Hi Friends,
My newest personal Inkscape challenge is to reproduce a bookcase in my living room, realistically, including to scale. The problem I've run into, rather early in the game, is that the measured dimensions will need to be somehow translated to account for perspective. It's somewhat of an isometric angle, but instead of 45/45, it's more like 60/30. I hope you understand what I mean by that.

I'm afraid that I will be stymied by the necessary calculations, so that it will probably end up somewhat close to scale. And of course there's no serious need to draw it to scale -- just that I was thinking of trying it. Anyway, I wonder if anyone knows, or knows where to find the proper ratios by which to convert the dimensions (H, W, D)?

Thinking out loud, I'm imagining a triangle with one angle 30 degrees and one side the width of the bookcase. OH! I think I just figured it out! Uummm....well, not quite. But I got this far:



This is the view from above looking down on the scene. Oh, oh, oh -- I think I've got it! I seem to recall (from somewhere long ago) that the formula for a right triangle is a^2 + b^2 = c^2. Oh, well that means that the angle is irrelevant. I have c. I can get a (or b). So then I can calculate b (or a).

Oh no, that's not right. I need my line of site to the corners.... This must be the right way to look at it....(literally ) :



The gray triangles created by the overlapping of the green and pink, are the triangles I need to calculate. So the length I need will be where the red ?s (question marks) are. But now I have no right triangles. I still know....theoretically....well, all the angles, theoretically. And one side length. But since it's not a right triangle, I don't know how to calculate. And I'm not even sure if the measurement CAN be calculated, having only 1 side's length and one corner's angle.

Any tips, clues, or comments?
Thanks

PS -- Just before I posted this I had a flash. I can get the length of the lines indicated by the black ?s (question marks). So that means I'll have 2 sides. I'm pretty sure the 3rd side can be calculated, but I don't know how.

I'm also not entirely sure this is the proper solution. Sorry for all the rambling and thinking out loud (I leave some of the rambling in case it might be instructional)

Thanks again

Probably a little more complex than what you're looking for (I certainly don't understand it at first blush) but here's a link about the math of perspective drawing:
http://www.math.utah.edu/~treiberg/Pers ... rspect.htm

I think your attempt has a fundamental flaw, which is that you have the object vanishing toward the viewer. Perspective drawing vanishes away from the viewing point.

You have made a good start, Brynn, and although it's not the method i would use, there's nothing wrong with it.

You are projecting onto a Picture Plane between you and the object, but note that
1) the actual dimensions of the object don't matter, just the proportions, and
2) the proportions of the object when projected onto the PP remain the same when the PP is moved back and forth, but change when the viewing distance is changed without moving the PP. There is no particular reason your PP has to touch the object you are rendering as it does now. Moving the PP is the same as zooming in or out, but changing the viewing distance actually changes your perspective.

Realistically, your viewing angle should be only 30° to 35° - that is - the angle you are drawing from the 'me' circle should be no more than this, and still enclose all the objects you want to render. To achieve this, you will have to move your 'me' back quite a bit. When adjusting viewing distance you might consider the height of your object as well. It might be better to think of a viewing cone rather than an angle - does your object's greatest dimension fit into a 35° cone? with some room to spare?

If you give a value to your viewing distance, you can use the law of sines to determine the projected lengths but let me emphasize here that it is ABSOLUTELY not necessary and a big waste of time to figure out these lengths; you can snap guides to the intersections of segments and proceed without doing any calculation at all.

These comments are based solely on my personal understanding and usage of perspective and are not backed by any citable source.

As druban has already mentioned, all of this could be done precisely and calculated but... ,)



On the left is a centered frontal view of the object (one point perspective) and on the right, ViewPoint is moved to the right, while object stands at the same position (two point perspective, so to say).

This way all you need to do is to draw plain top-view of the desired object (brown rectangle) in scale or in exact measures and then to position ViewPoint and the horizon. These would define the "working space" where positioning of the ViewPoint and the horizon does not really need to be calculated nor 100% precise, unless the steadiness of some building depends on it ,) Line which is touching the top corner of the "table" (under the horizon) is just an reference for a perspective transposing. Draw perpendicular guides from the object to the "reference line" and connect these intersection points with the "base" of the ViewPoint, which is laying on the horizon. Intersections of these guides and those drawn from the object to the ViewPoint would produce perspective view of the object. The same could be done on the one at the right but it is simpler to just use the angle O` to define VanishingPoint (angle O`` should lay on the horizon with one ray while intersection of the other ray and the ViewPoint would determine the position of the angle`s endpoint - VanishingPoint).

This way could be tricky when even simple curved lines (circle or arc) should be transposed since some more reference points on the original shape should be added (in order to define the ellipse which would be formed and in most cases center points should be added also).

llogg -- Yes, I think you're correct. "Perspective", in the sense that the road gets narrower, the further from viewer, was the wrong word to use. Truly, there would be some tiny amount of perspective, but negligable, when we're talking about 67 inches (or 49). I guess instead of "perspective" I should say "at an angle". When you said "vanishing towards the viewer" I think you misunderstood my illustration. The lines forming the cone, which converge on the circle labeled "me", are my lines of site. But definitely thank you for the link. I will look forward to exploring it.

druban wrote:Realistically, your viewing angle should be only 30° to 35° - that is - the angle you are drawing from the 'me' circle should be no more than this, and still enclose all the objects you want to render.

Is this according to some kind of artistic standard.....or limits of human vision....or....?? (I'm curious -- don't mean to sound rude.) The illustrations above are not to scale....the "me" circle really should be....about 50 to 100% further away. My viewing angle in reality, is....gosh, it's hard to say....I'd say 20 degrees, but certainly no more than 30. Maybe that explains why I had so much trouble with the angles in my "dictionary of Inkscape" image. It literally filled 45 degrees of my view! Actually even more than that, counting the lamp pole. Oh wait....the bookcase is much more taller than wide. It probably fills 45 degrees of the vertical view.

[quote-"druban"]....you can snap guides to the intersections of segments and proceed without doing any calculation at all.[/quote]
I don't quite follow that. But I'll give it some thought. Thanks druban

Maestral, I don't quite follow your illustrations. It could be you were mislead by my poor choice of words. Or maybe it will just take me some time studying the lines. I see you made some right triangles, using the line of sight....or what I think might be the line of site. Well, anyway, I will study them all more closely.

Thank you all for your comments. I'll put my brain power back on the problem, and post again if I get stuck

PS - I'll edit the title

[quote-brynn]
"PS -- Just before I posted this I had a flash. I can get the length of the lines indicated by the black ?s (question marks). So that means I'll have 2 sides. I'm pretty sure the 3rd side can be calculated, but I don't know how."

Perhaps? Try with the irregular ones

Oh wow -- that's it! Thanks Maestral

I won't be able to work on it, at the moment. But later I will....and also try to do it without calculating, using Guides. Thanks again

Edit
Ok, now it's later.

druban wrote:...you can snap guides to the intersections of segments and proceed without doing any calculation at all.

druban, do you mean if I draw the layout (like the ones above) to scale?

Edit #2
And later still....
I think I made it much more complicated than it needed to be. But I do think I've got it worked out.

Thanks again for everyone's help

brynn wrote:druban, do you mean if I draw the layout (like the ones above) to scale?

Yes, at least your viewing distance has to be in scale with your object...

Maestral wrote:Perhaps? Try with the irregular ones

That's the Law of Sines that i mentioned earlier - only made nice and painless to use! Nice find, Maestral.

^ I was surprised too when I found it since I still remember the cover page of my Logarithm tables ,)