Go Back   IceInSpace > Equipment > Astrophotography and Imaging Equipment and Discussions

Reply
 
Thread Tools Rating: Thread Rating: 2 votes, 5.00 average.
  #1  
Old 18-01-2013, 09:29 AM
Intergage's Avatar
Intergage (Matt)
Registered User

Intergage is offline
 
Join Date: Nov 2011
Location: Brisbane
Posts: 201
The 600 (500) rule when taking pictures of stars

Hey all,

I did it. I went out and got my 600D last night with a decent tripod and stuff but didn't get a new lens. Just got a single lens kit. Anyway, I was readying a blog post just before about taking pictures of the Milky Way (I wanted to learn the technical side not just 'Expose the ass off it!") and I found this post about the 600 Rule

http://blog.starcircleacademy.com/2012/06/600-rule/

Basically it states that the maximum length of an exposure with stars that doesnt result in star streaks is achieved by dividing the effective focal length of the lens into the number 600.

Apparently the maths doesn't work out 100% when you get up the full frame sensors though.
Reply With Quote
  #2  
Old 18-01-2013, 09:53 AM
azagil's Avatar
azagil (Aaron)
Registered User

azagil is offline
 
Join Date: Nov 2012
Location: Werribee (Melbourne)
Posts: 55
Hi Mate, I shot the Milky way from time to time and I use the rule of 600 on my D7000. (Small sensor) but as you can tell from my avatar it works pretty well but is only a rough guide so best to play around with it a little. But in saying that...600 is what I work off. :-). (If you do zoom in on your final image there is small trailing but normaly its fine unless you really go looking for it!!!
Reply With Quote
  #3  
Old 18-01-2013, 03:49 PM
Intergage's Avatar
Intergage (Matt)
Registered User

Intergage is offline
 
Join Date: Nov 2011
Location: Brisbane
Posts: 201
Quote:
Originally Posted by azagil View Post
Hi Mate, I shot the Milky way from time to time and I use the rule of 600 on my D7000. (Small sensor) but as you can tell from my avatar it works pretty well but is only a rough guide so best to play around with it a little. But in saying that...600 is what I work off. :-). (If you do zoom in on your final image there is small trailing but normaly its fine unless you really go looking for it!!!
Going to use the Rule for the first time tonight. Just started looking at the details for my camera
Reply With Quote
  #4  
Old 18-01-2013, 03:56 PM
Octane's Avatar
Octane (Humayun)
IIS Member #671

Octane is offline
 
Join Date: Dec 2005
Location: Canberra
Posts: 11,159
I used to do 700/focal_length on a full frame sensor. Seemed to work well.

H
Reply With Quote
  #5  
Old 18-01-2013, 04:26 PM
Intergage's Avatar
Intergage (Matt)
Registered User

Intergage is offline
 
Join Date: Nov 2011
Location: Brisbane
Posts: 201
Quote:
Originally Posted by Octane View Post
I used to do 700/focal_length on a full frame sensor. Seemed to work well.

H
I think I have a max exposure of about 5 - 6 seconds before I start getting real trails..

I was doing 15 second exposures last night and the trails weren't THAT bad.. When you zoomed in you saw them quite easily though.
Reply With Quote
  #6  
Old 19-01-2013, 08:47 AM
gregbradley's Avatar
gregbradley
Registered User

gregbradley is offline
 
Join Date: Feb 2006
Location: Sydney
Posts: 15,309
14mm widefield lens on a full frame goes about 30 seconds usually with no trouble. But you will find those stars further away from the suth celestial pole move faster (more angle to cover) than those closer to the celestial pole so you can get in a widefield image stars streaking in some parts of the image and tight in other parts.

The solution is use a widefield lens like 10mm on APS sensors or 14 on full frame, use your 30 seconds, use maxmium ISO before noise becomes too bad (probably around ISO3200) and long exposure noise reduction on. White balance may vary with cameras with Nikon being a little green/yellow and Canon being a little bluish but around 4200K should be close. Save as a RAW that way you can fiddle with the white balance mostly afterwards.

If you want longer than that you need a Vixen Polarie or similar product. Then you can get 2 or 4 minutes if you set it up just right and are within the weight contraints.

50mm in my experience is way too long a focal length to use without tracking then it is awesome. 24mm on full frame I would consider the cutoff point and is pushing it a bit.

The rule should really state - the longest focal length you can use at 30 seconds (anything less than 30 secs ISO3200 is going to be underexposed) is 24mm on full frame or 15 seconds on APS sensors without tracking.

I don't know why camera manufacturers of DSLRs have maximum time of 30 seconds before the stupid bulb setting has to be used. Its like they think noone in their right mind will need more than 30 seconds. Hey - we're being discriminated against!!

Greg.
Reply With Quote
  #7  
Old 20-01-2013, 12:43 AM
OzEclipse's Avatar
OzEclipse (Joe Cali)
Registered User

OzEclipse is offline
 
Join Date: Jun 2012
Location: Canberra, Australia
Posts: 948
Revised rule for DSLR's

These rules or formulae are based on a set of starting assumptions and have been kicking around for a long time.

In the old days of 35mm film where very fast films used for tripod astrophotography had very course grains and so a bigger movement could pass off as no drift. It was based on someone looking at a 8x10inch enlargement at normal reading distance perceiving trails as almost stationary. The eye can resolve about 1 arc minute so it can in fact resolve approximately 200 microns on a print which scales back to 25 microns on the film. However the number 700 corresponds to a trail length of about 50 microns. This is about 9 pixels on a modern DSLR. But today, we mostly look at screen based images and images that are highly reduced from full size.

The number is related to how much drift you find acceptable. So I've derived a revised formula, modernized for DSLRs.

max exposure time(s) = [14*N*P] / [FL*cos(d)]


where
N..........Number of pixels of drift
to work this out load any full frame image from your camera and scale it to the size you want to view on screen. Look at the scale percentage. Number of pixels drift = 100 / percentage

P..........Pixel size (microns) Most DSLR's are between 5 and 8 microns. You can look it up in your manual or just split the difference and make it 6.5

FL.........Focal length(mm)

cos(d)...cos of the declination. Use the declination of the stars in the field that are closest to the celestial equator in the camera field

Example : Using a 14mm lens, 5 pixels drift, on an APS-C sensor with 5.4m pixels pointed at the celestial equator can take a 27 sec exposure. Pointed at the SCP the lens will see 45 degrees each side of the pole extends the time to 38 seconds.

cheers

Joe

Last edited by OzEclipse; 20-01-2013 at 01:50 PM.
Reply With Quote
  #8  
Old 20-01-2013, 11:56 AM
mithrandir's Avatar
mithrandir (Andrew)
Registered User

mithrandir is offline
 
Join Date: Jan 2009
Location: Glenhaven
Posts: 4,161
Quote:
Originally Posted by OzEclipse View Post
max exposure time(s) = 14*N*P / FL*cos(d)
Joe, is that supposed to be read as:

(14 * N * P / FL) * cos(d)
or
14 * N * P / (FL * cos(d))

Associative rules for multiplication say the first but it would be good to be sure. I would write that as 14 * N * P * cos(d) / FL which is less likely to be misinterpreted.
Reply With Quote
  #9  
Old 20-01-2013, 12:58 PM
OzEclipse's Avatar
OzEclipse (Joe Cali)
Registered User

OzEclipse is offline
 
Join Date: Jun 2012
Location: Canberra, Australia
Posts: 948
Quote:
Originally Posted by mithrandir View Post
Joe, is that supposed to be read as:

(14 * N * P / FL) * cos(d)
or
14 * N * P / (FL * cos(d))

Associative rules for multiplication say the first but it would be good to be sure. I would write that as 14 * N * P * cos(d) / FL which is less likely to be misinterpreted.
Andrew,
Sorry about that :
It is:-
max exposure time(s) = [14*N*P] / [FL*cos(d)]

I've edited the original post to reflect this.

The cos(d) term is 1 for a declination of 0 so you can leave it out all together if you want to calculate for stars on the celestial equator.
Reply With Quote
  #10  
Old 20-01-2013, 11:54 PM
mithrandir's Avatar
mithrandir (Andrew)
Registered User

mithrandir is offline
 
Join Date: Jan 2009
Location: Glenhaven
Posts: 4,161
Quote:
Originally Posted by OzEclipse View Post
Andrew,
Sorry about that :
It is:-
max exposure time(s) = [14*N*P] / [FL*cos(d)]

I've edited the original post to reflect this.
It seemed to make more sense that way than the other so I thought I check it.

Quote:
The cos(d) term is 1 for a declination of 0 so you can leave it out all together if you want to calculate for stars on the celestial equator.
To make use of that you need to know the image center and FOV to work out "d".
eg APS-C long axis 23.3mm. At 11mm -> 93 FOV so pointed at dec -45 with that axis aligned n/s some stars are on the equator.
Reply With Quote
  #11  
Old 21-01-2013, 01:49 AM
OzEclipse's Avatar
OzEclipse (Joe Cali)
Registered User

OzEclipse is offline
 
Join Date: Jun 2012
Location: Canberra, Australia
Posts: 948
Quote:
Originally Posted by mithrandir View Post
It seemed to make more sense that way than the other so I thought I check it.

To make use of that you need to know the image center and FOV to work out "d".
eg APS-C long axis 23.3mm. At 11mm -> 93 FOV so pointed at dec -45 with that axis aligned n/s some stars are on the equator.
Andrew,

Yes that's right. That's why I say declination of the stars in the field closest to the equator rather than dec of the stars at the center of the field of view.

Joe
Reply With Quote
  #12  
Old 21-01-2013, 02:03 AM
AstroJunk's Avatar
AstroJunk (Jonathan)
Shadow Chaser

AstroJunk is offline
 
Join Date: Jan 2006
Location: Brisbane
Posts: 1,675
I'm still pretty new at this having similarly just bought a 650d, but I have already come up with my own simplified equation for exposure without star trails. Its easier to use that the others and much along with Gregs comments. Its:

30 seconds or less / conditions on the night.

The attached photos are using an 8mm lens and with a 30 second exposure. Everything looks great zoomed out, but I chose an area with the LEAST trailing to enlarge.

I have been reading the Nightscapes section avidly, and that seems to be the general concensus of exposure before having to resort to making a composite image with the stars tracked.

The 600 rule gives 8mm @ f4.5 = 16 seconds which is better but will require higher ISO and corresponding noise issues.

I haven't done Joes' maths, but of course the key factors are fl, pixel size (chip size is sort of irrelevant) and rate of movement of the field. I'll stick to my equation for the timebeing
Attached Thumbnails
Click for full-size image (TinnieAndTaurus.jpg)
113.5 KB177 views
Click for full-size image (longpixels.JPG)
113.2 KB155 views
Reply With Quote
  #13  
Old 21-01-2013, 06:07 PM
OzEclipse's Avatar
OzEclipse (Joe Cali)
Registered User

OzEclipse is offline
 
Join Date: Jun 2012
Location: Canberra, Australia
Posts: 948
Quote:
Originally Posted by AstroJunk View Post
I'm still pretty new at this having similarly just bought a 650d, but I have already come up with my own simplified equation for exposure without star trails. Its easier to use that the others and much along with Gregs comments. Its:

30 seconds or less / conditions on the night.

The 600 rule gives 8mm @ f4.5 = 16 seconds which is better but will require higher ISO and corresponding noise issues.

I haven't done Joes' maths, but of course the key factors are fl, pixel size (chip size is sort of irrelevant) and rate of movement of the field. I'll stick to my equation for the timebeing
Hi Jonathan,

In which universe does 600/8 = 16 ? Is this some sort of corollary to Douglas Adams 6x9=42?

None of these things are perfect, they are just a guide. My formula takes no account of the light spread of brighter stars which makes a trail longer but also wider nor does it give you trail free images. Of course if you zoom right in then N, the number of pixels drift has to be made smaller and your time becomes much shorter with commensurate increasing noise problems.

Tripod photography is always best expressed using a small print or small screen image to minimize visibility of trailing.

All these rules fall over when you go to these ultra wide and fisheye lenses. The focal lengths stated are obviously not the real focal lengths. Your lens is trailing more like a 12mm focal length lens than an 8mm.

Joe
Reply With Quote
  #14  
Old 21-01-2013, 07:32 PM
AstroJunk's Avatar
AstroJunk (Jonathan)
Shadow Chaser

AstroJunk is offline
 
Join Date: Jan 2006
Location: Brisbane
Posts: 1,675
Quote:
Originally Posted by OzEclipse View Post
In which universe does 600/8 = 16 ? Is this some sort of corollary to Douglas Adams 6x9=42?
I blame the PangalacticGargleblasters myself.

I had rather telescopically taken the aperture multiplied by the focal ratio to be the fl. As I said, real cameras and me are a very new thing! I'll google what those numbers actually mean.

To be fair though, I had done 600/8 = 75 and thought that was such an absurd number that it couldn't be true...
Reply With Quote
  #15  
Old 21-01-2013, 08:00 PM
CapturingTheNight's Avatar
CapturingTheNight (Greg)
Registered User

CapturingTheNight is offline
 
Join Date: Jul 2011
Location: Holbrook, NSW
Posts: 1,230
Ah yes the good old "600 Rule". I have a blog post here http://www.capturingthenight.com/blog/?p=12 on how I use it for my fixed tripod night sky scapes. Personally I use 500 or less when imaging. It's not perfect as stated, but it's a pretty good guide.

Quote:
Originally Posted by AstroJunk View Post
To be fair though, I had done 600/8 = 75 and thought that was such an absurd number that it couldn't be true...
With your new 650D being a non frame camera then you need to factor that into the equation as well Jonathon, but yeah you don't need to worry about the aperture. Canon's crop factor is 1.6, so with your 8mm it is effectively (for the purpose of the 600 Rule) 12.8mm (8 times 1.6). 600/12.8= 46.8 seconds maximum shutter speed before the stars trail too much to be really noticeable. As I said I tend to do the 500 rule or less. 500/12.8= 39 seconds but for that lens I would just stick it on 30 seconds.
Reply With Quote
  #16  
Old 21-01-2013, 08:04 PM
CapturingTheNight's Avatar
CapturingTheNight (Greg)
Registered User

CapturingTheNight is offline
 
Join Date: Jul 2011
Location: Holbrook, NSW
Posts: 1,230
Quote:
Originally Posted by OzEclipse View Post
These rules or formulae are based on a set of starting assumptions and have been kicking around for a long time.

In the old days of 35mm film where very fast films used for tripod astrophotography had very course grains and so a bigger movement could pass off as no drift. It was based on someone looking at a 8x10inch enlargement at normal reading distance perceiving trails as almost stationary. The eye can resolve about 1 arc minute so it can in fact resolve approximately 200 microns on a print which scales back to 25 microns on the film. However the number 700 corresponds to a trail length of about 50 microns. This is about 9 pixels on a modern DSLR. But today, we mostly look at screen based images and images that are highly reduced from full size.

The number is related to how much drift you find acceptable. So I've derived a revised formula, modernized for DSLRs.

max exposure time(s) = [14*N*P] / [FL*cos(d)]


where
N..........Number of pixels of drift
to work this out load any full frame image from your camera and scale it to the size you want to view on screen. Look at the scale percentage. Number of pixels drift = 100 / percentage

P..........Pixel size (microns) Most DSLR's are between 5 and 8 microns. You can look it up in your manual or just split the difference and make it 6.5

FL.........Focal length(mm)

cos(d)...cos of the declination. Use the declination of the stars in the field that are closest to the celestial equator in the camera field

Example : Using a 14mm lens, 5 pixels drift, on an APS-C sensor with 5.4m pixels pointed at the celestial equator can take a 27 sec exposure. Pointed at the SCP the lens will see 45 degrees each side of the pole extends the time to 38 seconds.

cheers

Joe
Very interesting read Joe. Thanks for posting. Might have to have a play around with your new formula one night....
Cheers
Greg
Reply With Quote
  #17  
Old 21-01-2013, 09:59 PM
AstroJunk's Avatar
AstroJunk (Jonathan)
Shadow Chaser

AstroJunk is offline
 
Join Date: Jan 2006
Location: Brisbane
Posts: 1,675
Quote:
Originally Posted by obsidianphotos View Post
Ah yes the good old "600 Rule". I have a blog post here http://www.capturingthenight.com/blog/?p=12 on how I use it for my fixed tripod night sky scapes. Personally I use 500 or less when imaging. It's not perfect as stated, but it's a pretty good guide.



With your new 650D being a non frame camera then you need to factor that into the equation as well Jonathon, but yeah you don't need to worry about the aperture. Canon's crop factor is 1.6, so with your 8mm it is effectively (for the purpose of the 600 Rule) 12.8mm (8 times 1.6). 600/12.8= 46.8 seconds maximum shutter speed before the stars trail too much to be really noticeable. As I said I tend to do the 500 rule or less. 500/12.8= 39 seconds but for that lens I would just stick it on 30 seconds.

Great explanation on your blog Greg - thanks for taking the time to put it together. I feel that the resolution of the chips is beginning to give some of these equations a hard time as even 30 seconds on the 18mp aps-c chip is pushing it let alone 45+ which would provide a clear streak.

However, I have a way to go yet before streaky images are my primary concern
Reply With Quote
  #18  
Old 21-01-2013, 10:13 PM
mithrandir's Avatar
mithrandir (Andrew)
Registered User

mithrandir is offline
 
Join Date: Jan 2009
Location: Glenhaven
Posts: 4,161
Quote:
Originally Posted by AstroJunk View Post
Great explanation on your blog Greg - thanks for taking the time to put it together. I feel that the resolution of the chips is beginning to give some of these equations a hard time as even 30 seconds on the 18mp aps-c chip is pushing it let alone 45+ which would provide a clear streak.
Try an APS-C with 24m 3.89um pixels.
With Joe's formula if I want less than 4px of trailing, at 14mm (effective 21.6mm) I get 90 FOV and can't go beyond 16 sec.
Reply With Quote
  #19  
Old 22-01-2013, 09:21 AM
OzEclipse's Avatar
OzEclipse (Joe Cali)
Registered User

OzEclipse is offline
 
Join Date: Jun 2012
Location: Canberra, Australia
Posts: 948
Quote:
Originally Posted by mithrandir View Post
Try an APS-C with 24m 3.89um pixels.
With Joe's formula if I want less than 4px of trailing, at 14mm (effective 21.6mm) I get 90 FOV and can't go beyond 16 sec.
The crop factor changes the 600 rule to the 400 rule when going from a full frame to APS-C.


However, the crop factor does not change the focal length, it changes the FOV. When using my formula, do not multiply FL x crop factor in my formula.

Also with 24MPx, you can get away with a lot more than 4 Px of trailing. It goes without saying that these fixed tripod images usually get shown a bit smaller. Let's say you can accept 4 Px trailing at the final displayed image and you are going to scale the image to 800x1200 Px(20% scale of the 4000x6000px original).

Then you can have N, the number of pixels as
4 x 100% / 20% = 20 Px

so your max exposure is

20Px * 4m * 14 8 = 140s

When I made the comment about Jonathan's lens working at 12mm, I meant that the pixel smear indicated that the lens was acting more like a 12mm indicating that the manufacturer may have deliberately understated the focal length for marketing purposes. They do this all the time.

If you think about it, there is 45-55mm between the back of the lens mount and the focal plane in most brands of DSLR so any focal length shorter than about 70mm is a marketing label and not a true statement of focal length.

Joe
Reply With Quote
  #20  
Old 22-01-2013, 10:06 AM
mithrandir's Avatar
mithrandir (Andrew)
Registered User

mithrandir is offline
 
Join Date: Jan 2009
Location: Glenhaven
Posts: 4,161
Quote:
Originally Posted by OzEclipse View Post
The crop factor changes the 600 rule to the 400 rule when going from a full frame to APS-C.
Joe, that must depend on the camera. Canon have a 1.6 crop. Sony have a 1.5 crop. I forget what Nikon is but it is a bit different again. So does 400 apply to Sony and should it be 375 for Canon?
Quote:
Originally Posted by OzEclipse View Post
However, the crop factor does not change the focal length, it changes the FOV. When using my formula, do not multiply FL x crop factor in my formula.
I added the chip size to my spreadsheet so it can calculate the "effective" focal length so I can use that in the FOV.

Quote:
Originally Posted by OzEclipse View Post
Also with 24MPx, you can get away with a lot more than 4 Px of trailing. It goes without saying that these fixed tripod images usually get shown a bit smaller. Let's say you can accept 4 Px trailing at the final displayed image and you are going to scale the image to 800x1200 Px(20% scale of the 4000x6000px original).
That's certainly true on IIS. 800x1200 is about as big as you can upload at any reasonable quality setting.

If today's predicted thunderstorms clear tonight I'll see what 80sec looks like at 14mm. (Jonathan is the one with the 8mm lens.)

The numbers show why the 150-500mm @500 has to go on my StarLapse. You don't get much light in 2.3sec.
Reply With Quote
Reply

Bookmarks

Thread Tools
Rate This Thread
Rate This Thread:

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT +10. The time is now 03:16 AM.

Powered by vBulletin Version 3.8.7 | Copyright ©2000 - 2019, Jelsoft Enterprises Ltd.
Advertisement
Bintel
Advertisement
NexDome Observatories
Advertisement
SkyWatcher Australia
Advertisement
Celestron Australia
Advertisement
OzScopes Authorised Dealer
Advertisement
Meade Australia
Advertisement
Lunatico Astronomical
Advertisement
Astronomy and Electronics Centre
Advertisement