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Unformatted text preview: DRAGONFLY BASELINE PERP‘ORL‘KIANCE ANALYSIS OLI\'IEH L. DE WECK .\ND CHARLES COLEMAN 1. DRAGONFLY B.\sL‘LtNL‘ ANALYSIS 1.1. Objective. The ohjet'tive of this lecture is to give you a. HHlt'lillg point. to establish a perfor
n'tatne baseline for the Dragonfly DesignBailrlFly Competition. First you need to establish a elear
baseline for the I'Nlﬂ't'lt‘tl perforniant'e of the amnorliﬁed aimraft. then you ran start with redesign
considerat.ions, il'aLlChtutliCS and optimization. '2. Cot‘oPtL‘I‘I'I'JOV Seotnno .—\.(.:0H‘]‘II.‘.I A rril.it'al analysis of the Dragonfly ("otnpetilion Storing Algorithm 1eveals that points ran he
gathered in three (listrete phases of the competition. :3. R/(? Moor.1. AmtRAFT Drsnm ELrhttiNTs The key elements of the ah‘t'raft that. will (1 mlrihnte are the "propulsion system". the “wing“ and
the "fuselage". 3.1. Propulsion System.
4. BASELINE. PAFLtt‘llZTERS Before we eat] s1.art. with analysis we need to establish a set. of Dragonfly haseline parameters.
A set is given here, but some of these \‘tllttes are just. estimates or guesses. You shoultl verify the
validity of the baseline parameters. TABLE 1. Dragonﬂy Baseline Parameters b wing span e18 in S wing area 4:30 [i113] 11' mass ("weight“) 15 [oz]
wing loading [71.8 [oz/soft] r. t'horrl It) [in] I overall lengl 11 33 [in]
Motor Craltpner Speed 1100 7.2 [\] E hattery t'harge 33“ tn:\h :3. STRAIGHT AND LEVEL FLIGHT The haseline \‘t‘lUtﬁTj' of Dragonfly is mtitnatml as follows: Kry Itrrntt'r and phrases. Baseline l't'l'fl’llllnllt‘t‘. 2 OLIVIER L. DE \VECK AND CHARLES COLEMAN 5.1. Velocity and Reynold’s Number.
(5.1) 'u = 20[ft/sec] = 20  0.6818[ft/sec][mph/ft/sec] m 13.64[mph] % 6.1[m/sec] Based on this estimate and some baseline parameters we can estimate the Reynold’s number
regime at which the aircraft is operating. (5.2) Re = &
,u
The Reynolds number, Re, is dimensionless, i.e. it is a ratio of two quantities with the same
unit. '0 = the relative velocity of the ﬂuid in [m/s]. 0 = The characteristic length, in this case the
wing chord. ,Ll. = The viscosity of the ﬂuid in [NS/m2]. The viscosity of air, also called the dynamic
viscosity of air is ,u=1.8 x 10—5 at 15C and atmospheric pressure at sea level. With p = 1.23 [kg/m3]
the density of the ﬂuid in [kg/m3]. Substituting in Equation (5.2), we obtain (5.3) Re 2 13.64[mph]  10[in] .780 = 106, 392
Substituting the numbers in SI units yields:
1.23  6.1 0.254
.4 . : — : 1 r
(5 ) Rc 1.810—5 00,876 Thus we can say that the Reynolds number is roughly Re 2 100k. This regime features both
pressure and friction drag. Most powered model R/ C planes operate in the Re 200,000 to well over
1,000,000 range. 5.2. Aspect Ratio. The aspect ratio AR is going to have a large effect on the induced drag of the
model aircraft. We estimate: 172 span [ml] 482
57 AR = — = — = =
( o) S wing area [in2] 450 5.12
We can put this aspect ratio in context with other types of aircraft: TABLE 2. Typical Aspect Ratios Application wing loading [oz/sqft] AR
High speed, highly maneuvrable 22—26 4—6
Moderate speed sport 16—22 68
Low speed trainer 1216 810
Slope gliders 1214 810
Soaring gliders 8—12 10—15 The corresponding wing loading during straight and level, unaccelerated ﬂight is found as: W 15[oz]
5.6 WL=—=—=4.8 ft
( ) S 450/144[sqft] [OZ/sq ]
5.3. Lift Coefﬁcient. We need to obtain an estimate of the Dragonﬂy lift coeﬂicient CL. We know
from the lift equation that: 1
(5.7) L = 510112615
In straight, level, unaccelerated ﬂight lift must equal weight , L = W, thus we can solve for CL
2 . l/V
5.8 C =
( ) L p025 Substituting the values for the known quantities on the right hand side yields:
2 15[oz]  0.030[kg/oz]  9.81[m/s2] (5.9) CL = .) . ~ 2
1.23  6.1— 3.125  0.093[m /sqft] Note that this lift coefficient is a rough estimate and only valid during cruise. m 0.63 IZPT 3 5.4. Drag Coefﬁcient. The drag coefficient is given as: Ci
7r  AR  P If we are considering an aircraft, we can think of the drag coefﬁcient as being composed of two
main components; a basic drag coefﬁcient which includes the effects of skin friction and shape (form),
C00 and an additional drag coefficient, CD, related to the lift of the aircraft. This additional source
of drag is called the induced drag and it is produced at the wing tips due to aircraft lift. The
induced drag coefﬁcient is equal to the square of the lift coefﬁcient CL divided by the quantity: pi
(3.14159) times the aspect ratio AR times an efﬁciency factor e. The aspect ratio is the square of the span divided by the wing area as computed above. Long,
slender, high aspect ratio wings have lower induced drag than short, thick, low aspect ratio wings.
Lifting line theory shows that the optimum (lowest) induced drag occurs for an elliptic distribution
of lift from tip to tip. The efficiency factor (6) is equal to 1.0 for an elliptic distribution and is some
value less than 1.0 for any other lift distribution. (Typical e = .7) From experience we estimate CD0 2 0.015. Substituting the values we already know we obtain: 0.63.2
.l = .l" —= .1 .2m .4
(51) CD UGO—P31451210 005+005 000 5.5. Drag Force. This allows us to compute the drag force that is acting on the Dragonfly. The
drag equation is: (5.10) CDZCDa‘l'CDiZCDa‘l' 1 (5.12) D : EpUZCDS
we now have estimates for all quantities on the RHS and can substitute.
(5.13) D = 0.5  1.23  6.12  0.040  3.125  0.093 = 0.27[N] N 0.97Ioz] Now, we are equipped with estimates of lift and drag. How does this help us find the endurance for
the design competition? 6. COMPETITION PERFORMANCE ESTIMATE We need to estimate the amount of power that is consumed by the Dragonfly during straight and
level flight. We know that the thrust is generated by the propulsion system, i.e. the combination of
battery—electrical D.C. motor—propeller. In straight, 1g, unaccelerated ﬂight we have (6.1) T = D i.e. thrust is qual to drag. Power is qual to force times velocity, i.e.
(6.2) P = T  v (6.3) P = D  v \Vhere does the power come from? Right, it is produced by the electrical motor, unfortunately there
are some losses and the efﬁciency, 17, is smaller than one. (6.4) P = Pm 17
We can now write the equation
(6.5) Pm ~17 : D ~12 [W] The efﬁciency of the motor must generally be determined experimentally. Some tests from previous
years show that: 17mm. N 0.66. Solving for the power delivered by the elctrical motor we get: D  v 0276.1
6.6 P = = — m 2; w
( l m n 0.66 O[ ] 4 OLIVIER L, DE \VECK AND CHARLES COLEMAN 6.1. DC Motor Model. In order to estimate the endurance, i.e. how long the battery charge will
last. we have to get an estimate of the current drawn from the battery. The motor power equation
is: (67) Pm : (E _ RZ ' 1, )(Im _ [0) [\Vl With E = 8.4 [W/A], R, = .357 [Q] [o = .72 [A] and Pm : 2.5 [W] we can solve for the motor current, by solving a quadratic equation for [m:
(6.8) [My m 1.04[A]
This is the current drawn by the DC. motor from the battery. Assuming that this current drawn
is accurate and that the charge of the battery pack is 3.5 [Ah] as advertised we get:
.35 1.04
This corresponds to a ﬂight time of 20 minutes. From experience we know that this is the right
order of magnitude. but somewhat optimistic. It is the goal of the baseline assessment to verify this
endurance during ﬂight test. (6.9) AT 2 = .33[h] 6.2. Competition Score. \Ve can obtain a rough estimate of the score at the design competition.
(1) Fly two laps. The distance between pylons is roughly 200 [ft]. Assume a radius of 60 [ft]
around each pylon. Thus the total distance is 4  200 + 4 7r  60 z 1500 [ft]. At 20 [ft/sec]
this should take ca. 1:15, i.e. get the maximum of 30 points.
(2) Crew Time. Assume that two people stop the aircraft after ground roll. One person loads
the eggs. Estimate ca. 15 [sec] crew time : — 15 points.
(3) Flight for endurance: Assume that we have loaded one egg. With an endurance of 20
minutes we would get 20  60 1 [pts/sec] = 1200 points
This would result in a hypothetical competition score of Points = 30 15 +1200=1215
points.
However, this analysis is ﬂawed. There are a number of assumptions that have to be corrected.
0 In the drag and lift calculation we must consider not only the weight of the empty Dragonﬂy,
but also the weight of the payload (eggs).
0 The motor efﬁciency 77 is a crucial number and it depends on the current. It might not be
valid to use the peak efﬁciency nmazc.
0 Note: Actual experience shows that ﬂight times are more on the order of 5 [min] rather than
20 [min]. Verify this in Johnson. Discuss and discover additional sources 01' energy loss.
0 Success in the competition is not only driven by eﬂicient aircraft design, but also by eﬂicient
operations and robustness. There is no use designing an extremely high AR wing if it will
crack during ﬂight. l\IlASS.~\CHUSE’["I‘S INSTITUTE OF TECHNOLOGY ...
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 Fall '05
 MarkDrela

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