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  #21  
Old 22-01-2013, 06:27 PM
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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).
If i ever take a half decent image, I'm gonna have it blown up to the size of a freekin' wall!
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  #22  
Old 22-01-2013, 10:50 PM
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OzEclipse (Joe Cali)
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Originally Posted by mithrandir View Post
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?.
Andrew,

You are reading way way too much into this. It's a very rough guideline not a precise calculation. For a 14mm lens the difference between using a factor of 400 and 375 is about 2s difference in exposure or 0.002 microns or about 1/2000 of one of your 4um pixels. It doesn't matter.

cheers

Joe
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  #23  
Old 22-01-2013, 11:38 PM
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Just to show that I do practice what I preach.

20s with a 10mm f5.6 full frame fisheye ISO 12500.

I only derived the formula last weekend so in this image taken winter 2011, I was just playing with using short exposures. Shorter than the 600/500/400 rule. I was playing around with different high ISO's trying to push the camera's limits.

Now with my formula I can maximize my exposure time minimize the ISO for a calculated number of pixels drift.

Joe
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Last edited by OzEclipse; 23-01-2013 at 10:17 PM.
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  #24  
Old 23-01-2013, 08:54 PM
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Thanks for your detailed research and answer, Joe. Very appreciated!

H
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  #25  
Old 05-12-2014, 10:37 PM
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Hi

I did the same calculation as OzEclipse but did not get the same result. Joe, could you explain how you made your calculation ? I explain the maths behind mine here (that's in french but Google can translate it).

No matter the crop factor, if you take into account the physics under the optical light path (Airy disk) and the CFA pattern of the DSLRs, you will arrive to the following rule that is a good compromise between star trails close to the equator and allowable exposure time :

Simplified NPF rule: t = [35*N + 30*p(µm) ] / f (mm)

Where :
- N is the aperture (and not the number of pixel drift as Joe's formula)
- f is the focal length in millimeters
- p is the pixel size, in microns

For example :
- 5D mk II (pixels=6.4µm) with a 16 mm lens open at 2.8 : t=[35x2.8+30x6.4]/16=18 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 2.8 : t=[35x2.8+30x4.7]/16=15 s
- Nokia Lumia 1020 (pixel=1.1µm), focal length=5.9 mm, aperture=2.2 : t=[35x2.2+30x1.1]/5.9=19 s (but this phone is not able to shoot such a long exposure !)

This simplified formula is based on declination of 60°. It is very close to the formula of OzEclipse except it adds N and P instead of multiplying them.

Should you need a more precise formula to take into account the declination of the stars, and really experience NO trail anywhere on your shot, then it becomes :

Full NPF rule : t=[17*N + 14*p(µm) + f(mm)/10 ] / [f(mm)*cos(declination)]

Therefore, using the same equipment and at the equatorial plan (cos(dec)=1) which is the most defavourable place :
- 5D mk II (pixels=6.4µm) with a 16 mm lens open at 2.8 : t=[17x2.8+14x6.4+16/10]/16=9 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 2.8 : t=[17x2.8+14x4.7+16/10]/16=7 s
- Nokia Lumia 1020 (pixel=1.1µm), focal length=5.9 mm, aperture=2.2 : t=[17x2.2+14x1.1+5.9/10]/5.9=9 s (but again, this phone is not able to shoot such a long exposure !)

Fred

Last edited by Fred_76; 06-12-2014 at 02:20 AM. Reason: error in letter
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  #26  
Old 12-12-2014, 03:54 PM
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Static tripod formula

Hi Fred

Nice piece of work.

Sorry only just saw your post

I did not take account of airy disc or seeing or bayer matrix effects directly however I would take it into account by common sense. I would not see the point of ever assigning a value of 1 to N. Even for no drift, I'd use a value of 1.5. I did not explain this clearly in my original post. When I do this and compare, there is only a slight deviation between results of our formulae for ultra wide lenses which I suspect would be almost un-noticeable. This deviation becomes smaller as the focal length increases.

Anything < 1.5 does not make sense. Even if you print at 300 dpi, I'd recommend N=3.

I'm pretty busy at the moment. If I can find my original derivation, I'll post a scan otherwise I can write it out again. It's very easy. I'll post my derivation soon as possible.

kind regards

Joe
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  #27  
Old 14-12-2014, 09:22 AM
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Derivation of the tripod rule

Hi Fred,
I have attached a PDF to this post with the formula derivation. Looking back at the original post and recalling my derivation of the formula, I was deriving a formula that made allowance for the fact that these days, we look at a lot of down-sized images on screen. Your formula makes no such allowance. Fixed tripod shots are never going to be perfect. Hence I introduced the single quantity N - number of pixels drift to allow for reduced image size viewing, and tolerance for non-round stars. Assuming a down-sized image, effects like seeing and airy disc don't need to be taken into account.

Your approach is different to mine. You seem to be looking at using short exposures as a quasi-high precision guiding for stacking. My feeling is that if high precision guiding and 100% viewing is the target, the photographer should probably be using a tracking mount! I was looking at wide angle reduced resolution images where airy discs and seeing effects could be disregarded. We have two different but equally valid approaches. I must admit that I did not explain the quantity N and considerations around its value carefully enough in my original post. The value of N should never be set to 1. The star image is always going to be spread at the sensel over >1 pixel by the bayer matrix, seeing etc depending on lens focal lengths and f ratios.


Your statement on your web page that

"Cette règle date du temps de l'argentique et certains photographes l'on améliorée en tenant compte de la déclinaison minimale contenue sur l'image. D'autres formules plus compliquées ont été écrites mais aucune ne donne de résultat convaincant dans tous les cas. Elles sont toutes basées sur des méthodes empiriques."

translates roughly to a statement that your formula is the only formula that works and no others found on the internet are accurate or "convincing" in all cases. They are all empirically based. I think your statement is a bit of a stretch. Both your derivation and mine are based on well-known celestial and optical theory. Neither of us have invented anything new. There is more than one way to arrive at the same result. I have included a plot of a comparison of the indicated exposure times from your formula and mine for lenses ranging from 5mm - 500 mm. At f2.8,f4,f5.6 the differences are negligible.

At f22, there is a significant difference between our results. However, If we restrict the comparison to practical uses not theoretical extremes ( no exposures at >f8 for a static shot, no phone cameras with 1 micron pixels, no >500mm focal lengths ) then the airy discs and seeing effects are always less than pixel dimension of 5.5 µm and the results of the two formulae are very similar as can be seen in the illustration below. IIS does not allow excel files to be posted so I have taken a screen shot of the results.

kind regards

Joe
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  #28  
Old 20-12-2014, 12:07 AM
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Hi Joe,

You are absolutely right. Both formulas are based on the same theory, and both of us found the "14" factor. The only difference is that I took into account the Airy disk size (hence the 17xA term in the "full NPF formula"). I agree with you that the seeing has a minor effect (f/10).

If we neglect the Airy ans seeing contribution, we both arrive to the same formula :

t = 14*N*p(µm) / [ F(mm)*cos(dec) ]

(my formula is with n=1)

Assume that the original picture (full size, out of the DSLR) has a width of Wo pixels. It is then reduced (for Internet) to a width of Wi pixels. We can calculate the coefficient n = Wo/Wi and use that in the formulas.

Example :
- photo shot with a 1000D, pixel size = 5.7 µm, images are 3888 px witdh
- published image is 1000 px wide
- lens is a Canon 17 F/4 has a focal length of 17 mm, aperture of 4

Exposure time (declination = 0°)
- Fred : t=[17*4+14*5.7+17/10]/17=8.8 s => rounded to 9 s
- Joe : t=[14*5.7*3888/1000]/17=18.3 s => rounded to 18 s

If we combine our works, we can derive a better formula :

t = [17*A+7*(1+n)*p(µm)]/ [ F(mm)*cos(dec) ]

where :
A is the aperture
F is the focal length in mm
p is the pixel physical size on the sensor, in µm
n, two cases :
- photo to be published on screen : N=nb pixels width original / nb pixels width on screen
- photo to be printed with a resolution of nDPI : N=nb pixels width original / (width of print in inches x nDPI)


and the example becomes :
- Fred+Joe : t=[17*4+7*(1+3888/1000)*5.7]/17=15.5 s => rounded to 16 s

Note that in the case one uses a real B&W sensor the "1+n" factor shall be replaced by n only as there is no more CFA matrix.

Concerning my statement (in french), I agree with you that I did'nt invent anything. However, I did an extended survey, 5 years ago when I first publish my formula (here), to find out what kind of "rule of the thumb" was used. The most used was the "500 or 600 rule" and many photographers were claiming they had problems with these rules with their DSLRs. They were proposing some modifications by taking the crop factor into account, or by adding more parameters. These solutions were OK for their equipment, but not for all cases. That is why I tried to understand the physics behind and derived my solution.

In fact I discovered the 500 rule was already used in the film epoch and was working pretty well. But at that time, the films had to be used a high ISO grade for night shoots, with a quite poor resolution (600 to 1000 dpi on the film) and today, the DSLRs have a far better resolution (more than 4 times).


But as you explained me in PM, these formulas are too complex for most photographers. I therefore propose a simplier formulation assuming the following :
- declination of about 45°
- allowable drift of about 1.5 pixels (n=1.5)

Therefore the formula becomes :

t = 25 x (N + p) / F, rounded to the highest available speed

where :
- N is the aperture
- p is the pixel width in µm
- F is the focal length in mm


again, in the above example, we get :

t = 25 x (4 + 5.7) / 17 = 14.3 s, rounded to 15 s

Last edited by Fred_76; 20-12-2014 at 12:27 AM.
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  #29  
Old 16-01-2015, 08:16 PM
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I tried the formula for the C/2014 Q2 Lovejoy comet.

My equipment is a Canon 500D (pixel size = 4.7 µm) with a Tamron 135 mm F/2.8 lens open at F/4. The comet, yesterday was at 13° declination.

The formula (in blue below) gives :

t = 25x(4.7+4)/[135xcos(13°)]=1.7s rounded to 2s

The stacking of 67 photos with 2 s exposures gives the following result (not enough to correctly see the comet but enough to imagine it) :

http://www.sahavre.fr/forum/download...=949&mode=view

The stars are round !
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  #30  
Old 21-01-2015, 10:36 AM
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hm, formula doesn't take into account resonant frequency and wind speed?

Interesting read and the technical refinements are interesting, just not too practical for Jo Public. There are many "Rules" in photography and too many people take them as LAW. They are all really general guidelines its up to the photographer to try them as a starting point if in a hurry and they will generally yield an acceptable shot (so you at least get something) but you then refine settings from there to get the shot you are after for the result you want. Understanding is often missing.

Streaky/blobby stars are not just a factor of exposure time, lens distortion/cleanliness, tripod sturdiness, internal/external vibrations also can contribute. I usually start with a test shot at the 500 rule exposure time.Then look at the shot on camera making sure my target is in shot (and should remain in shot for my session (if i'm not tracking) then pixel peep for streaking, if everything is pin sharp or slightly streaky in the middle of my target then I readjust exposure time one increment in either direction then take another test shot. Rinse and repeat until I find the longest exposure time to keep round stars, that's the time I use for that target, that session.

If lens distortion is noticeable I adjust aperture again testing to find optimal for what I want. Same with ISO to get cleanest, optimal exposure for sky conditions.

Moving parts such as shutter and mirror inside cameras can create vibrations which can limit the roundness of stars too. So understand your camera, some have features like mirror up to reduce this. Likewise wind, people moving around, nearby vehicles can all transmit vibrations that can result in star distortion in shots. If shooting on camera tripod, don't extend it fully if you don't have too, to make it more sturdy. Good tripods often have a hook or hanging point at the base of the central pillar so if you hang a bag or something heavy from it it can help dampen external vibrations.

Only takes a little time to get this sorted up front. If you're waiting for astronomical dark time to collect your shot set, then go through all this ahead of time during the twilight while you are waiting. At least you can get focus , shutter speed and tripod vibrations sorted, then do the rest for exposure at the right time. Don't forget to set up your intervalometer. Now you are set to take a set of shots you can start to process.

As far as I'm concerned the only rule you need to follow to get a good photograph is Understand How to take a Good photograph and Why you need to adjust settings. There is no magic button, You have to Learn so deal with it.

So, the 500 rule is just one starting factor for taking non-streaky stars shots but it gets you in the right ballpark, but you still need to find the optimal settings for the session and target if you want to get picky. This applies to any budget, you do NOT need expensive cameras or lenses and the rules change depending on What you are shooting: if you just want nice pretty stars in night landscape it will differ from shooting the moon which will differ from shooting the Orion Nebula and will differ from shooting a constellation. Basically it all depends. And it will change if you are doing photometry or want an artistic shot or just want your first shot. The Rules are are generally good starting points (Note the word: generally does NOT mean always) but from there you need to refine and learn.
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  #31  
Old 14-10-2015, 09:32 AM
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Sil, I fully agree with you.

To avoid miscalculations, for those who like formulas, I implemented the NPF rule in a form page. It is still in french but you can choose between 340 DSLR or Digital Backs from 15 brands (that's almost exhaustive) from 4/3" to medium format :

http://www.sahavre.fr/tutoriels/astr...file-d-etoiles
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