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Q: Height of a wave as a function of wind speed ( Answered ,   4 Comments )
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 Subject: Height of a wave as a function of wind speed Category: Science > Earth Sciences Asked by: david_j_kaplan-ga List Price: \$3.00 Posted: 22 Aug 2002 16:09 PDT Expires: 21 Sep 2002 16:09 PDT Question ID: 57585
 ```What is the root-mean-square height (in feet) of a wave in open ocean as a function of wind speed (in knots) ?```
 Subject: Re: Height of a wave as a function of wind speed Answered By: sublime1-ga on 22 Aug 2002 20:04 PDT Rated:
 ```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```
 david_j_kaplan-ga rated this answer: ```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 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 " Given that, it's especially useful to provide your search strategy, along with the answer. :) Other than that, great addition (multiplication?)! sublime1-ga```