500 Rule & NPF Calculator

Pick your sensor, megapixels, focal length, aperture and the declination of the sky you’re framing: get the longest exposure for pinpoint stars with the NPF rule, compare it with the 500 rule and see how many pixels the star would travel with each.

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Your camera, lens and sky

Sensor
Sensor resolution

The declination of the stars in your frame closest to the celestial equator. With a wide-angle lens the equator is almost always in shot: leave it at 0°.

NPF trailing allowed (k)
Classic rule to compare
Max exposure · NPF rule
0
0pixel pitch (µm)
0crop factor
0star trail (px)
0star diameter (px)
Typical setups

NPF vs the 500 rule

RuleExposureOn cameraTrailTrailing (k)
Full NPF (k = 1)
Simplified NPF

“On camera” is the next standard shutter speed down (in third stops), so you never go over. Trail: how far the star moves across the sensor during the exposure. Trailing (k): trail ÷ star radius in the NPF model; 1 = pinpoint, 2 = barely noticeable, 3 = noticeable.

How the 500 rule and the NPF rule are calculated

The 500 rule only looks at the equivalent focal length:

t=500focal length × crop factor

Frédéric Michaud’s NPF rule adds the aperture N, the pixel pitch p (in µm) and the declination δ. This is the full form, with the focal length f in mm:

t=k ×16.9·N + 13.7·p + 0.1·ff · cos δ

It comes from a physical model: a star is not a point on the sensor but a spot whose diameter adds up atmospheric turbulence (3″ of seeing), lens diffraction (4.47·λ·N, with λ = 550 nm) and the sensor’s demosaicing (2 pixels). Meanwhile the Earth’s rotation (one turn every 86,164 s relative to the stars) shifts it by f·cos δ ÷ 13,713 mm per second. At k = 1 the star moves by its radius and stays round; at k = 2, by one diameter (barely noticeable); at k = 3, by three radii.

The simplified NPF rule, (35·N + 30·p) ÷ f, is the one most calculators use: it amounts to k = 2 with the star on the celestial equator and no turbulence. The calculator’s table shows all three side by side and, for each, how many pixels the star travels and which k that equals.

Worked example: the 500 rule vs NPF with a 20 mm lens

A 20 mm lens at f/2.8 on a 24 MP full-frame camera (6.0 µm pixels), aimed at a patch of sky that includes the celestial equator (δ = 0°):

  • 500 rule: 500 ÷ 20 = 25 s. The star travels 6.1 pixels: trailing of k = 3.8.
  • Simplified NPF: (35 × 2.8 + 30 × 6) ÷ 20 = 13.9 s (13 s on camera).
  • Full NPF at k = 1: (16.9 × 2.8 + 13.7 × 6 + 0.1 × 20) ÷ 20 = 6.6 s (6 s on camera). The star is 3.2 pixels across and moves 1.6: its radius.

With the same lens on a 61 MP full-frame camera (3.76 µm pixels), the 500 rule still says 25 s, but now the star travels 9.7 pixels. Full NPF at k = 1 drops to 5.0 s.

500 rule or NPF rule: which to use

The 500 rule was born with film, which resolves far less than a modern sensor, and it ignores both megapixels and aperture. With today’s 3–6 µm pixels it leaves visible trails at 100 % view, more so the higher the camera’s resolution. Use NPF: k = 1 if you will enlarge or crop; k = 2 (or the simplified rule) for the web, social media and moderate prints, as Michaud suggests. His model assumes a lens with moderate aberrations and 3″ seeing: treat the result as a starting point and check your first frame zoomed in.

Methodology & sources

Every result is computed in your browser with the formula shown above — nothing is sent to a server. The numeric model is based on this data and these references:

Formula reviewed and verified against real-world cases.

500 rule and NPF rule FAQ

What is the 500 rule?
A rule from the film era: the longest exposure before stars trail is 500 divided by the equivalent focal length (focal length × crop factor). With a 20 mm lens on full frame it gives 25 seconds. Frédéric Michaud derives it from the 0.029 mm circle of confusion of 35 mm film: depending on declination you get between 398 and 562, and 500 is the middle value.
What is the NPF rule?
A rule by Frédéric Michaud of the Astronomical Society of Le Havre that accounts for the aperture (N), the pixel pitch (P) and the focal length (F). It works out how long the star can move before it leaves its own blur spot, which depends on atmospheric turbulence, lens diffraction and the sensor’s demosaicing. Its simplified form is (35·N + 30·p) ÷ f seconds.
Why does the NPF rule give shorter times than the 500 rule?
Because modern sensors have far smaller pixels than film grain. With a 20 mm lens at f/2.8 on full frame, the 500 rule gives 25 s on both a 24 MP and a 61 MP camera, but the star travels 6.1 pixels on the first and 9.7 on the second. NPF corrects for that: 6.6 s and 5.0 s for pinpoint stars (k = 1).
Which declination should I enter?
The one closest to the celestial equator among the stars in your frame, ignoring the sign: stars on the equator (0°) move fastest. A wide-angle frame almost always includes some, so 0° is the safe choice. Only with tighter framing away from the equator (say the Milky Way core at −29°, or the area around Polaris) can you expose for longer.
What is the difference between the full and the simplified NPF rule?
The simplified rule, (35·N + 30·p) ÷ f, assumes you accept the star moving one diameter (k = 2), that it sits on the celestial equator and that there is no turbulence. The full rule, k·(16.9·N + 13.7·p + 0.1·f) ÷ (f·cos δ), lets you choose the trailing you accept (k from 1 to 3) and the declination. At k = 2 and 0° they almost agree: 13.2 s against 13.9 s with a 20 mm f/2.8 lens and 24 MP.
How do I find my camera’s pixel pitch?
Divide the sensor width in millimetres by the image width in pixels and multiply by 1000. A 24 MP full-frame camera is 6000 pixels wide: 36 ÷ 6000 × 1000 = 6.0 µm. A 61 MP one is about 9566 wide: 3.76 µm. The calculator works it out from the megapixels; if you know your camera’s exact figure, pick “Pixel pitch” and type it in.