代写ENG5062、代写Python/Java编程语言

日期：2022-08-03 09:14

Degrees of MEng, BEng, MSc and BSc in Engineering

NAVIGATION SYSTEMS

(ENG5062 / ENG4184)

Thursday 17th December 2020

Release time: 09:00AM (GMT) for 2.5 hours

Exam duration: 2 hours to complete exam plus 30 mins for download/upload of

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)

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,

And the length of the road is L = 1126.46m. (1)

[8 marks]

(b) The vector we are interested in is the reflector position with respect to the antenna

resolved into antenna axes,