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

[math]\displaystyle{ PR^j }[/math]

[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 }[/math]

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

[math]\displaystyle{ R^j-D^j\simeq \sqrt{(x-x^j)^2+(y-y^j)^2+(z-z^j)^2}+c\,\delta t }[/math] [math]\displaystyle{ j=1,2,...,n~~~~ (n \geq 4) }[/math]

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

[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]

[math]\displaystyle{ {r}=[x,y,z]^T }[/math]

[math]\displaystyle{ \left \langle{a},{b}\right \rangle={a}^{t} \; {M} \; {'b}= \left[ \begin{array}{c} a_1,a_2,a_3,a_4 \end{array} \right] \left( \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1\\ \end{array} \right) \left[ \begin{array}{c} b_1\\ b_2\\ b_3\\ b_4 \end{array} \right] }[/math]

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

[math]\displaystyle{ PR^j }[/math]

[math]\displaystyle{ {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) }[/math]

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

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

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

[math]\displaystyle{ \langle {M}{g},{M}{h} \rangle=\langle {g},{h} \rangle }[/math]

[math]\displaystyle{ \Lambda= \frac{1}{2} \left \langle \left[ \begin{array}{c} {r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {r}\\ c\,\delta t\\ \end{array} \right] \right \rangle }[/math]

[math]\displaystyle{ \left \langle {B}^{-1} {1}, {B}^{-1} {1} \right \rangle \Lambda^2+ 2\left [ \left \langle {B}^{-1} {1}, {B}^{-1} {a} \right \rangle -1 \right ] \Lambda + \left \langle {B}^{-1} {a}, {B}^{-1} {a} \right \rangle =0 }[/math]