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gorann

Traveling close to M42

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11 hours ago, jetstream said:

Well the hand is just an easy way to reduce aperture and is not of uniform shape really so it illustrates 2 points.

The dimming effect will come from the reduced exit pupil that reducing the aperture causes at the same focal length of telescope. ie my 250mm dia 1200mm fl = f4.8 vs the masked 127mm dia fl 1200 = f9.4.

If we view using a 1mm exit pupil in the 250mm f4.8, this is a 4.8mm fl eyepiece.

Using the same eyepiece in the masked 127mm (250mm masked to 5") this results in an exit pupil of 4.8/9.4= .51mm exit pupil, and this will dim the image visually for sure.

Second- I find it intriguing that reducing aperture with different shaped masks (ie the hand) still results in the image form being retained at focus.

 

I would think that it matters which extended source you are talking about as to whether it appears brighter the closer we get.... galaxies are made up of many point sources and also diffuse nebula and dust and....

Sorry Gerry, just got round to seeing this. Work you know! Does get in the way somewhat!

I didn't appreciate there was a difference between imaging and viewing in that way. Thank you for that.

Also, I would agree that it depends on whether you are talking about galaxies or nebula etc. It's the nebula that intrigues me. I still can't work out where the optimum viewing point might be.

I'm thinking it must be at a point where the image 'fills the frame' for the eye. Any closer and you are capturing less and any further away and your not using the receptors 'to the max'.

This has all been so interesting.

cheers

gaj

 

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2 hours ago, Stu said:

The image scale must scale with distance. If I move closer to an object then it appears bigger to me. If I move from 1500 ly away from something to 500 ly away, it will appear bigger. The effect is obvious with other extended objects like planets. Mars for instance varies hugely in apparent size depending upon the relative orbital positions of Earth and Mars.

The object appears bigger, so it's light is more spread out. The telescope's apparent field of view will remain unchanged so you will be looking at a relatively smaller area, the object surface brightness will not increase.

One variable is whether you are in a position where the object fills the field of view or not. Before you reach that point, I agree that it will 'appear' to get brighter because it will appear larger in the fov. After that point my I would expect it to remain the same, the area you are looking at becomes smaller, the object brightness gets larger as it gets closer but the surface brightness remains the same.

The image scale does increase but not in the same proportion to the distance - twice as close will not make image twice as large theoretically. For this to be, the distance from the mirror to the image plane will have to be twice as long and we know from experience that we do not change focus by that much. In the 12" f4 vs. 6" f8 example we know that the image scale will scale proportionally if we changed the 12" to f8 and we will have similar image brightness with a larger image. But with your Mars example we should actually be able to prove this. We can probably get information on the angular diameter of mars when it is at oposition and when it is at it furthest from us and compare this with the corresponding distances. I will try...

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19 minutes ago, beka said:

The image scale does increase but not in the same proportion to the distance - twice as close will not make image twice as large theoretically. For this to be, the distance from the mirror to the image plane will have to be twice as long and we know from experience that we do not change focus by that much. In the 12" f4 vs. 6" f8 example we know that the image scale will scale proportionally if we changed the 12" to f8 and we will have similar image brightness with a larger image. But with your Mars example we should actually be able to prove this. We can probably get information on the angular diameter of mars when it is at oposition and when it is at it furthest from us and compare this with the corresponding distances. I will try...

Beka,

I am not sure what you mean by saying that the image scale does not increase in proportion to the distance. At least the diameter of a circle will double when you move to half the distance. The area will increase by the square of that. Is that what you mean? Look at this fancy drawing I just made:

FW and distance.jpg

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2 hours ago, Stu said:

Can I just add that this has been one of the most enjoyable and informative threads I've taken part in for quite some time. Great stuff ??

Yes I agree this is very interesting and full of information. Stu, I did  check out your links, the first is very informative and emphasises the importance of the exit pupil. I don't think I found anything that contradicts my opinion that an extended object will appear brighter as we get closer. The fact that the maximum brighss we can get is the same as that of the naked eye...  no denying the math! I imagine what this means is that for these extended objects it is just the magnification which makes them easier to see. But I  am arguing that effectively both eye and telescope will see a brighter images as we get closer. 

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53 minutes ago, beka said:

Yes I agree this is very interesting and full of information. Stu, I did  check out your links, the first is very informative and emphasises the importance of the exit pupil. I don't think I found anything that contradicts my opinion that an extended object will appear brighter as we get closer. The fact that the maximum brighss we can get is the same as that of the naked eye...  no denying the math! I imagine what this means is that for these extended objects it is just the magnification which makes them easier to see. But I  am arguing that effectively both eye and telescope will see a brighter images as we get closer. 

Beka, I think we are now basically at the same place. I have tried all along to be specific about the fact that I am referring to surface brightness rather than overall visibility.

Absolutely correct (and in line with my posts) that magnifying these objects make them easier to see, and that the effect of larger aperture is to allow more magnification to be applied whilst maintaining a sufficiently large exit pupil for the image to remain bright enough to see. What they cannot do is make the surface brightness brighter than naked eye, that is not possible. I'm glad you agree.

On your last sentence, I agree to an extent. As I mentioned earlier, I think the position changes when you get close enough for the object to fill either your visual or telescopic field of view. At that point, the object will no longer be growing in size as you approach, in fact you will be viewing less and less of the nebula. I think that this means the optimum position is where the nebula totally fills your field of view, beyond this point I THINK (emphasis on think!!) that it will reduce in surface brightness until it basically becomes imperceptible once you are in amongst it. This last point I am very much making assumptions on though so would welcome other views.

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2 hours ago, gorann said:

Beka,

I am not sure what you mean by saying that the image scale does not increase in proportion to the distance. At least the diameter of a circle will double when you move to half the distance. The area will increase by the square of that. Is that what you mean? Look at this fancy drawing I just made:

FW and distance.jpg

Hi gorann, I am saying exactly that the diameter will not double at half the distance! The situation is not exactly like in your diagram. If you half the distance to a far away object the change in diameter will be smaller than when you half it for a closer one. Or putting it in another  way, if the distance is halved the angular diameter in the sky (and so on the image) does not double - it is actually the tangent of the angular diameter that doubles, with the actual angle changing to a smaller degree. In your diagram if you draw lines from your eye to the ends of the more distant object, you will see that the angle for this object is not be half that of the closer one, and if you imagine very distant objects the difference in the angle will be even less.

Edited by beka

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20 minutes ago, beka said:

Hi gorann, I am saying exactly that the diameter will not double at half the distance! The situation is not exactly like in your diagram. If you half the distance to a far away object the change in diameter will be smaller than when you half it for a closer one. Or putting it in another  way, if the distance is halved the angular diameter in the sky (and so on the image) does not double - it is actually the tangent of the angular diameter that doubles, with the actual angle changing to a smaller degree.

Do you have any links or diagrams to explanations of this as I really don't understand it!

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12 minutes ago, Stu said:

Do you have any links or diagrams to explanations of this as I really don't understand it!

I  am on my phone now but I will draw a diagram on a computer and post.

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Beka's right you know. The angular diameter (theta) of an object  X high and Y away is given by tan theta = X/Y.

So when object is 1 light year away and 1 light year high, theta is 45 degrees.

When the object is 100 light years away, and 1 light year high, theta is 0.57 degrees NOT the 0.45 you might guess.

So a far object's angular size and area reduces much less than you expect when you move it a lot farther away.

 

Draw the second set of rays to the top and bottom of your far object. The angle between them will be more than half the angle between the two outer rays.

 

One demonstration of the effect is 'Dolly Zoom' as seen in Home Alone.

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1 hour ago, Stu said:

Do you have any links or diagrams to explanations of this as I really don't understand it!

Sorry Stu and others, I realized as I was drawing and doing the calculations that my reasoning was exactly the wrong way round. The image scale in relation to distance is not linear but it is for close objects that it does not work.  For distant objects the tangent and angle are almost directly proportional - thus no increase in image brightness... I was so sure :-/. But I learned a lot thanks everyone!

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Ahh, ok, now I understand :)

Thanks Stub Mandrel, that's clear now. ??

Its obviously a far more complex subject that we perhaps imagined but through this thread I think we've all learned a lot, I certainly have!

Great thread :)

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6 minutes ago, Stub Mandrel said:

Beka's right you know. The angular diameter (theta) of an object  X high and Y away is given by tan theta = X/Y.

So when object is 1 light year away and 1 light year high, theta is 45 degrees.

When the object is 100 light years away, and 1 light year high, theta is 0.57 degrees NOT the 0.45 you might guess.

So a far object's angular size and area reduces much less than you expect when you move it a lot farther away.

 

Draw the second set of rays to the top and bottom of your far object. The angle between them will be more than half the angle between the two outer rays.

 

One demonstration of the effect is 'Dolly Zoom' as seen in Home Alone.

Thanks Stub Mandrel, that is right but if you use different numbers say 50 light years and 100 light years for the same 1 light year sized object I realized that the angle will be almost exactly double and for larger distances becomes closer to exactly double!

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1 hour ago, beka said:

Thanks Stub Mandrel, that is right but if you use different numbers say 50 light years and 100 light years for the same 1 light year sized object I realized that the angle will be almost exactly double and for larger distances becomes closer to exactly double!

Yes, I see now - of course the extreme point is when you are AT the object and the distance is zero - the angular size becomes 180 degrees even if the object is tiny, and Tan theta becomes infinity!

Still this has been a fascinating discussion.

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I saw a strange example of this type of geometrical effect tonight, purely by coincidence.

We had a brief power it whilst I was getting my little daughter ready for bed. I turned my phone torch on, and we were playing a game with her shadow. I noticed that when I held the torch a long way from her, the shadow was about the same size as her. As I brought the torch closer, not much changed until I got significantly closer, then the shadow began to grow more and more quickly, becoming huge as the torch was very close.

A fun way to understand a similar concept to that which we've been discussing. :) 

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3 hours ago, Stub Mandrel said:

Beka's right you know. The angular diameter (theta) of an object  X high and Y away is given by tan theta = X/Y.

So when object is 1 light year away and 1 light year high, theta is 45 degrees.

When the object is 100 light years away, and 1 light year high, theta is 0.57 degrees NOT the 0.45 you might guess.

So a far object's angular size and area reduces much less than you expect when you move it a lot farther away.

 

Draw the second set of rays to the top and bottom of your far object. The angle between them will be more than half the angle between the two outer rays.

 

One demonstration of the effect is 'Dolly Zoom' as seen in Home Alone.

I am getting confused now by your's and Beka's geometry. Here is the same drawing with distances in light years. Are you saying that the proportion that an object will take of our visual field is not doubling when we get to half the distance? Can you make a drawing to explain?

FW from spaceship.jpg

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2 hours ago, gorann said:

I am getting confused now by your's and Beka's geometry. Here is the same drawing with distances in light years. Are you saying that the proportion that an object will take of our visual field is not doubling when we get to half the distance? Can you make a drawing to explain?

Your diagram should actually show the issue clearly.  Try measuring the angles from the observer to the top + bottom of the two 5 light year tall objects.  You should see that one angle is not double the other.

The reason for this (I think) is the middle of the field of view is significantly closer to the observer than the top + bottom edges of the field of view meaning that the centre area appears larger and therefore occupies more of the apparent angle.  Add to this the fact that as your diagram is drawn the top and bottom areas are viewed at an angle while the centre area is viewed straight on.

As the distance from the observer becomes much greater than the height of the object, then this effect will have less of an impact and halving the distance will be closer to doubling the apparent height of the object.

Cheers,
Chris

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WOW - I have just read two pages of posts that completely baffled me :sad:

Here is a challenge for those who know their DSOs. Is there a nebula of similar size to M42 that is a material distance farther away from us (2x, 3x or 4x) that would allow us to test the theories discussed in this thread by observation, both in terms of brightness and size of the object?

 

...and just to add a bit of scale to this discussion - I discovered today that the Trapezium cluster in Orion is 1.5 light years in diameter! Quite staggering when you think of how small it looks surrounded by that enormous cloud of gas :eek:

Edited by DRT

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9 hours ago, Stu said:

I saw a strange example of this type of geometrical effect tonight, purely by coincidence.

We had a brief power it whilst I was getting my little daughter ready for bed. I turned my phone torch on, and we were playing a game with her shadow. I noticed that when I held the torch a long way from her, the shadow was about the same size as her. As I brought the torch closer, not much changed until I got significantly closer, then the shadow began to grow more and more quickly, becoming huge as the torch was very close.

A fun way to understand a similar concept to that which we've been discussing. :) 

Your game demonstrates the concept perfectly!

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7 hours ago, DRT said:

WOW - I have just read two pages of posts that completely baffled me :sad:

Here is a challenge for those who know their DSOs. Is there a nebula of similar size to M42 that is a material distance farther away from us (2x, 3x or 4x) that would allow us to test the theories discussed in this thread by observation, both in terms of brightness and size of the object?

 

...and just to add a bit of scale to this discussion - I discovered today that the Trapezium cluster in Orion is 1.5 light years in diameter! Quite staggering when you think of how small it looks surrounded by that enormous cloud of gas :eek:

Hi DRT,

Rather than a nebula here is a link http://www.nakedeyeplanets.com/mars-oppositions.htm that shows Mars's angular size at different distances - for oppositions at different years.

If you take years 2012 and 2018 the figures are as follows:

Year     Distance(AU)    Angular Size

2012    0.6745              13.9"

2018    0.3862              22.4"

The change in distance is 1.7467 times while the change in apparent diameter is 1.6115 times - so for Mars the surface brightness should increase if our understanding is correct. Now I am looking forward to 2018!

 

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That's not where I was going with this, beka. Mars is extremely close to us and is not a diffuse cloud of gas and dust so I can't see how it could possibly prove or disprove whether or not a nebula increases in brightness as it gets closer to the viewing point.

Whilst brightness is obviously a factor, I am actually more interested in discovering the level of detail and depth of colour that the human eye could detect from being closer to one of these objects. Would it just be a slightly brighter and much lager greyish cloud or would it look anything like what we see in processed images? I suspect the former would be true.

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3 hours ago, beka said:

If you take years 2012 and 2018 the figures are as follows:

Year     Distance(AU)    Angular Size

2012    0.6745              13.9"

2018    0.3862              22.4"

The change in distance is 1.7467 times while the change in apparent diameter is 1.6115 times - so for Mars the surface brightness should increase if our understanding is correct. Now I am looking forward to 2018!

I think the numbers you quote have an error.  For a distance of 0.3862 AU and a diameter of 6778 Km I calculate the angular size to be 24.2", not 22.4".  This makes the change in apparent diameter 1.7410 which considering all the rounding errors in the calculation means there will be no significant increase in surface brightness.

Cheers,
Chris

Edited by cgarry

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56 minutes ago, cgarry said:

I think the numbers you quote have an error.  For a distance of 0.3862 AU and a diameter of 6778 Km I calculate the angular size to be 24.2", not 22.4".  This makes the change in apparent diameter 1.7410 which considering all the rounding errors in the calculation means there will be no significant increase in surface brightness.

Cheers,
Chris

Yes Chris, I took the wrong value of the diameter for 2020 - so no increase in brightness even at Mars distances. I also did the calculation thinking the site was wrong - the result of the calculation matches the 2018 value of the diameter given on the site.

Edited by beka

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Here's the wonderful sketch of Joe Strummer surfing on a spaceship, with the angles measured on it:

angles.thumb.jpg.b5134f5abe462b3ad2cea0c

 

OK there may be some minor inaccuracies, but 54 degrees is NOT twice 28 degrees. If another vertical is added a 2 1/2 light years the angle would be about 91 degrees MUCH less than twice 54 degrees.

 

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I spent some time today putting together a spreadsheet trying to calculate all this. I will post it up later for everyone to rip apart ;). Actually I've just found that with the new forum software, I've been able to attach it directly from google Drive on my phone. Amazing!

I reckon I've got the apparent size changing correctly with distance. I've used an equation I found for point sources to calculate the change in apparent magnitude. This probably works ok down to a certain distance but then things go a bit haywire!

Basically the surface brightness stays the same down to about 35 LY if I remember correctly, then starts to increase dramatically. This may be correct, or it may be as a result of using the point source formula to calculate the apparent magnitude.

Thoughts appreciated!!

M42 Calculations

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50 minutes ago, Stu said:

I spent some time today putting together a spreadsheet trying to calculate all this. I will post it up later for everyone to rip apart ;). Actually I've just found that with the new forum software, I've been able to attach it directly from google Drive on my phone. Amazing!

M42 Calculations

What file extesnion does it take? It hasn't got one and it doesn't appear to be csv or similar.

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