How the angle of the camera would have any influence? It would not.

What put me off initially was the rotation of the Earth around itself. That is because Earth is selfish.

Then I took a pencil.

Assume for a blink that the earth does not rotate, but it is just a dot, i.e. the observer is in the center of the Earth. You know exactly the movement of Earth and Mercury around the Sun. If you draw a line through the center of the two planets, this line will intersect (or not) the visible surface of the Sun in a point, which I formerly called "shadow". For the Earth observer, he does not see the Mercury, he only sees the Shadow describing a curve (in time) on the surface of the Sun, which is, for the purpose of this puzzle, just a fixed yellow disk. Earth and Mercury have different speeds, and different revolution planes. This makes the respective curve unique, because you know the approximate time when the transit took place (we see it as a line from west to south-east, but it is not a line, however I don't know what it is... how does one sees an ellipse if he rotates on another ellipse out of it? is that also an ellipse? or it is a parabola? hyperbola? some cycloid/trochoid/roulette curve?). Therefore, any point on this curve gives you the accurate time, assuming that you can "zoom in" to see exactly the coordinates of the point. Only at that particular time you would have _that_ position of Earth and _that_ position of Mercury to give you _that_ position of the shadow. No question about it.

One microsecond earlier, or later, the shadow would have moved a pico-pixel, because Mercury revolves faster than the Earth.

But now, you are not at the center of the Earth, you are on its surface, and the Earth has a bad and selfish habit of rotating around itself, moving you up and down (due to the fact that the rotation axis is inclined), which makes the shadow move in the opposite direction... down and up... like a swing balancer, anchored on Mercury, one end (you) moves up (as the earth rotates), the other end (the 'shadow' as defined above) moves down.

So, the curve that the shadow describes now is a screw, or [URL="https://en.wikipedia.org/wiki/Helix"]thread[/URL], overposed over the ellipse/line we see. But how long the transit takes? If longer than a day, then the thread [URL="https://en.wikipedia.org/wiki/Screw_thread#Lead,_pitch,_and_starts"]pitch over lead[/URL] is a small number, eventually under-unity -- in that case, we are "screwed", sorry for the pun, because different times of the day may give the same position of the dot (see projections of a helix). But if the transit is shorter than a day, which it is, then the earth rotation will only accentuate the inclination of the initial curve, a little bit.

So, a guy on point A on the Earth will see one screw, another guy on point B will see another screw. These screws intersect inevitably, because they all rotate around the same ellipse. Or line. Or throchoid. Or whatever that is. This means that there are points (zones) of "incertitude", guy in point A will see the shadow at moment X of time in the same place as guy on point B will see it at moment Y of time. So, one could not say where the guy was, unless the time is known.

We are screwed...

The camera angle I'm referring to is if it's been rotated about a line from it towards the Sun. If all you get is a picture of the Sun's disk with Mercury as a spot somewhere inside the disk but don't know which direction the line from the centre of the disk to the spot is relative to the plane of the Earth's orbit round the Sun how can you work out where the picture was taken from?

If you have two pictures at different times you can probably work out which line the observer was on. With three or more you might be able to get their location on the Earth.

Chris

Good point. It even didn't occur to me that the camera may not be perfectly horizontal :sad:

[QUOTE=LaurV;530956]<...snip...> [you are] in the center of the Earth <...snip...> [and] draw a line through the center of the two planets, this line will<...snip...>[/QUOTE]
...this line will go exactly through your a$$. (comment from a friend).

As I said, we are screwed... :rofl: