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The limit

How long you have to watch

Two constituents can only be separated once the record is long enough for them to drift a full cycle apart: T = 360° / |σᵢ − σⱼ|. The closer two speeds are, the longer the record has to be.

The one page where the answer is known in advance

When we pick the answer, how close does the solver get?

Every other number on this site is fitted from a real record, and there is no way to check it — the sea does not publish its own constants. So here it is the other way round: we chose four constituents, generated a 60-day record from them, added noise, and fitted it with exactly the solver every station page uses. What went in and what came back are side by side below.

What we put inM2 0.600 m @ 120° · S2 0.300 m @ 150° · K1 0.250 m @ 300° · O1 0.150 m @ 280°These four constants are invented. Not a place, not published values, and no other page reads them — the round numbers are chosen so the comparison is easy to make.

When we pick the answer, how close does the solver get?
Noise per readingWhat came backWorst errorPhase errorResidual RMS
M2 0.6000 · S2 0.3000 · K1 0.2500 · O1 0.1500000.0000
± 0.02 mM2 0.5987 · S2 0.3000 · K1 0.2509 · O1 0.14900.00130.26°0.0200
± 0.05 mM2 0.5968 · S2 0.2999 · K1 0.2523 · O1 0.14760.00320.66°0.0500
± 0.15 mM2 0.5903 · S2 0.2998 · K1 0.2570 · O1 0.14270.00971.93°0.1501

Why a full cycle

K1 and P1 differ by 0.0821°/h — less than a tenth of a degree. These four two-day snapshots are taken from different points across the 183 days this pair demands, and show what the criterion is waiting for.

  1. Day 0 — 0° apart: one wave with two names

    Day 0 — 0° apart: one wave with two names

    K1 - - P1 · 2 days

  2. Day 45.7 — 90° apart

    Day 45.7 — 90° apart

    K1 - - P1 · 2 days

  3. Day 91.3 — 180° apart: opposed, and this is where they become separable

    Day 91.3 — 180° apart: opposed, and this is where they become separable

    K1 - - P1 · 2 days

  4. Day 183 — back to 0°: as they began

    Day 183 — back to 0°: as they began

    K1 - - P1 · 2 days

What each length of observation buys

The standard set holds ten constituents. Here is how many of them an observation this long can actually separate — not how many a solver that does not check would hand back.
Constituents separable by length of observation
Time observedSeparableDropped
1 daysM2 M4 (2/10)S2 N2 K2 K1 O1 P1 Q1 MS4
7 daysM2 K1 M4 (3/10)S2 N2 K2 O1 P1 Q1 MS4
15 daysM2 S2 K1 O1 M4 MS4 (6/10)N2 K2 P1 Q1
29 daysM2 S2 N2 K1 O1 Q1 M4 MS4 (8/10)K2 P1
90 daysM2 S2 N2 K1 O1 Q1 M4 MS4 (8/10)K2 P1
183 daysM2 S2 N2 K2 K1 O1 P1 Q1 M4 MS4 (10/10)
365 daysM2 S2 N2 K2 K1 O1 P1 Q1 M4 MS4 (10/10)

The full ladder: every separation the standard set demands

Thirteen separations decide how long a survey has to run. The rest are met by any record of more than two days, and are listed below for completeness.
Minimum record length to separate each pair of constituents
PairSpeed difference (°/h)Minimum length (days)
K1 / P10.0821183
K2 / S20.0821183
M2 / N20.544427.6
O1 / Q10.544427.6
M4 / MS41.015914.8
M2 / S21.015914.8
O1 / P11.015914.8
K1 / O11.098013.7
K2 / M21.098013.7
N2 / S21.56039.6
P1 / Q11.56039.6
K2 / N21.64249.1
K1 / Q11.64249.1

The remaining separations, all under two days. A record too short to meet these is too short to fit at all — a single constituent needs one full period just to separate from the mean level.

Q1/Z₀ 1.1·K1/N2 1.1·N2/P1 1.1·K1/M2 1.1·O1/Z₀ 1.1·M2/P1 1.1·N2/O1 1.0·K1/S2 1.0·P1/Z₀ 1.0·K1/Z₀ 1.0·K1/K2 1.0·M2/O1 1.0·P1/S2 1.0·N2/Q1 1.0·K2/P1 1.0·M2/Q1 1.0·O1/S2 0.9·K2/O1 0.9·Q1/S2 0.9·K2/Q1 0.9·K2/M4 0.5·M4/S2 0.5·N2/Z₀ 0.5·K2/MS4 0.5·MS4/S2 0.5·M2/Z₀ 0.5·M2/M4 0.5·M4/N2 0.5·M2/MS4 0.5·S2/Z₀ 0.5·K2/Z₀ 0.5·MS4/N2 0.5·K1/M4 0.3·M4/P1 0.3·K1/MS4 0.3·M4/O1 0.3·MS4/P1 0.3·M4/Q1 0.3·MS4/O1 0.3·MS4/Q1 0.3·M4/Z₀ 0.3·MS4/Z₀ 0.3

The records on this site

The ladder above is universal. To watch it meet a real record — constituents dropping out and the condition number climbing as the window shortens — pick a station.