Boomerang Flight Dynamics

43 Pages·2018·2.056 MB·English

by  John C. Vassberg

AIAA 2012-2650
30th AIAA Applied Aerodynamics Conference
25 - 28 June 2012, New Orleans, Louisiana
Boomerang Flight Dynamics
John C. Vassberg∗
Long Beach, California 90803
Aerodynamics Lecture
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26 30th AIAA Applied Aerodynamics Conference
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K Abstract
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LI Aerodynamic characteristics and flight dynamics of boomerangs are investigated. A basic
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N aerodynamic model, developed in the 1960’s, is expanded upon using Blade Element Theory.
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K The new aerodynamic model is coupled with a gyroscope model for rudimentary analyses.
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d Some significant findings are made regarding the radius of a boomerang’s circular flight path,
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a the required inclination angle of its axis-of-rotation, its trim state, as well as its dynamic
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w stability. These discoveries provide a basic understanding of how the interplay between aero-
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dynamic forces and moments, and gyroscopic precession combine to return the boomerang to
its rightful owner by way of a circular flight path.
A traditional V-shaped boomerang design is developed as a case study for further detailed
analyses. Unsteady Reynolds-averaged Navier-Stokes solutions provide accurate aerodynamic
characteristics ofthe subjectboomerang. The high-fidelityaerodynamicmodeliscoupledwith
the equations of motion to provide accurate six-degree-of-freedom simulations of boomerang
flightdynamics. Boomerangorientationduringitsflighttrajectory isdescribedbytheclassical
Euler angles.
∗BoeingTechnical Fellow,FellowAIAA,BoomerangNovice
Copyright (cid:13)c 2012 by John C. Vassberg.
Published by the American Institute of Aeronautics and Astronautics with permission.
Copyright © 2012 by John C. Vassberg. Published by the American Institute of Aeronautics and Astronautics, Inc., with permission.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 1 of 43
Nomenclature
a,b,c,d Arbitrary Coefficients s Vector in Rotation Plane ⊥n
b Coordinate Vectors of Body Frame S Reference Area =RC
C Blade Reference Chord T Transformation Matrix
C 2D Lift Coefficient = lift TE Airfoil Trailing Edge
l qC
Cl0 2D Lift Coefficient at α=0 V Velocity
C 2D Lift-Curve Slope x Streamwise Cartesian Coordinate
lα
CL 3D Lift Coefficient = LqiSft y Lateral Cartesian Coordinate
CM 3D Rolling-Moment Coefficient z Vertical Cartesian Coordinate
CP Pressure Coefficient = P−q∞P∞
g Gravitational Acceleration ≃32 ft ND Non-Dimensionalized
sec2
50 i Coordinate Vectors of Inertial Frame 2D Two Dimensional
2-26 I Moment of Inertia 3D Three Dimensional
1
20 L Lift
4/6. L0 Basic Lift α Angle-of-Attack
DOI: 10.251 LLmEα AMLiifartsfosdiuleLetoadAinngglEed-ogfe-Attack πηθ 3EF.ru1a4lce1t5ri9oA2n6n5og4fl..e.B:laOdrebiRtaald-Piulsan=eRrInclination
g | M Rolling-Moment ρ Air Density =1.3755×10−6 slugs/in3
aa.or M0 Basic Rolling-Moment φ Euler Angle: Right Ascension
p://arc.ai Mnα NRooldliengV-eMctoomreonftBdoudeytForAamngele-of-Attack χχˆ LNeDssoRrootaftχioonral1;V=elomciinty(χ=,1)ωVR
15 | htt qr DRaydniaaml CicoPorredsinsuartee = 12ρV2 ψψc ECuulte-rOAffnAgnleg:leA=rgcousm−1e(n−tχo)f Perigee
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1, 2 R Reference Radius of Blade ω Rotational Velocity
er 2 Rp Radius of Circular Flight Path
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H o I. Introduction
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K
N
E
OL Boomerangs are deceptively simple-looking devices. But make no mistake, boomerang aerodynamics are
K
S rather complex. Not only is the flow unsteady and very three dimensional; it also reverses itself across
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O
H the wing such that the leading and trailing edges swap roles twice every cycle. Furthermore, the Reynolds
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K number is very low, (50,000).
S O
NI
K In spite of these challenges and by trial and error,Australian Aborigines managed to develop boomerang
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T technologysome10,000yearsago;seeLorenz,1 Mauro2 andHanson.3 Infact,severalcivilizationsdeveloped
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LIG straight-flying”boomerangs”forhuntingandwarmanymilleniaago. Agold-tipthrowing-stickwasfoundin
NG King Tutankhamun’s tomb. A boomerang was found in the sand dunes of The Netherlands that dates back
U
K 2,400 years. An exquisite mammoth-tusk throwing stick found in Poland is about 23,000 years old. This
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d b unique artifacthas a mass of about800grams,a spanof almost70cm, a chordof about6 cmand is 1.5 cm
e
ad thick. The ivoryof this relic has a density of about1,900kg-per-cubic-meter. These ancientthrowing sticks
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wnl have aerodynamic and gyroscopic properties that stabilize their flights and extend their range. However, it
o
D was only the Aborigines who createdthe returning boomerangwhich they used primarilyfor recreationand
sport, just as we do today. Indirectly, they also used the returning boomerang to aid in hunting fowl. Here,
they would throw a boomerang over a flock of ducks floating on a pond. The waterfowl would mistake the
boomerang for a hawk and would take flight, yet remain low to avoid the ”predator”. The low-flying game
would then be well within range of hunters waiting nearby. Another unusual, but very utilitarian, use of
the boomerang is when the Aborigines rubbed the leading edge against another piece of wood in a sawing
motion to ignite a fire.
Figure 1 providesa small sampling ofearly20th-centuryAboriginalboomerangs. The boomerangsin this
figure are probably straight-flying configurations. However, it is hard to know without throwing them - it
is possible they are of the returning type. Figure 2 illustrates a wide variety of modern-day boomerangs;
all of these examples are returning designs. Of particular note are the Tri-Fly and LED boomerangs, as
these are utilized by the author for flight tests and further analysis in this body of work. The middle V-
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 2 of 43
type boomerang of the modern Aboriginal group is similar to a design developed herein for the purpose of
high-fidelity analysis using computational fluid dynamics.
A number of boomerang world records as of June 2010 are included; see Wikipedia.4 The most number
of catches made within 5 minutes is 81 by Manuel Schutz, Switzerland, 2005. The quickest five-throw-catch
sequence is 14.6 seconds by Adam Ruhf, USA, 1996. The most number of consecutive catches is 2,251 by
Haruki Taketomi, Japan, 2009. The maximum time aloft in the unlimited class is 380.6 seconds by Billy
Brazelton, USA, 2010. The farthest distance out for a returning flight is 238 meters by Manuel Schutz,
Switzerland, 1999. The longest hand-held throw of any object was set in 2005 by David Schummy when he
threw a boomerang 427.2 meters (over a quarter-mile). All of these are incredible statistics; little did the
Aborigines know just how far their returning sticks could go.
This paper is organizedinthe followingmanner. SectionII introduces Euler’sclassicalorientationangles.
SectionIIIderivestheequationsofangularmotionassociatedwithboomerangdynamics. SectionIVprovides
0 a discussion regarding the kinematics of an over-hand boomerang throw. Section V revisits work done in
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26 the 1960’s regarding the aerodynamics of a boomerang. Section VI develops a new aerodynamic model
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01 for boomerangs based on blade element theory. Section VII develops general formulas for the radius of a
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4/6. boomerang’s circular flight path, as well as the rotation-plane inclination angle. Section VIII develops a
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25 generalconditionforstatictrimandprovesthatdynamicstabilityisanintrinsicpropertyoftheboomerang.
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OI: 1 SectionIXdevelopsatraditionalV-type boomerangdesignforfurther analysis. SectionXprovidesthe two-
D dimensional steady-state CFD results for the boomerang airfoil. Section XI discusses the three-dimensional
g | unsteady CFD results for the boomerang subjected to a fixed cyclic motion. Section XII provides a full 6-
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aa. DOF numericalsimulation of a boomerang in flight. Section XIII is a summary of conclusions. Throughout
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c. the text, references are givenas positive superscripts while footnotes are indicated by negative superscripts.
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p:// Tables of data are embedded within the text while all figures are appended to the end of the paper.
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1,
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mb II. Euler’s Classical Orientation Angles
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c
e
D
n The classicalEuleranglesφ, θ andψ areshowninFigure3. AlsodepictedinFigure3arethe inertialand
o
KTH body-fixed coordinate systems, labeled as (i1,i2,i3) and (b1,b2,b3), respectively. In celestial mechanics,
N the angles(φ, θ, ψ) are usedto specify the orientationofanorbit. They areknownasthe rightascensionof
E
OL the ascending node, the inclination of the orbital plane, and the argument of perigee, respectively. For the
K
GS boomerangproblem,asset-upherein, the axis ofrotationisaboutthe b3 vector,θ describes the bank angle
HO that the plane of rotation makes with the horizon, while the precession rate of b3 about i3 is given by φ˙.
KA Transformingfromtheinertialframetothebodyframecanbeaccomplishedwiththreerotationsinseries.
S
NI Payattentiontotheorderofrotationsasthisiscrucial. Firstrotatebyφaboutthe i3-axis,thenbyθ about
K
E the node vectornandfinally witha rotationofψ aboutthe b -axis. The complete transformationfromthe
T 3
A inertial frame to the body frame using this three-rotation sequence can be described as
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LI
NG Tib =R(ψ;b )R(θ;n)R(φ;i ). (1)
U 3 3
K
by Here, R(α;v) represents an elementary rotation of angle α about vector v. This sequence of rotations is
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de graphically illustrated by Figure 4. Completing the multiplication symbolized by Eqn (1), one obtains the
a
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nl transformation matrix
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cφcψ sφcθsψ sφcψ+cφcθsψ sθsψ
−
Tib = cφsψ sφcθcψ sφsψ+cφcθcψ sθcψ . (2)
 
− − −
sφsθ cφsθ cθ
 − 
 
Here,cindicatescosineandsindicatessine. SinceEqn(2)isapurerotation,itsinverseisjustitstranspose.
Hence, to transform from the body frame to the inertial frame, one uses
cφcψ sφcθsψ cφsψ sφcθcψ sφsθ
− − −
Tbi = sφcψ+cφcθsψ sφsψ+cφcθcψ cφsθ . (3)
 
− −
sθsψ sθcψ cθ
 
 
The next section discusses Euler’s equations of angular motion.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 3 of 43
III. Euler’s Equations of Angular Motion
Inthis section,webeginwiththe equationsofangularmotionasdefinedbyWiesel5 andthenadaptthem
forboomerangdynamics. WestartthediscussionwithaverygeneralizedformofEuler’sequationsofangular
motion,thenprogressthroughaseriesofsimplificationswhichretainonlyessentialelementsoftheequations
needed to describe a level circular flight path typical of a returning boomerang. Further simplifications
regardingmasspropertiesareintroducedwhichappropriatelycontinueto capturethe low-frequencycontent
of dynamic motion for boomerang flight. While this may filter or blur some of the high-frequency motion
within a spin cycle, it greatly reduces the complexity of the ensuing derivations presented throughout this
body of work.
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65 For the boomerang problem, the principal-axis frame naturally aligns with an appropriate body-fixed
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12- coordinate system. This is fortuitous as the equations of angular motion are simplest when expressed in
0
2
6. the principal-axis body frame because the off-diagonal terms of the moment-of-inertia matrix vanish. In
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51 the present work, the originof the principal-axis frame coincides with the center-of-mass of the boomerang.
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10. When the Euler angles are zeroed, the body and inertial frames are aligned; see initial state of Figure 4. In
OI: thisposition,theZ-axisistheaxis-of-rotation,Y isinthespanwisedirectionandX isstreamwise. Here,the
D
g | boomerangreferenceplanecoincides withthe X-Y plane whereZ =0, the boomerang’scenterline coincides
a.or with the X-axis, the right wing resides in the half-space where Y 0 and the boomerang upper-surface
c.aia resides in the Z 0 half-space. For this system, Euler’s equations of≥angular motion are given as:
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2015 | http:// MM21 == II2121 ωω˙˙21++((II1313−−II3232))ωω12 ωω33 (4)
er 21, M3 = I33 ω˙3+(I22−I11)ω1 ω2.
b
m Here, the ω-vector describes the angular rates in the principal body-axis frame. These equations provide
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c
De the time-rate-of-change of the ω-vector, yet only constitute one half of the rotational equations of motion.
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o Another set of equations is needed which link the angular rates of the body frame to the orientation angles
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KT of the body in the inertial frame. The remaining set of equations is given as:
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LE ω = φ˙ sinθ sinψ+θ˙cosψ
O 1
K
GS ω2 = φ˙ sinθ cosψ θ˙ sinψ (5)
O −
A H ω3 = ψ˙ +φ˙ cosθ.
K
S
NI These equations transform the ω-vector given by Eqn (4) to provide the time-rate-of-change of the Euler
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TE angles (φ˙,θ˙,ψ˙). In practice, these angular velocities are used to update the orientation angles (φ,θ,ψ) for
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G the body as a function of time.
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G
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U Axi-Symmetric Body
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y
b
d Ifthe bodyis axi-symmetricaboutoneofthe principalaxes,thenfurthersimplificationscanbe madeto the
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oa equations of angular motion. So how can a boomerang be axi-symmetric? This question is best answered
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w if one views the boomerang when it is spinning fast, rather than when statically displayed. Just as this
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motion blurs the visual of the boomerang to appear as a spinning disk, the moments of inertia can be time-
averaged in the X-Y plane. Although this simplification alters high-frequency content, it retains sufficient
mass properties to accurately describe low-frequency boomerang dynamics.
For the purpose of this discussion, let’s assume that the body is symmetric about the b -axis and a
3
rolling-moment exists purely about the n-axis. Let’s designate this torque about the nodal line as M . See
n
Figure 3 to get familiar with this situation. Note that the n-axis is perpendicular to the b -axis and thus
3
can be described as a linear combination of b and b .
1 2
n=cosψ b sinψ b (6)
1 2
−
Also define an s-axis which is perpendicular to both the n-axis and the b -axis.
3
s=sinψ b +cosψ b (7)
1 2
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 4 of 43
Note that the s-axis is not shown in Figure 3.
Because of symmetry, let’s define a moment of inertia (I ) that is constant for any radial in the (n,s)
rr
plane. Hence,
I I =I . (8)
rr 11 22
≡
Combining Eqns (4-8) gives:
M = I θ¨ + (I I )φ˙2sinθ cosθ + I ψ˙ φ˙ sinθ
n rr 33 rr 33
−
M = I φ¨sinθ + 2I θ˙φ˙ cosθ I θ˙ ψ˙ +φ˙ cosθ (9)
s rr rr 33
−
d (cid:16) (cid:17)
M = I ψ˙ +φ˙ cosθ .
3 33
dt
(cid:16) (cid:17)
Axi-Symmetric Disk
0
5
6
2 Ifanaxi-symmetricbodyisalsoplanarandhomogeneous(i.e.,adisk),thentheprincipalmomentsofinertia
2-
01 are simply:
2
6. 1
2514/ I33 = 2mR2 =2Irr, (10)
0.
1 where m is the mass of the disk and R its radius. Now, the equations of angular motion reduced to:
OI:
D 1
g | Mn = mR2 θ¨+φ˙2 sinθ cosθ+2ψ˙ φ˙ sinθ
or 4
aa. 1 h i
c.ai Ms = mR2 φ¨sinθ 2θ˙ψ˙ (11)
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p:// 1 hd i
5 | htt M3 = 2mR2 dt ψ˙ +φ˙ cosθ .
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0 h i
1, 2 For the moment, let’s assume that the angular motion is such that θ is constant, then θ˙ = θ¨= 0. Now,
2
er if the rolling-moment is purely about the n-axis, then Ms = 0. Let’s also assume that there is no moment
b
m being applied about the spin axis, i.e., M = 0. Under these conditions, the equations of angular motion
ce 3
De reduce further to:
n
TH o Mn = 1mR2 φ˙2 sinθ cosθ+2ψ˙ φ˙ sinθ
K 4
LEN 0 = φ¨ h i (12)
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SK 0 = ψ¨.
G
O
H One may recognize these equations as those that describe the motion of an inclined toy gyroscope, with
A
K its tip resting on a table. Here, the high spin rate of the gyroscope disk remains at a constant ψ˙, with the
S
NI gyroscope slowly rotating around its pivot-point-vertical-axis with a constant precession rate of φ˙, all the
K
E
A T while, maintaining a constant inclination angle θ. This precessional motion is driven by the moment Mn
G createdby the gyroscope’sweight, multiplied by the lateraloff-set distance between the pivot point and the
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G gyroscope’s center-of-mass. For boomerang dynamics, the moment (M ) is supplied aerodynamically.
N n
KU If Mn, θ and ψ˙ are known, then φ˙ can be determined using the quadratic formula applied to Eqn (12).
y
b
aded φ˙ = 1 ψ˙ ψ˙2+ 4cosθMn . (13)
wnlo cosθ "− ±r sinθmR2 #
o
D From experience, we know that the direction of the precession is the same as that of the disk spin. Hence,
the positive branch of Eqn (13) is physically correct. Further, if φ˙ ψ˙, then solution of Eqn (13) can be
≪
approximated by:
φ˙ =. 2Mn = Mn , (14)
sinθmR2ψ˙ sinθI ψ˙
33
where the second form is included for the more general case of constant θ motion. Eqn (14) can be directly
derivedfromEqn(12)bydroppingtheφ˙2 term;the intermediatestepofEqn(13)isincludedasaformality.
Eqns (12-14) will be used to develop a simplified analysis of a boomerang’s flight dynamics to determine
the radius of its circular flight path as well as other pertinent relationships.
The next section reviews the kinematics of an over-hand boomerang throw.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 5 of 43
IV. Kinematics of a Boomerang Throw
Abriefdiscussionregardingthe kinematicsofaboomerangthrowisincludedinthis section. Theprimary
objectiveistounderstandhowtheinitialconditionsoftheboomerang’skineticsaresomewhatpredetermined
by the kinematic motion of an over-handthrow.
Consider holding a basketball in the cup of your hand and throwing it under-handed as if bowling. The
natural motion associated with this toss has the basketball rolling off the fingertips just as the ball is being
released. This establishes both a forward velocity as well as a rotational velocity for the basketball. When
the basketball makes contact with the ground, its spin rate will remain essentially unchanged. Hence, the
natural kinematics of bowling a basketball creates an initial relationship between the ball’s velocity and its
spin rate.
.
ωR=V (15)
0
5 Here, V is the translational velocity of the basketball, ω is its rotational velocity and R its radius. While
6
2
2- the relationship defined in Eqn (15) is only an approximate one and the person bowling the basketball can
1
0
2 bias it one way or the other, it is quite difficult to violate it entirely.
6.
14/ Let’s now consider a boomerang throw and draw similarities with the above example. To start, one holds
5
2
0. a blade of the boomerang near its tip, grasping it in the palm of his hand. Because of the width of the
1
OI: hand, the center of this grasp is located around 85%-90% of the blade’s radius. Traditionally, a V-type
g | D configuration is held by the lifting blade, with the dingle blade-1 pointed forward at the far end of the
or boomerang. The boomerang throw proceeds in a manner analogous to throwing a small axe or a hatchet
a.
a
ai over-handed. The plane of rotation is nearly vertical, with only a slight lay-over angle. Hence, the axis of
c.
ar rotationis nearlyhorizontal. In termsofEuler’sorientationangles,this correspondsto aninitialinclination
p://
htt angle of the rotation plane in the range:
015 | 70◦ ≤θ0 ≤80◦. (16)
2
1, Since the boomerang blade must be firmly held, it does not roll out of the hand as efficiently as does
2
ber the finger-tip roll of the basketball. Hence, the relationship between the initial translational and rotational
m
ce velocities is slightly different than before. In the case of the boomerang throw, we draw from experience,
e
D
n make an educated guess, and assume this relationship to be:
o
H
T .
K ω R=0.85V . (17)
N 0 0
E
L
O Here, ω and V are the boomerang’s initial rotational and translationalvelocities, respectively. Also define
K 0 0
S
G theinitialrotationalvelocity(ω )ofEqn(17)inanondimensionalizedform;thisformwillbeusedthroughout
O 0
H the body of this work.
A
KNISK χ0 ≡ ωV00R =. 0.85 (18)
E
T
A The initial condition of Eqns (17-18) has the forward-going tip moving at 1.85V0 and the bottom tip also
G
LI moving forward but only at a speed of 0.15V0. In practice, a typical throw of a boomerang will yield a
G
N translational speed of about 60mph, with a rotational cadence of about 10Hz.
U
K
y
b
d Analysis of a Night Flight
e
d
a
o
nl In order to verify the assumption made by Eqns (17-18), we analyze a strobed image of a boomerang in
w
Do night flight;6 see Figures 5-6. In Figure 5, note the insert at the bottom-right corner. This insert shows
a close-up view of a three-bladed boomerang with an LED strobe light located on each blade at 75% of
its radius. Also in Figure 5 is a red rectangular box that traps a portion of the flight path approximately
perpendicular to the camera’s view. This segment of the flight path is depicted in Figure 6 and is further
analyzed. Since there are three strobed paths, focus on the one centered about the solid bold vertical aqua
line. To extract the relationship between translational and rotational velocities of the boomerang at this
stage of its flight, a hypocycloid is fit throughthe strobed curve of interest. The result is shown as the bold
gray line that fits the strobed-light path. The hypocycloid corresponding to this path is graphically given
by the very-large-radius circular contact surface depicted by the orange line at the bottom of the image
and the rolling wheel given by the yellow circle. The green circle in this figure represents the swept area of
the boomerang while the gray circle coincides with the location of the LED light. Finally, the ratio of the
-1Thedinglebladeistheblademost-influencedbythewakeoftheotherforaV-typeboomerang.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 6 of 43
radii of the gray-to-green circles provides a derived relationship of rotational-to-translational velocities for
this flight segment. In this particular case, the initial condition of Eqns (17-18) is exactly verified. In the
author’s experience, the translationalvelocity decays slightly faster than does the rotationalvelocity during
the course of a boomerang flight. Hence, the nondimensionalized rotational velocity (χ) of a boomerang
increases during the course of its mission, from start to finish, so we have:
dχ
>0. (19)
dt
ThispropertywillbeusedinSectionVIItoexplainwhytheinclinationangleofaproperly-designboomerang
reduces during flight.
Inducing a High-Spin-Rate Flight
0
5
6 Due to the natural kinematics of a boomerang throw, it is almost impossible to induce an extremely-high
2
12- rotationalvelocityatpointoflaunch. However,ifonethrowsaboomerangslightlydownward(likethrowing
0
2
6. a skipping stone) and bounces it off the ground, a high-spin-rate flight can be achieved. In essence, some of
4/
51 the translationalkinetic energygets impulsively convertedto rotationalkinetic energy. To establisharough
2
10. order-of-magnitude estimate, assume that the instantaneous bounce event conserves kinetic energy. Using
OI: the moment of inertia of a disk from Eqn (10), the states before and after the bounce are related as:
D
g |
or 1 1 . 1 1
a. mV2+ Iω2 = mV2+ Iω2
aia 2 0 2 0 2 1 2 1
http://arc. 2V022++(χω20RV)22 ==.. 22V+12χ+2(ωV12R.)2 (20)
5 | 0 0 1 1
1
0
1, 2 where R is the radius of the boomeran(cid:2)g, and(cid:3)subscripts(cid:2)0 and(cid:3)1 indicate the states before and after the
er 2 bounce event, respectively. With some rearrangement,substitution of Eqn (17) into Eqn (20) gives:
b
m
ce 2
n De ω1 2 . 2.7225−2 VV10
o = (21)
H ω  0.7225(cid:16) (cid:17) 
KT (cid:18) 0(cid:19)
N  
OLE 2 2.7225 V0 2 2
SK χ1 =. V1 − . (22)
HOG (cid:18)χ0(cid:19)  0.(cid:16)7225(cid:17) 
A  
SK  
NI Hence, if the bounce event reduces the translational velocity (V) by 50%, the rotational velocity (ω) will
K
E increasebyabout75%. Howeverinnondimensionalizedterms,theratiobetweenrotationalandtranslational
T
GA velocities change by a factor of about 3.5 due to the impulse of the bounce, to give χ1 =. 3.0.
GLI Thenextsectionprovidessomebackgroundonworkdoneinthefieldofboomerangflightdynamicsnearly
N
U 50 years ago.
K
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b
d
e
d
a
o
nl
w
o
D V. Felix Hess Model
In the 1960’s, the aerodynamics of boomerangs was studied by Felix Hess.7 Hess concluded that for
sufficientlyrapidrotation,thetime-averagedliftandrolling-momentofanadvancingboomerangaredirectly
proportional to ω2 and ωV, respectively. Hence,
L¯ =aω2 , M¯ =bωV, (23)
where a and b are constants of proportionality, ω is the rotational speed of the boomerang, and V is its
flight speed. Note that the coefficients a and b lump together both aerodynamic characteristics and mass
properties of a given boomerang. The units of a and b are lbs-sec2.
Hessutilizedhisaerodynamicmodel,givenbyEqn(23),inconjunctionwiththegyroscopemodelgivenby
Eqn(11),todeveloprepresentativeflightpathsforafamilyofboomerangs,simplybyvaryingthecoefficients
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 7 of 43
aandb. AdeficiencyofHess’aerodynamicsmodelisthatitcannotselftrim. Further,ifmanuallytrimmed,
it is neutrally stable. However from practice, we know that boomerangs are dynamically stable over a wide
varietyofshapesandflightconditions;theremustexistaninherentself-trimmingcapabilityintheirdesigns.
Hence, further workis needed to developan aerodynamics model that can accommodate this characteristic.
For easier comparison later, Hess’ model, given by Eqn (23), is reformulated. In the current work, the
rotational speed (ω) is nondimensionalized as follows.
ωR V
tip
χ= = , (24)
V V
where R is the radius of boomerang blades, and V is the blade tip speed due to rotation. We also refer to
tip
χ as the tip-speed ratio. Substitution of Eqn (24) into Eqn (23) gives Hess’ aerodynamic model in terms of
χ as:
a b
50 L¯ = χ2V2 , M¯ = χV2, (25)
26 R2 R
2- (cid:16) (cid:17) (cid:18) (cid:19)
01 where the bracketed terms are now the constants of proportionality.
2
4/6. Time-averagedrelationships for lift and rolling-moment will be significantly enhanced in the next section
1
25 using Blade Element Theory.
0.
1
OI:
D
g |
or VI. Blade Element Theory
a.
a
ai
c.
ar An aerodynamic model is developed herein using Blade Element Theory (BET). This model uses the
http:// lift-curve of a boomerang’s two-dimensional airfoil section and it is assumed that the boomerang operates
5 | at angles-of-attack close to zero.-2 Hence, the relationship between lift and alpha is essentially linear.
1
0
2
21, Cl(α)=Cl0+αClα. (26)
er
b
m Here, α is the angle-of-attack of the airfoil, C is its lift coefficient at α=0 and C is the slope of the lift
e l0 lα
c
De curve. These values will be determined numerically by running computational fluid dynamics (CFD) on a
on representative airfoil section of the boomerang blade over a range of angles-of-attack.
H
T Figure 7 provides a top-view schematic of a single, non-tapered blade that rotates about the Z-axis in
K
N a counter-clockwise manner. In this figure, R is the radius of the blade, C is its chord and the freestream
E
L
O velocity(V) is alignedwiththe X-axis. Also inthis schematic,ψ is the angularpositionof the blade, where
K
GS ψ = 0 represents the position when the advancing blade is perpendicular to the freestream flow. Vn is the
O
H local normal on-set velocity component to a blade section; it is a function of radial and angular positions
A
K and is given by:
S
NI V (r,ψ)=V cosψ+ωr. (27)
K n
E
A T Here, ω = ψ˙ is the rotational speed of the blade spinning about the Z-axis. In the present work, this
G
LI rotational speed is nondimensionalized as follows.
G
UN ωR
K χ= . (28)
y V
b
d
ade Figures 8-10 illustrate how Vn varies within the disk area swept by the blade for rotational speeds of
o
nl χ = (0.7, 1.0, 1.5), respectively. In these images, the freestream velocity is from left-to-right. Note that
w
Do Vn < 0 is depicted by the blue-shaded regions which only occur when 90◦ ψ 270◦. As an aside, the
≤ ≤
V =0 contour is a circle of diameter 1.
n χ
In general, the rotational plane of the blade can be at an angle-of-attack (α) to the freestream velocity
(V). The velocity component normal to the rotational plane (V ) can be approximated as:
z
.
V =V sinα=αV , α 1. (29)
z
| |≪
Asthebladesweepsinψ,theairfoilsectionsalongitsspanaresubjectedtoacyclicvariationofbothnormal
velocity (V ) and apparent angle-of-attack. Designate this effective angle-of-attack as α and approximate
n n
with:
α =tan−1 Vz =. Vz , V V . (30)
n z n
V V | |≪| |
(cid:18)| n|(cid:19) | n|
-2Actually,verylargeangles-of-attackexist,butonlyatverylowdynamicpressures,sotheerrorintroducedisminimal.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 8 of 43
The reason the absolute value of V is needed above is because, even when V is negative, the sign of lift
n n
due to angle-of-attackis solely determined by the sign of V . As will be seen later, this complicates matters
z
in the derivations for lift and rolling-moment due to angle-of-attack.
While all boomerangs have more than one blade, with blade element theory each are treated identically;
there is no distinction between the force and moments of the lifting and dingle blades. As a consequence,
derivingthecoefficientsofliftandrolling-momentforasinglebladeisequivalenttodoingsoforanN-bladed
configuration. For example, a 3-bladed boomerang will have 3 times the lift of one of its blades but its lift
will also be nondimensionalized by 3 times the reference area of a blade.
Lift Derivation
The next step is to developa relationshipfor the dimensionallift generatedby the blade as a function of its
angular position (ψ).
0
5
6
2
2- R 1
4/6.201 L(ψ) = Z0 2ρVn2Cl(αn)Cdr
51 1 R V
OI: 10.2 = 2ρCZ0 Vn2 (cid:20)Cl0+ |Vnz|Clα(cid:21) dr
g | D = 1ρC R V2C + V V C dr. (31)
aa.or 2 Z0 n l0 | n| z lα
ai (cid:2) (cid:3)
p://arc. gHievreen, tbhyeEinqtnegr(a2t6i)o,nbiustovneorwthaesspaafnunofcttihoenbolafdαe, ρainsdaiCr diesnscihtoy,rdCll(eαnng)thi.s thEeqn2sD(s2e6c-t3io0n)ahlalivfet caolseofficbieeennt
htt n
5 | incorporatedinto Eqn (31). By observation,note that the total lift of Eqn (31) can be split into a basic lift
1
20 and a lift due to angle-of-attack. To help keep the remaining derivations manageable, we will develop these
21, lifts separately, then combine later.
er
b
m
e
ec Basic Lift
D
n
o
H In the context of this work, basic lift is defined as that which is generated by an airfoil acting at no angle-
T
K of-attack to its onset flow. Starting with the first term of the integrand of Eqn (31), and substituting
N
LE Eqns (27-28), the basic-lift derivation continues as:
O
K
GS 1 R
HO L0(ψ) = ρCCl0 [V cosψ+ωr]2dr
KA 2 Z0
KNIS = 1ρV2CCl0 Rcos2ψ+2cosψχ r +χ2 r 2 dr
E 2 R R
GA T 1 Z0 (cid:16)1 (cid:17)R (cid:16) (cid:17)
LI = ρV2RCC cos2ψ+cosψχ+ χ2
UNG 2 l0(cid:20) 3 (cid:21)0
d by K = qSCl0 cos2ψ+cosψχ+ 13χ2 . (32)
de (cid:20) (cid:21)
a
o
wnl where q = 1ρV2 is the freestream dynamic pressure and S =RC is the blade planform area.
o 2
D To determine a time-averaged value for the basic lift, L (ψ) is integrated over ψ. However, since L is
0 0
symmetric about the Y-axis, this integration can be performed from 0 to π.
1 π
L¯ = L (ψ)dψ
0 0
π
Z0
π
1 1 1
= qSC (ψ+sinψcosψ)+χsinψ+ χ2ψ
l0
π 2 3
(cid:20) (cid:21)0
1 1
= qSC + χ2 (33)
l0
2 3
(cid:20) (cid:21)
ComparingEqn(33)withEqn(25)illustratesthatHess’aerodynamicmodelisinsufficientforsmallrotational
speeds where χ<1.5, even for basic lift.
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 9 of 43
Special Integration Topic
Before proceeding to the derivation of the lift due to angle-of-attack,it will be necessary to briefly touch on
a special integration topic. Consider a line segment on the interval [0,1] of the form y(x) = a+bx. If this
segment is either completely above or below the X-axis, then an integration of its absolute value is trivial.
However, let’s consider the situation where the line starts below the X-axis at x = 0 and ends above it at
x=1. For this case, it can be shown that:
1 a2 b
a+bx dx= +a+ , a 0 a b. (34)
| | b 2 ≤ ≤| |≤
Z0 (cid:20) (cid:21)
Here, a and b are arbitrary constants. Note that the first term in the square brackets of Eqn (34) is an
addition to the result one would get if the integrand had not been made an absolute value.
0
65 Lift due to Angle-of-Attack
2
2-
1
20 We now proceed with the derivation of lift due to angle-of-attack, starting with the second term of the
6.
4/ integrand of Eqn (31).
1
5
2
10. 1 R
org | DOI: Lα(ψ) = 12ρCZ0R|Vn|VzClαdr
c.aiaa. = 2ρCZ0 |V cosψ+ωr|αV Clαdr
p://ar = 1ρV2αC C R cosψ+χ r dr
htt 2 lα R
5 | Z0 (cid:12) (cid:16) (cid:17)(cid:12)
01 1 (cid:12) (cid:12)
21, 2 = qSαClα |cosψ(cid:12)+χη| dη (cid:12) (35)
er Z0
mb Here, η = r is the nondimensional spanwise coordinate along the blade. Because of the absolute-valued
ce R
De term inside the integral of Eqn (35), the following four cases exist for 0 ψ π.
on For 0 ψ π, and any value of χ>0, the term [cosψ+χη] 0 ove≤r the≤interval of integration, so:
H ≤ ≤ 2 ≥
T
K χ
N L (ψ)=qSαC cosψ+ . (36)
LE α lα 2
SKO For π <ψ π, and χ 1, the term [cosψ+χη] swithches sign, sio we apply Eqn (34) to get:
G 2 ≤ ≥
O
KA H Lα(ψ)=qSαClα cos2ψ +cosψ+ χ . (37)
S χ 2
NI (cid:20) (cid:21)
K
E For π <ψ cos−1( χ), and χ<1, the term [cosψ+χη] also switches sign, and again we have:
A T 2 ≤ −
LIG cos2ψ χ
NG Lα(ψ)=qSαClα χ +cosψ+ 2 . (38)
KU (cid:20) (cid:21)
by For cos−1( χ)<ψ π, and χ<1, the term [cosψ+χη]<0 over the interval, so:
d − ≤
e
ad χ
wnlo Lα(ψ)=qSαClα −cosψ− 2 . (39)
o
D h i
In order to determine a time-averaged value for lift due to angle-of-attack, L (ψ) is integrated over ψ
α
from 0 to π. It is possible to do this considering only two cases; one with χ 1 and the other with χ < 1.
≥
However, this does require a little manipulation of Eqns (36-39) when folded into the integration over ψ.
For χ 1,
≥
1 π χ π cos2ψ
L¯ = qSαC cosψ+ dψ+ dψ
α lα
π (Z0 h 2i Zπ2 (cid:20) χ (cid:21) )
π π
1 χψ 1
= qSαC sinψ+ + (ψ+sinψcosψ)
lα
π 2 2χ
((cid:20) (cid:21)0 (cid:20) (cid:21)π2)
χ 1
= qSαC + . (40)
lα
2 4χ
(cid:20) (cid:21)
Vassberg, Aerodynamics Lecture, New Orleans, Louisiana, June 2012 10 of 43