Light Gathering Power vs Area - Which one should I use?

Moon0326

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Hi!

I think I got the following equation for light gathering area from this forum.

pi*(14/2.8)^2 = 78.54

pi*(20/1.4)^2 = 641

300 / 14 = 21.4

300 / 20 = 15

78.54 * 21.4 = 1680

641 * 15 = 9615

When I compare two lens to see how much more light lens A gathers than lens B, should I use light gathering power or area?

Thanks!

Moon
 
Hi!

I think I got the following equation for light gathering area from this forum.

pi*(14/2.8)^2 = 78.54

pi*(20/1.4)^2 = 641

300 / 14 = 21.4

300 / 20 = 15

78.54 * 21.4 = 1680

641 * 15 = 9615

When I compare two lens to see how much more light lens A gathers than lens B, should I use light gathering power or area?

Thanks!

Moon
It depends on what you want to do.

Without tracking, you are limited by star trails, so faster wide is better.

A good exposure for Milkyway shots is 30s @f/1.4 = 120s @ f/2.8 and this will require tracking.

With tracking you get more detail with a long lens.
 
Hi!

I think I got the following equation for light gathering area from this forum.

pi*(14/2.8)^2 = 78.54

pi*(20/1.4)^2 = 641

300 / 14 = 21.4

300 / 20 = 15

78.54 * 21.4 = 1680

641 * 15 = 9615

When I compare two lens to see how much more light lens A gathers than lens B, should I use light gathering power or area?

Thanks!

Moon
It depends on what you want to do.

Without tracking, you are limited by star trails, so faster wide is better.

A good exposure for Milkyway shots is 30s @f/1.4 = 120s @ f/2.8 and this will require tracking.

With tracking you get more detail with a long lens.

--
cheers!
Gunn
-- Get a big lens and get closer™.
Etendue....learn it....faster wider is not always correct.

Also, to get an acceptable 30" untracked exposure requires at maximum, a 9mm lens on a FF camera.
 
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Thank you chris, but I don't understand it.

30s @ f/1.4? That will create startrail for sure even for 9mm lens (assuming there is f/1.4 lol). I try to use 300 or 200 rules.
 
Maximum exposure time for avoiding star trails for an untracked camera is the 10 rule:

10*(long side in mm) / (focal length in mm)

which means for a 14mm on a FF is 10x36/14 = 26 seconds.

And the absolute aperture rules ! The larger the lens, the more stars, so a 21mm f/2.8 on FF captures more light than a 14mm f/2.8 on an APS-C despite the aperture ratio is the same. The first is 7.5mm aperture and the other 5mm which captures (1.5)2 = 2.25 as much light which equals 0.9 stellar magnitude value.
 
I'm not sure if you actually tried that rule. 26 seconds obviously give you startrail. Even 20 seconds give you startrail. You need about 14 seconds to avoid startrail with 14mm f/2.8. And..all due respect, the question not about max shutter speed...
 
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To All, is my question vague? It looks like no one is understanding my question correctly.
 
This value of 26 is valid for full frame as I mentioned in the calculation. With APS-C it is indeed close to 14.
 
Please test it yourself. 26 never worked for me. 26 seconds is almost like using 400 rule. That does not work for a high megapixel camera. If you zoom 1:1, you will see startrail.
 
What matter most - the sky and the landscape or the sky only.

More aperture will always record fainter stars.

Wide and fast lenses are ideal in theory but vignetting, coma and astigmatism complicate matters in real life.

Longer lenses will record fainter stars (assuming similar f/numbers and larger aperture). But at the cost of reduced field of view, and reduced exposures if no tracking.

There is always a compromise between going wide and going faint when using the camera mounted on a fixed tripod. Several seem to favour a good 1.4/24mm lens to record as faint as possible without using a star tracker - a quite popular compromise.

There is no best lens in this case - all is down to individual preference!

Willing to use a star tracker?

In my case a tracked 15mm fish eye is used for maximum field of view and undistorted contellations - tracking will smear the landscape but record fainter stuff. If wanting a sharp landscape a double take will do - fixed landscape and tracked stars.

In my case a 35 mm lens is used for fainter stuff. Always tracking. Do panoramas if wanting something wider and fainter. Find using this lens a fine compromise between going wide and recording faint stuff - at least when having a star tracker at hand.

If faint stuff is wanted the CZ 2/135 and the 7DII will do a good job - tracked and used at full aperture. Recording 15-16 mag stuff is easy!
 
Thank you trollmannx, but I think you did not understand my question. I'm aware of all the things you explained.

I was wondering what is the best way to compare two lens. By 'compare', I mean..I want to say "Lens A gathers X times more lights than lens B".
 
I guess my question is too vague. No one understood my question or intention so far.

To moderator, please delete this post if possible.
 
Divide the area of the lenses.

A difference about 2.5 will correspond to 1 magnitude difference.

The magnitude series is approximately:

2.5 times area = 1 mag difference

6.3 times area = 2 mag

15.8 times area = 3 mag

39.8 times area = 4 mag

100 times area = 5 mag

For most lenses this will be correct for the central area. Vignetting will cause light loss towards the edge of the field.

My 2/35 mm lens will collect 11.6 times as much light as my 3.5/18 mm lens. This will correspond to a difference around 2.6 magnitudes.

A 2/35 mm lens will collect as much light as a 1.4/24 mm lens.

The 1.4/24 mm lens will not show as faint stars towards the edges as the 2/35 mm lens because more severe vignetting.
 
Moon, what was vague about your question was that it asked about two terms that are just different ways of saying the same thing, at least to us. The lens with the larger area of aperture will have the greater light-gathering power. So your question was somewhat like asking whether a lens is larger if it is measured in inches or in millimeters, or whether a glass will hold more Coke or more Pepsi.

It's possible that you're using the term "light gathering power" to describe something different from what we commonly use it to describe. The calculation with which you started your post was not one we commonly see here. It did appear, though, that the first two lines of it were efforts to determine the area of two lens apertures, but used the aperture diameters where the correct formula would use their radii.
 
Please test it yourself. 26 never worked for me. 26 seconds is almost like using 400 rule. That does not work for a high megapixel camera. If you zoom 1:1, you will see startrail.
I have tested this. But I am not a pixel peeper: when there are short trails at full 20MP resolution and not at 3MB I don't care.

Here two examples of non-tracked photos (but they are stacked):

Milky way, 13x30 seconds with Samyang 14 (the 'trails' in top right are a result of poor stacking).
Milky way, 13x30 seconds with Samyang 14 (the 'trails' in top right are a result of poor stacking).

Scorpius, 25x5 seconds with 85mm f/1.8
Scorpius, 25x5 seconds with 85mm f/1.8

--
Ricoh KR-5 ... Pentax ME Super ... Canon T90 ... ... ... 40d ... 7d ... 6d
 
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Those two images are really well controlled for 30 seconds exposure. I would not care about those tiny bit of startrail as well. All my friends and I can't get that clean and short trail with 30 seconds though.

I heard it also matters your location. I'm in north america. Where are you located? and do you have any information on it (though it's out of topic :) ).
 
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Thank you Lyle Aldridge.

I thought it was LGP then someone pointed out that those equations are actually for light gathering area. I could not really get much information on light gather area over the web. It looks like..I just have to work with the current equation.
 
Those two images are really well controlled for 30 seconds exposure. I would not care about those tiny bit of startrail as well. All my friends and I can't get that clean and short trail with 30 seconds though.

I heard it also matters your location. I'm in north america. Where are you located? and do you have any information on it (though it's out of topic :) ).
The first one I've taken in New Zealand and the second one in South Africa. I am from the EU.
But the location on Earth does not matter at all. The rotation rate is the same everywhere.
But the celestial location does matter. At the equator it is the fastest. The star trail speed is proportinal to the inverse of the cosine of the declination, so objects at 60 degrees (e.g. Crux, Cassiopeia, Big Dipper) declination move only half the speed compared to eqatorial objects (e.g. Orion).

I once shot the LMC (69 degrees declination) with 6 seconds with an 85mm on a crop camera and there were no discernible star trails.

According to the 10 rule it is 10x22.2/85 (it was a 7D which I had in 2013, hence the 22.2mm long side), which limits to 2.6 seconds, but divided by cos 69 degrees it is indeed 6 seconds. From the beautiful location of Laughing Waters. http://www.laughingwaters.co.za

Not optimally focused as I was less experienced in 2013. 12 frames stacked.
Not optimally focused as I was less experienced in 2013. 12 frames stacked.

EDIT: I found this one, also taken with the 7d 4 years ago, 15 seconds in a full moon night, single frame, well within the limit at declination 60 degrees.

Postprocessed with Photoshop, single frame.
Postprocessed with Photoshop, single frame.

--
Ricoh KR-5 ... Pentax ME Super ... Canon T90 ... ... ... 40d ... 7d ... 6d
 
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Thank you chris, but I don't understand it.

30s @ f/1.4? That will create startrail for sure even for 9mm lens (assuming there is f/1.4 lol). I try to use 300 or 200 rules.
Yes, I said:

"Without tracking, you are limited by star trails, so faster wide is better.

A good exposure for Milkyway shots is 30s @f/1.4 = 120s @ f/2.8 and this will require tracking."

Without tracking and at 24mm APSC @f1.4 I use 15s. With tracking I use f/2 and 60s.

I take it then you are talking about untracked, and you are trying to compare wide fast lenses. The Samyang 24/1.4 seems to be the best.
 

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