Boomerangs - Cryptography - Lecture Notes | MATH 0209A, Study notes of Cryptography and System Security

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What Makes Boomerangs Come Back?

Ernie Esser

UCLA

Parts of a Boomerang

[Hawes,

All About Boomerangs

]

Goals for the talk:

Discuss the lift equation, why a boomerang generates lift and how theselift forces produce torque on a boomerang

Explain gyroscopic precession and why torque on a boomerang makes itturn

Estimate the characteristic radius of a boomerang’s flight path

Explain a boomerang’s tendency to lay over as it flies

Comment on how to construct a boomerang

Discuss boomerang throwing and catching techniques

Aerodynamic Lift

An airfoil produces lift when the combined effects of its orientation (angle ofattack) and its shape cause oncoming air to be deflected downward.

Sketch of Streamlines [Acheson,

Elementary Fluid Dynamics

]

Boomerang wing cross section

Bevel on lifting arm

Intuition from Oversimplification

Downwash from a helicopter

Recall Newton’s Second Law: force

rate of change of momentum

where momentum

mass

velocity

mass of air deflected per unit time

ρU

change in deflected air’s vertical velocity

U

F

Lif t

ρU

(^2)

A more local viewpoint

Consider air following the top curveof the airfoil in two dimensions

airflow

direction of decreasing pressure

A downward force is required to accelerate the air this way.(think centripetal force

mv

2 r

for a spinning mass on a string)

This force comes from a difference in pressure, which is decreasing in thedirection of the force.Consequences:

Expect lower than normal pressure near top of airfoil

Expect tangential acceleration of oncoming air along top of airfoil

Another Viewpoint: Circulation

Under the previous assumptions, the solution to the equations for conservationof momentum and mass in the domain surrounding the airfoil is not unique. It is, however, determined by a quantity called circulation,

, that measures

the counterclockwise rotation of air about the airfoil.

C

u

ds

is computed by adding up the tangential component of velocity

u

along the

boundary of the airfoil given by the curve

C

Viscous Effects Determine Circulation

For an airfoil with a corner at the trailing edge, all solutions to the inviscidproblem have a singularity in the velocity there except for one. Adding eventhe tiniest amount of viscosity picks out the solution where a stagnation pointis at the corner. (Kutta Condition)

Zero Circulation

Negative Circulation

[Acheson,

Elementary Fluid Dynamics

]

Comments:

Equal transit time argument is false. Negative circulation implies airmoving even faster above airfoil.

Need to include effect of viscosity to correctly model dynamics.Conservation of angular momentum plus decreasing circulation requiresshedding of counterclockwise vortex.

Getting Back to Lift Equation

L

U

F

Lift

U L

and

F

Lif t

ρU

F

Lif t

ρU

(^2)

L

So for a wing with uniform crossection we get back the lift equation:

F

Lif t

ρU

(^2)

C

L

A

Note: If we measure or simulate

F

Lif t

we can compute

C

L

Reynolds Number

General characteristics of the flow can often be determined from the unitlessReynolds number

Re

defined by

Re

ρLU

μ

Where

ρ

density

L

characteristic length scale

U

typical flow velocity

μ

dynamic viscosity

Small

Re

smooth, steady flow

Large

Re

turbulent, unsteady flow, and thin boundary layer

Inviscid approximation can work when

Re

is large

Re

for boomerang is between

4

and

5

Re

for cruising jumbo jet wing is on the order of

7

Triangularization of Domain

[Used distmesh software by Persson and Strang from http://www-math.mit.edu/ persson/mesh/]

Simulation of

u

for max

U

in

Speed

u

|^

from

t

to

t

,^

F

Lif t

and

F

Drag

0

2

4

6

8

10

12

14

16

18

20

maxUin = 10 , t = 0.04 , Flift = 1.8094 , Fdrag = 0.

Re = 27000 , speed and Fnet/100 plotted

0

−0.01 −0.02 −0.03 −0.

0 0.04 0.03 0.02 0.

Simulation of

u

for max

U

in

Speed

u

|^

from

t

to

t

,^

F

Lif t

and

F

Drag

0

5

10

15

20

25

maxUin = 14.1421 , t = 0.04 , Flift = 2.8515 , Fdrag = 0.

Re = 38183.7662 , speed and Fnet/100 plotted

0

−0.01 −0.02 −0.03 −0.

0 0.04 0.03 0.02 0.

Simulation of

P

for max

U

in

Pressure

P

from

t

to

t

,^

F

Lif t

and

F

Drag

−

−

0

50

100

150

200

maxUin = 14.1421 , t = 0.04 , Flift = 2.8515 , Fdrag = 0.

Re = 38183.7662 , pressure and Fnet/100 plotted

0

−0.01 −0.02 −0.03 −0.

0 0.04 0.03 0.02 0.