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Abstract
One of the biggest problems of our time is the global warming. A direct result of thisphenomena is the melting of ice of the glaciers on the north- and the south pole. Asthis continues, the melted ice will contribute to an increase of the sea level, and maycause enormous natural disasters. To be able to prevent this, it’s important to study itsaffects. This reports contains a concept study of a Unmanned Aerial Vehicle, a UAV, seton the coast of Antarctica by the Australian owned base Davis Station to document thechanges and retracting of the glacier borderline. The purpose of the aircraft is to scouta pre-determined path whilst documenting the glaciers with photography from above.
Contents
Introduction 5
Background 6
1 Requirement specification 71.1 Mission profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.1 Takeoff and ascent . . . . . . . . . . . . . . . . . . . . . . . . 71.1.2 Steady state level flight . . . . . . . . . . . . . . . . . . . . . . 71.1.3 Descent and landing . . . . . . . . . . . . . . . . . . . . . . . 8
1.2 Weigth and geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Propulsion system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Power source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Perfomance analysis 92.1 Aerodynamics and control surfaces . . . . . . . . . . . . . . . . . . . . 9
2.1.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.2 Control surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 Phases of flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2.1 Takeoff and ascent . . . . . . . . . . . . . . . . . . . . . . . . 112.2.2 Steady state level flight . . . . . . . . . . . . . . . . . . . . . . 122.2.3 Descent and landning . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Propulsion system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Energy demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.5.1 Center of gravity . . . . . . . . . . . . . . . . . . . . . . . . . 162.5.2 Pitch stability and trim . . . . . . . . . . . . . . . . . . . . . . 16
3 Dimensioning 173.1 Weight and geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Electric engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Propeller . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4 Battery pack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.5 Charging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.6 Wing design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2
3.6.1 Profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.6.2 Aileron design . . . . . . . . . . . . . . . . . . . . . . . . . . 213.6.3 Ice protection system . . . . . . . . . . . . . . . . . . . . . . . 21
3.7 Tail design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.7.1 Horisontal stabilizer . . . . . . . . . . . . . . . . . . . . . . . 213.7.2 Vertical stabilizer . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.8 Center of gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.9 Optimisation of climb angle . . . . . . . . . . . . . . . . . . . . . . . 23
4 Avionics 264.1 Navigation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2 Flight Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.3 Observation equipment . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Summary 29
Appendix 32
3
Foreword
A special thanks would like to be denoted to our mentor Arne Karlsson for his help andguidance throughout this project. We would also like to thank the Royal Institute ofTechnology for the possibility to carry out this project. Lastly we would like to thankthe Royal Institute of Technology’s library for its contributions with information on thetopics in hand.
4
Introduction
Aerial vehicles play a huge role in todays society. They have become a great part in ev-erything from personal traveling to global warfare. Most aircrafts today are controlledby a pilot, and just like every other human being, a pilot makes errors, which may causedevastating results. For pilots in big commercial aircrafts, a mistake can lead to theirown and hundreds of people’s deaths, and for military pilots the cost can be both theirown lives and multi million dollar weaponry. So what if the human aspect of flight wasremoved? There is now a solution to this, and its called a UAV.
A UAV, or as it is commonly known, a drone, is an aerial vehicle that is controlledduring flight by a computor, i.e. no human pilot aboard. The way it is controlled is ei-ther autonomously or by a remote control from the ground or another vehicle. The firstUAV was simply a pack of unmanned, bomb-filled balloons, sent out by the Austriansas a way to attack Venice. The concept of a real unmanned aerial vehicle has long beensought out after. As technology got better, more advanced types of UAVs could be usedduring warfare, like during WWI.
Today the UAVs are most known for their participation in the US Military, where itis a great insurance for the concern of losing a valuable pilot for a dangerous mission.For very simple missions, like scouting an area where it can be very dangerous to bedue to the high enemy density in that specific area, the UAV is perfect. This way no pi-lot is harmed, but the mission can still executed and successful. The UAV is often usedoutside of the military aswell. It has a very wide range of use, as for example exploringoil-, gas-, and mineral sites, and is therefore used alot by private owned businesses.Another example is exploring just what this report is about, glaciers.
5
Background
The climate change is one of the biggest topics around the world in this day and age.Due to global warming, caused by greenhouse gases leaking into the atmosphere, dras-tic changes are occurring on earth. One of the biggest affects global warming has onthe earth is the melting of ice on the north and south pole, causing the sea level to rise.
One very important thing to do is to measure these changes, to better be able to un-derstand what is happening and the repercussions it has. This is the reason for thisproject. By doing a concept study of an unmanned aircraft with a mounted camera,we can plan a determined route for it to fly. Whilst flying, it will take pictures of theglacier below it. After completing its route, it will safely land on an airstrip and wait aset amount of hours until the next flight, while recharging. This will occur once a day,during a six month period. At the end of this half year, sufficient amount of data willhave been collected to be able to study the glaciar changes over time.
6
Chapter 1
Requirement specification
1.1 Mission profile
The aircraft will first takeoff and begin it’s ascent. Within 10 km, the aircraft mustreach an altitude of 1 km. The glacier that will be observed is about 80 km long, andthe aircraft will remain in steady state level flight until this entire distance has beencovered. The total mission distance is approximately 100 km.
Figure 1.1: Mission profile
1.1.1 Takeoff and ascent
During takeoff the aircraft should get airborn within 300 m. It should be able to startfrom ice aswell as dry ground. The aircraft should lift off at a speed that is sufficientenough to satisfy the lift to weight ratio. The aircraft should climb with a fix angleand speed, and since the aircraft will operate in the Antarctic region, there are no legalrestrictions to the rate of climb during ascent. However, the climb angle must be greatenough for the aircraft to reach cruising altitude within a distance of 10 km.
1.1.2 Steady state level flight
During steady state level flight the aircraft should maintain a cruising speed at 100-150km/h. To be able to clearly observe the changes of the glaciers, the aircraft should keep
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a cruising altitude of 1 km. This state of flight does not require any harsh manourversat all. The cruising range should be at least 80 km.
1.1.3 Descent and landing
The engine will be shut off to save energy and the aircraft will glide during the decent.The glide distance depends on the rate of decent, which is not restricted initially. Beforelanding, the speed will be reduced so that the aircraft can be able to touch down andbrake within 300 m. Just as for the the take-off process, the aircraft should manage toland on icy ground aswell as dry ground.
1.2 Weigth and geometry
The total take-off mass of the aircraft should not exceed 300 kg. The aircraft geometryshould be of conventional fashion, with a main wing attached to the body, and a tailwith a horisontal stabilizer and vertical stabilizer. The wing span as well as the totallenght of the aircraft should be less or equal to 8 m. The main body must be capaciousenough to carry the equipment required for the mission.
1.3 Propulsion system
The propulsion system in this case, will consist of an electric engine, and a propeller.The engine should be a brushless permanent magnet synchronous electric engine, and itshould be able to deliver the required amounts of power for the different flight phases.The engine mass should not exceed 50 kg. It should be able to operate in sub zerotemperatures without significantly losing performance. During steady state level flight,the engine should not operate above the continuous power setting.The propeller will be attached in a pusher-configuration at the rear of the aircraft. Theblade diameter of the propeller will be restricted to the start and landing phase, sincethe propeller must not hit the ground during rotation.
1.4 Power source
A direct current battery pack will be used as a power source. It should be powerfulenough to sufficiently fuel the aircrafts trip from start to landing. The battery pack mustbe rechargable, and be able to discharge and recharge enough times without signifi-cantly lowering the performance. The batteries should be able to withstand sub zerotemperatures, down to −20 ◦C. The total mass of the battery pack must not exceed 50kg.
8
Chapter 2
Perfomance analysis
2.1 Aerodynamics and control surfaces
2.1.1 Basics
As the aircraft is moving, the entire body is exposed to aerodynamic forces, changingthe motion of the aircraft. The wing is the most essential part of the aircraft. It generatesnext to all of the lift force, which makes it possible for the aircraft to get airborne. Thebody aswell as the wing also contributes to the drag force. The magnitude of the liftforce and the drag force depends on multiple variables, such as wing shape, air speed,air density, wing size, body shape and more. The drag force and the lift force can beexpressed as
D = CD · ρ · S ·v2
2(2.1)
L = CL · ρ · S ·v2
2(2.2)
where CD and CL are non-dimensional constants, determined by the shape of the wingand body, ρ is the density of the surrounding air, S is a reference area of the wing, and vis the velocity of the aircraft. A very common way of expressing CD is using the dragpolar
CD = CD0 + CDi = CD0 +K · CL2 (2.3)
whereCD0 is the zero-lift drag coefficient andCDi is the lift-dependent drag coefficient.CD0 can be expressed as
CD0 = CFe ·Swet
S(2.4)
where CFe is the equivalent skin-friction coefficient, Swet is the total wetted area of theentire aircraft. The drag-due-to-lift factor, K, is usually defined as
K =S
π · b2 · e0(2.5)
where b is the wingspan, and e0 is the Oswald efficiency factor.
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2.1.2 Control surfaces
Control surfaces on aircrafts controls the movement around the axes of motion, seeFigure 2.1.
Figure 2.1: Control surfaces and axes of motion
The ailerons control the movement around the roll axis, the rudder controls the move-ment around the yaw axis and the elevators control the movement around the pitch axis.These three surfaces are called primary control surfaces.There are also surfaces called secondary control surfaces such as flaps, slats, and airbrakes, which are mostly used on large and heavy aircrafts. The flaps and the slats areused to increase the camber and often the area of the wing, making it more effective atlow speed, thus creating more lift. Air brakes are used to increase drag.
Ailerons
When initially designing the ailerons, there are some rule of thumbs which can be used[18]. The aileron-to-wing-chord ratio, Ca/Cw, is usually about 15 to 25 percent, theaileron-to-wing-span ratio, ba/bw is about 20-30 percent. Also, the max deflection,Amax is usually around 20-25 degrees.
Rudders
In the design of the rudder, four parameters must be determined; rudder area, Sr, rudderchord, Cr, rudder span, br, and maximum rudder deflection, Rmax. Good inital valuesare Sr/Sv = 0.38, Cr/Cv = 0.42, and Rmax = 25 degrees. [18]
Elevators
In the design of the elevator, four parameters should be determined. They are elevatorplanform area, Se, elevator chord, Ce, elevator span, be, and maximum elevator deflec-tion, Emax. As a general guidance [18], the typical values for these parameters are as
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follows Se/Sh = 0.4−0.15, be/bh = 0.8−1, Ce/Ch = 0.2−0.4, andEmax,up = −25degrees, Emax,down = +20 degrees.
2.2 Phases of flight
2.2.1 Takeoff and ascent
During takeoff, the propeller must produce enough thrust power, Tprop, for the aircraftto accelerate from zero to the lift-off speed, vLOF , within a certain distance. The bal-ance equation during ground roll is
→ : Tprop −D − froll = m · a (2.6)
where m is the total mass of the aircraft, froll is the roll friction force and a is theaircraft’s acceleration.
Figure 2.2: Forces during takeoff
The propeller thrust force can be expressed as
Tprop =Peng · ηprop
v(2.7)
where Peng is the engine power and ηprop is the propeller efficiency. The roll frictionforce depends on the generated lift force, and on the roll friction coeffiecient, µroll.
froll = µroll · (W − L) (2.8)
where W is the total weight of the entire aircraft. The value of the roll friction coef-ficient depends on the contact materials. Eq. (2.6) is a non-linear ordinary differentialequation. It’s possible to obtain speed data, time data, distance data and more, by solv-ing this equation numerically in MATLAB using the function ODE45.
When the aircraft has reached its lift-off speed, it will leave ground and begin its tran-sition to climb. The lift-off speed, is set to be 20% greater than the stall speed, vs. The
11
stall speed is the absolute minimum speed an aircraft can fly at, and can be expressedas
Vs =
√2 ·W
ρ · CL,max · S(2.9)
where CL,max is the maximum value of of CL, which varies for different wing config-urations. Bacause of the short endurance of the transition to climb, no further perfor-mance analysis has been done regarding this flight phase.
Figure 2.3: Forces during ascent
The aircraft will keep a constant speed during the ascent, at a fixed angle of climb,γclimb, which is the angle between the horisontal axis and the aircraft’s flight path. Thebalance equations can be expressed as
↖: L−W cos(γclimb) = 0 (2.10)
↗: T −D −W sin(γclimb) = 0 (2.11)
The rate of climb, R/C, is the vertical climb speed and it can be expressed as
R/C = vclimb sin (γclimb) (2.12)
Since the air density is decreasing with an increase in altitude, and since the speed andangle of climb is required to be constant, the propeller thrust and the engine powerwill change with altitude. The times it takes for the aircraft to reach cruise altitude iscalculated by
∆tascent = t2 − t1 =
t2∫t1
dt =
h2∫h1
dh
R/C(2.13)
2.2.2 Steady state level flight
To maintain steady state level flight, it’s required that
↑: L−W = 0 (2.14)
→: Tprop −D = 0 (2.15)
12
The thrust vector isn’t actually completely parallell to the speed vector. However, theangle between them is often insignificantly small, and it can therefore be neglected inmost cases.
Figure 2.4: Forces during steady state level flight
The velocity at which the required engine power is at it’s minumum during steady statelevel flight is [14]
VPr,min =
√(2
ρ
)·√
K
3 · CD0·(W
S
)(2.16)
To prevent instability issues, the cruise speed if set to be 35 % greater than this.
2.2.3 Descent and landning
Figure 2.5: Forces during descent
During the descent, the engine will be shut off and the aircraft will glide it’s way down.As earlier mentioned, the air density will change with changes in altitude, and thiswill affect the angle of descent since the speed is required to be constant. The balanceequations for the descent are espressed as
↗: L−W cos(γdescent) = 0 (2.17)
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↖: D −W sin(γdescent) = 0 (2.18)
Because of the short endurance of the transition to landing, no further performanceanalysis has been done regarding this flight phase.
Figure 2.6: Forces during landing
For the landing phase, an analythical model can be applied to determine the perfor-mance. Two constants for the landing are calculated through the equations below.
Abreak = g · (−µroll,b) (2.19)
where µroll,b is the friction coefficient with breaked tyres.
Bbreak =g
W(0.5 · ρ · S · (CD − µroll,b · CL)) (2.20)
The speed at which the wheels begin to break, is usually lower than the touch downspeed. An assumption is this case is that the aircraft will begin to break one secondafter touch down, calling that thold, and that the break speed, vbreak, is equal to thetouch down speed, vTD.The total distance it takes for the aircraft to come to a stop is given by
Sland = Sroll + Sbreak = thold · vTD +1
2 ·Bbreakln
(1− Bbreak
Abreak· v2TD
)(2.21)
Just as for the take-off process, the performance of the landing process can also bedetermined by numerical calculations, using the function ODE45.
2.3 Propulsion system
For a propeller propulsion system, the main system parameter will be the power. Forthis project, an electric engine will be used. The shaft power generated by the electricengine, Peng, will be greater than the propeller thrust power, Pprop, due to the propeller
14
efficiency, ηprop. The relation between the propeller thrust power and the engine shaftpower is
Pprop = ηprop · Peng (2.22)
The propeller thrust power at steady state flight is Pprop = Tprop · v, which implies thatthe engine shaft power can be written as
Peng =Tprop · vηprop
(2.23)
The key feature of the electric engine is that the voltage is constant and the currentvaries. Given a steady voltage across the leads of a engine, the rotational frequency willremain constant regardless of load. The higher the power load on the engine, the morecurrent is drawn from the power source. A way of expressing the propeller efficiencyis by applying the Actuator Disk Theory [1]. This theory doesn’t take the rotationalfrequency into consideration. The efficiency can then be expressed as
ηprop =2
1 +√
1 +Tprop
Adisk·v2· ρ2
(2.24)
where Adisk is the propeller disk area. If the required thrust and propeller diameteris known at a certain airspeed and altitude, the propeller efficiency can now easily becalculated. The propeller tip Mach number should be kept below 1 to prevent high noiselevels and unexpected viscous effects. It can be expressed as
Mtip =
√(π ·D · n)2 + v2
a(2.25)
where n is the rotational frequency of the propeller, D is the diameter of the propellerand a is the speed of sound.
2.4 Energy demand
The aircraft will use a rechargeable battery pack as power source. It has to provide theengine and all the aircraft’s avionics with enough power to complete the mission withmargin. The total amount of energy (Joule) required to power the propulsion system forthe entire flight is calculated with
∆Eeng = ∆t ·n∑
i=1
(Pi,eng
ηeng
)(2.26)
where n is the number of flight phases. Every avionic component requires a specificvoltage and current. The energy amount required to power the avionics is
∆Eavionics = ∆t ·n∑
j=1
(Ij,req · Uj,req) (2.27)
15
where n is the number of avionic components. To get the total amount of energy re-quired for the entire trip is calculated by
∆Etot = ∆Eeng + ∆Eavionics (2.28)
The total energy consumption can easily be expressed in kilowatt hours kWh. One Whis equal to 3 600 J, which implies that,
kWhtot =∆Etot
3600000(2.29)
2.5 Stability
2.5.1 Center of gravity
Regarding flight stability, it’s very important that the center of gravity, CoG, is placedcorrectly. The CoG is often designated to be placed within the main wing’s chord. TheCoG can be expressed with the following coordinates below.
XCoG =
∑W (i) · x(i)
W(2.30)
YCoG =
∑W (i) · y(i)
W(2.31)
ZCoG =
∑W (i) · z(i)W
(2.32)
where W(i) is the weight of component i. The x-axis is placed along the roll axis, they-axis is placed along the pitch axis, and the z-axis is placed along the yaw axis.
2.5.2 Pitch stability and trim
To begin with, the aerodynamic center, a.c, of the main wing is the point at which thepitching moment coefficient for the main wing does not vary with lift coefficient (i.e.angle of attack). For thin symmertical airfoils, the a.c is usually placed around 1/4 ofthe chord. This is not entirely true for cambered airfoils, but it’s a good approximation.With that said, the pitching moment about the aircraft’s CoG can be expressed as
CM,CoG =xwc· CL,w −
xt · Shc · Sw
· CL,h + CM,a.c,w (2.33)
where xw is the distance from the main wing’s aerodynamic center back to the CoG, cis the chord of the wing, CL,w is the lift coefficient for the main wing, xt is the distancefrom CoG back to the a.c of the horisontal stabilizer, Sh is the stabilizer reference area,Sw is the main wing reference area, and CM,a.c,w is the wing pitching moment aboutits a.c. Trim can now be achieved by setting the incidence of the tail surface (whichadjusts its CL) to make CM,CoG = 0
16
Chapter 3
Dimensioning
3.1 Weight and geometry
The aircraft has been designed to meet many different requirements. The final wingspan has been set to six meters, and the fuselage’s length to four meters. The fuselagewill be made out of carbon fibre to minimize the weight while keeping some strenghtand flexibility to the structure. For the fuselage to be capacious enough, the front endhas been designed a little bit larger, with a maximum diameter of 0.5 m at the attachmentof the main wing. The landing gear will be a tricycle-configuration, with the backwheels positioned approximately 0.5 m behind the CoG, and the front wheel positionedin front of the wing. After take-off, the landing gear will retract to a slick positionoutside and along the fuselage. The final total weight of the aircraft is 134 kg.
3.2 Electric engine
For the propulsion system, a six phased synchronous brushless electric engine has beenchosen. Unlike brushed engines, brushless engines are purely inductive. This meansthat the engine life is limited primarily by the bearings, which makes it very reliable.The engine is provided by Joby Motors [10]. They primarily producine lightweight,ultra-efficient engines for use in electric aircrafts. The continuous engine power re-quired during steady state level flight has been calculated to 7 kW. The engine chosenhas a continuous shaft power of 14 kW, which is sufficiently enough.
17
Figure 3.1: J2 Engine
Construction InrunnerNominal voltage 100-600 [V]Poles 46Nominal RPM 2500Maximum RPM 3500Diameter 200 [mm]Mass 4000 [g]Length 75 [mm]Continuous Torque 53 [Nm]Continuous Shaft Power at Nominal RPM 14 [kW]Peak torque 80 [Nm]Peak Shaft Power at Nominal RPM (15s) 20.9 [kW]
Table 3.1: Engine specifications
The engine will be controlled by an Electronic Speed Controller, ESC. The functionof the ESC is to obtain flight status data from the flight control system, such as flightspeed, angles, and accelerations. With this data, the ESC will vary the current that isapplied to the engine, thereby controlling the torque for each point in time. No specificESC has been chosen for this engine.
3.3 Propeller
The propeller is provided by Pipistrel [9], which manufactures three different kinds ofpropellers, the LN, the BAM, and the VARLO, with four, three, and two blades respec-tively. They are designed to be used for ultra light and experimental aircrafts. The onechosen for this project is BAM. The layer and inside parts of this propeller are made of
18
composite materials, fibre reinforced plastics, which makes it very light and tough. Thebase of the blade is made of aluminium and stainless steel tube.
Figure 3.2: Propeller specification
3.4 Battery pack
The total energy requirements has been calculated to approximately 7 kWh for the entiretrip (see Appendix/Matlab code/169). The energy contribution from the avionics havebeen neglected, due to its insignificance. By knowing this energy amount, a batterypack can be chosen for the assignment. To extend the life cycle, the battery’s depthof discharge should not exceed 70%. In addition, due to unknown variable changeslike head wind, colder temperatures etc. an energy excess of an additional 20 % willbe required. By using these numbers, the final energy requirements is calculated toapproximately 12 kWh.One of the more optimal choices for a battery pack that will operate in a cold climateare battery packs using lithium polymere cells. These are among the most powerfulrechargeable batteries on the market due to their high ratio of dimensions to weight andcapacity, and are especially good during colder environments compared to other sorts.Lithium polymere cells have an volumetric energy density of 350Wh/L and gravimetricenergy density of 135Wh/kg [8]. To meet the required energy need for this project,with today’s polymere cells, the battery pack would need to have a volume of at least34 litres and a mass of 89 kg. However, with today’s rate of development, a presumptionis a 50% increase of the gravimetric energy density within 20-25 years. For the finalcalculations, battery mass is set to 50 kg.A battery pack for this application needs a Battery management system, BMS. The BMSworks by managing the different levels of the output, making sure it doesn’t operateoutside it’s safe zone, monitoring its state, controlling environment etc. No specificBMS has been choosen for this project.
19
Figure 3.3: Lithium polymere cell
3.5 Charging
The aircraft will take its place in the hangar after its route where it will be plugged inand recharged. Solar panels will be used for this purpose. The solar panels will be puton the roof of the hangar, and they must be dimensioned to be able to fully rechargethe batteries within its time frame. The solar panels are operating with their highestefficiency when facing the sun directly. This will be dealt with by using a small enginewith two axles, spinning and tilting to autonomously track the sun. The latest solarpanel on the market right now is from the Deutsch corporation Fraunhofer, which withworld breaking technology has an efficiency of 44.7 % [7]. A reasonable assumption isthat this number will rise to 60 % within 20 years. (6 % over an 8 year period, 12 %over 16 years, assuming linearity, most likely exponentially).Assuming regular solar power from the sun, at least a 1000 watts per square meter [6],and an minimum time of sun exposure of 8 hours a day [5], the time, t, to recharge thebattery with a solar panel of 4 square meters is
t =∆Etot
1000 · 0.6 · 4= 3h (3.1)
3.6 Wing design
3.6.1 Profile
The inital idea for the wing profile was to choose a cambered one with a high lift andrelatively low drag. Cambered profiles are used on most gliders and low speed aircrafts,so it seemed to be a good idea. In the end it came down to two different but similar pro-files; Clark-Y and the NACA 2412. To analyse the performance of these airfoils, the
20
platform XFLR5 [4] was used, which is an analysis tool for wings, planes and profiles.All calculations were made with a Reynolds nummer of 1.6 · 106. The final choice ofairfoil was the Clark-Y foil due to it’s high lift to drag ratio. It has a maximum thicknessof 11,7% at 30,4% of the chord. The angle of incidence was set to +5, 4 ◦ to maximizethe lift to drag ratio during steady state level flight.
Airfoil(
CLCD
)max
CL CD
Clark-Y 52,1 @ 5, 4 ◦ 0,98 0,019NACA 2412 49,8 @ 6, 9 ◦ 0,98 0,019
Table 3.2: Comparing airfoils for the main wing
Figure 3.4: The Clark-Y profile
3.6.2 Aileron design
The aileron-to-wing-chord ratio was set to 20%. With a wing chord of 0.8 m, the aileronchord was set to 0.16 m. The aileron-to-wing-span ratio was set to 25%. With a wingspan of 6 m, the final aileron chord length is 0.75 m. The max deflection was set to +-25 degrees.
3.6.3 Ice protection system
To keep ice from accumulating on the leading edge of the wing, a anti-ice protectionsystem is needed. For this aircraft, a passive system could be used. Such a systememploys hydrophobic surfaces that repells water. This surface can either be a coatingor some sort of fabric.
3.7 Tail design
3.7.1 Horisontal stabilizer
From the pitch stability analysis, the NACA 0012 was chosen as the horisontal stabilizer.To eliminate the pitch moment about the a.c for the tail, a symmetrical airfoil was
21
chosen. To fully satisfy the pitch moment about CoG, the angle incidence was chosento be +4 ◦. The final stabilizer chord was set to 0.45 m and the stabilizer span was setto 1.8 m.
Figure 3.5: NACA 0012 airfoil
Elevators
The elevator-to-stabilizer-span ratio was set to 0.9, giving a final elevator span of 0.81m. The elevator-to-stabilizer-chord ratio was set to 0.3, giving a final elevator chord of0.13 m. The maximum deflection was set to +-25 degrees.
3.7.2 Vertical stabilizer
For a Cessna 177, the planform area of the vertical stabilizer is 10% of the main wingplanform area. Also, the aspect ratio of the vertical stabilizer is about 1.71 [8]. Usingthese number as reference, gives a vertical stabilizer with a chord of 0.5 m and a spanof 0.85 m.
Rudder
The rudder-to-stabilizer-chord ratio was set to 0.40, and the rudder-to-stabilizer-arearatio was set to 0.40. Using these ratios, the rudder chord was set to 0.2 m and therudder span was set to 0.80 m.
3.8 Center of gravity
The center of gravity has been placed within the fuselage, where the chord of the wingis at it’s thickest. The ”free” masses (the avionics, battery pack, and engine) has beenplaced in order to achieve the required CoG. The landing gear configuration is assumedto have a horisontal CoG that coincides with the required one. Figure 3.6 and Table 3.3below shows the placement of the masses.
22
CoG x y zWings 0 0 0,12Fuselage 0,45 0 0Empennage 2,626 0 0,13Landing gear down 0 0 -0,7Landing gear up 0 0 -0,3Propeller 2,66 0 0Engine 1,66 0 0Battery pack -0,44 0 0Avionics -0,94 0 0Total 0 0 0
Table 3.3: Center of gravity
Figure 3.6: Figure showing placements of masses
3.9 Optimisation of climb angle
To minimize the energy consumption during ascent and steady state level flight, it isimportant to study the angle of climb. It is not completely obvious how the energyusage depends on the angle of climb, since the air density is changing with altitude,affecting the performance. A way of studying this is to let the climb angle vary, whilecalculating the total amount of energy required for each angle. The aircraft must reachan altitude of 1 km within a horisontal range of 10 km. This gives that the minimumclimb angle required is around 5.7 ◦. If the climb angle increases, the horisontal climb
23
distance decreases while the steady state level flight distance increases.
Figure 3.7: Optimisation of climb angle
To keep a constant climb angle and a constant speed during, the engine power is forcedto change with altitude. Figure 3.8 below presents the results from this optimisation.
Figure 3.8: Climb energy optimisation
At a climb angle of 19.3 degrees, the engine power exceeds the recommended continu-ous power of 14 kW. From Figure 3.8 it is clear that the best climb angle for this purpose
24
is tbe lowest possible. Therefore, the climb angle is set to 5.7 degrees. By using thisclimb angle compared to using 19.3, an energy save of 3.23% is obtained.
25
Chapter 4
Avionics
The word ”Avionics” comes from the words Aviation and Electronics. It is a collectiveword for all the electrical components in an aircraft such as wiring, navigation systems,flight controls and so on. The aircrafts mission is to scout a predetermined route. To beable to do this, the aircraft has to be equipped with the necessary electronics for it tocomplete it’s mission.
4.1 Navigation
For the aircraft to be able to find its location on the earth and make sure it correctlyfollows its route it needs to be equipped with a GPS. GPS stands for Global PositioningSystem and it works by uses three different orbiting satellites to pinpoint its location.The aircraft will use this technology to be able to steer its way through the predeter-mined route.
Figure 4.1: How a GPS works
26
4.2 Flight Control
For the aircraft to be able to steer its way it needs to be equipped with some sort of flightcontrol system. This system will make sure that the aircraft can turn left and right, andascend/descend at the required speeds for the aircraft to be able to complete its route.There are several different kinds of UAV navigation systems, used for different sizesand missions.
The flight control system picked for this UAV is called Proton [3]. The Proton weighsonly 50 grams, and contains a combined set of gyroscopes, accelerometers and GPS, allthere to control the flight on its course, whilst drawing only 0.7 W. The specificationsof the Proton are listed below.
Figure 4.2: Flight control system Proton
4.3 Observation equipment
When chosing a observation system for the aircraft it is important to have one with agyro stabilizer to make sure you get a steady focused picture. A gyro stabilizer is adevice for measuring orientation, meaning it can measure its position around a certainaxis. This way it knows if its wrongly positioned, and if it needs to adjust itself. Thestabilizer works by using a spinning wheel in which the axis is free to assume anyorientation without changing the axis of the wheel.The equipment used is called E-sky 02 and is produced by Dat Con [2]. This cameradevice is used by other UAVs in a wide range of fields, and is explained as a ”day andnight gyro stabilizer observation system capable of stabilizing demanding oscillations
27
up to approximately 45 ◦. It operates with a continous power of 3.6 W, and weighs only213 grams. There are two different E-sky models, either with thermal imaging, E-sky02, or without thermal imaging, E-sky 01. For this mission a thermal imaging system isof great use, hence the pick.
Figure 4.3: E-sky 02
28
Summary
The main goal of the project was to be able to plan a route for an unmanned aerial ve-hicle, UAV, along the coast of Antarctica, which has now been successfully done.The engine chosen for the project was a six phased synchronous brushless electric en-gine provided by Joby Motors, with a continuous shaft power of 14 kW, and a peak of20.9 kW.Along with the motor, a battery pack made of lithium polymere cells were chosen.These are some of the more powerful batteries on the market, and they work well underlow temperatures.For the main wing, a profile named Clark-Y was chosen, due to its high lift to dragratio.The propeller of the aircraft has been provided by the website pipistrel.si and is calledBAM. It has three blades and a diameter of 1.66 m.To be able to navigate its own course, the aircraft had been installed with a flight controlsystem called Proton. To be able to properly scout the area beneath, a camera calledE-Sky 02 had been installed, mounted right beneath the nose of the aircraft.When combining the complete aircraft into one piece, the center of gravity has beenplaced where the main wing chord is at its thickest.
The project will begin during the September month, due to longer days of the win-ter half year, meaning more sunshine for the solar cells. When the aircraft is on therunway, it has a 300 m runway ahead. During take-off, a continuous thrust power of15 kW is generated. The aircraft travels 82 m in almost 4.7 seconds, and takes off at aspeed of 25.6 m/s. During the ascent, the aircraft climbs at an angle of 5.7 degrees ata speed of 30.2 m/s, until reaching an altitude of 1 km. This takes approximately 5.3minutes.At this point, it continues to travel at approximately the same speed for 83.5 km, whichtakes about 47.9 minutes. After this phase, the aircraft starts to descend. During thisphase the engine is shut off, and the aircraft will glide a distance of 6.5 km, which takesabout 3.7 minutes. The aircraft will touch down with a speed of 25.6 m/s, where thebrakes will be applied after 1 second. The total distance from touch down to stop is119-232 m, depending on the ground conditions. At this point it will be brought to thehangar where it will recharge with the help of solar panels, making it ready for the nextdays flight. The entire project will continue once a day for a six month period. At theend of this period, the data of the glacier changes will be collected and analysed.
29
Bibliography
[1] http://web.mit.edu/16.unified/www/fall/thermodynamics/notes/node86.html.
[2] http://www.dat-con.com/sites/default/files/download/e-sky-2012.pdf, May 2014.
[3] http://www.uavnavigation.org/products/uav-autopilot-proton, May 2014.
[4] http://www.xflr5.com/xflr5.htm, March 2014.
[5] http://www.timeanddate.com/worldclock/astronomy.html?n=468, April 2014.
[6] http://en.wikipedia.org/wiki/sunlight, April 2014.
[7] http://www.ise.fraunhofer.de/en/press-and-media/press-releases/presseinformationen-2013/world-record-solar-cell-with-44.7-efficiency,April 2014.
[8] http://www.powerstream.com/li-pol.htm, May 2014.
[9] http://www.pipistrel.si, April 2014.
[10] http://www.jobymotors.com, April 2014.
[11] Arne Karlsson. The lift to drag ratio. 2004.
[12] Arne Karlsson. The aeroplane - some basics. 2012.
[13] Arne Karlsson. Steady climb performance with propeller propulsion. 2013.
[14] Arne Karlsson. Steady and level flight of an airplane with propeller propulsion.2013.
[15] Arne Karlsson. Cruise performance of airplanes with propeller propulsion. 2013.
[16] Arne Karlsson. How to estimate cd0 and k and the simple parabolic drag polar.2013.
[17] Arne Karlsson. Cruise performance. 2004.
[18] Mohammad Sadraey. Aircraft design: A systems enginering approach. 2012.
30
[19] D.P. Raymer. Aircraft design: A conceptual approach. 1992.
[20] Arne Karlsson. Aeroplane weight, balance and pitch stability. 2013.
31
Appendix
CAD-pictures
32
33
Mission data
Engine power Speed Time Distance γ
Start 15 kW 0-25.6 m/s 4.7 s 82/83 m * -Ascent 10.9 kW 30.2 m/s 5.3 min 10 km 5.7SSF 7 kW 29 m/s 47.9 min 83.5 km -Descent 0 kW 29 m/s 3.7 min 6.5 km -8.8Landing 0 kW 25.6-0 m/s 20.4/9.2 s * 232/119 m * -
* Ground condition ice/dry
34
Matlab code
1 close all2 clear all3 clc4
5 [m,g,Wto,b,c,S,stot,H,v_sound,rho,Swet,CL,CFe,...6 CD0,AR,e0,K,CD,eta_prop,eta_eng,...7 CL_max,v_stall,P_eng_1] = Initconstants();8
9 %% Weight estimation10
11 W_wings = 15; % Wings [kg]12 W_lg = 6; % Landing gear [kg]13 W_tail = 8; % Tail [kg]14 W_body = 30; % Fuselage [kg]15 W_prop = 6; % Propeller [kg]16 W_avion = 15; % Avioncics [kg]17 W_bat = 50; % Battery pack [kg]18 W_eng = 4; % Engine [kg]19
20 TotalW = (W_wings+W_lg+W_tail+W_body+W_prop+W_avion+W_bat+W_eng);21
22 disp(['Total mass of aircraft: ' num2str(TotalW) ' kg'])23 disp(' ')24
25 % Center of gravity26 Wi = [W_wings W_lg W_tail W_body W_prop W_avion W_eng W_bat];27 Xi = [ 0 0 -2.7 0 -3 0.5 2 0]';28 Zi = [0.15 -0.5 0.2 0 0 0 0 0]';29
30 Xcg = (Wi*Xi)/TotalW; % X-coordinate31 Zcg = (Wi*Zi)/TotalW; % Z-coordinate32
33 %% VELOCITY FOR MINIMUM POWER DURING SSF34
35 Vminpwr = ((2/rho(end))*((K/(3*CD0))ˆ(0.5))*(Wto/S))ˆ(0.5);36
37 %% TAKE OFF (Ground roll)38
39 % Solving ground roll time and distance with ODE4540
41 my_start = [0.025 0.05];42
43 index = 1;44 global mu_s45 for mu_s = my_start46 [t_s,z] = ode45(@startfil,[0 20],[0;10e-5]);47 s1 = z(:,1); % Distance48 v1 = z(:,2); % Velocity49 s1_spar(:,index)=s1;50 v1_spar(:,index)=v1;
35
51
52 index=index+1;53 end54
55
56
57 subplot(211)58 plot(t_s,s1_spar)59 title('Distance as function of time')60 xlabel('Time [s]'), ylabel('Distance [m]'), grid on61 subplot(212)62 plot(t_s,v1_spar)63 title('Velocity as function of time')64 xlabel('Time [s]'), ylabel('Velocity [m/s]'), grid on65 hold on66
67 [˜,j1] = min(abs(v1_spar(:,1)-1.2*v_stall)); % Find ...lift-off speed
68 [˜,j2] = min(abs(v1_spar(:,2)-1.2*v_stall)); % Find ...lift-off speed
69
70 E1 = (P_eng_1*t_s(j1))/(eta_eng); % Energy req to ...engine [J]
71
72 disp(' ')73 disp('TAKEOFF PERFORMANCE--------------------')74 disp(['Takeoff speed: ' num2str(1.2*v_stall) ' m/s'])75 disp(['Engine power: ' num2str(P_eng_1/1000) ' kW'])76 disp(['Ground roll time: ' num2str(t_s(j1)) ' seconds'])77 disp(['Ground roll distance: ' num2str(s1_spar(j1,1)) ' meters'])78 disp(['Ground roll time: ' num2str(t_s(j2)) ' seconds'])79 disp(['Ground roll distance: ' num2str(s1_spar(j2,2)) ' meters'])80
81 %% ASCENT ...----------------------------------------------------------------
82
83 s2hor = stot/10; % Horisontal distance [m]84 s2 = sqrt(Hˆ2+s2horˆ2); % Distance speed ...
direction [m]85 alpha = acosd(s2hor/s2); % Angle of attack ...
[degrees]86
87 H_as = linspace(0,1000,1000); % Cruise ...altitude SSF
88 s_as = H_as./sind(alpha); % Distance89 v_as = 1.35*sqrt(2*Wto*cosd(alpha)./... % Speed [m/s]90 (CL*rho(end)*S));91 T_as = CD*rho.*S*0.5.*(v_as.ˆ2)+Wto*sind(alpha); % ...
Propeller thrust [Nm]92 T_average = sum(T_as)/length(T_as); % Average ...
thrust93
94 ROC = v_as*sind(alpha); % Rate of ...climb
36
95
96 D_prop = 1.66; % ...Propeller diameter
97 Adisk = pi*(D_prop/2)ˆ2; % Disk area98 eta_prop_as = 2./(1 + sqrt(2*T_as./... % ...
Propeller efficiency99 (Adisk*(v_as.ˆ2).*rho) + 1));
100
101 P_as = (T_as.*v_as)./(eta_prop_as); % Engine ...power climb [W]
102 P_average = sum(P_as)/length(P_as);103 ds_as = diff(s_as(1:2));104 dt_as = ds_as./v_as;105 t2 = sum(dt_as);106 E2 = sum(P_as.*dt_as)/eta_eng; % Total ...
energy for climb107
108 disp(' ')109 disp('CLIMB PERFORMANCE--------------------')110 disp(['Rate of climb: ' num2str(ROC) ' m/s'])111 disp(['Angle of climb: ' num2str(alpha) ' degrees'])112 disp(['Speed: ' num2str(v_as) ' m/s'])113 disp(['Average engine power: ' num2str(P_average/1000) ' kW'])114 disp(['Average propeller thrust: ' num2str(T_average) ' Nm'])115 disp(['Time: ' num2str(t2/60) ' minutes'])116 disp(['Distance: ' num2str(s2hor) ' meters'])117
118 %% DESCENT ...---------------------------------------------------------------
119
120 Index = 1;121 for u = 0.1:0.01:30122 vdes = sqrt(Wto*sind(u)/(CD*rho(2)*S*0.5));123 Descentspeed(:,Index) = vdes;124 if abs(1.35*Vminpwr-vdes) <= 0.01125 k = [vdes u];126 u = 20;127 else128 end129 Index = Index+1;130 end131
132 s4 = H/sind(k(2)); % Distance [m]133 s4hor = s4*cosd(k(2)); % Horisontal ...
distance [m]134 t4 = s4/k(1); % Time [s]135 ROD = k(1)*sind(k(2)); % Rate of descent136
137 disp(' ')138 disp('DESCENT PERFORMANCE-------------')139 disp(['Speed: ' num2str(k(1)) ' m/s'])140 disp(['Rate of descent: ' num2str(ROD) ' m/s'])141 disp(['Angle of descent: ' num2str(k(2)) ' degrees'])142 disp('Engine power: 0 kW')
37
143 disp(['Time: ' num2str(t4/60) ' minutes'])144 disp(['Distance: ' num2str(s4hor) ' meters'])145
146 %% STEADY STATE FLIGHT ...---------------------------------------------------
147
148 s3hor = stot - s2hor - s4hor; % Distance [m]149 v3 = 1.35*Vminpwr; % Speed [m/s]150 T3 = CD*rho(end)*S*0.5*(v3ˆ2); % Propeller ...
thrust [N]151 P_eng_3 = (T3*v3)/(eta_prop); % Engine power [W]152 t3 = s3hor/v3; % Time [s]153 E3 = P_eng_3*t3/eta_eng;154
155 disp(' ')156 disp('SSF PERFORMANCE------------------')157 disp(['Speed: ' num2str(v3) ' m/s'])158 disp(['Engine power: ' num2str(P_eng_3/1000) ' kW'])159 disp(['Propeller thrust: ' num2str(T3) ' Nm'])160 disp(['Time: ' num2str(t3/60) ' minutes'])161 disp(['Distance: ' num2str(s3hor) ' meters'])162
163 disp(' ')164 disp(['Total mission time: ' num2str((t2+t3+t4)/60) ' minutes'])165
166 %% ENERGY ...----------------------------------------------------------------
167
168 Etotal = E1+E2+E3; % Total engine energy ...needed [J]
169 kWh = (Etotal/3600000); % Total kWh from battery170 disp(' ')171 disp(['Total kWh required from battery is: ' num2str(kWh)])172
173
174 %% LANDING ...---------------------------------------------------------------
175
176 my_break = [0.38 0.12]; % [Dry ground, Wet ice]177 A_land = g*(0-my_break);178 B_land = (g/m)*(0.5*rho(1)*S*(CD-my_break*CL));179 S_land = abs((1./(2*B_land)).*log(1-((B_land./A_land).*...180 ((1.2*v_stall)ˆ2))))+1*1.2*v_stall;181 t_stop = 1 + S_land./((1.2*v_stall)/2);182
183 index = 1;184 global mu_l185 for mu_l = my_break186 % Solving ground roll time and distance with ODE45187 [t_b,z] = ode45(@landfil,[0 30],[0;sqrt(2*Wto/(CL*rho(1)*S))]);188 s = z(:,1); % Distance189 v = z(:,2); % Velocity190 sspar(:,index)=s;191 vspar(:,index)=v;
38
192 [a,b] = max(sspar(:,index));193 in(:,index)=b;194 index=index+1;195 end196
197 figure(14)198 subplot(211)199 plot(t_b(1:in(1)),sspar(1:in(1),1),t_b(1:in(2)),sspar(1:in(2),2))200 title('Distance as function of time')201 xlabel('Time [s]'), ylabel('Distance [m]'), grid on202 legend('Dry ground','Wet ice')203
204 subplot(212)205 plot(t_b(1:in(1)),vspar(1:in(1),1),t_b(1:in(2)),vspar(1:in(2),2))206 title('Velocity as function of time')207 xlabel('Time [s]'), ylabel('Velocity [m/s]'), grid on208 legend('Dry ground','Wet ice')209 hold on210
211 [a1,j1] = min(abs(vspar(:,1)));212 [a2,j2] = min(abs(vspar(:,2)));213
214 disp(' ')215 disp('LANDING PERFORMANCE--------------------')216 disp(['Landing speed: ' num2str(sqrt(2*Wto/(CL*rho(1)*S))) ' m/s'])217 disp(['Time to stop (Dry ground): ' num2str(t_b(j1)) ' seconds'])218 disp(['Distance (Dry ground): ' num2str(sspar(j1,1)) ' meters'])219 disp(['Time to stop (Wet ice): ' num2str(t_b(j2)) ' seconds'])220 disp(['Distance (Wet ice): ' num2str(sspar(j2,2)) ' meters'])221
222
223 %% PROPELLER ...-------------------------------------------------------------
224
225 D_prop = 1.66; % Propeller diameter226 n_prop = (500:10:3500)/60; % Propeller cycles per second227 v_cruise = v3; % Cruise speed228 Mtip = sqrt(((pi*D_prop*n_prop).ˆ2)... % Tip mach number229 +v_cruiseˆ2)./v_sound(end);230
231
232 RPMnom = 2500; % Nominal RPM engine ...[rounds/min]
233 w = 2*pi*RPMnom/60; % Angular frequancy [rad/s]234 P_cruise = P_eng_3; % Cruise engine power [W]235 Moment = P_cruise/w; % Shaft momentum [Nm]236
237 % Recommended mach number at 0.72238
239 figure(2)240 plot(n_prop*60,Mtip), grid on241 title('Tip Mach number as function of RPM','fontsize',16)242 xlabel('RPM','fontsize',16), ylabel('Tip Mach number','fontsize',16)243
39
244 Adisk = pi*(D_prop/2)ˆ2; % Propeller disk area245
246 % Calculate propeller efficiency247 PE = P_eng_3;248 ind = 1;249 for v = 18:40;250 root = [PE/(Adisk*rho(end)*vˆ3) 0 4 -4];251 RO = abs(roots(root));252 Eff(ind) = RO(3);253 Thr(ind) = (Eff(ind)*PE)/v;254 ind=ind+1;255 end256
257 v=18:40;258 figure(3)259 plot(v,Eff), grid on260 legend('7 kW')261 xlabel('Airspeed [m/s]'), ylabel('Propeller efficiency [%]')262 figure(4)263 plot(v,Thr), grid on264 legend('7 kW')265 xlabel('Airspeed [m/s]'), ylabel('Propeller thrust [Nm]')266
267
268 %% PITCH STABILITY269
270 Cm_ac_w = 0.077; % Wing pitch moment about a.c [Nm]271 xw = 0.136; % Distance to c.g [m]272 xh = 2.5; % Distance to c.g [m]273 CL_h = 0.50; % Lift coefficient tail stabiliser274 Cm_ac_h = 0; % Tail pitch moment about a.c ...
(Symmetrical) [Nm]275
276 % Tail wing area to achieve pitch stability277 % NACA 0010 Angle of incidence 4 degrees.278
279 Sh = ((Cm_ac_w+Cm_ac_h+(xw*CL)/c)*c*S)/(xh*CL_h);280
281 % Tail stabilizer 0.9*0.45 on each side.282
283 %% CHARGING284
285 eta_panel = 0.6; % Efficiency solar panel286 E_sun = 1000; % Energy from sun [W/mˆ2]287 A = 4; % Total area of ...
panels [mˆ2]288 t_charg = Etotal/(E_sun*eta_panel*A*3600); % Charg time [hours]289
290 disp(' ')291 disp(['Charg time is: ' num2str(t_charg) ' hours']);292
293 %% OPTIMISATION ANGLE OF CLIMB294
295 disp(' '), disp('OPTIMISATION OF ANGLE---------------');
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296
297 H_ny = linspace(0,1000,1000); % Discrete ...altitude [m] % ...Import air density
298 alpha_vec = linspace(5.8,45,1000); % Define angle ...of climb
299 BOP = 0;300 for j = 1:length(alpha_vec)301 if BOP == 0;302 alpha_ny = alpha_vec(j); % Angle of climb ...
[deg]303
304 % Distance dimension (Constant for every alpha))305 s = H_ny./sind(alpha_ny);306
307 v1 = sqrt(2*Wto*cosd(alpha_ny)./(CL*rho(end)*S)); % Speed [m/s]308 T1 = CD*rho.*S*0.5.*(v1.ˆ2)+Wto*sind(alpha_ny); % Propeller ...
thrust [Nm]309
310 eta_prop = 2./(1 + sqrt(2*T1./... % Propeller ...efficiency
311 (Adisk*(v1.ˆ2).*rho) + 1));312
313 P1 = (T1.*v1)./(eta_prop); % Engine power ...climb [W]
314
315 ds = diff(s(1:2)); % Delta s [m]316 dt = ds./v1; % Delta t [s]317 E(j) = sum(P1.*dt)/eta_eng; % Climb energy [J]318
319 v2 = sqrt(2*Wto/(CL*rho(end)*S)); % Speed [m/s]320 T2 = CD(end)*rho(end)*S*0.5.*(v2ˆ2); % Propeller ...
thrust [Nm]321 eta_prop2 = 2/(1 + sqrt(2*T2/... & Propeller ...
efficiency322 (Adisk*(v2ˆ2).*rho(end)) + 1));323 P2 = (T2*v2)./(eta_prop2); % Engine power ...
climb [W]324 s2 = 10000-1000/tand(alpha_ny); % SSF distance [m]325 t2 = s2/v2; % SSF time [s]326
327 E2(j) = P2*t2/eta_eng; % SSF energy [J]328
329 Etot = E(j)+E2(j); % Total energy [J]330 Etotos(j) = Etot;331 eta_prop1(j,:) = eta_prop;332 P1_plot(:,j)=P1;333
334 if P1(:) >= 14000335
336 disp(['P_eng > 14 kW for ' num2str(alpha_ny) ' degrees.'])337 alpha_max = alpha_ny;338 Etot_max = Etot/10ˆ6;339 BOP = 1;
41
340 index = j;341 end342 end343 end344
345 disp(['You save ' num2str(100*((Etot_max/(Etotos(1)*10ˆ-6))-1))...346 ' percent of energy with angle 5.8 compared to ' ...
num2str(alpha_max)])347
348 alpha_vec = linspace(5.8,alpha_max,index);349 figure(5)350 plot(alpha_vec,Etotos/10ˆ6,alpha_vec,E/10ˆ6,alpha_vec,E2/10ˆ6)351 grid on352 %title('Energy optimisation','fontsize',18)353 xlabel('\gamma_{climb}','fontsize',18)354 ylabel('Energy use [MJ]','fontsize',18)355 leg=legend('Total energy','Climb energy','Steady state energy');356 set(leg,'FontSize',14)357
358 %% CONTROL SURFACES359
360 % Ailerons361 C_a = 0.20*c; % Aileron chord362 b_a = 0.25*(b/2); % Aileron span363
364 %Elevators365 C_e = 0.3*0.45; % Elevator chord366 b_e = 0.9*0.9; % Elevator span367
368 % Vertical stabilizer369
370 Av = 0.1*S; % VS area371 AR_v = 1.7; % VS Aspect ratio372 v_chord = 0.5; % VS chord373 v_span = 0.85; % VS length374
375 % Rudder376 S_r = 0.4*Av; % Rudder area377 C_r = 0.4*v_chord; % Rudder chord
1 function [m,g,Wto,b,c,S,stot,H,v_sound,rho,Swet,CL,CFe,...2 CD0,AR,e0,K,CD,eta_prop,eta_eng,...3 CL_max,v_stall,P_eng_1] = Initconstants()4
5 m = 134; % Total mass [kg]6 g = 9.82; % Oooooh gravity [m/sˆ2]7 Wto = m*g; % Total weigth [N]8 b = 6; % Wingspan [m]9 c = 0.8; % Width wing [m]
10 S = b*c; % Referencearea wing [mˆ2]11 stot = 100000; % Total distance [m]12 H = 1000; % Altitude [m]
42
13 [˜, v_sound, ˜, rho]...14 =atmosisa(linspace(0,1000,1000));15 Swet = 20; % Wet area (estimated) [mˆ2]16 CL = 0.98; % Lift coefficient (XFLR5)17 CFe = 0.0055; % Equivalent skin friction ...
coefficient18 CD0 = CFe*Swet/S; % Zero-lift drag coefficient19 AR = (bˆ2)/S; % Aspect ratio20 e0 = 0.7; % Oswald efficiency factor21 K = 1/(pi*AR*e0); % Drag due to lift factor K22 CD = CD0 + K*(CLˆ2); % Drag coefficient23 eta_prop = 0.75; % Efficiency factor propeller24 eta_eng = 0.95; % Efficiency engine25 my = 0.09; % Roll friction coefficient26 CL_max = CL; % No flaps, fix CL for wing27 v_stall = sqrt((2*m*g)/...28 (CL_max*rho(1)*S)); % Stall speed [m/s]29 P_eng_1 = 15000; % Takeoff engine power [W]30
31 end
1 function dz = landfil(˜,z)2
3 s = z(1);4 v = z(2);5
6 global mu_l7
8 [m,˜,Wto,˜,˜,S,˜,˜,˜,rho,˜,CL,˜,...9 ˜,˜,˜,˜,CD,˜,˜...
10 ,˜,˜,˜] = Initconstants();11
12 D = CD*rho(1)*0.5*(vˆ2)*S; % Drag [N]13 L = CL*rho(1)*0.5*(vˆ2)*S; % Lift [N]14
15 % Use friction force as long as tires are in contact16 % with ground17 if (Wto-L) > 018 f = mu_l*(Wto-L);19 else20 f = 0;21 end22
23 dz = zeros(size(z));24
25 dz(1) = z(2);26 dz(2) = -(D+f)/m;27
28 end
1 function dz = startfil(˜,z)
43
2
3 s = z(1);4 v = z(2);5
6 global mu_s7
8 [m,˜,Wto,˜,˜,S,˜,˜,˜,rho,˜,CL,˜,...9 ˜,˜,˜,˜,CD,eta_prop,˜...
10 ,˜,˜,P_eng_1] = Initconstants();11
12 D = CD*rho(1)*0.5*(vˆ2)*S; % Drag [N]13 T = (P_eng_1*eta_prop)/v; % Propeller thrust [N]14 L = CL*rho(1)*0.5*(vˆ2)*S; % Lift [N]15
16 % Use friction force as long as tires are in contact17 % with ground18 if (Wto-L) > 019 f = mu_s*(Wto-L);20 else21 f = 0;22 end23
24 dz = zeros(size(z));25
26 dz(1) = z(2);27 dz(2) = (T-D-f)/m;28
29 end
44
