(Mis-) Calculations
At chapter 05.04. ´Lift at Wings´ I described process of lift and made some calculations. As I don´t like that ´formula-caboodle´ I asked specialists to check these calculations. Past eight weeks that chapter was visited three thousand times and downloaded about one thousand times. However I got no hint for mistakes (even one value has one null too less), critic discussions came up only with some readers. ´Naturally´ my calculations were wrong and formula were used most fuzzy. These sections still are there unchanged, however at the following I will show most understandable and easy procedure of calculations and correct results.
 |
| Span |
80 m |
| Length / Height |
73 m / 24 m |
| Wing-Surface |
850 m^2 |
| Start-Weight |
560.000 - 590.000 kg |
| Payload |
83.000 - 152.000 kg |
| Fuel-Litre / -Weight |
310.000 L / 270.000 kg |
| Range |
10.000 - 15.000 km |
| Passengers |
500 - 850 |
| Freight-Volume |
1.000 m^3 |
| Optimum Speed |
0,85 Mach |
| Start- / Landing Speed |
270 km/h |
| Thrust (Start) |
310 - 340 kN |
| Noise Level |
100 dB |
One physician did not examine my calculations however gave me hints concerning formula and common opinions, e.g. lift of airplane would result by lowering corresponding masses of air and approved this by some data of new A380. So at the following are mentioned some figures of that impressive plane (some values rounded).
Data of A380
Eighty meters wide is that bird, seven 120-m^2-flats could be arranged at surface of wings, it weights as much as three hundred well build cars, each occupied with two or three passengers corresponds to its capacity. Like at land-vehicles, payload is relative small, only one seventh (or quarter at pure freight version), because that machine primary is a tanker lorry - nearby half of start weight is fuel.
More than 300.000 N thrust accelerate that airplane, which flies over quarter of globe at half of day by 850 km/h. Moving seven times faster than cars is expensive, each passenger ´consumes´ as much fuel as cars need (even there are four persons within car). Who believes it´s important coming to far continents within few hours, takes noise level of 100 dB easy.
Flying by itself is great, that A380 by itself is technical master piece - however it makes no sense for me. Nevertheless ´normal´ people seem not bothered by that stress and don´t suffer when ´flying off their souls´ and would need days to ´find back to themselves´. However that´s not subject here, because here are discussed only physical data and facts.
Classic Calculating
Previous specialist made following assumptions: mass m = 500 t (so 500.000 kg), wing surface S = 850 m^2, speed v = 100 m/s (so 360 km/h), density of air rho = 1 kg/m^3. Resulting is demanded force of lift A = m * g = 4.905.000 N. By classic formula for lift A = 0,5 * rho * v^2 * S * ca results lift-factor ca = 1,15. By data of wing further results demanded angles of attack alpha = 13,3 degrees (relative to angles at which wing produces null lift). Again results induced resistance W = 237.600 N, at previous speed to break by about 32.300 HP. Indeed, these results are calculated correct.
That expert now takes view of common understanding, that performance for production of downward-wind corresponds to force necessary for lifting airplane (by classic-mechanic reliable principle of actio = reactio). Based on angles of attack it´s easy to calculate, wings have to push down air by about 24 m/s. Only air direct downside of wing is accelerated that fast while air some more below is accelerated less. So this expert assumes average speed of downward-wind to be some 12 m/s.
Based on constant of impulses - airplane is affected by upward-impulse corresponding to downward-impulse of air - demanded downward speed needs movement of mass m = 415 t/s (some less than weight of airplane, because air is moved down some faster than by gravity g). Acceleration of that air-mass-flow requires performance of about 39.400 HP - and also these results are determined by rules of physics absolutely correct and accepted by general understanding.
Not to grasp
Verbal statements I made at diverse texts, now are covered by real data and I say thanks for that service. However I still am astonished, how such results - obviously theoretic correctly determined - are not checked with obvious realities. By previous calculations wing must push down air in extent of 415 t/s. These are 415.000 kg and by previous density rho = 1 kg/m^3 each kilogram corresponds with one cubic-meter. So volume of air of 415.000 m^3 should be accelerated downward by about 12 m/s.
At picture 05.12.02 schematic is sketched that volume. A380 shows span width of 80 m and at previous example plane flies by 100 m/s, so within one second covers most exactly football pitch (green). Previous volume results if 18-floor-building (blue walls) is constructed at total area, 52 m high. Goals like players at middle circle are drawn true to scale.
Upside right at this picture, also true to scale, wing (dark red) with its surface of 850 m^2 is drawn, so with 80 m span width and about 11 m length. Wing is drawn diagonal corresponding to previous angles of attack (yellow), where wing grasps layer of air (light red) of maximum 2,5 m height. Airplane moves length of wing within about one tenth of a second, pressing down volume of 80 * 11 * 2.5 = 2.200 m^3 air downwards by demanded acceleration. However, wing should also push remaining downside air of 80 * 11 * 49,5 = 43.500 m^3 within that tenth of second downward these 2,5 m completely (resp. also while one second whole volume of that ´church-tower-high-pitch-super-structure´).
That´s absolutely impossible in reality. Previous calculations are correct - however handling air like solid bodies (impulse onto (part-) surface immediately affects onto total mass of body). Air however is not to grasp likely, transmits pressure not pure mechanically, air is compressible and escapes immediately into areas of less pressure, by sound speed, without affecting corresponding (mechanic) counter-pressure. Wing just reaches not enough volume of air in order to produce downward winds in that extent demanded by impulse-constant.
Clutch at Straws
This mechanistic point of view reaches as far - and that opinion is wide spread - also air upside of wing is drawn down and behind wing goes on moving downward, so airplane correspondingly moves upward. At picture 05.12.03 at A likely naive is sketched: relation of that kind does not exist, because there are no ´green redirection-wheels´ hanging at heaven.
There is no possibility to move down that volume of air demanded for lift. At most would be possible to compress air by corresponding energy-input and to push up plane upon ´air-cushion´ B, again with corresponding mechanical energy-input. As long as plane rolls on ground or flies near ground, counter-pressure exists. Downward- resp. compression-pressure spreads by sound speed downward, however counter-pressure returns only by half of sound speed. Flying length of wing (previous 11 m) takes one tenth of second. Counter-pressure wanders some 15 m within that time-unit, so missing wing already by slow speed resp. height.
So within free space or at normal travel-speed, wing is affected only by upward-pressure resulting of air-layer directly grasped by wing, so here only that layer of maximum 2,5 m height, twentieth part of necessary air masses. In addition, that pressure naturally flows off aside and downside and backwards, so continuously energy-input is necessary, which continuously gets lost for system (like e.g. energy invested into bow-wave of ships spreads into ´infinity´).
Too less / too much Lift
For high resp. travel-speed, wings must show rather flat profile in order to produce sufficient lift and same time showing most less resistance. For slow resp. start speed, profiles must show stronger bending resp. at rear end of wings additional surfaces are extended. Nevertheless ´natural´ lift is not sufficient for tanked up airplane. After take-off, total plane is steep inclined, so previous wedge-shaped air-cushion of high density comes up. Mechanical pushing-upward upon that wedge is no original lift, which only comes up by accelerated speed at upside surfaces.
Like discussed at chapter 05.08. ´Airplane NT´ thus engines should suck-off air upside of wings and should not be arranged downside of wings, like common practise and also at previous A380. Dilemma however is, these engines would produce much too strong lift at high speed travel. Like described at mentioned chapter, thus engines should be installed backside of wings (at C) and if demanded, air only from upper side is sucked in, controlled by flaps. At normal flight or few demand for lift, air by parts could come from downside of wing (at D) - or airplanes should be designed just by New-Technology, see previous chapter.
Artificial Wind
Now however back to data of A380 and calculation of ´natural´ lift, i.e. only these lift-forces based on wing-profile by itself, resulting of normal molecular movement energy (plus some drive energy for balancing resistance). Starting point are previous data: mass m = 500 t = 500.000 kg resp. 5.000.000 N, surface of wing S = 850 m^2, speed v = 360 km/h = 100 m/s and density of air rho = 1 kg/m^3. Each square-meter of effective wing-surface thus must contribute lift force A = 5.000.000 / 850 = 5.882 N/m^2.
At chapter ´Lift at Wings´, based on movements of air particles and suction effects, I found flow of 45 m/s relative to wing at its upper side. That value could also be some higher because there was calculated without optimum angles of attack (about 3 degrees, for compensation of downside air, flowing upward at nose). In addition, calculations there assumed sound speed only with 300 m/s. Speed of these ´artificial winds´ about 45 to 50 m/s could be valid in general, so also at A380. So here is assumed, air downside of wing moves alongside by 100 m/s, air at upper sides however by 145 m/s.
Real Lift
Picture 05.12.04 visualizes following mode of calculation. Each blue cube represents one cubic-metre of air. At A this air is resting (V 0), normal atmospheric pressure of 1 bar exists resp. likely static pressure PS = 100.000 N/m^2 into any direction. This resting cubic-metre air shows no dynamic pressure (PD 0).
Below of wing (red) at B is drawn corresponding air mass, moving relative to wing by 100 m/s (V 100). This air shows dynamic pressure (flow-pressure, dam-up-pressure) at its right side (blue more dark), corresponding to known formula PD = 0.5 * rho * v^2. So flow-pressure at this downside surface of wing is PDU = 0.5 * 1 * 100^2 = 5.000 N/m^2. As sum of all pressures is constant, towards downside surface of wing affects reduced static pressure PSU = 100.000 - 5.000 = 95.000 N/m^2 (light blue).
Analogue mode of calculation is to use at upper side of wing with some faster relative speed of 145 m/s (V 145). Resulting is dynamic pressure PDO = 0.5 * 1 + 145^2 = 10.500 N/m^2 and corresponding static pressure PSO = 100.000 - 10.500 = 89.500 N/m^2. Difference of both static pressures is lift-force PA = 95.000 - 89.500 = 5.500 N/m^2 (and based on mutual dependence of pressures, also results directly as difference of both dynamic pressures). This lift-force of 5.500 N/m^2 is rather likely to necessary lift of 5.882 N/m^2 determined upside (especially as ´artificial wind´ of 45 m/s is assumed little bit too slow, e.g. 50 m/s would result lift-force of 6.325 N/m^2, so 443 N/m^2 more than necessary for horizontal flight).
Formalism becomes really simple if based only at undisputed fact of constant of all pressures. Downward-winds and mechanistic actio=reactio not at all are involved, only difference of static pressures raises airplanes. This formula is also valid, if not only that ´natural´ lift is used, but plane is pushed up by steep angles of attack. Air becomes dammed-up downside of wing, i.e. is pushed ahead by plane and air becomes compressed. If higher density and slower relative speed are used by that formula, increased lift-force for climbing-flight results.
By the way: naturally comes up downward flow behind wings, started by suction area upside of wing - as consequence and never ever as reason for lift. At the other hand: if angles of attack become too steep, flow at upper side becomes turbulent. As soon as suction there collapses, all lift-forces collapse too, i.e. pushing-downward air by itself never can result sufficient lift-forces. ´Natural´ Lift comes up only by pressure-differences and demands only few energy-input, while ´artificial´ lift of plane by mechanic pressing-upward demands huge energy-input and still is only available in addition to profile-based natural lift.
Correction
Critical is, these ´Bernoulli-rules´ are valid only for incompressible fluids, so usable for air only conditionally. At the other hand ´energy-constant-rule´ is valid and any particles movement all times can affect completely ahead or completely aside or diagonal with according components, so pressure constant should really exist.
However common formula do not pay attention to fact, flow initiated by suction is well ordered, particles fly nearby each other rather parallel, so affecting ´excessive´ dynamic pressure. Ordered flow alongside a surface protects better against atmospheric pressure as any flow produced by pressure - and common theories don´t mind that difference. So general calculation with pressure constant is not more wrong than paying no attention to specific suction effects
At chapter 05.04. ´Lift at Wings´ an example of airplane with these data was mentioned: wing surface 48 m^2, speed 150 m/s (half sound speed), additional flow at upper side 45 m/s (so air passes wing upside by 195 m/s). By previous mode of calculation thus results dynamic pressure at downside surface PDU = 0.5 * 1 * 150^2 = 11.250 N/m^2 and at upside surface PDO = 0.5 * 1 * 195^2 = 19.012 N/m^2. Difference is PA = 19.012 - 11.250 = 7.762 N/m^2, lift of total surface thus is PA = 7.762 * 48 = 372.576 N resp. sufficient for plane of about 37 t (instead of 20 t by calculation more wrong than right at that chapter). Following table shows these data at first row, below are shown data of some further situations (Remark: U means German ´unten´ = downside, O means German ´oben´ = upside).
A380 - Data
| Situation |
VU | VO | PDU | PDO | PA |
| 1. Example 05.04. | 150 | 195 | 11.250 | 19.012 | 7.762 |
| 2. A380 Start | 100 | 145 |
5.000 | 10.500 | 5.500 |
| 3. A380 Climbing | 200 | 250 |
20.000 | 31.250 | 11.250 |
| 4. A380 Optimum | 280 | 330 |
39.200 | 54.450 | 15.250 |
| 5. A380 Power | 300 | 330 | 45.000 | 54.450 | 9.450 |
| 6. A380 Limit | 320 | 330 | 51.200 | 54.450 | 3.250 |
| 7. Heli Mini | 0 | 30 | 0 | 450 | 450 |
| 8. Heli Normal | 0 | 45 | 0 | 1.000 |
1.000 |
| 9. Heli Load | 0 | 60 | 0 | 1.800 | 1.800 |
Second row of table shows data for lift of A380 with speed VU = 100 m/s, short time after start or at begin of climbing flight (analogue previous picture 05.12.04). At upper surface exists additional ´wind´ of 45 m/s, so VO = 145 m/s is assumed. Dynamic pressure at downside surface is PDU = 5.000 and at upper surface PDO = 10.500 N/m^2. Difference is lift PA = 5.500 N/m^2, i.e. additional pushing upward by steep angles of attack is merely necessary.
Third row of table shows data of further acceleration of A380, e.g. at 200 m/s and at upper surface now some increased flow of additional 50 m/s. Resulting is essential stronger lift PA = 11.250 m/s. So short time after first climbing phase, further raise demands much less energy-input.
Optimum speed of A380 is stated with 0,85 Mach, which means VU = 280 m/s (row 4), if sound speed is assumed with 330 m/s. Suction can not work faster than speed, so air moves alongside upper surface by maximum speed VO = 330 m/s. That speed is most economic for many airplanes and thus I expect, remaining 0,15 Mach resp. about 50 m/s are maximum speed of additional flow, which can be generated by suction at upper surface of wings. At this border of 0,85 Mach, A380 achieves its maximum lift PA = 15.250 N/m^2. That´s much too strong lift for conditions near ground, however just enough at level of 10 km height and low density rho = 0,4 kg/m^3 up there.
At row 5 situation is shown, where plane ´powers´ high speed of VU = 300 m/s. However that maximum VO = 330 m/s decreases lift PA = 9.450 N/m^2 dramatically, i.e. airplane can hold height only by mechanic pushing-upward. Row 6 marks approach towards limit, as speed of VU = 320 m/s produces only lift of PA = 3.250 N/m^2, so tending to zero - and it´s well known, no ´natural lift´ exists beyond sound speed.
Helicopter - Data
Previous results fit well to reality, so corresponding calculations are adequate also for determination of lift at helicopters, where ´artificial winds´ are produced alongside bended upper surfaces. At rows 7 to 9 of table is assumed, air is resting (VU = 0) at downside surface, while at upside surface flow VO of 30 or 45 or 60 m/s is generated.
At chapter 05.07. was presented a ´Suction-Helicopter´ with cap of 5 m diameter and thus with about 20 m^2 upside surface, however lift-forces exist at nearby double surface areas. As helicopter might weight or take load of 2 to 4 t, lift forces at size of 800 to 1800 N/m^2 are demanded. Previous table shows at row 8 and 9, this is achieved by flows of 45 to 60 m/s.
At previous chapter ´Spiral-Canal-Motor´ was proposed similar flight-unit and here at picture 05.12.05 practically combination of both conceptions schematic is drawn. Cabin is large enough e.g. with diameter of some 3 m, at the other hand that cap could well show diameter of about 6 m with round surface of 27 m^2. When flow moves only with 30 m/s, lift forces PA = 450 N/m^2 (row 7 of table) keep total weight of 1.200 kg in state of suspense.
Blades of normal helicopters reach out much wider than that cap. Blades of normal helicopters need much energy for whirl around and press down air masses, however no clear flows come up. Opposite, flows are established much more effective alongside wide bended surfaces, by much easier technology.
Spiral-Canal-Motor of this flight-unit could show diameter e.g. of 0.8 m (inclusive apron ca. 1.3 m) and thus circumference of about 2.5 m. Demanded speed of 30 to 60 m/s are achieved by 800 to 1.500 rpm. Canals of that pump are some 0.5 m long and at 1.500 rpm each rotor-canal glides over housing-canal within 8 milliseconds, so in average by 62.5 m/s, quite outside however by about 100 m/s - thus really sufficient speed is achieved.
At this motor, potential of normal molecular movements can transmit into ordered flows at its best and part of mechanical acceleration will only be one third. Correspondingly few drive is demanded - and if e.g. Tornado-Motor is used even no tanks are necessary. So these flight-units are much lighter to build than common helicopters and fly with much less HP and absolutely silent (in comparison to common helicopters or planes).
Result
Based on data of A380 that theory ´lifting-plane-mass by lowering-air-mass´ is disprove without any doubt. At its best only ´mechanical pushing-upward upon compressed air-cushion´ is possible - however only additional to ´original lift´. These natural lift-forces exclusively come up by suction-effect at upside surfaces, because generated flow affects less static pressure against upper sides than downside relative slow movement resp. atmospheric pressure of resting air affects upward. That mechanic pushing-upward involves heavy losses, while original lift only demands energy for overcoming resistance much weaker.
If one leaves (´impossible´) idea, air must be pressed down, but only optimum flows must be generated, quite new conceptions can be realized. All common considerations mostly are fixed on utilization of pressure and totally ignore enormous potential of normal molecular movement, easy to use however only by application of suction. Previous chapters offer diverse proposals concerning New Technology for building airplanes and helicopters as well.
Science finally must leave wrong understanding of limitations of energy constant and science of flows must consider specific differences between fluid- and solid-body-mechanics. If facts here described are integrated, real products can be build much more effective - and even flying could occur environmentally neutral.