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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.
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==
==Raising and resolution==
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Let <math>PR^j</math> the prefit-residual of satellite-<math>j</math>, computed from equation (1)
Let <math>PR^j</math> the prefit-residual of satellite-<math>j</math>, computed from equation (1)


<math>
::<math>
R^j=\rho^j+c(\delta t-\delta t^j)+T^j+\hat{\alpha}\, I^j+TGD^j+\mathcal{M}^j+{\boldsymbol \varepsilon}^j
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>
\qquad \mbox{(1)}</math>
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after removing all model terms  not needing the a priory knowledge of the receiver position:<ref group=“footnotes”>The tropospheric and ionospheric terms,  <math>T^j</math> and <math>\hat{\alpha} \,I^j</math>, can not be included, because the need to consider the satellite-receiver ray. Off course, after an initial computation of the receiver coordinates, the method could be iterated using the ionospheric and tropospheric corrections to improve the solution.</ref>
after removing all model terms  not needing the a priory knowledge of the receiver position:<ref group=“footnotes”>The tropospheric and ionospheric terms,  <math>T^j</math> and <math>\hat{\alpha} \,I^j</math>, can not be included, because the need to consider the satellite-receiver ray. Off course, after an initial computation of the receiver coordinates, the method could be iterated using the ionospheric and tropospheric corrections to improve the solution.</ref>


<math>
::<math>
PR^j\equiv R^j +c\,\delta t^j-TGD^j
PR^j\equiv R^j +c\,\delta t^j-TGD^j
\qquad \mbox{(2)}</math>
\qquad \mbox{(2)}</math>




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


<math>
::<math>
\begin{array}{r}
\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]
R^j-D^j\simeq \sqrt{(x-x^j)^2+(y-y^j)^2+(z-z^j)^2}+c\,\delta t\\[0.3cm]
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\end{array}
\end{array}
\qquad \mbox{(3)}</math>
\qquad \mbox{(3)}</math>


can be written as:
can be written as:


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




Developing the previous equation (4), it follows:
Developing the previous equation (4), it follows:


<math>\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>\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 \qquad \mbox{(5)}</math>




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</math>
</math>
</ref>
</ref>
the previous equation can be expressed in a more compact way as:
the previous equation (5) can be expressed in a more compact way as:


<math>\frac{1}{2} \left  \langle
::<math>\frac{1}{2} \left  \langle
\left[
\left[
\begin{array}{c}
\begin{array}{c}
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\right \rangle
\right \rangle
=0
=0
</math>
\qquad \mbox{(6)}</math>
 
 


The former equation can be raised for every satellite (or prefit-residual <math>PR^j</math>).
The former equation can be raised for every satellite (or prefit-residual <math>PR^j</math>).




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


<math>
::<math>
{\mathbf B}=
{\mathbf B}=
\left(
\left(
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\end{array}
\end{array}
\right)
\right)
\qquad \mbox{(5)}</math>
\qquad \mbox{(7)}</math>




Then, calling:
Then, calling:


<math>\Lambda=
::<math>\Lambda=
\frac{1}{2}
\frac{1}{2}
  \left  \langle
  \left  \langle
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\right]
\right]
\right \rangle
\right \rangle
</math>
\qquad \mbox{(8)}</math>
 




The four equations for pseudorange can be expressed as:
The four equations for pseudorange can be expressed as:


<math>{\mathbf a} -{\mathbf B}\,{\mathbf M} \left[
::<math>{\mathbf a} -{\mathbf B}\,{\mathbf M} \left[
\begin{array}{c}
\begin{array}{c}
{\mathbf r}\\
{\mathbf r}\\
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\end{array}
\end{array}
\right)
\right)
</math>
\qquad \mbox{(9)}</math>




from where:
from where:


<math>
::<math>
\left[
\left[
\begin{array}{c}
\begin{array}{c}
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\right]
\right]
={\mathbf M} {\mathbf B}^{-1} (\Lambda \; {\mathbf 1} + {\mathbf a})
={\mathbf M} {\mathbf B}^{-1} (\Lambda \; {\mathbf 1} + {\mathbf a})
\qquad \mbox{(6)}</math>
\qquad \mbox{(10)}</math>
 
 
 
Then, taking into account the following equality
 
:: <math>\langle {\mathbf M}{\mathbf g},{\mathbf M}{\mathbf h} \rangle=\langle {\mathbf g},{\mathbf h} \rangle \qquad \mbox{(11)}</math>,
 


and that


Then, taking into account the following equality <math>\langle {\mathbf M}{\mathbf g},{\mathbf M}{\mathbf h} \rangle=\langle {\mathbf g},{\mathbf h} \rangle</math>, and
::<math>
that
<math>
\Lambda=
\Lambda=
\frac{1}{2}
\frac{1}{2}
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\right]
\right]
\right \rangle
\right \rangle
</math>,  from the former expression, one obtains:
\qquad \mbox{(12)}</math>,   
 
 
from the former expression (10), one obtains:


<math>
::<math>
\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
\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>
\qquad \mbox{(13)}</math>
 


The previous expression (13) is a quadratic equation in <math>\Lambda</math> (note that matrix <math>{\mathbf B}</math> and the vector <math>\mathbf a </math> are also known) and provides two solutions, one of them is the searched solution


The previous expression is a quadratic equation in <math>\Lambda</math> (note that matrix <math>{\mathbf B}</math> and the vector {\mathbf a} are also known) and provides two solutions, one of them is the searched solution
::<math>
<math>
\left[
\left[
\begin{array}{c}
\begin{array}{c}
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c\,\delta t\\
c\,\delta t\\
\end{array}
\end{array}
\right]</math>.
\right]
\qquad \mbox{(14)}</math>.
 
The other solution is far from the earth surface.
The other solution is far from the earth surface.




==Generalisation to the case of <math>n</math>-measurements:==
==Generalisation to the case of <math>n</math>-measurements:==


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


<math>{\mathbf B}^T{\mathbf a} -{\mathbf B}^T {\mathbf B}\,{\mathbf M} \left[
::<math>
{\mathbf B}^T{\mathbf a} -{\mathbf B}^T {\mathbf B}\,{\mathbf M} \left[
\begin{array}{c}
\begin{array}{c}
{\mathbf r}\\
{\mathbf r}\\
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\right]
\right]
+\Lambda \; {\mathbf B}^T {\mathbf 1}=0
+\Lambda \; {\mathbf B}^T {\mathbf 1}=0
</math>
\qquad \mbox{(15)}</math>




where:
where:


<math>
::<math>
\left[
\left[
\begin{array}{c}
\begin{array}{c}
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\right]
\right]
={\mathbf M} ({\mathbf B}^T {\mathbf B})^{-1}{\mathbf B}^T(\Lambda \; {\mathbf 1} + {\mathbf a})
={\mathbf M} ({\mathbf B}^T {\mathbf B})^{-1}{\mathbf B}^T(\Lambda \; {\mathbf 1} + {\mathbf a})
</math>
\qquad \mbox{(16)}</math>




and then:
and then:


<math>
::<math>
\begin{array}{r}
\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 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
+ \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}
\end{array}
</math>
\qquad \mbox{(17)}</math>
 


==Notes==
==Notes==

Revision as of 09:01, 8 August 2011


FundamentalsFundamentals
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
Logo gAGE.png

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:[“footnotes” 1]

[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 \qquad \mbox{(5)} }[/math]


Then, calling [math]\displaystyle{ {\mathbf r}=[x,y,z]^T }[/math] and considering the inner product of Lorentz [“footnotes” 2] the previous equation (5) 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 \qquad \mbox{(6)} }[/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{(7)} }[/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 \qquad \mbox{(8)} }[/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) \qquad \mbox{(9)} }[/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{(10)} }[/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 \qquad \mbox{(11)} }[/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 \qquad \mbox{(12)} }[/math],


from the former expression (10), 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{(13)} }[/math]


The previous expression (13) is a quadratic equation in [math]\displaystyle{ \Lambda }[/math] (note that matrix [math]\displaystyle{ {\mathbf B} }[/math] and the vector [math]\displaystyle{ \mathbf a }[/math] 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] \qquad \mbox{(14)} }[/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 \qquad \mbox{(15)} }[/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}) \qquad \mbox{(16)} }[/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} \qquad \mbox{(17)} }[/math]


Notes

<references group="footnotes"/ >
Cite error: <ref> tags exist for a group named "“footnotes”", but no corresponding <references group="“footnotes”"/> tag was found