ORE 654Applications of Ocean Acoustics

Lecture 3aTransmission and attenuation

along ray paths

Bruce HoweOcean and Resources Engineering

School of Ocean and Earth Science and TechnologyUniversity of Hawai’i at Manoa

Fall Semester 2011

04/20/23 1ORE 654 L3

Prologue

• Newton’s “corpuscular energy” moving along “rays” helped describe propagation, reflection and refraction.

• In WWII – Ewing and Worzel discovered the SOFAR channel – sound fixing and ranging

• Perth-Bermuda 1960 shots

04/20/23 ORE 654 L3 2Ewing and Worzel, 1948

Transmission and attenuation along ray paths

• Energy transmission in ocean acoustics• Ray paths and ray tubes• Ray paths in refractive medium• Attenuation• The “SONAR Equation”; source level, sound

pressure level, transmission loss

04/20/23 ORE 654 L3 3

Energy transmission in ocean acousticsImpulse sources

• Explosions• Sparkers and air guns• Implosions (light bulbs)• Often milliseconds• Finite, discrete delta

function• Continuous – Direc delta

function

04/20/23 ORE 654 L3 4

δ f (t) = 1 / Δt 0 ≤ t ≤ Δt

dt

Δt0

Δt

∫ = 1

g(t)δ (t − t1)dt−∞

∞

∫ = g(t1)

Light bulb impulse source

• “100 W” light bulb• At 18.3 m depth

04/20/23 ORE 654 L3 5

Heard et al., 1997

Umbrella

Underwater crater:92 m across6 m deep

Yield: 8 kilotonsLocation: EniwetokDate: 8June 1958

Depth: 48 m

Nuclear testing

04/20/23 6ORE 654 L1

1 kiloton TNT = 4 × 109 J

Pressure, particle velocity, and intensity in a pulse

• Far field, spreading spherically from point source

• p0 at R0

• Waveform shape of outward traveling p same as p0; u follows same form

• Just time delay and amplitude change

04/20/23 ORE 654 L3 7

p(R, t) =p0 (t−R / c)R0

R

uR(R,t) ≈p

ρAc=

p0 (t−R / c)ρc

R0

R

iR(R,t) =puR =p20 (t−R / c)

ρAcR2

0

R2

Pressure in a pulse - 2

• Typical pulse shape – satisfies two conditions

• Step/delta with exponential decay, finite for t ≥ R/c

• τs – time to decay to 1/e

• tg – duration of integral or “gate time”; 98% correct for tg > 4τs

04/20/23 ORE 654 L3 8

p0 (t) =0 t < 0

p0 (t) dt0

∞

∫ =finite

p(R,t) =p0R0

Rexp −

1τ s

t−Rc

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

p0R0

Rexp −

1τ s

t−Rc

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥dt

Rc

Rc+tg

∫

=τ sp0R0

R1−exp −

tg

τ s

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢

⎤

⎦⎥

Particle velocity and intensity – transient signal

• Radial particle velocity and intensity in far field

• Message energy (Joules) pass through element ΔS at range R in gate open time tg, function of (t-R/c)

• Over full sphere ΔS = 4πR2

, total energy Em (J)• Energy flux at range R

transmitted during time interval pressure squared (tips) tg (J/m2)

04/20/23 ORE 654 L3 9

uR =p / (ρA / c) iR =puR

ΔEm =ΔS puR dt≈R2

0ΔSρAc0

∞

∫p0 (t−R / c)[ ]

2

R2 dtRc

Rc+tg

∫

ΔS=R2ΔΩ

Em ≈4πR2

0

ρAcp0 (t−R / c)[ ]

2dt

Rc

Rc+tg

∫

εg = puR dtRc

Rc+tg

∫ ≈tips[ ]R2

R20

ρAc

Power radiated by continuous wave signal

• Continuous wave sinusoid, peak value P0

• Source power over all angles is average energy over period T

• Same procedure for other signals

04/20/23 ORE 654 L3 10

p =P0ei2π ft

Π=4πR2

0

ρAcP0

2

2

P02

2=1T

P0 sin(2π ft) 2 dt0

T

∫ =Prms2

Ray paths and ray tubes

• Starting point -Spherical wave solution

• p0(t) is temporal function

• Time for message to arrive is R/c

• R0/R is spatial function• p(t,R) – output of

receiver at R

04/20/23 ORE 654 L3 11

p(t,R) =p0 (t−R / c)R0

R

Reflections along ray paths

• Reflected rays appear to come from image source• Pressure signal for bottom reflecting ray• Pressure signal for the bottom and surface reflected ray

04/20/23 ORE 654 L3 12

p(t) =p0 (t−Ra + Rb

c)

R0

Ra + Rb

R12

tpath =Ra +Rb

c

p(t) =p0 (t−Ra + Rb + Rc

c)

R0

Ra + Rb + Rc

R12R10

tpath =Ra + Rb + Rc

c

Multiple ray paths• Signal pressures

add vectorally and can constructively and destructively interfere at receiver

04/20/23 ORE 654 L3 13

Conservation of energy in ray tubes

04/20/23 ORE 654 L3 14

S1 =(ΔΩ)R12

S2 =(ΔΩ)R22

S1

ρA2c1p1(t−R1 / c1)[ ]

2dt

R1c1

R2

c1+tg

∫

=S2

ρA2c2p2 (t−R2 / c2 )[ ]

2dt

R2

c2

R2

c2+tg

∫

• Equality of energy• E1 = E2

Sound pressures in ray tubes• Sound pressures at 1 and 2

follow 1/R spherical divergence

• In homogenous media properties are constant, e.g., density and sound speed

• Change t-R/c to τ• Assume tg large enough to

include 1 and 2• Assume p1 proportional to p2,

so integrand must = 0• P changes as sqrt of area ratio,

or cross path ray tube diameter/scale

04/20/23 ORE 654 L3 15

τ1 = t − R1 / c; τ 2 = t − R2 / c

S1 p1(τ 1)[ ]2dτ

0

tg

∫ = S2 p2 (τ 2 )[ ]2dτ

0

tg

∫S1 p1(τ 1)[ ]

2− S2 p2 (τ 2 )[ ]

2⎡⎣

⎤⎦dτ0

tg

∫ = 0

p2 (τ 2 ) = +p1(τ 1)S1

S2

Ray paths in a refracting medium• Sounds speed / index of refraction usually function of space and time – can

be complicated• Use Snell’s Law to trace path of a small portion of a wave front –

determines direction• Ray is perpendicular to the local wave front• Gives direction and time• Valid for high frequencies:

– Changes in sound speed over scales large compared to wavelength– Water depth and range to receiver >> wavelength– Very good approximation when diffraction absent (wave effect)– Fails where rays cross (caustics), or shadow zones – use wave theory to patch

• Specular reflection at interfaces (mirror, flat)• Intensity losses along rays through geometric divergence (spherical

divergence modified by refraction), through absorption along paths, and reflections on interfaces

04/20/23 ORE 654 L3 16

Speed of sound - Seawater• From previous lecture• Sound speed (c or C m/s) is a complicated function of

temperature T °C, salinity S PSU, and pressure/depth z m

• Simple formula by Medwin (1975):

• Latest: http://www.teos-10.org/

04/20/23 ORE 654 L2 17

c = 1449.2 + 4.6T – 0.055T 2 + 0.00029T 3 + (1.34 – 0.010T )(S – 35) + 0.016z

Ocean stratification

• In general, sound speed = c(x,y,z,t)• Sound speed and other ocean properties such

as temperature and salinity (and pressure) have the largest variation in the vertical.

• Horizontal variations are typically smaller, over larger distances

• Assume c = c(z)

04/20/23 ORE 654 L3 18

Variation in sound speed – horizontal, vertical• Sargasso Sea at

750 m – main thermocline - +/- 5 m/s (1 °C)

• Perth to Bermuda – Blue 1,470 m/s, red 1,550 m/s

04/20/23 ORE 654 L3 19

Snell’s Law again

• Different layers• Define Ray parameter a – constant for one ray

no matter where along the ray you are• Start ray at z0 and c0. At depth z, angle is θz

and cz

04/20/23 ORE 654 L2 20

sinθ0

c0=

sinθ1

c1=

sinθz

cz

=a=constant

Formulating Ray integrals

• To determine ray paths (r,z) and travel time, must integrate, using Snell’s Law

• Dependent on z because c(z) only

• Differential distances dz, ds, dr and time dt

• Between “initial” and “final” positions

04/20/23 ORE 654 L2 21

ds =dz

cosθ

dt=ds

c(z)=

dzc(z)cosθ

dr =dztanθ

a=sinθ0

c0=constant

sinθ =ac(z)

cosθ = 1−a2c2 (z); ac(z) <1

tanθ =ac(z)

1−a2c2 (z)

tf −ti = dtzi

zf

∫ = dzzi

zf

∫1

c(z) 1−a2c2 (z)

rf −ri = drzi

zf

∫ = dzzi

zf

∫ac(z)

1−a2c2 (z)

dr

θds

dz

Ray tracing approximations

• Constant sound speed layers: – c(z) = cn for zn ≤ z ≤ zn+1

• Constant sound speed gradient layers, i.e., sound speed varies linearly in each layer:– c(z) = c(z1)+b(z-z1) for z1 ≤ z ≤ z2

– b = d(c(z))/dz

04/20/23 ORE 654 L3 22

Rays through constant sound speed layers• Real c(z) is continuous but

simplest solution - break into constant c layers

• Replace integrals with sums

Use Snell for next θ

04/20/23 ORE 654 L2 23

rf −ri = drzi

zf

∫ = dzzi

zf

∫ac(z)

1−a2c2 (z)

tf −ti = dtzi

zf

∫ = dzzi

zf

∫1

c(z) 1−a2c2 (z)sinθn =acn

cosθn = 1−(acn)2 ; acn <1

tanθn =acn

1−(acn)2

rN = (zn+1 −zn)tanθnn=0

N−1

∑

tN =(zn+1 −zn)cn cosθnn=0

N−1

∑

Rays through slowly changing sound speed layers

• Assume constant gradient in each layer

04/20/23 ORE 654 L2 25

c(z) =c(zn) +bn(z−zn) for zn ≤z≤zn+1

bn =c(zn+1)−c(zn)

zn+1 −zn

Rays through slowly changing sound speed layers

• Start with c(z) integrals

• discretize• change of

variables to simplify

04/20/23 ORE 654 L2 26

t f −ti = dtzi

zf

∫ = dzzi

zf

∫1

c(z) 1−a2c2 (z)

rf −ri = drzi

zf

∫ = dzzi

zf

∫ac(z)

1−a2c2 (z)

wn =z−zn +c(zn)bn

; dw=dz; c(w) =bnw

tn+1 −tn =dw

bnw 1−a2bn2w2( )

1/2wn

wn+1

∫

rn+1 −rn =abnwdw

1−a2bn2w2( )

1/2wn

wn+1

∫

Integral tables

04/20/23 ORE 654 L2 27

tn+1 −tn =dw

bnw 1−a2bn2w2( )

1/2wn

wn+1

∫

rn+1 −rn =abnwdw

1−a2bn2w2( )

1/2wn

wn+1

∫

dx

x (a2 −x2 )∫ =−1a

loga+ (a2 −x2 )

x

⎛

⎝⎜

⎞

⎠⎟

xdx

(a2 −x2 )∫ =− (a2 −x2 )

Rays through slowly changing sound speed layers

• Perform integrals• Sum over intervals to get total time and range

04/20/23 ORE 654 L2 28

tn+1 −tn =1bn

lnwn+1[1+ (1−a2bn

2wn2 )1/2 ]

wn[1+ (1−a2bn2wn+1

2 )1/2 ]=

1bn

lnwn+1(1+ cosθn)wn(1+ cosθn+1)

rn+1 −rn =1

abn

(1−a2bn2wn

2 )1/2 −(1−a2bn2wn+1

2 )1/2⎡⎣ ⎤⎦=1

abn

(cosθn −cosθn+1)

t= (tn+1 −tn)n=0

N−1

∑

r = (rn+1 −rn)n=0

N−1

∑

Rays through slowly changing sound speed layers• With linear sound

speed, ray paths are arcs of circles

04/20/23 ORE 654 L2 29

rn+1 −rn =1

abn

(cosθn −cosθn+1)

radius =1

abn

Rays through slowly changing sound speed layers

• Curvilinear path length• s = Rθ

04/20/23 ORE 654 L2 30

radius = Rn =1

abn

sn =Rn(θn+1 −θn)

Ray example - Arctic• Nearly isothermal• Linear sound

speed gradient = b ≈ 0.016 / s

04/20/23 ORE 654 L3 31

RayNorth Atlantic

• From Ewing and Worzel,1948.• Minimum at 1300 m

04/20/23 ORE 654 L3 32

RaysNorth Atlantic

• From Howe et al., 1987.• Minimum at 1300 m • 18°C mode water in upper layer• eigenrays

04/20/23 ORE 654 L3 33

RaysNorth Pacific

• From Dushaw et al., 1994.

• Minimum at 1000 m• Surface layer develops

as summer progresses• Transition from surface

reflecting to barely refracting

04/20/23 ORE 654 L3 34

Philippine Sea – 2009-2011

• ONR deep water acoustic propagation/tomography experiment

04/20/23 ORE 654 L3 35

Philippine Sea – profiles

04/20/23 ORE 654 L3 36

Philippine Sea - rays

04/20/23 ORE 654 L3 37

Philippine Sea – timefronts

04/20/23 ORE 654 L3 38

Attenuation• Seawater is a dissipative

propagation medium• Through viscosity or chemical

reactions to heat• Local amplitude decrease

proportional to the amplitude itself

• Acoustic pressure decreases exponentially with distance

• In terms of 1/e – neper/m• Then attenuation coefficient –

in terms of dB/km (relative power loss per km)

04/20/23 ORE 654 L3 39

dp =−αepdx

lnpp0

⎛

⎝⎜⎞

⎠⎟=αex

αe =1x

⎛⎝⎜

⎞⎠⎟ln

pp0

⎛

⎝⎜⎞

⎠⎟

p=p0e−αex

p(R,t) =p0 (t−R / c)R0

Re−αeR

α =1x

⎛⎝⎜

⎞⎠⎟10 log10

i1i2

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢

⎤

⎦⎥

α =1x

⎛⎝⎜

⎞⎠⎟

20 log10

p1

p2

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢

⎤

⎦⎥

1neper=8.68dB

8.68αe =α

Absorption losses in sea water: viscosities and molecular relaxation

• Coefficients of viscosity• μ dynamic or absolute

coefficient of shear viscosity – the viscosity =ratio shear stress to rate of strain

• Each component of stress is due to a shearing force parallel to area A, caused by velocity gradient (shear modulus)

• Bulk viscosity appears only in compressible media – i.e., acoustics – in compressible N-S, term proportional to rate of change of density – related to extensional modulus

04/20/23 ORE 654 L3 40

FxA

=μ∂u∂y

⎛⎝⎜

⎞⎠⎟

μ =dynamic viscosity

μb =bulk or volume viscosity

Absorption losses in sea water: Molecular relaxation

• Bulk viscosity – finite time for real fluid to respond to p change, or to relax back to a base state

• Ionic dissociation – activated, deactivated by condensation, rarefaction

• Magnesium sulfate and boric acid, even though minor parts of “salinity”

• Affects speed of propagation slightly (dispersive) – usually ignore this aspect

• Relaxation frequency• Damped oscillator

04/20/23 ORE 654 L3 41

c ≅cr

fr =1

2πτ r

αe =π fr / c( ) f 2

fr2 + f 2

α =Afr f 2

fr2 + f 2

α =Afr =constant for f ? fr

α =Afr

f2 : f 2 for f = fr

Molecular relaxation - 2

• Attenuation per wavelength

• For fresh water• Goes to zero very high

and very low frequency• Very high f – molecules

can’t respond• Very low – follow• Near fr – activated

molecules transfer energy from condensation to rarefaction – heat

04/20/23 ORE 654 L3 42

λ =c / f

αλ = Acfr( )f

fr2 + f 2

τ r =(4 / 3)μ + μ b

ρAc2

τ r = 2.1×10−12 s

f = 1 / (2π × 2.1×10−12 s) f = 1011Hz

Molecular relaxation - 2

• The fresh water curves• Relaxation frequency off

scale to right

04/20/23 ORE 654 L3 43

τ r = 2.1×10−12 s

f = 1 / (2π × 2.1×10−12 s) f = 1011Hz

Molecular relaxation in sea water

• During WWII, unexplained large attenuation at sonar frequencies – 20 kHz

• In 1950s – careful lab experiments by Leonard, Wilson and Bies

• Drive water filled sphere vs frequency - measure amplitudes of modes of oscillation and therefore damping, with various salts

• Determined magnesium sulfate, a relatively minor constituent by weight, was responsible

04/20/23 ORE 654 L3 44

Molecular relaxation in sea water

• Similar effects found by Mellen and Browning, 1970, for boric acid, moderated by pH

• Relaxation frequency ~ 1 kHz

04/20/23 ORE 654 L3 45

Molecular relaxation in sea water

• Final, including boric acid, magnesium sulfate and pure water

04/20/23 ORE 654 L3 46

α =A1P1 f1 f

2

f 2 + f12

+A2P2 f2 f

2

f 2 + f22

+ A3P3 f2

Molecular relaxation in sea water

• Boric acid• pH function of location• A1 in Pacific half of Atlantic

04/20/23 ORE 654 L3 47

α =A1P1 f1 f

2

f 2 + f12

+A2P2 f2 f

2

f 2 + f22

+ A3P3 f2

A1 =8.68

c10(0.78pH−5) dB km-1 kHz-1

P1 = 1

f1 = 2.8S

35⎛⎝⎜

⎞⎠⎟

0.5

10[(4 −1245 /(273+T )] kHz

Molecular relaxation in sea water

• Magnesium sulfate

04/20/23 ORE 654 L3 48

α =A1P1 f1 f

2

f 2 + f12

+A2P2 f2 f

2

f 2 + f22

+ A3P3 f2

A2 = 21.44S

c(1 + 0.025T ) dB km-1 kHz-1

P2 = 1 − 1.37 × 10−4 z + 6.2 × 10−9 z2

f2 =8.17 × 10[(8−1990 /(273+T )]

1 + 0.0018(S − 35)kHz

Molecular relaxation in sea water

• Pure water

04/20/23 ORE 654 L3 49

α =A1P1 f1 f

2

f 2 + f12

+A2P2 f2 f

2

f 2 + f22

+ A3P3 f2

A3 = 4.937 × 10−4 − 2.59 × 10−5T + 9.11 × 10−7T 2

− 1.50 × 10−8T 3 dB km-1 kHz2

P3 = 1 − 3.83 × 10−5 z + 4.9 × 10−10 z2

TL

04/20/23 ORE 654 L3 50

d

dR20 log10

R

R0

⎛

⎝⎜⎞

⎠⎟=

ddR

[α(R−R0 )]

Rt =8.68α

• Absorption dominant at long ranges, balance at R = Rt

Current interest in pH

• Change in sound absorption at 440 Hz due to CO2/pH

04/20/23 ORE 654 L3 51Hester et al., 2008

pH at ALOHA/HOT

• Variability? Time and space?

04/20/23 ORE 654 L3 52Duda, 2008

Borate absorption

• B(OH)3+ OH− ↔ B(OH)3 · OH− ↔ B(OH)4−

04/20/23 ORE 654 L3 53

Absorption - frequency

• Change in absorption with frequency vs frequency

• SNR(f,L) = SL − TLa(f,L) − TLg(L) − NL(f) − PG

• Big question is noise• Difficult!

04/20/23 ORE 654 L3 54

Scattering losses• Scattering of sound

out of the ray tube• “small” scale index of

refraction variation• Bubbles• Bodies – fish,

submarines, plankton, etc

• Address later

04/20/23 ORE 654 L3 55

The “SONAR Equation”: SL, SPL, and TL• Pressure can vary 10 orders

of magnitude• Use logarithmic scale –

decibel• Decibel = 10 times the log10

of the ratio of two powers• Ratio 10, 10 dB• Ratios – relative to 1 m and 1

μPa• Point source – spherical,

power Pac - Watts• At receiver, wave is plane,

intensity related to pressure• Transmitted acoustic

pressure brought back to R = 1 m

04/20/23 ORE 654 L3 56

I(R) =Pac

4πR2

I (R) =p2 (R)ρAc

p1m2 =

ρAc4π

Pac

p2 =p1m2 R0

2

R2 10−α(R−R0 )

energies/intensity/power dB = 10 log10

X1

X2

⎛

⎝⎜⎞

⎠⎟

pressure/velocity dB = 20 log10

x1x2

⎛

⎝⎜⎞

⎠⎟

re 1 μPa at 1 m

The “SONAR Equation”: SL, SPL, and TL• 1 bar = 1 atm = 105 Pa• 1 W = 170.8 db re 1 uPa at 1 m

04/20/23 ORE 654 L3 57

p1m2 =

ρAc4π

Pac

p2 =p1m2 R0

2

R2 10−α(R−R0 )

SPL(dB) =SL(dB)−TL(dB)

SL(dB) =10 log10 (Pac) +10 log10 (ρAc4π

) + 20 log10

106 μPa1Pa

⎛

⎝⎜⎞

⎠⎟

SL(dB) =10 log10 (Pac) + [(50.8 +120) =170.8]; re 1 μPa at 1 mTL(dB) =20 log10 (R / R0 ) +α(R−R0 )SPL(dB) =10 log10 (Pac) +170.8 −20 log10 (R / R0 )−α(R−R0 )

Re-cap: Transmission and attenuation along ray paths

• Energy transmission in ocean acoustics• Ray paths and ray tubes• Ray paths in refractive medium• Attenuation• The “SONAR Equation”; source level, sound

pressure level, transmission loss

04/20/23 ORE 654 L3 58