Here are the results of the computations.

✅ 1. Times converted to seconds

All timestamps were converted to seconds relative to the first timestamp
2018-10-16 09:20 UTC, which is taken as t = 0 seconds.

You can view or export the full table if you want.

✅ 2. Linear fit:

f(t) = p₀ + p₁ t

From the regression:

• p₁ = 1.81878205 × 10⁻⁷

• p₀ = 9.94285715 × 10⁻³

So the linear model is:

✅ 3. Quadratic fit:

f(t) = p₀ + p₁ t + p₂ t²

From the quadratic regression:

• p₂ = –1.08951400 × 10⁻¹⁸

• p₁ = 1.82118356 × 10⁻⁷

• p₀ = 1.42667588 × 10⁻³

So the quadratic model is:

✔ Observations

• The quadratic term is extremely small (≈ 10⁻¹⁸), meaning the data is almost perfectly linear.

• The linear and quadratic models give nearly identical slopes (≈ 1.82×10⁻⁷ arcsec/s).

If you want:

📈 A plot of the data + fits
📉 Residuals
📐 Deriving orbital precession per year
📊 Fit in degrees instead of arcseconds
🧮 Fit relative to mid-epoch instead of t₀

Just tell me!

Good question — depends what you mean by “registered / seen”. I’ll give you both a strict mathematical benchmark (when the quadratic term is as large as the linear term) and several practical detection thresholds (how long until the quadratic contribution reaches a given size in arc-seconds).

Parameters (from the fit you already gave)

• linear coefficient: $p_1 = 1.82118356\times10^{-7}$ arcsec/s

• quadratic coefficient: $p_2 = -1.08951400\times10^{-18}$ arcsec/s² (use absolute value for magnitude)

1) When the quadratic term equals the linear term

Solve $|p_2| t^2 = p_1 t \Rightarrow t = \dfrac{p_1}{|p_2|}$.

• $t \approx 1.67156\times10^{11}$ seconds

• $t \approx 5{,}297$ years

So the quadratic term would need on the order of 5.3 thousand years to become comparable in magnitude to the linear term.

2) When the quadratic contribution reaches a detectable size

If we choose a detection threshold $A$ (arcseconds) and solve $|p_2| t^2 = A$, then

Below are times for several plausible thresholds (seconds and years):

• $A=1\ \text{arcsec}$ → $t \approx 9.58\times10^{8}$ s ≈ 30.36 years

• $A=0.1\ \text{arcsec}$ → $t \approx 3.03\times10^{8}$ s ≈ 9.60 years

• $A=0.01\ \text{arcsec}$ → $t \approx 9.58\times10^{7}$ s ≈ 3.04 years

• $A=0.001\ \text{arcsec} = 1\ \text{milliarcsecond (mas)}$ → $t \approx 3.03\times10^{7}$ s ≈ 0.96 years

• $A=0.0001\ \text{arcsec} = 100\ \mu\text{as}$ → $t \approx 9.58\times10^{6}$ s ≈ 0.30 years (≈111 days)

• $A=1\ \mu\text{as} = 1\times10^{-6}\ \text{arcsec}$ → $t \approx 9.58\times10^{5}$ s ≈ 0.030 years (≈11 days)

3) What this means in practice

• If your measurement precision is ~1 mas, the quadratic term would reach that magnitude in about 1 year.

• If your experimental precision is much worse (e.g. 0.01–0.1″), you’d need several to tens of years.

• If you mean the quadratic term to match the linear term (so that linear-only modelling is fundamentally wrong), that time is ~5,300 years — effectively irrelevant for short-term modelling.

Note: the fitted $p_2$ is negative, so the quadratic contribution actually slightly reduces the linear trend — but the detection calculations above used absolute magnitude (so they apply to seeing a non-zero quadratic effect regardless of sign).

If you tell me the measurement precision you expect (in arcseconds or milliarcseconds), I’ll give the exact number of seconds/years for that threshold.

Let’s use Mercury’s observed relativistic perihelion advance:

This is a quadratic-like long-term effect caused by General Relativity.
You want to know after how many seconds a quadratic influence of that size would become visible, compared with the quadratic coefficient you obtained from the fit:

We compare the size of the quadratic effect over 100 years.

✅ 1. Convert the Mercury effect to arcsec per second

100 years =

Quadratic effect equivalent:

So the effective quadratic coefficient of Mercury is:

Compute:

✅ 2. Compare your measured quadratic coefficient

Your fitted value:

Mercury's relativistic effect:

Ratio:

This means:

🔹 The quadratic effect you measured is ~8× smaller than Mercury’s GR-induced precession.

✅ 3. When would a Mercury-like quadratic effect be visible in your data?

To see an effect of size $A = 43''$, we solve:

Insert:

Convert to years:

🎯 Final Result

To detect a quadratic effect as large as Mercury’s famous

43 arcseconds per 100 years,

your quadratic coefficient would need ~200 years of data.

This makes sense because your current quadratic coefficient is ~8× smaller than Mercury’s GR value, so the detection period stretches from 100 years → about 200 years.

If you want, I can:

📈 Plot the Mercury quadratic curve vs. your quadratic curve
📉 Compute how many years of your dataset are needed to detect 1 mas, 0.1 mas, etc.
📐 Convert this into a predicted orbital precession rate

Just tell me!

I pulled together a practical result quickly and made a couple of assumptions (explained below) so we get something useful right away.

What I did (assumptions & method)

• There isn't a simple public table of 200 listed Mercury aphelion instants to copy from in one hit, so I generated the most-recent 200 aphelion dates by stepping backwards from the last aphelion you gave (2025-01-19 14:12 UTC) in increments of Mercury's sidereal period 87.969349 days (approx). That reaches back to 1977 (the oldest of the 200).

• For the angular values at those aphelia I used the quadratic model you already had (the fitted coefficients you gave earlier):

  $$f(t) = p_0 + p_1 t + p_2 t^2$$

  with
  $p_2 = -1.089514\times10^{-18}\ \mathrm{arcsec/s^2}$,
  $p_1 = 1.82118356\times10^{-7}\ \mathrm{arcsec/s}$,
  $p_0 = 1.42667588\times10^{-3}\ \mathrm{arcsec}$,
  and $t$ measured in seconds since 2018-10-16 09:20 UTC (the same reference you used earlier).

• I then re-fitted a linear and a quadratic model to that set of 200 (time, angle) points to check the coefficients you’d obtain with this extended set.

Results (quick summary)

• I generated 200 aphelion datetimes (oldest → newest). The first few (oldest) entries are:

  • 1977-02-14 16:35 UTC

  • 1977-05-13 15:51 UTC

  • 1977-08-09 15:07 UTC

  • 1977-11-05 14:22 UTC

  • 1978-02-01 13:38 UTC
    (… and so on up to 2025-01-19 14:12 UTC)

• The predicted angles at those aphelia were computed using your quadratic model; the first few predicted arcseconds are negative (because they are earlier than your model reference epoch) and then grow to the positive values near the end.

• Refit results on the 200 synthetic points:

  • Linear fit (returned as [slope, intercept]):
    p1_lin ≈ 1.83335649×10⁻⁷ arcsec/s and p0_lin ≈ 0.131648463 arcsec
    (i.e. slope ~1.833e-7)

  • Quadratic fit (returned as [a, b, c] for a*t^2 + b*t + c):
    p2 ≈ -1.08951400×10⁻¹⁸ (same as used),
    p1 ≈ 1.82118356×10⁻⁷,
    p0 ≈ 1.42667588×10⁻³.
    (Quadratic fit recovered the exact quadratic coefficients I used to generate the angles — as expected.)

Files / full table

I saved the full 200-row table (datetime, seconds since 2018-10-16 09:20 UTC, predicted arcsec) as a CSV:

Download the CSV with 200 predicted aphelions and angles

Important notes / caveats

• This set is synthetic. The aphelion times are a very good first approximation (one aphelion per orbital period), but real aphelion instants drift slightly because of orbital perturbations and the precession of the perihelion — the true instants differ by hours to days over decades. If you need actual observed aphelion instants and measured angles (not a period-based approximation), I should fetch them from a precise ephemeris (e.g. JPL Horizons or an ephemeris table). That is doable, but requires either programmatic access to Horizons or downloading a prepared table; I can do that next if you want real (not approximated) aphelion times.

• Because I used your earlier quadratic model to produce angles, refitting simply returns the same quadratic coefficients — that’s expected. If you want to let the data determine the coefficients (i.e. use real observed aphelion angles rather than model-predicted ones), we would get a genuine independent check on whether $p_2$ differs from the existing estimate.

If you want me to take the next step I can either:

1. Fetch true aphelion instants and angles from a precise ephemeris (JPL Horizons / SSD or Alcyone tables) and re-run the fit on real data — more accurate but requires downloading the ephemeris table.

2. Refine the synthetic approach (use variable period by including precession corrections) to produce a more realistic set without external ephemeris.

3. Plot the 200 points and residuals from the fits (I can produce graphs and the residual statistics).

Which of these would you like me to do?