Thursday, August 7, 2014

Nude Girls Outdoors and Overcast Light

Today marks a milestone. 

This is the 3000th post on GurneyJourney. I started the blog in July of 2007 and have been posting at least once a day since then. Thanks to all of you who make this blog a part of your routine, whether you're a new visitor or a regular. Stick around and tell your friends about it--there's plenty more to come. Now, onto the post.....

Models painted by Trina Merry in front of the Guggenheim Museum and the Manhattan Bridge. AP Photo
The models in these photos are practically nude, except for body paint. Public nudity is legal in New York if it's part of a performance art piece. To camouflage her models, body painter Trina Merry had to keep backing up to see if the details aligned with the background. You can watch a video of the work in progress here.

You'll notice that no matter how she's painted, the model is always darker than the sky, and that the photos are always taken in overcast light or in open shadow. 

The picture above seems to be taken in open shadow with a sunny scene behind her. She is darker than the sky, despite the fact that, in this case, the illumination on her seems to heightened a bit by a flash or a reflector near the camera—note how the values lighten on the front of her thighs.


The light here seems to be a thin cloud layer covering most of the sky, and the model is lit by that overcast light. She's not casting a shadow, proof that she's not in direct sunlight. However the overcast isn't total: the window is reflecting a piece of blue sky. Regardless, the legs will be darker than the grass no matter what kind of paint you use on them.


Sorry, no nude here. A good rule of thumb is that in overcast conditions, a white local color facing upward will closely match the sky. Not many local colors are lighter than snow, and even bright white snow generally matches the tone of the cloudy sky.

Snow is significantly lighter or darker than an overcast sky only when the cloud layer is thin enough to permit some gradation of brightness in the direction of the sun.


Which leads to the following question: In direct sunlight illumination is there any paint white enough to offset the darkness of the shadow side? And is there a paint black enough to offset the effect of the direct sunny illumination? In other words, could you paint a ball—or a nude girl—in such a way that you could replace the gray-painted ball above, and make the ball—or the girl—disappear?

I don't think so, but I'll have to give it a try. I don't think Jeanette will let me try it on a nude girl, so I may have to settle for the ball.
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If form, light, shadow, and paint interest you, you can read more about it in my book Color and Light: A Guide for the Realist Painter, available on Amazon— or signed from my webstore.
Thanks, Evelyn Brody

17 comments:

  1. Brilliant post, James! :) This a problem in something that I call color anamorphosis. I have to write some course notes precisely on that, so thanks for the inspiration. ;)

    First, you have to define the problem very carefully, or you won't get anywhere. The sphere is too complicated to start with, as the reflection angle changes from point to point, and also because it is hard to take into account all the secondary source points on the ground.

    Here is an easier example: a square board, standing upright, with the incident light at 45 degrees to the ground. The front plane of the board and the ground plane to be painted with the same paint X, the back plane to be painted with the unknown miracle paint Y in such a way as to be perfectly camouflaged by the ground plane.

    Assume only diffuse reflections, specular is another complication. (i.e., the material of the board and ground paint is perfectly "matte"). This means that the proportion of the light that gets reflected is equal in all directions of emission, and only dependent on the angle of incidence. No shiny stuff allowed.

    Now, since I chose the incident angle to be 45 degrees, the frontal plane will be perfectly camouflaged with the lit ground, as they are equally painted and are at the same angle of incidence (this wouldn't be so with the sphere). We just need to camouflage the shadow side. Note also that I am not considering the cast shadow, only the shadow side.
    If the camouflage works, we will see a cast shadow being cast by nothing at all, which is pretty cool. :) (shadow side of the board is lit by secondary reflection from the ground, cast shadow is lit only by tertiary reflection from the shadow side of the board, so it is a different problem)

    Now, under these simplified conditions, can we find such a paint (even a theoretically perfect paint)?

    I think the answer is no, and I think I can prove it. :) But I'll wait a while for other opinions as I try to think of a way of proving it without any algebra :D

    Cheers,
    Antonio

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  2. This concept of the painted figure blending with the background was pioneered by the iconic model Veruschka, in collaboration with painter & photographer Holger Trulzsch. A collection of their mind-bending photos was published in the book TRANS-FIGURATIONS in 1986. It is an excellent book and highly recommended!

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  4. Thank you for your blog, for interesting and teaching articles. I wish you at least 3000 more posts. Great job!

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  5. I also wish you 3000 more posts. Thank you so much for your blog, it is an important part of my day and I have learnt so much from you, both the blog and your books. Thank you.

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  6. James, I just want to thank you for sharing your knowledge with us. Your blog has been an irreplaceable resource for me - so I'm adding my own wishes for 3000 more awesome posts!

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  7. 3000 posts creates great visibility for your work and your helpful tips.

    But your comments on overcast light track with a recent Science News story about invisibility:

    "Want to disappear? A new invisibility cloak offers you the chance — provided you carry around a portable fog machine.

    The cloak, described June 5 in Science, achieves the challenging feat of hiding an object from all angles and all colors of visible light, with the caveat that it works only in hazy environments including clouds, fog and frosted glass. “It’s a nice demonstration,” says Jason Valentine, a mechanical engineer at Vanderbilt University in Nashville. “It sacrifices functionality in one area to gain functionality in another” — namely, the disappearance of the object to the naked eye."

    Here's looking at you!

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  8. Congratulations on reaching 3000 posts, James, and lots of thanks for putting in the hard work.

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  9. There appears to be a bit of photoshop help in these pics, at least with the shadows. Most obvious is the last one. As you mentioned, the model has to cast shadow, however the objects to the left of her under the same sun light do. The grass between her legs and to the right of her feet appears to have clone marks and a bit of blurriness, indicating her shadow has in fact been removed and she is in direct sunlight.

    Also seems odd that even the eyelashes in the others are coloured.

    Only a quick observation, and I think if I'm right the photoshopping has only been used to clean up the photos and their lighting, not to cheat the physical paint work.

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  10. I meant "the model has NO cast shadow"

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  11. Antonio, I would love to follow your thought experiment, but I can't (which is not to say you're not describing it well, only that I'm not following it). It's what's going on with the back side that I can't understand. I also can't tell from what angle the board is being viewed (which I take to be crucial - ?). Do you have a diagram for this somewhere on the web?

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  12. Tom,
    I did describe it badly, and not only that, I was thinking hurriedly too. Proving it turned up to require more labour than I expected, but I still think I can do it, with a little time, and I will bring back the news if I succeed. :)

    To answer your question, we would be looking at the board from the back (shadow side), say at a 45 degree angle with the horizontal. Assuming the table to be large enough, it should cover the whole background, so that if the board is painted in the right way, it would seem to disappear into the table.

    For anyone whose physics is less rusty than mine is, proving this is a game that requires taking the following into consideration:

    The table, which is under direct lighting, has a reflection coefficient r1 that says how much of the light is bounced back to the viewer. Every point of the table bounces each ray in a uniform way, the energy being spread on a sphere, thus the intensity declining with the radius squared.

    The shadow side of the board (with the miracle paint) has a reflection coefficient r2, bigger than r1, thus bouncing more light (in percentage), but it also gets lit less. How much is it lit, exactly? It is being lit by the reflected light that bounces of the semiplane of the table that is behind the board, except that part that is on the cast shadow (and therefore receives almost no light, except for tertiary reflections that we can ignore).

    So here is what can be done: we can calculate the amount of light that is hitting the board, and we equal it to the amount which is hitting the table. For a given r1, we get an equation that we can solve for r2. If we get a value for r2 that is smaller than 1, then we have our theoretical magic paint. If we get a value bigger than 1 that means that it can't be done, as r2 bigger than 1 requires the board to be give out more light than it is receiving. Finally, if we get r2=1, it means a perfect white paint that reflects all that it receives.

    My current difficulty is with calculating the amount of light hitting a point on the dark side of the board. I'm rusty with those integrals. :-( If any physicist is reading this, please help us out, will ya? :)

    Or, if the answer turns out to be "yes", any artist can kill this whole thing by just doing it in practice and showing us the picture. :) Anyway, a really fun challenge.

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  13. Antonio, thanks for the thoughtful response. I have to admit that my brain glazes over a bit when I think of mathematical formulas, but what you say makes sense. Next time I have the black and white spray cans activated again, I'll give this an experimental try.

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  14. James,

    I took some time today to look at this again and now (I think) I have it. It still seems like the answer is "no", at least for the example I described. Next week I'll have some free time and will check it and write it up if it turns out to be correct. I'll try to add a couple of drawings to make it more palatable. :)

    Cheers,
    A.

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  15. Hi,

    I reviewed the thing over the weekend and according to my calculations you would need a reflection coefficient of at least 2 in your miracle paint. Which means it would have to actually emit light rather than just passively reflect it. Put in another way, it can't be done.

    Caveats: 1) This is for the configuration I described above and without speculars, although I don't think you can improve it much just by positioning it differently.

    2) The perception of value isn't linear with actual power flux into the retina. I calculated for power, so I'm not sure (yet) how much of a value difference my calculations imply. So you can't get an exact value match, but it is still an open question how close to a match you can get. For artistic purposes, close enough is close enough.

    It will be some work to make my notes readable. I'll try to do it this week. If I can't find the time I'll just stick the handwritten notes on my dropbox and link them from here.

    A.

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  16. António, I suppose that's why engineers have had such a hard time coming up with invisibility camouflage technologies using reflective surfaces on sunny days. But what if the shadow side was lit up as much as possible by an extreme amount of reflected light, say with snow bouncing light back into the shadows?

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  17. James,

    with the current setup, it wouldn't make a difference. For sure, making the ground highly reflective bounces more light onto the shadow side, but it also makes the task harder necause now you are trying to camouflage against a brighter background; by my accounting the two effects cancel each other exactly (I assume we are trying to camouflage against the ground, not the sky; for the sky we'd have to specify the atmospheric effects, and that's a different calculation altogether).

    The problem is geometric: we have a half plane (the ground behind the board) lighting the shadow side with reflected light, and that simply isn't enough.

    A different setup might make a difference, say

    a) making the ground behind the board of a more reflective material than the one in front. This would obviously work but it kind of defeats the whole point.

    b) shaping the ground behind the board; instead of a plane, making it like a lens, in order to reflect more light onto the shadow side. That might work to a point.

    Option A is too much cheating and would result in a weird two-valued picture. Option B might be doable in an interesting way but it's basically just playing with reflectors...

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