Bancroft Method

Fundamentals
Title	Bancroft Method
Author(s)	J. Sanz Subirana, J.M. Juan Zornoza and M. Hernández-Pajares, Technical University of Catalonia, Spain.
Level	Advanced
Year of Publication	2011

The Bancroft method allows obtaining a direct solution of the receiver position and the clock offset, without requesting any "a priori" knowledge for the receiver location.

Raising and resolution

Let $P{R}^j$ the prefit-residual of satellite-$j$, computed from equation (1)

$R^j={\rho }^j+c(\delta t-\delta t^j)+T^j+\hat{\alpha }{I}^j+TG{D}^j+{\mathcal{M}}^j+{\mathit{\epsilon }}^j\text{(1)}$

after removing all model terms not needing the a priory knowledge of the receiver position:

$P{R}^j\equiv R^j+c\delta t^j-TG{D}^j\text{(2)}$

Thence, neglecting the tropospheric and ionospheric terms, as well as the multipath and receiver noise, the equation (3)

$\begin{array}{r}{R}^j-D^j\simeq \sqrt{(x^j-x{)}^2+(y^j-y{)}^2+(z^j-z{)}^2}+c\delta t\\ j=1,2,...,n(n\ge 4)\end{array}\text{(3)}$

can be written as:

$P{R}^j=\sqrt{(x^j-x{)}^2+(y^j-y{)}^2+(z^j-z{)}^2}+c\delta t\text{(4)}$

Developing the previous equation (4), it follows:

$[{x^j}^2+{y^j}^2+{z^j}^2-{P{R}^j}^2]-2[x^{j}x+y^{j}y+z^{j}z-P{R}^{j}c\delta t]+[x^2+y^2+z^2-(c\delta t{)}^2]=0\text{(5)}$

Then, calling $\mathbf{r}=[x,y,z{]}^T$ and considering the inner product of Lorentz the previous equation (5) can be expressed in a more compact way as:

$\frac{1}{2}⟨\left[\begin{array}{c}{\mathbf{r}}^j\\ P{R}^j\end{array}\right],\left[\begin{array}{c}{\mathbf{r}}^j\\ P{R}^j\end{array}\right]⟩-⟨\left[\begin{array}{c}{\mathbf{r}}^j\\ P{R}^j\end{array}\right],\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]⟩+\frac{1}{2}⟨\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right],\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]⟩=0\text{(6)}$

The former equation can be raised for every satellite (or prefit-residual $P{R}^j$).

If four measurements are available, thence, the following matrix can be written, containing all the available information on satellite coordinates and pseudoranges (every row corresponds to a satellite):

$\mathbf{B}=\left(\begin{array}{cccc}{x}^1& y^1& z^1& P{R}^1\\ x^2& y^2& z^2& P{R}^2\\ x^3& y^3& z^3& P{R}^3\\ x^4& y^4& z^4& P{R}^4\end{array}\right)\text{(7)}$

Then, calling:

$\mathrm{\Lambda }=\frac{1}{2}⟨\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right],\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]⟩,\mathbf{1}=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],\mathbf{a}=\left[\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\end{array}\right]\text{being}{a}_j=\frac{1}{2}⟨\left[\begin{array}{c}{\mathbf{r}}^j\\ P{R}^j\end{array}\right],\left[\begin{array}{c}{\mathbf{r}}^j\\ P{R}^j\end{array}\right]⟩\text{(8)}$

The four equations for pseudorange can be expressed as:

$\mathbf{a}-\mathbf{B}\mathbf{M}\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]+\mathrm{\Lambda }\mathbf{1}=0,\text{being}\mathbf{M}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right)\text{(9)}$

from where:

$\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]=\mathbf{M}{\mathbf{B}}^{-1}(\mathrm{\Lambda }\mathbf{1}+\mathbf{a})\text{(10)}$

Then, taking into account the following equality

$⟨\mathbf{M}\mathbf{g},\mathbf{M}\mathbf{h}⟩=⟨\mathbf{g},\mathbf{h}⟩\text{(11)}$,

and that

$\mathrm{\Lambda }=\frac{1}{2}⟨\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right],\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]⟩\text{(12)}$,

from the former expression (10), one obtains:

$⟨{\mathbf{B}}^{-1}\mathbf{1},{\mathbf{B}}^{-1}\mathbf{1}⟩{\mathrm{\Lambda }}^2+2[⟨{\mathbf{B}}^{-1}\mathbf{1},{\mathbf{B}}^{-1}\mathbf{a}⟩-1]\mathrm{\Lambda }+⟨{\mathbf{B}}^{-1}\mathbf{a},{\mathbf{B}}^{-1}\mathbf{a}⟩=0\text{(13)}$

The previous expression (13) is a quadratic equation in $\mathrm{\Lambda }$ (note that matrix $\mathbf{B}$ and the vector $\mathbf{a}$ are also known) and provides two solutions, that introduced in expression (10) provides the searched solution:

$\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]\text{(14)}$.

The other solution is far from the earth surface.

Generalisation to the case of $n$-measurements:

If more than four observations are available, the matrix $\mathbf{B}$ is not square. However, multiplying by ${\mathbf{B}}^T$, one obtains (Least Squares solution):

${\mathbf{B}}^T\mathbf{a}-{\mathbf{B}}^T\mathbf{B}\mathbf{M}\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]+\mathrm{\Lambda }{\mathbf{B}}^T\mathbf{1}=0\text{(15)}$

where:

$\left[\begin{array}{c}\mathbf{r}\\ c\delta t\end{array}\right]=\mathbf{M}({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T(\mathrm{\Lambda }\mathbf{1}+\mathbf{a})\text{(16)}$

and then:

$\begin{array}{r}⟨({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{1},({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{1}⟩{\mathrm{\Lambda }}^2+2[⟨({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{1},({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{a}⟩-1]\mathrm{\Lambda }+\\ +⟨({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{a},({\mathbf{B}}^T\mathbf{B}{)}^{-1}{\mathbf{B}}^T\mathbf{a}⟩=0\end{array}\text{(17)}$