Wikipedia:Featured picture candidates/IsometricFlaw

Isometric projection flaw example

Image created by User:Algr. I really like it because besides looking nice, it also strongly contributes to the Isometric projection article, clearly pointing out a potential problem with the use of isometric projection pseudo-perspective, as well as the basis of some of M.C.Escher's artistic works.

Nominate and support. - AnonMoos 06:28, 15 August 2006 (UTC)[reply]
Strong support - per nom --GoOdCoNtEnT 07:34, 15 August 2006 (UTC)[reply]
Support this Escherian image. Even though PNG, no more information would be available if SVG (which, of course, would be better). --Janke | Talk 08:07, 15 August 2006 (UTC)[reply]
Oppose - Not striking, a fine image, but not an FP. HighInBC 12:41, 15 August 2006 (UTC)[reply]
Weak support - would prefer SVG. howch e ng {chat} 17:14, 15 August 2006 (UTC)[reply]
Support SVG edit 2. Nice job. howch e ng {chat} 21:01, 18 August 2006 (UTC)[reply]
- Comment: I have no idea whether the author's software would allow him to easily save to SVG, but if it becomes an issue, I could ask him... AnonMoos 19:19, 15 August 2006 (UTC)[reply]
  - I'm not the author, but I created an SVG version. Suggestions and feedback are welcome. Icey 19:22, 15 August 2006 (UTC)[reply]

Could you tidy up the corners please? Here's a blown-up screenshot showing what I'm talking about. —Keenan Pepper 02:12, 16 August 2006 (UTC)[reply]

Good catch. I am going to be ultra picky, but the edges should be cleaned too, there are some very slight misalignments. I believe it should not happen if done properly. --Bernard 02:24, 16 August 2006 (UTC)[reply]

Reply to Icey: a good start, but a few things should be better:

Your projection is not isometric. Measure the different lengths and you will see a difference. This holds for the original too, in a different way.
The shading on the narrow parts is not perfect, some are more in shadow than they should be.
The base of the red and blue pieces should not be a random curve but a half ellipse, and the lengths of both axis should be in the same ratio as the diagonals of the face (same is true for the original). Oh, they should be centered too.
The shading on the conical parts of the pieces should be a conical gradient, and should go in both directions (I must admit I have seen no way to do that in Inkscape. Well, at least it should pass through the top and be symmetric).

And also I think it would be better with a similar background gradient as in the original. In conclusion, oppose the svg for now, weak oppose the original (even this one should be fixed for the ellipse arcs and the isometry). --Bernard 02:15, 16 August 2006 (UTC). Support svg edit 2 --Bernard 22:04, 16 August 2006 (UTC)[reply]

Thanks for the feedback, I'll get those things sorted when I get home (12+ hours). Icey 07:37, 16 August 2006 (UTC)[reply]

Oppose both per Bernard. Hmm, maybe I'll make something for this with POV-Ray... —Keenan Pepper 02:25, 16 August 2006 (UTC)[reply]
Am I allowed to support my own picture? The hex measurements are off by less then one pixel at the size above - theoretically no image on a video screen could ever be perfectly isometric, so it is a question of under what context does the image fail? It turns out to be surprisingly difficult to make perfect hexagons on a program that doesn't directly support such geometry. Using a 3D image program definitely won't work unless it specifically supports isometric rendering. Any thoughts on how to calculate the right curve for the base? Flash gives me only X and Y pixel measurements, and no way to measure angles. Someone asked a while ago for SVG, but the software I tried downloading didn't work. Icey, If I send you the flash file, will that help? Algr
- You can do it in POV-Ray. It's just orthographic projection along the long diagonal of a cube. —Keenan Pepper 05:59, 16 August 2006 (UTC)[reply]
- Assuming that the pieces are cones (with horizontal cross-sections in the shape of a circle), the theoretical shape of the appearance of the bottom of the piece should be half of an ellipse which has its major axis and minor axis in a ratio of the square root of three to one.
  - Actually, it should be a bit longer than half of an ellipse: because of the conical shape, we can see a little behind. --Bernard 12:58, 16 August 2006 (UTC)[reply]
- My measurements on the original thumbnail are 28.7 upper right and 26.9 upper left, more than one pixel. It is about 172-167-164 on the large size. Not that serious, but you can't deny the problem Algr. --Bernard 12:58, 16 August 2006 (UTC)[reply]
  - I'm not denying the problem, I'm questioning it's relevance. Few photos would ever survive such scrutiny, nor would anyone expect them too. Algr
Alrighty, I've spent the last few hours completely remaking the image. I believe all the problems mentioned are now fixed. The only problem I have is that my knowledge of Inkscape is somewhat limited, so I don't know how to do the suggested shading on the pieces. I've just gone for some sphere's instead. As before, suggestions and feedback are welcome. Icey 20:28, 16 August 2006 (UTC)[reply]
Support SVG edit 2. —Keenan Pepper 23:09, 16 August 2006 (UTC)[reply]
Strong Support Contributes greatly to the linked article, and image quality great. There's no way you could understand the description of the limits discussed without an image like this. Prefer SVG edit 2. Jeeb 02:08, 18 August 2006 (UTC)[reply]
Support SVG edit 2. Good diagram, in SVG, and pleasing to the eye. --Tewy 20:44, 18 August 2006 (UTC)[reply]
There's one thing that bothers me, which I've illustrated here: http://img207.imageshack.us/my.php?image=dontgetitqf5.gif 71.124.216.138 16:38, 19 August 2006 (UTC)[reply]

Whoever you are, you bring up a good point. Can we please keep the boxes consistent, or is it meant to be like that? --Tewy 18:11, 19 August 2006 (UTC)[reply]

Aye, there's a perfectly logical explanation for that. *pauses* *looks round* Which, err, Algr will explain to you.
Essentially, I'm not sure, but if someone knows what it should be, I'll modify it to be correct. Icey

I think it's almost as consistent as it can be -- the thin layer is used when there is not a cube-thickness at hand, because having an abstract zero-height 2D plane instead of a thin 3D layer would look less real. So there's no need to insert a thin layer at the diagonal junction between the two cubes. The only possible real "inconsistency" is that the lower cube is placed on top of a thin layer, but the upper cube has an adjoining thin layer. However, doing it this way makes sure that the upper thin layer is exactly two cube-heights above the lower thin layer, so there's a good reason for it... AnonMoos 19:27, 19 August 2006 (UTC)[reply]

Cool. Thanks for explaining that. Icey 14:04, 20 August 2006 (UTC)[reply]

Promoted Image:IsometricFlaw_2.svg (+8/-2) Clear preference for edit 2 -- Moondigger 02:04, 25 August 2006 (UTC)[reply]