A coordinate-invariant model for deforming geodetic networks: understanding rank deficiencies, non-estimability of parameters, and the effect of the choice of minimal constraints

Abstract

By considering a deformable geodetic network, deforming in a linear-in-time mode, according to a coordinate-invariant model, it becomes possible to get an insight into the rank deficiency of the stacking procedure, which is the standard method for estimating initial station coordinates and constant velocities, from coordinate time series. Comparing any two out of the infinitely many least squares estimates of stacking unknowns (initial station coordinates, velocity components and transformation parameters for the reference system in each data epoch), it is proven that the two solutions differ only by a linear-in-time trend in the transformation parameters. These pass over to the initial coordinates (the constant term) and to the velocity estimates (the time coefficient part). While the difference in initial coordinates is equivalent to a change of the reference system at the initial epoch, the differences in velocity components do not comply with those predicted by the same change of reference system for all epochs. Consequently, the different velocity component estimates, obtained by introducing different sets of minimal constraints, correspond to physically different station velocities, which are therefore non-estimable quantities. The theoretical findings are numerically verified for a global, a regional and a local network, by obtaining solutions based on four different types of minimal constraints, three usual algebraic ones (inner or partial inner) and the lately introduced kinematic constraints. Finally, by resorting to the basic ideas of Felix Tisserand, it is explained why the station velocities are non-estimable quantities in a very natural way. The problem of the optimal choice of minimal constraints and, hence, of the corresponding spatio-temporal reference system is shortly discussed.

Appendices

Proof of Proposition 2

Let us first note that the observables are always estimable quantities (Dermanis 2003) and, therefore, the least squares estimates ${{ \hat{\mathbf{x}}}}_{i\vert S_{k} } (t_{k} )={{ \hat{\mathbf{x}}}}_{i0\vert S_{0} } +(t_{k} -t_{0} ){{ \hat{\mathbf{v}}}}_{i\vert S_{0} } +\mathbf{{ E}}_{i} {{ \hat{\mathbf{p}}}}_{k}$ of the observables ${{ {\hat{\mathbf{x}}}'}}_{i} (t_{k} )={{ {\hat{\mathbf{x}}}'}}_{i0\vert S_{0} } +(t_{k} -t_{0} ){{ {\hat{\mathbf{v}}}'}}_{i} +\mathbf{{ E}}_{i} {{ {\hat{\mathbf{p}}}'}}_{k}$ will be unique. To simplify the notation, we remove here the hats as a symbol for estimates and the network point index i. All initial coordinates and velocities refer to the same point and they are estimates, same as the transformation parameters $\mathbf{{ p}}_{k}$. We retain the index $S_{0}$ to denote the reference system to which initial coordinates and velocities refer.

Therefore, for any two sets of least squares solutions to the stacking problem ${\mathbf{x}}_{0\vert S_{0} }$, ${\mathbf{v}}_{S_{0} }$, $\mathbf{{ p}}_{k}$ and $\mathbf{{ {x}'}}_{0}$, $\mathbf{{ {v}'}}$, $\mathbf{{ {p}'}}_{k}$ it holds that

$$\begin{aligned} {\mathbf{x}}_{S_{k} } (t_{k} )= & {} {\mathbf{x}}_{0\vert S_{0} } +(t_{k} -t_{0} ){\mathbf{v}}_{S_{0} } +\mathbf{{ Ep}}_{k} =\mathbf{{ {x}'}}_{S_{k} } (t_{k} )\nonumber \\= & {} \mathbf{{ {x}'}}_{0} +(t_{k} -t_{0} )\mathbf{{ {v}'}}+\mathbf{{ E{p}'}}_{k} , \end{aligned}$$

(A1a)

$$\begin{aligned} \mathrm{or\, simply}&\nonumber \\ {\mathbf{x}}_{S_{k} } (t_{k} )= & {} {\mathbf{x}}_{S_{0} } (t_{k} )+\mathbf{{ Ep}}_{k} =\mathbf{{ {x}'}}_{S_{k} } (t_{k} )=\mathbf{{ {x}'}}(t_{k} )+\mathbf{{ E{p}'}}_{k},\nonumber \\ \end{aligned}$$

(A1b)

At the same time, we also have two different estimates of ${\mathbf{x}}(t_{k} )={\mathbf{x}}_{0} +(t_{k} -t_{0} ){\mathbf{v}}$, namely

$$\begin{aligned} {\mathbf{x}}_{S_{0} } (t_{k} )={\mathbf{x}}_{0\vert S_{0} } +(t_{k} -t_{0} ){\mathbf{v}}_{S_{0} } , \end{aligned}$$

(A2)

$$\begin{aligned} \mathbf{{ {x}'}}(t_{k} )=\mathbf{{ {x}'}}_{0} +(t_{k} -t_{0} )\mathbf{{ {v}'}}. \end{aligned}$$

(A3)

For $t_{k} =t_{0}$, Eq. (A1a) gives ${\mathbf{x}}_{S_{k} } (t_{0} )={\mathbf{x}}_{0\vert S_{0} } +\mathbf{{ Ep}}_{0} =\mathbf{{ {x}'}}_{0} +\mathbf{{ E{p}'}}_{0}$ and thus

$$\begin{aligned} \mathbf{{ {x}'}}_{0} ={\mathbf{x}}_{0\vert S_{0} } +\mathbf{{ E}}(\mathbf{{ p}}_{0} -\mathbf{{ {p}'}}_{0} ). \end{aligned}$$

(A4)

This means that $\mathbf{{ {x}'}}_{0}$ describes the same network shape as ${\mathbf{x}}_{0\vert S_{0} }$, but in a different reference system $\tilde{{S}}_{0}$, where $\mathbf{{ p}}_{0} -\mathbf{{ {p}'}}_{0}$ are the transformation parameters from $S_{0}$ to $\tilde{{S}}_{0}$. Setting $\mathbf{{ p}}_{0} -\mathbf{{ {p}'}}_{0} =\Delta \mathbf{{ p}}_{0}$, we may rewrite (A4) as:

$$\begin{aligned} \mathbf{{ {x}'}}_{0} ={\mathbf{x}}_{0\vert S_{0} } +\mathbf{{ E}}\Delta \mathbf{{ p}}_{0} \equiv {\mathbf{x}}_{0\vert \tilde{{S}}_{0} },\quad \Delta \mathbf{{ p}}_{0} =\mathbf{{ p}}_{0} -\mathbf{{ {p}'}}_{0} , \end{aligned}$$

(A5)

which is just the first of the results that we wanted to prove.

Note that this allows a direct interpretation of the different estimate $\mathbf{{ {x}'}}_{0}$ as a transformation ${\mathbf{x}}_{0\vert \tilde{{S}}_{0} }$ of the first estimate ${\mathbf{x}}_{0\vert S_{0} }$ from an original reference system $S_{0}$ to a new one $\tilde{{S}}_{0}$, with transformation parameters $\Delta \mathbf{{ p}}_{0}$ (for $S_{0} \rightarrow \tilde{{S}}_{0} )$.

The corresponding transformation of velocity components $\mathbf{{ v}}_{\tilde{{S}}_{0} } =(1+s_{0} )\mathbf{{ R}}({\varvec{ \uptheta }}_{0} ){\mathbf{v}}_{S_{0} } \approx (1+s_{0} )\left( {\mathbf{{ I}}-[\mathbf{{ \uptheta }}_{0} \times ]} \right) {\mathbf{v}}_{S_{0} }$ becomes in first-order approximation ${\mathbf{v}}_{\tilde{{S}}_{0} } \approx {\mathbf{v}}_{S_{0} } -[{\varvec{ \uptheta }}_{0} \times ]{\mathbf{v}}_{S_{0} } +s_{0} {\mathbf{v}}_{S_{0} }$ and since ${\varvec{ \uptheta }}_{0}$, $s_{0}$ as well as ${\mathbf{v}}_{S_{0} }$ are very small, the last two second terms may be neglected so that finally

$$\begin{aligned} {\mathbf{v}}_{\tilde{{S}}_{0} } \approx {\mathbf{v}}_{S_{0}}. \end{aligned}$$

(A6)

This allows us to compute an intermediate estimate

$$\begin{aligned} {\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )={\mathbf{x}}_{0\vert \tilde{{S}}_{0} } +(t_{k} -t_{0} ){\mathbf{v}}_{\tilde{{S}}_{0} } =\mathbf{{ {x}'}}_{0} +(t_{k} -t_{0} ){\mathbf{v}}_{S_{0} }. \end{aligned}$$

(A7)

Subtracting (A7) from (A3), we find that the two estimates differ by

$$\begin{aligned} \mathbf{{ {x}'}}(t_{k} )-{\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )=(t_{k} -t_{0} )(\mathbf{{ {v}'}}-{\mathbf{v}}_{S_{0} } ). \end{aligned}$$

(A8)

It is also possible to relate $\mathbf{{ {x}'}}(t_{k} )$ and ${\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )$ in a different way. From the first part of (A1a) and (A2), it follows that

$$\begin{aligned} {\mathbf{x}}_{S_{k} } (t_{k} )={\mathbf{x}}_{S_{0} } (t_{k} )+\mathbf{{ Ep}}_{k} , \end{aligned}$$

(A9)

while from the second part of (A1a), (A1b) and (A3), it follows that

$$\begin{aligned} {\mathbf{x}}_{S_{k} } (t_{k} )=\mathbf{{ {x}'}}(t_{k} )+\mathbf{{ E{p}'}}_{k}. \end{aligned}$$

(A10)

Subtracting (A10) from (A9) gives

$$\begin{aligned} \mathbf{{ {x}'}}(t_{k} )={\mathbf{x}}_{S_{0} } (t_{k} )+\mathbf{{ E}}(\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k} ). \end{aligned}$$

(A11)

On the other hand, subtracting (A2) from the second part of (A7) and taking (A5) into account gives

$$\begin{aligned}&{\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )-{\mathbf{x}}_{S_{0} } (t_{k} )=\mathbf{{ {x}'}}_{0} -{\mathbf{x}}_{0\vert S_{0} } =\mathbf{{ E}}\Delta \mathbf{{ p}}_{0},\quad \nonumber \\&\Delta \mathbf{{ p}}_{0} =[{\varvec{ \uptheta }}_{0}^{T} \,{\mathbf{d}}_{0}^{T} \,s_{0} ]^{T}. \end{aligned}$$

(A12)

Replacing ${\mathbf{x}}_{S_{0} } (t_{k} )={\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )-\mathbf{{ E}}\Delta \mathbf{{ p}}_{0}$ from (A12) into (A11) gives

$$\begin{aligned} \mathbf{{ {x}'}}(t_{k} )-{\mathbf{x}}_{\tilde{{S}}_{0} } (t_{k} )=\mathbf{{ E}}(\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k} -\Delta \mathbf{{ p}}_{0} ). \end{aligned}$$

(A13)

Since the left hand sides of (A8) and (A13) are equal, their right hand sides will also be equal giving

$$\begin{aligned} (t_{k} -t_{0} )(\mathbf{{ {v}'}}-{\mathbf{v}}_{S_{0} } )=\mathbf{{ E}}(\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k} -\Delta \mathbf{{ p}}_{0} ), \end{aligned}$$

(A14)

which must hold for all epochs $t_{k}$. The only term depending on $t_{k}$ in the right hand side of (A14) is $\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k}$, which must thus be of the linear form $\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k} =\mathbf{{ a}}+(t_{k} -t_{0} ){{ \dot{\mathbf{p}}}}$. This leads to

$$\begin{aligned} (t_{k} -t_{0} )(\mathbf{{ {v}'}}-{\mathbf{v}}_{S_{0} } )=\mathbf{{ E}}(\mathbf{{ a}}-\Delta \mathbf{{ p}}_{0} )+(t_{k} -t_{0} )\mathbf{{ E}}\dot{\mathbf{p}}, \end{aligned}$$

(A15)

a relation which holds for all $t_{k}$ only if $\mathbf{{ a}}-\Delta \mathbf{{ p}}_{0} =\mathbf{{ 0}}$ and $\mathbf{{ {v}'}}-{\mathbf{v}}_{S_{0} } =\mathbf{{ E}}{\dot{\mathbf{p}}}$. It follows that ${{ \mathbf{v}'}}={\mathbf{v}}_{S_{0} } +\mathbf{{ E}}\dot{\mathbf{p}}$, $\mathbf{{ a}}=\Delta \mathbf{{ p}}_{0}$, and $\mathbf{{ p}}_{k} -\mathbf{{ {p}'}}_{k} =\Delta \mathbf{{ p}}_{0} +(t_{k} -t_{0} ){{ \dot{\mathbf{p}}}}$. Therefore

$$\begin{aligned} \mathbf{{ {v}'}}={\mathbf{v}}_{S_{0} } +\mathbf{{ E}}\dot{\mathbf{p}}={\mathbf{v}}_{S_{0} } +[{\mathbf{x}}_{i0}^\mathrm{{ap}} \times ]{{ \dot{{\varvec{\uptheta }}}}}+{{ \dot{\mathbf{d}}}}+\dot{{s}}\,{\mathbf{x}}_{i0}^\mathrm{{ap}} , \end{aligned}$$

(A16)

$$\begin{aligned} \mathbf{{ {p}'}}_{k}= & {} \left[ {{\begin{array}{*{20}c} {{\varvec{ {\uptheta }'}}_{k} } \\ {\mathbf{{ {d}'}}_{k} } \\ {{s}'_{k} } \\ \end{array} }} \right] =\mathbf{{ p}}_{k} -\Delta \mathbf{{ p}}_{0} -(t_{k} -t_{0} ){{ \dot{\mathbf{p}}}}\nonumber \\= & {} \left[ {{\begin{array}{*{20}c} {{\varvec{ \uptheta }}_{k} -{\varvec{ \uptheta }}_{0} -(t_{k} -t_{0} ){{ \dot{\mathbf{\uptheta }}}}} \\ {{\mathbf{d}}_{k} -\mathbf{{ d}}_{0} -(t_{k} -t_{0} ){{ \dot{\mathbf{d}}}}} \\ {s_{k} -s_{0} -(t_{k} -t_{0} )\dot{{s}}} \\ \end{array} }} \right] , \end{aligned}$$

(A17)

and our proof has been completed.

Best fitting of initial transformation parameters between two sets of initial coordinates obtained using different minimal constraints in the stacking problem

The normal equations for the solution of the least squares problem (30) are

$$\begin{aligned} \left( {\sum \limits _{i=1}^n {\mathbf{{ E}}_{i}^{T} \mathbf{{ E}}_{i} } } \right) \mathbf{{ p}}_{0} =\sum \limits _{i=1}^n {\mathbf{{ E}}_{i}^{T} ({{ \hat{\mathbf{x}}}}_{i0}^{(2)} -{{ \hat{\mathbf{x}}}}_{i0}^{(1)} )} , \end{aligned}$$

(B1)

or explicitly

$$\begin{aligned}&\left[ {{\begin{array}{*{20}c} {{\mathbf{C}}-[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]^{2}} &{}\quad {-[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]} &{}\quad {\mathbf{{ 0}}} \\ {[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]} &{} 1 &{} {{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} } \\ {\mathbf{{ 0}}} &{}\quad {({{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}} &{}\quad {S^{2}+({{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}}} \\ \end{array} }} \right] \left[ {{\begin{array}{*{20}c} {{\varvec{ \uptheta }}_{0} } \\ {{\mathbf{d}}_{0} } \\ {s_{0} } \\ \end{array} }} \right] \nonumber \\&\quad =\left[ {{\begin{array}{*{20}c} {{\mathbf{h}}^{(1)}-\mathbf{{ h}}^{(2)}+[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]{{ \bar{\mathbf{x}}}}_{0}^{(1)} -[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]{{ \bar{\mathbf{x}}}}_{0}^{(2)} } \\ {{{ \bar{\mathbf{x}}}}_{0}^{(2)} -{{ \bar{\mathbf{x}}}}_{0}^{(1)} } \\ {s^{(2)}-s^{(1)}+({{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}{{ \bar{\mathbf{x}}}}_{0}^{(2)} -({{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}{{ \bar{\mathbf{x}}}}_{0}^{(1)} } \\ \end{array} }} \right] \end{aligned}$$

(B2)

where

$$\begin{aligned} {{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} =\frac{1}{n}\sum \limits _{i=1}^n {\mathbf{{ x}}_{i0}^\mathrm{{ap}} },\quad {{ \hat{{\bar{\mathbf{x}}}}}}_{0}^{(1)} =\frac{1}{n}\sum \limits _{i=1}^n {{{ \hat{\mathbf{x}}}}_{i0}^{(1)} },\quad {{ \hat{{\bar{\mathbf{x}}}}}}_{0}^{(2)} =\frac{1}{n}\sum \limits _{i=1}^n {{{ \hat{\mathbf{x}}}}_{i0}^{(2)} },\nonumber \\ \end{aligned}$$

(B3)

$$\begin{aligned} {\mathbf{C}}= & {} -\frac{1}{n}\sum \limits _{i=1}^n {[({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )\times ]^{2}},\nonumber \\S^{2}= & {} \frac{1}{n}\sum \limits _{i=1}^n {({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )}, \end{aligned}$$

(B4)

$$\begin{aligned} {\mathbf{h}}^{(1)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {[({\mathbf{x}}_{i0}^{\mathrm{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )\times ]({{ \hat{\mathbf{x}}}}_{i0}^{(1)} -{{ \bar{\mathbf{x}}}}_{0}^{(1)} )},\nonumber \\\mathbf{{ h}}^{(2)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {[({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )\times ]({{ \hat{\mathbf{x}}}}_{i0}^{(2)} -{{ \bar{\mathbf{x}}}}_{0}^{(2)} )}, \end{aligned}$$

(B5)

$$\begin{aligned} s^{(1)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}({{ \hat{\mathbf{x}}}}_{i0}^{(1)} -} {{ \bar{\mathbf{x}}}}_{0}^{(1)} ),\nonumber \\ s^{(2)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}({{ \hat{\mathbf{x}}}}_{i0}^{(2)} -} {{ \bar{\mathbf{x}}}}_{0}^{(2)} ). \end{aligned}$$

(B6)

It is relatively easy to verify that the solution to the normal equations is given by

$$\begin{aligned} {\varvec{ \uptheta }}_{0} ={\mathbf{C}}^{-1}({\mathbf{h}}^{(1)}-{\mathbf{h}}^{(2)}), \end{aligned}$$

(B7)

$$\begin{aligned} s_{0} =\frac{s^{(2)}-s^{(1)}}{S^{2}}, \end{aligned}$$

(B8)

$$\begin{aligned} {\mathbf{d}}_{0} ={{ \bar{\mathbf{x}}}}_{0}^{(2)} -{{ \bar{\mathbf{x}}}}_{0}^{(1)} -[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]{\varvec{ \uptheta }}_{0} -s_{0} {{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}}. \end{aligned}$$

(B9)

Best fitting of transformation parameter derivatives between two sets of velocities obtained using different minimal constraints in the stacking problem

By complete analogy, the solution to the least squares problem (31) is given by

$$\begin{aligned} {{ \dot{{\varvec{\uptheta }}}}}={\mathbf{C}}^{-1}({{ \dot{\mathbf{h}}}}^{(1)}-{{ \dot{\mathbf{h}}}}^{(2)}), \end{aligned}$$

(C1)

$$\begin{aligned} \dot{{s}}=\frac{\dot{{s}}^{(2)}-\dot{{s}}^{(1)}}{S^{2}}, \end{aligned}$$

(C2)

$$\begin{aligned} {{ \dot{\mathbf{d}}}}={{ \bar{\mathbf{v}}}}^{(2)}-{{\bar{\mathbf{v}}}}^{(1)}-[{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} \times ]{{ \dot{{\varvec{\uptheta }}}}}-\dot{{s}}{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} , \end{aligned}$$

(C3)

where

$$\begin{aligned} {{ \bar{\mathbf{v}}}}^{(1)}=\frac{1}{n}\sum \limits _{i=1}^n {{{ \hat{\mathbf{v}}}}_{i}^{(1)} },\quad {{\bar{\mathbf{v}}}}^{(2)}=\frac{1}{n}\sum \limits _{i=1}^n {{{ \hat{\mathbf{v}}}}_{i}^{(2)} }, \end{aligned}$$

(C4)

$$\begin{aligned} {{ \dot{\mathbf{h}}}}^{(1)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {[(\mathbf{{ x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )\times ]({{ \hat{\mathbf{v}}}}_{i}^{(1)} -{{ \bar{\mathbf{v}}}}^{(1)})},\nonumber \\{{ \dot{\mathbf{h}}}}^{(2)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {[({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )\times ]({{ \hat{\mathbf{v}}}}_{i}^{(2)} -{{ \bar{\mathbf{v}}}}^{(2)})} , \end{aligned}$$

(C5)

$$\begin{aligned} \dot{{s}}^{(1)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {({\mathbf{x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}({{ \hat{\mathbf{v}}}}_{i}^{(1)} -} {{ \bar{\mathbf{v}}}}^{(1)}),\nonumber \\ \dot{{s}}^{(2)}= & {} \frac{1}{n}\sum \limits _{i=1}^n {(\mathbf{{ x}}_{i0}^\mathrm{{ap}} -{{ \bar{\mathbf{x}}}}_{0}^\mathrm{{ap}} )^{T}({{ \hat{\mathbf{v}}}}_{i}^{(2)} -} {{ \bar{\mathbf{v}}}}^{(2)}). \end{aligned}$$

(C6)

Best fitting of initial transformation parameters and their derivatives between two sets of transformation parameter time series obtained using different minimal constraints in the stacking problem

The normal equations for the least squares solution of (32) are

$$\begin{aligned}&\left[ {{\begin{array}{*{20}c} {\mathbf{{ I}}_{3} } &{}\quad {(\bar{{t}}-t_{0} )\mathbf{{ I}}_{3} } \\ {(\bar{{t}}-t_{0} )\mathbf{{ I}}_{3} } &{}\quad {[s_{t}^{2} +(\bar{{t}}-t_{0} )^{2}]\mathbf{{ I}}_{3} } \\ \end{array} }} \right] \left[ {{\begin{array}{*{20}c} {\mathbf{{ p}}_{0} } \\ {{{ \dot{\mathbf{p}}}}} \\ \end{array} }} \right] \nonumber \\&\quad =\left[ {{\begin{array}{*{20}c} {{{ \bar{\mathbf{p}}}}_{k}^{(1)} \quad -{{ \bar{\mathbf{p}}}}_{k}^{(2)} } \\ {\mathbf{{ p}}_{t}^{(1)} \quad -\mathbf{{ p}}_{t}^{(2)} } \\ \end{array} }} \right] , \end{aligned}$$

(D1)

where

$$\begin{aligned} \bar{{t}}=\frac{1}{m}\sum \limits _{k=1}^m {t_{k} },\quad s_{t}^{2} =\frac{1}{m}\sum \limits _{k=1}^m {(t_{k} -\bar{{t}})^{2}} , \end{aligned}$$

(D2)

$$\begin{aligned} {{ \bar{\mathbf{p}}}}^{(1)}=\frac{1}{m}\sum \limits _{k=1}^m {{{ \hat{\mathbf{p}}}}_{k}^{(1)} },\quad {{ \bar{\mathbf{p}}}}^{(2)}=\frac{1}{m}\sum \limits _{k=1}^m {{{ \hat{\mathbf{p}}}}_{k}^{(2)} }, \end{aligned}$$

(D3)

$$\begin{aligned} \mathbf{{ p}}_{t}^{(1)} =\frac{1}{m}\sum \limits _{k=1}^m {(t_{k} -\bar{{t}}){{ \hat{\mathbf{p}}}}_{k}^{(1)} },\quad \mathbf{{ p}}_{t}^{(2)} =\frac{1}{m}\sum \limits _{k=1}^m {(t_{k} -\bar{{t}}){{ \hat{\mathbf{p}}}}_{k}^{(2)} }.\nonumber \\ \end{aligned}$$

(D4)

with solution

$$\begin{aligned} {{ \dot{\mathbf{p}}}}=\frac{1}{s_{t}^{2} }(\mathbf{{ p}}_{t}^{(1)} -\mathbf{{ p}}_{t}^{(2)} )-\frac{\bar{{t}}-t_{0} }{s_{t}^{2} }({{ \bar{\mathbf{p}}}}^{(1)}-{{ \bar{\mathbf{p}}}}^{(2)}), \end{aligned}$$

(D5)

$$\begin{aligned} \mathbf{{ p}}_{0}= & {} \frac{s_{t}^{2} +(\bar{{t}}-t_{0} )^{2}}{s_{t}^{2} }({{ \bar{\mathbf{p}}}}^{(1)}-{{\bar{\mathbf{p}}}}^{(2)})-\frac{\bar{{t}}-t_{0} }{s_{t}^{2} }(\mathbf{{ p}}_{t}^{(1)} -\mathbf{{ p}}_{t}^{(2)} )\nonumber \\ {}= & {} {{ \bar{\mathbf{p}}}}^{(1)}-{{\bar{\mathbf{p}}}}^{(2)}-(\bar{{t}}-t_{0} ){{ \dot{\mathbf{p}}}}. \end{aligned}$$

(D6)