# Simple and Exact Solutions to Position Calculation

#### Enrico Spinielli

#### 2022-06-02

Source:`vignettes/position-calculations.Rmd`

`position-calculations.Rmd`

This vignette contains solutions to various geographical position calculations. It is inspired and follows the 10 examples given at https://www.navlab.net/nvector/ .

Most of the content is based on (Gade 2010).

The color scheme in the Figures is as follows:

- \(\mathbf{\color{red}{Red}}\): Given
- \(\mathbf{\color{green}{Green}}\): Find this

### Example 1: A and B to delta

Given two positions \(A\) and \(B\), find the exact vector from \(A\) to \(B\) in meters north, east and down, and find the direction (azimuth/bearing) to \(B\), relative to north. Use WGS-84 ellipsoid.

#### Solution

Transform the positions \(A\) and \(B\) to (decimal) degrees and depths:

```
# Position A:
lat_EA <- rad(1)
lon_EA <- rad(2)
z_EA <- 3
# Position B:
lat_EB <- rad(4)
lon_EB <- rad(5)
z_EB <- 6
```

**Step 1**: Convert to n-vectors, \(\mathbf{n}_{EA}^E\) and \(\mathbf{n}_{EB}^E\)

```
(n_EA_E <- lat_lon2n_E(lat_EA, lon_EA))
#> [1] 0.99923861 0.03489418 0.01745241
(n_EB_E <- lat_lon2n_E(lat_EB, lon_EB))
#> [1] 0.99376802 0.08694344 0.06975647
```

**Step 2**: Find \(\mathbf{p}_{AB}^E\) (delta decomposed in E). WGS-84 ellipsoid is default

```
(p_AB_E <- n_EA_E_and_n_EB_E2p_AB_E(n_EA_E, n_EB_E, z_EA, z_EB))
#> [1] -34798.44 331985.66 331375.96
```

**Step 3**: Find \(\mathbf{R}_{EN}\) for position \(A\)

```
(R_EN <- n_E2R_EN(n_EA_E))
#> [,1] [,2] [,3]
#> [1,] -0.0174417749 -0.0348995 -0.99923861
#> [2,] -0.0006090802 0.9993908 -0.03489418
#> [3,] 0.9998476952 0.0000000 -0.01745241
```

**Step 4**: Find \(\mathbf{p}_{AB}^N = \mathbf{R}_{NE} \mathbf{p}_{AB}^E\)

```
# (Note the transpose of R_EN: The "closest-rule" says that when
# decomposing, the frame in the subscript of the rotation matrix that is
# closest to the vector, should equal the frame where the vector is
# decomposed. Thus the calculation R_NE*p_AB_E is correct, since the vector
# is decomposed in E, and E is closest to the vector. In the above example
# we only had R_EN, and thus we must transpose it: base::t(R_EN) = R_NE)
(p_AB_N <- base::t(R_EN) %*% p_AB_E %>%
as.vector())
#> [1] 331730.23 332997.87 17404.27
```

The vector \(\mathbf{p}_{AB}^N\) connects A to B in the North-East-Down framework. The line-of-sight distance, in meters, from A to B is

```
(los_distance <- norm(p_AB_N, type = "2"))
#> [1] 470356.7
```

while the altitude (elevation above the horizon), in decimal degrees, is

**Step 5**: Also find the direction to \(B\) (azimuth),
in decimal degrees, relative to true North

### Example 2: B and delta to C

A radar or sonar attached to a vehicle \(B\) (**B**ody coordinate frame) measures
the distance and direction to an object \(C\).

We assume that the distance and two angles (typically bearing and elevation relative to \(B\)) are already combined to the vector \(\mathbf{p}_{BC}^B\) (i.e. the vector from \(B\) to \(C\), decomposed in B).

The position of \(B\) is given as \(\mathbf{n}_{EB}^E\) and \(z_{EB}\), and the orientation (attitude)
of \(B\) is given as \(\mathbf{R}_{NB}\) (this rotation matrix can be found from
roll/pitch/yaw by using `zyx2R`

).

Find the exact position of object \(C\) as n-vector and depth (\(\mathbf{n}_{EC}^E\) and \(z_{EC}\)), assuming Earth ellipsoid with semi-major axis \(a\) and flattening \(f\).

For WGS-72, use \(a = 6378135~\mathrm{m}\) and \(f = \dfrac{1}{298.26}\).

#### Solution

```
p_BC_B <- c(3000, 2000, 100)
# Position and orientation of B is given:
(n_EB_E <- unit(c(1, 2, 3))) # unit() to get unit length of vector
#> [1] 0.2672612 0.5345225 0.8017837
z_EB <- -400
(R_NB <- zyx2R(rad(10),rad(20),rad(30))) # the three angles are yaw, pitch, and roll
#> [,1] [,2] [,3]
#> [1,] 0.9254166 0.01802831 0.3785223
#> [2,] 0.1631759 0.88256412 -0.4409696
#> [3,] -0.3420201 0.46984631 0.8137977
# A custom reference ellipsoid is given (replacing WGS-84):
# (WGS-72)
a <- 6378135
f <- 1 / 298.26
```

**Step 1**: Find \(\mathbf{R}_{EN}\)

```
(R_EN <- n_E2R_EN(n_EB_E))
#> [,1] [,2] [,3]
#> [1,] -0.3585686 -0.8944272 -0.2672612
#> [2,] -0.7171372 0.4472136 -0.5345225
#> [3,] 0.5976143 0.0000000 -0.8017837
```

**Step 2**: Find \(\mathbf{R}_{EB}\) from \(\mathbf{R}_{EN}\) and \(\mathbf{R}_{NB}\)

```
(R_EB <- R_EN %*% R_NB) # Note: closest frames cancel
#> [,1] [,2] [,3]
#> [1,] -0.3863656 -0.9214254 0.04119242
#> [2,] -0.4078587 0.1306225 -0.90365318
#> [3,] 0.8272684 -0.3659411 -0.42627939
```

**Step 3**: Decompose the delta vector \(\mathbf{p}_{BC}^B\) in E

```
(p_BC_E <- R_EB %*% p_BC_B) # no transpose of R_EB, since the vector is in B)
#> [,1]
#> [1,] -2997.828
#> [2,] -1052.696
#> [3,] 1707.295
```

**Step 4**: Find the position of \(C\), using the functions that goes from one
position and a delta, to a new position

```
l <- n_EA_E_and_p_AB_E2n_EB_E(n_EB_E, p_BC_E, z_EB, a, f)
(n_EB_E <- l[['n_EB_E']])
#> [1] 0.2667916 0.5343565 0.8020507
(z_EB <- l[['z_EB']])
#> [1] -406.0072
```

Convert to latitude and longitude, and height

```
lat_lon_EB <- n_E2lat_lon(n_EB_E)
(latitude <- lat_lon_EB[1])
#> [1] 0.9307209
(longitude <- lat_lon_EB[2])
#> [1] 1.107728
# height (= - depth)
(height <- -z_EB)
#> [1] 406.0072
```

### Example 3: ECEF-vector to geodetic latitude

Position \(B\) is given as an “ECEF-vector” \(\mathbf{p}_{EB}^E\) (i.e. a vector from E, the center of the Earth, to \(B\), decomposed in E).

Find the geodetic latitude, longitude and height (`latEB`

, `lonEB`

and `hEB`

),
assuming WGS-84 ellipsoid.

Position \(B\) is given as \(\mathbf{p}_{EB}^E\), i.e. “ECEF-vector”

```
(p_EB_E <- 6371e3 * c(0.9, -1, 1.1)) # m
#> [1] 5733900 -6371000 7008100
```

#### Solution

Find n-vector from the p-vector

```
l <- p_EB_E2n_EB_E(p_EB_E)
(n_EB_E <- l[['n_EB_E']])
#> [1] 0.5170890 -0.5745433 0.6344439
(z_EB <- l[['z_EB']])
#> [1] -4702060
```

Convert to latitude and longitude, and height

```
lat_lon_EB <- n_E2lat_lon(n_EB_E)
(latEB <- lat_lon_EB[1])
#> [1] 0.6872888
(lonEB <- lat_lon_EB[2])
#> [1] -0.8379812
# height (= - depth)
(hEB <- -z_EB)
#> [1] 4702060
```

### Example 4: Geodetic latitude to ECEF-vector

Find the ECEF-vector \(\mathbf{p}_{EB}^E\) for the geodetic position \(B\) given as latitude \(lat_{EB}\), longitude \(lon_{EB}\) and height \(h_{EB}\).

#### Solution

**Step 1**: Convert to n-vector

```
(n_EB_E <- lat_lon2n_E(lat_EB, lon_EB))
#> [1] 0.99923861 0.03489418 0.01745241
```

**Step 2**: Find the ECEF-vector p_EB_E

```
(p_EB_E <- n_EB_E2p_EB_E(n_EB_E, -h_EB))
#> [1] 6373290.3 222560.2 110568.8
```

### Example 5: Surface distance

Given two positions \(A\) \(\mathbf{n}_{EA}^E\) and \(B\) \(\mathbf{n}_{EB}^E\), find the surface distance \(s_{AB}\) (i.e. great circle distance). The heights of \(A\) and \(B\) are not relevant (i.e. if they don’t have zero height, we seek the distance between the points that are at the surface of the Earth, directly above/below \(A\) and \(B\)). Also find the Euclidean distance (chord length) \(d_{AB}\) using nonzero heights.

Assume a spherical model of the Earth with radius \(r_{Earth} = 6371~\mathrm{km}\).

Compare the results with exact calculations for the WGS-84 ellipsoid.

#### Solution

```
n_EA_E <- lat_lon2n_E(rad(88), rad(0));
n_EB_E <- lat_lon2n_E(rad(89), rad(-170))
r_Earth <- 6371e3
```

##### Spherical model

The great circle distance is given by equations (16) in (Gade 2010) (the \(\arccos\) is ill conditioned for small angles; the \(\arcsin\) is ill-conditioned for angles near \(\pi/2\), and not valid for angles greater than \(\pi/2\)) where \(r_{roc}\) is the radius of curvature, i.e. Earth radius + height:

\(\begin{align} s_{AB} & = r_{roc} \cdot \arccos \!\big(\mathbf{n}_{EA}^E \boldsymbol{\cdot} \mathbf{n}_{EB}^E\big)\\ & = r_{roc} \cdot \arcsin \!\big(\big|\mathbf{n}_{EA}^E \boldsymbol{\times} \mathbf{n}_{EB}^E\big|\big) \tag{16} \end{align}\)

The formulation via \(\operatorname{atan2}\) of equation (6) in (Gade 2010) is instead well conditioned for all angles:

\(s_{AB} = r_{roc} \cdot \operatorname{atan2}\big(\big|\mathbf{n}_{EA}^E \boldsymbol{\times} \mathbf{n}_{EB}^E\big|, \mathbf{n}_{EA}^E \boldsymbol{\cdot} \mathbf{n}_{EB}^E\big) \tag{6}\)

```
(s_AB <- (atan2(base::norm(pracma::cross(n_EA_E, n_EB_E), type = "2"),
pracma::dot(n_EA_E, n_EB_E)) * r_Earth))
#> [1] 332456.4
```

The Euclidean distance is given by

\(d = r_{roc} \cdot \big| \mathbf{n}_{EB}^E - \mathbf{n}_{EA}^E \big|\)

```
(d_AB <- base::norm(n_EB_E - n_EA_E, type = "2") * r_Earth)
#> [1] 332418.7
```

### Example 6: Interpolated position

Given the position of \(B\) at time \(t_0\) and \(t_1\), \(\mathbf{n}_{EB}^E(t_0)\) and \(\mathbf{n}_{EB}^E(t_1)\).

Find an interpolated position at time \(t_i\), \(\mathbf{n}_{EB}^E(t_i)\). All positions are given as n-vectors.

#### Solution

Standard interpolation can be used directly with n-vector as

\[ \mathbf{n}_{EB}^E(t_i) = \operatorname{unit}\Bigg(\mathbf{n}_{EB}^E(t_0) + \frac{t_i − t_0}{t_1 − t_0} \Big(\mathbf{n}_{EB}^E(t_1) − \mathbf{n}_{EB}^E(t_0)\Big)\Bigg) \]

```
n_EB_E_t0 <- lat_lon2n_E(rad(89.9), rad(-150))
n_EB_E_t1 <- lat_lon2n_E(rad(89.9), rad(150))
# The times are given as:
t0 <- 10
t1 <- 20
ti <- 16 # time of interpolation
```

Using the expression above

```
t_frac <- (ti - t0) / (t1 - t0)
(n_EB_E_ti <- unit(n_EB_E_t0 + t_frac * (n_EB_E_t1 - n_EB_E_t0) ))
#> [1] -0.0015114993 0.0001745329 0.9999988425
```

and converting back to longitude and latitude

```
(l <- n_E2lat_lon(n_EB_E_ti) %>% deg())
#> [1] 89.91282 173.41322
(latitude <- l[1])
#> [1] 89.91282
(longitude <- l[2])
#> [1] 173.4132
```

### Example 7: Mean position (center/midpoint)

Given three positions \(A\), \(B\), and \(C\) as n-vectors \(\mathbf{n}_{EA}^E\), \(\mathbf{n}_{EB}^E\), and \(\mathbf{n}_{EC}^E\), find the mean position, \(M\), as n-vector \(\mathbf{n}_{EM}^E\).

Note that the calculation is independent of the depths of the positions.

#### Solution

The (geographical) mean position \(B_{GM}\) is simply given equation (17) in (Gade 2010) (assuming spherical Earth)

\[ \mathbf{n}_{EB_{GM}}^E = \operatorname{unit}\Big( \sum_{i = 1}^{m} \mathbf{n}_{EB_i}^E \Big) \tag{17} \]

and specifically for the three given points

\[ \mathbf{n}_{EM}^E = \mathrm{unit}\Big(\mathbf{n}_{EA}^E + \mathbf{n}_{EB}^E + \mathbf{n}_{EC}^E \Big) = \frac{\mathbf{n}_{EA}^E + \mathbf{n}_{EB}^E + \mathbf{n}_{EC}^E}{\Big | \mathbf{n}_{EA}^E + \mathbf{n}_{EB}^E + \mathbf{n}_{EC}^E \Big| } \] Given the three n-vectors

```
n_EA_E <- lat_lon2n_E(rad(90), rad(0))
n_EB_E <- lat_lon2n_E(rad(60), rad(10))
n_EC_E <- lat_lon2n_E(rad(50), rad(-20))
```

find the horizontal mean position

```
(n_EM_E <- unit(n_EA_E + n_EB_E + n_EC_E))
#> [1] 0.38411717 -0.04660241 0.92210749
```

and convert to longitude/latitude

```
(l <- n_E2lat_lon(n_EM_E) %>% deg())
#> [1] 67.236153 -6.917511
(latitude <- l[1])
#> [1] 67.23615
(longitude <- l[2])
#> [1] -6.917511
```

### Example 8: A and azimuth/distance to B

Given a position \(A\) as n-vector \(\mathbf{n}_{EA}^E\), an initial direction of travel as an azimuth (bearing), \(\alpha\), relative to north (clockwise), and finally the distance to travel along a great circle, \(s_{AB}\) find the destination point \(B\), given as \(\mathbf{n}_{EB}^E\).

Use Earth radius \(r_{Earth}\).

In geodesy this is known as “The first geodetic problem” or “The direct geodetic problem” for a sphere, and we see that this is similar to Example 2, but now the delta is given as an azimuth and a great circle distance. (“The second/inverse geodetic problem” for a sphere is already solved in Examples 1 and 5.)

#### Solution

Given the initial values

```
n_EA_E <- lat_lon2n_E(rad(80),rad(-90))
azimuth <- rad(200)
s_AB <- 1000 # distance (m)
r_Earth <- 6371e3 # mean Earth radius (m)
```

**Step 1**: Find unit vectors for north and east as per equations (9) and (10)
in (Gade 2010)

$$ \[\begin{align} \mathbf{k}_{east}^E & = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \times \mathbf{n}^E \tag{9} \\ \mathbf{k}_{north}^E & = \mathbf{n}^E \times \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \times \mathbf{n}^E \tag{10} \end{align}\] $$

```
k_east_E <- unit(pracma::cross(base::t(R_Ee()) %*% c(1, 0, 0) %>% as.vector(), n_EA_E))
k_north_E <- pracma::cross(n_EA_E, k_east_E)
```

**Step 2**: Find the initial direction vector \(d_E\)

**Step 3**: Find \(\mathbf{n}_{EB}^E\)

Convert to longitude/latitude

```
(l <- n_E2lat_lon(n_EB_E) %>% deg())
#> [1] 79.99155 -90.01770
(latitude <- l[1])
#> [1] 79.99155
(longitude <- l[2])
#> [1] -90.0177
```

### Example 9: Intersection of two paths

Define a path from two given positions (at the surface of a spherical Earth), as the great circle that goes through the two points.

Path A is given by \(A_1\) and \(A_2\), while path B is given by \(B_1\) and \(B_2\).

Find the position C where the two great circles intersect.

#### Solution

```
n_EA1_E <- lat_lon2n_E(rad(50), rad(180))
n_EA2_E <- lat_lon2n_E(rad(90), rad(180))
n_EB1_E <- lat_lon2n_E(rad(60), rad(160))
n_EB2_E <- lat_lon2n_E(rad(80), rad(-140))
# These are from the python version (results are the same ;-)
# n_EA1_E <- lat_lon2n_E(rad(10), rad(20))
# n_EA2_E <- lat_lon2n_E(rad(30), rad(40))
# n_EB1_E <- lat_lon2n_E(rad(50), rad(60))
# n_EB2_E <- lat_lon2n_E(rad(70), rad(80))
```

Find the intersection between the two paths, \(\mathbf{n}_{EC}^E\)

```
n_EC_E_tmp <- unit(pracma::cross(
pracma::cross(n_EA1_E, n_EA2_E),
pracma::cross(n_EB1_E, n_EB2_E)))
```

\(\mathbf{n}_{{EC}_{tmp}}^E\) is one of two solutions, the other is \(-\mathbf{n}_{{EC}_{tmp}}^E\). Select the one that is closest to \(\mathbf{n}_{EA_1}^E\), by selecting sign from the dot product between \(\mathbf{n}_{{EC}_{tmp}}^E\) and \(\mathbf{n}_{EA_1}^E\)

Convert to longitude/latitude

```
(l <- n_E2lat_lon(n_EC_E) %>% deg())
#> [1] 74.16345 180.00000
(latitude <- l[1])
#> [1] 74.16345
(longitude <- l[2])
#> [1] 180
```

### Example 10: Cross track distance (cross track error)

Path A is given by the two positions \(A_1\) and \(A_2\) (similar to the previous example).

Find the cross track distance \(s_{xt}\) between the path A (i.e. the great circle through \(A_1\) and \(A_2\)) and the position \(B\) (i.e. the shortest distance at the surface, between the great circle and \(B\)).

Also find the Euclidean distance \(d_{xt}\) between \(B\) and the plane defined by the great circle.

Use Earth radius \(6371~\mathrm{km}\).

#### Solution

Given

```
n_EA1_E <- lat_lon2n_E(rad(0), rad(0))
n_EA2_E <- lat_lon2n_E(rad(10),rad(0))
n_EB_E <- lat_lon2n_E(rad(1), rad(0.1))
r_Earth <- 6371e3 # mean Earth radius (m)
```

Find the unit normal to the great circle between n_EA1_E and n_EA2_E as shown in the Figure 11.

Find the great circle cross track distance

Find the Euclidean cross track distance

```
(d_xt <- -pracma::dot(c_E, n_EB_E) * r_Earth)
#> [1] 11117.79
```

### Example 11: Cross track intersection

Path A is given by the two positions \(A_1\) and \(A_2\) (similar to the previous example).

Find the cross track intersection point \(C\) between the path A (i.e. the great circle through \(A_1\) and \(A_2\)) and the position \(B\), i.e. the shortest distance point at the surface, between the great circle and \(B\).

#### Solution

Given (note that \(B\) doesn’t necessarily need to lie in between \(A_1\) and \(A_2\) as per Figure above)

```
n_EA1_E <- lat_lon2n_E(rad(0), rad(3))
n_EA2_E <- lat_lon2n_E(rad(0),rad(10))
n_EB_E <- lat_lon2n_E(rad(-1), rad(-1))
```

Find the normal to the great circle between n_EA1_E and n_EA2_E:

Find the intersection points (one antipodal to the other):

Choose the one closest to B:

Convert to longitude/latitude

```
(l <- n_E2lat_lon(n_EC_E) %>% deg())
#> [1] 0 -1
(latitude <- l[1])
#> [1] 0
(longitude <- l[2])
#> [1] -1
```

## References

Gade, Kenneth. 2010. “A Nonsingular Horizontal Position Representation.” *The Journal of Navigation* 63 (03): 395–417. https://www.navlab.net/Publications/A_Nonsingular_Horizontal_Position_Representation.pdf.