UNIVERSITY OF GLASGOW
Degrees of MEng, BEng, MSc and BSc in Engineering
NAVIGATION SYSTEMS
(ENG5062 / ENG4184)
Thursday 17th December 2020
Release time: 09:30AM (BST) for 4 hours
This exam should take you: 2 hours to complete
However, you have a 4 hours window to download/complete/upload your submission
Attempt ALL THREE Questions.
The numbers in square brackets in the right-hand margin indicate the marks allotted to the
part of the question against which the mark is shown.
These marks are for guidance only.
A FORMULA SHEET IS PROVIDED AT THE END OF PAPER
A calculator may be used. Show intermediate steps in calculations.
Page 1 of 17 Continued overleaf…
Q1.
(a) The following coordinates were recorded using a hand-held GNSS at the beginning
and end of Sauchiehall Street in Glasgow. Calculate the length of the road.
Buchanan Galleries – 55 51′51′′N 4 15′11′′W
Charing Cross – 55 51′58′′N 4 16′14′′W
(8)
(b) A radar drone surveillance system is used to detect and track small airborne vehicles
operating close to restricted sites such as airports, military bases and nuclear power
plants. The radar antenna is mounted on a two-axis gimbal that allows it to rotate in
both azimuth and elevation. Each drone within range will reflect a small portion of
the radar energy from favourably aligned reflector points on the fuselage.
Using the navigation kinematic notation, show that the velocity of a reflector point p
on the fuselage as seen from the antenna and resolved into antenna axes is given by,
r˙aap = v
a
eb vaea +Cab beblbbp aea
(
raeb raea +Cab lpbp
)
where, F b,F a are free to rotate with respect to F e and lbbp is the position of the
fuselage reflector with respect to the body axes. (12)
oi,e xe
ye
ze
reea
reeb
reab = r
e
eb reea
xa
ya
za
xb yb
zb
p
lbbp
Figure Q1: Coordinate representation of UAV and radar antenna.
[20 Marks]
Page 2 of 17 Continued overleaf…
Solution:
To solve this problem we need to obtain the location of each end in Cartesian coordi-
nates using,
xeeb = (RE(Lb) + hb) cosLb cosλb
yeeb = (RE(Lb) + hb) cosLb sinλb
zeeb =
[
(1 e2)RE(Lb) + hb
)
sinLb
There are a few variables we need to find first. The geodetic latitudes and longitudes
need to be converted to their decimal equivalents from deg/min/sec format. Using the
standard conversion,
Longitude (deg) Latitude (deg)
Buchanan Galleries -4.2531 55.8642
Charing Cross -4.2707 55.8663 (1)
(Note 1
2
point for each correct coordinate).To ensure correct calculation, the meridian
radius of curvature need to be calculated,
RE(L) =
R0√
1 e2 sin2 L
Using the latitudes for each end we get,
Buchanan Galleries
RE(L) =
6378137√
1 0.08182 sin2(55.8642)
= 6.392814× 106m
Charing Cross
RE(L) =
6378137√
1 0.08182 sin2(55.8663)
= 6.392814× 106m (2)
(Note 1 point for each correct radius of curvature). We are now able to calculate the
Cartesian position for each end of the road, assuming hb = 0m.
Page 3 of 17 Continued overleaf…
Buchanan Galleries
xeeb = (6392814) cos(55.8642) cos( 4.2531) = 3577488m
yeeb = (6392814) cos(55.8642) sin( 4.2531) = 266047m
zeeb =
[
(1 0.0818192)6392814] sin(55.8642) = 5255972m
Charing Cross
xeeb = (6392814) cos(55.8663) cos( 4.2707) = 3577213m
yeeb = (6392814) cos(55.8663) sin( 4.2707) = 267132m
zeeb =
[
(1 0.0818192)6392814] sin(55.8663) = 5256103m
(3)
(Note 1
2
point for each correct coordinate).
As this is a relatively short distance, there is no need to consider curvature of the earth
and so to calculate distance, use the simple Euclidian distance.
dist(a, b) = ∥b a∥ (1)
The vector from ’a’ to ’b’ is then,
b a =