Where 15.04 arcseconds a second comes from
The Earth turns once relative to the stars in a sidereal day: 23 hours, 56 minutes and about 4 seconds, or 86,164.09 seconds. The four-minute difference from a solar day is the Earth having moved along its orbit, so it has to turn a little further to bring the sun back to the same place, and the stars do not care about that.
One turn is 360 degrees, which is 1,296,000 arcseconds. Divide by 86,164.09 seconds and you get 15.0411 arcseconds per second. That is the rate at the celestial equator. Away from it, the star traces a smaller circle, and the radius of that circle scales with the cosine of the declination, so the linear rate on the sky scales the same way. At declination 60 the cosine is one half and the drift is 7.52 arcseconds a second. At declination minus 60 it is the same, because cosine does not care about the sign.
The rules of thumb, and what they leave out
The old rule divides a number by the focal length and gives you seconds. The number is 500 in the most common version, 600 or 400 or 300 in others, and there is a more elaborate variant that pulls in pixel size and aperture. All of them are conventions rather than measurements, which is why this page asks you which one you compare against instead of picking one.
What they leave out is the two things that actually decide the answer. Pixel size sets how much sky each pixel covers, so the same angular smear is invisible on a coarse sensor and obvious on a fine one. Declination scales the drift by up to a factor of nothing. A rule that ignores both was written for a print viewed at arm length from a 35 mm negative, and it does about as well as you would expect when applied to a 3.76 micron pixel viewed at 100 percent.
| Situation | What the divisor rule misses |
|---|---|
| Dense sensor, short lens | Reads generous — the smear the rule allows is several pixels wide |
| Target near the pole | Reads conservative — the cosine has removed most of the drift |
| Frame destined for a big print or a crop | The rule assumed a viewing distance you are not using |
| Target rising in the east or setting in the west | Refraction near the horizon adds drift the rule never modelled |
Deciding your own trailing budget
The calculator asks how many pixels of smear you will accept, and that is the only genuinely subjective input. One pixel is strict and gives round stars at any enlargement. Two to three is where most people land for images that will be viewed whole rather than pixel-peeped. Beyond that you are photographing short star trails, which is a legitimate thing to do on purpose and an annoying thing to discover afterwards.
One useful way to set it: take the star width you actually record on a tracked frame, in pixels, and allow a smear well under that. A star that is already three pixels across from seeing and optics is not visibly damaged by half a pixel of drift, and is clearly damaged by three. Trailing that is small compared with the star you were going to get anyway is trailing nobody will see.
Periodic error is a sine wave you only see part of
A worm-driven mount repeats a small tracking error once per turn of the worm. Model that as a sine wave with a peak-to-peak amplitude and a period, and the question becomes how much of the swing a sub exposure of a given length can catch. A sub shorter than half the period catches part of the wave; the worst placement is the one straddling the steepest section, and the excursion works out as the peak-to-peak figure times the sine of pi times the sub length divided by the period. Once the sub reaches half a period it can span a full peak to a full trough and sees the whole swing.
That model is deliberately simple. Real mounts have harmonics from other gears in the train, and usually a slow drift underneath from alignment and flexure. It is enough to answer the practical question, which is whether shortening the sub buys anything: at a tenth of the worm period a sub sees about 31 percent of the peak-to-peak error, at a quarter it sees 71 percent, and past half it sees all of it. Halving a sub that was already short of a quarter period roughly halves the error it catches.
Polar alignment error turns into drift and rotation
If the mount axis misses the celestial pole by some angle, the point the mount tracks circles the point you wanted, once per sidereal day. The linear drift rate is the offset multiplied by the rotation rate in radians per second, which is 2 pi over 86,164. Five arcminutes of error works out at about 0.022 arcseconds a second, or 6.6 arcseconds over a five minute sub — enough to elongate stars visibly at most imaging scales, and enough to walk a target across the frame over a night.
The same error also rotates the field, which is what guiding cannot fix. Guiding holds one star still; a rotating field turns everything else around it, so the stars at the corners smear in arcs while the guide star sits perfectly. That rotation depends on where the target sits in altitude and azimuth as well as on the alignment error, which is why it is not computed here — the honest version needs the pointing, not just the misalignment. What is worth knowing is that it exists and that it sets a ceiling on sub length that no amount of guiding raises.
Questions people ask
How long can I expose without a tracker?
Long enough for a star to move less than your trailing budget across the sensor. Work the image scale first — 206.265 times pixel size in microns over focal length in millimetres — then divide the smear you will accept, in arcseconds, by the drift rate. The drift rate is 15.041 arcseconds a second times the cosine of the declination. At 135 mm with 4.3 micron pixels the scale is 6.57 arcseconds per pixel, so aimed at declination 20 where the sky drifts 14.13 arcseconds a second, one and a half pixels of trailing arrives in about seven tenths of a second. That is far shorter than any divisor rule suggests, and it is why the rule and the derivation disagree so sharply on short lenses.
Is the 500 rule accurate?
It is a convention that predates digital sensors, and it carries neither pixel size nor declination, which are the two things that decide the answer. On a dense modern sensor it typically allows several pixels of smear rather than one, which looks fine in a web-sized image and obvious at full resolution. On a target near the celestial pole it is far more conservative than it needs to be. This page shows the rule you enter beside the derived figure and reports how many pixels of trailing the rule is actually permitting at your setup.
Why does declination change the exposure I can use?
A star at declination zero traces the largest possible circle as the sky turns, so it moves the fastest. A star at declination 60 traces a circle half the radius and moves at half the linear rate. A star at the celestial pole traces almost no circle at all and barely moves. The scaling is exactly the cosine of the declination, and it works identically north and south of the equator because the cosine of a negative angle equals the cosine of the positive one.
Does shortening my subs help with periodic error?
Up to a point. A sub much shorter than the worm period only catches part of the swing, and the fraction it catches is the sine of pi times the sub length over the period. At a tenth of a period you see about 31 percent of the peak-to-peak error; at a quarter, 71 percent; at half or more, all of it. So going from a five minute sub to a one minute sub on an eight minute worm cuts the error caught by roughly half. Below that the returns flatten and read noise starts to dominate instead.
Will guiding fix a poor polar alignment?
It fixes the drift and not the rotation. Guiding holds the guide star in place, which removes the slow translation the misalignment causes, so stars near the guide star stay round. What it cannot remove is field rotation: the whole frame turns slowly about the guided point, so corner stars draw arcs while the centre is sharp. That rotation depends on where the target sits in the sky as well as on how far the mount axis misses the pole, and the only fix is a better alignment or shorter subs.