# Bow Wave and Kelvin Wake

Discussion in 'Hydrodynamics and Aerodynamics' started by WilliamP, Oct 21, 2012.

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### WilliamPNew Member

I suspect I'm on a 2+2=5 with this, but if wake leaves a boat at angle of around 19 degrees does it help to have a bow angle smaller than this so that the wake doesn't run along the hull?

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It doesn't have to be "hull shape". Any pressure field that is moving has the same effect and same classic Kelvin wave pattern.

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### WilliamPNew Member

That's why I was thinking that if the bow angle was less than 19 degrees the wake might diverge from the hull. If the hull angle was greater the wake would not diverge from the hull and would rub along the hull increasing drag. So maybe this is a critical bow angle, below which there is a step change reduction in drag?

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No.

It can be a single point, and has the same kelvin wave pattern. There was a hull desigmed and made in the late 1800s with such a hull shape, extremely fine.....I can't recall its name name, but can dig it out for you, if you're interested. Didn't work.

Reduction of drag is a different matter. That is influence far more by the length displacement ratio.

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### daiquiriEngineering and Design

My understanding is that Kelvin's theorem considers the ship as a tiny pressure point on the surface. Like when you look at it from an airplane flying high above. Hence, the resulting wave pattern and the characteristic 19.5° angle are relative to the far-field of the ship's wake. Plus, being a pressure-point, it is independent from the hull shape.
Am I correct in this one, AH?

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### WilliamPNew Member

Thanks Ad Hoc. I agree the 19.5ish degrees is independent of the hull shape, but even though the theory may assume the ship is a point, doesn't the wake in reality get formed by the bow and stern waves? I was just wondering whether *if* the bow wave forms a wake pattern that is at 19.5 degrees, then can you keep the hull "inside" the wake by beeping the half angle of entry below 9.75 degrees? Many dinghies, for example have a half entry of about 10-11 degrees. If they were pinched in to be less than 9.75 would there be a significant drag reduction? Obviously a narrow bow is generally lower drag, but is there a cliff edge redduction at this ange?

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In a word, yes.

As noted before, this has been tried before.

The problem with an ever decreasing angle of entrance is that this results in a greater curvature required to bring the lines, or slope, back to zero at the max beam. What I mean by this is, if you image in looking in plan view. Take a block of wood, say 10m by 2m. If you make the angle of entrance so fine, where does this line intersect the hull beam, that being the 2.0m, or rather 1.0m either side of the centreline? So, at the usual 19.47 degrees intersects at 2.83m aft of the FP, this being at the angle of entrance intersecting with the max beam line, as a straight line.

Yet if you now have this at 9.75 degrees, the location moves aft to 5.82m. But, where is the volume??....so to gain this back, you need to create shape, otherwise you’ll have a triangular plan form hull. You can add a “hollow” into the shape to help bring the intersection fwd.

So, the lower the angle of entrance the further and further aft this intersection becomes. Which means there comes a point where you really need the volume in your hull ie beam. So this intersection is brought fwd to help create the volume; a hollow in the lines if you like. But, where this line intersects with the max beam has an increase in its slope (the hollow) simply to get back to the max beam further fwd rather than further aft…which means an increase in the pressure disturbance, ergo an increase in drag not lower.

You can ask Leo ( http://www.boatdesign.net/forums/profile/leo-lazauskas.html ) for a more analytical solution if you want one. He's good for it

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### Leo LazauskasSenior Member

A more accurate representation of the Kelvin wake is shown in the attached
graphic from Newman's "Marine Hydrodynamics".

Note how the Kelvin angle is defined. It is not, as depicted incorrectly in Ad
Hoc's graphic and many other places, the line through the intersection of
the crests of the transverse and diverging wave systems.

In thin-ship theory the flow is simulated with a distribution of sources and
sinks distributed on the vertical centreplane. Other theories distribute
sources and sinks over the actual hull surface, and sometimes the actual
free-surface.

In the equivalent theory for hovercraft, planing hulls, and other travelling
pressure distributions, a distribution of pressure points on the undisturbed
free surface is used.
These pressure points are equivalent to horizontal dipoles.

To simulate cambered hulls, vertical horseshoe vortices can be used.

The simple singularities found in aerodynamics (i.e. when there is no free
surface) each have a "Havelock" equivalent, i.e. Havelock sources, Havelock
dipoles, Havelock vortices etc.
These singularities are the same as their aerodynamic counterparts but they
automatically satisfy the linearised free-surface boundary condition.

Last edited: Aug 12, 2015
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### daiquiriEngineering and Design

Talking about triangular hull forms and wave resistance, did you see this paper: http://www.din.unina.it/HSMV 2011 Proceedings/html/Papers/28.pdf ?
They claim some pretty much incredible gains over more traditional hull forms. Though it's hard for me to believe that the seakeeping gets improved that much, in particular with quartering or following waves, they have these towing-test data and I have just an opinion...

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### daiquiriEngineering and Design

"Principle of Naval Architecture", Tupper&Rawson's "Basic Ship Theory" and Molland's "Maritime Engineering Reference Book" define it just like shown in Ad Hoc's picture...
So you say that it is mathematically defined as an angle to the line connecting intersections of divergent waves and their tangents set at an angle of 35°16' to the centreline? But why, for God sake???

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### Leo LazauskasSenior Member

You need to go through the relatively simple derivation, but the mathematics does work out exactly.
See:
http://www.wikiwaves.org/
and especially:
http://www.wikiwaves.org/Ship_Kelvin_Wake

The Kelvin angle is equal to arctan(1/(2^1.5)) = 19.47122... degrees,
which occurs when the wave angle theta is equal to
theta = arcsin(1/sqrt(3)) = 35.264... degrees.

The situation is a bit more complicated in finite depth water.

The incorrect graphs seem impossible to eradicate. It's like the myth of flowing glass windowpanes

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### daiquiriEngineering and Design

Let's say that we defined the Kelvin angle as in the "wrong" illustrations, the one given by AH. Do you know perhaps what would be it's value, according to that definition?

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### Leo LazauskasSenior Member

I don't know if there is a precise simple formula as for the Kelvin angle and
the wave angle (theta_K) at which it occurs.

The correct definitions are not that complicated, and are very useful
because they allows us to separate wave resistance due to the transverse
wave system and the diverging wave system.

Transverse waves are those for -theta_K < theta < theta_K
Diverging waves are those with |theta| > theta_K.
The "incorrect" angle would not allow such an easy distinction.

It is also possible to derive that waves along the cusp line (theta = theta_K)
decay like 1/cuberoot(x) and hence they persist downstream for much
longer than transverse waves, which decay like 1/sqrt(x).
Waves along the line at the "incorrect angle" would decay like 1/sqrt(x)
because they are inside the cusp line.

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Not according to all my reference books:

Again, according to these reference books, it can be shown to be 35.267 degrees mathematically by superposition of parallel systems of wave crests. With the cusp line making an angle arcsin (1/3) = 19.47 degrees.

See above

You'll have to include Prof Whittaker in that too. Whom I consider probably one of the, if not the world's leading expert on wash.

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### Leo LazauskasSenior Member

Yes, the diagrams in your old references, and many others are incorrect and

Now that I see what they are calculating I concede that they give the correct angles. But their diagrams are wrong.
You can tell Whittaker he forgot about the phase shift.

Newman is correct, and the diagram was reproduced in
Faltinsen's "Hydrodynamics of High-Speed Marine Vehicles".

Here's some evidence that there are many textbooks and
papers with the wrong format.

From an article by L. J. Doctors in the Australian Naval
Architect, Vol. 6, No. 3, Nov. 2002, p 26.

"The specific analysis of water waves generated by a moving
disturbance has been presented by many authors. The works
of both Wehausen and Laitone (1960) and Stoker (1966) are
excellent in that the wave pattern generated by a point
source has been clearly described. It is noteworthy that
these two references are the only two papers that I possess
in which it is properly stated that the transverse waves
do not truly meet the divergent waves on the Kelvin line.
This is because there is a phase difference of one quarter
of a period between them. The majority of publications show
the two wave systems meeting at a cusp on the Kelvin line,
which is not correct."

You have your favourites. I have mine.

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