Mercury's transit
 

Will the transit of Mercury next month be visible to the unaided eye ?
I would be using the glasses I used for the last solar eclipse a short time ago
and, of course, assuming no clouds. I have seen video on youtube, and it looks like one would
need at least good binoculars. TIA

[QUOTE=tServo;528217]Will the transit of Mercury next month be visible to the unaided eye ?
I would be using the glasses I used for the last solar eclipse a short time ago
and, of course, assuming no clouds. I have seen video on youtube, and it looks like one would need at least good binoculars. TIA[/QUOTE]It's undoubtedly not visible to my unaided eye. I'm severely myopic.

Assuming you have typical vision, you can probably resolve something about an arc minute across. The diameter of the sun and that of Mercury, in kilometers, are easily available from Wikipedia. I'll throw in the angular diameter of the sun as seen from the earth for free --- it's 30 arc minutes to an adequate degree of accuracy. To adequate accuracy, the relative distances between the sun, Mercury and the Earth can also be obtained from Wikipedia.

All the rest is left as a simple exercise in geometry for the reader. You didn't expect me to do [B]all[/B] your work for you, did you?

Just to be clear, I do know the answer.

[QUOTE=tServo;528217]Will the transit of Mercury next month be visible to the unaided eye ?
I would be using the glasses I used for the last solar eclipse a short time ago
and, of course, assuming no clouds. I have seen video on youtube, and it looks like one would
need at least good binoculars. TIA[/QUOTE]

Looking into the sun is fraught with danger, more so using binoculars! Do you want to be blinded? I suggest projecting the sun's image onto a a piece of paper, maybe with a simple pinhole camera.

[QUOTE=paulunderwood;528221]Looking into the sun is fraught with danger, more so using binoculars! Do you want to be blinded? I suggest projecting the sun's image onto a a piece of paper, maybe with a simple pinhole camera.[/QUOTE]True. However, if you use a filter suitable for the task, it's fine.

I've observed the sun many times through 15x80 binoculars, including the 1999 98% partial eclipse visible from Oxford.

If you don't know what constitutes a suitable filter, don't observe the sun through a telescope or binoculars.

[QUOTE=paulunderwood;528221]I suggest projecting the sun's image onto a a piece of paper, maybe with a simple pinhole camera.[/QUOTE]

That worked for Venus, but Mercury is too small. Its angular diameter will be five times smaller than Venus, so the dot will have 25 times less area.

Compare these images:

[url]https://en.wikipedia.org/wiki/Transit_of_Venus#/media/File:20040608_Venus_Transit.JPG[/url] (Venus)

[url]https://en.wikipedia.org/wiki/Transit_of_Mercury#/media/File:Transit_Of_Mercury,_May_9th,_2016.png[/url] (Mercury)

[QUOTE=GP2;528224]That worked for Venus, but Mercury is too small. Its angular diameter will be five times smaller than Venus, so the dot will have 25 times less area.

Compare these images:

[url]https://en.wikipedia.org/wiki/Transit_of_Venus#/media/File:20040608_Venus_Transit.JPG[/url] (Venus)

[url]https://en.wikipedia.org/wiki/Transit_of_Mercury#/media/File:Transit_Of_Mercury,_May_9th,_2016.png[/url] (Mercury)[/QUOTE]

I see. Or rather I did not! The image of Mercury's transit was indistinguishable between the dots of filth on on my 1080p monitor, :smile:

[QUOTE=xilman;528220]IAll the rest is left as a simple exercise in geometry for the reader. You didn't expect me to do [B]all[/B] your work for you, did you?
[/QUOTE]

You can also look up on heavens-above that the Mercury diameter is [SPOILER]9,95 arc seconds[/SPOILER] on Nov 11th.
But it also says the Mercury distance is 0.676 AU that day, and wikipedia informs us Mercury's radius is 2439.7 km, so we can double check:

[SPOILER]Mercury diameter: 3600 arc seconds / degree * 2 * arctan( 2439700 m / (0.676 AU * 149597870700 m/AU)) ~ 9.95 arc seconds[/SPOILER]

The Sun distance is 0.9899949 AU that day and the Sun radius is 696,342 km:
[SPOILER]Sun diameter: 60 arc minutes / degree * 2 * arctan( 696342000 m / (0.9899949 AU * 149597870700 m/AU)) ~ 32.33 arc minutes[/SPOILER]

On May 9, 2016 (previous transit) I was in Escalante, Utah. I didn't even know it was going to happen. The National Park Service visitor center had a little viewing setup (maybe a small Celestron with appropriate filter?) and it was an interesting experience to see this.

Here is the [URL="https://en.wikipedia.org/wiki/Transit_of_Mercury#/media/File:Transit_of_Mercury,_2006-11-08_2.jpg"]Wikipedia[/URL] image of that event.

During the 2012 Venus transit, I was barely able to see Venus through just eclipse shades. I had to really squint to see it. And I couldn't see any of the sunspots.

Once I projected it through binoculars onto a piece of paper, both Venus and the sunspots were clear as day.

No way in hell would I be able to see Mercury - which would be even smaller than the sunspots.

[QUOTE=Mysticial;528327]Once I projected it through binoculars onto a piece of paper, both Venus and the sunspots were clear as day.[/QUOTE]

I tried that method with Mercury in 2016 and couldn't see it. It worked with Venus in 2004 and 2012.

The dot for Mercury is so small that you probably need the focus to be exact, and that's hard to do when you're projecting onto paper.

When I was a kid, sunglasses were referred to as smoke glasses in Persian (now they are called Sun glasses). At the time, people would use the black residue of smoke from a candle (Soot) to darken transparent glass to look at a solar eclipse. I assume this is the origin of the prank were someone is given eye glasses which leave dark rings around the eyes when removed.
In any case if you use smoked glass you still risk eye injury if you don't smoke the glasses dark enough to be safe.

Sic transit gloria Monday,
 

Completely clouded out.

Exactly what one should expect in this ${DEITY}-forsaken country.

[QUOTE=xilman;530322]Completely clouded out.

Exactly what one should expect in this ${DEITY}-forsaken country.[/QUOTE]Edmond Halley had a similarly gloomy view of the weather on the island of St. Helena, where he spent two years observing the heavens. In [url=https://halleyslog.wordpress.com/2014/03/28/return-to-st-helena/]Return to St Helena[/url] we find[quote]In November 1677 Halley wrote to Jonas Moore that:
[indent]such hath been my ill fortune, that the Horizon of this Island is almost always covered with a Cloud, which sometimes for some weeks together hath hid the Stars from us, and when it is clear, is of so small continuance, that we cannot take any number of Observations at once; so that now, when I expected to be returning, I have not finished above half my intended work; and almost despair to accomplish what you ought to expect from me.[/indent][/quote]

He was, however, able to observe the transit of Mercury on 28 October 1677.

[QUOTE=xilman;530322]Completely clouded out.

Exactly what one should expect in this ${DEITY}-forsaken country.[/QUOTE]

Partially clouded out here in desert, but the best pictures were the ones with the clouds, actually, as there were no interesting sunspots. Nor any uninteresting sunspots, even. Cellphone camera again, so I'm sure you'll see many higher resolution images on the internet.

Better than waiting until 2032.

Norm

[QUOTE=xilman;530322]Sic transit gloria Monday[/QUOTE]
Correct is: sic transit Gloria Gaynor. Or, in this case, sic transit gloria alien, aaa... sorry, mercurian...

[YOUTUBE]ybXrrTX3LuI[/YOUTUBE]

[QUOTE=Spherical Cow;530349] the best pictures were the ones with the clouds, [/QUOTE]
That photo is not good, because the sun has a black dot in the middle.... :cmd:
I somehow imagine some communist censor publishing it in a scientific magazine, or a newspaper, but not before fixing the inaccuracy of the film...

Now, joking apart, assuming I have an infinite-resolution of that photo, could I guess/calculate your exact location on earth, where the photo was taken, without knowing the time it was taken at? (In theory this would be possible, I think)

[QUOTE=LaurV;530359]That photo is not good, because the sun has a black dot in the middle.... :cmd:
I somehow imagine some communist censor publishing it in a scientific magazine, or a newspaper, but not before fixing the inaccuracy of the film...

Now, joking apart, assuming I have an infinite-resolution of that photo, could I guess/calculate your exact location on earth, where the photo was taken, without knowing the time it was taken at? (In theory this would be possible, I think)[/QUOTE]

Hmmm. Interesting; I'll have to think that through- my first guess would have been that we would need to know the time to calculate the location on earth. Plus, we would probably have to make assumptions about distortion of the image. The optics are not likely perfectly aligned- this is a cellphone mounted tentatively over the eyepiece, with cellophane tape and rubber washers as spacers; a [URL="https://en.wikipedia.org/wiki/Rube_Goldberg"]"Rube Goldberg"[/URL] approach (. (Although the cellophane tape has been carefully calibrated...). Guess I better measure the solar image first, to see if its actually circular.

Norm

[QUOTE=LaurV;530359]Now, joking apart, assuming I have an infinite-resolution of that photo, could I guess/calculate your exact location on earth, where the photo was taken, without knowing the time it was taken at? (In theory this would be possible, I think)[/QUOTE]I think the best you could expect is to narrow it down to a line crossing the Earth. To pinpoint the precise position on the line you need the exact time. Fortunately the exif metadata does have the time (but it doesn't have the GPS coords), and assuming the time is suitably accurate then you could compute the position.

[QUOTE=LaurV;530359]That photo is not good, because the sun has a black dot in the middle.... :cmd:
I somehow imagine some communist censor publishing it in a scientific magazine, or a newspaper, but not before fixing the inaccuracy of the film...

Now, joking apart, assuming I have an infinite-resolution of that photo, could I guess/calculate your exact location on earth, where the photo was taken, without knowing the time it was taken at? (In theory this would be possible, I think)[/QUOTE]

Because of the principle of the diffraction of light, there isn't any such thing as infinite resolution.

[QUOTE=PhilF;530387]Because of the principle of the diffraction of light, there isn't any such thing as infinite resolution.[/QUOTE]Are you sure about that?

I'm not.

[QUOTE=xilman;530392]Are you sure about that?

I'm not.[/QUOTE]

A good discussion of diffraction limit can be found here:

[url]https://books.google.com/books?id=1g_OBQAAQBAJ&pg=PA282&lpg=PA282&dq=lens+%22infinite+resolution%22&source=bl&ots=7NqAoSBuhc&sig=ACfU3U08Ba8Fja1j5wHyGoqdund3B7vJew&hl=en&sa=X&ved=2ahUKEwir2riSweXlAhVPqZ4KHWSqC6kQ6AEwDHoECAcQAQ#v=onepage&q=lens%20%22infinite%20resolution%22&f=false[/url]

Video from NASA's SDO:
[url]https://vimeo.com/372671661[/url]

[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

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: