Turbo sail - 100 knots?

[SailDesign]

Anton,
What gives ice its advantage is the fact that it is hard. It cannot be pushed out of the way, but the force of trying to do so melts it and the "Boat" (skate) hydroplanes on this tiny amount of water, with little friction.
Steve

[dimitarp]

How to achieve the 100 knots on water?. Is it possible to run with 100 knots on water?
To answer of these questions or similar questions we have to use some theoretical knowledge of aerodynamics and hydrodynamics. Here are some empirical considerations.
If we accept that exist only a sail ( wing ) and there are no other resistances in air and water then the speed of a ship is defined from:

Vs=Vw* L/D


Where: Vs - is velocity of ship (boat, windsurf), m/s ;
Vw – is velocity of wind, m/s;
L – is lift of sail, N;
And D – is drag of sail, N.

L/D ratio is a important characteristic of every sail and wing. For monofoil sail the maximum L/D is about 12; for doublefoil sail – 15 and for wing – 18 .
Let us now to add the resistances in air nad water, the equation for ship velocity will be:

Vs= (Vw/(1+m*Aa/As+n*Aw/As))*L/D


Where m – is ratio of resistance in air to drag of sail;
n – is ratio of resistance in water to drag of sail;
Aa – is all projection area of things that generate air resistance in direction of apparent wind.
Aw – is contact area of ship to water.
and As – is sail area

It is difficulty to estimate factors m and n theoretically. But if make some admissions they can be defined. The admissions are:
1. The wind is uniform;
2. The all things that generate the air resistance have a cylindrical form without gap and sharp angles;
3. All resistances in water are transform to a resistance of a flat with area Aw to water. This mean that water resistance is transform to only viscose resistance;
4. For large Re the resistances do not influence from speed.

Then m is about 10 and n is about 20.

I take part in a project that aim to estimate the possibility of achieving 60 knots in sailing on water. The first stage of this project is to define the m and n experimentally. The web site of this project now in constructing and will coming soon. The project is open for funds and advices from all. Now I will shortly present the second stage of the project.

Example:
Let us have a windsurf with Aw=0.8m2 , double sail with As=8m2 and crew with Aa=0.8m2 , then :

Vs = 3.75xVw. For achieving 100 knots the wind speed must be about 13.3 m/s. For this speed wind will be never uniform and heel moment will be very large. That is mean the crew (one man) have to will be with fine skill and must compensate the heel moment ( of about 30000 N.m). For compensating the heel moment we will use central boat fin and other fin.
But why the skill is so important?
The answer of this question lies in dynamics of windsurf. The acceleration of a ship is define from:

Fx+Fd = +- M*(dVs/dt)

where Fx – is the general tractive force, N
Fd –s the general resistance force, N
M – is general mass of ship, kg
t – is time, s
+ note acceleration
- note decelerate.

When the crew is made a mistake Fd is increase fast and velocity fall, but when every think is all right Fx increase much slowly from Fd about 10 times.

My conclusion is. The speed of 100 knots is attainable but we need from:

1. Special boat,
2. Special rig (probably wing)
3. Special conditions ( flat water and moderate uniform wind)
4. And Special crew.
Is it possible?

[frosh]

Hi Dimitarp, just when I thought that this was dead and sunk, as a serious proposition, you come along with some complex mathematics and show us that in fact 100kts is possible!
However even with my level of maths, which is probably less than yours, I can see some fatal flaws in your presentation.
You make an initial assumption that there is only a wing on this boat to create drive, and suffer drag, nothing else???
We havn't been able to do this yet, and I understand that hydrofoils are a type of wing but you do not include these in your calculations, do you?
Then there are those ratios, m and n. They make no real sense. If m is the ratio of resistance in air to drag of sail, these two items appear to me to be two items that have a fairly similar meaning. To make mathematical sense the two items need to be completely separate. The problem with n is even worse.
The ratio of resistance in water to drag of the sail. Is the sail drag occurring when it is in the air or water? If part of the sail is in the water, then again the two items have fairly similar meanings and can't form a mathematical ratio.
Then we have an assumption that m=10 and n=20. So if we simplify, and say the ratio m in air is 1, then the ratio n in water is only 2.
I must be missing something here! Jet aircraft wings can go through air at Mach 3.3 approx. 2000 knots. Accordingly then should a wing be able to driven through water at 1000 knots?
Please explain a bit better as you have lost some of us along the way!

A foiler team from Australia including Sean Langman, Andy Dovell, John Biddlecombe and Rohan Veal believe they have a design for crunching the 50 knot mark on supercavitating foils.
More here: http://www.rohanveal.com/home.html

[dimitarp]

Originally Posted by frosh

Hi Dimitarp, just when I thought that this was dead and sunk, as a serious proposition, you come along with some complex mathematics and show us that in fact 100kts is possible!

Hi,
The resistance in air and water do not defines only from fluid characteristic but also it defines from area that in contact in fluid and from shape of body. The m parameter mean that the resistance in air from all other things ( exception of sail) is m times greater than restance from sail with same area. The n parametar mean that the resistance in water is n times greater than resistance from sail with same area.
The hydrofoil I do not comment at all. I think that at speed 100 knots it is anpossible to compensate the heel moment. The heel moment will be so great so can not be compensate with wight of sailor.
At high Mah number there are another effects and lift to drag ratio is fall to about 3

[marshmat]

I see a few iffy things in the mathematics mentioned earlier. The resistances are defined in very unconventional ways that do not bear a clear relationship to known formulae.
For instance, drag force (with no free surface) is normally parametrized as:
Fd = 1/2 * Cd * rho * V^2 * A
for drag coefficient Cd defined in terms of area A, at speed V in a fluid of density rho. Cd IS dependent on Reynolds number. I would like to see the calculations expressed in accordance with accepted engineering practice and with a little less hand-waving.
13.3 m/s of wind is about 48 km/h, or a little under 25 knots. Are you saying this boat would travel at over four times the speed of the wind driving it?
-Added- > By the way, the Mach number ( Ma = v / a ) is an indicator of compressible flow character and is only important if it is greater than about 0.3. For an object in air moving at less than about 100 metres per second, Ma is irrelevant.

[frosh]

Hi Dimitarp, I think that partly it is a language difficulty but are you now backing down from your earlier claim that 100 knots or even 60 knots is possible by a sail boat? Matt quite rightly points out that your mathematical equations are "unconventional" to say the least! "Yellow Pages Endeavor" solved the problem of lack of RM by a regular solo sailor but still has not achieved 47 knots even in the new model "Macquarie Innovations". You can find all the info on these at: http://www.macquarie.com.au/speedsailing/background.htm

[tspeer]

Originally Posted by dimitarp

...
Let us now to add the resistances in air nad water, the equation for ship velocity will be:

Vs= (Vw/(1+m*Aa/As+n*Aw/As))*L/D


Where m – is ratio of resistance in air to drag of sail;
n – is ratio of resistance in water to drag of sail;
Aa – is all projection area of things that generate air resistance in direction of apparent wind.
Aw – is contact area of ship to water.
and As – is sail area
...

This is not true. The correct relationship is:

Vs = Vw * sin(gamma - beta) / sin(beta) = Vw * [sin(gamma)/tan(beta) - cos(gamma)]
beta = arctan(Ds/Ls) + arctan(Dh/Lh)

gamma = course relative to the true wind
beta = angle between apparent wind and boat's course
Ds = aerodynamic drag
Ls = aerodynamic lift
Dh = hydrodynamic drag
Lh = hydrodynamic lift

Even if beta is a small angle, the course gamma will not be a small angle. For small beta and a beam reach (gamma = 90 deg), the speed becomes:

Vs = Vw / (Ds/Ls + Dh/Lh)

Notice that this has absolutely nothing to do with projected areas at all, and it does not reduce down to the same expression when you substitute in the ratios m and n.

Here's a real-world example of the kind of sailing performance you're talking about. The current landsailing speed record is 116 mph, or right at 100 kt, set in winds of 25 - 30 mph. This is a boatspeed/windspeed ratio of about 4. So beta had to be no greater than 14.5 degrees. This is what you have to shoot for if you want to go 100 kt on water.

Say, beta = 14 degrees just to provide a little margin. The optimum angle to the true wind will be 104 degrees [sin(gamma-beta)=1], just below a beam reach. In order to establish some requirements, arbitrarily divide beta into its aerodynamic and hydrodynamic drag angles:

beta_s, beta_h, Ls/Ds, Lh/Dh
7, 7, 8.1, 8.1
5, 9, 11.4, 6.3
3, 11, 19.1, 5.1
1, 13, 57.3, 4.3

The best Standard Class sailplanes achieve L/D's on the order of 40, so an aerodynamic drag angle of 2 - 3 degrees is probably the best you can hope for. Call it an L/D of 20 for everything in the air. This means that you have to achieve L/D's for the hull & foils of at least 5. At 100 kt.

The lift on the hull will be approximately equal to the lift on the sails, and the drag of the hull will be equal to the drag necessary to support the weight plus the drag from providing the lift to oppose the side force on the sail.

Dh/Lh = (W/Ls) * 1/(Lv/Dv) + 1/(Lf/Df)
W = weight of the boat + vertical load from the sails
Lv/Dv = L/D of vertical support
Lf/Df = L/D of foils providing horizontal lift

This is where sail area comes in. If the L/D's can be maintained, then a high ratio of sail loading to weight (Ls/W) will improve the performance. If the vertical and horizontal L/D's are the same and the sail-force to weight ratio is about 1, then an L/D around 10 from the vertical foils is in order.

At 100 kt, you're way past the subcavitating regime, so you will have to use supercavitating foils to provide side force. Or fully ventilated angled planing surfaces. Getting an L/D of 10 out of these is not going to be easy.

With regard to compensating for heeling and pitching moments, this will dictate much of the craft's configuration. Iron Duck was a semi-symmetrical craft with a symmetrical wing section, but angled wheel fairings and wheel placement. It could sail on both tacks, but sailed better on one tack. Past sailing record breakers, like Crossbow, Yellow Pages, and MacQuarie Innovation are asymmetrical single-tack craft.