### Ultimate sailing for oldies part 6

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**Sun Mar 08, 2020 6:56 pm**Ultimate sailing for oldies Part VI

Diamond planform wing

The double size hangglider layout is rejected because of the weight of the rigid members (100ft). So my attention turned to my paraglider which weighs only 10lbs (dry). This fails due to my inability to shorten the lines to get the canopy into the ground affect region without ruining its aerodynamic shape. In any event, it does not respond well when wet.

The project is morphing into how to make the most compact flying machine, and how to build a hanglider weighing as little as a paraglider. The weight and strength problems arise from the long wingspan and chord, which must be supported by spars. The double-size hangglider had a wingspan of 34 ft and a chord of 25ft. The paragilder has a wingspan of 30 ft and a maximum chord of 9.5ft.

The most compact wing planform is square, if one rejects the bizzare idea of a round wing. Assuming that the required horizontal projected area is sufficient to lift the machine and pilot, the wing’s requirements are lightness and indestructability. That means a fabric wing. I propose to use a B&Q tarpaulin. It is strong, impervious to water, easy to source, has strong metal eyelets at the corners where all the forces concentrate, and has reinforced edges. It must be see-through since it surrounds the pilot like a shroud. If a tarpaulin can lift a person, think what could be achieved with real aerofoil!

Imagine a Viking ship’s square sail, then turn it 90° so that the sale is horizontal. It needs a line to a spar to each corner for the sail to be rigid. The minimum number of spars is two, both diagonally to the four corners. A 400 sq ft square sail would require two 28.28 ft long spars, which is too long and heavy (or weak). A 225 sq ft sail with 15ft sides would need two poles only 21.21ft long.

The weight of the machine is as follows:-

Inflatable hulls 2 x 6.5lbs 13 lbs

4.5ft mast 2lbs

Tarpaulin 8lbs

Campari Catapult deck 29lbs

Pilot (the witches weight) 101lbs

Poles (say) 16lbs

Hapa (floating: nil) 0lbs

The lift formula is 176lbs

Lift = weight = 176lbs = CL p/2 SV2 = 2 x 0.0024/2 x 225 x 18ft/sec 2

=12.3 mph apparent wind.

The lift formula for a square wing has the addition of Aspect ratio/2 + Aspect ratio = 1/2+1 = 1/3, so the lift is only one third of 176lbs, so a square sail is rejected. This lift penalty is due to the two wingtips forming such a large part of the aerofoil, i.e. there are two very large wingtip vortices which result in a large loss of lift at the wingtips.

This penalty does not exist with delta wings, however low their aspect ratio, because the wingtips form such a small area of the wing. If the square wing is rotated 45° it becomes a diamond shape. I assume that this is in effect a delta wing.

In order to support the wing the two spanwise poles will stick out both sides of the tiny Campari deck. The reason why yachts do not have “bowsprints” athwartships is because they could not come alongside. We in AYRS are not troubled by such practicalities!

A high sail camber produces good lift coefficients, so I will have a large billow in the tarpaulin. The maximum chord is 21.21 ft so the mean Reynolds Number is Re = 6250 x 21.21/2 x 18ft/sec = 1,193,062. This is high compared with manpowered aircraft so it would insure that the profile drag of the wing would be lower than in manpowered aircraft (Manpowered Flight, p132).

The centre of effort of my hangglider, which is diamond shape, is 50% chord. If incorrect the pilot can easily change his seating place since the trampoline has been removed and his seat can be placed anywhere along the deck.

The wingtips of this ekranoplane contribute little lift so they could be cut off, resulting in the tapered wings usual on ekranoplanes. The constuctual complexity of this means I omit this refinement.

During the taxiing takeoff run the wintips can be one foot under water, thereby contributing a useful bit of drogue to reduce leeway and making the sea a huge weightless wingtip plate to suppress the wingtip vortices. It may have the effect of a wing of infinite span, and therefor no induced drag at all!

The weakest points structurally are the two unsupported poles. They are 21.21ft long, i.e. the diagonals across the 15ft by 15ft square. In order to decrease the length of the chordwise pole while still retaining the 225 sq ft area I have elongated the sail into a lozenge shape spanwise. The maximum chord is now 16ft 3”. The aspect ratio is increased to 3.41, which reduces induced drag. Stability is increased due to the wingtips being further apart and it looks more aeroplane-like. The two spansize poles are each only 13 ft 10” long, and therefor stronger than the 16ft 3” pole, and separate from each other because they supply the anhedral. My wasp hangglider canopy is 24ft maximum chord and 18.5 ft span, so is no use here.

The induced drag coefficient formula is:-

KCL2

Π A

The constant (K) is 1.12 for a delta wing. The higher the CL the higher the coefficient of induced drag because the latter is caused by the former. The higher the aspect ratio, the lower the induced drag, so 1.12 x 22 /π x 3.41 = 4.48/10.712856 = 0.4182

This CD1 ,is reduced by 80% when flying at one foot wingtip height, viz

Wing height/wingspan = 1ft/27.7ft = 80% (appendix 4)

CD1 0.4182x 20% = 0.0836

CD (other) = 0.0220

CD (all) =0.1056

Lift/drag ratio = CL / CD(all) = 2/0.1056 = 19 to 1, which is excellent. With lift-induced drag decreased by 80% it follows that lift is increased by 80% since they are different sides of the same coin. So the lift formula becomes : Lift = Weight = 176lbs = 2 x 0.0024/2 x (225 + 80% x 225) x 13.46 ft/sec2 =9MPH.

If the machine is drifting downward we must add the leeway speed to get the true windspeed needed on takeoff.

Checking the mean Reynolds Number we get:- 6250 x 16.25 ft/2 x 13.46 ft/sec = 683,516, which is higher than Puffin II and well within the range of other manpowered aircraft aerofoils.

This shown that the project is well worth pursuing, so I recommend it to the House!

R.E. Glencross

12 January 2019

Diamond planform wing

The double size hangglider layout is rejected because of the weight of the rigid members (100ft). So my attention turned to my paraglider which weighs only 10lbs (dry). This fails due to my inability to shorten the lines to get the canopy into the ground affect region without ruining its aerodynamic shape. In any event, it does not respond well when wet.

The project is morphing into how to make the most compact flying machine, and how to build a hanglider weighing as little as a paraglider. The weight and strength problems arise from the long wingspan and chord, which must be supported by spars. The double-size hangglider had a wingspan of 34 ft and a chord of 25ft. The paragilder has a wingspan of 30 ft and a maximum chord of 9.5ft.

The most compact wing planform is square, if one rejects the bizzare idea of a round wing. Assuming that the required horizontal projected area is sufficient to lift the machine and pilot, the wing’s requirements are lightness and indestructability. That means a fabric wing. I propose to use a B&Q tarpaulin. It is strong, impervious to water, easy to source, has strong metal eyelets at the corners where all the forces concentrate, and has reinforced edges. It must be see-through since it surrounds the pilot like a shroud. If a tarpaulin can lift a person, think what could be achieved with real aerofoil!

Imagine a Viking ship’s square sail, then turn it 90° so that the sale is horizontal. It needs a line to a spar to each corner for the sail to be rigid. The minimum number of spars is two, both diagonally to the four corners. A 400 sq ft square sail would require two 28.28 ft long spars, which is too long and heavy (or weak). A 225 sq ft sail with 15ft sides would need two poles only 21.21ft long.

The weight of the machine is as follows:-

Inflatable hulls 2 x 6.5lbs 13 lbs

4.5ft mast 2lbs

Tarpaulin 8lbs

Campari Catapult deck 29lbs

Pilot (the witches weight) 101lbs

Poles (say) 16lbs

Hapa (floating: nil) 0lbs

The lift formula is 176lbs

Lift = weight = 176lbs = CL p/2 SV2 = 2 x 0.0024/2 x 225 x 18ft/sec 2

=12.3 mph apparent wind.

The lift formula for a square wing has the addition of Aspect ratio/2 + Aspect ratio = 1/2+1 = 1/3, so the lift is only one third of 176lbs, so a square sail is rejected. This lift penalty is due to the two wingtips forming such a large part of the aerofoil, i.e. there are two very large wingtip vortices which result in a large loss of lift at the wingtips.

This penalty does not exist with delta wings, however low their aspect ratio, because the wingtips form such a small area of the wing. If the square wing is rotated 45° it becomes a diamond shape. I assume that this is in effect a delta wing.

In order to support the wing the two spanwise poles will stick out both sides of the tiny Campari deck. The reason why yachts do not have “bowsprints” athwartships is because they could not come alongside. We in AYRS are not troubled by such practicalities!

A high sail camber produces good lift coefficients, so I will have a large billow in the tarpaulin. The maximum chord is 21.21 ft so the mean Reynolds Number is Re = 6250 x 21.21/2 x 18ft/sec = 1,193,062. This is high compared with manpowered aircraft so it would insure that the profile drag of the wing would be lower than in manpowered aircraft (Manpowered Flight, p132).

The centre of effort of my hangglider, which is diamond shape, is 50% chord. If incorrect the pilot can easily change his seating place since the trampoline has been removed and his seat can be placed anywhere along the deck.

The wingtips of this ekranoplane contribute little lift so they could be cut off, resulting in the tapered wings usual on ekranoplanes. The constuctual complexity of this means I omit this refinement.

During the taxiing takeoff run the wintips can be one foot under water, thereby contributing a useful bit of drogue to reduce leeway and making the sea a huge weightless wingtip plate to suppress the wingtip vortices. It may have the effect of a wing of infinite span, and therefor no induced drag at all!

The weakest points structurally are the two unsupported poles. They are 21.21ft long, i.e. the diagonals across the 15ft by 15ft square. In order to decrease the length of the chordwise pole while still retaining the 225 sq ft area I have elongated the sail into a lozenge shape spanwise. The maximum chord is now 16ft 3”. The aspect ratio is increased to 3.41, which reduces induced drag. Stability is increased due to the wingtips being further apart and it looks more aeroplane-like. The two spansize poles are each only 13 ft 10” long, and therefor stronger than the 16ft 3” pole, and separate from each other because they supply the anhedral. My wasp hangglider canopy is 24ft maximum chord and 18.5 ft span, so is no use here.

The induced drag coefficient formula is:-

KCL2

Π A

The constant (K) is 1.12 for a delta wing. The higher the CL the higher the coefficient of induced drag because the latter is caused by the former. The higher the aspect ratio, the lower the induced drag, so 1.12 x 22 /π x 3.41 = 4.48/10.712856 = 0.4182

This CD1 ,is reduced by 80% when flying at one foot wingtip height, viz

Wing height/wingspan = 1ft/27.7ft = 80% (appendix 4)

CD1 0.4182x 20% = 0.0836

CD (other) = 0.0220

CD (all) =0.1056

Lift/drag ratio = CL / CD(all) = 2/0.1056 = 19 to 1, which is excellent. With lift-induced drag decreased by 80% it follows that lift is increased by 80% since they are different sides of the same coin. So the lift formula becomes : Lift = Weight = 176lbs = 2 x 0.0024/2 x (225 + 80% x 225) x 13.46 ft/sec2 =9MPH.

If the machine is drifting downward we must add the leeway speed to get the true windspeed needed on takeoff.

Checking the mean Reynolds Number we get:- 6250 x 16.25 ft/2 x 13.46 ft/sec = 683,516, which is higher than Puffin II and well within the range of other manpowered aircraft aerofoils.

This shown that the project is well worth pursuing, so I recommend it to the House!

R.E. Glencross

12 January 2019