Bancroft Method

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

The Bancroft method allows to obtain 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 [math]\displaystyle{ PR^j }[/math] the prefit-residual of satellite-[math]\displaystyle{ j }[/math], computed from equation (1)

[math]\displaystyle{ R^j=\rho^j+c(\delta t-\delta t^j)+T^j+\hat{\alpha}\, I^j+TGD^j+\mathcal{M}^j+{\boldsymbol \varepsilon}^j \qquad \mbox{(1)} }[/math]

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

[math]\displaystyle{ PR^j\equiv R^j +c\,\delta t^j-TGD^j \qquad \mbox{(2)} }[/math]

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

[math]\displaystyle{ \begin{array}{r} R^j-D^j\simeq \sqrt{(x-x^j)^2+(y-y^j)^2+(z-z^j)^2}+c\,\delta t\\[0.3cm] j=1,2,...,n~~~~ (n \geq 4)\\ \end{array} \qquad \mbox{(3)} }[/math]

can be written as:

[math]\displaystyle{ PR^j = \sqrt{(x-x^j)^2+(y-y^j)^2+(z-z^j)^2}+c \, \delta t \qquad \mbox{(4)} }[/math]

Developing the previous equation (4), it follows:

[math]\displaystyle{ \left[{x^j}^2+{y^j}^2+{z^j}^2-{PR^j}^2 \right]-2 \left[x^j x+y^j y+z^j z-{PR^j} \; \right] + \left[x^2+y^2+z^2-(c\,\delta t)^2 \right]=0 }[/math]

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

[math]\displaystyle{ \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right] \right \rangle - \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle + \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle =0 }[/math]

The former equation can be raised for every satellite (or prefit-residual [math]\displaystyle{ PR^j }[/math]).

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):

[math]\displaystyle{ {\mathbf B}= \left( \begin{array}{cccc} x^1&y^1&z^1&PR^1\\ x^2&y^2&z^2&PR^2\\ x^3&y^3&z^3&PR^3\\ x^4&y^4&z^4&PR^4\\ \end{array} \right) \qquad \mbox{(5)} }[/math]

Then, calling:

[math]\displaystyle{ \Lambda= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle \; , \; {\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] \; \mbox{being} \; \; a_j= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right] \right \rangle }[/math]

The four equations for pseudorange can be expressed as:

[math]\displaystyle{ {\mathbf a} -{\mathbf B}\,{\mathbf M} \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] +\Lambda \; {\mathbf 1}=0\;\;,\;\;\;\; \mbox{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) }[/math]

from where:

[math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] ={\mathbf M} {\mathbf B}^{-1} (\Lambda \; {\mathbf 1} + {\mathbf a}) \qquad \mbox{(6)} }[/math]

Then, taking into account the following equality [math]\displaystyle{ \langle {\mathbf M}{\mathbf g},{\mathbf M}{\mathbf h} \rangle=\langle {\mathbf g},{\mathbf h} \rangle }[/math], and that [math]\displaystyle{ \Lambda= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle }[/math], from the former expression, one obtains:

[math]\displaystyle{ \left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf 1} \right \rangle \Lambda^2+ 2\left [ \left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf a} \right \rangle -1 \right ] \Lambda + \left \langle {\mathbf B}^{-1} {\mathbf a}, {\mathbf B}^{-1} {\mathbf a} \right \rangle =0 \qquad \mbox{(7)} }[/math]

The previous expression is a quadratic equation in [math]\displaystyle{ \Lambda }[/math] (note that matrix [math]\displaystyle{ {\mathbf B} }[/math] and the vector {\mathbf a} are also known) and provides two solutions, one of them is the searched solution [math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] }[/math]. The other solution is far from the earth surface.

Generalisation to the case of [math]\displaystyle{ n }[/math]-measurements:

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

[math]\displaystyle{ {\mathbf B}^T{\mathbf a} -{\mathbf B}^T {\mathbf B}\,{\mathbf M} \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] +\Lambda \; {\mathbf B}^T {\mathbf 1}=0 }[/math]

where:

[math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] ={\mathbf M} ({\mathbf B}^T {\mathbf B})^{-1}{\mathbf B}^T(\Lambda \; {\mathbf 1} + {\mathbf a}) }[/math]

and then:

[math]\displaystyle{ \begin{array}{r} \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1} \right \rangle \Lambda^2+ 2\left [ \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle -1 \right ] \Lambda +\\[0.3cm] + \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle =0 \end{array} }[/math]

Notes

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