## Ultimate Sailing for Oldies

Including submissions to John Hogg Prize, Howard Fund, and Catalyst
R Glencross
newcomer
Posts: 7
Joined: Wed Dec 06, 2017 11:59 am

### Ultimate sailing for oldies part 2

Ultimate Sailing for Oldies, Part 2

The seaplane’s lift has been calculated above.
The craft’s flying capabilities and course can be computed if we know the aerodynamic and hydrodynamic drag and the true wind speed and direction. With regards to the aerodynamic drag, the vast majority of this is induced drag, because of the low aspect ratio and delta wing planform. This is the downside of the huge lift vortices that occur at the wingtips as experienced by crabclaw rigs and the late lamented Concord airliner.

The formula for the coefficient of induced drag is:-

CD, = 1.12 Cl2
Π A
For the 1.12 see appendix 5 (Manpowered Flight, page 51). The wing tapers to a point. The Cl2 is included because the induced drag is induced by lift, so the higher the Cl the higher the higher the CD,. π creeps into many equations for reasons which defeat this student of ‘O’ level maths. (Editor – can you help here?). (No - Ed). The aspect ratio is included in order to punish low aspect-ratio aerofoils like this one. The wingtip region has a greater effect on airflow round the wing on a low aspect ratio wing than on a high aspect- ratio wing i.e. the wingtip vortices are larger on a low aspect ratio wing.

So we get:-

CD, = 1.12 x 12 = 1.12 = 0.133
Π x 2.67 = 8.39

The remaining drag is the form drag, skin friction and parasitic drag. These form only small proportion of the total drag in a low aspect ratio craft. The most draggy manpowered aircraft mentioned by Sherwin is the Ottawa craft with a CD of 0.011. It was streamlined whilst my pilot and undercarriage are not streamlined so the figure will be in excess of 0.011,say 0.022. The Ottawa craft is also the manpowered aeroplane with the highest Reynolds Number (Re), at 900,000. The lower the Re the greater the drag, so let us compute the Re for the hanglider boat. The formula for computing Re is :-
Re =PCV
M
P is the density of the air, C is the wing’s mean chord in feet, V is airspeed in ft/sec, and m is the coefficient of friction of air, i.e. the viscosity of air,
P = 6250 in air (see manpowered flight page 132).
M
So the Re of the hanglider boat is:-
Re = 6250 x 24ft span x 22 ft/sec = 1,650,000.
2
which is reassuringly high.
So we have nothing to worry about concerning low Reynolds numbers.

As a further feasibility check on the figures, let us compute the lift/drag ratio of the hanglider in free flight outside the ground effect region. Cl is unity, CDi is 0.133 and CD (other) is 0.022, so total drag coefficient is 0.155

Lift = Cl = 1 = 6.45 to 1
Drag = CD all = 0.155
which is within the bounds of possibility, but seems optimistic. I suspect that, pace Abbott and Boenhoff, the cruising Cl outside the ground effect region is in excess of unity. If it is 2 we get CDi = 1.12 x 22 = 4.48 = 0.534
Π x 2.67 8.39
Add CD (other) of 0.022 we get CD of 0.556, so
Lift = Cl = 2 = 3.6 to 1 which looks much more feasible.
Drag = CD all = 0.556

The biggest unknown in this experiment is what is the Cl of a hanglider. I can find nothing about this in the hangliding and paragliding magazines. They do not seem interested in the maths. Anyway, Cl alters with angle of attack, and also with speed. In order to adopt a conservative approach I will assume Cl is unity, except on takeoff. (otherwise I cannot take off).

During takeoff Cl = 2, so we get

CDl = 1.12 X 22 = 4.48 = 0.534
π x 2.67 8.39

Induced drag is reduced by 70% at a wingtip height of one foot (see appendix 4) so:-

CD1 = 30% x 0.534 = 0.160
CDother = = 0.022
CD all = = 0.182
So the lift/drag ratio of the hanglider boat on takeoff is:-
L = Cl = 2 =11 to 1
D CD all 0.182
This seems perfectly feasible since the advertised lift/drag ratio in free flight is five to one and we would expect a lift/drag ratio higher than that in the ground effect region.

In order for the machine to fly in any direction other than directly downwind the hangglider needs to produce a side force. This is achieved by banking the hangglider while at the same time deploying the hapa from drogue mode to hapa mode on the side opposite to the direction of bank. Let us assume an angle of bank of 10 degrees (I assume that it will not sideslip). See appendix six. This results in the vertical component of lift being reduced by approximately 4%, so the airspeed will have to be increased in order to maintain the vertical lift of 250 lbs, so:-
Vertical component of lift = 250 lbs = 1 x 0.0024 x 207 sqft x 31.7 ft/sec2
2
=Airspeed of 21.6 mph.

The hanglider’s top penetration speed is 25 mph.

At an angle of bank of 10° the side force that is produced is 45lbs (see appendix 7). The hapa formula in hapa mode is therefore:-
Thrust H= 45lbs = 1x 1.956 x 18sqft x 1.6 ft sec2
2
= 1.09 mph water speed.

So the hapa produces the required amount of thrust or resistance when pulled through the water at 1.09 mph. A smaller hapa would produce this same amount of thrust at a higher speed, but how would I take off in the first place when I need a larger area at that moment?

When the hangglider banks, the amount of ground effect is reduced on the upper wingtip due to its increased height. The mean wingtip height is now

1ft + 5 ft = 3 ft
2

So the ground effect is mean wing height = 3ft = 0.125
Wingspan 24 ft

On the graph at appendix 4 we see that the induced drag coefficient is reduced to 53% of the free flight figure. (On takeoff it was reduced to 30% of the free flight figure.)

So, with this new induced drag coefficient the air drag is as follows:-
CD1 = 53% x 0.133 = 0.070
CDother = = 0.022
CD all = = 0.092
l/d = Cl = 1 = 10.87 to 1
Cd all 0.092

This enables us to compute the drag angle. Cotangent 10.87 is about 5°, so drag angle is 5°.

The hapa has a tested drag angle of 45° so β° is 5°+ 45° = 50°. This gives the triangle of velocities as per appendix 8. This shows that there is no point in attempting a flight in less than a 21 mph wind.

A greater angle of bank would require greater airspeed in order to maintain the 250 lbs of vertical lift. Drag would also increase as the raised wingtip would receive less ground effect. It could also sideslip. A less draggy 18 square foot hapa would increase ground speed and require less true wind speed.

The real answer is to employ a lighter pilot, but volunteers are hard to find!

In the next part of my article I intend to compute the aircraft’s performance assuming a lighter (still elderly) pilot, a Cl of 2 for the hangglider, and a more efficient hapa.

As a yacht it is a disaster, requiring a 21 mph wind to achieve a water speed of one mile per hour. But as an aeroplane it is the lowest, slowest, lightest, cheapest, greenest, smallest, funnest, safest, most novel manned aircraft in the world, and I recommend it to the House!

Roger Glencross

22 January 2017

R Glencross
newcomer
Posts: 7
Joined: Wed Dec 06, 2017 11:59 am

### Ultimate sailing for oldies part 3

Ultimate Sailing for Oldies part III

Let us take the most optimistic view of Ultimate Sailing in order to check whether the game is worth the candle. Let us assume that an elderly lightweight pilot is found. Let us also assume that the coefficient of lift of a first generation hanglider is two.

Abbott and Doenhoff’s textbook “Theory of Wing Sections” was completed in June 1958, from the original 1949 edition. The authors “considered best to expedite republication by foregoing extensive revision” As a result there is no mention of single sided or flexible wings which were just emerging.

However there is a reference to Francis Rogollo, the inventor of the Rogollo hangglider. It refers to his work in designing the NACA 23030 double sided wing section. This section has a maximum coefficient of lift of 1.9 before the inevitable stall. (See appendix 9). This is with a slotted flap, a high lift device. There is no scope for sophisticated high lift devices on a simple hangglider, but perhaps the hangglider itself is a high lift device!? This strengthens my belief that a hang-glider can have a coefficient of lift of two. Perhaps AYRS members have some information on this?

The task of improving the lift/drag ratio of my crudely constructed 18 sqft flexible fabric hapa should be simple. The improved hapa should be rigid and thus heavier than my present fabric hapa. It may weigh as much as a small dinghy. In fact, a small unmanned dingy might do! A simple test, towing a dinghy behind a motorboat on two lines of unequal length, i.e. in hapa mode, would show the angle to the side that the dinghy hapa could achieve. Let us assume that a lift/drag ratio of two is achieved, as opposed to unity on my fabric hapa.

It would be unlikely that a volunteer could be found to man this dinghy hapa to deploy the two hapa lines under load! He would be at the mercy of the amateur hangglider pilot who would be in the last stages of terror!

The weight breakdown of the equipage is as follows. (The hapa’s weight is excluded because it is supported by its water displacement) :-

Lbs
Aquaviatrix (the witch’s weight) 101
Hangglider with aluminum poles 45
Less reduction for bamboo poles (21)
Campari Catapult undercarriage 55
All – up weight 180 Lbs

Assuming an angle of bank of only 5° the higher wingtip is 3ft above sea-level (appendix 6A), so the average wing height is 3ft + 1ft =2ft
2
So the ground effect is mean wing height = 2ft =0.08333
Wing span 24ft

We see on the graph at appendix 4 that the induced drag coefficient is reduced to 45% of the free flight figure. The induced drag coefficient is:-

CD1 = 1.12x22 = 4.48 = 0.534
Π x 2.67 = 8.39

With ground effect we get
CD1 = 45% x 0.534 =0.240

CD other = 0.022

CD all = 0.262

The lift/drag ratio of the hangglider is:-
L = Cl = 2 = 7.6 to 1
D CD all 0.262

Cotangent 7.6 is 7.5 degrees, so the aerodynamic part of the experiment has a drag angle of 7.5 degrees.

The horizontal projected area of the wing is reduced by approximately 2% due to the angle of bank of five degrees. So the true wing area of 216 sq ft is reduced to 211 sq ft. So the lift formula is:-

Lift = weight = 180lbs = 2 x 0.0024 x 211 x 18.852 ft/sec = airspeed of 12.8 mph.
2
The speed range for this hangglider is 15 to 25 mph But a lightweight pilot like ours could fly at a speed lower that 15 mph. The speed range given is when flown as a glider, whereas I am abusing the machine as a kite. A glider is powered solely by gravity, which is a weak force. (ignoring thermals). A kite, however, is powered by the wind, the force of which can frequently exeed the force of gravity. (Editor, can you calculate what windspeed equals the force of gravity?)

With an angle of bank of five degrees we get the triangle of forces per appendix 7 A. This shows a side force on the hapa of 15lbs. The hapa lift formula is :-
Side force or hapa thrust = 15lbs = 1 x 1.956 x 18 x 0.92 ft/sec2
2
= 0.62 mph water speed.

So the hapa produces the required amount of thrust of 15lbs when pulled through the water at 0.62 mph. The lift co-efficient of unity should not be confused with the lift/drag ratio of the hapa of two. The former informs the quantity of lift and the latter informs the direction of the lift.

We can now prepare the new triangle of velocities (appendix 8A). The drag angle of the aerodynamic part is 7.5 degrees (see above). The lift/drag ratio of the hapa is two to one. Cotangent 2 is 26.5 degrees, so the beta angle is 7.5 degrees plus 26.5 degrees equal 34 degrees. The triangle of velocities shows that the machine will fly in a true windspeed of only 12.2 mph!

The forces on the machines can be adjusted by altering the angle of bank, therely giving control over speed and course, within limits.

The main attraction of this experiment is that all the hardware is to hand:- the hangglider canopy comes courtesy of John Smith and Bob Date of the Bristol Mob, the A frame is salvaged from my old hangglider, the undercarriage is a veteran of many previous experiments and I possess a large stock of bamboo. All that is wanting is an airy, agile, aged Aquarviatrix……….

Roger Glencross

February 2017

R Glencross
newcomer
Posts: 7
Joined: Wed Dec 06, 2017 11:59 am

### ultimate sailing for oldies part 4

Ultimate Sailing for Oldies part IV

In part III I assumed that the hanglider- boat had an all-up weight of 180 lbs, that the hanglider had a lift co-efficient of two, the 18 sq. ft. hapa had a lift/drag ratio of two, the hapa had a lift co-efficient of unity and, the hang-glider had a lift/drag ratio of 7.6 in ground effect.

The craft must have a reasonable range of flying ability in order to be worth constructing. The flying range which I am proposing is as per appendix 8B. I have taken three situations: at a VT/VAW angle of 30° (triangle ABC), at a VT/VAW angle of 45° (triangle ABD) and at a VT/VAW angle of 90° (triangle ABE). These are all in a true windspeed of 12.2mph, the minimum windspeed which permits normal flight. The range of apparent windspeeds from 18.2 mph to 21.5 mph are within the speed range of this hanglider, which has a top airspeed of 25 mph when flown as a glider. When misused as a kite, it could no doubt go faster than 25 mph, until something broke!

There is no reason to believe that the beta angle of 34° would alter at different speeds because lift/drag ratios are fairly consistent over quite a wide speed range. Any course greater than 90° between VT/VAW would require a beta angle of less than 34°, so would be impossible with a hapa and a hangglider with these lift/drag ratios.

At top air speed (triangle ABD) the hangglider experiences an airspeed of 21.5 mph (31.5 ft./sec). But it cannot fly outside of the ground effect region, so the pilot must pull in the control bar to reduce the angle of attack, i.e. reduce the co-efficient of lift to, say 0.72, as follows:-

Weight = lift=180lbs = 0.72 X 0.0024 X 211 sq. ft. X 31.5ft/sec2.
2
Whatever the hapa’s water speed, the thrust or lift which it produces must remain constant i.e. not increase with speed. This is because the weight of the flying machine remains constant, the lift remains constant (i.e. the machine does not leave the ground effect region), the lift/drag ratio of the hangglider remains constant over the speed range, therefor the drag remains the same, the angle of bank of 5° remains constant, therefor the side force of 15lbs remains constant. So the thrust required of the hapa must remain at 15lbs at all water speeds ;(Action and reaction are equal and opposite).

The hapa lift formulae for water speeds of 11 mph (triangle ABC), 15.5 mph (triangle ABD) and 22 mph (triangle ABE) are as follows for a constant hapa lift of 15 lbs:-

11 mph LH = 15lbs = 1 X 1.956 X 0.0589 sq. ft. X 16.137ft/sec2
2
15.5 mph LH = 15lbs = 1 X 1.956 X 0.0297 sq. ft. X 22.7385ft/sec2
2
22 mph LH = 15lbs = 1 X 1.956 X 0.0147 sq. ft. X 32.274ft/sec2
2
So the working hapa area has to automatically reduce from 18 sq ft (as required for takeoff) to 2.1 square inches at 22 mph! So the necessarity of a variable geometry hapa is established.

Various ways of dodging this immense problem come to mind. The hangglider could increase its angle of bank, and thereby increase the side force. At an angle of bank of 60° while still maintaining a vertical lift force of 180lbs, it would produce a side force of 316 lbs when flying at 18.2 mph (appendix 7B) But at 22 mph hapa speed we get the hapa lift formula as:
LH = 1 X 1.956 X 18 sq. ft. X 32.274 ft./sec2 = 18,462 lbs.!
2
One could increase the flying height and experience stronger winds due to the wind velocity gradient but that would loose ground effect. Also, the horizontal components of the hapa’s lift (its motive power), would reduce, resulting in speed reducing, whilst doing nothing to reduce the total hapa drag, thus reducing its effective lift/drag ratio. I assume that the hapa will remain vertical. If one wished to reduce the effective hapa area by making it take a less vertical attitude, one would need a line to the deepest wingtip of the hapa. Dragging lines, even fared lines through the water at speed is a definite no-no. Lines vibrate at speed so as to resemble a wide plank being dragged through the water wide side first!

So I see no way of avoiding the need for a variable geometry hapa. One can easily envisage an object which becomes larger when under load e.g. stretching, overlapping shutters, springs etc. but I need an object which becomes smaller under load. One can envisage a ladderfoil as per Williwaw, but the reduction in submerged area would have to be 98% at top speed. The present mark 1,2 and 3 hapas do rise out of the water at speed to a small extent, but not sufficiently to increase the flyer’s speed range noticeably. In any case their takeoff speed is too fast for my purpose.

One could use drag devises e.g. buckets, which emerge from the water one by one as the amount of resistance required is reduced, but one would need a great many buckets. Alternatively the craft could be secured to an anchored buoy before takeoff, thereby having in effect a drogue of infinite area (dangerous, because one cannot release energy from the system in a gust), release the line under great load, shoot sideways with the hapa and hope for the best! But the best small hapa in the world would only produce a little lift at very low speeds, so the equipage would probably come to an abrupt halt.

A drawstring bag becomes smaller when pulled, but cannot open out again automatically. It could be spring loaded to make it open out when the load is reduced.

In order to take off in the first place, the hapa will have to adopt drogue mode. The hanglider will have to face into wind and drift downwind. When the resistance of the hapa equals the drag of the hangglider the equipage will lift off. The lift of the hangglider in order ot take off must equal or exceed its weight, i.e. 180 lbs. This occurs in an apparent wind of 12.8 mph. The aerodynamic drag when in drogue mode is 180 = 25lbs, rather larger than the 15lbs drag when
7.2
flying in hapa mode .

Drogues have a drag co-efficeint of 1.2, rather larger than the lift co-efficient of unity which I have assumed for hapa mode.
The drogue formula is:-
25lbs = 1.2 X 1.956 X 18 X 1.09 ft/sec2
2

So the required quantity of drogue resistance is reached when the hapa is being dragged downwind at 0.74 mph. This speed must be added to the required apparent windspeed of 12.8 mph to compute the true windspeed required for takeoff of 13.54 mph. So a slightly higher true windspeed is required at takeoff than is required for normal flight, in order to overcome the low speed hump. This low speed hump problem continues until the hapa is travelling at one or two miles per hour, thus requiring it to travel in “demi-drogue” mode at that speed. This requirement, and the need to change from drogue mode to hapa mode in a controlled way, may not happen automatically, but may require the hapa to be controlled by the pilot at all times.

Whoever said that there was no further scope for amateur inventors in the field of yachting! The design of a variable geometry hapa would make an excellent AYRS project and I urge the AYRS committee to offer a prize to the member (or non-member) who successfully builds one.

Roger Glencross
November 2017

R Glencross
newcomer
Posts: 7
Joined: Wed Dec 06, 2017 11:59 am

### ultimate sailing for oldies part 5

Ultimate Sailing for Oldies part V

At the AYRS all-day seminars in November 2017 and January 2018 various comments, not to say objections, were raised re windpowered flight for Oldies. In true New Scientist style I will now correct the errors which have occurred in previous editions of Ultimate Sailing for Oldies.

Fred Ball carried out stress tests on my bamboo poles which proved that they were too weak. So I would build the hangglider with carbon fibre poles, which are not only stronger but lighter. It was also pointed out that I would need more lift than I had computed, thus requiring a higher windspeed or larger wing. I investigated the latest state-of-the-art hanggliders: the combat GT glider weighs 79lbs, over three times the weight of my hangglider. It has only 145 sq ft sail area, two-thirds of mine. The vital statistic in this project is a high lift/drag ratio, but the Combat GT is no better in this respect than models which went before it. It merely has a higher top speed, which is not the object of my experiment. It costs £3,500!

So rejecting a faster wing leaves the need for a larger wing. AYRS members thought that there would be difficulty recruiting an Aquaviatrix weighing 101lbs: also more force might be needed to overcome hydrodynamic drag during takeoff and the windspeed would be lower at sea level. The thought arose of using not one but two hanggliders. The trigger for this notion was that many older fliers posses first generation hanggliders which they would never dream of flying again, but which need a good home.

A biplane layout could be built compact and therefore strong, and would be spectacular and fun. But the benefit of the all-important ground effect would occur only under the lower wing. There would be interference effects between the wings which would reduce lift. There would be greatly increased induced drag due to four, instead of two wingtips. Thus there would be a reduction in lift/drag ratio. This must be kept high to permit low-speed flight. So I reject the biplane layout.

The hanggliders could be set up one behind the other. But that would mean the rear glider flying in turbulent airflow. Also, the pilot would not be under the centre of lift as he operates one or other of the gliders so I reject this setup.

That leaves the hanggliders side by side. This has the advantage of increasing the aspect ratio and therefore the lift/drag ratio. The gliders could share a leading edge spar and control frame, so the total weight would be less than double the weight of one glider.

With two gliders one has the option of introducing dihedral. This produces extra stability because the glider wallows in the air rather than balancing on it. The side elevation would give the necessary sideforce to permit sailing. This area could be increased if required by increasing the angle of dihedral or by erecting a sail vertically on top of the gliders.

However, the raised wingtips would loose the all-important ground effect, so I propose anhedral instead, thereby bringing both wingtips close to the water’s surface, thus creating a sailing ekranoplane or wing in ground effect! Small floats or amas on the wingtips would add stability before and during takeoff and enable the central undercarriage to be reduced in size to an ama, resulting in a saving in weight. Happily, AYRS publication 126 “Low Flying Boats” provides the necessary formulae to design the ekranoplane.

Other suggested ways of increasing lift were: awaiting stronger winds, but they are seldom available when required and they produce bigger waves; flying higher to use the wind velocity gradient, but that would loose ground effect.

Another error pointed out concerned Edmond Bruce’s work in Ayrs 61 “Sailing Analyses.” I had assumed a hapa lift coefficient of unity. Bruce states (page 27) that a foil’s Cl at maximum lift/drag ratio is 0.40 regardless of the NACA section, even with a flat plate. Since it is necessary for the hapa to have the highest lift/drag ratio possible, and since the lift\drag ratio which I have posited of two to one is optimistic, I must deploy the hapa at its best lift/drag ratio. Therefor the hapa Cl of 0.40 must be substituted for unity.

But this figure is only correct at a high Reynolds Number (Re), which means six million, and at an aspect ratio of six to one. The formula for Re is:
1 metre chord X 1 metre per second speed x 1,000,000 =Re. In other words a foil of one metre chord travelling at one metre per second has a Re of one million. So, referring to appendix 8B, triangle ABF at hapa speed of 2MPH (equals approximately one metre per second) we get:
6m chord x 1m/sec x 1,000,000 = 6,000,000
So the hapa must be 19.5 feet wide! In order to reduce the required width, let us assume 4MPH (= approximately 2m/sec):-
3m chord x 2m/sec x 1,000,000 = 6,000,000
That is, 9.75 ft chord. It would be difficult to make a hapa 9.75 ft wide which could be pulled at 22MPH (appendix 8B triangle ABE).

But how does the hapa get up to 4 MPH in the first place? The answer is a tow-up. Readers may feel that this makes a farce of the whole project, if they have not already come to that conclusion! But no! As demonstrated on film at the AYRS winter seminars, ice yacht races start by the sailor pushing his yacht, then jumping aboard when it is up and running. I assumed that this was because the coefficient of static resistance is much higher than the coefficient of sliding resistance. But I am informed that it is in order to pump the sail up. No one considers ice yachting a farce and my craft requires this initial boost for the very same reason.

The hapa needs an aspect ratio of six to one per Bruce. With a chord of 9.75ft that means a draught of 58.5.ft and an area of 570 sq ft! Is the project doomed? No! In AYRS publication 117 “Natural Aerodynamics” page 40, Ian Hannay states “An aspect ratio of ten to one will feel to nature the same as an aspect ratio of one to ten The maximum drag reduces below an aspect ratio of one. The drag force is along the line of flow so that an aspect ratio of ten to one looks the same to the drag flow as 1/10. It is just rotated 90°,thus below an aspect ratio of one the induced drag reduces”. Thus a javelin (AR 1/100) will always travel further than a discus of the same weight.

So, with an aspect ratio of 1/6 we have a span of 9.75ft = 1.625 ft and an
area of 15.84 sq ft. 6
At 4 MPH (5.868 ft/sec) we get this hapa lift:-
Lift = 0.4 x 1.956 x 15.84 sq ft x 5.868 ft/sec2 = 213lbs. But the side force required
2
is only 15lbs (appendix 7A), so the hapa is far too large.

However all is not lost. In a letter from the author of AYRS publication “Power from the Wind”, Amic informed me in the context of a NACA 0012 foil (in air, but the fluid is irrelevant at the same Re), at a Re of only 440,000 it had a maximum Cl of 0.95 at an aspect ratio of only 3.5. A flat plate section has a maximum L/D ratio of 6.7 at Re = 6,000,000 and AR = 6 per Bruce. Cl and L/D ratio will be lower at lower Re and at lower AR. We can do better than a flat plate by using a segment of a circle of various thicknesses. The hapa must be symmetrical in the fore and aft sense because it must shunt.

Let us assume that C l deteriorates to 0.10 at low Re and low AR and L/D goes down to two. With a hapa of 1.3 ft chord, 1.93 ft projected span, 2.5sq ft projected area and a projected aspect ratio of 1.49 to one we obtain the required force of 15 lbs at 8 ft/sec, as follows:
Lift – 0.10 x 1.956 x 2.5 sq ft x 8ft/sec2 =15lbs.
2
So at a hapa speed of only 5.4 MPH sufficient thrust is produced to propel the machine, which is in effect weightless since the lift equals the weight. Up to a water speed of 5.4 MPH a tow-up is required. Tow-up also avoids the need for a drogue, which would have to somehow automatically morph into a hapa. The Re at 1.3 ft chord (0.4 M) and speed of 8ft/sec (2.43m/sec) is:
0.4 x 2.43 x 1,000,000 = Re 972,000, well above Amic’s 440,000 figure.

There is no difficulty in getting sufficient lift in water because it is 815 times thicker than air, and salt water is thicker than fresh. The problem of getting the hapa to reduce in area as it gets faster (variable geometry) can be left to practical experimentation on water. The main problem is getting enough aerodynamic lift for the machine to take off while travelling slowly at sea level. This is the problem which I will now address.

The amount of anhedral should be the minimum possible which still gives enough headroom to permit the pilot to operate the A frame. Per appendix 2 that means a 7 ft distance from the kingbolt to the water surface. The reason for this is that it must produce the maximum amount of lift and therefore must have the maximum projected horizontal area. It may be impractical to join together two identical hanggliders so a bespoke glider of twice the area of a first generation hangglider may have to be manufactured.

A structure built twice as large as a similar structure will not weigh twice the smaller structure’s weight but considerably more than twice the weight (the square cube law). But there will be savings in weight (a single A frame, a smaller central ama) to counter this, but also the additional weight of the two wingtip floats, a strut to prevent the glider falling on the pilot’s head, and a horizontal strut at sea level joining the wingtips to the central ama to keep the glider rigid. I will take a very optimistic view of the weight so that if the figures show that even this unfeasably light craft cannot take off then I know that the game is not worth the candle.
So estimated weight is :- lbs
Aquaviatrix (the witch’s weight) 101
Hanggliders 2 x 24lbs 48
Central ama plus extras 55
204 lbs

Scaling up the hangglider per appendix 2 we get:
Span 34 ft
length 25.5ft
Area 432sq ft (34ft x 25.5 ft)
2
Appendix 2 A shows that the anhedral angle is 134° (67° x 2) and the projected horizontal area is 31.6 ft span x 25.5 ft length ÷ 2 = 403ft. The lift formula is :-
Lift = weight = 204lbs = Cl p Sv2 = 2 x 0.0024 x 403 x 14.5 ft/sec2
2 2
So takeoff speed is 10 MPH (apparent wind).

AYRS 126 “low flying boats” page 11 mentions the problem of dealing with changes of the position of the centre of lift and the associated pitch stability problem when a wing moves in and out of the ground effect. I do not intend my craft to leave the ground effect region, so that will not be a problem.

Checking the Reynolds Number in air:-
6250 x chord (ft) x velocity (ft.sec)
6250 x 25.5 x 14.5 ft/sec = 2.310,937 which is reassuringly high.

Chuck Bixel states (p38 AYRS 126) “I’m not sure the Reynolds Numbers are a good means of interpreting very low aspect ratio lift data”. And on page 39: “For very low aspect ratio double-wing designs, Reynolds Number scaling formulae may not be quite the same as for conventional wings.” He states (pg 42) “airspeeds over 100 kts are required before substantial ground effect reactions can occur”. He states (p 47) that a high angle of attack and a very low aspect ratio are needed to produce large ground effect, with the trailing edge very close to the ground. These claims are contradicted by Keith Sherwin in his book “Man Powered Flight”

Page 49 states, in the context of the Flarecraft Corporation’s sales speil, that “the pressure under the wing increases with speed”. If this means that slow speed equals poor ground effect it is not evidenced in Keith Sherwin’s graph in appendix four.

Page 14 states “in ground effect, the wing performs as if its aspect ratio were larger than that calculated from the planform of the wing”. But this does not apply to wings that taper to a point, like mine, so I have ignored this. Page 15 states “If the angle of attack is held constant, in ground effect the lift co-efficient, and thus the lift, will be larger than in free flight. But Hoerner (p 17) indicates that for Cl greater than 1.5 ground effect decreases. Wingtip plates are said to increase lift by 20%.

I assume that both wingtips have a height over the water of one foot, as a margin of safety and to skim any waves. The coefficient of ground effect is:-
CD = 1.12 x 22 = 4.48 = 0.534 per Ultimate Sailing for Oldies part two.
π x 2.67 8.39

Wing height over projected wing span is 1 ft i.e. 0.03, which shows (appendix 4)
31.6ft
that induced drag is reduced by the maximum possible (80%). So CDl =20% x 0.534 = 0.107 Other drag is more or less proportional to sail area, so 2 x 0.022 = 0.044. So total drag coefficient is
CD1 0.107
CD other 0.044
CD all 0.151

Lift/drag ratio of the machine is:-
Cl = 2 = 13.25 to 1
CDall 0.151

Contangent 13.25 is about 4.5 degrees. The hapa L/D ratio is assumed to be 2 to 1. Cotangent 2 is 26.5 so beta angle is 4.5 ° + 26.5° = 31°. That gives us the triangles of velocities per appendix 8c.

The machine must have a side force if it is to sail. Once the ekranoplane has taken off it is, in effect, weightless, since its lift equals its weight. Thus the side force required, which must equal the hapa force, is small.

The hapa experiences hydrodynamic resistance and overcomes it with a lift/drag ratio of 2 to 1, which ensures that most of the hapa force is working in the required direction. The flying ekranoplane experiences no hydrodynamic forces, but it does experience aerodynamic drag. This is the only force that the hapa’s lift must equal and oppose (once acceleration has ceased).

This aerodynamic drag should not be confused with that aerodynamic drag which works at right angles to the ekronoplane’s lift force i.e. in exactly the opposite direction of flight. The drag force which the hapa must oppose is the drag force from, in effect, the vertical sail atop the ekranoplane. This can either be an actual sail on top of the machine or the machine’s side elevation if it is sufficient. If it is insufficient it can be increased by altering the anhedral angle.

This project is inexpensive, does not require extreme conditions, is not re-inventing the wheel, not too large, low tech, requires no hard-to-obtain facilities like runways, etc, is safe, silent, green, suitable for the amateur and I recommend it to the House!

Roger Glencross
December 2018

R Glencross
newcomer
Posts: 7
Joined: Wed Dec 06, 2017 11:59 am

### Ultimate Sailing for Oldies

Introduction

Now that Ultimate Sailing has been achieved by the French, using hapas inspired by Didier Costes and designed and manufactured by Stephane Rousson, it is opportune to upgrade Professor J. G. Hagedoorn’s booklet “Ultimate Sailing, introducing the hapa”. I propose to move from Hagedoorn’s theory to the practice of existing aerofoils and hapas.

Hagedoorn considered using the 242 square foot Notre Dame Para-foil of aspect ratio of two as his aerofoil. He did not design or build a hapa that was stable at speed. He computed the triangle of velocities (true windspeed, apparent windspeed and water speed) for notional kite/hapa combinations with lift/drag ratios of kite and hapa of five to one each, on a towline slope of three in ten. He also computed the air speeds and water speeds for a notional range of kite/hapa combinations with lift/drag ratios from two to ten for both elements, with pull over mass on the towline of from one to twelve, all at a towline slope of three in ten.

But the practice is somewhat different from Hagedoorn’s mathematical models! Kites parafoils and hanggliders have progressed greatly since his 1960’s designed Notre Dame Para-foil. However kites and hapas with lift/drag ratios of five to one, let alone ten to one, have not been realized. Also the towline slope posited of three in ten is yet to be proved and may not be constant.

Let us consider existing hapas (The marks are my own).
• Mark 1 0.833 square foot projected area Aspect ratio 8.5 to 1
• Mark 2 1.3 square foot projected area Aspect ratio 4.46 to 1
• Mark 3 1 square foot projected area Aspect ration 5.7 to 1
• Fabric 18 square foot projected area Aspect ratio 0.42 to 1
• Fabric 3.67 square foot projected area Aspect ratio 0.27 to 1
The Mark 1,2 and 3 hapas seem to have drag angles varying with speed from 25° to 35° , i.e. lift/drag ratios from two to 1.4. The Mark 3 hapa worked well at October 2016 Speedweek. I do not know the lift/drag ratios of inflatable kites, but as kitesurfers are not very close-winded, even when using efficient boards and skegs, I suspect they are low.

In order to lift the weight of a person one must have sufficient vertical projected kite area and sufficient apparent wind. An enormous kite is dangerous. The vertical projected kite area is reduced by the semicircular shape and the far from vertical slope of the kite lines of kitesurfer kites. Thus all kitepowered man lifting has succeeded only at relatively high water speed. Until now! Happily Mark 1,2 and 3 hapas cope well at such speeds, but unhappily they confine ultimate sailing to the Young, the Athletic, the Strong and the Light in Weight.

What is left for us Oldies? I am not concerned with how close-winded I may fly, but only with flying! So I am only concerned with lift, not drag. I plan to deploy my 18 square foot fabric hapa initially in drogue mode, which would produce a drag co-efficient of 1.2 (see Ian Hannay's Natural Aerodynamics, AYRS 117 page 41). For the aerofoil I would use a Skyhook 111A hangglider of 216 square foot wing area, aspect ratio 2.63 to 1, which flies at an airspeed between 15 to 25 MPH (22 to 37 feet per second).

The advantage of a hangglider over kites is that the flier has control over the angle of attack in a hangglider via the A frame control bar, provided the Aquaviator’s arms are long enough and he has the strength to push forward hard enough. This ability to produce “superlift” i.e. lift above the sustainable figure, for a short time before the flow collapses, is necessary for the craft to take off. See AYRS 117 page 39.

Abbott and von Doenhoff tell us that a lift co-efficient of unity is the best that one can normally expect at an acceptable lift/drag ratio. So with the Skyhook 111A hangglider we get:-
Lift = CL × ½ ×Air density × wing area × velocity2
The minimum airspeed is 22 ft/sec so we get:-
Lift = 1 × ½×0.0024 × 216 × 22 ×22 = 125lbs
But the all-up weight is 250lbs so we either need to operate at a CL of 2 (“superlift”) or increase the flying speed to over 31 ft/sec (nearly 22 mph), which is a little high for my liking.

Hence we have a need for superlift by pushing forward the A-frame control handle, effecting a larger angle of attack, and producing a lift co-efficient of 2.0. A “B” bar would be better, enabling the Aquaviator to pull it back to his stomach while still enabling him to push the bar further out.

In practice not all of the 250lbs lift will available to lift the Aquaviator, due to the positive angle of the hapa line with the horizon. But the effect of this is not significant, in my opinion.

The drogue must produce sufficient resistance to resist the aerofoil from merely blowing downwind and losing its apparent wind of 22 ft/sec. The hangglider will have a lift/drag ratio of less than its advertised five to one ratio due to its enhanced angle of attack on takeoff, so let us say three to one. So the required resistance is 250 lbs ÷ 3 = 83lbs.

The 18 sq ft drogue has a drag co-efficient of 1.2 (AYRS 117 page 41), so it will develop a resistance of 83lbs at a water speed given by:-
83lbs = 1.2 × ½×1.956 × 18 × VW2 ft/ sec
or VW = 1.98 ft/sec i.e 1.3 MPH.
This nearly 2ft/sec water speed downwind must be added to the aerofoil’s air speed of 22 ft/sec, making a required true windspeed on 24 ft/see i.e. 16 MPH.

Before the “superlift” collapses the drogue must be deployed into hapa mode while the hangglider reduces its angle of attack as it picks up speed and becomes less downwinded. The lift/drag ratio of the 18sqft fabric hapa is poor, that is about one to one, a drag angle of 45°, but as I am only concerned with flying and not with travelling close-winded or achieving a fast ground speed, all that is required from the hapa is sufficient resistance to stop too much leeway. Thus I will maintain sufficient apparent wind to fly.

The hangglider would fly close to the sea surface so ground effect would reduce the induced drag. This is the main component of drag in a low aspect ratio aerofoil with a lift co-efficient as high as two due to the wingtip eddies, i.e. vortex lift, which are so marked in delta wing aircraft. Ground effect also increases the amount of lift generated. The quantity of ground effect depends on the height of the wing tips divided by the length of the wingspan. The lower the wingtip height and the lower the aspect ratio the more the benefit of the ground effect is felt.

The wing span is 24 feet and I compute the wingtip height at one foot. This gives a 70% reduction in induced drag. (see K. Sherwin “Manpowered Flight” page 53). I have not included this welcome bonus in my figures, but not because of any lack of faith in the benefits of ground effect.
If in the process of the above project I invent the slowest, lowest, most inefficient manned aircraft ever, so much the better!

What could possibly go wrong?
Roger Glencross
October 2016

Appendix 1: Weight breakdown
Pilot 150
Hangglider (dry) 45
Campari Catapult dinghy as undercarriage 55
All-up weight 250lbs