An 80-minutes star trail exposure vs. a 20-seconds exposure of the Grand Tetons ~ © Royce Bair |
Longer exposures cause the stars to "trail" |
Equatorial Mounts: When astro-photographers take time exposures of just the stars, they use an equatorial mount and a tracking mechanism that keeps the stars in sync with the earth's rotation. Using this system, exposures can be extremely long without blurring the stars or causing them to "trail". If any of the earth's landscape is included in the photo, it is the landscape that would now become blurred during the exposure.
Sharp stars AND landscapes: Because astro-landscape or "NightScape" photos require both the stars and the earth to remain sharp during the time exposure, the length of the exposure must be short enough so that the stars do not appear to rotate or trail due to the earth's rotation.
For this reason, lenses with a longer focal length (more telephoto) will have faster star movement or trailing than lenses with a shorter focal length (more wide angle). Using simple math, amateur astronomers developed a formula called the "600" rule to determine the maximum exposure times for various lenses mounted to 35mm film cameras. The 600 Rule formula says that 600 divided by the focal length of the lens (in millimeters) equals the maximum allowable exposure time in seconds. Example: a camera using a 24mm wide angle lens should use a maximum exposure time of 25 seconds (600 / 24 = 25).
< an 8-sec exposure and a 15-sec shot enlarged to 100% > 8"x10" prints look great using the 600 Rule, but not at 24"x30" |
A simple chart: Both rules or formulas are based on the full-frame 35mm (24x36mm) film format, so the chart below has a column for both full-frame sensor cameras (and their lenses) and a column the smaller APS-C sensor cameras (and their lenses). Users of the four-thirds (4/3) format should use the APS-C column. Instructions: Pick the column that describes your camera system. Find the number of millimeters that is closest to your lens, then find the maximum exposure time in seconds to the right in either the 450 (less star movement) or the 600 column. (To print this chart, select the blue text with your cursor and any instructions you want to include. Copy and paste the text into your favorite word processor. Change the text to a mono-space font, i.e. and "Courier", so the columns will be properly aligned, and print.)
"450" "600"
Rule Rule
Full-Frm APS-C* Maximum Maximum
Sensor Sensor Exposure Exposure
Camera Camera Time in Time in
Lenses Lenses Seconds Seconds
8mm 5.3mm 56 75
9mm 6.0mm 50 67
10mm 6.7mm 45 60
11mm 7.3mm 41 55
12mm 8.0mm 38 50
13mm 8.7mm 35 46
14mm 9.3mm 32 43
15mm 10.0mm 30 40
16mm 10.7mm 28 37
17mm 11.3mm 26 35
18mm 12.0mm 25 33
19mm 12.7mm 24 32
20mm 13.3mm 23 30
21mm 14.0mm 21 29
22mm 14.7mm 20 27
23mm 15.3mm 20 26
24mm 16.0mm 19 25
25mm 16.7mm 18 24
26mm 17.3mm 17 23
28mm 18.7mm 16 21
30mm 20.0mm 15 20
32mm 21.3mm 14 19
35mm 23.3mm 13 17
40mm 26.7mm 11 15
45mm 30.0mm 10 13
50mm 33.3mm 9 12
55mm 36.7mm 8 11
60mm 40.0mm 8 10
65mm 43.3mm 7 9
70mm 46.7mm 6 9
80mm 53.3mm 6 7
90mm 60.0mm 5 7
100mm 66.7mm 5 6
120mm 80.0mm 4 5
135mm 90.0mm 3 4
150mm 100.0mm 3 4
175mm 116.7mm 3 3
200mm 133.3mm 2 3
*Although the Canon APS-C sensor is slightly smaller than the Nikon APS-C sensor (1.6X factor vs. 1.5X factor), I did not feel the difference is significant enough to warrant a 5th column.
Notes: If your ISO is already to the limits, use the 600 column times, or go even a little longer if you have too (ugly noise is worse than having elongated stars). For the least star movement and highest enlargements, use the 450 column. Choose the shutter speed time that best fits your camera. For instance, if you're using a 24mm lens on a full-frame sensor camera, the 450 column says to exposure for 19 seconds. However, the closest setting on you shutter speed dial is 20 seconds. And, if you want even better enlargement quality (and your ISO isn't already max-ed out beyond your tastes), go one shutter speed setting lower to 15 seconds—you'll be surprised at the star movement difference, and the improvement in quality. Use some practical sense, too. For instance, many 15mm wide angle lenses made for a full-frame camera are "fisheye" lenses, with a view angle of 180º. Many 14mm wide angle lenses, made for a full-frame camera, have a view angle of 114º, making their view narrower than a 15mm! The 450 column recommends a maximum exposure time of 32 seconds for the 14mm, whereas the 15mm recommendation is 30 seconds. In reality, because the 14mm has a narrower field of view, an exposure time of 20 to 25 seconds will give much better results.
Normal and telephoto lenses: As you approach the normal and medium telephoto focal lengths, star movement becomes even more apparent. Although the 450 column says I can get by with a 9 second exposure (10 seconds on your shutter speed dial), the sharpness of the stars and planets in the photo below was greatly improved by going to only 5 seconds.
Morning twilight: Venus & Jupiter within the constellation Taurus (50mm f/1.4 lens @ f/3.5 • 5 seconds • ISO 5000) |
This is so helpful; thank you! And I love both photos. Each has a unique flavor very appealing to me.
ReplyDeleteI see a lot of you're (beautiful) pictures on 30 seconds.
ReplyDeleteWhy is that? if 15 seconds prevents many star stripes trailing?
ps I am a big fan of you're work
Many of my early photos were taken with a Canon EF 15mm f/2.8 Fisheye lens (for a full-frame camera). 30 seconds is the recommended exposure time for that lens, even with the 450 Rule :)
DeleteHi its very intersting ,I've always used the 500 rules, I might need a remote timer now ;) here is another interseting article
ReplyDeletehttp://starcircleacademy.com/2012/06/600-rule/
Thanks! I just tried to take star pictures with my 200mm lens and was surprised to see long star trails. But then, I remembered that a telephoto would make a difference. Many of the standard articles presume the use of a wide lens,, so it was nice to find the confirmation and a rule of thumb to use. Much easier than trying to figure out the degrees of movement of the stars & then calculating the effect of different length telephotos!
ReplyDeleteUltimately the sensor geometry (size and maximum pixels) dictates what "rule of thumb" to use. Using a 1.6 crop sensor and making an image the same size as a full frame sensor will result in a streak that is 1.6 times as long from the cropped camera.
ReplyDeleteA 5D Mark II at 16mm results in stars that "streak" after 5.3 seconds as can be observed by "pixel peeping". After 28 seconds, that star streak will be about 5 pixels long on the 5D II at 16mm.
On the Nikon D800 - also a full frame camera - using a 16mm lens shows streaks that begin after 4.2 seconds. On the D800 that same 28 second exposure will result in a 6.6 pixel streak of the star.
The calculations above are "worst case" - a star at the celestial equator, but wide angle lenses typically include this area of the sky.
The famous "600 rule" is quite out of date, now, people are amending it by reducing this number to 500, 450 or even 185... It is because it misses many of the fundamentals of the optic.
ReplyDeleteNo 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 valid for any crop factor (because it takes into account the pixels width) and gives an optimum between star trails close to the equator and allowable exposure time :
t(s) = [35 x N + 30 x p(µm) ] / f (mm)
Where :
- A is the aperture
- f is the focal length in millimeters
- p is the pixel size, in micro-meters
- t is the exposure, in seconds
If you don't know the pixel size of your sensor, just divide the width in mm of your sensor by the number of pixels across, and multiply the result by 1000. An APS-C sensor is about 22.2 mm wide, a full frame is about 26 mm wide.
For example :
- 5D mk II (pixels=6.4µm) with a 16 mm lens open at 1.8 : t=[35x1.8+30x6.4]/16=16 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 1.8 : t=[35x1.8+30x4.7]/16=13 s
This formula is based on declination of 60°.
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 :
t=[17 x N + 14 x p(µm) + f(mm)/10 ] / [f(mm) x 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 1.8 : t=[17x1.8+14x6.4+16/10]/16=8 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 1.8 : t=[17x1.8+14x4.7+16/10]/16=6 s
Fred
For the D800, with the same 16 mm/1.8, the exposure time would be :
ReplyDelete- 13 s for the "simple NPF rule"
- 6 s for the "complete NPF rule"
The assumption that this "Rule" applies equally at all latitudes does not seems right. Surely, an object at the equator travels further in the same amount of time than one halfway to the poles and surely this should affect things shouldn't it?
ReplyDeleteI found your this post while searching for information about blog-related research ..
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