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Hey mister, what power is your scope?

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#1 Seaquel47

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Posted 21 January 2021 - 12:05 PM

Occasionally when I do public outreach using a camera on the scope, I get this question.  When using eyepieces I know how to answer the question.  With a camera I don't know the effective magnification of the setup.  Of course if zoomed on the screen that changes the equation.  I currently use an ASI294MC Pro on a TV85 with a 0.8 reducer/flattener for an effective 480mm focal length.  How do you calculate the effective magnification?



#2 Jim Davis

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Posted 21 January 2021 - 12:21 PM

Occasionally when I do public outreach using a camera on the scope, I get this question.  When using eyepieces I know how to answer the question.  With a camera I don't know the effective magnification of the setup.  Of course if zoomed on the screen that changes the equation.  I currently use an ASI294MC Pro on a TV85 with a 0.8 reducer/flattener for an effective 480mm focal length.  How do you calculate the effective magnification?

You don't. It doesn't make sense. The image on the camera sensor is tiny. How much larger your computer makes it is how much magnification you have. A longer focal length will make a bigger image. An eyepiece literally makes that image bigger. The closer you are to it, the bigger it gets. That's why shorter eyepieces have higher magnification. So you can measure the difference in size.

Edit: I guess, if you wanted to, you could measure the image on your screen with a ruler, and divide it by the size of the camera sensor.


Edited by Jim Davis, 21 January 2021 - 12:24 PM.

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#3 eyeoftexas

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Posted 21 January 2021 - 12:21 PM

Just use your best W.C. Fields voice and say, "Go away kid, you bother me".  [kidding, of course]


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#4 slepage

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Posted 21 January 2021 - 12:25 PM

I explain it like this if I was using your setup - 

 

A telescope's magnification power is based off of two things.  First is the focal length of the scope and second is the focal length of the eyepiece.  Since we are not using an eyepiece the power of the scope cannot be calculated,  only the focal length.  So what we are using tonight is equivalent to a 480 mm telephoto lens.

 

Hope this helps,

Steve


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#5 sg6

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Posted 21 January 2021 - 12:36 PM

Answer:

No magnification as such, the image size created is dependant of the focal length and the angular size of the object. Good old Pythagoras and Trigonometry apply and what you need.

 

That usually makes them hesitate, at least for a few years.

And it is correct.

 

Could throw in the image size is:

The Tangent of the objects subtended angular size multiplied by your scopes focal length.

 

That most probably will add a year or two more to the hesitation.

 

Just be sure to get across that there is no magnification as such.


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#6 psugrue

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Posted 21 January 2021 - 12:42 PM

 

Edit: I guess, if you wanted to, you could measure the image on your screen with a ruler, and divide it by the size of the camera sensor.

From a practical point of view, I don't think viewing the image on a 17 inch laptop or a 90 inch tv or at your local movie theater, increases the magnification. Yes the planet Saturn will be bigger when you measure it with a yard stick but it's still only being represented by the same number of pixels in the grid on the sensor that cover that portion of the image. Imagine if Saturn was represented by a 10 by 10 grid of 100 pixels (yep pretty bad image) but viewing on a larger screen does not help. The same holds true for any size grid. However very high def images would benefit from larger screens in order to do them justice.

 

As always please remember, this is just my opinion and I am an Idiot! 

 

Regards,

 

Patrick 


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#7 Steve C.

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Posted 21 January 2021 - 01:02 PM

I've gone round and round on this in my mind, and while it's true that image scale on a sensor is not equal to magnification in a eyepiece, I think it's valid to say something like, "this image on the screen is roughly similar to what you would see at 100x (or whatever) in an eyepiece." The person unexperienced with optics is looking for an answer that they can understand.  There's a time for exactitude and a time for when it's confusing.


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#8 Voyageur

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Posted 21 January 2021 - 01:09 PM

I've gone round and round on this in my mind, and while it's true that image scale on a sensor is not equal to magnification in a eyepiece, I think it's valid to say something like, "this image on the screen is roughly similar to what you would see at 100x (or whatever) in an eyepiece." The person unexperienced with optics is looking for an answer that they can understand.  There's a time for exactitude and a time for when it's confusing.

I like your response. Might depend on who’s asking, though. For a nice person at a star party where I’m doing public outreach, it’s a good reply. But then you might get the guy who says, “Is that all? I grabbed a scope at Costco last week that will go to 675x.”


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#9 alphatripleplus

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Posted 21 January 2021 - 01:32 PM

Moving from EAA to Equipment for a potentially better fit.



#10 nic35

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Posted 21 January 2021 - 02:10 PM

FWIW, your rig has an FOV of 2.29x1.56 degrees.  This is roughly the same size as a 13mm Nagler, (2.22 degree FOV).  So, 37x.  But it is an apples to oranges comparison, as others have said. 

 

john


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#11 Seaquel47

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Posted 21 January 2021 - 02:16 PM

John (nic35) thanks I agree it is an apples and oranges thing but I like your comparison to a 13mm Nagler.  I think a casual telescope user can relate to that.



#12 Jim Davis

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Posted 21 January 2021 - 02:24 PM

FWIW, your rig has an FOV of 2.29x1.56 degrees.  This is roughly the same size as a 13mm Nagler, (2.22 degree FOV).  So, 37x.  But it is an apples to oranges comparison, as others have said. 

 

john

Yes, since a 25mm orthoscopic would also give you 2.2 degrees FOV, at 19x. Since eyepieces can have various apparent fields of view, it would depend on that. Magnification is an angular measurement, which you are not getting in a picture.


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#13 LittleEd

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Posted 21 January 2021 - 03:07 PM

There is, of course, an actual answer.

 

It's a combination of the image scale (focal length/sensor combination), the size of the monitor the image is being displayed on, and how close they are to the image (yes, when you get closer the image is bigger - that's also magnification). The problem with giving them an actual answer (assuming you went through all the effort of that calculation) is that it's a rather abstract number that doesn't really inform them much.

 

What might be more informative for them is if you have handy a paper plate  (or other similar flat circular object) that is the size the moon would appear if being imaged in exactly the same way. Then there's something they're familiar with that they can compare to what they're seeing.


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#14 RichA

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Posted 21 January 2021 - 03:29 PM

Occasionally when I do public outreach using a camera on the scope, I get this question.  When using eyepieces I know how to answer the question.  With a camera I don't know the effective magnification of the setup.  Of course if zoomed on the screen that changes the equation.  I currently use an ASI294MC Pro on a TV85 with a 0.8 reducer/flattener for an effective 480mm focal length.  How do you calculate the effective magnification?

Why not lie to impress them?  IF the average claim for power for a 60mm crapfractor is 450x, then go by diameter and tell them your TV85 is 638x.


Edited by RichA, 21 January 2021 - 08:42 PM.

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#15 psugrue

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Posted 21 January 2021 - 03:44 PM

And even if it is true that technically, there is no answer. I am not convinced of that yet. From a practical perspective, we could view an eye chart with a specific camera and move the eye chart back until we can only read down to the 20/20 line. Then swap out eyepieces until we can only read down to the 20/20 line. Divide the focal length by the eyepiece. Boom! Equivalent magnification. No long exposure or stacking allowed.

 

What do you think?    


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#16 GaryShaw

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Posted 21 January 2021 - 06:04 PM

If one purpose of conducting ‘Outreach’ viewing sessions is to foster an interest, appreciation and an awareness of Astronomy in the general public, then smug or dismissive answers will have the opposite effect and are totally the wrong way to go.

 

I’d hope the response would begin with a direct answer to the question in words the questioner can relate to. If appropriate, based on who the questioner is, there could be a short follow up explaining that it’s tricky to give an exact magnification because the camera can zoom and has lots of potential magnifications. 

 

IMO

Gary


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#17 WadeH237

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Posted 21 January 2021 - 06:06 PM

Yes the planet Saturn will be bigger when you measure it with a yard stick but it's still only being represented by the same number of pixels in the grid on the sensor that cover that portion of the image.

How is this any different from what an eyepiece does?  I would maintain that "increasing magnification" by showing an image on a bigger monitor is exactly the same thing as increasing magnification by swapping in a shorter focal length eyepiece on a visual scope.

 

When you swap out eyepieces for one with higher magnification, you are literally increasing the magnification on the same image formed by the scope at the focal plane.

 

As for an answer to the question in the title, I usually ask if they are familiar with camera lenses.  If they are (and they usually are), then I state the characteristics of the scope as if it were a camera lens.  For example, my C14 is a 3900mm prime lens at F/11.


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#18 WadeH237

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Posted 21 January 2021 - 06:22 PM

And even if it is true that technically, there is no answer. I am not convinced of that yet. From a practical perspective, we could view an eye chart with a specific camera and move the eye chart back until we can only read down to the 20/20 line. Then swap out eyepieces until we can only read down to the 20/20 line. Divide the focal length by the eyepiece. Boom! Equivalent magnification. No long exposure or stacking allowed.

 

What do you think?    

I think that the scope produces an image at the focal plane, whose size is determined solely by the focal length of the telescope.  Once you add a camera, you are capturing the image at a certain image scale, which is measured in arc seconds per pixel, which is a measure of the area of sky captured on each pixel.  The field of view is then determined by the number of pixels in the sensor.

 

There is literally nothing inside of an astrophotography rig that magnifies the image.

 

You are not the first person to try an coerce a definition of "magnification" onto an astrophotography rig - but it just does not apply.  This is very different from a visual telescope, where the eyepiece is literally a magnifying glass.


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#19 kvastronomer

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Posted 21 January 2021 - 06:35 PM

Can we use scope's highest useful magnification (HUM) as the answer for such question?


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#20 WadeH237

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Posted 21 January 2021 - 06:50 PM

Can we use scope's highest useful magnification (HUM) as the answer for such question?

If someone asks the question, then they have at least some level of interest.

 

I generally try to figure out what kind of answer they would understand, and then try to get them as close an answer as possible.  For astrophotography, I try and express it in terms of terrestrial photography.  For visual, I usually try and explain the telescope's magnification is not as important as how much light it can bring to your eye.  Most everyone has seen pictures of the Andromeda Galaxy, so it's a good reference.  People are often stunned when I explain that the Andromeda Galaxy is actually six times wider in the sky than the full moon, and that the reason that you don't see it is because it's too faint.  This often leads to more outreach conversation.

 

I would hesitate to use highest useful magnification as an answer for a couple of reasons.  First, it's just a rule of thumb, and you'll often find that seeing conditions have a bigger impact on how far you can go, than the optics do.  Next, I think that people, by nature, want a single fact or number that they can use to compare different things - but I don't think that reality can be nicely packaged up that way.  As above, if it seems that they have enough interest, I would rather explain what makes a good telescope, instead of just rattling off a number. 


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#21 KBHornblower

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Posted 21 January 2021 - 07:22 PM

The screen on my iPad subtends about 30 degrees at a normal viewing distance.  If I were using it for outreach viewing with a telescope, and the imaging setup just filled it with the Moon, I would tell spectators that the magnification is about 60x if anyone asks.  Having seen this thread, if I were using a different image scale, I would make sure I had a corresponding answer before the event.  If I were projecting it onto a big screen I would do this in terms of the angle subtended by the screen as seen from the middle of the spectator gathering.  My hunch is that most spectators would take my word for it at that point.  If someone asks for technical reasoning I would respond with a detailed explanation as needed.


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#22 dghent

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Posted 21 January 2021 - 07:52 PM

At least you get asked what the power is. All I seem to get asked is how "far" mine can "see".


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#23 RobertMaples

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Posted 21 January 2021 - 08:34 PM

The screen on my iPad subtends about 30 degrees at a normal viewing distance.  If I were using it for outreach viewing with a telescope, and the imaging setup just filled it with the Moon, I would tell spectators that the magnification is about 60x if anyone asks.  Having seen this thread, if I were using a different image scale, I would make sure I had a corresponding answer before the event.  If I were projecting it onto a big screen I would do this in terms of the angle subtended by the screen as seen from the middle of the spectator gathering.  My hunch is that most spectators would take my word for it at that point.  If someone asks for technical reasoning I would respond with a detailed explanation as needed.

waytogo.gif When someone not familiar with telescopes ask "how much power?" or "how much magnification?" generally all they want to know is how much bigger it makes the object appear.


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#24 RichA

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Posted 21 January 2021 - 08:47 PM

"How much power?  My 9" reflector?  4000x."

-Sir William Herschel

 

According to one source.



#25 rhetfield

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Posted 21 January 2021 - 08:49 PM

I would give magnification based on what an eyepiece 2 pe with the same fov would be. In my case, most smaller sensors end up the same fov as my 5mm ep - therefore 130x. As good of an answer as any.
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