[QUOTE=PhilF;530393]A good discussion of diffraction limit can be found here[/QUOTE]May I suggest that you research the term "super-resolution"? Here is just [URL="https://en.wikipedia.org/wiki/Super-resolution_microscopy"]one link[/URL].

In a closely related field, image deblurring, I have some practical experience. In [URL="https://britastro.org/node/19348"]this example[/URL] and [URL="https://britastro.org/node/19566"]this[/URL] the blurring is caused by the convolution of the point images with the diffraction limit of the optical train and the refraction of the atmosphere between the stars and the telescope.

Contrary to popular opinion, image reconstruction is limited primarily by lack of knowledge of the optical transfer function and by noise in its multitudinous forms.

Here's an exercise for you to perform with the aid of something like Mathematica or Octave. The diffraction pattern of a point source viewed through a circular aperture is a [URL="https://en.wikipedia.org/wiki/Diffraction#Circular_aperture"]messy function[/URL], [I]I[/I], parameterized by the radius of the aperture, [I]d[/I], in units of the wavelength of the light [I]lambda[/I] (assumed monochromatic for this exercise) used to form the image. Compute and plot the image of two identical point sources (i.e delta functions) separated by, say, 0.1 [I]lambda/d[/I] radians, which is well below the diffraction limit. The image is just the sum of two [I]I[/I] functions.

Next, compute and plot the difference of the two functions. You will find that it is not zero. The reason you can say that is because you know the point spread function perfectly and your have computed its convolution with the two delta functions to an accuracy much greater than the noise level which, in the example, is pure quantization noise.

If, as a follow-up exercise, I were to give you the sum of two such functions corresponding to an image made with the same equipment but no information about the separation of the two possibly dissimilar delta functions, you should be able to perform least-squares fitting, say, to estimate the unknown values, [i]specifically because[/i] you know the extent to which the data is not modelled by a single [I]I[/I] function.

[QUOTE=retina;530373]Fortunately the exif metadata[/QUOTE]
No metadata. I have no idea what a file is and I can not open the file or save it, I only have the image on screen and can do nothing with it beside zooming into it infinitely and measuring distances on it in some arbitrary units (say, solar diameters) with infinite precision. Can I pinpoint exactly on the earth where the photo was taken from? I don't know the time the photo was taken, but I can calculate the position of celestial objects at a given time with infinite precision (which is actually not far from the truth, in reasonable limits, this is what a GPS is doing).

Then, my feeling is that I can pinpoint the exact position, and also calculate the time the photo was taken, very accurate, from all the puzzle (in fact, the two are dependent, one would give the other). The key here may be the fact the Earth and Mercury orbits are not perfect co-planar, and the black dot is not on the solar equator. And you will have to prove me wrong...

Edit: And yes, retina, I was thinking EXACTLY about pinpointing where your lair is! :razz:

[QUOTE=LaurV;530458]No metadata. I have no idea what a file is and I can not open the file or save it, I only have the image on screen and can do nothing with it beside zooming into it infinitely and measuring distances on it in some arbitrary units (say, solar diameters) with infinite precision.
Edit: And yes, retina, I was thinking EXACTLY about pinpointing where your lair is! :razz:[/QUOTE]

Oops- Sorry; was going to post the EXIF for you. Here we go-

Norm

And, as I think about it- the image posted was cropped, so the size is no longer what is shown in the EXIF file. The important info is the time (UT minus 7 hours), and since I have an app on that phone that synchronizes with "atomic" time, from the National Bureau of Standards I think, the time should be good within a second or two, at most.

[QUOTE=Spherical Cow;530469]Oops- Sorry; was going to post the EXIF for you.[/QUOTE]
Haha, I don't know if this was a joke or you took my poor sentence in the total opposite way. My meaning was that I should only use the picture, and no metadata, nor file, nor time information. I am talking about a hypothetical situation in which I have a very high resolution bitmap with your picture, I am trying here to make a puzzle which I have the feeling it can be positively answered, but I don't know how to solve.

Gnats. I was hoping that with the image and EXIF file alone, you would show me how to calculate where the picture was taken from, street address, computer IP address, make and model of my car, and where I left my car keys.

I'll keep working on the puzzle, too.

Norm

[QUOTE=LaurV;530458]Edit: And yes, retina, I was thinking EXACTLY about pinpointing where your lair is! :razz:[/QUOTE]You will have to wait for me to post a photo of some celestial objects taken from my lair. But don't hold your breath waiting, I don't have a camera.

[size=1]And, no, I still think you need the time stamp, else you'll only narrow it to a line, not a point.[/size]

[QUOTE=retina;530485][SIZE=1]And, no, I still think you need the time stamp, else you'll only narrow it to a line, not a point.[/SIZE][/QUOTE]
That would be true if the Mercury and the Earth would have coplanar orbits. But they are not perfectly coplanar, and you see Mercury at some distance below the Sun's equator. My felling is that knowing the approximate time, and having this distance, you can compute the exact time. The rest is just a simple geometry problem. But reality is more complex. Assuming the Sun is fixed, you have Mercury rotating around it on an ellipse, say a perfect one, then the Earth rotating around the Sun on a (perfect) ellipse, but slightly out of plane, then the Earth rotating around itself in a totally out of plan perfect circle (i.e. the point you are on, describes a circle, but in fact, that is a "screw", because the earth itself moves). The "shadow" of Mercury on the Sun from your point of view (I use quotes, because the real shadow is on the other side, because the light comes from the Sun) will describe a "screw" on the Sun's surface, which has different position/coordinates on the Sun's surface. The question is if you are on point A on Earth, and I am on point B on Earth (reasonably close to A), how much "your screw" will intersect with "my screw" :razz: on the Solar surface. If they intersect only in few points, we may talk. If they intersect a lot... bummer...

[QUOTE=LaurV;530821]That would be true if the Mercury and the Earth would have coplanar orbits. But they are not perfectly coplanar, and you see Mercury at some distance below the Sun's equator. My felling is that knowing the approximate time, and having this distance, you can compute the exact time. The rest is just a simple geometry problem. But reality is more complex. Assuming the Sun is fixed, you have Mercury rotating around it on an ellipse, say a perfect one, then the Earth rotating around the Sun on a (perfect) ellipse, but slightly out of plane, then the Earth rotating around itself in a totally out of plan perfect circle (i.e. the point you are on, describes a circle, but in fact, that is a "screw", because the earth itself moves). The "shadow" of Mercury on the Sun from your point of view (I use quotes, because the real shadow is on the other side, because the light comes from the Sun) will describe a "screw" on the Sun's surface, which has different position/coordinates on the Sun's surface. The question is if you are on point A on Earth, and I am on point B on Earth (reasonably close to A), how much "your screw" will intersect with "my screw" :razz: on the Solar surface. If they intersect only in few points, we may talk. If they intersect a lot... bummer...[/QUOTE]You know where you are. The orbits of the earth and of Mercury are known to very high precision and the data is freely available.

Why don't you just compute the track of Mercury across the Sun from a sampling of locations and times and see which matches the observation?

You very much overestimate my abilities...

But would you know exactly which angle the camera was at when it took the picture? Unless Mercury was exactly in the centre of the Sun's disk you wouldn't be able to work out a location without that.

Of course you can safely assume it was taken from somewhere where the transit was visible. Just as a post saying someone was clouded out says they were in such a place.

Chris