# 600 Rule?

You may have heard it elsewhere as the “600 rule”.  I first heard about the rule while visiting the Looney Bean in Bishop, California in 2008.  Five photographers sitting in a coffee shop pouring over their laptops reviewing what they recently bagged are bound to start talking.  It was my good fortune that one of those present was the very talented Brenda Tharp who first quoted the 600 rule to me.

I, however, have repeated the rule as the “500 Rule” because I think 600 is overly optimistic.  What is the rule?  The rule states that the maximum length of an exposure with stars that doesn’t result in star streaks is achieved by dividing the effective focal length of the lens into the number 600.  A 50mm lens on a 35 mm camera, therefore would allow 600 / 50 = 12 seconds of exposure before streaks are noticeable.  That same 50 mm lens on a 1.6 crop factor camera would only allow 7.5 seconds of exposure.

## But Wait. The Rule Isn’t All That Great!

The real number is quite subjective.  A little math reveals that on the Canon 5D Mark II (a full frame camera), with a 16mm lens a pin point star on the celestial equator moves from one pixel* to the next in 5.3 seconds.  But the 600 rule would allow 37 seconds of exposure and the 500 rule 31 seconds.  Both rules will produce streaks on the sensor! The visibility of those streaks will depend on the finished print size and viewing distance.  Print it large and stand close and the streaks will be obvious.

So what does a 30 second exposure look like at the pixel level:

Clearly those stars are streaking across about 5 pixels* just as the math would bear out.

What is going on here?  The Canon 5D Mark II images are 5634 x 3753 pixels* from a sensor that measures 36 x 24 millimeters. Dividing 36 by 5,634 reveals that the distance from the center of one pixel* to the next is a scant 0.00639 millimeters (or 6.4 microns).

The formula for calculating the distance in millimeters (d) that a star travels across a sensor due to the earths rotation looks like this:

d = t * f / 13750

Where t is time in seconds, f the effective focal length and 13750 is, well 13750.  Is the math scaring you a bit… don’t worry… we’re almost done. Earlier we calculated the pixel* to pixel distance as 0.00639, what we want to find is how long (t) it takes for a star to move that far on the sensor.

0.00639 = t * f / 13750

Solving for f = 16mm we get a t value of 5.3 seconds as I asserted earlier.

But how does that calculate out on a different sensor, the Canon 50D, for example?

The Canon 50D has 4770 pixels across 25.1 mm or an inter-pixel* distance of 0.0053 millimeters.  Substituting into the earlier equation we find that a star marches across a pixel on the 50D with the same 16mm lens in 2.83 seconds.  With a 50mm lens on the same camera… the bad news is the star is speeding from one pixel* to the next in less than a second!

What does an image look like with a 30 second exposure at 16mm on a full frame camera? Remember that the streaks will be 40% longer on the cropped Canon 50D.

30 Second Exposure – a close look shows elongated stars.

Scaled down to only 16% of the original image size or seen from a distance no streaking is obvious! We will try not to twitch knowing – because we pixel peeped – that the stars are really dashes not nice round pinpricks of light. And indeed only the eagle eyed are likely to notice the dash-like nature of the stars until the photo is printed large, say at 20 x 30 inches.

## What Can We Conclude?

1. Streaking starts a LOT sooner than any rule you may have learned.
2. The time it takes to streak depends on the inter-pixel* distance (sensor density / mm) and the focal length.
3. How much streaking to allow depends on your aesthetic tolerances.
4. You can not get more or brighter stars by exposing longer; starlight has already given up on one pixel* and moved on to the next in just a few seconds.
5. The longer the focal length, the more impossible it becomes to prevent streaking.
6. Gaps in your star trails may be unavoidable if the inter-shot delay (normally 1 second) is long enough to skip pixels*.

## Final Note

I carefully added asterisks* to every location where I wrote the word “pixel” in a way that might imply your camera collects light in pixels. You might be wondering why I did that. The answer is: your sensor is comprised of sensels, not pixels. It takes 4 to 9 sensels to create a single pixel depending on the de-mosaic-ing algorithm your camera uses. Maybe you aren’t that picky, but I didn’t want to hear complaints from the purists.

I particularly relish this epiphany because I reported long ago that “longer exposures do not result in more stars“.  I just never got around to doing the extra bit of math – or the experiments – to prove out my assertions.

## Real Final Note

A commenter has rightfully taken me to task by pointing out that the perception of a streak is dependent on many things other than just the actual sensor values recorded. In particular, if the image is not enlarged much some streaking will be scarcely or completely unnoticeable because the feature will be too small for the eye to perceive.  The problem with this assertion is that it assumes a lot of preconditions: e.g. how large the print is, how far from the print a viewer stands, and the subjective experience of the viewer.  My real world experience has led me to conclude that it is a reasonable goal to keep the streaking to 2 to 3 pixels or less because that will provide the greatest possible usable magnification (finished viewing size).  There would be no point to collecting a high megapixel image if you can not produce a print proportionately larger or more detailed than a lower megapixel image!

Here is an example that makes my point. I love this image captured on a Canon 5D Mark II. When printed at 20×30″ and viewed at 4 feet there is some streaking. Perhaps only a critical eye would notice, but even an untrained eye will notice when viewed from two feet away.

## 28 thoughts on “600 Rule?”

1. Bob Stothfang

Quote: “You can not get more or brighter stars by exposing longer; starlight has already given up on one pixel* and moved on to the next in just a few seconds.”

This statement makes complete sense to me if the camera is stationary. If I place my camera on a “perfectly aligned tracking equatorial mount” then theoretically a star would expose the same sensor throughout the whole exposure. Can I get more or brighter stars in that case?

1. Steven Christenson

Yes, Bob. That is exactly one reason for using an equatorial mount. Beware, however, as you’re likely to discover that brighter stars are easy to “blow out”. That won’t matter much unless you’re eager to capture their color. By the way you played right into my plan since my earlier articles talk about astrophotography, and the most recent series about capturing the Milky Way also talks about tracked captures.

2. Harry

Clear and useful, thanks. I just do not get why you would still use 500 (instead of 600) as the magic number at all?
As solving for the 5D the above gives about 88, are you taking the 4-9 sensels per pixels into account to get to 500?

It might be nice to create a simple list of magic numbers per camera/sensor; the above equation might be a bit overwhelming for some, but, say, 100/f=t is a lot easier to remember and use.
Thanks!

1. Steven Christenson

Harry, Good questions. I actually started working on embedding an app in the page that would calculate the ideal time. The math is easy, but I’ve been too busy with other things – and it is painful to collect accurate data for every camera.

I don’t recommend the 500 rule. I was revolting against the 600 rule which I found always streaked too much for my tastes. But as I state in the article, 500 is too much! The problem is that the “correct rule” depends on the camera and the desired aesthetic. Moreover, I also notice that people seldom understand or accommodate the effect of the “crop factor”.

The point of this article is to show that there is a calculable maximum time, and everything beyond the maximum is an aesthetic judgment. E.g. for me, a smear of 5 pixels is too much, but I think a 2 or 2.5 pixel smear would normally be quite tolerable in nearly all display or print modes for an image.

The Canon 5D has a pitch of 8.2 microns, so with a 16mm lens after 7.04 seconds a star at the celestial equator will have moved onto the next “pixel”. Allowing a dash of 2 to 2.5 pixels per star, the exposure time for the 5D to not noticeably streak is therefore 14 to 17 seconds. The rule 250/f would work for the 5D Whereas the same results on the 5DII would require a 210/f rule and on the 50D that rule would be 110/f.

2. Gilles

I vote for Harry’s “100 rule”, that says: “10 s for 10 mm”.
It’s easy to remember, and matches quite well my experience.

3. Steven Christenson

It may not be obvious from the article, but those who are plagued by gaps in their star trails may find that “doing the math” reveals the source of the problem. If, for example there is a three second delay between exposures (see How Long is a 30 Second Exposure), then that may well be enough time to skip all or part of a pixel.

4. Tibi

The relevant maximum streak length that you should allow is almost never a single pixel (or sensel, for that matter). The relevant quantity is the ‘circle of confusion’, which under usual viewing conditions is about 30 microns on a full-format camera. This is quite close to 4-5 pixels on the 5D, so the 30 second exposure from the “500 rule” doesn’t sound bad. Of course, it’s not adequate under all conditions, but it’s a good rule of thumb.
http://en.wikipedia.org/wiki/Circle_of_confusion

1. Steven Christenson

Thank you for you comments, Tiberiu, though I’m not sure I agree. Circle of Confusion describes perceived sharpness and in particular why blurring occurs when something imaged spans more than one “pixel”. I have a 20×30 inch print of a Bristlecone Pine Tree with a Milky Way behind it. Even though it was taken at 30 seconds or less with a wide field on a full frame camera (well under the 600 or 500 rules), at a distance of 3 feet the stars are noticeably dashes rather than dots and because the angular speed of the sky is not uniform**, the dashes are not of equal length across the field. Of course I’m being picky.

Your comment does, however, underscore my argument why the “600 (or 500) rule” can not be blindly applied to all cameras because of the large differences in sensor geometry.

**The stars nearer to the celestial equator move farther across the sensor than those near the celestial poles in any unit of time.

1. Tibi

Steven, the calculation I alluded to has a number of assumptions, as outlined for example in the Wikipedia article. That said, unless you have superhuman vision, you won’t be able to see details much smaller than about 0.2 mm at a distance of 25 cm, or about 0.7 mm at 3 feet. Assuming a focal length of 10 mm on a full-frame camera and the maximal possible angular velocity (which is 1/13750 rad/s), stars take about 40 seconds to move that much.

Now, of course, some of these numbers are somewhat subjective, and I’m not trying to tell you what you can or cannot see in a picture that you’re looking at and I’ve never seen. Like all rules in photography, this “6(5)00 rule” can’t be more than a rule of thumb, and experimentation should help you decide what works for you. My only point was that in most circumstances, there’s no reason to reduce the shutter speed to the point where the streaks are shorter than single pixels (or sensels).

1. Steven Christenson

Doing the math: a 6 pixel streak when enlarged to 20 x 30 inches (508 x 762 mm) becomes a 0.81 mm feature on the print of an uncropped image – larger than the “theoretical” maximum. The feature would become even larger if the 3753×5616 image was cropped. My vision is not great by the way, and I can clearly see dashes not dots for stars. Is the streaking painfully obvious? no, but this may well be a case of reality trumping theory. Order my print and I’ll bet you’ll notice, too. (at 20×30 inches)

In summary:

1. Longer exposures don’t gain any more stars. Once stars have moved on to the next pixel(*) no additional photons at that site are being counted.
2. A full frame sensor and a 10mm lens is not typical. There are almost no full frame lenses at speeds of f/2.8 or better. Indeed nearly all 10mm lenses are for APS-C (not full frame) sensors where the geometry is less favorable.

1. Tibi

Steven, if you take a picture with a 50MP camera instead of 20MP, the maximum exposure time to avoid star trails will stay exactly the same, although the length of the trails in pixels will change; that’s because the limit is given by human visual acuity, and a pixel in a 20MP picture is already much smaller than the circle of confusion (CoC). Of course, the CoC will change in the obvious way if you crop the picture, or if you pixel-peep, or look at a huge print from a very small distance. But sensor resolution won’t change it as long as it’s above a certain threshold (which is probably in the range of 5-10 megapixels). If the camera is not full frame, you just need to multiply the focal length by the crop factor before plugging it into the “X/f” rule.

“Longer exposures don’t gain any more stars.”
Even if that were true, you might still want a longer exposure to get a nicer sky color, as you do in the picture you posted above — very nice picture, by the way.

But also, while it’s true that the intensity of each star pixel no longer increases once the streak grows longer than a pixel, when you look at a print-out, you don’t see individual pixels. Your eye averages the intensity over an area approximately given by the CoC. As such, any trail that’s shorter than the CoC will look like a point whose intensity is proportional to the length of the trail. Here’s a figure that can help you test that. If you look at it from far enough away, some of the trails will look like points of different intensities, although all the pixels of all the trails are exactly as bright.

That means longer exposure does buy you more intensity for stars, and implicitly, it can increase the number of stars that are visible in the picture, as long as the trails stay shorter than the CoC.

2. Steven Christenson

I don’t see any night/star shots in your portfolio, so I’d like to encourage you to try some experimentation to see if your actual results match the theoretical values as closely as you think.

Your theoretical arguments concerning CoC seem sound except that my experience tells me otherwise. Once a visual threshold is crossed it is not a matter of apparent sharpness. Though my eye might perceive a dash as a brighter element that element is no longer star-like.

Also, be careful about assuming that the sensor density doesn’t affect the outcome. It does – though in an important different way then you’ve noted. Since the denser sensor has less area per pixel* every measured value is SMALLER than it would be on a less dense sensor, and thus all values are closer to the native sensor noise level. This reduction decreases the usable dynamic range just as any underexposure does.

While the dense sensor has more adjacent pixels that somewhat offset the reduction in contrast, it doesn’t really buy much else. Assuming only the sensor density is increased the star streaks now cover more pixels. As a result one CANNOT make a larger image from the greater density image unless we move people proportionately farther away. My clients hang 11×14 or 20×30 or 40×60 prints on their walls but they don’t stand farther away to view them!

Even ignoring the streaking issue in any night sky image increasing exposure time is usually not an effective solution. A longer exposure results in more accumulated skyglow and thus a reduction in the overall contrast between the stars and the sky. I found this out the hard way when trying to collect meteors. Long exposures miss fewer meteors, but they also allow the sky glow to overwhelm the dimmer meteors.

5. Enrico

That math is all great but practically you end up using around 30s shutter speed anyway (for Milky Way and meteor shots) with a wide angle lens (20′s and lower) the only thing you may change is the aperture and ISO. With 35mm and longer effective focal length it’s difficult to compose Milky Way shots or catch meteors.
That means you will always have star streaking except you use a camera tracking device.

1. Steven Christenson

As a result of my work, I’ve found that you can get good results with lower shutter speeds (15 seconds to 20 seconds). It means more processing, but it is nigh impossible to remove streaks after the shot as compared to boosting contrast and lightness on a shorter shot. Of course it’s important to have a camera capable of good high-ISO performance.

For example I shot this at 25 seconds, 15mm rather than the 30 seconds to which I had become accustomed. 25 seconds at 15mm meant 1.5 pixels less streaking.

At this size the streaks are not at all obvious but when printed large you will definitely notice.

6. Steve Jones

Trying to contact privately, because I can’t find a post on it. Did you ever find a way to shoot trails without gaps…without the extra work in Photoshop?

1. Steven Christenson

Yes I have, Steve. However the technique/method is something I do not plan to publish on my blog – perhaps in my book. It’s too valuable, I think to have others appropriate it for their own blogs.

7. Gilles

Following the discussion started by Fernando on the main page, I had a closer look at this page, and as a scientist I regret some loose wording in the way you present the central formula:
- The distance travelled by a star on the sensor depends on the actual focal length of the lens, not its effective focal length (the laws of optics don’t know about our preference for the “full frame” sensor). The crop factor was added after the fact in the equation, to take into account the extra magnification that will be required for a smaller sensor when displaying the image. You’re most probably right to do so, but then d is not “the distance … that a star travels across a sensor”, as stated in the text. It’s an effective distance if you like, obtained after the camera sensor has been stretched to the size of your reference sensor. Knowing the physical size of a pixel on the sensor, the number of pixels crossed by a star is obtained from the physical distance travelled (computed without the crop factor). Then, the acceptable number of pixels, for a streak to be unnoticed, does depend on the crop factor (amongst other things).
- As you know, stars have different motions depending on their declination. There is a factor cos(δ) in the general equation, that was here implicitly chosen to be equal to one. So d is not the distance travelled by “a star”, but by a star at the equator. It is also the maximum distance travelled by any star (from which you can derive the maximum exposure time, as wanted). This is clarified elsewhere on the page, but regrettably not when the equation is introduced.
- The number “13750″ is actually an approximation of 86400/2π, the inverse of which has a simple physical meaning: it’s the angular frequency of stars at the equator (in radians per seconds).

1. Steven Christenson

You are correct! I tried to avoid all of the details that seem irrelevant. I clearly stated that the remarks addressed a star at the celestial equator because it’s the worst case that matters. The star declination is irrelevant for nearly all wide field landscape astrophotography where a field of view of 60 to 120 degrees includes stars at nearly every declination. Based on the questions I get, it’s clear the average reader doesn’t think about (or want to think about) how the direction they aim their camera will affect the outcome.

I did make an unstated assumption which I should have set forth: regardless of the size of the sensors used to obtain the image in the end we will be comparing finished images of the same size viewed from the same distance. The “effective distance” as you’ve called it brings all that together.

8. Jay Kenyson

I have a Nikon D8oo with a Nikkor 14-24mm F2.8 wide angle lens
Could you advise me in what settings I would need to achieve good
results ??

1. Steven Christenson

@Jay, Unfortunately, I’m too lazy to look up the specifications of the D800 sensor and you haven’t specified what focal length you will use, how big you plan to make your finished image or how far you’ll stand away from it. But don’t worry, as I noted in the article the length of exposure is about aesthetics. Start with the 500 rule, check your results and then change your settings as necessary.

Here is some quick help: 500 / 14 = 35; while 500 / 24 = 20 seconds.

9. Jenn Grover

This is a fascinating discussion. Thanks for sharing your findings. I am a novice at shooting astrophotography and have been following the 600 rule but not thrilled with streaking. I am going to try a little experimenting with your formula. shooting with the 5D III it might buy me a little higher ISO.

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