Transcript paper airplanes aerodynamics

Particle Motion
Refresher on translational kinematics
and kinetics.
Linear momentum, Newton’s laws,
coordinate frames.
Examples from aircraft performance,
orbital mechanics, and launch vehicle
trajectories.
Newton’s Second Law
~f = m~a
• Applicable to mass particles with constant
mass and zero volume – extensible to rigid
bodies
• Not as simple as it looks
– Determination of the force usually requires
free-body diagram – can be extremely
complicated
– Determination of the acceleration sometimes
involves complicated kinematics
– Acceleration is second derivative of position
vector with respect to time
Reference Frame & Position Vector
^
~r = x^i + y^j + z k
Position vector
must be from
inertial origin to
mass particle
^
k
z
~r
x
^i
m
y
^j
If expressed in
terms of inertial
frame components,
then differentiation
is easy
Inertial Origin
• An inertial origin is a point that is not
accelerating with respect to any other
inertial origin
– Alternatively, an inertial origin is a point for
which Newton’s laws are applicable
– There is no known inertial origin, but for most
problems an origin can be found that is
“inertial enough”
– For some problems, an Earth-fixed reference
point is sufficient, whereas for others, the
rotation of the Earth must be considered
Inertial Reference Frame
• An inertial reference frame is a set of three
unit vectors that are mutually
perpendicular, with their origins at a single
inertial origin, and whose directions
remain fixed with respect to inertial space
– Alternatively, an inertial reference frame is a
frame for which Newton’s laws are applicable
– Usually the d&c analyst must determine the
simplest frame that is “inertial enough”
– What about these vehicles?
• Paper airplane, Cessna 152, B2, container ship,
Space Shuttle, Apollo 11, Cassini, Rama
Reference Frames
• A reference frame is a set of three mutually
perpendicular (orthogonal) unit vectors
• Typical notations include
ˆi ˆj kˆ , Iˆ Jˆ K
ˆ , eˆ eˆ eˆ , bˆ bˆ bˆ
1 2 3
1 2 3
• Typical reference frames of interest for vehicles
include
–
–
–
–
–
–
ECI (Earth-centered inertial)
Perifocal (Earth-centered, orbit-based inertial)
ECEF (Earth-centered, Earth-fixed, rotating)
Orbital (Earth-centered, orbit-based, rotating)
Wind (vehicle-centered, rotating)
Body (vehicle-fixed, rotating)
Earth-Centered Inertial (ECI)
• Also called “Celestial
Coordinates”
• The I-axis is in vernal
equinox direction
• The K-axis is Earth’s
rotation axis,
perpendicular to
equatorial plane
• The J-axis is in the
equatorial plane and
finishes the “triad” of unit
vectors
• The IJ-plane is the
equatorial plane
K̂
Ĵ
Î
Towards Sun at
vernal equinox
Perifocal Frame
• Earth-centered, orbitbased, inertial
• The P-axis is in periapsis
direction
• The W-axis is
perpendicular to orbital
plane (direction of orbit
angular momentum

vector, r  v)
• The Q-axis is in the
orbital plane and
finishes the “triad” of
unit vectors
Q̂
Ŵ
P̂
Q̂
P̂
Orbital Frame
• Similar to “roll-pitchyaw” frame, for
spacecraft
• The o3 axis is in the
nadir direction
• The o2 axis is in the
negative orbit normal
direction
• The o1 axis completes
the triad, and is in the
velocity vector direction
for circular orbits

v
ô1

 r ô3
ô 2
 ŵ
Body-Fixed Frame
• Applicable to any type
of vehicle
• Typically denoted using
“b” unit vectors
For spacecraft:
• The b3 axis is in the
nadir direction
• The b2 axis is in the
negative orbit normal
direction
• The b1 axis completes
the triad, and is in the
velocity vector direction
for circular orbits
ô1
b̂1
b̂ 3
ô3
b̂ 2
b̂1
ô1
b̂ 3
ô3
b̂ 2
ô 2
ô 2
Vector, Frame, and Matrix Notation
Symbol
n ~v o
^i ; ^i ; ^i
1
2
F
i
no
^i
vi
vb
Rbi
µ
~
!
~ bi
!
! bi
b
! bi
a
3
Meaning
vector, an abstract mathematical object with direction and length
the three unit base vectors of a reference frame
n
o
the reference frame with base vectors ^i1 ; ^i2 ; ^i3
a column matrix whose 3 elements are the unit vectors of Fi
a column matrix whose 3 elements are the components
of the vector ~v expressed in Fi
a column matrix whose 3 elements are the components
of the vector ~v expressed in Fb
rotation matrix that transforms vectors from Fi to Fb
a column matrix whose 3 elements are the Euler
angles µ1 , µ2 , µ3
an angular velocity vector
angular velocity of Fb with respect to Fi
angular velocity of Fb with respect to Fi expressed in Fb
angular velocity of Fb with respect to Fi expressed in Fa
Position, Velocity and Acceleration Vectors
^
~r = x^i + y^j + z k
d
^
~v = ~r = ~r_ = x_^i + y_^j + z_ k
dt
^
~a = ~v_ = x
Ä^i + yÄ^j + zÄk
^
k
z
~r
x
^i
m
y
^j
Derivatives are
simple because
unit vectors are
constant, in
direction and
magnitude
Application of
~f = m~a
Need to know the components of the force vector
~f
)
^
= fx^i + fy^j + fz k
³
´
^
= m x
Ä^i + yÄ^j + zÄk
fx
fy
fz
= mÄ
x
= mÄ
y
= mÄ
z
Three second-order ordinary
differential equations.
Linearity and coupling depend
on nature of forces.
Simple Feedback Control Forces
Suppose the ¡
forces take
¡ the following form
fx
=
fy
=
fz
=
kpx x
kdx x_
pz
dz
¡k y ¡ k y_
py
dy
¡k z ¡ k z_
Here the “k” terms are feedback gains, whose values are
selected
kp by the d&c analyst.
The k terms are proportional to the position errors, and
d
the
terms are proportional to the velocity errors, or to
the derivatives of the position errors, hence the controller is
a “PD” controller
Equations of Motion with PD Control
First, define some new variables, termed the states
x_ 1 = x4
x1 = x
x2 = y
x_ 2 = x5
x3 = z
x_ 3 = x6
1 ¡
x4 = x_
x_ 4 = fx =m =
( kpx x1 ¡ kdx x4 )
m
x5 = y_
1 ¡
x_ 5 = fy =m =
( kpy x2 ¡ kdy x5 )
x6 = z_
m
#
1 ¡
¡k x )
x
_
=
f
=m
=
(
k
x
6
z
pz 3
dz 6
x 2 R6
m
x_ = f (x)
x is the state vector x_ = f (x; x0 ; k)
)
Linear System of Equations (PD Control)
• Because of the unique form of these
equations, they can be written in matrix
form
– Each “right-hand side” appears as a summation
of constants multiplying states, with the states
always appearing linearly
– Thus, the equations comprise a system of
linear, constant-coefficient ordinary differential
equations
x_ = f (x) = Ax
Linear System (PD) (continued)
Simple rearrangement of the equations of motion
leads to the block matrix form:
2
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
6
6
6
x_ = 6
¡k =m
6 ¡k =m
6
px
dx
¡
¡k =m
4
0
kpy =m
0
dy
¡
¡k =m
0
0
kpz =m
0
0
dz
·
¸
0
1
=
¡ diag(k )=m ¡ diag(k )=m x
p
d
= Ax
Because A is constant, the system is easily solved:
x(t) = eAt x0
3
7
7
7
7x
7
7
5
PD Example Plots
The nature of the PD controller is that it
causes all of the states to approach zero
asymptotically; thus it is a stable controller
Note poor quality of these graphs. Not acceptable
for technical presentations. How to improve?
PD Example Exercise
• The lecture notes include the Matlab code
for numerical integration of this system of
equations
– Implement the Matlab code and carry out
some simple experimentation with varying
gains and initial conditions
– Compare the numerically integrated solution
to the exact solution on the previous slide
• We will return to the closed form solution
later in the course
Polar Coordinates, Rotating Reference Frame
~r = r e
^r
~v = r^
^_ r
_ er + r e
Convenient to visualize
^r
e
^µ
e
^j
~r
µ
frame origin at particle, but
position vector must be
from inertial origin.
Unit vectors of rotating
reference frame are constant
in length, but not in
m
direction
^i
Unit vectors are orthogonal,
in spite of distortion that
appears in figure!
Rate of Change of the Unit Vectors
• The unit vectors change because the angle 
changes
• Denote small change in time t and corresponding
change in angle  with t and 
• Consequently, the required derivatives can be
written as
¢^
er
¢µ
_e
_e
^r = lim
^µ = µ^
= lim
e
µ
¢t!0 ¢t
¢t!0 ¢t
_e
^_ µ = ¡µ^
e
r
Exercise: convince yourself
~v = r_ e
^r + rµ_e
^µ
³
´
³this limit is correct
´
~a =
^r + rµÄ + 2r_ µ_ e
^µ
rÄ ¡ rµ_2 e
Orbital Mechanics Application
• When the net force is easily decomposed into
radial and transverse components, polar
coordinates are quite useful
• The most common example is the simple twobody problem, where the force is the inversesquare universal gravitational law
m
~r
M
~f = ¡ GMm e
^r
r2
where G is the universal
gravitational constant
Simple Orbit Example
Suppose we select units suchmthat GM=1; then the
¡ e
~
^r
f
=
force is
r
r2
Application of Newton’s 2nd law leads to the two
scalar nonlinear ordinary differential
1 equations:
rÄ ¡ rµ_2
= ¡
rµÄ + 2r_ µ_
= 0
r2
Why are these equations nonlinear?
Exercise: derive these equations
Rewrite as
x_ = f (x)
Define new state variables
x_ 1 = x3
x1 = r
x2 = µ
x_ 2 = x4
)
x3 = r_
x4 = µ_
#
x 2 R4
1
¡
x_ 3 = x1 x2
4
x2
1
2x3 x4
¡
x_ 4 =
x1
x_ = f (x)
Note the nonlinearities, as well as the singularity
for zero radius
Orbital Mechanics Example (cont)
• We can identify a special solution to the
equations of motion
• Suppose x1=constant (circular orbit), then x3=0,
so x4=constant also (constant angular rate)
• From the d.e. for x3 (which must be zero), we can
x3 x2 = 1
deduce that
1 4
• This result is a form of Kepler’s Third Law, which
relates the orbital period to the orbital radius –
the “1” is a result of our choice of units such that
GM=1
Exercise: verify the equivalence between
this result and Kepler’s Third Law
Some Numerical Results
Plot generated using Matlab code from lecture notes.
Particle Motion
… so far …
f=ma
Reference frames, inertial origins and
frames
Position, velocity, acceleration
State vector form of EOM
“PD” control example
Rotating reference frame, polar
coordinates
Simple orbital motion
Normal-Tangential Coordinates
• For a circular orbit, the radial-transverse frame
and coordinates are also normal and tangential
to the spacecraft flight path, respectively
• We can also define a frame that has these
properties for non-circular orbits, but the
gravitational force is more complicated in such a
frame
• This “normal-tangential” frame is well-suited for
problems involving aircraft trajectories, since lift
and drag are aerodynamic forces defined in the
normal and tangential directions, respectively
Normal-Tangential Coordinates
~v = v e
^t
~a = v_ e
^t + v e
^_ t
For aircraft, the velocity
^n
e
^j
m
appears in the aerodynamic
forces, and these forces are
defined in the normal and
tangential directions
^t
e
µ
Unit vectors of rotating
reference frame are constant
in length, but not in
~r
direction
^i
Flight path
Rate of Change of the Unit Vectors
• The unit vectors change because the angle 
changes
• Denote small change in time t and corresponding
change in angle  with t and 
• Consequently, the various derivatives can be
written as
^_ t
e
^_ n
e
~v
~a
=
=
=
=
^n
µ_e
¡µ_e
^t
^t
ve
_e
v^
_ et + v µ^
n
Exercise: convince yourself …
Note that if  is increasing, then the
radius of curvature of the flight
^n path
e
emanates from the positive
direction, whereas if  is decreasing,
then the radius of e
emanates
^curvature
n
from the negative direction.
The normal acceleration (and hence
the net normal force) is in the same
sense.
Normal-Tangential Coordinates
• The speed can generally be written in terms of
_ and the radius of curvature as
thevangular
= rc µrate
• Hence the acceleration vector can be written as
~a = v^
_ et +
v2 e
^n
rc
• This acceleration vector can be used in a
straightforward way in developing the equations
of motion for an aircraft modeled as a point
mass moving in a vertical plane
Simplest Aircraft Translational Motion
The four forces:
Lift L, perpendicular to flight path
Drag D, parallel to flight path
Weight W, toward center of Earth
Thrust T, generally inclined wrt flight path
^n
e
L
aT T
D
W
a

^t
e
Application of Newton’s 2nd Law
^t : T cos ®T ¡ D ¡ W sin µ = mv_
e
v2
¡
^n : T sin ®T + L W cos µ = m
e
rc
• The simplest problem is the case of straight, level,
) rflight,
1
non-accelerating
for aT=0
=
c
– Straight ) µ = 0
– Level
– Equations simplify:
¡
^t : T D
e
^n : L ¡ W
e
= 0
= 0
Thrust Required for Straight, Level Flight
• Solve the thrust=drag, lift=weight
equations
obtain required thrust
T =to
W=(L=D)
R
• Obvious conclusion: required thrust is
minimized when L/D is maximized
• Further analysis requires stating lift and
drag in terms of velocity and dynamic
pressure
q = 1 ½v 2
2
• Dynamic pressure:
where  is atmospheric density
Lift and Drag
• The lift force is
written in terms of
dynamic pressure, a
characteristic area,
and a dimensionless
variable called the lift
coefficient
L = qSCL
• Similarly, the drag
force is
D = qSCD
• Keep in mind that
dynamic pressure
1
2is
q = ½v
2
Drag Coefficient
Drag Polar
• The drag coefficient can
be written as
C2
CD = CD;0 +
L
¼eAR
• The CD,0 term is the zerolift drag term, or parasite
drag coefficient
CD
Drag Polar
• The term involving the lift
coefficient is called the
induced drag
• The term AR is the aspect
ratio, AR = b2/S
• The term 0 < e < 1 is the
Oswald efficiency factor
CL
Thrust Required
2
Putting itTall =
together,
qSC we+canWwrite
R
D;0
qS¼eAR
Why are lift and drag coefficients higher at low
speeds and lower at high speeds?
Lagrange’s Equations
• Brief treatment applicable only to systems of n
particles subject only to conservative forces
• Basic idea:
– Write down the position vectors of the n particles in
terms of independent generalized coordinates
– Differentiate the position vectors to get velocity
vectors in terms of generalized velocities
– Write down the kinetic energy, T
– Write down the potential energy, V
– Form the Lagrangian, L=T-V
– Take derivatives of the Lagrangian to get the
differential equations of motion
Lagrange’s Equations (2)
• The position vectors must be written in terms of
independent generalized coordinates, denoted qk,
k=1,…,N
– For example, for a particle with unconstrained motion in three
dimensions, one might use q1=x, q2=y, q3=z
– For a particle constrained to move on a circle in a plane (e.g., a
pendulum), there is only one degree of freedom, so one might
use q1=
– In general, the position vectors are written as
~ri = ~ri (q)
~vi = ~r_ i (q; q)
_
– Consequently, the velocity vectors are
Lagrange’s Equations (3)
• The potential energy depends only on the
positions
V = Vof(~rthe
) =particle
V (q)
i
¢ ~v =on
• TheT kinetic
velocities
n m ~
_
= 1 §energy
vdepends
T (q;
q)
2
i=1
i i
i
• Thus the Lagrangian depends on
generalized
and
_ velocities
L = T ¡coordinates
V = L(q; q)
Lagrange’s Equations (4)
• Take the derivatives
of the Lagrangian:
¡
d @L
@L = 0; k = 1; ¢ ¢ ¢
dt @q_k
@qk
;N
• This step gives N second-order ordinary
differential equations describing the motion of
the n particles
• If there were non-conservative forces acting on
the system of particles, then the right-hand sides
of these equations would have non-zero terms,
called generalized forces.
Lagrange’s Equations Example
Consider a single particle of mass m connected to a linear spring with
spring constant k, suspended in a constant gravitational field with
acceleration g, and constrained to move in a vertical plane, i.e., an
elastic planar pendulum. Assume, for simplicity, that the unstretched
length of the spring is zero.
Use polar coordinates, measured from the fixed end of the spring. The
position of the
respect to this inertial origin is
~rparticle
= r^
ewith
r
The potential energy associated with the spring is
Vs = kr2 =2
The potential energy¡associated with gravity is
Vg =
where
µ
mgr cos µ
is the angle between
^r
e
and the downward direction
Lagrange’s Equations Example (2)
~v = ~r_ = r_ e
^r + rµ_e
^µ
³
´
1
1
T = m~v ¢ ~v = m r_ 2 + r2 µ_2
2
2
³
´ 1
1
L = T ¡ V = m r_ 2 + r2 µ_2 ¡ kr2 + mgr cos µ
2
2
All that remains is to take the required derivatives of this scalar function with
respect to the generalized coordinates and velocities, obtaining the two secondorder di®erential equations of motion:
mÄ
r ¡ mrµ_2 + kr ¡ mg cos µ
mr2 µÄ + 2mrr_ µ_ + mgr sin µ
= 0
= 0
An interesting exercise is to re-derive these equations by direct application of
Newton's Second Law.