What is the root-mean-square height (in feet) of a wave in open ocean
as a function of wind speed (in knots) ?

Hi david...

Here's what I found:

There's an educational site run by Nasa, showing
radar images which correlate wind speed and 
wave height. You can view it at:
http://eospso.gsfc.nasa.gov/eos_edu.pack/p14.html

Referring to the radar image on the page, it says
"In this image, the strongest winds (about 15 
meters per second, or 54 kilometers per hour)."
and
"In this image, the highest waves occur in the
Southern Ocean, where waves up to 6 meters high
...are found"
referring to the same area.

A little math tells us that 54kph = 33.5556 mph,
33.5556 mph = 29.1262608knots and
6 meters = 19.686 ft. So, in this study,
over a ten day period, winds of about 29 knots
produced waves of up to 19+ feet in height.

Here's a page for converting between knots,
or kts, to mph and back:
http://www.srh.noaa.gov/fwd/media/appendix/windf.htm

The next page on our tour is one which displays the
Beaufort Scale of Wind Force and Probable Wave Height:
http://marine.cwb.gov.tw/CWBMMC/windwaveE.html

It gives probable and maximum wave heights for various
winds, in both knots and meters/sec. The probable
heights can be taken as 'average' heights. You will
need to convert their figures, in meters of height,
into feet, at this ratio: 1 meter = 3.281 ft.
Using this, you can see that the maximum wave height
they have listed, which is 16 meters, would be 52.496
feet high, in winds of 64-71 knots, or about 80 mph.

Here's a page which allows you to convert 'average'
and 'peak' values to root mean square, or rms values:
http://www.forcontrols.com/html/formulas.html

So, for the 'strong gale' listing on the chart, a 
wind of 41-47 knots, or about 44 knots, would produce
a peak height of 10 meters, or 32.81 feet, with a 
rms values of 23.204 feet (32.81/1.414).


Searches done, via Google:

"wave height" "wind speed"
://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=%22wave+height%22+%22wind+speed%22

"knots to mph"
://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=%22knots+to+mph%22

"calculating root mean square"
://www.google.com/search?num=50&hl=en&lr=&ie=UTF-8&oe=UTF-8&q=%22calculating+root+mean+square%22&btnG=Google+Search

I hope this satisfies your interest, and
please feel free to post a request for 
clarification if necessary, before you
rate the answer.

sublime1-ga

---

Request for Answer Clarification by david_j_kaplan-ga on 23 Aug 2002 06:38 PDT

I asked for a formula, theoretical or empircal, relating wind speed to
rms height.  You send me a series of websites that to a degree related
these variables.

The only site that you sent me that related some version of these
variables, in the form of a table, was the one titled "Beaufort Scale
of Wind Force and Its Probable Wave Height".  got the address of this
particual site by putting "wave height" and " wind speed" into google.

What I was looking for a function with the units built-in that I could
insert into a computer program without doing the interpolation myself.

david_j_kaplan-ga

---

Clarification of Answer by sublime1-ga on 23 Aug 2002 09:28 PDT

david...

Thank you for the opportunity to clarify my answer.

While such a simplified formula might be theorized, the 
calculations are, in reality, quite complex, and require
that wave height be sampled over a period of at least
10 minutes in order to account for the many variables.

Surflink, a resource for surfers seeking large waves,
notes the following in regards to its calculations:
"The reports draw from 30 sources both out at sea as
well as land based in order to obtain current swell,
tide and wind information. This information is sent
through a 100+ step formula that contains 11 
different geographic characteristics for each break"
http://www.surflink.com/pwr/reportinfo.htm

The DNV Course in Ocean Engineering site, here:
http://www.dnv.com/ocean/course.htm
goes into great details about the mathematics
involved, and the factors taken into account,
such as whether the wind is moving in the 
same direction as the wave, thus driving and
amplifying it.

This page introduces the subject of wind/wave
calculations, from the simple to the complex:
http://www.dnv.com/ocean/bk/c/a34/s0.htm

Arriving at the formula on the following page:
http://www.dnv.com/ocean/bk/c/a34/s5.htm#E3473

That being formula 3.4.73:
which I cannot reproduce in any way than this:
H sub s approx 5.74 cdot 10 sup -4 V sqrt{V over g} t sup 2/3

I can continue to search when I return later
today, but I wanted you to know that I was
working on it, and that the complexity of the
subject makes it seem unlikely that a simple
formula exists, since even the complex math
cannot produce numbers without considerable
amounts of error:

"Expected error in the wave height estimates
may be of order 15%. The time of blowing 
needed to obtain this wave condition is 12.9 
hours. This duration is realistic and does 
also support the initial choice of 10 hours 
duration used to transform the wind speed."
http://www.dnv.com/ocean/bk/s123.htm


Searches done, via Google:

"wave height" "wind speed" + formula
://www.google.com/search?num=50&hl=en&lr=&ie=UTF-8&oe=UTF-8&q=%22wave+height%22+%22wind+speed%22+%2B+formula&btnG=Google+Search

sublime1-ga

I commend sublime1-ga for his or her answer which I take to be
"Arriving at the formula on the following page:
http://www.dnv.com/ocean/bk/c/a34/s5.htm#E3473 
formula 3.4.73: 
which I cannot reproduce in any way than this: 
H sub s approx 5.74 cdot 10 sup -4 V sqrt{V over g} t sup 2/3 

You will remember that I specified the wave to be in open ocean.  That
gets rid of some of the complexity (but not enough).

This was an experiment to see what would happen if an apparently
simple question was asked about a complex topic.

Thanks for a really good attempt.

david...

Yes, on first glance it did appear that there might
be a simple solution, yet upon researching the 
answer, I realized I was "in over my head", as it
were... :)  I'm glad you allowed me a second chance
to look into it, since it was a very educational
experience for me to learn about this complex topic.

sublime1-ga

It is a simple solution, just not a simple search <grin>

Here is the solution

h = RMS wave height
H = Significant wave height
g = gravity
v = adjusted wind speed

From 
http://www.utexas.edu/research/mopro/chapter12/chapter12-9.htm

We find that H = 4*h

And from  http://www.care.auckland.ac.nz/~cchr010/courses/maritime/mre06.pdf

we find that  (g*H)/v^2 = 0.2433

solving for h, we find h = 4*(0.2433*v^2)/g

note that this formula is only valid if a long enough time period has
lapsed for the waves to adjust to the wind speed.  If you want a finer
time resolution you'll need more information and more complex formulas
(available in the pdf file.).

LetterRip

A further note...

your units for the formula should be consistent, otherwise your
dimensionless constant will be wrong.

LetterRip

LetterRip...

You said:
"It is a simple solution, just not a simple search <grin>"

Given that, it's especially useful to provide your search
strategy, along with the answer.  :)

Other than that, great addition (multiplication?)!
 
sublime1-ga