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 Boat Design Forums Wave height calculations

#1
06-02-2009, 10:35 PM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
Wave height calculations

Does anyone have a simple equation to estimate wave height based upon fetch, depth and wind speed? Thanks in advance.

Chris
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Chris Tucker Marine Design
www.ctmd.com.au


#2
06-02-2009, 11:58 PM
 MikeJohns Senior Member Join Date: Aug 2004 Rep: 2011 Posts: 3,008 Location: Australia
Chris
Oceanography is the key. Usually the predicitions are quite complex spectral approaches and it's not trivial, I'm not sure what you want to do with it but for a simple approach have a look at this:

http://www4.ncsu.edu/eos/users/c/cek...r10/part2.html

Cheers
#3
06-03-2009, 12:10 AM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
Thanks Mike, In this case its very symplistic. Ie what approximate wave height can a developed expect in their "man made" lakes (Dams)
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Chris Tucker Marine Design
www.ctmd.com.au
#4
06-03-2009, 12:12 AM
 Ad Hoc Naval Architect Join Date: Oct 2008 Rep: 2358 Posts: 4,615 Location: Japan
Is there such a thing as "simple" in wave analysis??

Fetch and depth will affect the results, but if you ignore these for now there are 2 basic formula you can apply.

For fully developed sea
Hs= 0.025V^2 (ft) - height
Ts = 0.64V (s) - period

where V = wind speed in knots

In "nearly developed seas"

Hs = 0.023V^2
Ts = 0.45V

These are from Bretschneider and Sverdrup and Munk, but produces wide variations, despite being well used.

However the classic and now widely used method of Pierson-Moskowitz has:

Hs^2 = 3.5V^4/10^4 (ft)

as THE equation to use.
#5
06-03-2009, 12:40 AM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
Thanks,

I have these but when the fetch is approx 20m and the depth is 1.6m so they won't work. Basically the question I've been asked is "If we want to place some Art in the middle of our lake, what height should it stand above the surface to avoid waves hitting the bottom in normal wind speeds"

The info I've been given is.

Lake is approx 40x40 square.
depth is approx 1.6m throughout
they define "normal wind" as up to 25 knots.
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Chris Tucker Marine Design
www.ctmd.com.au
#6
06-03-2009, 12:51 AM
 Ad Hoc Naval Architect Join Date: Oct 2008 Rep: 2358 Posts: 4,615 Location: Japan
CTMD

ah...now you have supplied more information than "...Does anyone have a simple equation to estimate wave height based upon fetch, depth and wind speed? Thanks in advance...". It does help when asking a question to provide the details from the start!

The fetch you describe is not correct. Because the fetch of 20m assumes that there is only 20m in all direction, ie it is inside a box of 20x20m or a large tube 20m in diameter.

What fetch is there, ie complete unobstructed land/surface in each direction of the 'lake' (more like a big pond really)...there must be more than 20m. Unless it is a pond in a prison cell!!!!

Ops hit sned before i forgot to add....Because your open fetch of 20m of water would behave differently if the wind could only approach from 10m compared to 1km across it, as well as buildings being near by cretaing 'funnels' etc. Personally i wouldn't worry too much about 20m, just assume a height of 1.0m..wont get worse than that!
#7
06-03-2009, 01:04 AM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
That's more info than I have, lets assume a worst case of needing a wave height in the middle of a 40X40 "lake" with at least 1000m of clear land in any direction. it will eventually be reduced but its a paddock with roads at present.
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Chris Tucker Marine Design
www.ctmd.com.au
#8
06-03-2009, 01:19 AM
 Ad Hoc Naval Architect Join Date: Oct 2008 Rep: 2358 Posts: 4,615 Location: Japan
CTMD

well, the problem comes in that for such a small body of water, there is no "real validated" data for 'simple' formula. You would ahve to estimate the energy in the waves system and hence find the approximate sea spectrum that could be developed over the area...not so simple and straight forward when all you ahve is 40x40 stretch of water...you'd ahve to assume a certain probability distribution, is it regular or irregular, one-dimensional, etc etc

However to 'assume' lots again you could use

E=rho x g x H^2 x length/ 8

where H= 2r = wave height

which is Nm per width of wave...but this is based upon trochoidal and subtrochoidal theory....and on it goes....so, as i said before, not so simple for a simple one line formula
#9
06-03-2009, 01:25 AM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
I figured I'd have to do it the hard way, but sometimes you hope someone else knows more than you do, to save you some time. The whole thing is a "favour" for a client who has just signed up for a new power cat design.
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Chris Tucker Marine Design
www.ctmd.com.au
#10
06-03-2009, 01:29 AM
 Ad Hoc Naval Architect Join Date: Oct 2008 Rep: 2358 Posts: 4,615 Location: Japan
CTMD

we always look for short cuts, but sometimes, the hard way is the best way..especially in the absence of any 'real data'.

As I said, for such a small body of water i wouldn't worry about it. You're not going to get swells or rough waves...with a depth of 1.6m you're not going to get anything above 1.0m at an absolute extreme worst case possible...so just place it where common sense dictates rather than formulae based upon the ocean/sea waves not ponds!
#11
06-03-2009, 01:32 AM
 CTMD Naval Architect Join Date: Dec 2007 Rep: 117 Posts: 196 Location: Melbourne, Aus
That was my answer too, I suggested that if they were really worried they could place a gradiated post in the middle of the lake and take some photos on the next windy day.
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Chris Tucker Marine Design
www.ctmd.com.au