12 replies to this topic

[PlanetNamek]

Just a quick question here. Without going all over the sky and seeing what magnitude stars you can and can't see with your scope is there a way to calculate it using your telescopes specs? I am pretty sure 5' scope can see up to about 12 magnitude I believe? Thanks a lot for the help.

[cliff mygatt]

Here is a good chart that should answer your questions. Good Luck and welcome to Cloudy Nights!


Aperture Resolving Light Max
Power Gathering Magnitude
Power

50mm 2.32" 51x 10.3
60mm 1.93" 73x 10.7
80mm 1.45" 131x 11.3
90mm 1.29" 165x 11.5
100mm 1.16" 204x 11.8
125mm 0.93" 319x 12.3
150mm 0.77" 459x 12.7
200mm 0.63" 816x 13.3
235mm 0.49" 1127x 13.6
280mm 0.4" 1600x 14.5
355mm 0.3" 2500x 15

[tezster]

Telescope limiting magnitude calculator

[David Knisely]

The exact magnitude you will be able to reach with your telescope will depend on many things, so it is not some hard and fast number. It depends on aperture, conditions, magnification, seeing, and the sensitivity of the observer's eye. To a rough approximation, a formula like this might be somewhat close, but even it isn't all that accurate:

LM = 5*log(d) + 10.3

where d is the aperture of the telescope in inches and log is the base 10 logarithm. Thus, a 10 inch telescope could be able to get down to around magnitude 15.3 or so. However, there are times when even this isn't very accurate. At lower powers, you may not be able to go quite as faint, as the skyglow at that power level can make the really faint stars harder to see. There will also be many nights when the conditions just won't let you go quite that faint. There are even times when you might exceed the above formula's magnitude limit. For example, on one really superb night from my driveway, I hit about magnitude 15.6 with a 9.25 inch SCT at around 480x. The star was not visible quite all the time, but it sort of faded in and out. The best way to find out how faint you can go is by actually observing certain areas of the sky that have calibrated star fields with stars of precisely known magnitude to use as a reference. Also, using high power (30x per inch of aperture or higher) will allow you to catch the fainter stars, although that power may not be optimum for some other objects. Clear skies to you.

[Starman1]

Looking at the html code for that calculator, I see several errors in percentages and secondary obstruction figures.
Here is one that has been corrected to reflect contemporary scopes, coatings, and diagonal sizes:
http://www.scopecity...-calculator.cfm

[David Knisely]

Here is a good chart that should answer your questions. Good Luck and welcome to Cloudy Nights!


Aperture Resolving Light Max
Power Gathering Magnitude
Power

50mm 2.32" 51x 10.3
60mm 1.93" 73x 10.7
80mm 1.45" 131x 11.3
90mm 1.29" 165x 11.5
100mm 1.16" 204x 11.8
125mm 0.93" 319x 12.3
150mm 0.77" 459x 12.7
200mm 0.63" 816x 13.3
235mm 0.49" 1127x 13.6
280mm 0.4" 1600x 14.5
355mm 0.3" 2500x 15

These magnitude figures tend to be quite a bit on the conservative side (by a full magnitude at least). I have seen the 13th magnitude star that sits next to the Ring Nebula in an 80mm aperture, and regularly get down to 15th magnitude in my 10 inch (250 mm) Newtonian. Clear skies to you.

[GlennLeDrew]

It's important to at least pass on to the beginner the following:

Through a telescope at lowest useable magnification, where the exit pupil equals the eye's fully dilated iris, the limiting magnitude will essentially scale as the ratio of the areas objective : eye pupil.

For example, let's suppose the observer's pupil opens to 6mm, and at the time the NELM is 5.5 magnitude. A 100mm scope collects (100 / 6)^2 = 278X more light than the eye. This is equivalent to LOG(278) * 2.5 = 6.1 magnitudes. At lowest power, the sky brightness in the eyepiece equals the sky brightness as seen by the unaided eye. And so the faintest star seen at lowest power is 5.5 + 6.1 = 11.6 magnitude.

As magnification is increased, the exit pupil shrinks and so the sky darkens. This has the effect of increasing contrast for stars (NOT extended objects!), bringing ever fainter ones into visibility. By the time the exit pupil has shrunk to about 1mm, stars some 1.5-2 magnitudes fainter than visible at lowest power will be seen. Further shrinking of the exit pupil only expands the now resolved Airy disk, with the star behaving like an extended object.

Further refinements of the lowest power magnitude limit should include transmission losses. For example, if the transmission efficiency of the system above were to be 85%, we would compute the ratio of light gain as (100 / 6)^2 * 0.85 = 236.

And for scopes having a central obstruction, the fraction of light blocked thereby should be subtracted. For example, an 8" SCT having a 67mm central obstruction has subtracted (67 / 203)^2 = 0.109 the admitted light, leaving 1 - 0.109 = 0.891 of the light to pass through the system.

And if that SCT (with diagonal and eyepiece) has a transmission efficiency of, say, 87% (losses due to coating inefficiencies and absorption in glass), the total efficiency equals 0.87 * 0.891 = 0.775, or 77.5%.

[Starman1]

It's worth noting that the calculator to which I linked in the earlier post has adjustments for the following:
aperture
magnification
visual magnitude limit
observer's age
telescope type (taking transmission into account)
optics cleanliness
color of star
distance from zenith
extinction coefficient (0.15 mag. in pristine site, up to 0.3-0.5mag in city with some pollution
seeing (size of star image or closest star resolvable if image size not estimable)
experience in viewing.

That will give very different results depending on what you plug in. If you're not familiar with the calculator, try plugging in different values for each parameter and see what happens. It's, uh, illuminating.

[David Knisely]

Looking at the html code for that calculator, I see several errors in percentages and secondary obstruction figures.
Here is one that has been corrected to reflect contemporary scopes, coatings, and diagonal sizes:
http://www.scopecity...-calculator.cfm

I tried that one and it way overshot the magnitude limit for my 9.25 inch SCT the night I went to 15.6 unless I used a 0.3 extinction coeficient. Most of the time, I have trouble passing 15.0 in the SCT. Indeed, in my 14 inch Dob, I can go to 16th magnitude or a few tenths of a magnitude past that on a really good night, but not a great deal fainter. My old "off the cuff" formula yields a figure of 16.0, so it at least gives me something halfway close to reality, even with the variables mentioned earlier. Again, many of these formulae offer ball-park estimates but little else of significant accuracy. If a person needs to know how faint they are going, it is best to do observations of calibrated star fields. Clear skies to you.

[Tony Flanders]

Through a telescope at lowest useable magnification, where the exit pupil equals the eye's fully dilated iris, the limiting magnitude will essentially scale as the ratio of the areas objective : eye pupil.

For example, let's suppose the observer's pupil opens to 6mm, any at the time the NELM is 5.5 magnitude. A 100mm scope collects (100 / 6)^2 = 278X more light than the eye. This is equivalent to LOG(278) * 2.5 = 6.1 magnitudes.

That's an excellent rule of thumb -- and just plain common sense, too. However, I would say that although it applies beautifully to binoculars, you should subtract about 0.3 or 0.4 magnitude for a telescope. That's roughly the difference between one-eyed and two-eyed viewing.

On the other hand, it's hard to estimate either naked-eye or telescopic limiting magnitude within 0.2, so what's 0.3 or 0.4 between friends?

[Starman1]

Looking at the html code for that calculator, I see several errors in percentages and secondary obstruction figures.
Here is one that has been corrected to reflect contemporary scopes, coatings, and diagonal sizes:
http://www.scopecity...-calculator.cfm

I tried that one and it way overshot the magnitude limit for my 9.25 inch SCT the night I went to 15.6 unless I used a 0.3 extinction coeficient. Most of the time, I have trouble passing 15.0 in the SCT. Indeed, in my 14 inch Dob, I can go to 16th magnitude or a few tenths of a magnitude past that on a really good night, but not a great deal fainter. My old "off the cuff" formula yields a figure of 16.0, so it at least gives me something halfway close to reality, even with the variables mentioned earlier. Again, many of these formulae offer ball-park estimates but little else of significant accuracy. If a person needs to know how faint they are going, it is best to do observations of calibrated star fields. Clear skies to you.

Assumption is that the SCT is a new one, with 96% reflective coatings on the two internal mirrors, 98.5% transmission on each side of the corrector, and a 92% semi-enhanced star diagonal. You can change the html code if your scope is different.

All of the figures obtained are based on Schaefer's work, which utilizes "10% visible" as the dividing line between visible and not visible and that is a tad more difficult than most viewers prefer.

It predicts a range for me from magnitude 14 at my home to 17.6 in a pristine sky with my 12.5"

I've reached somewhere between 17.1 and 17.3 a few times with my 12.5", which is within the margin of error. I reached magnitude 15.6 with my 8" SCT a decade ago under perfect skies (roughly mag. 22 skies), with 15.1 my average at Mt. Pinos (not a pristine site, but dark).

I used the star charts in Roger Clark's book as test clusters because they are all high in the sky for northern observers. His charts don't go deep enough for the 12.5", so I use other fields.

An extinction of 0.15 magnitude is nearly perfection. 0.3 magnitudes is far more common, and I've seen 0.5 magnitudes in the desert when there was dust in the air, though that was a severe case. If your SQM is one of the narrow ones, you can take a reading at +30 degrees and get a direct reading.

My complaints with the calculator relate to the necessity of putting in an NELM figure, which, as we have discussed, is VERY personal vision acuity-related, and the fact that increasing age always gives a lower limit. It's obvious that if the NELM stays the same but your pupil diameter diminishes with age that your vision is getting more sensitive to light as you age. That is contrary to most experience though, so I back into the NELM using an SQM reading, using the formula:
NELM = [(SQM-8.89)/2] + 0.5 This would yield an NELM of 7.05 under perfect skies, even though some individuals go deeper than that with stars. Myself, I would barely reach mag. 7 under mag.22 skies, so I might fit in quite well with the calculator's predictions, while others do not.

The test chart around M57 is a good one for summer around here, and I find the mag.16.8 star in between IC1296 and M57 to usually be visible with direct vision part of the time and averted vision 100% of the time.

Here's an extinction calculator you might like:
http://www.orion-dru.../extinction.htm

[BrooksObs]

I always find it mildly amusing to peruse these well meaning threads purporting to define the magnitude limits for various aperture instruments. The honest fact is that no formula and particularly none of the almost pointlessly complex ones being derived these days have serious and practical application for 99.9% of visual observers. They are based on the results of a select few highly experienced individuals typically working with outstanding instruments under truly excellent skies. Unless that also describes your own situation take any predictions derived from them with not just a few grains of salt, but about 2 pounds there of.

There are so many individually unknown/undefined variables involved, not the least of which is today's hugh range in sky brightness from light pollution and probably an even greater range in observer skill and experience that utterly precludes any proposed formula from having a true "accuracy" of greater than less than one full magnitude and in some situations even twice that.

This comes from someone with nearly 60 years of observing experience and participation in a number of the studies from which the complex modern formulae were derived. So, if you really what to have any true idea of how faint a star your scope will reveal the only honest way to determine that figure is to go out and look for yourself.

BrooksObs

[Starman1]

I always find it mildly amusing to peruse these well meaning threads purporting to define the magnitude limits for various aperture instruments. The honest fact is that no formula and particularly none of the almost pointlessly complex ones being derived these days have serious and practical application for 99.9% of visual observers. They are based on the results of a select few highly experienced individuals typically working with outstanding instruments under truly excellent skies. Unless that also describes your own situation take any predictions derived from them with not just a few grains of salt, but about 2 pounds there of.

There are so many individually unknown/undefined variables involved, not the least of which is today's huge range in sky brightness from light pollution and probably an even greater range in observer skill and experience that utterly precludes any proposed formula from having a true "accuracy" of greater than less than one full magnitude and in some situations even twice that.

This comes from someone with nearly 60 years of observing experience and participation in a number of the studies from which the complex modern formulae were derived. So, if you really what to have any true idea of how faint a star your scope will reveal the only honest way to determine that figure is to go out and look for yourself.

BrooksObs

Whereas that is true in general (after all, how many observers really try to determine the limits of their observations with charts and the like?), it is not true for the Bogen calculator based on Schaefer's work.

That calculator (the one linked) DOES take into account experience, light pollution, dirt on the optics, how high the object being viewed is, and magnification, just to mention some of the factors allowed.

When I type in my worst case scenario, which is my backyard in LA, I get a limit of mag. 12.2 for my 12.5" scope.

When I type in the best case scenario, which is a pristine site in Utah or Arizona, I get a limit, on the same scope, of mag.17.9!

That's a huge range, and probably reflective of what someone could experience. The typical observer, in the typical environment, will likely exceed the former and not reach the latter.

Most of the time when a beginner asks this question they want to know how faint a deep-sky object they can see, based upon the listed magnitude of the object. As you and I know, any calculator that merely predicts limiting magnitude for stars will fail miserably on DSOs, and then you get involved with a discussion of total integrated magnitude versus surface brightness magnitude and attempt to explain brightness gradients, the inaccuracy of listed magnitudes, etc.

And every attempt at trying to derive a "Visibility Index" to predict the visibility of a particular DSO in a particular scope has failed.

And yet, where stars are concerned, I have come very close to exactly the magnitudes predicted using 5", 6", 8", and 12.5" scopes. IF you know how to input the correct values for each field, the prediction is uncannily accurate, and many other CNers have found that true as well.

So what is the value of a calculator for the beginner merely asking how deep he can go? Not much. And that's why I basically agree with your points.
The best advice is the advice given to me a long long time ago: "Don't assume anything can't be seen until you can't see it. And then, try to go back to see it again when conditions are better." I noticed my observing log, when I sort by magnitude, has tons of 16th magnitude stuff "seen" and lots of 13th magnitude stuff "not found". The "proof is in the pudding".