Simple and Exact Solutions to Position Calculation

In nvctr: The n-vector Approach to Geographical Position Calculations using an Ellipsoidal Model of Earth

knitr::opts_chunk$set(
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  comment = "#>",
  out.width = "220px",
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library(png)
library(nvctr)

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 [@gade2010].

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 {#example-01}

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.

knitr::include_graphics("ex1img.png")

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))
(n_EB_E <- lat_lon2n_E(lat_EB, lon_EB))

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))

Step 3: Find $\mathbf{R}_{EN}$ for position $A$

(R_EN <- n_E2R_EN(n_EA_E))

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())

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"))

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

(elevation <- atan2(-p_AB_N[3], p_AB_N[2]) %>% deg())

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

(azimuth <- atan2(p_AB_N[2], p_AB_N[1]) %>%   # positive angle about down-axis
  deg())

Example 2: B and delta to C {#example-02}

A radar or sonar attached to a vehicle $B$ (Body 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 $f = \dfrac{1}{298.26}$.

knitr::include_graphics("ex2img.png")

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
z_EB <- -400
(R_NB <- zyx2R(rad(10),rad(20),rad(30))) # the three angles are yaw, pitch, and roll

# 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))

Step 2: Find $\mathbf{R}{EB}$ from $\mathbf{R}{EN}$ and $\mathbf{R}_{NB}$

(R_EB <- R_EN %*% R_NB) # Note: closest frames cancel

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)

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']])
(z_EB <- l[['z_EB']])

Convert to latitude and longitude, and height

lat_lon_EB <- n_E2lat_lon(n_EB_E)
(latitude  <- lat_lon_EB[1])
(longitude <- lat_lon_EB[2])

# height (= - depth)
(height <- -z_EB)

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.

knitr::include_graphics("ex3img.png")

Position $B$ is given as $\mathbf{p}_{EB}^E$, i.e. "ECEF-vector"

(p_EB_E <- 6371e3 * c(0.9, -1, 1.1)) # m

Solution

Find n-vector from the p-vector

l <- p_EB_E2n_EB_E(p_EB_E)
(n_EB_E <- l[['n_EB_E']])
(z_EB <- l[['z_EB']])

Convert to latitude and longitude, and height

lat_lon_EB <- n_E2lat_lon(n_EB_E)
(latEB  <- lat_lon_EB[1])
(lonEB <- lat_lon_EB[2])

# height (= - depth)
(hEB <- -z_EB)

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 $h_{EB}$.

knitr::include_graphics("ex4img.png")

Solution

lat_EB <- rad(1)
lon_EB <- rad(2)
h_EB <- 3

Step 1: Convert to n-vector

(n_EB_E <- lat_lon2n_E(lat_EB, lon_EB))

Step 2: Find the ECEF-vector p_EB_E

(p_EB_E <- n_EB_E2p_EB_E(n_EB_E, -h_EB))

Example 5: Surface distance {#example-05}

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.

knitr::include_graphics("ex5img.png")

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 [@gade2010] (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 [@gade2010] 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))

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)

Elliptical model (WGS-84 ellipsoid)

The distance between $A$ and $B$ ca be calculated via geosphere package

geosphere::distGeo(c(0, 88), c(-170, 89))

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.

knitr::include_graphics("ex6img.png")

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) ))

and converting back to longitude and latitude

(l  <- n_E2lat_lon(n_EB_E_ti) %>% deg())
(latitude  <- l[1])
(longitude <- l[2])

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.

knitr::include_graphics("ex7img.png")

Solution

The (geographical) mean position $B_{GM}$ is simply given equation (17) in [@gade2010] (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))

and convert to longitude/latitude

(l  <- n_E2lat_lon(n_EM_E) %>% deg())
(latitude  <- l[1])
(longitude <- l[2])

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.)

knitr::include_graphics("ex8img.png")

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 [@gade2010]

$$ \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 $d_E$

d_E <- k_north_E * cos(azimuth) + k_east_E * sin(azimuth)

Step 3: Find $\mathbf{n}_{EB}^E$

n_EB_E <- n_EA_E * cos(s_AB / r_Earth) + d_E * sin(s_AB / r_Earth)

Convert to longitude/latitude

(l  <- n_E2lat_lon(n_EB_E) %>% deg())
(latitude  <- l[1])
(longitude <- l[2])

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.

knitr::include_graphics("ex9img.png")

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 $\mathbf{n}{EA_1}^E$

n_EC_E <- sign(pracma::dot(n_EC_E_tmp, n_EA1_E)) * n_EC_E_tmp

Convert to longitude/latitude

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

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 $6371~\mathrm{km}$.

knitr::include_graphics("ex10img.png")

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 \@ref(fig:solution-10-fig).

c_E <- unit(pracma::cross(n_EA1_E, n_EA2_E))

knitr::include_graphics("solution10img.png")

Find the great circle cross track distance

(s_xt <- (acos(pracma::dot(c_E, n_EB_E)) - pi / 2) * r_Earth)

Find the Euclidean cross track distance

(d_xt <- -pracma::dot(c_E, n_EB_E) * r_Earth)

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 $B$.

knitr::include_graphics("ex11img.png")

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:

n_EN_E <- unit(pracma::cross(n_EA1_E, n_EA2_E))

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

n_EC_E_tmp <- unit(
  pracma::cross(
    n_EN_E,
    pracma::cross(n_EN_E, n_EB_E)
  )
)

Choose the one closest to B:

n_EC_E <- sign(pracma::dot(n_EC_E_tmp, n_EB_E)) * n_EC_E_tmp

Convert to longitude/latitude

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

References

nvctr documentation built on Oct. 28, 2020, 5:07 p.m.