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Various magnifications replicates what visual "distance" from the moon?

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#1 miniqtone

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Posted 12 June 2019 - 10:58 AM

If I were to view the moon at 200x, what distance from the moon would that simulate via the naked eye?

 

For example imagine being on Apollo 11 flying to the moon and you are 50,000 miles from the moon. How much magnification via an Earthbound scope would be needed to replicate that view?

 

Is there any sort of chart that shows these simulated distances relative to magnification?


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#2 Michael Covington

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Posted 12 June 2019 - 11:03 AM

At 200x, it would seem to be 1/200 as far away as it really is.

 

It is 240,000 miles away.   So at 200x it would seem to be 12,000 miles away.

 

 

You don't need a chart.  All you need is division.


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#3 miniqtone

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Posted 12 June 2019 - 11:09 AM

At 200x, it would seem to be 1/200 as far away as it really is.

 

It is 240,000 miles away.   So at 200x it would seem to be 12,000 miles away.

 

 

You don't need a chart.  All you need is division.

Why, of course. Thanks for encouraging me to put my minute thinking-cap on this morning! :)


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#4 miniqtone

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Posted 12 June 2019 - 03:43 PM

OK here's another question. 

 

Back to Apollo 11; when they were 12,000 miles from the moon, from there how long did it take them to achieve lunar orbit? 



#5 jks2000

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Posted Yesterday, 02:31 PM

I think 200x means 200 times bigger, not 200 times closer. The trigonometry means moving to 1/200 the distance increases the apparent size by less than 200x. No I can't remember the formulas! But there is an online calculator at https://planetcalc.com/1897/ that lets you play with calculating different apparent sizes from different distances and physical sizes etc. Assuming their math checks out anyway.

#6 Michael Covington

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Posted Yesterday, 02:38 PM

The relation is exact at the center of the field. You will find a tiny difference over appreciable fields because we're actually dealing with tan(theta) rather than theta.  But the difference is tiny.

For a field 1 degree in diameter (0.5 degree in semidiameter), the angle itself and its tangent in radians differ by a factor of 0.99997. 

Most telescope fields are much smaller than that.

 

Designers of wide-angle lenses, on the other hand, have to think about this.



#7 jks2000

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Posted Yesterday, 03:08 PM

Makes sense. Like i said, I remember about the math but not exactly how to do it. :)

But at 200x the moon way exceeds the eyepiece tfov you're talking about. I dropped some rough ballpark figures for distance and size into that calculator tool, and it spit out something like 1/200th distance enlarges the apparent size of the image about 170x. Does that check out?

#8 Michael Covington

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Posted Yesterday, 04:10 PM

Makes sense. Like i said, I remember about the math but not exactly how to do it. smile.gif

But at 200x the moon way exceeds the eyepiece tfov you're talking about. I dropped some rough ballpark figures for distance and size into that calculator tool, and it spit out something like 1/200th distance enlarges the apparent size of the image about 170x. Does that check out?

Not at all.  What assumptions is that calculator making?  And what do we mean by image?  I assume we're talking about the image as seen in the telescope vs. an equal apparent area of sky without a telescope.



#9 jks2000

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Posted Yesterday, 05:05 PM

Interesting. I asked it for the angular size of an object 2200 miles in diameter at a distance of 250,000 miles. It gave 30 minutes 15 sec, which is about right. Then I changed the distance to 1250 miles (1/200) and it gave an angular size of 82 deg and change. (165x the original size).I think we were just approaching the OP question differently, I was picturing looking out the window of an orbiting spacecraft at 1/200 the distance, not just magnifying what was in the eyepiece. I agree completely with you that if you're just looking at whatever fits in the eyepiece, the relationship is essentially linear. Sorry if I made a hash of what I was saying. All that being said, hopefully the math that calculator tool is using wasn't sending me on a snipe hunt. :)

#10 Michael Covington

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Posted Yesterday, 05:14 PM

Yes, you are looking at the angular size of a large object.  At 82 degrees angular size, the tangent is *appreciably* different from the angle itself.  I was thinking about what is actually seen in the telescope -- the craters look as though they are 200 times closer.   Compare a 200x telescope to a 1x telescope 200 times closer to the surface, but with the same field of view.  

 

When you consider an object filling an appreciable part of the sky, trigonometry and perspective do get involved.




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