(5) a Point of view (VR)
(4) looking at 3d modeling software (screen)
(3) a 2d representation of 3d models (graph paper)
(2) a 1d representation of 2d models. (Drawing paper)
(1) a 1d model (finger paints)
Also, great video with very concise info!
How do you arrive at the 83 millimeters? When you’re setting up the 90 degree points from each vanishing point.
83 mm is the radius of the circle, and the radius of the circle is 90 degrees. So for this circle, 90 degrees is 83 mm (on a line that passes through the center). If your circle was 100mm, you'd use 100mm as your 90 degree measurement to place the distance from a VP to it's opposite horizon line.
@@Jasoncm I had the same question as the original person about how you arrived at this, and I must honestly tell you that I found this explanation incomprehensible.
"and the radius of the circle is 90 degrees"??? Relative to what?
I feel like you're using mm and radius interchangeably here, but one is a measurement of length (i.e. the radius is the length of a ray emanating from the centerpoint of a circle out to its perimeter, or half the diameter), and the other is a measurement of an angle (implying two lines orginating at a common point). There's a piece of important info missing. Not trying to be difficult here---I would really like to understand this and keep going with your videos. But you've lost me very early on with this one point that Luke brought up about where 83mm is coming from (in the video you said centimeters). Any further clarification would be most welcome. Thank you.
@@anahata2009 Hi! Thanks for explaining your question. It's tricky, and I'll try to do it in a comment, but if that doesn't work, maybe a little video answer would be best. It's about the "cone of vision," or "what width of scene can you see in the picture?" In this perspective system, you are seeing a 180 degree field of vision. So the center of the circle is looking straight ahead, the top of the circle is straight up (90degrees away from straight ahead), the rightmost edge of the circle is straight to the viewer's right (90 degrees away from straight ahead) etc. The whole circumference of the circle represents 90 degrees away from looking straight ahead. So in this system, the radius of the circle represents a 90 degree change in the direction you're looking (as long as it's on a straight line that passes through the center of the circle).
Let me know if this doesn't clear things up!
@@Jasoncm Hi Jason. Thanks for taking the time to try to further clarify.
I'm still bewildered, like some crucial piece of information to orient myself is missing . . .but maybe getting closer? I understand the cone of vision representing everything in front of the viewer from left to right (arc of 180 degrees), and up and down. So when you say "The whole circumference of the circle represents 90 degrees away from looking straight ahead" I guess what you're describing is essentially a flat plane (picture a round plate) that is oriented perpendicular to the viewer's straight-ahead line of vision? Maybe?
Even if I've understood that correctly, what you said next is still confusing to me:
"in this system, the radius of the circle represents a 90 degree change in the direction you're looking (as long as it's on a straight line that passes through the center of the circle)" There are an extraordinary number of angles such a 'straight line through the center of the circle" on that plane could pass through. Are you talking about a straight line on the aforementioned perpendicular plane at the viewer's eye level, paralell to the ground?
And I still don't know where a measurement of 88mm comes into play . . .
A little video answer would be great, but at this point I feel like I'm going to need an elaborate 3D animation to untwist my neurons. ;-)
@@anahata2009 exactly! And that round picture plate, your eye itself is right at the center of it, because that 90 degree cone (or 180, depending on if you're thinking radius or diameter) lets you look straight up and straight down and fully to the left and fully to the right, just not behind you.
Wonderful. Thank you
Very cool!
Thanks! You know cool, so I'm flattered!
@@Jasoncm all of your videos are so good. I realized I should comment more and tell you how much I love them.
Nice demonstration of how to construct a curvilinear perspective, and super cool that you're constructing a 6-point perspective; I haven't seen that before!
However, a small detail (that is somewhat involved to explain)-it seems to me like you're almost using a stereographic projection but not quite. The fact that you're drawing circles makes it seem like you're using a stereographic projection, because all lines becomes circular arcs (or lines) in that projection. However, the fact that you equate a right angle with a distance on the paper, as you do at 5:52, makes it seem like you're using an equidistant projection, as in an equidistant projection, a right angle will always have the same length as long as it either intersects COV or is part of a line that does, while in a stereographic projection it will be differently long on the paper depending on where it is located (even if it intersects COV in both cases) since it is not a distance-preserving projection. In your case, specifically, the distance on the paper should be slightly shorter between VP1 and the point where the horizon intersects the line that intersects VP1 and COV, compared with the radius of the 90 degree cone of vision (because the half-way point between VP1 and the horizon is more centered, i.e. closer to COV, than a half-way point between COV and the 90-degree cone of vision, and objects become larger in the projection the farther away from COV they are). (Thus, if you make them equally long, that is an error that could potentially lead to inconsistencies later down the road.)
Apart from that, it seems like you have worked out a very rigorous way to construct the perspective!
That's really interesting stuff. There's some expediency to this system to make it more practical for creating drawings and paintings. It's never been intended to be mathematically precise. I don't know that I'd give up the arcs -- having to plot sine curve sections instead of being able to use a compass I think would make this process so onerous I'd never use it. I wonder if there's a relatively elegant way to plot the distance between VPs other than equidistant. Thanks for writing, I enjoy talking about this stuff.
@@Jasoncm I completely get that, and what you do still seems to work very well for you which is the most important thing. I also admit that I don't know how to find out where the horizon arc should be located in a stereographic projection in any easy way (without using a calculator). The only way I have found to actually work in a stereographic projection when drawing is by using a Wulff net (a.k.a. stereonet), but from what I have seen those always have one vanishing point in COV. And of course, even if you could get a similar net where no vanishing point is in COV (which I think should be possible), having to use a computer to print a visual guide on a separate piece of paper would still be very cumbersome and make the process more complicated. I think you're process is elegant in its (relative) simplicity.
@@kristoferkrus thanks! It's largely taken from Flocon and Barre's book "Curvilnear Perspective"
@@Jasoncm I created a proof-of-concept for a curvilinear perspective based on the stereographic perspective in GeoGebra here if you are interested, with the code buexkeus (I couldn't paste the URL without TH-cam removing my comment, but if you are familiar with GeoGebra, you may be able to figure it out). You can move the blue points to change the perspective, and you can move around and zoom in/out like in Google Maps. You can also show some of the supporting shapes I used to create the perspective by going to algebra mode (by pressing the calculator) and checking the shapes you want to show, but I have used some constructions that are not pure straightedge and compass constructions, which may make it slightly more cumbersome if you want to draw in on paper.
@@kristoferkrus I don't know geogebra I'm afraid. I went to geogebra and put buexkeus in the search and got nothing. Any hints to help me find your stuff?
Problem: find the arc that passes through three specified points.
Solution: three non-colinear points describe a triangle. Construct the perpendicular bisectors of each side of the triangle (method demonstrated in video). The intersection is called the circumcenter of the triangle. The circumcenter is the same center as the circle that inscribes the triangle. Therefore, the circumcenter is the center of the desired arc, and the radius is the distance between that center and any of the specified points.
Follow-up... *why* isn't it exact? Do we need to take any considerations for the angles between the vanishing points? vp1, vp2 seem to be chosen somewhat arbitrarily. Next step... how do we fill in a grid? Equidistant spacing along the first two axes we drew, same as in three point perspective to find the intersections on the surface of the sphere, and then draw arcs of increasing curvature passing through the poles orthogonal to the corresponding vp?
@@innovationsanonymous8841 there are some great tricks for equal spacing in 5-point, and some of them extrapolate pretty well to six-point. VP 1 and VP two are chosen in one sense arbitrarily, but it's really to give the angle of view you want. If your up VP is close to the edge of the circle, you're looking a little up, if it's closer to the center, you're looking WAY up.
@@innovationsanonymous8841 why isn't it exact? I honestly don't know. Flocon and Barre said it wasn't exact, and I read through their explanation years ago. I remember being convinced at the time, but I don't remember the exact explanation. You can probably find a pdf of their book "curvilinear perspective" if you want to read it. I THINK if you were to make this space with sine curves instead of arcs, it might be more exact.
excellent!
❤
Thank you so much
You're most welcome!
Thank you so much.
I'm glad you liked it
the video❌
THE MUSIC✅
Learning from someone that says it's a pain and complicated can't be a good thing if rather hear it's fun and easy so I'm not stressed out by a teacher that sounds stressed out
Lots of drawing stuff is fun and easy. Some stuff is complicated and has a lot of steps but is still worth it
Does 7 points perspective possible 🤨
7 points or more is definitely possible. First I should say that you could easily have 7 points or more in a standard 2-pt perspective drawing if there are multiple objects in the scene that don't line up with the main scene. If you have chairs around a circular table, each chair will use separate vanishing points. But that's not what I'm talking about here, I'm just talking about the primary scene's vanishing points. Each vp represents a direction. So you could have vps for up, down, left, right, forward, backward (that's six) and then another point for forward again, and up again, etc and your directions would show up multiple times in the same drawing (maybe with changes, maybe the same)
What is 90 degrees in terms of mm. I didnt understand... what do you exactly mean by that, what am i supposed to do? Will you please help. Where did that 83 mm come from? What's the measurement i shall take and from where. Please clarify it. You should be specific in terms of geometry, this is really not a great way to explain the geometrical measurments and the rays you are drawing... i hope you understand.
So, the main circle that circumscribes the drawing represents a 90 degree cone of vision. So the radius of that circle is 90 degrees. So whatever size you draw your circle, the radius is your 90 degree measurement (that can only be applied with a ruler if it passes through the center of the circle, otherwise the line of measurement would curve and the measurement would change.
Really upset with your tutorial, not done man... sorry
This is Part 1. There's more