Find someone else to answer that question. I have a method to show you, if you want to hear about it fine, drop all the other conditions and let's just stick to this method. You say you want simple and then you insist on trying to complicate everything.
Come on, get on with it will you!
Ok then let's get on with it. Bit by bit. one small piece at a time with explanations of how and why, before we even try to move on.
Off you go then.
OK, so I think (difficult to be sure) that you are fine with step 1 - two people taking simultaneous photos of the moon. You'll have to correct me if I'm wrong about that.
Step 2.
First part is to figure out the linear size of a pixel on the camera's sensor. The specs for some cameras quote that figure directly, but if not, take the physical width of the sensor (usually given in mm) and divide by the number of horizontal pixels. Since this number is a fraction of a mm, you can convert to µm if you want as the numbers are more convenient.
Second part of this is to work out the angle represented by one pixel. This depends on how much you are zoomed in. The more you are zoomed in, the smaller this is.
The calculation is pixel size / focal length of lens. The result is in radians, so either keep it in radians or convert to degrees, or since the numbers are very small, convert to arcminutes or arcseconds for convenience. The units for pixel size and focal length have to match so either use mm for both or µm for both.
That's step 2. OK with this or do you have questions? Please let's keep this on track so questions should be about the method and not diving off into some other topic entirely.
Ok, so just to be clear about this camera and pixel stuff, before we go any further....how much...in size... do the pixels change from your standpoint to, say.....10 miles away full zoom?
Once you answer that can you then equate that to 240,000 miles away zoom, as your moon apparently is....or am I not getting this?
Not sure I entirely understand the question, but I'll have a go at explaining this a bit more, see if it helps.
This example image shows the moon taken with a zoom lens at 400mm, 500mm and 1000mm settings.
The image is 440px wide and let's assume that's not been cropped, so is the whole width of the sensor. It's a real low res camera in this example. Let's assume the sensor is a DX format, so say 24mm wide.
Each pixel is therefore 24/440 mm wide (approx 0.066mm). That pixel size is fixed. It can't change no matter what zoom setting you use. It's a physical pixel on the sensor.
What does change as you zoom is the angular size of each pixel. That is to say the amount of sky each pixel captures. Zoom in and you are capturing a smaller part of the sky, i.e. more detail, with each pixel. To calculate that value, divide pixel size by focal length. So for example for the 400mm zoom, that's 0.066/400 = 0.000136364 radians or 0.007813061 degrees. Do the same calculation for the others and you get 0.006250449 (500mm zoom) and 0.003125224 (1000mm zoom).
Notice how the angular pixel size has gone down from roughly 0.007, to 0.006 to 0.003 as the focal length has increased.
Now we can use these to work out the angular width of the moon. Just multiply the angular size of a pixel by the number of pixels.
At 400mm, that's 0.007813061 x 65 = 0.508 degrees.
At 500mm, that's 0.006250449 x 81 = 0.506 degrees.
At 1000mm, that's 0.003125224 x 160 = 0.500 degrees.
There's a little bit of variation because the image is very low res so the pixel widths aren't that accurate, but the point is, it's saying the moon is 1/2 degree wide no matter whether it's zoomed in or zoomed out. Which is correct.
Once we've done this pixel size calculation, if we know the focal length we're using we can convert pixel distances on the image to angular distances. That's the point of this step.
Ok, so what is the width of the moon?
Ok as in ok with this step, no more questions on the method?
Honestly, why not try giving clear and simple unambiguous answers. All I need to know at each stage is, are you clear about the method so far or do I need to explain further.
The moon is approximately 1/2 degree wide.
Ok, carry on.
Step 3:
Take the two photos. Use some photo editing software, such as PhotoShop. Overlay the images. Slide and rotate the two moon images until they sit on top of each other with all the features (craters etc.) lining up.
Remember that both images also include at least one nearby bright star (let's call this our reference star).
The reference star will appear to have shifted position, so the merged image will contain two images of the reference star. In fact all the stars in the image will all appear to have shifted by the same amount in the same direction.
Since we know the stars haven't moved, we conclude the moon must have (or at least appeared to).
How much did the moon appear to shift between the two images? Well if we measure the distance (in pixels) between the two different positions of the reference star, then that must be how much the moon has shifted.
Think of it this way. Put a chair next to a table in an otherwise featureless room. Take a photo, centred on the chair. Move the chair away from the table. Take another photo, again centred on the chair. The two photos suggest the table has been moved, but we know it's the other way around. By measuring the amount the table appeared to move, we then know how much the chair actually moved.
We now know by how many pixels the moon apparently shifted. Since we also have a method to convert pixel distance to angular distance, we can now calculate the angle the moon has shifted by.
That's step 3. OK with this step?
NB: to avoid me having to ask you to clarify your answer to this, if/when you are OK with this step, how about you just say "Yes, I'm OK with this step" or something very similar. Obviously if you have questions, go ahead and ask.
Not really ok, no.
What would happen if the point of light ( your star) in your picture, near your moon is very close to your moon and looking like it does to you, rather than what you're told in terms of light years away?
How does this marry up with pixels?
It's a fair question. If the star was close then it too would apparently move position. If the star was the same distance as the moon, then it would move with it and the relative distance between the two would not change. Since stars are sometimes occulted by the moon (i.e. the moon passes in front of a star) rather than the other way around, we at least know that the stars are further away than the moon.
Now consider two observers at the same latitude. We know that Polaris is less than a degree from due north and that doesn't vary no matter where you observe it from. Similarly for our two observers at the same latitude, Polaris is always the same altitude. What this means is that for these two observers, Polaris is completely fixed in place, no matter how far apart the observers are.
We can then determine the positions of all the other stars relative to Polaris and we find these relative positions are also fixed. The positions of all these fixed objects in the sky are given coordinates analogous to latitude and longitude. These are right ascension (RA) and declination (DEC). You can look these coordinates up in an atlas.
If you want to find some fixed object in the night sky, find an identifiable bright star nearby, point your telescope at it and then alter the telescope settings to match the known RA/DEC coordinates of your bright star. Then point the telescope to the RA/DEC coordinates of the object you are trying to find and if your telescope is properly set up, it should be right there in the viewfinder. This is how we find things in the night sky and demonstrates that the fixed objects are indeed fixed and don't change position no matter where the observer is.
If stars shifted their positions for different observers, then RA/DEC coordinates would vary for each observer and everyone would need their own personalised atlas.
A good way to imagine this is to pretend (note - this is pretend) that there is an invisible, absolutely huge sphere with the earth at the centre. All the stars are nailed to the inside of this sphere and it rotates around an axis once a day. The moon and planets move relative to this sphere, the stars do not.
Since we know the stars' positions are fixed for all observers, the stars provide a fixed background and therefore it has to be the moon whose apparent position has changed and not the star.