# User talk:RHB100

I have contributed several sections to the GPS article in order to provide a basic understanding of global positioning. In some cases these contributions have included almost the entire section. In other cases only part of the section. I have also contributed to other articles including trilateration RHB100 (talk) 20:37, 19 July 2010 (UTC)

## Introduction of Robert H. Biggadike

Hello, I am Robert H. Biggadike. I am a Licensed Professional Engineer in the field of Control System Engineering. I hold advanced engineering degrees from the University of Arkansas and UCLA in addition to an undergraduate engineering degree from Arkansas. In addition to my technical interests, I am interested in libertarianism and capitalism. I am originally from Newport, Arkansas. I now live in California.

BSME & MSEM University of Arkansas
Degree of Engineer UCLA (majoring in rigid and elastic body dynamics) RHB100 (talk) 19:43, 28 July 2009 (UTC)

I am a philanthropist. My engineering career has been so successful that with savings and investments I have become a multi-millionaire. I am a benefactor of the University of Arkansas, who has made an estate gift commitment valued at more than \$7.8 million to the University of Arkansas College of Engineering as documented at [[1]]

## Contributions of Robert H. Biggadike

This is a work in progress. I have been making contributions to Wikipedia since 27 June 2008. I am now (19 July 2010) attempting to make a list of some of my more important contributions along with references to the edits for substantiation. ~~

My contribution to the GPS article include the sections on computation of GDOP (general dilution of precision) and a section on derivation of the equations for computing GDOP. These are new sections which I created from scratch. This can be confirmed by observing on the revision page the edits by RHB100 on 10 December 2008 at 1:28, the edits of 10 December 2008 from 19:54 till 21:43, and the edit on 11 December 2008 at 21:30 for the GDOP computation. The derivation of the GDOP computation equations was done with the edit of 11 December 2008 at 22:09.

I have also contributed to the sections on Timeline and Modernization, Basic concepts of GPS including its subsections, the subsections of Communication, a very significant portion of the sections, Equations and Methods of solution, almost all of Multidimensional Newton-Raphson, the section, Signal arrival and measurement, and the section, Carrier phase tracking and measurement (surveying).

## Possible Error in Wikipedia Article: GPS

Hi! I replied on my talk page, to keep the discussion in one place. Basically I think you are right and the article has it backwards. This would be a good thing to fix, LouScheffer (talk) 20:12, 27 June 2008 (UTC)

## Using the C/A Code

Use "["[#multi_nr|Multidimensional Newton Raphson]"]" Multidimensional Newton Raphson to link to this target.
Put "<"span id="multi_nr"></span">" in the target.
"["http://en.wikipedia.org/w/index.php?title=User_talk:RHB100&action=edit&section=4#multi_nr Multi Newton Raphson"]" to same article does not work.
<A=HREF="http://en.wikipedia.org/w/index.php?title=User_talk:RHB100&action=edit&section=4#multi_nr">Multi Newton Raphson Test</A> to same article does does not work

I contributed the following paragraph to this section of the GPS article.

When pseudoranges are determined for four satellites, an estimate of the GPS receiver position can be made. Trilateration is used to determine the two points of intersection of three sphere surfaces corresponding to three satellites. The surface of the sphere corresponding to the fourth satellite or the surface of the earth is used to determine which of the two intersections provides a valid estimate of GPS receiver position. The valid estimate is the point closest to the surface of the sphere corresponding to the fourth satellite or the surface of the earth. It is likely the surfaces of the three spheres intersect since the circle of intersection of the first two spheres is normally quite large and thus the third sphere surface is likely to intersect this large circle. It is very unlikely that the surface of the sphere corresponding to the fourth satellite will intersect either of the two points of intersection of the first three since any clock error could cause it to miss intersecting a point. However the distance from the valid estimate of GPS receiver position to the surface of the sphere corresponding to the fourth satellite can be used to compute a clock correction. Let R4 denote the distance from the valid estimate of GPS receiver position to the fourth satellite and let P4 denote the pseudorange of the fourth satellite. Let DA=R4-P4. Then the quotient, DEL_T= DA/(speed of light), provides an estimate of UTC - (time indicated by the receiver's on-board clock) and the GPS receiver clock can be moved forward if DEL_T is positive or backwards if DEL_T is negative.

Improvements desired for "Using the C/A Code" section include adding figures of 2 and 3 spheres intersecting. In this section or possibly in the trilateration article the method of formulating the equations of the spheres in the planar coordinate system required for trilateration should be explained.

## Last Version Before Lou Scheffer

Hi! Neither your description, nor mine, describes how GPS receivers actually do the calculation. Furthermore, we both are describing how the pseudoranges are used, and how clock corrections are made. So the main difference is the explanation, I think. My personal description is that equations and variables help little in a case like this - the folks who want equations can derive them from the text, and other who do not want equations or variable definitions do not need to wade through them.
I propose to put both versions on the talk page for GPS, and let others comment. LouScheffer (talk) 02:18, 7 July 2008 (UTC)

To Lou Scheffer: All right, we will see if anything useful can be learned from this. I have used very few equations and the ones that I have are very simple. Variables just provide a name so as to avoid excessive wordiness. 

## TOA (Spheres) is Better than TDOA (Hyperboloids)

There are references to using hyperboloids in the Wikipedia GPS article. Analytic Geometry tells us that the locus of points which have a constant difference in distance from two satellite positions lie on a surface called a hyperboloid. Since this surface can be computed based on the difference in time of arrival at the GPS Receiver of messages from different satellites, it suggests that this data would reduce the error due to the GPS Receiver Clock. However judging from GPS literature it seems that spheres based on TOA are used more often. I think there is a good reason for this. It is because you can get a very good clock correction using trilateration with four spheres. This negates the time difference advantage of using hyperboloids in my opinion.RHB100 (talk) 21:39, 9 July 2008 (UTC)

Crazy Software Productions (talk) 19:58, 3 September 2008 (UTC) Using pseudorange calculations (sometimes this is called hyperboloid calculation) can be used for positional calculations. This method converges fast, it is not influenced by the clockerror and with a close first estimate only one iteration is needed. One of the effects of pseudorange calculation is that the the clockerror is also calculated in the process. The solutions I have seen up to now based on spheres have never been complete. The suggested clock correction in the GPS mainarticle is not always in the correct direction and is often far of by amount. So the method described is not a completely working solution. Then the amount of calculation needed is far more than is needed for pseudorange calculations.

Pseudorange calculations converges very fast. An example with (0, 0, 0) (centre of the earth) as a first estimate gave the following 'errors' (this is independend of the timing error):
0 first estimate, error is 6372836 meters
1 first iteration, error is 1718732 meters
2 second iteration, error is 108173 meters
3 third iteration, error is 333 meters
4 fourth iteration, error is 0.002695871 meters (2 millimeters)
5 fifth iteration, error is 2.8 E-9 meters.

So in most situations only one iteration will be sufficient.

The above numbers where calculated using an implementation of the pseudorange calculation in a spreadsheat, only using + - * / and the squareroot. (Spreadsheat is available).

The number of calculations needed for one iteration for the pseudorange calculation can be as low as 69 additions/subtractions, 42 multiplications, 22 divisions and 4 squareroots. Because only one iteration is needed in most situations, this is thus the amount of calculation needed for one positioning. I have not analysed the trilateration but having seen parts of the algoritm I estimate that the amount of calculations for one iteration does exceed the above numbers and the number of iterations needed is larger as wel.

So my conclusion is that TDAO is far better than TOA. Based on the amount of calculation needed and based on the number of iterations needed. (But because I haven't seen the complete algoritm or implementation of TOA a real comparison can not be made). If you can show that TOA does produce better numbers than the above I will be very interrested.

The reason why: "However judging from GPS literature it seems that spheres based on TOA are used more often." Is because this explanes the physics of the GPS system very well. The signals are spread in spheres (and not hyperboloids). And would the signals be timed in such a way that they arrive at the receivers end at the same time, the spheres exactly describe the situation at that point in time. But the picture is draw based on the knowledge where the point of the receiver is. It's much more difficult to explain the process if within that explanation the location of the receiver must be left unknown until the end of the explanation. Where those explanations very often fail is the explaining why a fourth satellite is needed. So sometimes this is described for picking the correct point of two. Or sometimes it is described that it's used to correct the clock. But with pseudorange calculation the whole calculation is based on the information of four satellites, the calculation of the clock error is just one of the results.

Crazy Software Productions (talk) 19:58, 3 September 2008 (UTC)

## Basic Concept of GPS

To describe the basic concept of how a GPS receiver works, the errors are at first ignored. Using messages received from four satellites, the GPS receiver is able to determine the satellite positions and time sent. The x, y, and z components of position and the time sent are designated as ${\displaystyle \left[x_{i},y_{i},z_{i},t_{i}\right]}$ where the subscript i denotes which satellite and has the value 1, 2, 3, or 4. Knowing the indicated time the message was received ${\displaystyle \ tr_{i}}$, the GPS receiver can compute the indicated transit time, ${\displaystyle \left(tr_{i}-t_{i}\right)}$. of the message. Assuming the message traveled at the speed of light, c, the distance travelled, ${\displaystyle \ p_{i}}$ can be computed as ${\displaystyle \left(tr_{i}-t_{i}\right)c}$. Knowing the distance from GPS receiver to a satellite and the position of a satellite implies that the GPS receiver is on the surface of a sphere centered at the position of a satellite. Thus we know that the indicated position of the GPS receiver is at or near the intersection of the surfaces of four spheres. In the ideal case of no errors, the GPS receiver will be at an intersection of the surfaces of four spheres. The surfaces of two spheres if they intersect in more than one point intersect in a circle. A figure, Two Sphere Surfaces Intersecting in a Circle, is shown below depicting this which hopefully will aid the reader in visualizing this intersection.

Two Sphere Surfaces Intersecting in a Circle

The article, trilateration, shows mathematically how the equation for a circle is determined. A circle and sphere surface in most cases of practical interest intersect at two points, although it is conceivable that they could intersect in 0 or 1 point. Another figure, Surface of Sphere Intersecting a Circle (not disk) at Two Points, is shown below to aid in visualizing this intersection. Again trilateration clearly show this mathematically. The correct position of the GPS receiver is the one that is closest to the fourth sphere. This paragraph has described the basic concept of GPS while ignoring errors. The next problem is how to process the messages when errors are present.

Surface of Sphere Intersecting a Circle (not disk) at Two Points

Let ${\displaystyle \ b}$ denote the clock error or bias, the amount by which the receiver's clock is slow. The GPS receiver has four unknowns, the three components of GPS receiver position and the clock bias ${\displaystyle \left[x,y,z,b\right]}$. The equation of the sphere surfaces are given by:

OLD DESCRIPTION OF TRILATERATION AND TIME ERROR COMPUTATION The above paragraph has provided an overview of the process for computing position and clock correction. For the interested reader this paragraph provides a little more detail on how the spheres intersect, where trilateration is used, the computation of receiver position, clock correction and the iteration process. Utilizing the pseudoranges determined for four satellites, an estimate of the GPS receiver position is made. For the ideal case when the pseudoranges are correct the GPS Receiver must be on the surface of each of the corresponding spheres and therefore must be at one of the intersections of these sphere surfaces. Trilateration is used to determine the two points of intersection of three sphere surfaces corresponding to three satellites. The surface of the sphere corresponding to the fourth satellite or the surface of the earth is used to determine which of the two intersections provides a valid estimate of GPS receiver position. The valid estimate is the point closest to the surface of the sphere corresponding to the fourth satellite or the surface of the earth.

It is likely the surfaces of the three spheres intersect since the circle of intersection of the first two spheres is normally quite large and thus the third sphere surface is likely to intersect this large circle. It is very unlikely that the surface of the sphere corresponding to the fourth satellite will intersect either of the two points of intersection of the first three since any clock error could cause it to miss intersecting a point. However the distance from the valid estimate of GPS receiver position to the surface of the sphere corresponding to the fourth satellite can be used to compute a clock correction. Let ${\displaystyle \ r_{4}}$ denote the distance from the valid estimate of GPS receiver position to the fourth satellite and let ${\displaystyle \ p_{4}\ }$denote the pseudorange of the fourth satellite. Let ${\displaystyle \ da=r_{4}-p_{4}}$. Note that ${\displaystyle \ da}$ is the distance from the computed GPS receiver position to the surface of the surface of the sphere corresponding to the fourth satellite. Thus the quotient, ${\displaystyle \ b=da/c\ }$, provides an estimate of
(correct time) - (time indicated by the receiver's on-board clock)
and the GPS receiver clock can be moved forward if ${\displaystyle \ b}$ is positive or backwards if ${\displaystyle \ b}$ is negative.

The above procedure has not taken into account the change in pseudoranges resulting from the correction to the GPS receiver clock. When the magnitude of Δt is small this may be adequate. However when Δt is large an iterative procedure should be used. The pseudoranges should be recomputed using the updated GPS receiver clock and a new valid estimate of GPS receiver position should be computed as described above. A new value of Δt should then be computed. These iterations should be continued until the magnitude of Δt is sufficiently small.

## Methods of Solution witn No More Than Four Satellites

Two of the most important methods of computing GPS receiver position and clock bias are (1) trilateration followed by one dimensional numerical root finding and (2) multidimensional Newton-Raphson. These two methods along with their advantages are discussed.

• Solve by trilateration followed by one dimensional numerical root finding[1]. This method involves using Trilateration to determine the intersection of the surfaces of three spheres. It is clearly shown in trilateration that the surfaces of three spheres intersect in 0, 1, or 2 points. In the usual case of two intersections, the solution which is nearest the the surface of the sphere corresponding to the fourth satellite is chosen. The surface of the earth can also sometimes be used instead, especially in the case of civilian GPS receivers since it is illegal in the United States to track vehicles of more than 60,000 feet in altitude. The bias, ${\displaystyle \ b}$ is then computed based on the distance from the solution to the surface of the sphere corresponding to the fourth satellite. Using an updated received time based on this bias, new spheres are computed and the process is repeated. One advantage of this method is that it involves one dimensional as opposed to multidimensional numerical root finding.

It is likely the surfaces of the three spheres intersect since the circle of intersection of the first two spheres is normally quite large and thus the third sphere surface is likely to intersect this large circle. It is very unlikely that the surface of the sphere corresponding to the fourth satellite will intersect either of the two points of intersection of the first three since any clock error could cause it to miss intersecting a point. However the distance from the valid estimate of GPS receiver position to the surface of the sphere corresponding to the fourth satellite can be used to compute a clock correction. Let r4 denote the distance from the valid estimate of GPS receiver position to the fourth satellite and let p4 denote the pseudorange of the fourth satellite. Let da = r4 - p4. Then the quotient, Δt = da / c, provides an estimate of UTC - (time indicated by the receiver's on-board clock) and the GPS receiver clock can be moved forward if Δt is positive or backwards if Δt is negative.

The above procedure has not taken into account the change in pseudoranges resulting from the correction to the GPS receiver clock. When the magnitude of Δt is small this may be adequate. However when Δt is large an iterative procedure should be used. The pseudoranges should be recomputed using the updated GPS receiver clock and a new valid estimate of GPS receiver position should be computed as described above. A new value of Δt should then be computed. These iterations should be continued until the magnitude of Δt is sufficiently small.

• Utilize multidimensional Newton-Raphson[1]. Linearize around an approximate solution say ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ from iteration k, then solve four linear equations derived from the quadratic equations above to obtain ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$. The radii are large and so the sphere surfaces are close to flat.[2][3] This near flatness may cause the iterative procedure to converge rapidly in the case where ${\displaystyle \ b}$ is near the correct value and the primary change is in the values of ${\displaystyle x,y,\;and\;z}$, since in this case the problem is merely to find the intersection of nearly flat surfaces and thus close to a linear problem. However when ${\displaystyle \ b}$ is changing significantly, this near flatness does not appear to be advantageous in producing rapid convergence, since in this case these near flat surfaces will be moving as the spheres expand and contract. This possible fast convergence is an advantage of this method. Also it has been claimed that this method is the "typical" method used by GPS receivers.[4]. This disadvantage of this method is that according to [1], "There are no good general methods for solving systems of more than one nonlinear equations."
• Other methods include: Solve for the intersection of the expanding signals form light cones in 4-space cones, Solve for the intersection of hyperbloids determined by the time difference of signals received from satellites utilizing multilateration, *Solve the equations in accordance with[5][6].

## Mathematics of Method Involving Multidimensional Newton-Raphson to Solve GPS Equations

Let ${\displaystyle x,y\ and\ z\;}$ denote the true coordinates of GPS receiver position at time, ${\displaystyle \ t}$. Let ${\displaystyle \ b}$ denote the unknown clock error or bias, the amount by which the receiver's clock is slow. Let the coordinates of each satellite, and the time the message was sent, be ${\displaystyle \left[x_{i},y_{i},z_{i},t_{i}\right]}$, let the GPS clock's indicated received time be ${\displaystyle tr_{i}\;for\ i=1,2,3,4}$ and c be the speed of light. The pseudorange is computed as ${\displaystyle p_{i}=\left(tr_{i}-t_{i}\right)c}$. Assume the message travels at the speed of light, then the pseudorange satisfies the equation,

1. :${\displaystyle p_{i}={\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2})}}-bc,\ for\;i=1,2,3,4.}$, When an approximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ rather than the exact solution, ${\displaystyle \left[x,y,z,b\right]}$ is used in equation 1, there is a residual, ${\displaystyle f_{i}^{(k)}}$. Transforming ${\displaystyle \ p_{i}}$ to the right hand side of the equation there results,
2. :${\displaystyle f_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2})}}-b^{(k)}c-p_{i},\ for\;i=1,2,3,4}$. A solution will have been found when ${\displaystyle f_{i}^{(k)}}$ is zero or sufficiently close to zero for ${\displaystyle \ i=1,2,3,4}$. In order to linearize equation 2 the partial derivatives are computed as
3. ${\displaystyle {\frac {\partial f_{i}^{(k)}}{\partial x^{(k)}}}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial y^{(k)}}}={\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial z^{(k)}}}={\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}},and\;{\frac {\partial f_{i}^{(k)}}{\partial b^{(k)}}}=-c}$ where ${\displaystyle R_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2})}}}$. Linearizing equation 2 about the apprximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ there results
4. ${\displaystyle f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}}x^{(k)}+{\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}}y^{(k)}+{\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}}z^{(k)}-b^{(k)}c-p_{i}\ for\;i=1,2,3,4.}$. In order to drive ${\displaystyle f_{i}^{(k+1)}}$ closer to zero choose the values ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ such that
5. ${\displaystyle -f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{p_{i}}}(x^{(k+1)}-x^{(k)})+{\frac {(y^{(k)}-y_{i})}{p_{i}}}(y^{(k+1)}-y^{(k)})\;+}$ ${\displaystyle {\frac {(z^{(k)}-z_{i})}{p_{i}}}(z^{(k+1)}-z^{(k)})-(b^{(k+1)}-b^{(k)})c-p_{i}}$. That is choose the values ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ such that the residual in equation 2 changes by approximately ${\displaystyle \ -f_{i}^{(k)}}$.
Let ${\displaystyle \Delta x^{(k+1)}\ =\ (x^{(k+1)}-x^{(k)}),\ \Delta y^{(k+1)}\ =\ (y^{(k+1)}-y^{(k)}),\ \Delta z^{(k+1)}\ =\ (z^{(k+1)}-z^{(k)}),\ and}$ ${\displaystyle \Delta b^{(k+1)}\ =\ (b^{(k+1)}-b^{(k)}).}$ Substituting and transposing ${\displaystyle \ p_{i}}$ to the left hand side of the equation, there results
6. ${\displaystyle -f_{i}^{(k)}+p_{i}={\frac {(x^{(k)}-x_{i})}{p_{i}}}\Delta x^{(k+1)}+{\frac {(y^{(k)}-y_{i})}{p_{i}}}\Delta y^{(k+1)}\;+}$ ${\displaystyle {\frac {(z^{(k)}-z_{i})}{p_{i}}}\Delta z^{(k+1)}-\Delta b^{(k+1)}c}$. Equations 6 provide a set of four linear equations in four unknowns, the delta terms. They are in a form for solution. After a solution is found, the ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ are evaluated using,
7. ${\displaystyle x^{(k+1)}\ =\ x^{(k)}\ +\ \Delta x^{(k+1)},y^{(k+1)}\ =\ y^{(k)}\ +\ \Delta y^{(k+1)},z^{(k+1)}\ =\ z^{(k)}\ +\ \Delta z^{(k+1)}}$, and ${\displaystyle b^{(k+1)}\ =\ b^{(k)}\ +\ \Delta b^{(k+1)}}$. Then set ${\displaystyle \ k=k+1}$ in equations 2 through 6, plug the ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ from equations 7 into equations 2, set ${\displaystyle \ k=k+1}$ in equations 7, and reevaluate the residuals in equations 2. This procedure is repeated until the residuals are sufficiently small in magnitude.

### Multidimensional Newton-Raphson for GPS

The equations used in multidimensional Newton-Raphson for GPS are stated below. For another derivation of these equations see [7] or for an online derivation see Navigation Solution in GPS World by Torralbas, Roberto and Alvarez Jose Manuel. For an understanding of multidimensional Newton-Raphson as a general numerical method, see [1].

Let ${\displaystyle x,y\;}$ and ${\displaystyle z\;}$ denote the true receiver position coordinates. Let ${\displaystyle \ b}$ denote the unknown clock error or bias. Let the coordinates of each satellite, and the time the message was sent, be ${\displaystyle \left[x_{i},y_{i},z_{i},t_{i}\right]}$, let the GPS clock's indicated received time be ${\displaystyle tr_{i}\;{\mbox{for}}\ i=1,2,3,4}$ and c be the speed of light. The pseudorange is computed as:

${\displaystyle p_{i}=\left(tr_{i}-t_{i}\right)c}$

Assume the message travels at the speed of light, then the pseudorange satisfies the equation:

${\displaystyle p_{i}={\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}-bc,}$ ${\displaystyle \;\;{\mbox{for}}\;i=1,2,3,4.\qquad (1)}$

When an approximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ rather than the exact solution, ${\displaystyle \left[x,y,z,b\right]}$ is used in equation 1, there is a residual, ${\displaystyle f_{i}^{(k)}}$. Transforming ${\displaystyle \ p_{i}}$ to the right hand side of the equation produces,

${\displaystyle f_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2}}}-b^{(k)}c-p_{i},\;\;}$
${\displaystyle {\mbox{for}}\;i=1,2,3,4.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;\;\;(2)}$

A solution will have been obtained when ${\displaystyle f_{i}^{(k)}}$ is zero or sufficiently close to zero for ${\displaystyle \ i=1,2,3,4}$.

In order to linearize equation 2, the partial derivatives are computed as:

${\displaystyle {\frac {\partial f_{i}^{(k)}}{\partial x^{(k)}}}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial y^{(k)}}}={\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}},}$
${\displaystyle {\frac {\partial f_{i}^{(k)}}{\partial z^{(k)}}}={\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial b^{(k)}}}=-c\qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;(3)}$

where

${\displaystyle R_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2}}}}$.

Linearizing the right hand side of equation 2 about the approximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ there results

${\displaystyle f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}}x^{(k)}+{\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}}y^{(k)}+{\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}}z^{(k)}-b^{(k)}c-p_{i}+\epsilon _{i},\;\;}$
${\displaystyle {\mbox{for}}\ i=1,2,3,4\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (4)}$

where ${\displaystyle \ \epsilon _{i}}$ is the residual due to linearization, which is in addition to the residual, ${\displaystyle \ f_{i}^{(k)}}$, due to an approximate solution.

In order to drive ${\displaystyle \ f_{i}^{(k+1)}}$ closer to zero choose the values ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ such that

${\displaystyle -f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{R_{i}}}(x^{(k+1)}-x^{(k)})+{\frac {(y^{(k)}-y_{i})}{R_{i}}}(y^{(k+1)}-y^{(k)})\;+}$
${\displaystyle {\frac {(z^{(k)}-z_{i})}{R_{i}}}(z^{(k+1)}-z^{(k)})-(b^{(k+1)}-b^{(k)})c-p_{i}.\ {\mbox{for}}\ i=1,2,3,4\qquad (5)}$

That is choose the values

${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$

such that the residual in equation 2 changes by approximately ${\displaystyle \ \ -f_{i}^{(k)}}$ as shown below.

Let

${\displaystyle \Delta x^{(k+1)}\ =\ (x^{(k+1)}-x^{(k)}),\Delta y^{(k+1)}\ =\ (y^{(k+1)}-y^{(k)}),}$
${\displaystyle \Delta z^{(k+1)}\ =\ (z^{(k+1)}-z^{(k)}),\Delta b^{(k+1)}\ =\ (b^{(k+1)}-b^{(k)}).}$

Substituting and transposing ${\displaystyle \ \ p_{i}}$ to the left hand side of the equation, there results

${\displaystyle {\begin{Bmatrix}-f_{1}^{(k)}+p_{1}\\-f_{2}^{(k)}+p_{2}\\-f_{3}^{(k)}+p_{3}\\-f_{4}^{(k)}+p_{4}\\\end{Bmatrix}}={\begin{bmatrix}{\frac {(x^{(k)}-x_{1})}{R_{1}}}&{\frac {(y^{(k)}-y_{1})}{R_{1}}}&{\frac {(z^{(k)}-z_{1})}{R_{1}}}&-c\\{\frac {(x^{(k)}-x_{2})}{R_{2}}}&{\frac {(y^{(k)}-y_{2})}{R_{2}}}&{\frac {(z^{(k)}-z_{2})}{R_{2}}}&-c\\{\frac {(x^{(k)}-x_{3})}{R_{3}}}&{\frac {(y^{(k)}-y_{3})}{R_{3}}}&{\frac {(z^{(k)}-z_{3})}{R_{3}}}&-c\\{\frac {(x^{(k)}-x_{4})}{R_{4}}}&{\frac {(y^{(k)}-y_{4})}{R_{4}}}&{\frac {(z^{(k)}-z_{4})}{R_{4}}}&-c\\\end{bmatrix}}{\begin{Bmatrix}\Delta x^{(k+1)}\\\Delta y^{(k+1)}\\\Delta z^{(k+1)}\\\Delta b^{(k+1)}\\\end{Bmatrix}}\qquad \qquad \qquad \;\;\;\;\;\;(6)}$

Equations 6 provide four linear equations in four unknowns, the delta terms. They are in a form for solution. Using the values of ${\displaystyle \ \;\Delta x^{(k+1)},\Delta y^{(k+1)},\Delta z^{(k+1)}}$ and ${\displaystyle \;\Delta b^{(k+1)}}$ determined by this linear equation solution,

${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$

is evaluated using:

${\displaystyle \ x^{(k+1)}\ =\ x^{(k)}\ +\Delta x^{(k+1)},\ y^{(k+1)}\ =\ y^{(k)}\ +\Delta y^{(k+1)},}$
${\displaystyle \ z^{(k+1)}\ =\ z^{(k)}\ +\Delta z^{(k+1)},\ b^{(k+1)}\ =\ b^{(k)}\ +\Delta b^{(k+1)}.\qquad \qquad \;\;\;\;\;\;(7)}$

Then set ${\displaystyle \ k=k+1}$ in equations 2 through 6, plug the terms

${\displaystyle \left\{x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right\}}$

from equations 7 into equations 2, set ${\displaystyle \ k=k+1}$ in equations 7, and reevaluate the residuals in equations 2. This procedure is repeated until the residuals are sufficiently small in magnitude.

### Multidimensional Case for more than four satellites

The equations used in the case of more than four satellites are stated below. For another derivation of these equations see [8] or for an online derivation see Navigation Solution in GPS World by Torralbas, Roberto and Alvarez Jose Manuel.

Let ${\displaystyle x,y\;}$ and ${\displaystyle z\;}$ denote the true receiver position coordinates. Let ${\displaystyle \ b}$ denote the unknown clock error or bias. Let the coordinates of each satellite, and the time the message was sent, be ${\displaystyle \left[x_{i},y_{i},z_{i},t_{i}\right]}$, let the GPS clock's indicated received time be ${\displaystyle tr_{i}\;{\mbox{for}}\ i=1,2,3,4....n}$ and c be the speed of light. The pseudorange is computed as:

${\displaystyle p_{i}=\left(tr_{i}-t_{i}\right)c}$

Assume the message travels at the speed of light, then the pseudorange satisfies the equation:

${\displaystyle p_{i}={\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}-bc,}$ ${\displaystyle \;\;{\mbox{for}}\;i=1,2,3,4.\qquad (1)}$

When an approximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ rather than the exact solution, ${\displaystyle \left[x,y,z,b\right]}$ is used in equation 1, there is a residual, ${\displaystyle f_{i}^{(k)}}$. Transforming ${\displaystyle \ p_{i}}$ to the right hand side of the equation produces,

${\displaystyle f_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2}}}-b^{(k)}c-p_{i},\;\;}$
${\displaystyle {\mbox{for}}\;i=1,2,3,4,...n.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;\;\;(2)}$

A solution will have been obtained when ${\displaystyle f_{i}^{(k)}}$ is zero or sufficiently close to zero for ${\displaystyle \ i=1,2,3,4,...n}$.

In order to linearize equation 2, the partial derivatives are computed as:

${\displaystyle {\frac {\partial f_{i}^{(k)}}{\partial x^{(k)}}}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial y^{(k)}}}={\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}},}$
${\displaystyle {\frac {\partial f_{i}^{(k)}}{\partial z^{(k)}}}={\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}},{\frac {\partial f_{i}^{(k)}}{\partial b^{(k)}}}=-c\qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;(3)}$

where

${\displaystyle R_{i}^{(k)}={\sqrt {(x^{(k)}-x_{i})^{2}+(y^{(k)}-y_{i})^{2}+(z^{(k)}-z_{i})^{2}}}}$.

Linearizing the right hand side of equation 2 about the approximate solution, ${\displaystyle \left[x^{(k)},y^{(k)},z^{(k)},b^{(k)}\right]}$ there results

${\displaystyle f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{R_{i}^{(k)}}}x^{(k)}+{\frac {(y^{(k)}-y_{i})}{R_{i}^{(k)}}}y^{(k)}+{\frac {(z^{(k)}-z_{i})}{R_{i}^{(k)}}}z^{(k)}-b^{(k)}c-p_{i}+\epsilon _{i},\;\;}$
${\displaystyle {\mbox{for}}\ i=1,2,3,4\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (4)}$

where ${\displaystyle \ \epsilon _{i}}$ is the residual due to linearization, which is in addition to the residual, ${\displaystyle \ f_{i}^{(k)}}$, due to an approximate solution.

In order to drive ${\displaystyle \ f_{i}^{(k+1)}}$ closer to zero choose the values ${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$ such that

${\displaystyle -f_{i}^{(k)}={\frac {(x^{(k)}-x_{i})}{R_{i}}}(x^{(k+1)}-x^{(k)})+{\frac {(y^{(k)}-y_{i})}{R_{i}}}(y^{(k+1)}-y^{(k)})\;+}$
${\displaystyle {\frac {(z^{(k)}-z_{i})}{R_{i}}}(z^{(k+1)}-z^{(k)})-(b^{(k+1)}-b^{(k)})c-p_{i}.\ {\mbox{for}}\ i=1,2,3,4\qquad (5)}$

That is choose the values

${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$

such that the residual in equation 2 changes by approximately ${\displaystyle \ \ -f_{i}^{(k)}}$ as shown below.

Let

${\displaystyle \Delta x^{(k+1)}\ =\ (x^{(k+1)}-x^{(k)}),\Delta y^{(k+1)}\ =\ (y^{(k+1)}-y^{(k)}),}$
${\displaystyle \Delta z^{(k+1)}\ =\ (z^{(k+1)}-z^{(k)}),\Delta b^{(k+1)}\ =\ (b^{(k+1)}-b^{(k)}).}$

Substituting and transposing ${\displaystyle \ \ p_{i}}$ to the left hand side of the equation, there results

${\displaystyle {\begin{Bmatrix}-f_{1}^{(k)}+p_{1}\\-f_{2}^{(k)}+p_{2}\\-f_{3}^{(k)}+p_{3}\\-f_{4}^{(k)}+p_{4}\\\end{Bmatrix}}={\begin{bmatrix}{\frac {(x^{(k)}-x_{1})}{R_{1}}}&{\frac {(y^{(k)}-y_{1})}{R_{1}}}&{\frac {(z^{(k)}-z_{1})}{R_{1}}}&-c\\{\frac {(x^{(k)}-x_{2})}{R_{2}}}&{\frac {(y^{(k)}-y_{2})}{R_{2}}}&{\frac {(z^{(k)}-z_{2})}{R_{2}}}&-c\\{\frac {(x^{(k)}-x_{3})}{R_{3}}}&{\frac {(y^{(k)}-y_{3})}{R_{3}}}&{\frac {(z^{(k)}-z_{3})}{R_{3}}}&-c\\{\frac {(x^{(k)}-x_{4})}{R_{4}}}&{\frac {(y^{(k)}-y_{4})}{R_{4}}}&{\frac {(z^{(k)}-z_{4})}{R_{4}}}&-c\\\end{bmatrix}}{\begin{Bmatrix}\Delta x^{(k+1)}\\\Delta y^{(k+1)}\\\Delta z^{(k+1)}\\\Delta b^{(k+1)}\\\end{Bmatrix}}\qquad \qquad \qquad \;\;\;\;\;\;(6)}$

Equations 6 provide four linear equations in four unknowns, the delta terms. They are in a form for solution. Using the values of ${\displaystyle \ \;\Delta x^{(k+1)},\Delta y^{(k+1)},\Delta z^{(k+1)}}$ and ${\displaystyle \;\Delta b^{(k+1)}}$ determined by this linear equation solution,

${\displaystyle \left[x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right]}$

is evaluated using:

${\displaystyle \ x^{(k+1)}\ =\ x^{(k)}\ +\Delta x^{(k+1)},\ y^{(k+1)}\ =\ y^{(k)}\ +\Delta y^{(k+1)},}$
${\displaystyle \ z^{(k+1)}\ =\ z^{(k)}\ +\Delta z^{(k+1)},\ b^{(k+1)}\ =\ b^{(k)}\ +\Delta b^{(k+1)}.\qquad \qquad \;\;\;\;\;\;(7)}$

Then set ${\displaystyle \ k=k+1}$ in equations 2 through 6, plug the terms

${\displaystyle \left\{x^{(k+1)},y^{(k+1)},z^{(k+1)},b^{(k+1)}\right\}}$

from equations 7 into equations 2, set ${\displaystyle \ k=k+1}$ in equations 7, and reevaluate the residuals in equations 2. This procedure is repeated until the residuals are sufficiently small in magnitude.

## Extracting Navigation Message Using Demodulation and C/A Gold Code

Demodulating and Decoding GPS Satellite Signals

The figure depicts the processing of the signals broadcast by the satellites using the Coarse Acquisition Code. The broadcast signal is demodulated with the 1575.43 MHz L1 carrier. All of the satellite signals are modulated with the L1 carrier. Thus there is a need to separate the signals after demodulation. This is done by modulo 2 addition of the Gold code corresponding to satellite n1, satellite n2, ... nk where k is the number of channels in the GPS receiver. The result of these modulo 2 additions are the 50 bps navigation messages from satellites n1, n2 ... nk. The Gold codes used in GPS are a sequence of 1023 bits with a period of one millisecond. The Gold codes have the property of low cross correlation so that it is unlikely that one satellite signal will be misinterpreted as another. Also the Gold codes have good autocorellation properties. [9]

There are 1025 different Gold codes of length 1023 bits, but only 32 are used. These Gold codes are quite often referred to as pseudo random noise since they contain no message. However, this may be misleading to some readers since they are actually deterministic sequences. Gold_code provides a discusion as does [10].

## Changes of 17 August 2008

Reorganized sections. Moved template to Technical versus Encyclopedic in discussion. Added statements about technical nature of Processing the Navigation Message.

## Basic concept of GPS operation

A GPS receiver calculates its position by carefully timing the signals sent by the constellation of GPS satellites high above the Earth. Each satellite continually transmits messages containing the time the message was sent, a precise orbit for the satellite sending the message (the ephemeris), and the general system health and rough orbits of all GPS satellites (the almanac). These signals travel at the speed of light through outer space, and slightly slower through the atmosphere. The receiver uses the arrival time of each message to measure the distance to each satellite, from which it determines the position of the receiver (conceptually the intersection of spheres - see trilateration[11] ) The resulting coordinates are converted to more user-friendly forms such as latitude and longitude, or location on a map, then displayed to the user.

As the reader may have noticed trilateration can result in two solutions.

It is therefore necessary to explain to the reader how this ambiguity can be resolved. There are different methods for accomplishing this.

It might seem that three satellites would be enough to solve for a position, since space has three dimensions. However, a three satellite solution requires the time be known to a nanosecond or so, far better than any non-laboratory clock can provide. Using four or more satellites allows the receiver to solve for time as well as geographical position, eliminating the need for a very accurate clock. In other words, the receiver uses four measurements to solve for four variables: x, y, z, and t. While most GPS applications use the computed location and not the (very accurate) computed time, the time is used in some GPS applications such as time transfer and traffic signal timing.

Although four satellites are required for normal operation, fewer may be needed in some special cases. If one variable is already known (for example, a ship or plane may already know its altitude), a receiver can determine its position using only three satellites. Also, in practice, receivers use additional clues (doppler shift of satellite signals, last known position, dead reckoning, inertial navigation, and so on) to give degraded answers when fewer than four satellites are visible.[12][13][14]

A GPS receiver calculates its position by carefully timing the signals sent by the constellation of GPS satellites high above the Earth. Each satellite continually transmits messages containing the time the message was sent, a precise orbit for the satellite sending the message (the ephemeris), and the general system health and rough orbits of all GPS satellites (the almanac). These signals travel at the speed of light through outer space, and slightly slower through the atmosphere. The receiver uses the arrival time of each message to measure the distance to each satellite thereby establishing that the GPS receiver is approximately on the surfaces of spheres centered at each satellite. This information is then used to estimate the position of the GPS receiver as the intersection of sphere surfaces. The resulting coordinates are converted to a more convenient form for the user such as latitude and longitude, or location on a map, then displayed.

It might seem that three satellites would be enough to solve for a position, since space has three dimensions. However it turns out that three spheres usually intersect in two points. Thus the sphere corresponding to the fourth satellite is needed to determine which intersection is the GPS receiver position. Also a fourth satellite is needed to correct the GPS receiver clock by a method which will be described below. While many GPS applications have no particular use for the computed time, it is used in some GPS applications such as time transfer, and it is the only variable of interest in some applications, such as traffic signal timing.

Although four satellites are required for normal operation, fewer may be needed in some special cases. If one variable is already known (for example, a sea-going ship knows its altitude is 0), a receiver can determine its position using only three satellites.

I notice you keep complicating the section headings in Global Positioning System which were quite concise before. Granted, the titles chosen weren't perfect, but changing "Demodulation" to "Demodulation and decoding satellites signals" is excessively verbose. The short form surely hinted sufficiently at the section's subtopic. See WP:HEAD which references WP:MOS#Article titles.2C headings and sections which prefers short section names. —EncMstr (talk) 21:42, 22 August 2008 (UTC)

I was trying to save you some reading, but since you didn't see the headings and sections portion, you should just read all of the Manual of Style. Either that, or read some featured articles and you'll see that brevity is valued. Here are some good examples of technical articles: HTTP cookie, Search engine optimization, Automatic number plate recognition, Plug-in hybrid, and Construction of the World Trade Center. —EncMstr (talk) 01:48, 23 August 2008 (UTC)

## Picking one of two solutions

Hi!

You seem very sure that one of the uses of the fourth satellite is to pick among one of two solutions. But I think this is (almost) never needed, and never used in practice, since the two solutions are always widely separated, with only one near the Earth's surface. If you can find a technical reference that shows that any GPS receivers do this (other than perhaps some specialized receivers for use in deep outer space), I'll be very surprised. Otherwise I think we should state the method of disambiguation that is really used, and not a theoretical use.

This is part of a more general comment. There is a very fine line between a simplified explanation, and an explanation simplified so far it is not longer correct, and could mislead the reader. My personal belief is that we should err on the side of a correct explanation, and not talk down to the reader. This is because one of the main targets is someone who wants to learn more, and they should not find that the detailed information contradicts the simple stuff they read first. Such contradictions make them suspicious of the information as a whole... LouScheffer (talk) 12:15, 24 August 2008 (UTC)

Hi, The 3 satellite solutions you referenced works by assuming 0 altitude, not by assuming an accurate clock. This would give very bad results in practice. It also explains why 4 are needed to determine altitude - with 3 they just assume the altitude is 0. This page also says: "But more to the point, using only three satellites inevitably involves timing error that leads to inaccurate measurements. If you add a fourth satellite, its sphere will not intersect at the same point as the other three. That permits the receiver to figure out the timing error and factor it into the calculations of the other spheres." This I think reinforces the point that the main (and in practice only) reason for satellite 4 is to reduce clock errors. LouScheffer (talk) 20:36, 25 August 2008 (UTC)

## Demodulation and decoding

Demodulating and Decoding GPS Satellite Signals using the Coarse/Acquisition Gold code.

Since all of the satellite signals are modulated onto the same L1 carrier frequency, there is a need to separate the signals after demodulation. This is done by assigning each satellite a unique pseudorandom sequence known as a Gold code, and the signals are decoded, after demodulation, using modulo 2 addition of the Gold codes corresponding to satellites n1 through nk, where k is the number of channels in the GPS receiver and n1 through nk are the pseudorandom numbers associated with the satellites. The result of these modulo 2 additions are the 50 bps navigation messages from satellites n1 through nk. The Gold codes used in GPS are a sequence of 1023 bits with a period of one millisecond. These Gold codes are highly mutually orthogonal, so that it is unlikely that one satellite signal will be misinterpreted as another. As well, the Gold codes have good auto-correlation properties..[15]

There are 1025 different Gold codes of length 1023 bits, but only 32 are used. These Gold codes are quite often referred to as "pseudo-random noise" since they contain no data. However, this may be misleading since they are actually deterministic sequences.

If the almanac information has previously been acquired, the receiver picks which satellites to listen for by their PRN numbers. If the almanac information is not in memory, the receiver enters a search mode and cycles through the PRN numbers until a lock is obtained on one of the satellites. To obtain a lock, it is necessary that there be an unobstructed line of sight from the receiver to the satellite. The receiver can then acquire the almanac and determine the satellites it should listen for. As it detects each satellite's signal, it identifies it by its distinct C/A code pattern.

For each sattelite to which it listens, the receiver must calculate the time the message was sent. Transmission of each 30 second navigation message frame begins precisely on the minute or half minute as indicated by the satellite's atomic clock according to Satellite message format. In particular the first frame of every navigation message begins precisely on the minute or half minute indicated by the satellite atomic clock. Correcting these times by satellite's clock offset previously obtained gives a set of times differing by 30 seconds at which time the message was sent according to receiver clock. Since 30 seconds corresponds to a distance of about 9 million kilometers, the receiver can select which one corresponds to the time at which the message was sent.

The receiver also measures the received time for each satellite. As a first step in doing this, the receiver uses the C/A Gold code with the same PRN number as the satellite to compute an offset from the time sent that generates the best correlation. Call this offset, O. The true offset giving the best estimate of message transit time differs from O by an approximately integral number of milliseconds since the satellite repeats the broadcast of the PRN sequence each millisecond. The offset, O, is computed in a trial and error manner. The 1023 bits of the satellite PRN signal are compared with the receiver PRN signal. If correlation is not achieved, the 1023 bits of the sattelite PRN signal is shifted by one bit relative to the receiver PRN signal and again compared with this signal. This process is repeated until correlation is achieved.

Next the clock adjustment data in the navigation message is used to recalculate the time sent. Then the orbital position data, or ephemeris, from the Navigation Message is used to calculate precisely where the satellite was at the start of the message. A more sensitive receiver will potentially acquire the ephemeris data more quickly than a less sensitive receiver, especially in a noisy environment.[16]

This computation of sent time, received time, etc. is repeated for each satellite to which the receiver listens.

## Basic concept of GPS

A GPS receiver calculates its position by carefully timing the signals sent by the constellation of GPS satellites high above the Earth. Each satellite continually transmits messages containing the time the message was sent, precise orbital information (the ephemeris), and the general system health and rough orbits of all GPS satellites (the almanac). These signals travel at the speed of light[17]. The receiver measures the transit time of each message and uses these transit times along with the speed of light to compute the distance to each satellite. These distances are used to compute two estimates of receiver positions (conceptually the intersection of sphere surfaces—see trilateration[18]). It uses additional known information, such as being on board a near Earth vehicle, to determine which of the two is the valid estimate of position.[19] The receiver clock is then corrected as discussed below to get a much more accurate estimate of position. The final calculated position is displayed as

1) latitude, longitude, and possibly altitude or

2) location on a map and possibly altitude.

It might seem that three satellites would be enough to solve for a position, since space has three dimensions. However, a three-satellite solution requires that time be known to a nanosecond or so—far more accurate than a non-laboratory clock provides. Using four (or more) satellites allows the receiver to solve for time as well as geographical position, eliminating the need for a very accurate clock. In other words, the receiver uses four measurements to solve for four variables: x, y, z, and t. While most GPS applications use the computed location only and effectively hide the very accurately computed time, the time is used in a few GPS applications such as time transfer and traffic signal timing.

Although four satellites are required for normal operation, fewer may be needed in special cases. If one variable is already known (for example, a ship or plane may have known elevation), a receiver can determine its position using only three satellites. Some GPS receivers may use additional clues or assumptions (such as re-using the last known altitude, dead reckoning, inertial navigation, or including information from the vehicle computer) to give degraded answers when fewer than four satellites are visible.[20][21][22]

## Phase and Frequency Tracking

according to [23]

, Chapters 7 and 8 of

[24]

and

[25].

## Geometric dilution of precision computation (DOP)

As a first step n computing DOP, consider the unit vectors from the receiver to satellite i.

${\displaystyle {\frac {(x_{i}-x)}{R_{i}}},{\frac {(y_{i}-y)}{R_{i}}},and{\frac {(z_{i}-z)}{R_{i}}}}$ where
${\displaystyle R_{i}={\sqrt {(x_{i}-x)^{2}+(y_{i}-y)^{2}+(z_{i}-z)^{2}}}}$ and where ${\displaystyle \ x,y,and\ z}$ denote the position of the receiver and ${\displaystyle \ x_{i},y_{i},and\ z_{i}}$ denote the position of satellite i. Formulate the matrix A as
${\displaystyle A={\begin{bmatrix}{\frac {(x_{1}-x)}{R_{1}}}&{\frac {(y_{1}-y)}{R_{1}}}&{\frac {(z_{1}-z)}{R_{1}}}&c\\{\frac {(x_{2}-x)}{R_{2}}}&{\frac {(y_{2}-y)}{R_{2}}}&{\frac {(z_{2}-z)}{R_{2}}}&c\\{\frac {(x_{3}-x)}{R_{3}}}&{\frac {(y_{3}-y)}{R_{3}}}&{\frac {(z_{3}-z)}{R_{3}}}&c\\{\frac {(x_{4}-x)}{R_{4}}}&{\frac {(y_{4}-y)}{R_{4}}}&{\frac {(z_{4}-z)}{R_{4}}}&c\end{bmatrix}}}$

The first three elements of each row of A are the components of a unit vector from the receiver to the indicated satellite. The elements in the fourth column are c where c denotes the speed of light. Formulate the matrix, Q, as

${\displaystyle Q=\left(A^{T}A\right)^{-1}}$

This computation is in accordance with "Section 1.4.2 of PRINCIPLES OF SATELLITE POSITIONING" where the weighting matrix, P, has been set to the identity matrix.

The elements of the Q matrix are designated as

${\displaystyle Q={\begin{bmatrix}d_{x}^{2}&d_{xy}^{2}&d_{xz}^{2}&d_{xt}^{2}\\d_{xy}^{2}&d_{y}^{2}&d_{yz}^{2}&d_{yt}^{2}\\d_{xz}^{2}&d_{yz}^{2}&d_{z}^{2}&d_{zt}^{2}\\d_{xt}^{2}&d_{yt}^{2}&d_{zt}^{2}&d_{t}^{2}\end{bmatrix}}}$

The Greek letter ${\displaystyle \sigma }$ is used quite often where we have used d. However the elements of the Q matrix do not represent variances and covariances as they are defined in probability and statistics. Instead they are strictly geometric terms. Therefore d as in dilution of precision is used. PDOP, TDOP and GDOP are given by

${\displaystyle PDOP={\sqrt {d_{x}^{2}+d_{Y}^{2}+d_{Z}^{2}}}}$,
${\displaystyle \ TDOP={\sqrt {d_{t}^{2}}}}$, and
${\displaystyle GDOP={\sqrt {PDOP^{2}+TDOP^{2}}}}$ in agreement with "Section 1.4.9 of PRINCIPLES OF SATELLITE POSITIONING".

The horizontal dilution of precision, ${\displaystyle HDOP={\sqrt {d_{x}^{2}+d_{Y}^{2}}}}$, and the vertical dilution of precision, ${\displaystyle \ VDOP={\sqrt {d_{z}^{2}}}}$, are both dependent on the coordinate system used. To correspond to the local horizon plane and the local vertical, x, y, and z should denote positions in either a North, East, Down coordinate system or a South, East, Up coordinate system.RHB100 (talk) 22:36, 11 December 2008 (UTC)

### Derivation of DOP Equations

The equations for computing the geometric dilution of precision terms has been described in the previous section. This section describes the derivation of these equations. The method used here is similar to that used in "Global Positioning System (preview) by Parkinson and Spiker"

Consider the position error vector, e, defined as the vector from the indicated postion to the true position of the receiver.  :${\displaystyle \mathbf {e} =e_{x}{\hat {x}}+e_{y}{\hat {y}}+e_{z}{\hat {z}}}$ where bold denotes a vector and ${\displaystyle {\hat {x}},\ {\hat {y}},\ and\ {\hat {z}}}$ denote unit vectors along the x, y, and z axes respectively. Let ${\displaystyle \ e_{t}}$ denote the time error, the true time minus the receiver indicated time. Assume that the mean value of the three components of e and ${\displaystyle \ e_{t}}$ are zero.

${\displaystyle A\ {\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\\e_{t}\end{bmatrix}}={\begin{bmatrix}{\frac {(x_{1}-x)}{R_{1}}}&{\frac {(y_{1}-y)}{R_{1}}}&{\frac {(z_{1}-z)}{R_{1}}}&c\\{\frac {(x_{2}-x)}{R_{2}}}&{\frac {(y_{2}-y)}{R_{2}}}&{\frac {(z_{2}-z)}{R_{2}}}&c\\{\frac {(x_{3}-x)}{R_{3}}}&{\frac {(y_{3}-y)}{R_{3}}}&{\frac {(z_{3}-z)}{R_{3}}}&c\\{\frac {(x_{4}-x)}{R_{4}}}&{\frac {(y_{4}-y)}{R_{4}}}&{\frac {(z_{4}-z)}{R_{4}}}&c\end{bmatrix}}\ {\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\\e_{t}\end{bmatrix}}={\begin{bmatrix}e_{1}\\e_{2}\\e_{3}\\e_{4}\end{bmatrix}}\ (1)\ where\ e_{1},\ e_{2},\ e_{3},\ and\ e_{4}}$ are the errors in pseudoranges 1 through 4 respectively. This equation comes from linearizing the equation relating pseudoranges to receiver position, satellite positions, and receiver clock errors as shown in [2] .

Multiplying both sides by ${\displaystyle \ A^{-1}\ }$ there results

${\displaystyle \ {\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\\e_{t}\end{bmatrix}}=A^{-1}{\begin{bmatrix}e_{1}\\e_{2}\\e_{3}\\e_{4}\end{bmatrix}}\ (2)}$ . Transposing both sides
${\displaystyle \ {\begin{bmatrix}e_{x}&e_{y}&e_{z}&e_{t}\end{bmatrix}}={\begin{bmatrix}e_{1}&e_{2}&e_{3}&e_{4}\end{bmatrix}}\left(A^{-1}\right)^{T}\ (3)}$ .

Post multiplying the matrices on both sides of equation (2) by the corresponding matrices in equation (3), there results

${\displaystyle \ {\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\\e_{t}\end{bmatrix}}{\begin{bmatrix}e_{x}&e_{y}&e_{z}&e_{t}\end{bmatrix}}=A^{-1}{\begin{bmatrix}e_{1}\\e_{2}\\e_{3}\\e_{4}\end{bmatrix}}{\begin{bmatrix}e_{1}&e_{2}&e_{3}&e_{4}\end{bmatrix}}\left(A^{-1}\right)^{T}\ (4)}$ .

Taking the expected value of both sides and taking the non-random matrtices outside the expectation operator, E, there results

${\displaystyle \ E\left({\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\\e_{t}\end{bmatrix}}{\begin{bmatrix}e_{x}&e_{y}&e_{z}&e_{t}\end{bmatrix}}\right)=A^{-1}\ E\left({\begin{bmatrix}e_{1}\\e_{2}\\e_{3}\\e_{4}\end{bmatrix}}{\begin{bmatrix}e_{1}&e_{2}&e_{3}&e_{4}\end{bmatrix}}\right)\left(A^{-1}\right)^{T}\ (5)}$ .

Assuming the pseudorange errors are uncorrelated and have the same variance, the covariance matrix on the right side can be expressed as a scalar times the identity matrix. Thus

${\displaystyle {\begin{bmatrix}\sigma _{x}^{2}&\sigma _{xy}^{2}&\sigma _{xz}^{2}&\sigma _{xt}^{2}\\\sigma _{xy}^{2}&\sigma _{y}^{2}&\sigma _{yz}^{2}&\sigma _{yt}^{2}\\\sigma _{xz}^{2}&\sigma _{yz}^{2}&\sigma _{z}^{2}&\sigma _{zt}^{2}\\\sigma _{xt}^{2}&\sigma _{yt}^{2}&\sigma _{zt}^{2}&\sigma _{t}^{2}\end{bmatrix}}=\sigma _{R}^{2}\ A^{-1}\left(A^{-1}\right)^{T}=\sigma _{R}^{2}\ \left(A^{T}A\right)^{-1}\ (6)\ since\ A^{-1}\left(A^{-1}\right)^{T}\left(A^{T}A\right)=I}$
${\displaystyle \ Note:\left(A^{-1}\right)^{T}=\left(A^{T}\right)^{-1}\ since\ I=\left(AA^{-1}\right)^{T}=\left(A^{-1}\right)^{T}A^{T}}$

Substituting for ${\displaystyle \left(A^{T}A\right)^{-1}=Q}$ there follows

${\displaystyle {\begin{bmatrix}\sigma _{x}^{2}&\sigma _{xy}^{2}&\sigma _{xz}^{2}&\sigma _{xt}^{2}\\\sigma _{xy}^{2}&\sigma _{y}^{2}&\sigma _{yz}^{2}&\sigma _{yt}^{2}\\\sigma _{xz}^{2}&\sigma _{yz}^{2}&\sigma _{z}^{2}&\sigma _{zt}^{2}\\\sigma _{xt}^{2}&\sigma _{yt}^{2}&\sigma _{zt}^{2}&\sigma _{t}^{2}\end{bmatrix}}=\sigma _{R}^{2}\ {\begin{bmatrix}d_{x}^{2}&d_{xy}^{2}&d_{xz}^{2}&d_{xt}^{2}\\d_{xy}^{2}&d_{y}^{2}&d_{yz}^{2}&d_{yt}^{2}\\d_{xz}^{2}&d_{yz}^{2}&d_{z}^{2}&d_{zt}^{2}\\d_{xt}^{2}&d_{yt}^{2}&d_{zt}^{2}&d_{t}^{2}\end{bmatrix}}\ (7)}$

From equation (7), it follows that the variances of indicated receiver position and time are

${\displaystyle \sigma _{rc}^{2}=\sigma _{x}^{2}+\sigma _{y}^{2}+\sigma _{z}^{2}=\sigma _{R}^{2}\left(d_{x}^{2}+d_{y}^{2}+d_{z}^{2}\right)=PDOP^{2}\sigma _{R}^{2}}$ and
${\displaystyle \sigma _{t}^{2}=\sigma _{R}^{2}d_{t}^{2}=TDOP^{2}\sigma _{R}^{2}}$ .

The remaining position and time error variance terms follow in a straightforward manner.

## Carrier phase tracking (surveying)

Utilizing the navigation message to measure pseudorange has been discussed. Another method that is used in GPS surveying applications is carrier phase tracking. The period of the carrier frequency times the speed of light gives the wave length, which is about 0.19 meters for the L1 carrier. With a 1% of wave length accuracy in detecting the leading edge, this component of pseudorange error might be as low as 2 millimeters. This compares to 3 meters for the C/A code and 0.3 meters for the P code.

However, this 2 millimeter accuracy requires measuring the total phase, that is the total number of wave lengths plus the fractional wavelength. This requires specially equipped receivers. This method has many applications in the field of surveying.

We now describe a method which could potentially be used to estimate the position of receiver 2 given the position of receiver 1 using triple differencing followed by a mathematical technique called least squares. A detailed discussion of the errors is omitted in order to avoid detracting from the description of the methodology. In this description differences are taken in the order of differencing between satellites, differencing between receivers, and differencing between epochs. This should not be construed to mean that this is the only order which can be used. Indeed other orders of taking differences are equally valid.

The satellite carrier total phase can be measured with ambiguity as to the number of cycles as described in CARRIER PHASE MEASUREMENT and CARRIER BEAT PHASE. Let ${\displaystyle \ \phi (r_{i},s_{j},t_{k})}$ denote the phase of the carrier of satellite j measured by receiver i at time ${\displaystyle \ t_{k}}$. This notation has been chosen so as to make it clear what the subscripts i, j, and k mean. In view of the fact that the receiver, satellite, and time come in alphabetical order as arguments of ${\displaystyle \ \phi }$ and to strike a balance between readability and conciseness, let ${\displaystyle \ \phi _{i,j,k}=\phi (r_{i},s_{j},t_{k})}$ so as to have a concise abbreviation. Also we define a three functions, :${\displaystyle \ \Delta ^{r},\Delta ^{s},and\Delta ^{t}}$ which perform differences between receivers, satellites, and time points respectively. Each of these functions has a linear combination of three subscripted variable as its argument. These three functions are defined with the following six equations.

${\displaystyle \ \Delta ^{r}(\alpha _{i,j,k})=\alpha _{i+1,j,k}-\alpha _{i,j,k}}$ ,
${\displaystyle \ \Delta ^{s}(\alpha _{i,j,k})=\alpha _{i,j+1,k}-\alpha _{i,j,k}}$ ,
${\displaystyle \ \Delta ^{t}(\alpha _{i,j,k})=\alpha _{i,j,k+1}-\alpha _{i,j,k}}$ . and also
${\displaystyle \ \Delta ^{r}(\alpha _{i,j,k}\pm \beta _{l,m,n})=\Delta ^{r}(\alpha _{i,j,k})\pm \Delta ^{r}(\beta _{l,m,n})}$ ,
${\displaystyle \ \Delta ^{s}(\alpha _{i,j,k}\pm \beta _{l,m,n})=\Delta ^{s}(\alpha _{i,j,k})\pm \Delta ^{s}(\beta _{l,m,n})}$ ,
${\displaystyle \ \Delta ^{t}(\alpha _{i,j,k}\pm \beta _{l,m,n})=\Delta ^{t}(\alpha _{i,j,k})\pm \Delta ^{t}(\beta _{l,m,n})}$ ,

Receiver clock errors can be approximately eliminated by differencing the phases measured from satellite 1 with that from satellite 2 at the same epoch as shown in BETWEEN-SATELLITE DIFFERENCING. This difference is designated as ${\displaystyle \ \Delta ^{s}(\phi _{1,1,1})=\phi _{1,2,1}-\phi _{1,1,1}}$

Double differencing can be performed by taking the differeces of the between satellite difference observed by receiver 1 with that observed by receiver 2. The satellite clock errors will be approximately eliminated by this between receiver differencing. This double difference is designated as ${\displaystyle \ \Delta ^{r}(\Delta ^{s}(\phi _{1,1,1})=\Delta ^{r}(\phi _{1,2,1}-\phi _{1,1,1})}$ .

Triple differencing can be performed by taking the difference of double differencing performed at time ${\displaystyle \ t_{2}}$ with that performed at time ${\displaystyle \ t_{1}}$. This will eliminate the ambiguity associated with the integral number of wave lengths in carrier phase provided this ambiguity does not change with time. Thus the triple difference result has eliminated all or practically all clock bias errors and the integer ambiguity. Also errors associated with atmospheric delay and satellite ephemeris have been significantly reduced. This triple difference is designated as ${\displaystyle \ \Delta ^{t}(\Delta ^{r}(\Delta ^{s}(\phi _{1,1,1})))}$ .

Triple difference results can be used to estimate unknown variables. For example if the position of receiver 1 is known but the position of receiver 2 unknown, it may be possible to estimate the position of receiver 2 using least squares. Although the errors in the triple difference results have been reduced there is still some residual error. Thus triple differencing should be performed at several time pairs to increase estimate accuracy. This will result in an over determined system. To get estimates for an over determined system, least squares can be used. The least squres procedure determines the position of receiver 2 which best fits the observed triple differences under the criterion of minimizing the sum of the squares of the differences between the measured triple differences and the triple differences computed with the estimated position of receiver 2.

### Problem description

The receiver uses messages received from satellites to determine the satellite positions and time sent. The x, y, and z components of satellite position and the time sent are designated as [xi, yi, zi, si] where the subscript i denotes the satellite and has the value 1, 2, ..., n, where n ≥ 4. When the time of message reception indicated by the on-board receiver clock is , the true reception time is t = - b, where b is the receiver's clock offset from the much more accurate GPS system clocks employed by the satellites. The receiver clock offset is the same for all received satellite signals (assuming the satellite clocks are all perfectly synchronized). The message's transit time is - b - si. Assuming the message traveled at the speed of light, c, the distance traveled is ( - b - si) c.

For n satellites, the equations to satisfy are:

${\displaystyle (x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}={\bigl (}[{\tilde {t}}-b-s_{i}]c{\bigr )}^{2},\;i=1,2,\dots ,n}$

or in terms of pseudoranges, ${\displaystyle p_{i}=\left({\tilde {t}}-s_{i}\right)c}$, as

${\displaystyle {\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}+bc=p_{i},\;i=1,2,...,n}$ .[26][27]

Comparison of these equations with the Equations in R3 section of Sphere in which ${\displaystyle (x-x_{i})}$ corresponds to ${\displaystyle (x-x_{0})}$, ${\displaystyle (y-y_{i})}$ corresponds to ${\displaystyle (y-y_{0})}$, ${\displaystyle (z-z_{i})}$ corresponds to ${\displaystyle (z-z_{0})}$, and ${\displaystyle {\bigl (}[{\tilde {t}}-b-s_{i}]c{\bigr )}}$ corresponds to ${\displaystyle r}$ shows that these equations are spheres as documented in Sphere.

Since the equations have four unknowns [x, y, z, b]—the three components of GPS receiver position and the clock bias—signals from at least four satellites are necessary to attempt solving these equations. They can be solved by algebraic or numerical methods. Existence and uniqueness of GPS solutions are discussed by Abell and Chaffee.[28] When n is greater than 4 this system is overdetermined and a fitting method must be used.

With each combination of satellites, GDOP quantities can be calculated based on the relative sky directions of the satellites used.[29] The receiver location is expressed in a specific coordinate system, such as latitude and longitude using the WGS 84 geodetic datum or a country-specific system.[30]

#### Problem description - Talk page

Here is a quote from this document: "The expression under the square root sign is the true range to the satellite. It is actually a representation of the sphere centered on coordinates x,y,z, the position of the satellite."

The equations, ${\displaystyle (x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}={\bigl (}[{\tilde {t}}-b-s_{i}]c{\bigr )}^{2},\;i=1,2,\dots ,n}$ represent spheres centered on the satellites with coordinates ${\displaystyle x_{i},y_{i},z_{i},\;i=1,2,\dots ,n}$ and with radii given by ${\displaystyle {\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}=[{\tilde {t}}-b-s_{i}]c,\;i=1,2,...,n}$. [31]

Well at least we are discussing the issues. That seems to indicate some progress. I agree with Woodstone that a pseudorange is just a distance and that the basic equations given above describe spheres (=spherical surfaces) for each fixed value of the (unknown) clock bias b. I agree with Fgnievinski that the discussion should be restricted to the geometrical three-dimensional space -- no 4D space-time. I think that we should recognize b as the clock bias. And that there is one and only one clock bias. Therefore we should talk about one and only one value of b. There is no reason to drag the reader through what may be envisioned as happening in an iterative procedure with b changing. The pseudorange, ${\displaystyle p_{i}}$ is the measured quantity, equal to ${\displaystyle \left({\tilde {t}}-s_{i}\right)c}$ where ${\displaystyle \left({\tilde {t}}\right)}$ denotes the indicated time of reception, ${\displaystyle \left(s_{i}\right)}$ denotes the time of transmission and ${\displaystyle c}$ denotes the speed of light.. But the pseudorange typically contains large errors because of the large velocity of light along with the fact that there are inaccuracies in the receiver clock. The corrections to the pseudoranges are given by :${\displaystyle {\bigl (}-bc{\bigr )},\;i=1,2,\dots ,n}$ where b is the clock bias at least to a highly accurate approximtion. The distances from receiver to satellites (at least to an accurate approximation are ${\displaystyle {\bigl (}\left({\tilde {t}}-s_{i}\right)c-bc{\bigr )},\;i=1,2,\dots ,n}$ . The squares of these distances are equal to the sum of the squares of the three components of the distances from receiver to the satellites. Thus these equations appear to be describing the surfaces of spheres as pointed out by Woodstone.

Here are some statements on which we may be able to agree. (1)The material in The Mathematics of GPS BY Richard Langley down through the section titled, Linearization of the pseudorange equations, provides a correct description of the GPS mathematical problem, its solution, and its geometry. (2) The GPS mathematical problem is most clearly envisioned as a problem involving four unknowns in three dimensional space. (3) The problem description section should include the equations to be solved and the properties of the solution. Details of the solution method should be hidden from the reader in this section with only links or names of the solution method. We should not drag the reader through what may be envisioned as occurring in the solution process such as b, the clock bias, taking on different values in an iterative solution method. This allows the reader to concentrate on the meaning of the equations and the properties of the solution without the distraction of computer algorithms. The mathematics of GPS can be better understood when you think just about that. Then in a section devoted to solution methods if included, the reader can think exclusively about the solution method and better understand that. RHB100 (talk) 21:19, 29 June 2015 (UTC)

### Geometric interpretation

The GPS equations can be solved by numerical and analytical methods. Geometrical interpretations can enhance the understanding of these solution methods.

#### Spheres

The solution of the equations in the problem description section result in corrected distances from satellites to receiver, ${\displaystyle {\bigl (}[{\tilde {t}}-b-s_{i}]c{\bigr )},\;i=1,2,\dots ,n}$. These distances represent the radii of spheres, each centered on one of the transmitting satellites. The solution for the position of the receiver is then at or near the intersection of the surfaces of four or more of these spheres. [32] [26] [27] This near intersection of four or more sphere surfaces only exists when the clock bias, b, is at least approximately correct. It should be kept in mind that this near intersection of the surfaces of four or more sphere surfaces in no way contradicts that the receiver is at the near intersection of hyperboloid surfaces as stated below.

#### Hyperboloids

If the distance traveled between the receiver and satellite i and the distance traveled between the receiver and satellite j are subtracted, the result is ( - si) c - ( - sj) c, which only involves known or measured quantities. The locus of points having a constant difference in distance to two points (here, two satellites) is a hyperboloid (see Multilateration). Thus, from four or more measured reception times, the receiver can be placed at the intersection of the surfaces of three or more hyperboloids.[28][33]

#### Spherical cones

The solution space [x, y, z, b] can be seen as a four-dimensional geometric space. In that case each of the equations describes a spherical cone,[34] with the cusp located at the satellite, and the base a sphere around the satellite. The receiver is at the intersection of four or more of such cones.

### Solution methods

#### Least squares

When more than four satellites are available, the calculation can use the four best, or more than four simultaneously (up to all visible satellites), depending on the number of receiver channels, processing capability, and geometric dilution of precision (GDOP).

Using more than four involves an over-determined system of equations with no unique solution; such a system can be solved by a least-squares or weighted least squares method.[26]

${\displaystyle \left({\hat {x}},{\hat {y}},{\hat {z}},{\hat {b}}\right)={\underset {\left(x,y,z,b\right)}{\arg \min }}\sum _{i}\left({\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}+bc-p_{i}\right)^{2}}$

#### Iterative

Both the equations for four satellites, or the least squares equations for more than four, are non-linear and need special solution methods. A common approach is by iteration on a linearized form of the equations, (e.g., Gauss–Newton algorithm).

The GPS system was initially developed assuming use of a numerical least-squares solution method—i.e., before closed-form solutions were found.

#### Closed-form

One closed-form solution to the above set of equations was developed by S. Bancroft.[27][35] Its properties are well known;[28][33][36] in particular, proponents claim it is superior in low-GDOP situations, compared to iterative least squares methods.[35]

Bancroft's method is algebraic, as opposed to numerical, and can be used for four or more satellites. When four satellites are used, the key steps are inversion of a 4x4 matrix and solution of a single-variable quadratic equation. Bancroft's method provides one or two solutions for the unknown quantities. When there are two (usually the case), only one is a near-earth sensible solution.[27]

When a receiver uses more than four satellites for a solution, Bancroft uses the generalized inverse (i.e., the pseudoinverse) to find a solution. However, a case has been made that iterative methods (e.g., Gauss-Newton algorithm) for solving over-determined non-linear least squares (NLLS) problems generally provide more accurate solutions.[37]

Leick et al. (2015) states that "Bancroft’s (1985) solution is a very early, if not the first, closed-form solution."[38] Other closed-form solutions were published afterwards,[39][40] although their adoption in practice is unclear.

## References

1. ^ a b c d Press, Flannery, Tekolsky, and Vetterling 1986, Numerical Recipes, The Art of Scientific Computing (Cambridge University Press).
2. ^ Noe, P.S.; Myers, K.A. (March 1976). "A Position Fixing Algorithm for the Low-Cost GPS Receiver". IEEE Transactions on Aerospace and Electronic Systems. AES-12 (2): 295 – 297.
3. ^ Richard Langley (July/August 1991). "The Mathematics of GPS" (PDF). GPS World. Check date values in: |date= (help)
4. ^ Jay Farrell, Matthew Barth (1999). The global positioning system and inertial navigation. McGraw-Hill. ISBN 007022045X., p. 145.
5. ^ Bancroft, S. (1985). "An Algebraic Solution of the GPS Equations". Aerospace and Electronic Systems, IEEE Transactions on: 56––59.
6. ^ Krause, L.O. (1987). "A Direct Solution to GPS-Type Navigation Equations". Aerospace and Electronic Systems, IEEE Transactions on. AES-23 (2): 225–232. doi:10.1109/TAES.1987.313376. Unknown parameter |month= ignored (help)
7. ^ Langley, R. B., "The Mathematics of GPS," GPS World, Vol. 2, No. 7, July/August 1991, pp. 45-50
8. ^ Langley, R. B., "The Mathematics of GPS," GPS World, Vol. 2, No. 7, July/August 1991, pp. 45-50
9. ^ "[George, M., Hamid, M., and Miller A. "Gold Code Generators in Virtex Devices" (PDF).
10. ^
11. ^
12. ^ Georg zur Bonsen, Daniel Ammann, Michael Ammann, Etienne Favey, Pascal Flammant (2005-04-01). "Continuous Navigation Combining GPS with Sensor-Based Dead Reckoning". GPS World.
13. ^ "NAVSTAR GPS USER EQUIPMENT INTRODUCTION" (PDF). US Government., Chapters 7 and 8.
14. ^ "GPS Support Notes" (PDF).
15. ^ "[George, M., Hamid, M., and Miller A. "Gold Code Generators in Virtex Devices" (PDF).
16. ^ "AN02 Network Assistance" (HTML). Retrieved 2007-09-10.
17. ^ The speed of light varies slightly between the partial vacuum of space and the denser atmosphere, however a receiver can approximate these effects. Once a rough position is determined, some receivers carefully compute the amount of atmosphere the signal traveled through and adjust the signal timing accordingly.
18. ^ HowStuffWorks: How GPS Receivers Work accessed May 14, 2006.
19. ^ The (usually erroneous)second position is typically tens of thousands of miles above the earth.
20. ^ Georg zur Bonsen, Daniel Ammann, Michael Ammann, Etienne Favey, Pascal Flammant (2005-04-01). "Continuous Navigation Combining GPS with Sensor-Based Dead Reckoning". GPS World.
21. ^ "NAVSTAR GPS USER EQUIPMENT INTRODUCTION" (PDF). US Government., Chapters 7 and 8.
22. ^ "GPS Support Notes" (PDF).
23. ^ Georg zur Bonsen, Daniel Ammann, Michael Ammann, Etienne Favey, Pascal Flammant (2005-04-01). "Continuous Navigation Combining GPS with Sensor-Based Dead Reckoning". GPS World.
24. ^ "NAVSTAR GPS USER EQUIPMENT INTRODUCTION" (PDF). US Government.
25. ^ "GPS Support Notes" (PDF). January 19, 2007. Retrieved 2008-11-10.
26. ^ a b c section 4 beginning on page 15 GEOFFREY BLEWITT: BASICS OF THE GPS TECHNIQUE
27. ^ a b c d "Global Positioning Systems" (PDF). Archived from the original (PDF) on July 19, 2011. Retrieved October 15, 2010. Cite error: Invalid <ref> tag; name "Bancroft" defined multiple times with different content (see the help page).
28. ^ a b c Cite error: The named reference Abel1 was invoked but never defined (see the help page).
29. ^ Dana, Peter H. "Geometric Dilution of Precision (GDOP) and Visibility". University of Colorado at Boulder. Retrieved July 7, 2008.
30. ^ Peter H. Dana. "Receiver Position, Velocity, and Time". University of Colorado at Boulder. Retrieved July 7, 2008.
31. ^ Richard Langley (1991). "The Mathematics of GPS" (PDF). GPS World.
32. ^ Langley, R. B., "The Mathematics of GPS," GPS World, Vol. 2, No. 7, July/August 1991, pp. 45-50
33. ^ a b Cite error: The named reference Fang was invoked but never defined (see the help page).
34. ^
35. ^ a b Bancroft, S. (January 1985). "An Algebraic Solution of the GPS Equations". IEEE Transactions on Aerospace and Electronic Systems. AES-21: 56–59. Bibcode:1985ITAES..21...56B. doi:10.1109/TAES.1985.310538.
36. ^ Chaffee, J. and Abel, J., "On the Exact Solutions of Pseudorange Equations", IEEE Transactions on Aerospace and Electronic Systems, vol:30, no:4, pp: 1021-1030, 1994
37. ^ Sirola, Niilo, "Closed-form Algorithms in Mobile Positioning: Myths and Misconceptions", Proceedings of the 7th Workshop on Positioning, Navigation and Communication 2010 (WPNC'10), Dresden, March 2010. Retrieved November 10, 2014
38. ^ doi:10.1002/9781119018612.ch6
39. ^ Alfred Kleusberg, "Analytical GPS Navigation Solution", University of Stuttgart Research Compendium,1994
40. ^ Oszczak, B., "New Algorithm for GNSS Positioning Using System of Linear Equations," Proceedings of the 26th International Technical Meeting of The Satellite Division of the Institute of Navigation (ION GNSS+ 2013), Nashville, TN, September 2013, pp. 3560-3563.

## G-force

Remove-This commonly used term is a common error. An encyclopedia should not repeat common misconceptions. G-force was not natural to Isaac Newton who regarded force and acceleration as two completely different things.

BSME & MSEM University of Arkansas
Degree of Engineer UCLA (majoring in rigid and elastic body dynamics) RHB100 (talk) 19:51, 28 July 2009 (UTC)
You have placed a Dubious on G-force but no arguement. Would you like to open discussion? Is your point the statement above? If so how do we handle the fact that the term is in almost universal usage, and not just by laypeople? We cannot say it doen't exist. It is valid and useful to say my airplane is certified to +6g/-3g, it's even got a g recorder to measure it. Ex nihil (talk) 23:58, 4 August 2009 (UTC)

## Re: Why is PRN 08 not listed in article, List of GPS satellite launches

Because I couldn't find it in the first pass. It's in there now. -- 23:37, 17 August 2009 (UTC)

## If ice did not float, bodies of water would freeze from the bottom to the top - Unjustified Speculation

If ice did not float, bodies of water would freeze from the bottom toward the top[1]. Sinking ice brings about two effects at the surface. (1) An increased temperature difference between the water and the atmosphere due to warmer water replacing the sinking ice. This effect would cause greater heat transfer from the water to the atmosphere. (2) A reduction of thermal conductivity across the layer formerly occupied by the ice since as shown in Selected properties of fresh water, the thermal conductivity of water is lower than that of ice by almost a factor of four near freezing temperature. This effect would cause the heat flow from the water to the atmosphere to decrease. If the former effect is predominant then there will be an increased heat flow from the surface to the atmosphere. If it is also assumed that any change in the heat flow from the earth into the body of water as a result of the ice falling to the bottom is small compared to the heat transfer at the surface then in accordance with the First law of thermodynamics, there would be increased cooling of the body of water under the condition of ice sinking over the condition of reality, floating ice. Thus under these assumptions, freezing from the bottom all the way to the top would be more likely if ice did not float.

This freezing of the body of water could kill fish and other creatures not resistant to freezing temperatures. Density of ice increases slightly with decreasing temperature and has a value of 0.9340 g/cm³ at −180 °C (93 K).[2]

References

1. ^ Water, Water by Neil deGrasse Tyson
2. ^ Lide, D. R., ed. (2005). CRC Handbook of Chemistry and Physics (86th ed.). Boca Raton (FL): CRC Press. ISBN 0-8493-0486-5.

## Cross reference of PRN's to Blocks

 Cross reference of PRN's to Blocks H denotes healthy and in orbit, U denotes unhealthy, unusable, or decommisioned and not assigned to another block. PRN 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 BLOCK IIA H H H H H H U H H H H H BLOCK IIR H H H H H H H H H H H H BLOCK IIR-M U H H H H H H H

## Geoid non-rotating gravitational potential in progress

From the vector diagram, using bold to denote vectors it follows that

R = r1 + r ,
r = R - r1 , and
r = |R - r1| . Squaring and then taking the square root,
r = ${\displaystyle {\sqrt {R^{2}-2\mathbf {R\cdot r1} +r1^{2}}}\ .}$ Factoring out R.
r = ${\displaystyle R{\sqrt {1-{\frac {2\mathbf {R\cdot r1} }{R^{2}}}+\left({\frac {r1}{R}}\right)^{2}}}\ .}$ Since we are interested in the case in which r > 0, we can invert both sides to get
${\displaystyle r^{-1}=R^{-1}\left(1-{\frac {2\mathbf {R\cdot r1} }{R^{2}}}+\left({\frac {r1}{R}}\right)^{2}\right)^{-{\frac {1}{2}}}\ .}$
${\displaystyle r^{-1}=R^{-1}\left(1-2cos(b){\frac {r1}{R}}+\left({\frac {r1}{R}}\right)^{2}\right)^{-{\frac {1}{2}}}\ .}$

Consider the function,

${\displaystyle f(x)=\left(1-2cos(b)x+x^{2}\right)^{-{\frac {1}{2}}}\ .}$
${\displaystyle f'(x)=\left(cos(b)-x\right)\left(1-2cos(b)x+x^{2}\right)^{-{\frac {3}{2}}}\ .}$
${\displaystyle f''(x)=-\left(1-2cos(b)x+x^{2}\right)^{-{\frac {3}{2}}}}$ ${\displaystyle +3\left(cos(b)-x\right)^{2}\left(1-2cos(b)x+x^{2}\right)^{-{\frac {5}{2}}}\ .}$
Expanding f(x) in a Taylor series about x = 0, we get,
${\displaystyle f(x)=1+x\ cos(b)+x^{2}{\frac {3cos(b)^{2}-1}{2}}+\ ...\ .\ }$ Utilizing the definition of Legendre Polynomials we get,
${\displaystyle f(x)=P_{0}(cos(b))+x\ P_{1}(cos(b))+x^{2}\ P_{2}(cos(b))+\ ...\ }$ where the ${\displaystyle \ P_{n}(cos(b))}$ are Legendre Polynomials of order n with argument, cos(b). The appearance of the first three terms of the series for f(x) suggests that the infinite series might be of similar form. On page 454 of AEM by C. R. Wylie, Jr., Theorem 2 which states that ${\displaystyle \left(1-2XZ+Z^{2}\right)^{-{\frac {1}{2}}}\ =\sum _{n=0}^{\infty }Z^{n}P_{n}(X)\ }$, is proven. Applied to the notation used here Theorem 2 states that ${\displaystyle \left(1-2cos(b)x+x^{2}\right)^{-{\frac {1}{2}}}\ =\sum _{n=0}^{\infty }x^{n}P_{n}(cos(b))\ .}$ If we let ${\displaystyle x={\frac {r1}{R}}}$ then we get the result,
${\displaystyle {\frac {1}{r}}={\frac {1}{R}}\sum _{n=0}^{\infty }\left({\frac {r1}{R}}\right)^{n}P_{n}(cos(b))\ .}$

Consider the function,

${\displaystyle f({\frac {x}{R}})=\left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {1}{2}}}\ .}$

${\displaystyle f'({\frac {x}{R}})=\left(cos(\beta )-\left({\frac {x}{R}}\right)\right)\left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {3}{2}}}}$

${\displaystyle f''({\frac {x}{R}})=-\left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {3}{2}}}}$ ${\displaystyle +3\left(cos(\beta )-\left({\frac {x}{R}}\right)\right)^{2}\left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {5}{2}}}\ .}$

### Earth Precession

From the vector diagram, using bold to denote vectors it follows that

${\displaystyle \mathbf {R} =\mathbf {\rho } +\mathbf {r} }$ ,
${\displaystyle \mathbf {r} =\mathbf {R} -\mathbf {\rho } }$, and
${\displaystyle r=|\mathbf {R} -\mathbf {\rho } |}$. Squaring and then taking the square root,
${\displaystyle r={\sqrt {R^{2}-2\mathbf {R\cdot \rho } +\rho ^{2}}}\ .}$ Factoring out R.
${\displaystyle r=R{\sqrt {1-{\frac {2\mathbf {R\cdot \rho } }{R^{2}}}+\left({\frac {\rho }{R}}\right)^{2}}}\ .}$ Since we are interested in the case in which r > 0, we can invert both sides to get

## GPS pictures

I've improved the diagramm for the GPS articles, as replacement for the raster picture. You removed the picture because there was no bus (from change description). Is the picture OK now ? Does it needs some correction ? Other pictures (I've made) have also been removed (for other reasons). I tried to make better picture (vector drawings), but with errors apparently. Could you also please contact the author of the picture to correct them , rather then just removing them. Thanks --Tsaitgaist (talk) 11:39, 10 February 2010 (UTC)

I've also update this picture as replacement for this picture. The polar axis are now vertical.--Tsaitgaist (talk) 13:58, 10 February 2010 (UTC)

Now this picture looks more like this one. --Tsaitgaist (talk) 14:09, 10 February 2010 (UTC)

However, I still disagree with you regarding, File:Gps error diagram.svg. This drawing looks more like a flow chart than a geometric figure and gives the false impression that the errors are all in the same direction and thus additive in my opinion. RHB100 (talk) 21:10, 10 February 2010 (UTC)

I'm happy to correct it, but I don't understand the error. Should it be horizontal rather then vertical ? Should the bubbles go up ? Are there missing or too many arrows ? I read the 2 diagrams the same way, but I'm just a drawer, there might be a technical aspect I don't see.--Tsaitgaist (talk) 14:24, 12 February 2010 (UTC)

Why don't you go ahead and add the figure, File:Lat 2spheres 2.svg, since we both agree that this is a good figure, the color serves to distinguish the 2 spheres and the readers may appreciate a little color in the figures. After that we can discuss the figure on which we disagree. RHB100 (talk) 22:22, 12 February 2010 (UTC) RHB100 (talk) 22:25, 12 February 2010 (UTC)

I was just too lazy :p. I waited for all the drawings to be ok to change them. I've change the correct ones. We can discuss to correct and enhance the last picture to fit the article --Tsaitgaist (talk) 10:21, 13 February 2010 (UTC)

Why don't you go ahead and add the figure, File:Lat 2spheres 2.svg, since we both agree that this is a good figure, the color serves to distinguish the 2 spheres and the readers may appreciate a little color in the figures. After that we can discuss the figure on which we disagree. RHB100 (talk) 22:22, 12 February 2010 (UTC)

I was just too lazy :p. I waited for all the drawings to be ok to change them. Now it's done. We can discuss to correct and enhance the picture to fit the article --Tsaitgaist (talk) 10:21, 13 February 2010 (UTC)

I don't think there is anything wrong with File:Gps error diagram.jpg. I think it shluld bel left as it is. Can you find anything wrong with it and if so what? RHB100 (talk) 19:49, 15 February 2010 (UTC)

The current picture is just of poor quality (the picture as file, not the information it contains). The size of the canvas it too big for the diagram (a lot of blank space), you can also see compression artifacts around all the diagram. A vector graphic file has also more benefits : it's scalable (if you zoom you don't loose quality, the diagram quality remains), it's easily modifiable by others (change the position/color/... of the objects), it's easy to translate (for wikis in other languages). There is no drawback. SVG is a lot more adapted for diagrams. I'm just vectorizing the pictures I see in the articles I read to enhance the quality of them. --Tsaitgaist (talk) 21:58, 16 February 2010 (UTC)

I created the diagram to make the effect of errors more understandable. I can increase from the thumbnail to full size with no noticeable loss of quality. I am not sure you fully understand the reason why it is drawn as it is. Therefore if you want to draw it as a .svg file, it is necessary that you draw just as it is. RHB100 (talk) 22:52, 16 February 2010 (UTC)

Now it's like the original. I do not understand everything in the diagram. Here the aspects it don't understand :

1. why do the bubbles go up
2. is it important that it goes from left to right (the arrow already indicates the direction)
3. why is the arrow between the two first bubbles longer then the two last. Is this error more important
4. why does "numerical error" have an arrow and not "effect of incorrect pseudo range ..."

Because I thought these were not important details, I erased them in the first pictures, so to not have the other readers ask the same questions and get confused. --Tsaitgaist (talk) 10:21, 17 February 2010 (UTC)

I think you have made valuable contributions to the GPS article with the figures you have already added. However, I ask that you leave the error diagram. File:Gps error diagram.jpg, as it is. I think the time spent for you to continue asking questions and for me to try to answer then is not worth it. I am sure that you have the talents to make improvements in many other areas. In summary you have made valuable contributions with improvement of GPS figures. But let's just leave the error diagram as it is. RHB100 (talk) 20:45, 17 February 2010 (UTC)

## Gravitational potential expanded in series of Legendre polynomials

For the continuous case the potential, V at point b is computed as an integral over the distributed mass,

${\displaystyle V=-\int {\frac {G}{r}}\ dm=-\int {\frac {G}{|\mathbf {R} -\mathbf {x} |}}\ dm,}$

where dm equals density times differential volume, where x is a vector from the center of mass to the differential element of mass, where r is a vector from the differential element of mass to point b i.e. ${\displaystyle \mathbf {r} =\mathbf {R} -\mathbf {x} \,}$, , and where R is a vector from the center of mass to point b as shown in figure.

Diagram showing vectors, R, x, and r

Expanding the denominator by taking the square root of the square, carrying out the dot product, and factoring R out of the denominator we get the more useful expression,

${\displaystyle V=-\int {\frac {G}{\sqrt {R^{2}-2\mathbf {R} \cdot \mathbf {x} +x^{2}}}}\,dm\ =}$
${\displaystyle -{\frac {1}{R}}\int G\,/\,{\sqrt {1-2cos(\beta ){\frac {x}{R}}+\left({\frac {x}{R}}\right)^{2}}}\,dm.\ }$ V can be expanded in a series of Legendre Polynomials

Consider the function,

${\displaystyle f({\frac {x}{R}})=\left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {1}{2}}}\ .}$
Expanding ${\displaystyle f({\frac {x}{R}})}$ in a Taylor series about ${\displaystyle \left({\frac {x}{R}}\right)=0}$, we get the approximation,
${\displaystyle f({\frac {x}{R}})\approx 1+\left({\frac {x}{R}}\right)\ cos(\beta )+\left({\frac {x}{R}}\right)^{2}{\frac {3cos(\beta )^{2}-1}{2}}\ .\ }$

The coefficients are Legendre polynomials with argument ${\displaystyle cos(\beta )}$:

${\displaystyle f({\frac {x}{R}})\approx P_{0}(cos(\beta ))+\left({\frac {x}{R}}\right)\ P_{1}(cos(\beta ))+\left({\frac {x}{R}}\right)^{2}\ P_{2}(cos(\beta ))\ }$.

The appearance of the first three terms of the Taylor series expansion for ${\displaystyle f({\frac {x}{R}})}$ suggests that the infinite series might follow the same pattern. On page 454, section 10.8 of [1] Theorem 2 which states that ${\displaystyle \left(1-2XZ+Z^{2}\right)^{-{\frac {1}{2}}}\ =\sum _{n=0}^{\infty }Z^{n}P_{n}(X)\ }$, is proven. Also see Applications of Legendre polynomials in physics. Applied to the notation used here Theorem 2 states that

${\displaystyle \left(1-2cos(\beta )\left({\frac {x}{R}}\right)+\left({\frac {x}{R}}\right)^{2}\right)^{-{\frac {1}{2}}}\ =\sum _{n=0}^{\infty }\left({\frac {x}{R}}\right)^{n}P_{n}(cos(\beta ))\ .}$

We thus get the result,

${\displaystyle V=-{\frac {G}{R}}\int \sum _{n=0}^{\infty }\left({\frac {x}{R}}\right)^{n}P_{n}(cos(\beta ))\,dm\ {\rm {,}}}$

where the ${\displaystyle P_{n}(cos(\beta ))}$ are Legendre Polynomials of degree n and argument ${\displaystyle cos(\beta )\,.}$ This is a form which is quite useful for applications as shown in section 11.7 Gravitational Potential ... of [2] . Taking integrals of terms in the series we get:

${\displaystyle V=-{\frac {G}{R}}\int (1+\left({\frac {x}{R}}\right)\ cos(\beta )+\left({\frac {x}{R}}\right)^{2}{\frac {3cos(\beta )^{2}-1}{2}}\ dm+...}$

Note that ${\displaystyle \int x\ cos(\beta )dm=0.}$ This is true since ${\displaystyle x\ cos(\beta )}$ is the component of the location with respect to the point cm of the differential element of mass along the direction of R as can be seen from the figure. But the point cm is the location of the center of mass. Therefore we get the desired result that ${\displaystyle \int x\ cos(\beta )dm=0\ }$ and that ${\displaystyle \int \left({\frac {x}{R}}\right)\ cos(\beta )dm=0\ }$ since R is invariant over the integral. We therefore get

${\displaystyle V=-{\frac {GM}{R}}-{\frac {G}{R}}\int \left({\frac {x}{R}}\right)^{2}{\frac {3cos(\beta )^{2}-1}{2}}dm+...}$

RHB100 (talk) 20:14, 7 April 2010 (UTC)

References

1. ^ C. R. Wylie, Jr. 1960,Advanced Engineering Mathematics (McGraw-Hill Book Company)
2. ^ Leonard Meirovitch 1970, Methods of Analytical Dynamics (McGraw-Hill Book Company)

## Spherical, earth fixed coordinates (in progress)

In order to put in a form useful for determining the gravitational potential of bodies which are nearly spherical such as the earth and the moon, x and R are expressed in spherical coordinates. Consider the spherical coordinates angles of x and R with respect to the u, v, w coordinate system.

The Cartesian coordinate system, u, v, w is earth centered, earth fixed. The u and v axes are in the equatorial plane with w along the polar axis positive from the earth center toward the North pole so as to form a right handed orthogonal system.

Let u', v', w' denote a second coordinate system which is displaced from u, v, w by a rotation through the angle, λ (longitude) about the w axis with the positive direction defined by the right hand rule as shown in the figure.

Now consider a third coordinate system u", v", w" which is displaced from u', v', w' by a rotation through the angle Φ (latitude) about the negative v' axis. When the angles λ and Φ have been chosen so that the final direction of u" correspond with the direction of R then R, λ, and Φ are the spherical coordinates of R. Similarly we define x, λ', and Φ' as the spherical coordinates of x. Expressing x and R in terms of the unit vectors, ${\displaystyle {\hat {\mathbf {w} }},{\hat {\mathbf {v} }},and\ {\hat {\mathbf {u} }}}$ we get

${\displaystyle \mathbf {R} =R\left[{\hat {\mathbf {w} }}\ sin(\phi )+{\hat {\mathbf {v} }}\ cos(\phi )sin(\lambda )+{\hat {\mathbf {u} }}\ cos(\phi )cos(\lambda )\right]}$ and

${\displaystyle \mathbf {x} =x\left[{\hat {\mathbf {w} }}\ sin(\phi ')+{\hat {\mathbf {v} }}\ cos(\phi ')sin(\lambda ')+{\hat {\mathbf {u} }}\ cos(\phi ')cos(\lambda ')\right]}$ .

Taking the dot product and normalizing there results,

${\displaystyle {\frac {\mathbf {x} \cdot \mathbf {R} }{xR}}=cos(\theta )=sin(\phi )sin(\phi ')\ +\ cos(\phi )cos(\phi ')\left[sin(\lambda )sin(\lambda ')\ +\ cos(\lambda )cos(\lambda ')\right]}$ .

Thus using the cosine angle sum identity,

${\displaystyle cos(\theta )=sin(\phi )sin(\phi ')\ +\ cos(\phi )cos(\phi ')cos(\lambda -\lambda ')}$ .

RHB100 (talk) 02:13, 27 March 2010 (UTC)

Hello, RHB100. This message is being sent to inform you that there currently is a discussion at Wikipedia:Wikiquette alerts regarding an issue with which you may have been involved. Thank you. Ozob (talk) 05:54, 13 March 2010 (UTC)

## A vector does not designate a point

There are many things wrong with the writing of Gravitational potential. One place we can start is by pointing out that a vector does not designate a point. A mistake of referring to x as a point, even though x is a vector.

A vector has only a direction and a magnitude. It does not have a location. A vector beginning at the point (x,y,z) = (0,0,0) and ending at the point, (x,y,z) = (2,0,0), is exactly the same as a vector beginning at the point (x,y,z) = (5,0,0) and ending at the point, (x,y,z) = (7,0,0). This is true since they both have the same magnitude and direction.

Thus a vector certainly does not designate a point and the writing of Gravitational potentia shows a failure to understand very basic fundamentals on the nature of vectors and points. RHB100 (talk) 20:41, 7 April 2010 (UTC)

## ANI notice

Hello. This message is being sent to inform you that there currently is a discussion at Wikipedia:Administrators' noticeboard/Incidents regarding an issue with which you may have been involved. Thank you. Ozob (talk) 23:31, 13 March 2010 (UTC)

I'd appreciate if you would cease with putting entire posts in bold text. Thanks. Gerardw (talk) 23:19, 15 March 2010 (UTC)

OK, I have removed most of bold from a paragraph on the administration page at [http://en.wikipedia.org/wiki/Wikipedia:Administrators%27_noticeboard/Incidents#User:RHB100 . RHB100 (talk) 23:54, 15 March 2010 (UTC)

## Edit warring at Gravitational potential

It looks to me that you broke the WP:Three revert rule by making four reverts in 24 hours starting with your 01:09 edit of 18 March on this article. You are advised to stop warring on this article or you may be blocked from editing Wikipedia. If you cannot reach agreement on a contested issue, follow the steps of Dispute resolution, which might include a Request for comment. EdJohnston (talk) 01:06, 19 March 2010 (UTC)

## Preliminary computations

The section Derivation pointed out that the coordinate system in which the sphere centers are designated must be such that (1) all three centers are in the plane, Z = 0, (2) the sphere center, P1, is at the origin, and (3) the sphere center, P2, is on the X axis. In general the problem will not be given in a form such that these requirement are met. This section discusses the coordinate system translation and rotations necessary to meet these requirements.

Assume the original coordinate system in which the centers are expressed is X0, Y0, Z0. X0(I), Y0(I), Z0(I) denote the coordinates for circle centers P1, P2, and P3 for I =1,2,3 respectively and similarly for X1(I), Y1(I), Z1(I) and the other coordinate systems discussed below. We assume at the outset that X0(1), Y0(1), and Z0(1) are not colinear.

Form a new coordinate system X1, Y1, Z1 which is translated from X0, Y0, Z0 by -X0(1), -Y0(1), -Z0(1). After this translation the requirement that P1 be at the origin will have been met since

X1(1) = 0,
Y1(1) = 0, and
Z1(1) = 0.

Next form a new coordinate system, X2, Y2, Z2 which is rotated through λ about the positive Z1 axis where

λ = atan2(Y1(2),X1(2)) if either Y1(2) or X1(2) is not zero. Otherwise set λ equal to zero.
For I=1,2,3 set
X2(I) = X1(I)*COS(λ) + Y1(I)*SIN(λ),
Y2(I) = -X1(I)*SIN(λ) + Y1(I)*COS(λ), and
Z2(I) = Z1(I). As a result of this coordinate system rotation through angle λ, the Y component of P2 is now zero, that is Y2(2) = 0.

Now form an X3, Y3, Z3 coordinate system which is rotated through φ about the -Y2 axis where

${\displaystyle \phi =arcsin\left({\frac {Z2(2)}{\sqrt {X2(2)^{2}+Z2(2)^{2}}}}\right)}$ if Z2(2) is non-zero. Otherwise set φ equal to zero.
For I=1,2,3 set
X3(I) = X2(I)*COS(φ) + Z2(I)*SIN(φ),
Y3(I) = Y2(I), and
Z3(I) = -X2(I)*SIN(φ) + Z2(I)*COS(φ). As a result of this coordinate system rotation through angle φ, the Z component of P2 is now zero, that is Z3(2) = 0. Thus P2 is on the X3 axis.

Finally form a new X4, Y4, Z4 coordinate system which is rotated through the angle θ about the negative X3 axis where

θ = atan2(Z3(2), Y3(2) provided either Z3(2) or Y3(2) is non-zero. Otherwise set θ equal to zero.
For I=1,2,3 set
X4(I) = X3(I)
Y4(I) = Y3(I)*COS(THETA) + Z3(I)*SIN(THETA)
Z4(I) = -Y3(I)*SIN(THETA) + Z3(I)*COS(THETA)

Now compute the intersections of the three sphere surfaces as described in the section, Derivation. Then express the solutions in the X0, Y0, Z0 coordinate system using the inverse transformations.

RHB100 (talk) 02:48, 17 April 2010 (UTC)

## Show Hide

See [Usage of Show Hide]

Title text here

Body text line 1
Body text line 2
Body text line 3

        FUNCTION DACOMP ( BIAS )
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
C        1         2         3         4         5         6         7
C 3456789012345678901234567890123456789012345678901234567890123456789012
C
DOUBLE PRECISION PRANGE(3), ERRMAT(3), PRTEMP(3)
DOUBLE PRECISION ERRTOT(2), SATTIM(3), RSAT(5,3), R4(2), DAA(2)
COMMON /COMMDA/ RSAT, PRANGE, SATTIM, PRANTI

DO 10 I=1,3
PRTEMP(I) = PRANGE(I) + BIAS
10      CONTINUE
PTMTMP = PRANTI + BIAS

CALL TRILAT( RSAT(1,1), RSAT(1,2), RSAT(1,3), PRTEMP, NSOLNS)

DO 40 I=4,5 ! Compute errore for each of 2 solutions
DO 35 J=1,3   ! Compute distances to J=1,2,3 satellites
SQUARE = 0.0
DO 33 K=1,3 ! X, Y, Z components of error
SQUARE = SQUARE + (  RSAT(I,K) - RSAT(J,K)  )**2
33              CONTINUE
ERRMAT(J) = PRTEMP(J) - SQRT( SQUARE )
35          CONTINUE
ERRTOT(I-3) = SQRT( ERRMAT(1)**2 +  ERRMAT(2)**2 +
2       ERRMAT(3)**2 )
40      CONTINUE

IF( (ERRTOT(1) .GE. 1.E-6) .OR. (ERRTOT(2) .GE. 1.E-6)  ) THEN
WRITE(6,*) ' Excessive trilateration error '
WRITE(6,*) ' ERRTOT = ', (ERRTOT(I), I=1,2)
IPRINT = 1
END IF

DO 50 I=4,5 ! Distance to 4th sphere for solutions in 4 & 5
SQUARE = 0.0
DO 47 J=1,3 ! X, Y, Z components
SQUARE = SQUARE + (  RSAT(I,J) - SATTIM(J)  )**2
47          CONTINUE
R4(I-3)  = SQRT( SQUARE )
DAA(I-3) = R4(I-3) - PTMTMP
50      CONTINUE

IF( ABS(DAA(1)) .LE. ABS(DAA(2)) ) THEN
DA = DAA(1)
ISOLN = 1
ELSE
DA = DAA(2)
ISOLN = 2
END IF

C       WRITE(6,*) ' R4 = ', (R4(I), I=1,2)
C       WRITE(6,*) ' DAA = ', (DAA(I), I=1,2)
C       WRITE(6,*) ' Solution ', ISOLN, ' is closest.', ' DA = ', DA
DACOMP = DA
RETURN
END


RHB100 (talk) 19:42, 29 April 2010 (UTC)

### More Show Hide

RHB100 (talk) 21:38, 29 April 2010 (UTC)

All== GPS ==

Please let me know when you take a pause in editing GPS. I'll finish my changes after that. (I don't have many more at the moment.) It appears we're stepping on each other which is a waste of time. Cheers! Lfstevens (talk) 22:19, 12 July 2010 (UTC)

All yours. Thanks for your patience. Lfstevens (talk) 02:16, 13 July 2010 (UTC)

## Article Feedback Tool, Version 5

Hey; I've moved your question to the talkpage and replied to it there. Thanks! Okeyes (WMF) (talk) 20:08, 27 February 2012 (UTC)

## Beam Theory

In the figure on Bending of an Euler-Bernoulli beam, the x and z direction of a right handed coordinate system are shown. Since ${\displaystyle \mathbf {e_{z}} \times \mathbf {e_{x}} =\mathbf {e_{y}} }$ where ${\displaystyle \mathbf {e_{x}} }$ , ${\displaystyle \mathbf {e_{y}} }$ , and ${\displaystyle \mathbf {e_{z}} }$ are unit vectors in the direction of the x, y, and z axes respectively, the y axis direction is into the figure.

Because of the fundamental importance of this equation in engineering, we will provide a short derivation. The length of the neutral axis in the figure is ${\displaystyle \rho \ d\theta }$. The length of a fiber with a radial distance, e, below the neutral axis is ${\displaystyle \left(\rho +e\right)\ d\theta }$. Therefore the strain of this fiber is ${\displaystyle {\frac {\left(\rho +e-\rho \right)\ d\theta }{\rho \ d\theta }}={\frac {e}{\rho }}\ .}$ The stress of this fiber is ${\displaystyle E\ {\frac {e}{\rho }}}$ where E is the modulus of elasticity in accordance with Hooke's Law. The differential force vector, ${\displaystyle d\mathbf {F} \ ,}$ resulting from this stress is given by, ${\displaystyle d\mathbf {F} =E\ {\frac {e}{\rho }}\ dA\ \mathbf {e_{x}} \ .}$ This is the differential force vector exerted on the right hand side of the section shown in the figure. We know that it is in the ${\displaystyle \mathbf {e_{x}} }$ direction since the figure clearly shows that the fibers in the lower half are in tension. ${\displaystyle dA}$ is the differential element of area at the location of the fiber. The differential bending moment vector, ${\displaystyle d\mathbf {M} \ ,}$ associated with ${\displaystyle d\mathbf {F} \ ,}$ is given by

${\displaystyle d\mathbf {M} =-e\mathbf {e_{z}} \times d\mathbf {F} \ =-\mathbf {e_{y}} E\ {\frac {e^{2}}{\rho }}\ dA\ .}$ This expression is valid for the fibers in the lower half of the beam.

The expression for the fibers in the upper half of the beam will be similar except that the moment arm vector will be in the positive z direction and the force vector will be in the -x direction since the upper fibers are in compression. But the resulting bending moment vector will still be in the -y direction since ${\displaystyle \mathbf {e_{z}} \times -\mathbf {e_{x}} =-\mathbf {e_{y}} \ .}$ Therefore we integrate over the entire cross section of the beam and get for ${\displaystyle \mathbf {M} }$ the bending moment vector exerted on the right cross section of the beam the expression

${\displaystyle \mathbf {M} =\int d\mathbf {M} =-\mathbf {e_{y}} {\frac {E}{\rho }}\int {e^{2}}\ dA=\ -\mathbf {e_{y}} {\frac {EI}{\rho }}}$ where ${\displaystyle I}$ is the area moment of inertia . From calculus, we know that when ${\displaystyle {dz \over dx}}$ is small as it is for an Euler-Bernoulli beam. ${\displaystyle {\frac {1}{\rho }}={d^{2}z \over dx^{2}}}$ . Therefore
${\displaystyle \mathbf {M} =\ -\mathbf {e_{y}} EI{d^{2}z \over dx^{2}}}$ .

A number of different sign conventions can be found in the literature on the bending of beams and care should be taken to maintain consistency.[1] In this article, the sign convention has been chosen so the coordinate system is right handed. Forces acting in the positive ${\displaystyle x}$ and ${\displaystyle z}$ directions are assumed positive. The sign of the bending moment is positive when the torque vector associated with the bending moment on the right hand side of the section is in the positive y direction (i.e. so that a positive value of M leads to a compressive stress at the bottom cords). The sign of the shear force has been chosen such that it matches the sign of the bending moment.

## Binomial coefficient

The article at present is either deliberately designed to confuse readers or it is so poorly written that it confuses the reader as much as if it had been deliberate. The statement above implying that the formula, ${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\mbox{for }}\ 0\leq k\leq n}$ defining the value of the binomial coefficient where the notation, ${\displaystyle n!}$, called n factorial is defined by ${\displaystyle n!=n(n-1)(n-2)...1}$ with ${\displaystyle 0!=1}$, is of less importance than Pascal's triangle and such is a very ignorant and stupid statement.

## 3RR

Your recent editing history at Binomial coefficient shows that you are currently engaged in an edit war. Being involved in an edit war can result in you being blocked from editing—especially if you violate the three-revert rule, which states that an editor must not perform more than three reverts on a single page within a 24-hour period. Undoing another editor's work—whether in whole or in part, whether involving the same or different material each time—counts as a revert. Also keep in mind that while violating the three-revert rule often leads to a block, you can still be blocked for edit warring—even if you don't violate the three-revert rule—should your behavior indicate that you intend to continue reverting repeatedly.

To avoid being blocked, instead of reverting please consider using the article's talk page to work toward making a version that represents consensus among editors. See BRD for how this is done. You can post a request for help at a relevant noticeboard or seek dispute resolution. In some cases, you may wish to request temporary page protection. - DVdm (talk) 21:12, 12 August 2012 (UTC)

## File:Sphere clock cor.jpg listed for deletion

A file that you uploaded or altered, File:Sphere clock cor.jpg, has been listed at Wikipedia:Files for deletion. Please see the discussion to see why it has been listed (you may have to search for the title of the image to find its entry). Feel free to add your opinion on the matter below the nomination. Thank you. Sfan00 IMG (talk) 20:40, 2 August 2013 (UTC)

## File:Gps error diagram.jpg listed for deletion

A file that you uploaded or altered, File:Gps error diagram.jpg, has been listed at Wikipedia:Files for deletion. Please see the discussion to see why it has been listed (you may have to search for the title of the image to find its entry). Feel free to add your opinion on the matter below the nomination. Thank you. Sfan00 IMG (talk) 09:19, 3 August 2013 (UTC)

## Basic concept of GPS 3 October 2013

A GPS receiver calculates its position by precisely timing the signals sent by GPS satellites high above the Earth. Each satellite continually transmits messages that include

• the time the message was transmitted
• satellite position at time of message transmission

The receiver uses the messages it receives to determine the transit time of each message and computes the distance to each satellite using the speed of light. Each of these distances and satellites' locations define a sphere. The receiver is on the surface of each of these spheres when the distances and the satellites' locations are correct. These distances and satellites' locations are used to compute the location of the receiver using the navigation equations. This location is then displayed, perhaps with a moving map display or latitude and longitude; elevation or altitude information may be included, based on height above the geoid (e.g. EGM96). Many GPS units show derived information such as direction and speed, calculated from position changes.

In typical GPS operation, four or more satellites must be visible to obtain an accurate result. Four sphere surfaces typically do not intersect. [a] Because of this, it can be said with confidence that when the navigation equations are solved to find an intersection, this solution gives the position of the receiver along with the difference between the time kept by the receiver's on-board clock and the true time-of-day, thereby eliminating the need for a very large, expensive, and power hungry clock. The very accurately computed time is used only for display or not at all in many GPS applications, which use only the location. A number of applications for GPS do make use of this cheap and highly accurate timing. These include time transfer, traffic signal timing, and synchronization of cell phone base stations.

Although four satellites are required for normal operation, fewer apply in special cases. If one variable is already known, a receiver can determine its position using only three satellites. For example, a ship or aircraft may have known elevation. Some GPS receivers may use additional clues or assumptions such as reusing the last known altitude, dead reckoning, inertial navigation, or including information from the vehicle computer, to give a (possibly degraded) position when fewer than four satellites are visible.[2][3][4]

References

1. ^ Cite error: The named reference Gere was invoked but never defined (see the help page).
2. ^ Georg zur Bonsen, Daniel Ammann, Michael Ammann, Etienne Favey, Pascal Flammant (April 1, 2005). "Continuous Navigation Combining GPS with Sensor-Based Dead Reckoning". GPS World. Archived from the original on November 11, 2006.
3. ^ "NAVSTAR GPS User Equipment Introduction" (PDF). United States Government. Chapter 7
4. ^ "GPS Support Notes" (PDF). January 19, 2007. Archived from the original (PDF) on March 27, 2009. Retrieved November 10, 2008.

## Accusations of blatant editorializing made by DVdm

I, RHB100, have been accused of blatant editorializing by DVdm. This accusation was made twice at [3] and also at the talk page, [4] subsection, Solution based on intersection of at least four spheres, not TDOA and not Multilateration , where I quote

Wikipedia is not in the business of alerting readers. This is a blatant example of wp:EDITORIALIZING. I have removed it. - DVdm (talk) 18:47, 26 January 2015 (UTC)

About all I said was "The reader should note the equations to be solved are not the equations for a hyperboloid discussed above" as can be seen on the diff page, [5] .

I think to accuse me of blatant editorializing for the simple remark I made is absolutely ridiculous and highly irresponsible on the part of DVdm. I have asked DVdm what specifically I wrote that he thinks is blatant editorializing. He has not been able to come up with a single remark that I made which is blatant editorializing. I claim that I have made no statement that is blatant editorializing. DVdm has been unable to refute this. I am completely innocent of the accusations made by DVdm .

Elsewhere on the GPS talk page section, DVdm indicates he thinks wp:EDITORIALIZING applies to remarks made on the talk page. DVdm is doing a lot of damage and his failure to understand wp:EDITORIALIZING shows that he is unqualified to edit. He should be blocked from editing until he can demonstrate that he is competent. RHB100 (talk) 21:49, 27 January 2015 (UTC)

## Navigation equations as of 3 October 2013

The receiver uses messages received from satellites to determine the satellite positions and time sent. The x, y, and z components of satellite position and the time sent are designated as [xi, yi, zi, ti] where the subscript i denotes the satellite and has the value 1, 2, ..., n, where ${\displaystyle n\geq 4.}$ When the time of message reception indicated by the on-board clock is ${\displaystyle \,{\tilde {t}}_{\text{r}}}$, the true reception time is ${\displaystyle \,t_{\text{r}}={\tilde {t}}_{\text{r}}+b}$ where ${\displaystyle \,b}$ is receiver's clock bias (i.e., clock delay). The message's transit time is ${\displaystyle \,{\tilde {t}}_{\text{r}}+b-t_{i}}$. Assuming the message traveled at the speed of light, ${\displaystyle \,c}$, the distance traveled is ${\displaystyle \,\left({\tilde {t}}_{\text{r}}+b-t_{i}\right)c}$. Knowing the distance from receiver to satellite and the satellite's position implies that the receiver is on the surface of a sphere centered at the satellite's position with radius equal to this distance. Thus the receiver is at or near the intersection of the surfaces of the four or more spheres. In the ideal case of no errors, the receiver is at the intersection of the surfaces of the spheres.

The clock error or bias, b, is the amount that the receiver's clock is off. The receiver has four unknowns, the three components of GPS receiver position and the clock bias [x, y, z, b]. The equations of the sphere surfaces are given by:

${\displaystyle (x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}={\bigl (}[{\tilde {t}}_{\text{r}}+b-t_{i}]c{\bigr )}^{2},\;i=1,2,\dots ,n}$

or in terms of pseudoranges, ${\displaystyle p_{i}=\left({\tilde {t}}_{\text{r}}-t_{i}\right)c}$, as

${\displaystyle p_{i}={\sqrt {(x-x_{i})^{2}+(y-y_{i})^{2}+(z-z_{i})^{2}}}-bc,\;i=1,2,...,n}$ .

These equations can be solved by algebraic or numerical methods.

## Notice of Edit warring noticeboard discussion

Hello. This message is being sent to inform you that there is currently a discussion involving you at Wikipedia:Administrators' noticeboard/Edit warring regarding a possible violation of Wikipedia's policy on edit warring. Thank you. siafu (talk) 02:31, 23 June 2015 (UTC)

See Wikipedia:Administrators' noticeboard/Edit warring#User:RHB100 reported by User:Siafu (Result: ). People say you are conducting a long-term war that resumes every 6-12 months. It will be to your advantage to reply there and promise to stop warring at Global Positioning System. Otherwise a long block from Wikipedia is possible. The best way for us to deal with this permanently is either (a) you promise to wait for consensus before reverting again, or (b) an admin blocks your account indefinitely. Thank you, EdJohnston (talk) 02:57, 23 June 2015 (UTC)
If your response at WP:AN3 is truly the best you can do, I'm planning to go ahead with an indefinite block, if I am the closing admin. Thank you, EdJohnston (talk) 03:52, 23 June 2015 (UTC)

I have told you that I have concluded that it appears useless to try to educate these people on how GPS works. What more do you want me to do. I did nothing but try to make the GPS article correct. I didn't know that normal editing was counted as reverts. Other people were far more guilty of edit warring than me. Is there anyway to change an incorrectly written article. I will keep in mind that what I thought were normal edits is sometimes considered a revert. I will refrain from making changes until I understand the difference between normal editing and reverts better. RHB100 (talk) 04:26, 23 June 2015 (UTC)

## You've accepted a voluntary restriction at Global Positioning System

Per this AN3 closure (permalink) you've agreed to make no further edits to Global Positioning System unless you first make a proposal on the talk page and it gets consensus there. Let me know if you have any questions. For whatever reason, nobody else is agreeing with your changes, so the consensus is against you. On Wikipedia the correct behavior is to back away from the situation and stop insisting that your version is the one that belongs. Thanks, EdJohnston (talk) 16:42, 23 June 2015 (UTC)

## Re:GPS page protection

I've left a notice on the talk page but I feel that the best course of action at the moment would be to let it expire, my reasons being that in the spirit of AGF we need to try and let the public edit again, and more covertly those who move first to tweak the article's information are usually the people who cause the editing disruption, which may be valuable information to have if this goes to mediation or arbitration. If it does turn out that the page is not ready for general editing (IE if the edit warring flares up again) I'll re-protect the page, otherwise a wait-and-see approach ehre would probably be best for all. TomStar81 (Talk) 18:01, 6 July 2015 (UTC)

## AN-notice

This message is being sent to inform you that there is currently a discussion at Wikipedia:Administrators' noticeboard regarding an issue with which you may have been involved. Thank you.

## August 2015

Please do not insert text that I wrote, with my signature on it, in to a different page on Wikipedia, as you did at WP:AN, without putting it in quotes or in a quote block. The way you did this makes it look like I put that text there. Please go read Wikipedia:Copying within Wikipedia. Kendall-K1 (talk) 02:28, 9 August 2015 (UTC)

## ArbCom elections are now open!

Hi,
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## Quaternions

Quaternions have advantages over Euler angles and direction cosines. In formulating the nonlinear differential equations of motion of a vehicle, the dynamics engineer is faced with the choice of what to use for vehicle attitude description. Euler angles have been widely used in the past. However there is a singularity in the Euler angle derivatives when the pitch angle is 90 degree. Also trigonometric operations are required to compute the coordinate transformation matrix as a function of the Euler angles. Direction cosines could also be used but there are 9 direction cosines compared to only 4 quaternions. Quaternions have the advantage that there are no singularities in the quaternion derivatives with respect to time and that the coordinate transformation matrix can be computed from the quaternions with only multiplications, additions, and subtractions (no trigonometric operations).

Before getting into a discussion of quaternions for performing rotations some difinitions are necessary. Historically quaternions, denoted here by q, were conceived by Hamilton as like extended complex numbers.

${\displaystyle \mathbf {q} =s+a\mathbf {i} +b\mathbf {j} +c\mathbf {k} =\lbrack s,\mathbf {v} \rbrack }$

s id the scalar part and ${\displaystyle a\mathbf {i} +b\mathbf {j} +c\mathbf {k} =\mathbf {v} }$ is the vector part of the quaternion. A 3 dimensional vector is represented in the 4 dimensional space of quaternions as a quaternion with zero real part. The ${\displaystyle \mathbf {i} }$ , ${\displaystyle \mathbf {j} }$ , and ${\displaystyle \mathbf {k} }$ obey the following multiplication laws:

{\displaystyle {\begin{alignedat}{2}\mathbf {ij} &=\mathbf {k} ,&\mathbf {ji} &=\mathbf {-k} ,\\\mathbf {jk} &=\mathbf {i} ,&\mathbf {kj} &=\mathbf {-i} ,\\\mathbf {ki} &=\mathbf {j} ,&\mathbf {ik} &=\mathbf {-j} ,\\\mathbf {i} ^{2}&=\mathbf {j} ^{2}&=\mathbf {k} ^{2}&=-1\end{alignedat}}}

Note that multiplication is not commutative for ${\displaystyle \mathbf {i} }$ , ${\displaystyle \mathbf {j} }$ , and ${\displaystyle \mathbf {k} }$. Multiplication of quaternions is defined as follows:

${\displaystyle \mathbf {q} \mathbf {q'} =\lbrack s,\mathbf {v} \rbrack \lbrack s',\mathbf {v'} \rbrack =}$ ${\displaystyle ss'+s\mathbf {v'} +s'\mathbf {v} +\mathbf {v} \times \mathbf {v'} -\mathbf {v} \cdot \mathbf {v'} }$ .[1][2]

A rotation of vector ${\displaystyle \mathbf {p} }$ through an angle of θ around a unit vector, ${\displaystyle \mathbf {u} }$ to obtain vector ${\displaystyle \mathbf {p'} }$ can be achieved by performing the following multiplication

${\displaystyle \mathbf {p'} =\mathbf {q} \mathbf {p} \mathbf {q} ^{-1}}$ , where
${\displaystyle \mathbf {u} =(u_{x},u_{y},u_{z})=u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} }$ where ${\displaystyle {\hat {\mathbf {x} }}}$ , ${\displaystyle {\hat {\mathbf {y} }}}$ , and ${\displaystyle {\hat {\mathbf {z} }}}$ are unit vectors in the directions of the x, y, and z axes of a three dimensional Cartesian coordinate system ,
where ${\displaystyle \mathbf {p} =(p_{x},p_{y},p_{z})=p_{x}\mathbf {i} +p_{y}\mathbf {j} +p_{z}\mathbf {k} }$ and
where ${\displaystyle \mathbf {q} }$ is a quaternion defined by an extension of Euler's formula through the equation,
${\displaystyle \mathbf {q} =e^{{\frac {\theta }{2}}{(u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )}}=\cos {\frac {\theta }{2}}+(u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\sin {\frac {\theta }{2}}}$

The inverse of ${\displaystyle \mathbf {q} }$, ${\displaystyle \mathbf {q} ^{-1}}$ is given by

${\displaystyle \mathbf {q} ^{-1}=e^{-{\frac {\theta }{2}}{(u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )}}=\cos {\frac {\theta }{2}}-(u_{x}\mathbf {i} +u_{y}\mathbf {j} +u_{z}\mathbf {k} )\sin {\frac {\theta }{2}}}$.

${\displaystyle \mathbf {q} }$ is a unit quaternion as is ${\displaystyle \mathbf {q} ^{-1}}$ . The rotation is clockwise if our line of sight points in the same direction as u.

Verifying that the rotation of ${\displaystyle \mathbf {p} }$ about ${\displaystyle \mathbf {u} }$ is through angle ${\displaystyle \theta }$,

${\displaystyle \mathbf {p'} =q\mathbf {p} q^{-1}=\left(\cos {\frac {\theta }{2}}+\mathbf {u} \sin {\frac {\theta }{2}}\right)\,\mathbf {p} \,\left(\cos {\frac {\theta }{2}}-\mathbf {u} \sin {\frac {\theta }{2}}\right)}$
{\displaystyle {\begin{aligned}\mathbf {p'} &=\mathbf {p} \cos ^{2}{\frac {\theta }{2}}+(\mathbf {u} \mathbf {p} -\mathbf {p} \mathbf {u} )\sin {\frac {\theta }{2}}\cos {\frac {\theta }{2}}-\mathbf {u} \mathbf {p} \mathbf {u} \sin ^{2}{\frac {\theta }{2}}\\[6pt]&=\mathbf {p} \cos ^{2}{\frac {\theta }{2}}+2(\mathbf {u} \times \mathbf {p} )\sin {\frac {\theta }{2}}\cos {\frac {\theta }{2}}-(\mathbf {p} (\mathbf {u} \cdot \mathbf {u} )-2\mathbf {u} (\mathbf {u} \cdot \mathbf {p} ))\sin ^{2}{\frac {\theta }{2}}\\[6pt]&=\mathbf {p} (\cos ^{2}{\frac {\theta }{2}}-\sin ^{2}{\frac {\theta }{2}})+(\mathbf {u} \times \mathbf {p} )(2\sin {\frac {\theta }{2}}\cos {\frac {\theta }{2}})+\mathbf {u} (\mathbf {u} \cdot \mathbf {p} )(2\sin ^{2}{\frac {\theta }{2}})\\[6pt]&=\mathbf {p} \cos \theta +(\mathbf {u} \times \mathbf {p} )\sin \theta +\mathbf {u} (\mathbf {u} \cdot \mathbf {p} )(1-\cos \theta )\\[6pt]&=(\mathbf {p} -\mathbf {u} (\mathbf {u} \cdot \mathbf {p} ))\cos \theta +(\mathbf {u} \times \mathbf {p} )\sin \theta +\mathbf {u} (\mathbf {u} \cdot \mathbf {p} )\\[6pt]&=\mathbf {p} _{\bot }\cos \theta +(\mathbf {u} \times \mathbf {p} )\sin \theta +\mathbf {p} _{\|}\end{aligned}}}

where ${\displaystyle \mathbf {p} _{\bot }}$ and ${\displaystyle \mathbf {p} _{\|}}$ are the components of v perpendicular and parallel to u respectively. This is the formula of a rotation by θ around the u axis.

Let ${\displaystyle \mathbf {q} =E_{0}+E_{1}\mathbf {i} +E_{2}\mathbf {j} +E_{3}\mathbf {k} }$ denote a quaternion and let ${\displaystyle \mathbf {\omega } }$ denote the array of components of vehicle angular velocity vector expressed in body fixed coordinates with P, Q, R along x, y, z. Then it follows that ${\displaystyle {\frac {d\mathbf {q} }{dt}}={\frac {1}{2}}\mathbf {q} \mathbf {\omega } }$. [3] Thus ${\displaystyle {\frac {d\mathbf {q} }{dt}}={\frac {1}{2}}\left(\lbrack E_{0},\mathbf {v} \rbrack \lbrack 0,\omega \rbrack \right)}$ = ${\displaystyle {\frac {1}{2}}\left(E_{0}\mathbf {\omega } +\mathbf {v} \times \mathbf {\omega } -\mathbf {v} \cdot \mathbf {\omega } \right)}$ where ${\displaystyle \mathbf {v} =E_{1}\mathbf {i} +E_{2}\mathbf {j} +E_{3}\mathbf {k} }$

${\displaystyle {\frac {d\mathbf {q} }{dt}}={\frac {1}{2}}\lbrace E_{0}\mathbf {\omega } +\left(RE_{2}-QE_{3}\right)\mathbf {i} -\left(RE_{1}-PE_{3}\right)\mathbf {j} +\left(QE_{1}-PE_{2}\right)\mathbf {k} -Pe1-QE_{2}-RE_{3}\rbrace }$
${\displaystyle {\frac {d\mathbf {q} }{dt}}={\frac {1}{2}}\lbrace \left(PE_{0}-QE_{3}+RE_{2}\right)\mathbf {i} +\left(+PE_{3}+QE_{0}-RE_{1}\right)\mathbf {j} +\left(-PE_{2}+QE_{1}+RE_{0}\right)\mathbf {k} -Pe1-QE_{2}-RE_{3}\rbrace }$, thus it follows that
${\displaystyle {\dot {E_{0}}}={\frac {1}{2}}\left(-PE_{1}-QE_{2}-RE_{3}\right)}$
${\displaystyle {\dot {E_{1}}}={\frac {1}{2}}\left(+PE_{0}-QE_{3}+RE_{2}\right)}$
${\displaystyle {\dot {E_{2}}}={\frac {1}{2}}\left(+PE_{3}+QE_{0}-RE_{1}\right)}$
${\displaystyle {\dot {E_{3}}}={\frac {1}{2}}\left(-PE_{2}+QE_{1}+RE_{0}\right)}$

The equations for the quaternion derivatives with respect to time are as shown below. [4] If we replace ${\displaystyle E_{0}}$ ,${\displaystyle E_{1}}$ ,${\displaystyle E_{2}}$ , and ${\displaystyle E_{3}}$ in the above 4 equations with ${\displaystyle e_{1}}$ ,${\displaystyle e_{4}}$ ,${\displaystyle e_{3}}$ , and ${\displaystyle e_{2}}$ respectively and then reorder we can see that the equations below are the same as those above with different names for the variables involved.

${\displaystyle {\dot {e_{1}}}={\frac {1}{2}}\left(-Pe_{4}-Qe_{3}-Re_{2}\right)}$
${\displaystyle {\dot {e_{2}}}={\frac {1}{2}}\left(-Pe_{3}+Qe_{4}+Re_{1}\right)}$
${\displaystyle {\dot {e_{3}}}={\frac {1}{2}}\left(+Pe_{2}+Qe_{1}-Re_{4}\right)}$
${\displaystyle {\dot {e_{4}}}={\frac {1}{2}}\left(+Pe_{1}-Qe_{2}+Re_{3}\right)}$

Comparing these equations with equations (77) in section V show that they agree. [5] This provides an important verification of the correctness of the above equations.

For any unit quaternion ,${\displaystyle \mathbf {q} ={\frac {\cos(\alpha )}{2}}+\mathbf {\hat {u}} {\frac {sin(\alpha )}{2}}}$ , and vector, ${\displaystyle \mathbf {v} }$, ${\displaystyle \mathbf {q^{*}} \mathbf {v} \mathbf {q} }$ can be interpreted as rotating the coordinate frame through the angle ${\displaystyle \alpha }$ about axis ${\displaystyle \mathbf {\hat {u}} }$ while v is not rotated at all.[3] When the coordinate frame is rotated ${\displaystyle \mathbf {i} }$, ${\displaystyle \mathbf {j} }$, and ${\displaystyle \mathbf {k} }$ are rotated so that they become unit vectors along the axes of the rotated reference frame. It thus follows that if ${\displaystyle \mathbf {v} =X\mathbf {i} +Y\mathbf {j} +Z\mathbf {k} }$ is a vector expressed in inertial or fixed coordinates, ${\displaystyle \mathbf {q} =s+a\mathbf {i} +b\mathbf {j} +c\mathbf {k} }$ then the vector, ${\displaystyle \mathbf {v'} }$ expressed in body fixed coordinates is given by ${\displaystyle \mathbf {v'} =(s-a\mathbf {i} -b\mathbf {j} -c\mathbf {k} )(X\mathbf {i} +Y\mathbf {j} +Z\mathbf {k} )(s+a\mathbf {i} +b\mathbf {j} +c\mathbf {k} )}$ or