File talk:Douma bodies 252pm.jpg

A few things about this photo and this scene:

First, prior thoughts, from Monitor blog:

Site 1: Men and boys, dirt floor (mosque?) courtyard, unwrapped then wrapped - canopy cover, west or southwest wall is blue. No gate at the corner.Body count: considering 6 photos at different times, there are at least 24 (probably 25+) in the nearer row, at least 17 in the back row. So at least 41 total. Five are clearly boys (considered here), others might be.

...

Some notes then:

1) The bodies don't seem to display rigor mortis, which means they either died less than 2 hour ago, or about a day ago.

2) The blood on them all seem fairly dry. No one's pouring blood, and only the worst injuries seem to be smearing at all.

3) There are at least 41 bodies, just of men and boys, who already died from their injuries, stopped bleeding, and got fairly dry ... no more than 90 minutes after the alleged attack. In all my research, I'm still not sure if that's possible, but it definitely seems unlikely. There was no rubble really to pull them from, and no sign they were under it either. Just standing around, hit by shrapnel, died and collected quickly, on this scale?

4) The injuries they supposedly died from are generally unclear. At least one has a missing leg, one a nasty chest wound, another a head wound, and some are covered up. But dozens of men and boys are seen, totally intact, just some combination of peppered and ripped-up a bit, bleeding from random spots, smoky by not dusty,

5) Pants torn a lot with both the men and the boys, and with the men but not the boys, belts get undone and pants get pulled halfway down more often than not. This an happen by accident, or be a sign of disrespect for the dead. (that is, it might suggest these were people the rebels disliked, government loyalist or non-Sunni families)

6) Again, this is a segregated scene, but in all other scenes and records, there's still no sign of more than 3 women and about that many girls killed, to what seems about a dozen boys and about a hundred men./ Were they spared the way bombs sometimes do, randomly? Or were they "spared" the way Islamist war booty sometimes is?

Next, refinement of the 90 minutes part. Attack time appears a bit earlier than I thought (" somewhere between 1:25 and 1:45")... maybe closer to 1 PM or even a bit earlier. Activists told HRW it was at "about noon," but I think more like 1 (earlier photo analysis pending).

Exact time for this photo: no later than 2:52 is pretty vague, when there's sunlight showing an exact time. I'm rusty, or feel rusty, but the elevation angle might be 67 degrees? That would be about 1:10 PM! I've asked Petri Krohn elsewhere, opening the question here too, for him or whoever. --Caustic Logic (talk) 13:25, 3 September 2015 (UTC)

And here's a crop with the best sun info: --Caustic Logic (talk) 23:31, 3 September 2015 (UTC)


 * I think the line from the left hand corner of the metal box to corner of its shadow is perpendicular to the view from the camera. Using Gimp I rotated the photo -5.44 degrees to make it horizontal. I then rotated it to -26.67 degrees to get the line drawn by the shadow vertical. This would mean that the sun is at 21.23 degrees from the vertical, i.e. an elevation of 68.77 degrees. -- Petri Krohn (talk) 07:35, 4 September 2015 (UTC)


 * Using the ESRL solar calculator with the arrow set to the central mosque in Douma I get the time 13:12:30 for the elevation 68.77 on August 16th. -- Petri Krohn (talk) 07:54, 4 September 2015 (UTC)


 * Awesome, Petri, thanks! I need to learn some new tricks, one of these days. But we get about the same thing. Somewhat by coincidence - checking back my 67 degrees is reached only at 1:30, not 1:10. You get a bit higher though, which means earlier (with a slow change for moderate differences). You're probably closer. But small differences in elevation are big enough in time it's safer to say in the 1:00-1:30 range, with a reliable reading around 112 (not likely the equivalent span before noon).


 * The significance, again, is this is only x minutes after the explosions at the markets. That still needs set. --Caustic Logic (talk) 09:51, 4 September 2015 (UTC)


 * Notes after re-examining: I still see 5 degrees rotation, but 5.5 is close enough. The main thing I wonder about is how the angle the wall is seen from effects the angle we measure. I made a flat triangle 80px by 250px, about the area covered, drew two lines, 69 and 70 degrees. Did what you did with the cooler, thinking the very corner of the shadow is probably invisible, but best. So I got a bit shallower, 68. I'd say between all this 67 is an outlier and 68-69 is the right range to call. Time range for that is 1:09:33 to 1:20 on the dot. --Caustic Logic (talk) 10:23, 4 September 2015 (UTC)

3D
I decided I could tackle the 3-D aspect, or get a start, and refine the elevation reading. A flat reading, even rotated, isn't quite right. The apparent shadow angle will be offset in 3-D space, with the distance 'back' coming through partly as 'up,' etc. Here, the blue bar is the imaginary pole at the corner of this object, and the yellow triangle is the patch of light it would block, in a plane that here is seen at an angle, green line as its base. Note a flat measure will either use an angle less than 90, or you're not using the right line. dark green = true triangle base, light green = simple horizontal/possibly apparent triangle base. Because of foreshortening, the sliver of difference between them appears very slight. But in real space, which is what matters ... the difference is slight, but significant in context.

How long is this true base? next graphic - Looking at the vertical shadow at the corner of the two walls, the line of light travel between those orange lines (wall corner and shadow edge) is dark green because it'll be the same angle as our base. Foreshortening makes it hard to be sure, but I estimate about 3 units back for every 8 to the left (8:3 - the white bars between the orange). It might be closer to and even 45 degrees, but I don't think fully 8:8.

Apparent triangle vs. real triangle will have a visual:real base proportion, depending, of 29:29 (if there was no offset), 29:32 (if a 3:8 angle), less than 29:37 (if 45 degrees). (visual measure of # of 16" units...) Triangles with these bases shown in that order, each with its overhead shadow offset illustration to explain. Left to right, the elevation angle = time for each: That last is not likely, an upper roof. I think the first one (about Petri's measure) is out, but barely. The middle one is close - range centered just later than it - 68 high, I'd say down to 63 is reasonable = almost exactly 1:20-1:59, with 1:35 - 1:45 is maybe the best range.
 * 68.5 = 1:15:40
 * 66.25 = 1:36:30.
 * 60.5 = 2:16:38

Point stands as I see it. Any other thoughts, Petri, anyone? --Caustic Logic (talk) 11:59, 8 September 2015 (UTC)


 * Can we assume all the pilasters are equally thick and at equal distance from the face of the wall? We could compare the width of the the shadow (on the left) to the width of the part of the pilaster that is exposed (on the right). We can correct for the different distances by dividing by the height of the fence from the top of the rusticated part to the top of the fence.
 * "Can we assume all the pilasters are equally thick and at equal distance from the face of the wall?" I don't think so. I'm not sure what you're suggesting even. (BTW, I hope I wasn't too confusing above - I'm pretty sure I get it, but it's hard to explain). Degree of foreshortening vs. actual pixels would say how deep that little triangle is. I'm for ranges anyway, though. This one's bit broad, could stand narrowing... --Caustic Logic (talk) 10:50, 9 September 2015 (UTC)


 * Things that could be done: if we knew the distance between pillars (from sat image of matched site) I could skew a whole wall to accurate props and measure shadow widths. Extrusion could be measured... less well. It will still be rough. But any flat rectangular surface, skewed to its true known proportions, should have be readable like that. But that may be unneeded. We'll see. --Caustic Logic (talk) 12:30, 9 September 2015 (UTC)


 * P.S. - As the lens is rectilinear parallel surfaces are "compressed" the same way. The relationship of the height to width of a rectangle remains the same wherever in the photo it is, left, right or center. -- Petri Krohn (talk) 08:39, 9 September 2015 (UTC)
 * I agree with Petri. As all vertical lines in the image are parallel (they don't converge to a vanishing point), the plane of the camera image is vertical, and so is the corner of the cooler.  If the ground is horizontal, the right angle between the shadow line and the cooler implies that the the plane of the triangle formed by the cooler and the shadow line is parallel to the camera image.  Think of an ideal pinhole camera. So the solar elevation can be read directly.  Pmr9 (talk) 09:52, 9 September 2015 (UTC)


 * Okay, I see two smart people saying something I'm not sure I get, and not sure is even relevant. Thinking about an ideal pinhole camera drew up a blank card saying insert info about ideal pinhole camera here. I can be lame that way. The difference isn't huge, as I read it, but that imaginary sundial direction (green line) wouldn't be and doesn't come out quite visual-horizontal from the corner, just foreshortened like it is. That was my point. Is it even challenged? I'm happy to take that if I can see why. It makes for earlier and better argument, obviously. --Caustic Logic (talk) 10:50, 9 September 2015 (UTC)


 * I do not know if either of you understood me. We can use the height of the wall as an unit length; call it "1 meter", "1 yard", or "1 wall". We then measure the width of the shadow on the left (~ 0.06 walls) and the pilaster on the right (~ 0.08 walls). The triangle 0.06 / 0.08 will give us the angle to the sun. What we measure on the photograph is not the actual width as we are not looking at the wall or shadow at a right angle. But because the surfaces are parallel and the lens is rectilinear and not some fisheye lens, both measurements are "compressed" by the same proportion. (This all relies on the assumption that back wall and the side wall have similar construction.) -- Petri Krohn (talk) 14:31, 9 September 2015 (UTC)
 * First, it's not quite rectilinear anyway, from perspective or lens, I don't know. Rotated so ground is horizontal, the top of the wall isn't quite parallel. Next, I don't know why we need wall height first, nor what ~o,o6 means besides left wall, but just on that side, measuring the distance from pilaster face to the wall's face (depth of triangle) vs. shadow width gives the angle. That was my point to start. I just don't know still how to measure both at the true size (corrected for perspective). And actual units don't matter, so long as we can be accurate about the same units on both walls - it's a simple proportion thing. If you know how, let me know. I'll ponder on it at work tonight. --Caustic Logic (talk) 23:04, 9 September 2015 (UTC)


 * I think what I said above about skewing walls might provide the answer. It looks to me like the pilasters are meant to be the same size all around, or same width anyway. And they seem to be square. So once each wall is skewed right to where the pillars are the same width, the "back stance" and "left" distance of the shadow will be at the same scale, proportion can be set. Back with Photoshop work when it's ready. --Caustic Logic (talk) 10:01, 10 September 2015 (UTC)
 * Bummer. That would work, I think, IF done right. I have a confusing graphic I don't feel like showing that apparently wasn't done right - a wrong presumption or wrong method - seems to show 5 units back to 3 over, to a standard pilaster with of 16-17 units. That gives a triangle 29:45 from apparent, massively foreshortened, yielding elevation 54, time 2:52 or so. But the picture was published a good half hour earlier, so that's not a good center time. And it just looks way steeper than anything close to 45 degrees. Eh, I'm taking a break. --Caustic Logic (talk) 10:54, 10 September 2015 (UTC)
 * Google Sketchup Make (http://www.dummies.com/how-to/content/how-to-set-up-fr-photomatching-in-google-sketchup.html) looks like a useful (free) tool for building a 3D model from a single photo of a scene in which you can assume some angles to be right angles. I think the base angle of the right-angled triangle formed by the cooler and its shadow is by far the best way to estimate the solar elevation: indirect estimates based on shadows on walls will introduce additonal uncertainties.  But to estimate the solar azimuth relative to the courtyard, you need to estimate the angle between the camera image plane and one of the walls.  Pmr9 (talk) 19:59, 10 September 2015 (UTC)
 * I hate new things, and it might say the same (right or wrong) but I should try that this weekend. Glad you agree on that corner - I used wall angles at first and it seemed about the same, but unsure what that angle really is, looking for sundial vertical, space, then clear shadow. Tried pilaster top to shadow, but none were very clear (odd floor) --Caustic Logic (talk) 22:09, 10 September 2015 (UTC)

I didn't download anything new, but thought out how to measure the angle of the cooler's shadow across the ground, in itself. decide both the pillars and spans between are standard with, skew both walls so they share proportions, draw red rectangles one span and one pillar wide, extend the lines out along the ground plane, in red. Then I do the same for the cooler, in orange. Each direction considered from its correct angle, at the same proportions, yields this handy overhead grid (right, bottom).

It's still not exact, but the best method yet. I measure where the shadow falls on the outer orange line, zoomed way in, and draw that angle into the grid in our green. As we can see, it's not much different from the light green apparent line. In fact, sampled onto my original three options from above (right, top), it almost matches the second one perfectly. That was my "what it looks like to me" angle, and it seems to be just about as right as this method.

So with no need for further calculations, we can say the best single time reading is:
 * 66.25 degrees solar elevation = 1:36:30 PM.
 * I'd say it's more likely to be a hair later, best range, simplified: 1:30 to 1:45. --Caustic Logic (talk) 10:32, 14 September 2015 (UTC)

One thing I should have done earlier is get an enhanced view of that shadow edge. Levels adjustment pops out lighter edges of the shadow and shows it extends another pixel back from the evident corner - shifts it a bit closer to the third option. And looking at my grid graphic, it doesn't seem to match this or even the photo really - the gap before the east wall is too wide. The shadow slopes back a bit further, not sure how much. --Caustic Logic (talk) 11:36, 24 September 2015 (UTC)

Azimuth?
Another thing about this green line is shouldn't that be our solar azimuth (compass direction to the sun)? If we knew how far it reaches from the corner, and I estimate (anywhere from 3:8 to unsure) from - whatever. I get measures from known directions on a placed location where we know the angle of each wall, etc. I can say that right hand wall face south, probably a bit southeast. But some nexus of az+elev, that helping narrow down the spot, something in this will pull it all together. I'm not quite in top form here. Anyone else get the spark? --Caustic Logic (talk) 12:22, 9 September 2015 (UTC)

Above was the best yet method for reading time by elevation, but there's a breakthrough on location that lets us measure the azimuth - Petri decided both body collection scenes are at Annatr school, here on Wikimapia, in the northern enclosure (see . I think he's right, definitely on the later collection (west wall) and probably on the earlier one we're studying (southeast corner). This school is about 550 meters due north of impact #4, east of the others, 700 meters from the furthest (park market).

If so, we have our wall orientation, about one degree clockwise from true east-west. This lets me take the last best measure of the triangle base as solar azimuth (from the red ground grid). This gives an azimuth: 201.5, which means it's only 1:09:56. Elevation would be 69.98. That's quite early, but line set-up was never exact. Fudging a bit, 204.5 seems reasonable = about 1:14:40 elevation 68.6. Could be later, or earlier, though earlier seems unlikely.

From elevation, I got a best reading of about 1:37. Fact is, I don't have the kind of precision needed to really pin it down for sure, but we have a range of, say, 1:10-1:40 and we're back to range that includes times probably before the attack. Azimuth is preferable to elevation - though neither is exact, the range should favor the earlier reading - a good center estimate might 1:18 +/- 8 min. --Caustic Logic (talk) 01:05, 20 September 2015 (UTC)


 * As noted above, the grid wasn't set up perfectly, and the shadow stretches back further, closer to the east wall than shown. So, without saying how much, the azimuth will be a bit later than the 201.5. Going back to the three options shown above, I'd say it's between 2 and 3 but closer to 2. Finally putting these into the mep (turned 90, +1 for actual alignment) for azimuth, I get


 * option 2 = 203 degrees = app 1:12
 * option 3 = 213 deg = app 1:29
 * a good guess in this range is 207 = 1:18:45. --Caustic Logic (talk) 12:11, 24 September 2015 (UTC)

Calculation
Now that we have a location and we can see that the east and south walls are (near enough) exactly at right angles to each other, I've had a go at the geometry of the shadow on the back wall. I think this is the best measurement to use for an accurate reading - we can't accurately determine the direction of the cooler shadow.

Write


 * alpha = angle of shadow of south wall on east wall
 * phi = angle of solar azimuth relative to south wall (add one degree to get true azimuth)
 * theta = angle of solar elevation
 * beta = angle between image plane of camera and east wall

Expressions below are in the R language

We can calculate theta from alpha and phi


 * theta = atan( tan(alpha) * cos(phi) ), where atan is the inverse tangent function

We can't directly measure alpha as the image plane of the camera is not parallel to the east wall. The relation of the true angle alpha to the angle alphafalse measured on the image is


 * alpha = atan( tan(alphafalse) / cos(beta) )

Substituting this into the equation above, with angles in degrees (rather than radians), we have


 * theta = 180 * atan(tan(alphafalse * pi/180) * cos(phi * pi/180) / cos(beta * pi/180)) / pi

We can see in this expression that to get the elevation theta, we correct the measured shadow angle alphafalse by multiplying it by cos(phi) / cos(beta). The measured shadow angle (about 67 degrees) is very close to the solar elevation because the corrections for azimuth and camera angle almost cancel out: the plane of the camera image contains the solar azimuth.

To use this expression, we have to estimate beta. Extend the lines formed by the top and bottom of the south wall to the left vanishing point, and extend the lines formed by the top and bottom of the east wall to the right vanishing point. The line joining the two vanishing points is the camera horizon. Measure (a) the distance from the southeast corner to the left vanishing point, and (b) the distance from the southeastcorner to the right vanishing point. The angle between the camera image plane and the east wall is


 * 180 * atan(b/a) / pi

I tried this, with not very accurate drawing tools and got beta = 13.7 degrees. Using alpha' = 67, beta = 13.7, phi = 23.5 (204.5 - 1 - 180, relative to the wall), we get theta = 64.7, which is too low (should be 68.6). More accurate measurement of alpha' would help, but I think this is more consistent with an earlier time (phi close to beta, i.e. azimuth in image plane). Pmr9 (talk) 23:11, 21 September 2015 (UTC)


 * P, thank you so much for this. I hope you can continue to a useable range. Because: I'm not inclined to the math part. This is a chance to follow and learn. But, I'm pressed, other research pendin ... was going to try, but just called in for unexpected work ... seals the deal for me. I'm sitting it out for now ... but next few days, maybe. --Caustic Logic (talk) 09:27, 22 September 2015 (UTC)


 * Okay, I came back to this. I get some of it already. 13.7 degrees from parallel to the east wall sounds about right. As for the shadow angle, you'd correct that using Beta to be the true angle, I suppose, because ignoring the distortion and measuring straight from the pixels seen would give too steep of an angle. I'm not sure about wall shadows though. The sundial method is a vertical line's shadow on a horizontal plane. Here it's the shadow of a horizontal line (top of wall) on a vertical one (the wall/pilaster), and I'm not sure what in the 3-D mix that would measure. BUT, it did and does seem at least close to the elevation angle, so maybe that does work fine. (??)--Caustic Logic (talk) 10:48, 24 September 2015 (UTC)


 * UPDATE: The problem with my earlier calculations was that I couldn't get values of elevation and azimuth that were consistent with the solar calculator. The azimuth estimates indicated an earlier time than the elevation estimates  I've had another go.

With more careful drawing of the vanishing points, I estimate beta, the angle between the camera image plane and the east wall, as 11 degrees. With more careful measurement, I estimate alphafalse, the measured elevation angle of the shadow of the top of the south wall on the east wall as 70 degrees (magnification shows another shadow (?cable) just above the true one). This agrees with a measurement based on the elevation of the shadow on the cooler door (71 degrees), though the cooler door is not quite parallel with the east wall. Putting these together, I can plot a curve for the possible values of solar elevation theta and azimuth (181 + phi) that matches the curve obtained by putting a range of time values into the solar calculator. This curve isn't sensitive to the exact value of beta: that is, it's still a good fit to the solar calculator even if the value of beta is as high as 14 degrees.

To read the time from this curve, we have to put in a value of phi (angle between solar azimuth and the east wall). I think the most accurate azimuth is read from the shadow of the yellow pole to the right of the cooler. If the shadow line were parallel with the image horizon, this would imply that the image plane was parallel with the azimuth. The shadow line is very slightly anticlockwise of the image plane, implying that the solar azimuth is very slightly anticlockwise (viewed from above) of the image plane. So we can put an upper bound on phi as beta, and an upper bound on the solar azimuth as (181 + beta). If my estimate of beta as 11 degrees is correct, this gives azimuth 192, solar elevation 69.8. If beta is 14 degrees, we have theta=195, solar elevation 69.5. This gives an upper limit on the time as 13:00. I'll upload some figures and code to show the calculation Pmr9 (talk) 12:40, 27 September 2015 (UTC)


 * Following somewhat ... this seems like a pretty sound process, but 1:00 or before seems so early it raises problems. Can that be only 11-14 degrees the cooler shadow slants from the perpendicular to the south wall? I was seeing 22 at least. A graphic might help. But if nothing else I can cite your range and mine and between them we're clear it's ... so early it raises problems, one way or the other. --Caustic Logic (talk) 23:54, 27 September 2015 (UTC)


 * I've added a PDF of the photo with vanishing points drawn in but the original is a drawing in .ODG (open document graphic) format that the wiki won't let me upload.  I've also uploaded a figure that shows how the geometry-derived curve of solar elevation against azimuth, based on the measured shadow angle alphafalse and the image plane angle beta agrees with the curve obtained from putting a few values into the solar calculator.   Below is a script in the R language that generates this curve - again I can't upload it as a file.  The key point is that the geometry-derived curve has to fit the curve from the solar calculator: otherwise something is wrong. To get values that The vertical lines in the photo are not quite parallel, implying that the image plane is tilted very slightly back from the vertical.  Strictly the calculation should allow for this, but the correction is probably very small.
 * (hint: I've taken screen captures of Open Office-generated images and saved those as png/jpg and uploaded that - preview quality anyway - will do with your PDF image if that's okay. Optional display, if it's adequate) --Caustic Logic (talk) 11:25, 30 September 2015 (UTC)

alpha.false <- 69.8 # angle of shadow on cooler door

beta <- 180 * atan(45/236) /pi # estimate from inverse tangent of ratio of distances to left and right VPs

beta <- 14 # more conservative estimate (consistent with later time)

phi <- 181:220 - 181 # limits of plausible range

theta <- 180 * atan(tan(alpha.false * pi/180) * cos(phi * pi/180) / cos(beta * pi/180)) / pi

png("Douma_solar.png")

plot(phi+181, theta, xlim=c(180,220), ylim=c(64, 71), col="blue", xlab="Solar azimuth", ylab="solar elevation", main="Relation of solar elevation to azimuth", )

points(208.03, 68.05, col="red")

points(194.98, 69.62, col="red")

points(194.98, 69.62, col="red")

points(213.51, 67.03, col="red")

points(184.49, 70.15, col="red")

legend(x=c(180, 180), y=c(66, 65), legend=c("curve calculated from image plane and east wall shadow angle", "curve from NOAA solar calculator"),       col=c("blue", "red"), pch=c("o", "o"))

dev.off

Pmr9 (talk) 21:22, 29 September 2015 (UTC)

Videos
These videos are from the same place: -- Petri Krohn (talk) 02:06, 17 September 2015 (UTC)
 * https://www.youtube.com/watch?v=q2R2Pseg7YM
 * https://www.youtube.com/watch?v=UMFrqsvE__8

Update: There is one more video from the same place. This one is from the same corner with the yellow steel structure. Details, like the checkered tiling of the floor allow us to conclude all these videos + photo are from the same courtyard. -- Petri Krohn (talk) 01:59, 19 September 2015 (UTC)
 * https://www.youtube.com/watch?v=ba0vDB5GieM (In playlist)
 * Mirror: https://www.youtube.com/watch?v=4s8dvbk37TQ


 * Further, Orient News 1:02 scene really clarifies that point. --Caustic Logic (talk) 23:29, 19 September 2015 (UTC)

Location?
I think the place is the northern half of Annatr School / مدرسة العناتر (دمشق الكبرى) which is here on Wikimapia. -- Petri Krohn (talk) 18:45, 17 September 2015 (UTC)
 * I think you hit the nail on the head, for the other spot the bodies were seen. The little square reed-roofed (snack shack?) is a good fit, there's a gate in this west-facing wall, etc. But that's site 2, tile floor. This photo is at site 1, dirt floor, with the right-hand wall facing south or I think southeast, the left-hand wall then east or northeast. I think it'll be hard to place, haven't really tried (priorities). I was sure they were different places, but it hit me they could be different walls of the same place. But no such walls here, so it is different. One more thing learned. --Caustic Logic (talk) 10:20, 18 September 2015 (UTC)
 * This photo is taken in the southeast corner of the northern enclave of the school compound. In fact there seem to be two totally separate schools or whatever public buildings in the same block, but Wikimapia marks them as one. -- Petri Krohn (talk) 21:46, 18 September 2015 (UTC)
 * Ah, the other corner - could be, if the tile stops. Google Earth shows a blue shape along that south wall that changes size, could be the blue tarp wall we see (until 2013, was a second red-roof shack). Overhanging tree at corner, check. Cooler, unclear - tile seems continuous in 2012 and before, and now it also seems pretty continuous color all across. So, not convinced but it could be. Worth checking what the azimuth would be. will be back. --Caustic Logic (talk) 22:53, 18 September 2015 (UTC)
 * Basic measure, taking the walls as square to north (pretty darn close) with shadow graphic rotated in, says directly azimuth 200.5, which means it's only 1:08:20. Elevation would be 69.09. I doubt this is right - it should probably be at least after the attack news. Unless this was a White Helmets whistleblower photo. --Caustic Logic (talk) 23:05, 18 September 2015 (UTC)
 * (that complicates stuff above, needs addressed there, end side-track) --Caustic Logic (talk) 13:10, 19 September 2015 (UTC)


 * Petri, I'm coming around. the third video you shared, DE1 on my list, does show the same dirt-floor scene and also clearly enough shows similar tile stopping along its edge, continuing to the west. A clear video showing both scenes would help but I'm pretty sure this is all the same spot. I'm digesting and working it in. --Caustic Logic (talk) 13:10, 19 September 2015 (UTC)


 * Annatr school ( مدرسة العناتر ) video, minor damage in 2012 to southern part (girls school?), helps little. I was hoping to find a news story that anyone (opp. would say regime, and prob. not these days) had used the school as a prison. That happens, and if it's the case here, might suggest they were only trucked out that day, not in and then out. --Caustic Logic (talk) 23:22, 18 September 2015 (UTC)

Date?
We now know the time and place of the photo. We would still need to verify it is from the same date. It could as well be some old White Helmets photo that was just released on August 16th to illustrate the massacre.

We would need to show that the bodies are the same or we see the same people wearing the same clothes. I think the man in the shirt with yellow stripes is also seen at the end of Orient News video, this time wearing a White Helmets vest on top of the shirt. It is also possible that the man seen at 40 seconds into the video wearing a key chain and a watch is in the photo with his head covered by the WH logo.

There are changes in the trash and junk surrounding the bodies. Also, the sunshade tarp is broken in the photo but seems to give a rectangular shadow in the video. -- Petri Krohn (talk) 02:58, 20 September 2015 (UTC)


 * My feeling is no way is all this recycled. It's good to be thorough, and I think the one guy is probably the same. I may check more later. Bodies are the same ... as what? The ID videos? Those could be recycled too. I've kept an eye out for any match, but not really closely and didn't notice any yet.


 * There was a similar batch of bodies, I think a smaller and bloodier batch, in the same or very similar place, in Twitter photos of the 12th, after another market massacre. I saved a few links. One anyway (copy) Pretty sure I've seen that recycled as the 16th, but it seems to be a different array. And of course I've seen no photoas matching the scene under study at any prior time. So best guess is it's a new scene that day (in case we get no further). --Caustic Logic (talk) 23:37, 20 September 2015 (UTC)


 * Here is a video from the August 12th incident showing the same bodies as in the tweeted photo. This is well before noon, so it is unlikely to be the same as our photo. The bodies are already in "professional" plastic body bags with zippers and peek windows. Maybe they were bought in this way. Most of the August 16 victims are wrapped in simple white sheets.
 * I tried to look for something earlier. Here is one video June 16th that may be from the same place. -- Petri Krohn (talk) 01:25, 21 September 2015 (UTC)