astronomy - What accounts for the discrepancies in my calculations of year lengths?

A common exercise in many introductory astronomy texts is to use the lengths of various kinds days to calculate the approximate length of the corresponding year.

For example, ratio $k$ of the length of the sidereal day to the mean solar day $D$, can be used to estimate the length of the tropical year, $a_t$ from¹

$$ a_t \approx \left(k-1\right)^{-1}$$

and, similarly, the ration of the length of stellar day, $k'$, along with the established value for general precession, can be used to estimate the length of the sidereal year²

$$ a_s \approx \left(Pk'-1\right)^{-1}$$

where $P \approx 0.999999894$.

But both of the values I arrive at using these calculations differ from established values. The value I get for $a_t$ is off by about a quarter of a second per year,³ while my value for $a_s$ is off by nearly 20 minutes.⁴

What accounts for these differences? Specifically,

1. Is my reasoning sound, or have a made an error in my thinking or calculations?


2. In particular, do my methods fail to incorporate some phenomena that should be accounted for, or fail to remove some that should be omitted?


3. If there are such omitted or wrongly included phenomena, do they correspond to the differences I'm seeing between my computed values and the established ones; for example, could I use those differences to determine some constant associated with the phenomena?


4. Are these differences just a result of the fact that each of these established values (for $a_t$, $a_s$, $k$, $k'$, and $D$) are determined independently, so that it is too much to expect them to "line up" any better than they do in these results?

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^{Please forgive the extensive footnotes, but I want to be absolutely sure that my reasoning and calculations are, at least, sound.}

^{1: Given the average angular velocity, $\dot{\alpha }$, of the meridian with respect to the ecliptic coordinate system, it must be the case that in a mean solar day, the sun has passed through an angle in that system of$$\theta =\dot{\alpha} D-24^h$$so that it will take $$\frac{24^h}{\theta}=\left( \frac{\dot{\alpha }}{24^h}D-1\right)^{-1}$$days for the sun to complete the circuit of the ecliptic that determines the mean tropical year. And since by definition$$\dot{\alpha } = \frac{24^hk}{D}$$the length of the tropical year is$$\left(k-1\right)^{-1}$$}

^{2: Similarly, from the average angular velocity of the meridian with respect to fixed stars along the ecliptic, $\dot{\sigma }$, we can calculate the angular progress of the sun against fixed stars over the course of a mean solar day$$\phi =\text{ }\dot{\sigma} D- 360{}^{\circ}$$and since this progress exactly corresponds to the angular progress of the earth on its orbit with respect to fixed stars (i.e. excluding precession of the ecliptic) it will take$$\frac{360{}^{\circ}}{\phi }=\left( \frac{ \dot{\sigma }}{360{}^{\circ}}D-1\right)^{-1}$$mean solar days for a complete circuit of that orbit, which is the definition of the sidereal year; where $\dot{\sigma }$ consists of two components: the angular velocity of the earth on its axis which is by definition exactly $360{}^{\circ}$ per stellar day, $D/k'$, and the rate that that axis precesses relative to fixed stars$$\dot{\sigma } = \frac{360{}^{\circ}k'}{D}-\frac{50.28792\text{''}}{a_j}=\frac{360{}^{\circ}k'}{D} P$$where, where $a_j$ is a julian year and thus$$P\approx 0.999999894$$so that it will take$$\left(Pk'-1\right)^{-1}$$days for the sun to complete full circuit against the background stars, which is the definition of the duration of the sidereal year.}

^{3: Using the IERS values for $D$ and $k$ I arrive at a value for $a_t$ which is too small by $0.265284\text{ s}$.}

^{4: Using the IERS values for $D$ and $k'$ and the rate of precession, I arrive at a value for $a_s$ which is too large by $18.7156\text{ min}$, while, confusingly, ignoring the effect of precession (by setting $P=1$) produces a closer value, that comes up short by $1.6943\text{ min}$, which is more than an order of magnitude smaller ($4.1478\text{"}$) than the expected effect of omitting precession over the course of a sidereal year ($50.2888\text{"}$).}

^{Note that I'm pretty sure my discrepancies are not a result of daily variations (due to the obliquity of the ecliptic and the eccentricity of the Earth's orbit) in the values of ecliptic velocities, since these all average out over the course of a year to exactly the values used.
For example, the daily motion of the celestial sphere is around the celestial pole, not the ecliptic pole, so even if the velocity of the meridian around the pole is constant, the velocity with respect to the ecliptic will vary as the meridian sweeps through the ecliptic at varying declinations. Though the average ecliptic velocity over a complete circuit of the ecliptic will be $\dot{\alpha}$, the meridian completes less than a full circuit of the ecliptic in completing a full equatorial circuit (due to precession), and more than a complete circuit in a mean solar day (due to the sun's movement). However both of these omitted and added segments of the ecliptic follow the sun in the course of a year through the full range of declinations, so that over the course of a tropical year the average is $\dot{\alpha}$. So in the calculations for the tropical year, it is the fact that things average out for complete equatorial circuits that lets me use $24^h$ as a constant in the ecliptic system (as in the first equation in note 1), while it is the fact that things average out for mean solar days that let me apply $\dot{\alpha}$ to $D$.}