As for the speed of sound in water:
Speed of Sound in Water
R. J. Wilkes, 11/5/97
The literature on sound speed in water (freshwater and seawater) is rather confusing. There are 6 primary references in the literature (refs. [1]-[6] below). Treating them in chronological order, their content can be summarized as follows:
Greenspan and Tschiegg (1959): Tables of c vs T for distilled water, 0 < T< 100 deg C; no pressure is given and it is presumably standard atmospheric P. Also given is a ploynomial fit with standard deviation 0.026 m/sec or 17 ppm.
Wilson (1960a): Tables of c for seawater, -3 < T < 30 deg C; 1.033 (=1 atm)< P < 1000 kg/cm2; 33 < S < 37 ppt. Note that P is absolute pressure. A polynomial fit is reported with standard deviation 0.22 m/sec.
Wilson (1960b): Tables of sound speed in distilled water, for 0 < T < 100 degC, 14.7 < P < 14,000 psia (ie, 1~1000 atm). The accuracy is 0.1 ppt.
Wilson (1960c): A brief note giving an updated fit equation covering a wider range of parameters (especially salinity): -4 < T < 30 deg C, 1.0 < P < 1000 kg/cm2, 0 < S < 37 ppt.
Del Grosso (1974): Provides a fit formula and tables of differences relative to results of previous formulas. Ranges are not given explicitly but tables in the paper cover 0 < T < 35 degC, 0 < S < 43 ppt, 0 < P < 1000 kg/cm2. Note that he uses gauge pressure, ie P=0 corresponds to 1 atm = 1.033 kg/cm2 absolute.
Chen and Millero (1977): Tables and fit for 0 < T < 40 deg C, 0 < P < 1000 bars, 5 < S < 40 ppt. Note pressure is gauge pressure (P=0 corresponds to 1 atm). Standard deviation of fit is 0.19 m/sec.
Attached is source code for Fortran-77 functions to compute the sound velocity for specified S,T,P using the fits described in refs. [1], [4] (which covers [2] and [3]), [5] and [6] within their valid ranges. The functions have been arranged for uniform input (T in degC, P in kg/cm2 absolute, S in ppt) and return c=-1 if input parameters are out of the range of validity for the fit coded. Naturally there is no warranty express or implied for the use of these functions!
Following the source code is a table of c vs T according to the various authors, for fresh water, and one point for seawater at T=1 degC, P=500 kg/cm2 abs, S=35 ppt.
References
[1] M. Greenspan and C. Tschiegg, (1959), JASA 31:75.
[2] W. Wilson, (1959), JASA 31:1067.
[3] W. Wilson, (1960a), JASA 32:641.
[4] W. Wilson, (1960b), JASA 32:1357.
[5] V. Del Grosso, (1974), JASA 56:1084.
[6] C. Chen and F. Millero, (1977), JASA 62:1129.
Fortran-77 functions to calculate sound speed in water
double precision function greensp(T,P,S)
implicit double precision (a-z)
c calculate c(m/sec) in fresh water at 1 atm given t(degC)
c according to m. greenspan and c. tschiegg, JASA 31:75 (1959)
if (s.gt.0.or.p.gt.1.033) then
greensp=-1.0
return
endif
c=1402.736
c=c + (5.03358)*t +(-0.0579506)*t**2
# + (3.31636e-04)*t**3 + (-1.45262e-06)*T**4
# + (3.0449e-09)*t**5
greensp=c
return
end
double precision function wilson(t,p,s)
c find c(m/sec) in water with (T(degC), P(kg/cm2 abs), S(ppt))
c according to wilson, JASA 1960
implicit double precision (a-z)
logical notok
c test for in-range
wilson=-1.0
notok=.false.
if ((t.lt.(-4.0)).or.(t.gt.30.0)) notok=.true.
if ((s.lt.0).or.(s.gt.37.0)) notok=.true.
if ((p.lt.0).or.(p.gt.1000.0)) notok=.true.
if (notok) return
c ok
c0 = 1449.14
ct1= 4.5721e00
ct2=-4.4532e-02
ct3=-2.6045e-04
ct4= 7.9851e-06
cp1=1.60272e-01
cp2=1.0268e-05
cp3=3.5216e-09
cp4=-3.3603e-12
dcs=1.39799*(S - 35.0)+1.69202e-03*(S - 35.0)**2
cstp1=-1.1244e-02
cstp2=7.7711e-07
cstp3=7.7016e-05
cstp4=-1.2943e-07
cstp5=3.1580e-08
cstp6= 1.5790e-09
cstp7=-1.8607e-04
cstp8=7.4812e-06
cstp9=4.5283e-08
cstp10=-2.5294e-07
cstp11=1.8563e-09
cstp12=-1.9646e-10
dct=t*(ct1+t*(ct2+t*(ct3+t*ct4)))
dcp=p*(cp1+p*(cp2+p*(cp3+p*cp4)))
dcstp=(s-35.0)*(cstp1*t + cstp2*t**2 + cstp3*p
# + cstp4*p**2 + cstp5*p*t + cstp6*p*t**2)
# + p*(cstp7*t + cstp8*t**2 + cstp9*t**3)
# + p*p*(cstp10*t + cstp11*t**2)
# + p**3*cstp12*t
c=c0+dct+dcp+dcs+dcstp
wilson=c
return
end
double precision function delgros(t,pa,s)
implicit double precision (a-z)
logical notok
c returns c(m/s) in water with (T(deg C), P(kg/cm2 abs), S(ppt))
c according to V. Del Grosso, JASA 56:1084 (1974)
c convert kg/cm2 absolute to gauge pressure
P=Pa-1.033
c test for in-range
delgros=-1.0
notok=.false.
if ((t.lt.0).or.(t.gt.35.0)) notok=.true.
if ((s.lt.0).or.(s.gt.43.0)) notok=.true.
if ((p.lt.0).or.(p.gt.1000.0)) notok=.true.
if (notok) return
c ok
c0=1402.392
ct1= 0.501109398873E+01
ct2= -0.550946843172E-01
ct3=+0.221535969240E-03
cs1= 0.132952290781E+01
cs2= +0.128955756844e-03
cp1= 0.156059257041e00
cp2= +0.244998688441e-04
cp3= -0.883392332513e-08
c1= -0.127562783426e-01
c2= +0.635191613389e-02
c3= +0.265484716608e-07
c4= -0.159349479045e-05
c5= +0.522116437235e-09
c6= -0.438031096213e-06
c7= -0.161674495909e-08
c8= +0.968403156410e-04
c9= +0.485639620015e-05
c10= -0.340597039004e-03
dct=t*(ct1 +t*(ct2 + t*ct3))
dcs=s*(cs1 + s*cs2)
dcp=p*(cp1 + p*(cp2 + p*cp3))
dcstp=c1*t*s + c2*t*p + c3*(t*p)**2 + c4*t*p**2
# + c5*t*p**3 + c6*p*t**3 + c7*(s*p)**2
# + c8*s*t**2 + c9*t*p*s**2 +c10*t*s*p
c=c0+dct+dcs+dcp+dcstp
delgros=c
return
end
double precision chenmil(t,p0,s)
c * sound speed according to Chen and Millero (1977) JASA,62,1129
implicit none
double precision s,t,p0
double precision a,a0,a1,a2,a3,b,b0,b1
double c,c0,c1,c2,c3,p,sr,d,sv
c convert from absolute to bars
p = p0 - 1.033
c test for in-range
wilson=-1.0
notok=.false.
if ((t.lt.0.0).or.(t.gt.40.0)) notok=.true.
if ((s.lt.5.0).or.(s.gt.40.0)) notok=.true.
if ((p.lt.0).or.(p.gt.1000.0)) notok=.true.
if (notok) return
c ok
sr = sqrt(s)
d = 1.727e-3 - 7.9836e-6 * p
b1 = 7.3637e-5 + 1.7945e-7 * t
b0 = -1.922e-2 - 4.42e-5 * t
b = b0 + b1 * p
a3 = (-3.389e-13 * t + 6.649e-12) * t + 1.100e-10
a2 = ((7.988e-12 * t - 1.6002e-10) * t
# + 9.1041e-9) * t - 3.9064e-7
a1 = (((-2.0122e-10 * t + 1.0507e-8) * t
# - 6.4885e-8) * t - 1.2580e-5) * t + 9.4742e-5
a0 = (((-3.21e-8 * t + 2.006e-6) * t
# + 7.164e-5) * t -1.262e-2) * t + 1.389
a = ((a3 * p + a2) * p + a1) * p + a0
c3 = (-2.3643e-12 * t + 3.8504e-10) * t - 9.7729e-9
c2 = (((1.0405e-12 * t -2.5335e-10) * t
# + 2.5974e-8) * t - 1.7107e-6) * t + 3.1260e-5
c1 = (((-6.1185e-10 * t + 1.3621e-7) * t
# - 8.1788e-6) * t + 6.8982e-4) * t
+ 0.153563
c0 = ((((3.1464e-9 * t - 1.47800e-6) * t
# + 3.3420e-4) * t - 5.80852e-2) * t
# + 5.03711) * t + 1402.388
c = ((c3 * p + c2) * p + c1) * p + c0
chenmil = c + (a + b * sr + d * s) * s
return
end
Sample output
Fresh water, surface
T,degC Greenspan Wilson DelGrosso
0 0.14027E+04 0.14024E+04 0.14024E+04
1 0.14077E+04 0.14074E+04 0.14073E+04
2 0.14126E+04 0.14122E+04 0.14122E+04
3 0.14173E+04 0.14169E+04 0.14169E+04
4 0.14220E+04 0.14216E+04 0.14216E+04
5 0.14265E+04 0.14261E+04 0.14261E+04
6 0.14309E+04 0.14306E+04 0.14305E+04
7 0.14352E+04 0.14350E+04 0.14348E+04
8 0.14395E+04 0.14392E+04 0.14391E+04
9 0.14436E+04 0.14434E+04 0.14432E+04
10 0.14476E+04 0.14475E+04 0.14472E+04
11 0.14515E+04 0.14514E+04 0.14511E+04
12 0.14553E+04 0.14553E+04 0.14550E+04
13 0.14591E+04 0.14591E+04 0.14587E+04
14 0.14627E+04 0.14628E+04 0.14624E+04
15 0.14662E+04 0.14664E+04 0.14659E+04
16 0.14697E+04 0.14699E+04 0.14694E+04
17 0.14731E+04 0.14734E+04 0.14727E+04
18 0.14764E+04 0.14767E+04 0.14760E+04
19 0.14795E+04 0.14800E+04 0.14792E+04
20 0.14827E+04 0.14831E+04 0.14823E+04
Seawater, p=500, s=35
T,degC Greenspan Wilson DelGrosso
1 -0.10000E+01 0.15364E+04 0.15358E+04
In other words, the speed of sound in water is roughly 1500 meters per second; that is 4920 feet per second or 3356 miles per hour. --Boris
Yes, that's about right. For an easier way to do it, see: http://www.npl.co.uk/npl/acoustics/techguides/soundseawater/content.html. It provides several alternative equations -- just press the "Interactive version" link next to your preferred one and punch in the parameters, it'll calculate the speed of sound for you.
I got 3,300 miles per hour using one equation.
So I'm very skeptical of the claims in the article. At the very least, it would take ENORMOUS forces to shove aside water fast enough to travel at 3000+ mph in the water. Not only is water quite heavy, by volume, but it's practically incomprehsible -- unlike air, you can't just "squeeze" it out of your way, you have to shove it aside, plus all the water that's all around it. That's why jumping off a bridge into a body of water is almost always fatal -- at high speeds, water is about as "hard" as concrete.