GC3RVE3 Triangulation Calculation (Wherigo Cache) in Ohio, United States created by Landondena

GC3RVE3 ▼

A cache by Landondena Message this owner

Hidden : 8/3/2012

Difficulty:

Terrain:

Size: Size: micro (micro)

Join now to view geocache location details. It's free!

Watch

How Geocaching Works

Related Web Page

Please note Use of geocaching.com services is subject to the terms and conditions in our disclaimer.

Geocache Description:

There is only one container and that is the final.
This is rated a 5 terrain because you have a 10% chance of needing gear, one depends upon luck of draw. 4 difficulty due to the math/mapping
The basic concept of the wherigo is a Reverse Geocache
You get 10 attempts before you will get a new location.

Do not access the final from I75!!

Trilateration

In geometry, trilateration is the process of determinating absolute or relative locations of points by measurement of distances, using the geometry of circles, spheres or triangles. In addition to its interest as a geometric problem, trilateration does have practical applications in surveying and navigation, including global positioning systems (GPS). In contrast to triangulation it does not involve the measurement of angles.

In two-dimensional geometry, when it is known that a point lies on two curves such as the boundaries of two circles then the circle centers and the two radii provide sufficient information to narrow the possible locations down to two. Additional information may narrow the possibilities down to one unique location.

In three-dimensional geometry, when it is known that a point lies on three surfaces such as the surfaces of three spheres then the centers of the three spheres along with their radii provide sufficient information to narrow the possible locations down to no more than two. If it is known that the point lies on the surface of a fourth sphere then knowledge of this sphere's center along with its radius is sufficient to determine the one unique location.

This article describes a method for determining the intersections of three sphere surfaces given the centers and radii of the three spheres.

Derivation

The intersections of the surfaces of three spheres is found by formulating the equations for the three sphere surfaces and then solving the three equations for the three unknowns, x, y, and z. To simplify the calculations, the equations are formulated so that the centers of the spheres are on the z=0 plane. Also the formulation is such that one center is at the origin, and one other is on the x-axis. It is possible to formulate the equations in this manner since any three non-colinear points lie on a unique plane. After finding the solution it can be transformed back to the original three dimensional Cartesian coordinate system.

We start with the equations for the three spheres:

$r_1^2=x^2+y^2+z^2$ $r_2^2=(x-d)^2+y^2+z^2$ $r_3^2=(x-i)^2+(y-j)^2+z^2$

We need to find a point located at (x, y, z) that satisfies all three equations.

First we subtract the second equation from the first and solve for x:

$x=\frac{r_1^2-r_2^2+d^2}{2d}$ .

We assume that the first two spheres intersect in more than one point, that is that

$d-r_1 < r_2 < d+r_1$ .

In this case substituting the equation for x back into the equation for the first sphere produces the equation for a circle, the solution to the intersection of the first two spheres:

$y^2+z^2=r_1^2-\frac{(r_1^2-r_2^2+d^2)^2}{4d^2}$ .

Substituting : $y^2+z^2=r_1^2-x^2$ into the formula for the third sphere and solving for y there results:

$y=\frac{r_1^2-r_3^2-x^2+(x-i)^2+j^2}{2j}=\frac{r_1^2-r_3^2+i^2+j^2}{2j}-\frac{i}{j}x$ .

Now that we have the x- and y-coordinates of the solution point, we can simply rearrange the formula for the first sphere to find the z-coordinate:

$z=\pm \sqrt{r_1^2-x^2-y^2}.$

Now we have the solution to all three points x, y and z. Because z is expressed as the positive or negative square root, it is possible for there to be zero, one or two solutions to the problem.

This last part can be visualized as taking the circle found from intersecting the first and second sphere and intersecting that with the third sphere. If that circle falls entirely outside or inside of the sphere, z is equal to the square root of a negative number: no real solution exists. If that circle touches the sphere on exactly one point, z is equal to zero. If that circle touches the surface of the sphere at two points, then z is equal to plus or minus the square root of a positive number.

Preliminary and final computations

The Derivation section 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 requirements are met.

This problem can be overcome as described below where the points, P1, P2, and P3 are treated as vectors from the origin where indicated. P1, P2, and P3 are of course expressed in the original coordinate system.

$\hat e_x = \frac{ P2 - P1 }{ \| P2 - P1 \| }$ is the unit vector in the direction from P1 to P2. $i = \hat e_x \cdot ( P3 - P1 )$ is the signed magnitude of the x component, in the figure 1 coordinate system, of the vector from P1 to P3. $\hat e_y = \frac{ P3 - P1 - i \; \hat e_x}{ \| P3 - P1 - i \; \hat e_x \| }$ is the unit vector in the y direction. Note that the points P1, P2, and P3 are all in the z=0 plane of the figure 1 coordinate system.

The third basis unit vector is $\hat e_z = \hat e_x \times \hat e_y$ . Therefore,

$d = \| P2 - P1 \|$ the distance between the centers P1 and P2 and $j = \hat e_y \cdot ( P3 - P1 )$ is the signed magnitude of the y component, in the figure 1 coordinate system, of the vector from P1 to P3.

Using $i, \; d$ and $j$ as computed above, solve for x, y and z as described in the Derivation section. Then

$\vec p_{1,2} = P1 + x \ \hat e_x + y \ \hat e_y \ \pm \ z \ \hat e_z$

gives the points in the original coordinate system since $\hat e_x, \; \hat e_y$ and $\hat e_z$ , the basis unit vectors, are expressed in the original coordinate system.

In geometry, trilateration is the process of determinating absolute or relative locations of points by measurement of distances, using the geometry of circles, spheres or triangles. In addition to its interest as a geometric problem, trilateration does have practical applications in surveying and navigation, including global positioning systems (GPS). In contrast to triangulation it does not involve the measurement of angles.
In two-dimensional geometry, when it is known that a point lies on two curves such as the boundaries of two circles then the circle centers and the two radii provide sufficient information to narrow the possible locations down to two. Additional information may narrow the possibilities down to one unique location.
In three-dimensional geometry, when it is known that a point lies on three surfaces such as the surfaces of three spheres then the centers of the three spheres along with their radii provide sufficient information to narrow the possible locations down to no more than two. If it is known that the point lies on the surface of a fourth sphere then knowledge of this sphere's center along with its radius is sufficient to determine the one unique location.
This article describes a method for determining the intersections of three sphere surfaces given the centers and radii of the three spheres.

Derivation

We start with the equations for the three spheres:

$r_1^2=x^2+y^2+z^2$ $r_2^2=(x-d)^2+y^2+z^2$ $r_3^2=(x-i)^2+(y-j)^2+z^2$

We need to find a point located at (x, y, z) that satisfies all three equations.

First we subtract the second equation from the first and solve for x:

$x=\frac{r_1^2-r_2^2+d^2}{2d}$ .

We assume that the first two spheres intersect in more than one point, that is that

$d-r_1 < r_2 < d+r_1$ .

In this case substituting the equation for x back into the equation for the first sphere produces the equation for a circle, the solution to the intersection of the first two spheres:

$y^2+z^2=r_1^2-\frac{(r_1^2-r_2^2+d^2)^2}{4d^2}$ .

Substituting : $y^2+z^2=r_1^2-x^2$ into the formula for the third sphere and solving for y there results:

$y=\frac{r_1^2-r_3^2-x^2+(x-i)^2+j^2}{2j}=\frac{r_1^2-r_3^2+i^2+j^2}{2j}-\frac{i}{j}x$ .

Now that we have the x- and y-coordinates of the solution point, we can simply rearrange the formula for the first sphere to find the z-coordinate:

$z=\pm \sqrt{r_1^2-x^2-y^2}.$

Now we have the solution to all three points x, y and z. Because z is expressed as the positive or negative square root, it is possible for there to be zero, one or two solutions to the problem.

Preliminary and final computations

The third basis unit vector is $\hat e_z = \hat e_x \times \hat e_y$ . Therefore,

Using $i, \; d$ and $j$ as computed above, solve for x, y and z as described in the Derivation section. Then

$\vec p_{1,2} = P1 + x \ \hat e_x + y \ \hat e_y \ \pm \ z \ \hat e_z$

gives the points in the original coordinate system since $\hat e_x, \; \hat e_y$ and $\hat e_z$ , the basis unit vectors, are expressed in the original coordinate system.

This Wherigo has been tested on these Devices: IPhone 3G, Garmin Colorado, Garmin Oregon

This Wherigo does not work on these Devices: No Known Devices

If your device is not listed, it has not been tested on, if you run this wherigo on a non-listed device please state that in your log so I can add it to the list.

DOWNLOAD the Wherigo cartridge here.

A Wherigo capable smartphone or GPS (like a Colorado or Oregon) is required to run the cartridge. See the Wherigo site for details.

Additional Hints (Decrypt)

[HIDE] 3 srrg yrsg sebz pragre

Decryption Key

A|B|C|D|E|F|G|H|I|J|K|L|M
-------------------------
N|O|P|Q|R|S|T|U|V|W|X|Y|Z

(letter above equals below, and vice versa)

Watch

Geocache Description:

Preliminary and final computations

Preliminary and final computations

Attributes

Inventory

Bookmark Lists

Triangulation Calculation Wherigo Cache

Watch

Geocache Description:

Preliminary and final computations

Preliminary and final computations

Attributes

Inventory

Bookmark Lists

Treasures