Ultimate sailing for oldies VII
Vertical sail for progress across water
In order for the wind powered hapa-driven ekranoplane to fly it first needs a vertical component of area in the wing, I.e. a sail equivalent. Is the vertical projection of the anhedral (windward side only) sufficient to drive the machine across the water?
In order to compute the required vertical area needed at a given airspeed one needs to know the lift co-efficient of the sail and its lift/drag ratio. Available data is very suspect, and anyway CL and L/D ratio alter with changes in apparent wind speed, sail twist, angle of attack and angle of bank.
I have it on good authority that a sail CL of 0.8 is reasonable. This is when the sail is being employed as an aerofoil. It does not apply to a spinnaker going downwind, a pure drag device, an arial drogue.
The L/D ratio of the ekranoplane is 19 to 1 (see ultimate sailing VI) but that is in the vertical plane. In vertical yacht sails it seems that their sails have an L/D ratio of one to one, as per the diagram on page 48 of AYRS publication 61. I confirmed with the 101 year old author in February 2019 that this was as accurate drawing and not merely diagrammatic.
The main resistance which the sail of the airborne ekranoplane must equal is the hapa’s hydrodynamic resultant force. The best hapa lift/drag ratio which we can hope for at present (2019) is two to one. The vertical projection of the ekranoplane’s wing will produce aerodynamic drag. But this will be insignificant compared with the hapa’s hydrodynamic drag because water is nearly one thousand times denser than air. So I have ignored it.
The ekranoplane will require tow-up by motor boat until the hapa can produce the required lift to take over. This occurs at the modest water speed of 3.2 MPH so I am not re-inventing parascending.
At lower water speeds than 3.2 MPH there is not enough hapa lift produced by the small (say 1.3 sq ft) hapas which are currently available. One could have a bit of fun with an 18 sq ft drogue drifting slowly downwind. That would take off at a water speed of 0.66 ft/sec (0.45 MPH), thus:-
Resistance = ekranolane’s drag = weight ÷ L/D ratio = 176 ÷ 19 = 9.26 lbs.
9.26lbs = 1.2 x 1.956 ÷ 2 x 18sqft x 0.66ft/sec 2
I see no way of the hapa being deployed at the correct speed. At the moment of the hapa replacing the drogue the hapa’s position would be:-
Hapa thrust = 9.26 lbs = 0.4 x 1.956 ÷ 2 x 54 sqft x 0.66 ft/sec2.
The hapa is 1.3 sqft, not 54 sqft. The CL of 0.4 is as per Edmund Bruce (AYRS 61 pg 27), being the CL for best L/D ratio, and we need best L/D ratio. Maybe the drogue could collapse gradually.
The vertical projected area of the ekranoplane (windward side only) is max-chord x max height ÷ 2 = 21.21 ft x4 ft ÷2 = 42.42 sqft. The craft will experience a small sideforce at low air speed which will increase as VAW increases. But kiteboats (for this is such) do not experience capsizing moments. Instead the kiteline (i.e. hapa line) merely takes up a lower horizontal angle. If any residual sideforce is experienced it can be countered by weight shift by the pilot.
I have prepared diagrams of the craft in ground effect and out of ground effect. Outside ground effect the L/D ratio deteriorates greatly (see below) Anhedral is still preferable even outside ground effect because the pilot must be somewhere above the water, yet below the wing to avoid being top heavy. Also, the wings have a tendency of rising when they experience lift and the wingtips kissing above the craft. This can be prevented with little weight by the tension in a line joining the wingtips. With dihedral it would require a strong, therefore heavy and draggy spar in torsion for the full length of the wingspan.
Once the craft is up in the air and largely or completely outside the ground effect region it can bank, giving it the ability to manoeuvre. It can also experience stronger winds at height. The hapa’s change of shunt is accomplished in the same way that kitesurfers change shunt. The modest tow-up water speed at lift-off and the modest required true wind speed at which this occurs (see diagram) is not dangerous. It remains to be seen by practical experiment whether the hapa will automatically take over from the towboat.
I am not too worried about the hapa not producing the required amount of lift. A small increase in water speed or an increase in hapa area (e.g using two or more hapas), would solve that. The main problem is not the hydrodynamics but in the aerodynamics. Let us assume no ground affect and no lift increase due to ground effect . The CD will be 0.4182 per Ultimate Sailing VI, CD other remaines 0.0220 so CD all is 0.4402. L/D ratio is CL ÷ CD all = 2 ÷ 0.4182 = 4.78 to 1.
The point at which the resultant sail force and resultant hapa force are equal and opposite occurs at a force of 13.19 lbs, as follows:- Resultant sail force = 13.19lbs = 0.8 x 0.0024 ÷2 x 42.42 sqft x 18 ft/sec2
Resultant hapa force = 13.19lbs = 0.4 x 1.956 ÷2 x 1.3 sq ft x 5.1ft/sec2 . This is that part of the hapa force working directly opposing the sail force.
The reason why an apparent windspeed of 18 ft/sec is chosen is because that is the velocity at which the machine takes off, viz:-
Lift = weight = 176lbs = 2 x 0.0024 ÷ 2 x 225 x 18ft/sec2 (see ultimate sailing VI).
The diagrams appear to show that ground effect is not vital. But in view of the optimistic figures adopted for lift co-efficient and all-up weight, it is still required.
So at a VAW of 12 mph and a tow-up water speed of 3.2 mph per the enclosed diagrams of forces and velocities there is not much danger. It seems eminently achievable if ground effect can be obtained, so I recommend the project to House!
Ultimate sailing for oldies part 7
Including submissions to John Hogg Prize, Howard Fund, and Catalyst
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