Helping out a numpty!

[bingevader]

1 hour ago, ollypenrice said:

He's here, on top of the Rocher St Michel above Orpierre. He's reflecting on the width of the rocky ridge which leads off it and wondering if it might somehow be magnified...

 

Now that I'm down I simply agree that the apparent size goes as the inverse square of the distance. In the other thread (about 'magnification' in imaging) my point was that in visual astronomy we have something to magnify, namely the image projected onto the retina. We still have that here, an image of part of the moon, so yes, we can reasonably state a 'distance equivalence' for a magnification by a telescope. I can't see anything wrong with that. It won't work for stars because any angular size they possess in amateur telescopes is an artefact of the optics. They should be points. (Can we see points? I've no idea!)

Olly

Thanks for the clarification.

Took me a while to realise that was an arrow pointing to your location, not a stick-man of you after a nasty bicycle accident. ;D

[Ricochet]

12 hours ago, furrysocks2 said:

What have I got wrong here, then...?

Well...

12 hours ago, furrysocks2 said:

I went on the assumption that at 1x magnification, the moon would have an apparent angular diameter of 0.5183 degrees. At mag 2, this doubles, at 10x mag, it has an apparent angular diameter of 5.183, etc.

You have used the idea that Magnification = X / Y where X is the apparent diameter in the telescope and Y the apparent diameter with the naked eye, which will be fine for small angles, however, according to Wikipedia:

Quote

For optical instruments with an eyepiece, the linear dimension of the image seen in the eyepiece (virtual image in infinite distance) cannot be given, thus size means the angle subtended by the object at the focal point (angular size). Strictly speaking, one should take the tangent of that angle (in practice, this makes a difference only if the angle is larger than a few degrees). Thus, angular magnification is given by:

where  is the angle subtended by the object at the front focal point of the objective and  is the angle subtended by the image at the rear focal point of the eyepiece.

Now in my understanding "subtends" means edge to edge, i.e the angular diameter of the moon, but we've got tan functions in there which is going to limit us to angles <90° which I'm sure is something that going to upset all those people who have bought 100° eyepieces. As a result I'm going to take the unusual step of quoting an equation and then immediately insisting it must be wrong. Specifically, I'm going to declare that it must relate to the angle from one edge to the optical axis, i.e the angular radius of the moon, which will then give us angular diameters between 0° and 180°. For the sake of keyboard typing sanity I will rewrite the above equation as :

M = tan (X) / tan (Y)

Where X is the apparent radius through the telescope and Y through the naked eye.

So the angular radius of the magnified image must be:

X = atan (M * tan Y) [1]

12 hours ago, furrysocks2 said:

I then took "apparent diameter / 2 = tan ( radius / distance )" and rearranged to "distance = radius / atan( apparent size / 2 )"

This is also incorrect. The standard equation is:

tan Ø = opposite / adjacent 

and substituting in your terms we get:

tan (apparent diameter / 2) = radius / distance

so

distance = radius / tan (apparent diameter / 2) = radius / tan (apparent radius) [2]

Using equations [1] and [2] allows us to create the following table:

The pink columns show the discrepancy between using equation [1] to calculate the apparent diameter in the eyepiece with just multiplying the magnification by half a degree. A bit of testing next time there's a full moon and clear sky is needed to test whether this correlates to what we actually see through an eyepiece (wide AFoV/high magnification needed to differentiate!).

The yellow columns show the calculated "apparent distance" we get by taking the apparent radius determined in eqn [1] and plugging it into eqn [2] (left column). The right column shows what we get if we simply divide the distance to the moon by the magnification.

I'm not entirely sure what the point was, but I've made the angles tend to 180° and the apparent distances now match.

[MarsG76]

I like to think that magnification looks like the amount of times closer you are to the objects.. so for example, 384,000km looks like 3,840km at 100X... etc..

[furrysocks2]

1 hour ago, Ricochet said:

Well...

Thanks for taking that on, love it!

So I screwed up on two counts in the first table. I think I got there in the end but I need to go back and check when I'm more awake.

[JOC]

I wouldn't have thought that the size of image projected onto the retina was even relevant.  It is my understanding of the way the eye works that the thing governing what the eye sees is the light entering the pupil which of course is smaller or bigger depending on how light or dark it is.  I then understood that the eyes lens then focuses those light 'rays' (lets not get tangled up with wave, particle duality theory) directly onto the optic nerve where it leaves the eye (assuming your eyeball is spherical).  So I would have though it was the size of image that hits the pupil that was relevant as surely there is only a maximum size of image which can pass through the pupil and thus be focussed by the eyeball lens onto tbe optic nerve?  Or should I not try and make sense of these rhings at nearly 11pm?

[furrysocks2]

Just now, JOC said:

Or sbould I not try and make sense of tbese rhings at nearly 11pm?

I shouldn't still be trying... :/

[furrysocks2]

7 hours ago, ollypenrice said:

... apparent size goes as the inverse square of the distance

...until you get really close? I hope I've understood this.

[furrysocks2]

3 hours ago, Ricochet said:

X = atan (M / tan Y) [1]

Typo: "X = atan(M * tan Y)"

 

Thanks for the clarifications - think I got it.... "X times magnification" is always "X times closer"... it's the "X times bigger" that is non-linear and only when you start getting really close...?

Apologies for the goose chase. Off to sleep and forget this all happened.

[Ricochet]

1 hour ago, furrysocks2 said:

Typo: "X = atan(M * tan Y)"

 

Thanks for the clarifications - think I got it.... "X times magnification" is always "X times closer"... it's the "X times bigger" that is non-linear and only when you start getting really close...?

Apologies for the goose chase. Off to sleep and forget this all happened.

Thanks for the correction, I've amended my post. Thankfully that was what I used in the spreadsheet so everything still worked  

I think the confusion with the "times bigger" is because we're dealing with angular size and not a linear size. Put a protractor on the bottom edge of a piece of paper, then extend each major line to the top edge of the paper. Then measure the distance between each set of lines along the top edge. 

[bingevader]

Oh goodness! I'm sorry if I kept anyone up!

I now have a very thorough answer to my question.

Thank you all very much!

[furrysocks2]

48 minutes ago, bingevader said:

Oh goodness! I'm sorry if I kept anyone up!

I now have a very thorough answer to my question.

Thank you all very much!

Ignore most of what I said, the goose was chasing a herring.