_{1}

^{*}

We consider, compare, and contrast various aspects of aerodynamic and ballistic flight. We compare the energy efficiency of aerodynamic level flight at a given altitude versus that of ballistic flight beginning and ending at this same altitude. We show that for flights short compared to Earth’s radius, aerodynamic level flight with lift-to-drag ratio
L/D > 2 is more energy-efficient than ballistic flight, neglecting air resistance or drag in the latter. Smaller
L/D suffices if air resistance in ballistic flight is not neglected. For a single circumnavigation of Earth, we show that aerodynamic flight with
L/D > 4π is more energy-efficient than minimum-altitude circular-orbit ballistic spaceflight. We introduce the concept of gravitational scale height, which may in an auxiliary way be helpful in understanding this result. For flights traversing N circumnavigations of Earth, if
then even minimum-altitude circular-orbit ballistic spaceflight is much more energy-efficient than aerodynamic flight because even at minimum circular-orbit spaceflight altitude air resistance is very small. For higher-altitude spaceflight air resistance is even smaller and the energy-efficiency advantage of spaceflight over aerodynamic flight traversing the same distance is therefore even more pronounced. We distinguish between the energy efficiency of flight per se and the energy efficiency of the engine that powers flight. Next we consider the effects of air density on aerodynamic level flight and provide a simplified view of drag and lift. We estimate the low-density/high-altitude limits of aerodynamic level flight (and for comparison also of balloons) in Earth’s and Mars’ atmospheres. Employing Mars airplanes and underwater airplanes on Earth (and hypo-thetically also on Mars) as examples, we consider aerodynamic level flight in rarefied and dense aerodynamic media, respectively. We also briefly discuss hydrofoils. We appraise the optimum range of air densities for aerodynamic level flight. We then consider flights of hand-thrown projec-tiles that are unpowered except for the initial throw. We describe how aerodynamically efficient ones (
*i.e.*, with large
L/D) such as Frisbees, Aerobies, and boomerangs not only can traverse record horizontal distances, but (along with discuses) also can—since lift exceeds weight at achievable throwing speeds—maintain altitude farther if thrown horizontally against the wind than with it. Then we compare the energy efficiency of surface transportation versus that of both aerodynamic and ballistic flight.

We consider, compare, and contrast various aspects of aerodynamic and ballistic flight. In Section 2 we compare the energy efficiency of aerodynamic level flight at a given altitude versus that of ballistic flight beginning and ending at this same altitude. We show that for flights short compared to Earth’s radius, aerodynamic level flight with lift-to-drag ratio L / D > 2 is more energy-efficient than ballistic flight, neglecting air resistance or drag in the latter. If air resistance in ballistic flight is not neglected, then smaller L / D suffices for short aerodynamic level flight to be more energy-efficient than short ballistic flight. For a single circumnavigation of Earth (the longest possible flight whose purpose is to reach a destination on Earth, with the destination being the starting point after a round-the-world trip), we show that aerodynamic flight with L / D > 4 π is more energy-efficient than minimum-altitude single-circular-orbit ballistic spaceflight, neglecting the very small air resistance in the latter. We introduce the concept of gravitational scale height, which may in an auxiliary way be helpful in understanding this result. If the very small air resistance at minimum circular-orbit spaceflight altitude is not neglected, then L / D very slightly smaller than 4 π suffices for single-circumnavigation aerodynamic flight to be more energy-efficient than minimum-altitude single-circular-orbit ballistic spaceflight. For flights traversing N circumnavigations of Earth, owing to air resistance or drag being very small even at minimum circular-orbit ballistic spaceflight altitude, L / D must exceed 4 π N for aerodynamic flight to be more energy-efficient than minimum-altitude circular-orbit ballistic spaceflight. But L / D ≈ 100 may represent the practicable limit that can be achieved even with the most advanced aerodynamic technology. Hence if N ≫ 1 even minimum-altitude circular-orbit ballistic spaceflight is much more energy-efficient than aerodynamic flight. For higher-altitude spaceflight drag is even smaller and the energy-efficiency advantage of spaceflight over aerodynamic flight traversing the same distance is therefore even more pronounced. In Section 2.4 we distinguish between the energy efficiency of flight per se and the energy efficiency of the engine that powers flight. In Section 3 we consider the effects of air density on aerodynamic level flight and provide a simplified view of drag and lift. We estimate the low-density/high-altitude limits of aerodynamic level flight (and for comparison also of balloons) in Earth’s and Mars’ atmospheres. Employing Mars airplanes and underwater airplanes on Earth (and hypothetically also on Mars) as examples, we consider aerodynamic level flight in rarefied and dense aerodynamic media, respectively. We also briefly discuss hydrofoils. We appraise the optimum range of air densities for aerodynamic level flight. In Section 4 we consider flights of hand-thrown projectiles that are unpowered except for the initial throw. We describe how aerodynamically efficient ones (i.e., with large L / D ) such as Frisbees, Aerobies, and boomerangs not only can traverse record horizontal distances, but (along with discuses) also can—since lift exceeds weight at achievable throwing speeds—maintain altitude farther if thrown horizontally against the wind than with it. In Section 5 we compare the energy efficiency of surface transportation versus that of both aerodynamic and ballistic flight. Brief concluding remarks are given in Section 6. Footnotes provide supporting references and some auxiliary information. Supplementary Notes, providing more comprehensive auxiliary information concerning topics discussed in the main text and/or in the cited references including some additional supporting references, are given in the Appendix.

In this paper we define “aircraft” as any type of aerodynamic flying machine—for example man-made airplane or sailplane, bird, or flying insect in forward or hovering flight, including hand-thrown aircraft (e.g., discus, Frisbee, Aerobie, boomerang, etc.). Underwater airplanes, which we will briefly consider in Section 3.3.2, and hydrofoils, which we will briefly consider in Section 3.4, should be classified as aircraft because their lift obtains aerodynamically rather than via buoyancy, even though the density of their aerodynamic medium (water) is ≈ 800 times that of air at sea level. By lifting the hull out of water into air, drag on the hull of a hydrofoil at any given speed is reduced ≈ 800 times; only the wings need suffer water resistance as opposed to air resistance. Except for a few very brief parenthetical remarks concerning hovering flight, in this paper we consider only aircraft that obtain their lift by virtue of their translational forward motion, i.e., translational-lift aircraft—for example man-made airplanes and sailplanes, birds and insects that obtain their lift by virtue of their translational forward motion, underwater airplanes, hydrofoils, and hand-thrown translational-lift aircraft (e.g., discuses, Frisbees, Aerobies, boomerangs, etc.). Except for a brief consideration of underwater airplanes in Section 3.3.2, and a brief consideration in Section 3.4 and occasional other even briefer remarks concerning hydrofoils, from among translational-lift aircraft we consider only aerial translational-lift aircraft, which obtain their lift by virtue of their translational forward motion through air—for example man-made airplanes and sailplanes, birds and insects that obtain their lift by virtue of their translational forward motion, and hand-thrown translational-lift aircraft (e.g., discuses, Frisbees, Aerobies, boomerangs, etc.).^{1 }

In this paper our main goal is to elucidate more conceptually than mathematically some fundamental ideas concerning energy efficiency and a number of other aspects of aerodynamic versus ballistic flight, and to provide comparison with surface transportation. Some elucidations of this type have, of course, been provided elsewhere. But to the best knowledge of the author, this paper provides a range of such elucidations, and some new ones, not found elsewhere, at least not in a single work. We do not attempt the mathematically complex and detailed fully-quantitative analyses based on rigorous application of fluid dynamics, e.g., computational fluid dynamics, as is required in the actual design of aircraft, or the analyses required in the actual design of spacecraft or surface vehicles. Thus our analyses are qualitative to semiquantitative.

We compare the energy efficiency of aerodynamic level flight at a given altitude versus that of ballistic flight beginning and ending at this same altitude, at first neglecting air resistance or drag in the case of ballistic flight. We define a short flight as one traversing horizontal distance X much smaller than Earth’s radius R ≐ 6370 km . [The dot-equal sign ( ≐ ) means “very nearly equal to”.] Hence for a short flight Earth’s curvature can be neglected. Let h A be the altitude above mean sea level of aerodynamic level flight, and also the beginning or initial altitude h B , initial = h A and ending or final altitude h B , final = h A of ballistic flight. (The subscript A denotes aerodynamic flight and the subscript B denotes ballistic flight.) Hence this altitude is at radial distance r A = R + h A from Earth’s center. For simplicity we let h A ≪ R as is the case for all aerodynamic flight and h B ≪ R as is the case for low-altitude short ballistic flight (and even for minimum-altitude circular-orbit ballistic spaceflight).

By elementary Newtonian mechanics [

E B , short = 1 2 m g X . (1)

The local acceleration due to gravity is g and drag is neglected.

An aerodynamic level flight of an aircraft of mass m and weight mg subject to aerodynamic lift L and aerodynamic drag D traversing any horizontal distance X, short or long, costs energy

E A = D X = D L L X = D L m g X = m g X L / D . (2)

Of course, for any flight, short or long, “level” and “horizontal” mean that h A and therefore r A = R + h A is constant, a long flight following the curvature of Earth.

The first step of Equation (2) is justified because to maintain aerodynamic level flight through horizontal distance X air resistance (most generally, fluid resistance) or drag D must be overcome through distance X. The third step of Equation (2) is justified because to maintain aerodynamic level flight lift L must equal the weight mg of an aircraft [^{2} The lift-to-drag ratio L / D of a translational-lift aircraft is the ratio (horizontal distance traversed) ¸ (altitude lost) in gliding flight without engine power relative to the air (also relative to the ground if the wind is calm) [^{2} For any translational-lift aircraft L / D is maximized if the aircraft is flown at its most energy-efficient angle of attack. In this paper unless otherwise noted we always assume L / D to be thus maximized. [For auxiliary information, which may also be helpful in some cases wherein Refs. [^{2} are cited later in this paper, see Supplementary Notes 1-6. Also, we discuss these points more thoroughly in Section 3. Wherever helpful, refer to Decker, J.S. (2014) See How It Flies at https://www.av8n.com/how/.]

Comparing Equations (1) and (2), if L / D > 2 , then E A < E B , short . Hence short aerodynamic level flight with L / D > 2 is more energy-efficient than short ballistic flight beginning and ending at the same altitude, neglecting air resistance in the latter. This requirement L / D > 2 is met by practically all aircraft (including all birds and flying insects), even by aerodynamically inefficient ones [

By Equations (1) and (2), if air resistance in short ballistic flight can be neglected, then both E B , short and E A are directly proportional to mg. Hence reducing mg reduces the energy cost of both short ballistic flight and short aerodynamic level flight traversing given horizontal distance X equally and in direct proportion to the reduction in mg, but does not alter the ratio of energy costs between these two modes of flight. If air resistance in short ballistic flight cannot be neglected, then reducing mg reduces the energy cost of short ballistic flight less than in direct proportion to the reduction in mg.

Most typically, mg is reduced by reducing m. But we can also consider reduction of g. Two examples: (i) Aerodynamic level flight on Mars is at lower g. (Of course, for Mars R ≐ 3390 km .) (ii) An aircraft of mass m a fraction f ( 0 < f < 1 ) of whose weight mg is offset by buoyancy can be construed as either being of effective mass m ( 1 − f ) in a gravitational field g or as being of mass m in a gravitational field of effective strength g ( 1 − f ) . Such partial offset of weight by buoyancy obtains, for example, for a dirigible or blimp that relies on buoyancy for only part of its lift, with the balance obtaining aerodynamically, or for a hydrofoil that cruises so slowly that it must rely on buoyancy for a non-negligible part of its lift.

For numerical examples, consider a 100 km = 10 5 m aerodynamic level flight at a given altitude h A (within the troposphere) with L / D = 20 versus a 100 km = 10 5 m ballistic flight launched at a 45˚ angle, beginning and ending at this same altitude h B , initial = h B , final = h A . Let the mass of both the aerodynamic and ballistic craft be 1000 kg. Assume that the engines powering both the aerodynamic and ballistic craft are 25% efficient. Neglect air resistance in the ballistic flight. Then by elementary Newtonian mechanics [

E B , short ϵ = 1 0.25 × 1 2 m g X J = 2 m g X J = 2 × 1000 × 9.8 × 10 5 J = 1.96 × 10 9 J .

And by Equation (2),

E A ϵ = 1 0.25 × m g X L / D J = 4 m g X 20 J = 1000 × 9.8 × 10 5 J 5 = 1.96 × 10 8 J .

Since typical hydrocarbon aviation fuels yield ≈ 4.4 × 10 7 J / kg ,^{3} » 45 kg of fuel would be required for the ballistic flight, and » 4.5 kg of fuel would be required for the aerodynamic level flight. (This yield of ≈ 4.4 × 10 7 J / kg is per kg of fuel alone, not counting the O_{2} required to burn it.) Not neglecting air resistance in the ballistic flight, the energy-efficiency advantage of the aerodynamic level flight would exceed this 10:1 ratio. Engine efficiency is discussed in more detail, and distinguished from the energy efficiency of flight per se, in Section 2.4. In any case, as per the numerical examples given immediately above, the energy that must be supplied to an engine whose efficiency is ϵ in order to facilitate aerodynamic flight requiring energy E A is of course E A / ϵ . And likewise in the case of ballistic flight it is of course E B / ϵ .

Now let us consider the longest possible flight whose purpose is to reach a destination on Earth—a single circumnavigation of Earth. The destination is thus the starting point after a round-the-world trip. For circumnavigation X = 2 π r , X being the horizontal distance traversed given flight at constant h and hence also at constant r = R + h , following Earth’s curvature.

In this case, by elementary Newtonian mechanics minimum-energy ballistic spaceflight obtains at the lowest-altitude circular orbit at which air resistance or drag is negligible, i.e., at h B ≈ 100 mi ≈ 160 km or equivalently at r B = R + h B ≈ 6370 km + 160 km = 6530 km .^{4} By elementary Newtonian mechanics a circular orbit requires speed v orbit = ( G M / r B ) 1 / 2 , where M is Earth’s mass (not to be confused with the mass m of a spacecraft).^{4} Also by elementary Newtonian mechanics the energy of a spacecraft of mass m at rest at Earth’s surface is its gravitational potential energy E pot ( R ) = − G M m / R . Its gravitational potential energy at r B = R + h B is E pot ( h B ) = − G M m / ( R + h B ) , its orbital kinetic energy at r B = R + h B is E kin , orbit ( h B ) = G M m / [ 2 ( R + h B ) ] = m v orbit 2 / 2 , and its total (potential + kinetic) energy in circular orbit at r B = R + h B is E total ( h B ) = − G M m / [ 2 ( R + h B ) ] .^{4} In this Section 2.2 we consider only minimum-altitude ( h B ≈ 100 mi ≈ 160 km ) circular spaceflight orbits for which h B ≪ R and hence v orbit = ( G M / r B ) 1 / 2 ≐ ( G M / R ) 1 / 2 . If h B ≪ R , then by elementary Newtonian mechanics the total energy cost of establishing orbit and hence of ballistic circumnavigation of Earth at r B = R + h B is^{4}

E total ( h B ) = E B , circumnavigation = − G M m 2 ( R + h B ) − ( − G M m R ) = G M m R − G M m 2 ( R + h B ) = G M m R − G M m 2 R ( 1 + h B R ) ≐ G M m R − G M m 2 R ( 1 − h B R ) = G M m R [ 1 − 1 2 ( 1 − h B R ) ] = G M m R ( 1 − 1 2 + h B 2 R ) = G M m R ( 1 2 + h B 2 R ) = G M m 2 R ( 1 + h B R ) ≐ G M m 2 R = 1 2 m × G M R 2 × R ≐ 1 2 m g R = 1 2 m g × 2 π R 2π = m g × 2 π R 4π ≐ m g X 4 π . (3)

The first step in the last line of Equation (3) is justified because at all r ≥ R , g = G M / r 2 , and specifically at r = R , g = G M / R 2 . (We are primarily interested in the magnitude of g, so we always take g as positive.)

For all r ≥ R circular orbital speed at r is v orbit = ( G M / r ) 1 / 2 and escape velocity from r is v escape = ( 2 G M / r ) 1 / 2 .^{4} Thus energy cost for escape from r to infinity can be written as^{4}

E escape = 1 2 m v escape 2 = 1 2 m 2 G M r = m G M r = m G M r 2 r = m g r at all r ≥ R = m g R at r = R . (4)

The energy cost establishing of circular orbital speed at r, which equals the circular-orbital kinetic energy at r, is half as great:^{4}

E kin , orbit = 1 2 m v orbit 2 = 1 2 m G M r = 1 2 m G M r 2 r = 1 2 m g r at all r ≥ R = 1 2 m g R at r = R . (5)

Thus at all r ≥ R , E escape from radial distance r from Earth’s center to infinity equals the energy required for lifting through vertical distance r from radial distance r to radial distance 2r from Earth’s center if g had remained constant and equal to its value at r rather than decreasing with increasing distance from Earth. And E kin , orbit for circular orbit at r is half as great. Hence also at all r ≥ R the free-fall velocity through vertical distance r from radial distance 2r to radial distance r from Earth’s center if g had been constant and equal to its value at r equals the escape velocity from radial distance r from Earth’s center. Thus the “gravitational scale height” at all r ≥ R is equal to r itself—specifically, at r = R it is equal to R itself [^{4}

The “ r = 0 gravitational scale height” corresponding to escape through a borehole from the center r = 0 of a spherical gravitator of uniform density (which Earth is not) is 3/2 times that from R. Hence the energy cost for escape through a borehole from the center r = 0 of a spherical gravitator of uniform density is 3/2 times that from R. This is perhaps most easily understood if one observes that within a uniform-density spherical gravitator, i.e., at 0 ≤ r ≤ R , g ∝ r . Therefore the average value of g ( r ) within R equals g ( R ) / 2 . Thus the portion of an escape trip from r = 0 within R contributes R / 2 to the “ r = 0 gravitational scale height”, the portion at and beyond R contributes R itself, total 3 R / 2 . Thus the portion of an escape trip from r = 0 within R costs 1/2 as much energy (1/3 of the total) as the portion at and beyond R (2/3 of the total). Thus E escape ( r = 0 ) = 3 2 E escape ( r = R ) . Escape velocity from R is ( 2 G M / R ) 1 / 2 ;^{4} since v escape ∝ E escape 1 / 2 escape velocity through a borehole from the center r = 0 of a spherical gravitator of uniform density is v escape ( r = 0 ) = ( 3 2 ) 1 / 2 v escape ( r = R ) = ( 3 G M / R ) 1 / 2 .

Of course M can be construedas the mass of any spherically-symmetrical gravitator for which Newtonian gravitational theory is sufficiently accurate and Einstein’s General Relativity is not required, not necessarily Earth (the spherically-symmetrical gravitator taken to be necessarily of uniform density only in our discussion of r = 0 gravitational scale height). But in this Section 2.2 our main focus is on comparison of single-circumnavigation aerodynamic flight versus single-circumnavigation minimum-altitude circular-orbit ballistic spaceflight about Earth. (For auxiliary information concerning the concept of scale height see Supplementary Note 7.)

Since even minimum-altitude circular-orbit ballistic spaceflight must be above any appreciable atmosphere, r B = R + h B for a single-orbit spaceflight must exceed r A = R + h A for a single-circumnavigation aerodynamic flight. But since for simplicity we let the spaceflight orbit be a minimum-altitude circular one, h B ≪ R and hence slightly more strongly not only h A ≪ R but also h B − h A = r B − r A ≪ R . The concept of gravitational scale height introduced in the immediately preceding paragraph may help clarify why the strong inequality h B ≪ R ensures that the result derived for E total ( h B ) = E B , circumnavigation in Equation (3) is sufficiently accurate for our purposes.^{4}

Thus if h B is minimum-orbital altitude then h B − h A = r B − r A can be neglected compared to R, slightly more so compared to r A = R + h A , and slightly more so yet compared to r B = R + h B . Hence with negligible error we can set r A ≐ r B ≐ ( r A + r B ) / 2 = r ≐ R . Then for the longest possible (single-circumnavigation) aerodynamic level flight at altitude h A = r A − R ( h A < h B ≪ R ) whose purpose is to reach a destination (the starting point after a round-the-world trip) on Earth, Equation (2) remains valid if we set X = 2 π r A ≐ 2 π R . Hence

E A , circumnavigation = m g X L / D = m g × 2 π r A L / D ≐ m g × 2 π R L / D . (6)

Comparing Equations (3) and (6), if L / D > 4 π then E A , circumnavigation < E B , circumnavigation . Hence single-circumnavigation aerodynamic level (fixed-altitude) flight with L / D > 4 π is more energy-efficient than single-circumnavigation minimum-altitude circular-orbit ballistic spaceflight, neglecting the very small air resistance in the latter. This requirement L / D > 4 π is met by many, perhaps most, but not all aircraft. It is met by soaring birds such as albatrosses [

Neglecting air resistance in minimum-altitude ballistic spaceflight, for flights of intermediate length (X ranging from much smaller than R to approaching 2 π R ), the minimum value that L / D must exceed for aerodynamic level flight at altitude h A ≪ R to be more energy-efficient than ballistic flight beginning and ending at this same altitude h B , initial = h B , final = h A increases monotonically from 2 towards 4 π as X increases from very small values towards 2 π R . Not neglecting air resistance in ballistic flight, for flights of intermediate length (X ranging from much smaller than R to approaching 2 π R ), the minimum value that L / D must exceed for aerodynamic level flight at altitude h A ≪ R to be more energy-efficient than ballistic flight beginning and ending at this same altitude h B , initial = h B , final = h A increases monotonically from 2 n ( X ) towards 4 π n ( X ) as X increases from very small values towards 2 π R : n ( X ) < 1 but increases monotonically towards very nearly 1 as X increases from very small values towards 2 π R . Air resistance is very small in minimum-altitude circular-orbit ballistic flight ( X = 2 π R ), and hence also 1 − n ( X ) is very small, i.e., n ( X ) is very nearly 1 if X = 2 π R . (Since if X = 2 π R minimum-energy ballistic spaceflight is a circular orbit just above appreciable atmosphere at altitude h B , it cannot begin and end at the altitude h A of aerodynamic level flight, but h B − h A ≪ R . )

Generalizing the third-to-last paragraph of Section 2.1 in light of this Section 2.2, by Equations (1)-(3) and (6), if air resistance in ballistic flight can be neglected, then for flights traversing any given horizontal distance X, short or long, both E B and E A are directly proportional to mg. (Of course, a long horizontal flight follows Earth’s curvature at fixed altitude above mean sea level.) Hence reducing mg reduces the energy cost of both ballistic flight and aerodynamic level flight traversing any given horizontal distance X, short or long, equally and in direct proportion to the reduction in mg, but does not alter the ratio of energy costs between these two modes of flight. If air resistance in ballistic flight cannot be neglected, then reducing mg reduces the energy cost of any ballistic flight, short or long, less than in direct proportion to the reduction in mg. (The second-to-last paragraph of Section 2.1 requires no modification in light of Section 2.2.)

For an N-circular-orbit minimum-altitude ballistic spaceflight traversing distance X = 2 π r B N = 2 π ( R + h B ) N ≐ 2 π R N , whose purpose is typically scientific study of Earth as opposed to reaching a destination on Earth, in almost all cases N ≫ 1 . Also X = 2 π r A N = 2 π ( R + h A ) N ≐ 2 π R N for N-circumnavigation aerodynamic level (fixed-altitude) flight. By Equation (2), for N-circumnavigation aerodynamic level (fixed-altitude) flight, E A increases linearly with X = 2 π r A N = 2 π ( R + h A ) N ≐ 2 π R N . By contrast, for minimum-altitude circular-orbit ballistic spaceflight, irrespective of X = 2 π r B N = 2 π ( R + h B ) N ≐ 2 π R N , E B remains fixed at the value given by Equation (3) for N = 1 . For, even at minimum-orbital spaceflight altitude, air resistance is almost negligible, i.e., space is almost frictionless; thus the energy cost of launching a spacecraft is one-time. Hence for flights traversing N circumnavigations of Earth L / D must exceed 4 π N if aerodynamic level (fixed-altitude) flight is to be more energy-efficient than minimum-altitude circular-orbit ballistic spaceflight. But for even the best currently existing sailplanes L / D values are between 70 and 80 [^{4} In return for a one-time energy expenditure, X spaceflight / 2 π R → ∞ even for minimum-altitude circular-orbit spaceflight, X spaceflight / 2 π R → ∞ even more strongly for high-altitude, say geosynchronous, circular-orbital spaceflight, and X spaceflight / 2 π R → ∞ even more strongly yet for spaceflight exceeding escape velocity. This of course is simply owing to space being essentially frictionless, and increasingly frictionless with increasing altitude, thus allowing spacecraft but not aircraft to take full advantage of Newton’s first law of motion (inertia).^{5} The energy cost of speed in spaceflight is one-time; the energy cost of speed in aerodynamic flight is never-ending.^{5} Spaceflight is thus the only mode of transportation that can achieve ∞ mi / gal = ∞ km / l of fuel (or the equivalent thereof)—Spaceship Earth (whose fuel for its orbital and rotational motions was part of the solar nebula’s kinetic energy) is a good example.^{5} To save time in spaceflight continuous energy expenditure can be employed, for example employing solar, laser, or on-board nuclear energy. But in spaceflight continuous energy expenditure buys acceleration; in aerodynamic flight it buys only (constant) speed.^{5}

The energy efficiency of flight per se should not be confused with the energy efficiency of the engine that powers flight. If an aircraft engine is a heat engine operating in a cycle with heat input at temperature T hot and heat exhaust at temperature T cold , then its thermodynamic efficiency even assuming perfect reversibility is limited by the Carnot bound 1 − ( T cold / T hot ) .^{6} Of course any real cyclic heat engine is less than perfect and hence its actual thermodynamic efficiency is less than the Carnot^{6} bound. But the actual thermodynamic efficiency of any real cyclic heat engine under any given conditions, while less than the Carnot bound 1 − ( T cold / T hot ) for any given T cold / T hot , nonetheless, all other things being equal, like the Carnot^{6} bound increases monotonically with decreasing T cold / T hot . For example, the Curzon-Ahlborn efficiency at maximum power output assuming endoreversibility (irreversible heat flows directly proportional to finite temperature differences but otherwise reversible operation), 1 − ( T cold / T hot ) 1 / 2 [^{7} like the Carnot efficiency 1 − ( T cold / T hot ) ,^{6} increases monotonically with decreasing T cold / T hot [^{6,7} Both the Carnot and Curzon-Ahlborn [^{6,7} Unlike the Carnot efficiency, the Curzon-Ahlborn [^{7}) Thus if an aircraft engine is a cyclic heat engine, then this engine will operate most efficiently at the altitude where the atmosphere is coldest, most typically at or near the tropopause, but commonly as close to the surface as is safe in polar regions in winter. This of course assumes that the engine, if air-breathing and operating at altitude, is supercharged, and that the supercharger requires only a very small fraction of the engine's power output. The oxygen available to an air-breathing engine is of course directly proportional to air density ρ , so intuitively it would seem that so would be the engine’s power output. But actually with increasing altitude h A the power output of an unsupercharged air-breathing engine decreases slightly faster than ρ .^{8} Even a decrease in power output directly proportional to ρ in almost all cases more than offsets any increase in power output owing to increased thermodynamic efficiency on account of lower temperatures that typically obtain at higher altitudes.^{6,7} (A real cyclic heat engine may be difficult to start in cold weather, and its efficiency immediately at starting may be reduced by the high viscosity of still-cold lubricants, but upon attaining steady-state it will operate more efficiently than in hot weather.) In any case, the energy that must be supplied to an engine whose efficiency is ϵ in order to facilitate aerodynamic flight requiring energy E A is of course E A / ϵ . And likewise in the case of ballistic flight it is of course E B / ϵ .

Of course, neither nonheat engines such as electric motors and birds’ flight muscles nor noncyclic (necessarily single-use) heat engines such as rockets are limited ultimately by the Carnot bound, nor are they limited at maximum power output assuming endoreversibility by the Curzon-Ahlborn bound. (Even if a rocket engine is refurbished, each launch represents a separate single use.) Their Carnot efficiencies and even their Curzon-Ahlborn efficiencies can in principle approach 100% irrespective of T cold / T hot and indeed of temperature at all. But, more often than not, in practice these engines face other limitations. Electric motors typically in practice rather than merely in principle exceed 90% efficiency. But birds’ flight muscles are in practice typically considerably less efficient, typically in the range of 25% to 40%. And even if noncyclic, single-use rocket heat engines can in practice rather than merely in principle approach 100% efficiency irrespective of T cold / T hot and indeed of temperature at all, typically most of their work output must be expended in accelerating exhaust gases, with only a small fraction available for accelerating payloads.^{9}

We should note that: (a) Even if rocket heat engines operated in a cycle, owing to their very small T cold / T hot ratio their Carnot and even Curzon-Ahlborn bound would be nearly 100%. But it would still obtain that typically most of their work output must be expended in accelerating exhaust gases, with only a small fraction available for accelerating payloads.^{9} (b) Not all rocket engines are heat engines, not even noncyclic ones. For example, ion-drive rocket engines are not heat engines, not even noncyclic ones, and hence (like noncyclic heat engines) are not limited ultimately by the Carnot bound, nor at maximum power output assuming endoreversibility by the Curzon-Ahlborn bound. Because of their high exhaust speeds less mass need be exhausted to achieve a given spacecraft speed, and hence less of their work output need be expended on the exhaust.^{9} (c) Nonrocket spacecraft propulsion,^{9 }e.g., via solar sails or laser sails, or via the EM, MEGA or related drives if they are verified,^{9} is not limited ultimately by the Carnot bound, nor at maximum power output assuming endoreversibility by the Curzon-Ahlborn bound.

As an aside, we mention that the ratio of the Curzon-Ahlborn efficiency to the Carnot efficiency, [ 1 − ( T cold / T hot ) 1 / 2 ] ÷ [ 1 − ( T cold / T hot ) ] , decreases monotonically with increasing T cold / T hot from unity in the limit T cold / T hot → 0 to 1/2 in the limit T cold / T hot → 1 . Thus this ratio is never greater than 1 or less than 1/2. The former limit is obvious. The latter limit is most easily verified by setting T cold / T hot = 1 − δ : in the limit T cold / T hot → 1 ⇔ δ → 0 with the help of the binomial theorem [ 1 − ( 1 − δ ) 1 / 2 ] ÷ [ 1 − ( 1 − δ ) ] = [ 1 − ( 1 − 1 2 δ ) ] ÷ δ = 1 2 δ ÷ δ = 1 2 .

Of course, determination of how closely any given engine approaches to its theoretical maximum efficiency requires detailed analyses of the properties of that particular engine (e.g., bypass ratio of jet engines, battery and circuit design for electric motors, metabolic chemistry of birds’ flight muscles, etc.) We have not considered such detailed analyses in this Section 2.4.

Drag is given by [

D = 1 2 C D A frontal , geom ρ v 2 = 1 2 A frontal , eff ρ v 2 , (7)

where C D is the coefficient of drag, A frontal , geom is an aircraft’s geometrical frontal cross-sectional area, A frontal , eff is its effective frontal cross-sectional area (the frontal cross-sectional area that it effectively presents with respect to air resistance or drag), ρ is the air density, and v is the airspeed (also the ground speed if the wind is calm) of flight. [Note: The symbol A denoting surface area should not be confused with the subscript A, introduced in the first paragraph of Section 2.1, denoting aerodynamic flight (as opposed to ballistic flight, denoted by the subscript B).] Thus the coefficient of drag C D is given by

C D = A frontal , eff A frontal , geom . (8)

(See Supplementary Note 8.)

But more complete definitions of A frontal , eff and hence of C D are required. The drag owing to an aircraft’s frontal cross-sectional area per se is pressure drag. Pressure drag times frontal cross-sectional area is the force that an aircraft must impart to push air in front out of its way, and to overcome the pull of the partial vacuum behind it ensuing because adjacent air cannot move in behind it instantaneously. But there are two other components of drag: induced drag—a penalty that must be paid for lift (see Supplementary Notes 6 and 9), and skin-friction drag—owing to the viscosity of air rubbing against surfaces parallel to the airflow (see Supplementary Note 10). The contributions of induced drag and skin-friction drag are included along with that of pressure drag within A frontal , eff and hence within C D . Thus our more complete definitions of A frontal , eff and hence of C D : A frontal , eff is an aircraft’s effective frontal cross-sectional area—the frontal cross-sectional area that it would effectively present with respect to total air resistance or drag—as if total drag had been subsumed within pressure drag, i.e., as if induced drag and skin-friction drag had been converted to and incorporated within pressure drag. Since A frontal , geom is fixed for any given aircraft flying at any given angle of attack, by Equations (7) and (8) thus construing induced drag and skin-friction drag as incorporated within pressure drag modifies C D in direct proportion to the modification of A frontal , eff from its value with respect to pressure drag alone. Accordingly, we thus construe Equations (7) and (8) as if induced drag and skin-friction drag are converted to and incorporated within pressure drag. Drag is a complex phenomenon [^{10}) But regardless of classification scheme we construe A frontal , eff and hence C D as noted immediately above, i.e., as if all drag is subsumed within pressure drag. This implies subsuming all of the complexities within C D : with that understood, Equations (7) and (8) are correct as written. With good aerodynamic design, at or near the angle of attack at which L / D is maximized, C D ≪ 1 and hence A frontal , eff = C D A frontal , geom ≪ A frontal , geom (see Supplementary Notes 4 and 5). (In the case of hovering flight, as of a helicopter, or of a hovering hummingbird or insect, even if the aircraft executes no horizontal motion the revolving airfoils do and hence experience drag. The revolving airfoils’ geometrical area A frontal , geom with respect to drag is their geometrical frontal cross-sectional area, their effective area A frontal , eff with respect to drag is their effective frontal cross-sectional area; v with respect to drag is the root-mean- square average, taken over the geometrical frontal cross-sectional area of the airfoils, of airspeeds of the airfoils, be they blades of a helicopter, or wings of a hovering hummingbird or insect. Of course, if a hovering-flight aircraft also executes horizontal motion then its entire structure contributes to both A frontal , geom and A frontal , eff .)

Similarly (see Supplementary Notes 1-6 and 8), lift [

L = m g = 1 2 C L A wing , geom ρ v 2 = 1 2 A wing , eff ρ v 2 , (9)

where C L is the coefficient of lift, A wing , geom is an aircraft’s geometrical wing area, and A wing , eff is its effective wing area—the area that it effectively presents with respect to downward deflection of air required by Newton’s third law of motion^{11} as the price for the upward force that is its lift. The first step of Equation (9) recognizes that to maintain aerodynamic level flight L must equal the weight mg of an aircraft. We use “wing area” as shorthand for an aircraft’s entire lifting-surface area. A typical airplane obtains most but not all of its lift from its wings; its fuselage and elevators contribute some lift. The spectrum of airplane design ranges from flying wings with little or no fuselage to lifting bodies that are fuselage with little or no wing.^{12} Flying wings are based on the principle that the wing has a higher L / D ratio than any other part of an airplane, while lifting bodies seek to avoid structural stresses on wings, especially at high airspeeds.^{12} Thus the coefficient of lift C L is given by

C L = A wing , eff A wing , geom . (10)

Lift is a complex phenomenon, perhaps even more so than drag [^{11} must be obeyed. With that understood, Equations (9) and (10) are correct as written. (There does seem to be universal agreement that some elements of explanations of lift are incorrect, e.g., the “equal-transit-time” element: see Supplementary Note 1.) With good aerodynamic design, at or near the angle of attack at which L / D is maximized, C L is usually at least a significant fraction of unity and in some cases only a little smaller than unity, and hence A wing , eff = C L A wing , geom is usually at least a significant fraction of A wing , geom and in some cases almost as large as A wing , geom (see Supplementary Notes 4 and 5). At larger angles of attack C L not uncommonly exceeds unity, but at the expense of smaller L / D (see the last four paragraphs of Section 3.1.2 and Supplementary Notes 6 and 9). [In the case of hovering flight, as of a helicopter, or of a hovering hummingbird or insect, the wings are the revolving airfoils. The revolving airfoils’ geometrical area with respect to lift is their geometrical wing (not frontal) area, their effective area with respect to lift is their effective wing (not frontal) area; v with respect to lift is the root-mean-square average, taken over the geometrical wing area of the airfoils, of airspeeds of the airfoils, be they blades of a helicopter, or wings of a hovering hummingbird or insect.]

The forms of Equations (7)-(10), in particular the ρ v 2 functional dependency in Equations (7) and (9) that is paramount for both drag and lift, can perhaps in some measure be most easily physically understood via the following very simplified qualitative to semiquantitative arguments. First, note that both drag and lift are forces, and that ρ v 2 A is the only combination of ρ , v, and A that has dimensions of force, or equivalently that ρ v 2 is the only combination of ρ and v that has dimensions of force per unit area (= pressure). Construe, as discussed three paragraphs previously, induced drag and skin-friction drag as converted to and incorporated within pressure drag, so that total drag is construed as pressure drag. Pressure drag times effective frontal cross-sectional area A frontal , eff in aerodynamic level flight is the horizontal force—the horizontal momentum per unit time t—that an aircraft must impart to push air in front out of its way, and to pull air into the trailing partial vacuum behind it. Per unit time t a volume of air ~ v t A frontal , eff and hence a mass of air ~ ρ v t A frontal , eff must thus be given speed typically comparable to v, but at any rate at least approximately proportional to v. Thus to maintain constant horizontal forward airspeed v in the face of drag D an aircraft must impart to air horizontal force equal to D, in accordance with^{11}

D = horizontal force imparted to air = horizontal momentum imparted to air per unit time t ~ ( mass of air given speed ~ v per unit time t ) × v t ~ ( ρ v t A frontal , eff ) × v t = A frontal , eff ρ v 2 ~ C D A frontal , geom ρ v 2 . (11)

The upward force of lift L is by Newton’s third law of motion^{11} equal to the downward force—the downward momentum per unit time t—that an aircraft’s wings must impart to air. By similar reasoning as we employed with respect to drag, the wings must impart to air downward force equal to L, in accordance with^{11}

L = m g = downward force imparted to air = downward momentum imparted to air per unit time t ~ ( mass of air impelled downward per unit time t ) × v t ~ ( ρ v t A wing , eff ) × v t = A wing , eff ρ v 2 ~ C L A wing , geom ρ v 2 . (12)

The first step of Equation (12) recognizes that to maintain aerodynamic level flight L must equal the weight mg of an aircraft [^{11}, and hence are rigorously correct. Approximations are employed only in the last two lines of Equations (11) and (12).

If C L is independent of ρ and of v, as is typically approximately true at or near the most energy-efficient angle of attack and hence at or near airspeeds v corresponding to maximization of L / D , then induced drag, as pressure drag, is approximately proportional to ρ v 2 , and hence can be incorporated within Equations (7) and (8) via a simple approximately additive contribution to C D (see Supplementary Notes 4, 5, 6, and 9, especially Supplementary Note 9). But skin-friction drag is in general not even approximately proportional to ρ v 2 : skin-friction drag is a function of the coefficient of viscosity μ as well as of ρ , v, and A, and hence has dimensions of force even though it is not expressible as the combination ρ v 2 A , or equivalently dimensions of force per unit area (= pressure) even though per unit area it is not expressible as the combination ρ v 2 (see Supplementary Note 10). But skin-friction drag is nevertheless typically incorporated within Equations (7) and (8) via a contribution to the functional dependence of C D on ρ and on v. This is in accordance with Equations (7) and (8) being construed as if both induced drag and skin-friction drag are converted to and incorporated within pressure drag, as discussed in the second paragraph of this Section 3.1.1.

Thus for both drag and lift the ρ v 2 functional dependency is paramount [

But this minimum energy E A must be expended faster and hence the power P A required for maximally energy-efficient aerodynamic level flight increases with decreasing ρ (and hence with increasing h A ), because it is necessary to fly at faster v in thinner air to maintain L equal to the weight mg of an aircraft and hence to maintain aerodynamic level flight despite smaller ρ . By differentiating Equation (2) with respect to time t [or simply dividing Equation (2) by t given steady aerodynamic level flight], we obtain, for the power P A required to maintain aerodynamic level flight at speed v of an aircraft of mass m and weight mg subject to aerodynamic lift L and aerodynamic drag D:

P A = ∂ E A ∂ t = D ∂ X ∂ t = D v = D L L v = D L m g v = m g v L / D instantaneously = E A t = D X t = D v = D L L v = D L m g v = m g v L / D given steady flight . (13)

The fifth steps of both lines of Equation (13) recognize that to maintain aerodynamic level flight L must equal the weight mg of an aircraft [^{13} Obviously aerodynamic level flight is possible if and only if the maximum available power exceeds, or at the very least equals, the required power P A .^{13} Minimum required power P A = E A / t allows an aircraft to maintain aerodynamic level flight for the maximum possible time (maximum endurance); maximum energy efficiency E A / X = P A / v (which we always assume in this paper unless otherwise noted) allows an aircraft executing aerodynamic level flight to traverse the maximum possible distance. (See Supplementary Notes 5 and 6, and the references cited therein.)

Because it is necessary to fly at faster v in thinner air to maintain L equal to the weight mg of an aircraft and hence to maintain aerodynamic level flight despite smaller ρ , an aerodynamic level flight traversing given horizontal distance X requires less time t = X / v in thinner air, so the energy cost E A = P A t = P A X / v of aerodynamic level flight traversing given X, or equivalently E A per given X, i.e., E A / X = P A / v , is at least approximately independent of ρ (and hence of h A ). Focusing on the paramount ρ v 2 functional dependency of aerodynamic lift and drag [

In the four immediately preceding paragraphs we did not explicitly consider the effects of changing mg. But (recall the third-to-last and second-to-last paragraphs of Section 2.1 and the last paragraph of Section 2.2) this should be explicitly considered. Irrespective of the value of mg, it is still true, in accordance with Equation (2), that E A = D X = m g X ÷ ( L / D ) , and in accordance with Equation (13), that P A = E A / t = m g v ÷ ( L / D ) . Thus, all other things being equal, E A is directly proportional to mg. But, all other things being equal, P A is not directly proportional to mg, because the airspeed v required to maintain aerodynamic level flight increases with increasing mg. By Equations (9), (12), and (13) we have for aerodynamic level flight

ρ v 2 ∝ m g ⇒ v ∝ ( m g ρ ) 1 / 2 ⇒ P A = m g v L / D ∝ m g ( m g ρ ) 1 / 2 L / D = ( m g ) 3 / 2 ρ 1 / 2 ( L / D ) . (14)

We note that Equation (14) is consistent with E A being directly proportional to mg in accordance with Equation (2). For in accordance with Equation (14) E A = P A t = P A X / v = m g X ÷ ( L / D ) , in agreement with Equation (2).

Lift and drag are in general not exactly proportional to ρ v 2 , because C L and C D are in general not strictly constant [

It should be noted that, in the design of aircraft, even departures from the paramount ρ v 2 functional dependency of lift and/or drag smaller than those discussed in Items (a), (b), and (c) of the immediately preceding paragraph can yield modest but still significant improvements in aircraft energy efficiency [^{14} See the two immediately following paragraphs.

Not uncommonly, aerodynamic level flight at a given v is more energy-efficient at lower ρ and hence at higher h A . This can obtain despite the required increase in angle of attack to above that which maximizes C L / C D and hence L / D , consequently decreasing C L / C D and hence L / D , as the penalty for increasing C L itself and hence L itself sufficiently to maintain L = m g in the face of decreased ρ at fixed v [recall Equations (9) and (10)]: up to a limit, C L / C D and hence L / D decreases more slowly with the required increase in angle of attack than ρ decreases.^{14} Of course this obtains only up to a limit: with continued increase in angle of attack C L / C D and hence L / D decreases ever more rapidly until stalling occurs. Higher-altitude aerodynamic level flight while maintaining the (smaller) angle of attack that maximizes L / D would be at sufficiently faster v to decrease the flight time t more than it increases the required power P A , and hence would increase energy efficiency E A / X = P A t / X = P A / v even more. But the engine(s) may not be capable of the required increase in P A . Even if they are, in some cases increased v may be detrimental [e.g., owing to encountering shock-wave drag if Mach 1 is approached too closely (see Supplementary Note 12), or to excessive aerodynamic heating]. A specific example of this (owing to encountering shock-wave drag if Mach 1 is approached too closely) is discussed in the immediately following paragraph.

As a specific example, the energy efficiency (distance X per unit of fuel relative to the air, also relative to the ground if the wind is calm) of older commercial jet airliners was ≈43% higher at 35,000 ft to 40,000 ft (≈5 mi/100 lb fuel) than at 20,000 ft (≈3.5 mi/100 lb fuel) [^{7} 1 − ( T C / T H ) 1 / 2 and an atmosphere if not identical with then at least close to the U. S. Standard Atmosphere [^{6} 1 − ( T C / T H ) the figure is only ≈5%, but the Curzon-Ahlborn [^{7} is a more realistic estimate for real-world engines.] Thus improved aerodynamic efficiency rather than improved engine efficiency must have contributed a factor of ≈1.43/1.07 ≈ 1.34 to the improved energy efficiency at 35,000 ft to 40,000 ft over and above that at 20,000 ft. Perhaps improved L / D ratios at given angles of attack at lower ρ and hence at higher h A might contribute somewhat. But the major contribution to this improved energy efficiency—the only contribution if, as is usually at least approximately true, L / D ratios at given angles of attack are independent of ρ and hence of h A —is the reduced power P A and hence reduced energy E A required to traverse a given distance X at a given v at lower ρ and hence at higher h A . This is not uncommon, despite the required increase in angle of attack to above that which maximizes C L / C D and hence L / D , consequently decreasing C L / C D and hence L / D , as the penalty for increasing C L itself and hence L itself sufficiently to maintain L = m g in the face of decreased ρ at fixed v [recall Equations (9) and (10)]: up to a limit, C L / C D and hence L / D decreases more slowly with the required increase in angle of attack than ρ decreases.^{14} Of course this obtains only up to a limit: with continued increase in angle of attack C L / C D and hence L / D decreases ever more rapidly until stalling occurs. Higher-altitude aerodynamic level flight while maintaining the (smaller) angle of attack that would ordinarily maximize L / D would ordinarily also be at sufficiently faster v to decrease the flight time t more than it increases the required power P A , and hence would ordinarily increase energy efficiency E A / X = P A t / X = P A / v even more. But the engine(s) may not be capable of the required increase in P A . Even if they are, in this case increased v would be sufficiently close to Mach 1 to encounter shock-wave drag and consequently increased D and thus decreased L / D even at the optimum angle of attack, and hence also an increase in P A required to overcome the shock-wave drag over and above that owing to increased v. [See Item (c) of the first paragraph of this Section 3.1.2 and Supplementary Note 12.]

But, again, in this paper we focus mainly on the paramount ρ v 2 functional dependency of both lift and drag, which is the first-order dependency upon ρ and upon v. Considerations of departures from the first-order ρ v 2 functional dependency [^{14} but we do not attempt them in this paper: our analyses are qualitative to semiquantitative. Thus we conceal the difficult and complex physics underlying departures from the first-order ρ v 2 functional dependency [^{14} such as discussed in the second, third, and fourth paragraphs thereof (see also Supplementary Notes 5, 6, 8-12, and 14), the ρ v 2 functional dependency of both drag and lift is the paramount, first-order, functional dependency [

If ρ is so small that even the minimum airspeed required for aerodynamic level flight equals or exceeds the speed v orbit , Earth = ( G M Earth / r ) 1 / 2 ≐ ( G M Earth / R Earth ) 1 / 2 ≈ 8 × 10 3 m / s required for minimum-altitude circular-orbit ballistic spaceflight about Earth,^{4} then all flight about Earth must be ballistic rather than aerodynamic. We now estimate how small ρ must be and how high the altitude in Earth’s atmosphere must be for this to obtain. As discussed in Section 3.1, aerodynamic forces of lift and drag are typically, at least approximately, proportional to ρ v 2 [^{15}

ρ min , abs , Earth ≈ ρ 0 , Earth [ v min ( ρ 0 , Earth ) v orbit,Earth ] 2 ≈ ρ 0 , Earth [ v min ( ρ 0 , Earth ) 8 × 10 3 m / s ] 2 . (15)

Thus within the approximation that ρ v 2 is a conserved quantity, independent of ρ (and hence of flight altitude on either Earth or Mars) and of v, ρ min , abs , half , x = 4 ρ min , abs , all , x : with the help of centrifugal force the absolute lower limit of air density ρ min , abs , x for aerodynamic level flight is 4 times that without its help. Thus within this approximation as per the last six paragraphs of Section 3.1.1 the power required for aerodynamic level flight at the absolute lower limit of air density ρ min , abs , x , and hence also (since this power must be frictionally dissipated) the consequent frictional aerodynamic heating, is less by a factor of (1/4)^{1/2} = 1/2 with the help of centrifugal force than without its help. But, at any rate, in Section 3.2 and the last three paragraphs of Section 3.3.1, we give only in-the-ballpark estimates.

The minimum airspeeds v m i n ( ρ 0 , Earth ) required for aerodynamic level flight at sea-level air density ρ 0 , Earth ≈ 1 kg / m 3 of any aircraft obtain for the lightest unmanned model airplanes, which are limited to indoor flights, and to outdoor flights only if the wind is calm. (For hovering flight these minimum airspeeds are the root-mean-square average, taken over the geometrical area of the airfoils, of airspeeds of the airfoils, be they blades of a helicopter, or wings of a hovering hummingbird or insect.) These minimum airspeeds v m i n ( ρ 0 , Earth ) at sea-level air density ρ 0 , Earth ≈ 1 kg / m 3 are somewhat less than 1 m/s, or ≈10^{−4} of v orbit,Earth ≈ 8 × 10 3 m / s required for minimum-altitude circular-orbit ballistic spaceflight about Earth,^{4} say, v m i n ( ρ 0 , Earth ) ≈ 0.8 m / s . [By Equations (9) and (10), if C L ≈ 1 , taking ρ = ρ 0 , Earth ≈ 1 kg / m 3 and g = 9.8 m / s 2 , v ≈ 10 − 4 v orbit,Earth ≈ 0.8 m / s corresponds to a wing loading of

m g / A wing , geom ≈ 1 2 ρ 0 , Earth v 2 ≈ ( 1 2 × 1 × 0.8 2 ) N / m 2 ≈ 0.3 N / m 2 or

m / A wing , geom ≈ 0.03 kg / m 2 . Thus, putting v min ( ρ 0 , Earth ) ≈ 0.8 m / s in Equation (15), if ρ ≲ ρ min , abs , Earth ≈ ( 10 − 4 ) 2 ρ 0 , Earth = 10 − 8 ρ 0 , Earth ≈ 10 − 8 kg / m 3 , corresponding to altitudes h ≳ h A , max , abs , Earth ≈ 130 km ≈ 430000 ft above sea level in Earth’s atmosphere [^{15} Hence all flight about Earth must then be ballistic rather than aerodynamic.^{15 }

This estimate of the low-density/high-altitude limit for aerodynamic level flight about Earth is an ultimate limit that neglects all practical difficulties. To re-emphasize, the most obvious and most general of practical difficulties are the required power P A and especially the consequent frictional aerodynamic heating, the latter being equal to P A because P A is ultimately thermally dissipated via frictional aerodynamic heating. Other practical difficulties, which we do not consider, include the reduction of the power available from air-breathing engines with decreasing ρ and hence with increasing h A (obviously aerodynamic level flight is possible if and only if the maximum available power exceeds, or at the very least equals, the required power P A ),^{13} and practical difficulties that are specific for given types of aircraft, e.g., maximum airspeeds for propeller airplanes,^{16} and minimum and maximum airspeeds for various types of jets.^{17} By the last five paragraphs of Section 3.1.1, P A , and hence also the consequent frictional aerodynamic heating and the required rate of heat dissipation, is, at least approximately, proportional to ρ − 1 / 2 [^{18} Perhaps a reasonable estimate of the practical low-density/high-altitude limit of aerodynamic level flight in Earth’s atmosphere, even if attainable only by the lightest unmanned model airplanes, is ρ min , prac , Earth ≈ 10 − 3 ρ 0,Earth ≈ 10 − 3 kg / m 3 , corresponding to an altitude of h A , m a x , prac,Earth ≈ 180000 ft ≈ 55 km . (This is also approximately the practical low-density/high-altitude limit of balloons in Earth’s atmosphere as of this writing.^{18})

Note that, by Equation (2) and the last six paragraphs of Section 3.1.1 (neglecting exceptions to the paramount ρ v 2 functional dependency as per Section 3.1.2), the energy E A required for aerodynamic level flight of an aircraft of weight mg traversing given distance X is not greater at the practical—or even the ultimate—low-density/high-altitude limit of aerodynamic level flight than at sea level. Difficulties arise only because E A must be expended faster and hence thermally dissipated faster (frictional aerodynamic heating!) in thinner air— P A = ∂ E A / ∂ t ∝ ρ − 1 / 2 .

Atmospheric density at low altitudes on Mars, ρ 0 , Mars ≈ ( 1 / 70 ) kg / m 3 , is ≈1/70 of ρ 0, Earth ≈ 1 kg / m 3 as obtains on Earth [

Recalling the third and fourth paragraphs of Section 3.1.2, we also note that aerodynamic level flight at a given v can be more energy-efficient at lower ρ as on Mars. This can obtain despite the required increase in angle of attack to above that which maximizes C L / C D and hence L / D , consequently decreasing C L / C D and hence L / D , as the penalty for increasing C L itself and hence L itself sufficiently to maintain L = m g in the face of decreased ρ [recall Equations (9) and (10)]: up to a limit, C L / C D and hence L / D decreases more slowly with the required increase in angle of attack than ρ decreases.^{14} Of course this obtains only up to a limit: with continued increase in angle of attack C L / C D and hence L / D decreases ever more rapidly until stalling occurs. Aerodynamic level flight at lower ρ as on Mars while maintaining the (smaller) angle of attack that maximizes L / D would be at sufficiently faster v to decrease the flight time t more than it increases the required power P A , and hence would increase energy efficiency E A / X = P A t / X = P A / v even more. But the engine(s) may not be capable of the required increase in P A . Even if they are, in some cases increased v may be detrimental (recall the third and fourth paragraphs of Section 3.1.2).

What is more, owing to Mars’ weaker gravity, atmospheric density decreases more slowly with increasing altitude on Mars than on Earth, even in the face of the higher molecular weight and lower temperature of Mars’ atmosphere. The scale height [

R ≈ 1 70 e − − h 11 km e − − h 8.5 km = 1 70 e h ( 1 8.5 km − 1 11 km ) ≈ 1 70 e h 37 km . (16)

At h ≈ 157 km , R ≈ 1 , i.e., density is approximately equal in Mars’ and Earth’s atmospheres. At all higher altitudes, Mars’ atmosphere is denser than Earth’s, and in increasing ratio R with increasing altitude h.

Thus the practical low-density limit of aerodynamic level flight in Earth’s atmosphere ρ m i n , prac,Earth ≈ 10 − 3 ρ 0, Earth ≈ 10 − 3 kg / m 3 corresponds (solving 10 − 3 ≈ 1 70 e − h / 11 km ) to an altitude of h ≈ 29 km ≈ 95000 ft in Mars’ atmosphere. This is also approximately the practical low-density/high-altitude limit of balloons in Mars’ atmosphere. The practical low-density limit of buoyant flight ρ ≈ 10 − 3 kg / m 3 is equal in Earth’s and Mars’ atmospheres because lower g on Mars reduces both weight and buoyancy by the same factor of ≈0.38. Indeed for this reason it is ρ ≈ 10 − 3 kg / m 3 in any atmosphere of any planet. This is strictly true for buoyant flight via vacuum airships, which obtain buoyant lift equal to atmospheric density ρ per unit volume of (essentially) perfect vacuum, or ( 1 − F ) ρ per unit volume of imperfect or partial vacuum (e.g., hot-air balloons) of density F ρ (of course 0 < F < 1 ). Depressurized (as opposed to hot-air) vacuum airships are more practicable on Mars than on Earth: owing to the higher molecular weight and lower temperature of Mars’ atmosphere, its ratio of pressure to density is smaller than that of Earth’s atmosphere. See: Clarke, J.-P.; Rimoli, J.; Gloyd, J. T.; Logarzo, H.; Kraus, J. (2018) Evacuated Airship for Mars Missions (Georgia Tech Air Transportation Laboratory) at https://ntrs.nasa.gov/search.jsp?R=20180006789 (left-click on “View Document”) and Vacuum airship (most recently revised in 2019) at https://www.wikipedia.org. For buoyant flight via lifting gas of density F ρ (of course 0 < F < 1 ) buoyant lift equals ( 1 − F ) ρ per unit volume of lifting gas. Since for a lifting gas F equals the ratio of the molecular weight of the lifting gas to that of the ambient atmosphere, unlike for vacuum airships (or partial-vacuum airships such as hot-air balloons) F is larger for a given lifting gas in a higher-molecular-weight atmosphere such as Mars’ atmosphere than in a lower-molecular-weight one such as Earth’s. (Obviously, 0 < F ≪ 1 is required if the practical low-density limit of buoyant flight ρ ≈ 10 − 3 kg / m 3 is to be attained.)

But the practical high-altitude limit h A , m a x , prac,Mars of aerodynamic level flight on Mars is much higher than this, because we still must consider one important thing that is unequal in favor of Mars—Mars’ weaker gravity: g on Mars is ≈0.38 of g on Earth. By Equation (14) the power P A , Mars required for aerodynamic level flight in Mars’ atmosphere at the altitude where 0.38 3 / 2 / ρ m i n , prac,Mars 1 / 2 = 1 3 / 2 / ρ m i n , prac,Earth 1 / 2 = 1 / ρ m i n , prac,Earth 1 / 2 ≈ 1 / ( 10 − 3 kg / m 3 ) 1 / 2 , i.e., where ρ m i n , prac,Mars = 0.38 3 ρ m i n , prac,Earth ≈ 0.38 3 × 10 − 3 kg / m 3 ≈ 5 × 10 − 5 kg / m 3 , equals the power P A , Earth required for aerodynamic level flight in Earth’s atmosphere at the altitude h A , m a x , prac,Earth ≈ 180000 ft ≈ 55 km where ρ ≈ ρ m i n , prac,Earth ≈ 10 − 3 kg / m 3 , our estimate of the practical low-density/high-altitude limit of aerodynamic level flight in Earth’s atmosphere (recall the third paragraph of Section 3.2). Thus probably a reasonable estimate of the practical low-density/high-altitude limit of aerodynamic level flight in Mars’ atmosphere, even if attainable only by the lightest unmanned model airplanes (recall Section 3.2), is ρ m i n , prac,Mars ≈ 5 × 10 − 5 kg / m 3 , which corresponds(solving 5 × 10 − 5 ≈ 1 70 e − h A , m a x , prac,Mars / 11 km ) to an altitude of h A , m a x , prac,Mars ≈ 62 km ≈ 204000 ft in Mars’ atmosphere—slightly higher than h A , m a x , prac,Earth ≈ 180000 ft ≈ 55 km . Low g on Mars, ≈0.38 of g on Earth, implies a reasonable estimate of ρ m i n , prac,Mars being lower than ρ m i n , prac,Earth . This lower required density, combined with g on Mars ≈ 0.38 of g on Earth rendering the scale height [^{18} the former is considerably higher than the latter in Mars’ atmosphere. As noted in the immediately preceding paragraph, the practical high-altitude limit of buoyant flight corresponds to ρ ≈ 10 − 3 kg / m 3 on Mars as on Earth because lower g on Mars reduces both weight and buoyancy by the same factor of ≈0.38—indeed for this reason it is ρ ≈ 10 − 3 kg / m 3 in any atmosphere of any planet.)

Now let us consider the ultimate, as opposed to the practical, low-density/high-altitude limit of aerodynamic level flight in Mars’ atmosphere—corresponding to the speed of aerodynamic level flight of the lightest unmanned model airplanes equaling minimum-altitude circular-orbital speed on Mars, neglecting all practical considerations, the most obvious and most general of which are power requirements and frictional aerodynamic heating. Consider indoor flight of the lightest unmanned model airplanes on Mars with indoor air density maintained equal to that at sea level on Earth, i.e., ρ 0, Earth ≈ 1 kg / m 3 . Since g on Mars is ≈0.38 of g on Earth, by the first line of Equation (14) the speed required to maintain indoor aerodynamic level flight of the lightest unmanned model airplanes on Mars at indoor air density ρ 0, Earth ≈ 1 kg / m 3 is v m i n , Mars ( ρ 0, Earth ) ≈ 0.38 1 / 2 ≈ 0.62 of the ≈0.8 m/s required on Earth, i.e., ≈0.5 m/s. The speed v orbit , Mars = ( G M Mars / r ) 1 / 2 ≈ ( G M Mars / R Mars ) 1 / 2 ≈ 3.6 km / s required for minimum-altitude circular-orbit ballistic spaceflight on Mars is ≈0.45 times v orbit,Earth ≈ 8 × 10 3 m / s .^{4} Thus for Mars’ atmosphere, applying and slightly modifying Equation (15) yields^{15}

ρ min,abs,Mars ≈ ρ 0,Earth [ v min,Mars ( ρ 0,Earth ) v orbit,Mars ] 2 ≈ ρ 0,Earth [ 0.62 v min,Earth ( ρ 0,Earth ) 0.45 v orbit,Earth ] 2 ≈ ρ 0,Earth [ 0.62 ( 0.8 m / s ) 0.45 ( 8 × 10 3 m / s ) ] 2 ≈ ρ 0,Earth ( 0.62 0.45 × 10 − 4 ) 2 ≈ 1.9 × 10 − 8 ρ 0,Earth ≈ 1.9 × 10 − 8 kg / m 3 ≈ 1.9 ρ min,abs,Earth ≈ 1.3 × 10 − 6 ρ 0,Mars . (17)

Solving 1.3 × 10 − 6 ≈ e − h A ,max,abs,Mars / 11 km yields h A , m a x , abs,Mars ≈ 149 km ≈ 490000 ft —slightly higher than h A , m a x , abs,Earth ≈ 130 km ≈ 430000 ft . The two reasons for this high value h A , m a x , abs,Mars ≈ 149 km ≈ 490000 ft on Mars are g on Mars being ≈0.38 of g on Earth and Mars’ atmospheric scale height [

It is perhaps worthwhile to re-emphasize that, by Equation (2), the last six paragraphs of Section 3.1.1, the last paragraph of Section 3.2, and the first paragraph of this Section 3.3.1 (neglecting exceptions to the paramount ρ v 2 functional dependency as per Section 3.1.2), the energy E A , Mars required for aerodynamic level flight of an aircraft of mass m traversing given distance X is not greater at the practical—or even ultimate—low-density/high-altitude limit of aerodynamic level flight on Mars than at low altitudes on Mars: E A , Mars ≈ 0.38 of E A , Earth required to traverse X at sea level or any higher altitude on Earth. Difficulties arise only because E A must be expended faster and hence thermally dissipated faster (frictional aerodynamic heating!) in thinner air on Mars as on Earth as anywhere— P A = ∂ E A / ∂ t ∝ ρ − 1 / 2 .

Now let us consider, even if only hypothetically as a thought experiment, a fully-submerged underwater airplane on Earth, even though no such craft actually exists and one is extremely unlikely to ever be built. (An underwater airplane, which cruises fully submerged, should not be confused with a hydrofoil, which typically except for its wings cruises above water.) Let our underwater airplane be of density ( N + 1 ) × 10 3 kg / m 3 , the volume of all of its solid parts be V m^{3}, and hence its mass be m = ( N + 1 ) × 10 3 V kg . Let us compare our underwater airplane to a standard (atmospheric) airplane on Earth, of identical size and shape, but of density N × 10 3 kg / m 3 and hence of mass m = N × 10 3 V kg . Our underwater airplane’s extra 10^{3} kg/m^{3} of density offsets the buoyancy provided by water ( ρ water ≈ 10 3 kg / m 3 ), so that its underwater flight can be evenhandedly compared with atmospheric flight of our standard (atmospheric) airplane. Thus our underwater airplane’s extra 10^{3} kg/m^{3} of density renders water its aerodynamic medium as opposed to its buoyant medium, consistently with air being our standard (atmospheric) airplane’s aerodynamic medium. This facilitates an evenhanded comparison between our standard (atmospheric) airplane’s aerodynamic level flight in air and our underwater airplane’s aerodynamic level flight in water. We take N sufficiently larger than the density of air at sea level, ρ 0, Earth ≈ 1 kg / m 3 , that the buoyancy provided by air for our standard (atmospheric) airplane can be neglected (this easily obtains, because all solids are hundreds to thousands of times as dense as air at sea level), and we assume high enough Reynolds numbers for both airplanes that skin-friction drag is small compared to pressure drag for both airplanes (see Supplementary Note 10). Water is ≈ 800 times as dense as air at sea level. Thus, by the last six paragraphs of Section 3.1.1, our underwater airplane need fly only ≈ 800 − 1 / 2 ≈ 1 / 28 as fast as our standard (atmospheric) airplane at or near sea level to sustain level flight, and by Equation (14) requires only ≈ 800 − 1 / 2 ≈ 1 / 28 as much power to sustain level flight as our standard (atmospheric) airplane at or near sea level. But our underwater airplane takes ≈ 800 1 / 2 ≈ 28 times as long to traverse a given horizontal distance X as our standard (atmospheric) airplane. Hence by Equation (2) and the last six paragraphs of Section 3.1.1, our Earth underwater airplane requires the same energy E A as our standard (atmospheric) Earth airplane to traverse a given horizontal distance X.

Recalling the last six paragraphs of Section 3.1.1 and that g on Mars is ≈0.38 of g on Earth, if atmospheric pressure on Mars was high enough for liquid water to exist, then both the required flight speed and the required power for aerodynamic flight of a Mars underwater airplane would by Equation (14) be ≈ 0.38 1 / 2 ≈ 0.62 of the values required for an identical Earth underwater airplane, both underwater airplanes being of density ( N + 1 ) × 10 3 kg / m 3 and mass m = ( N + 1 ) × 10 3 V kg . (Atmospheric pressure on Mars was high enough for liquid water to exist in the past [^{19}) Also, by Equation (2) and the last six paragraphs of Section 3.1.1, a Mars underwater airplane requires the same energy E A , Mars to traverse given horizontal distance X as a standard (atmospheric) Mars airplane identical except for being of density N × 10 3 kg / m 3 and hence of mass m = N × 10 3 V kg . By Equation (2) and comparison with Section 3.3.1, this energy E A , Mars is ≈0.38 times E A , Earth required to traverse the same horizontal distance X by an identical Earth underwater airplane of density ( N + 1 ) × 10 3 kg / m 3 and mass m = ( N + 1 ) × 10 3 V kg , or by a standard (atmospheric) Earth airplane identical except for being of density N × 10 3 kg / m 3 and hence of mass m = N × 10 3 V kg .

Hydrofoils should be classified as aircraft because their lift obtains primarily if not essentially entirely aerodynamically rather than via buoyancy, even though the density ρ of their aerodynamic medium (water) is ≈800 times that of air at sea level. By lifting the hull out of the water into the air, drag on the hull at any given speed is reduced ≈800 times; only the wings need suffer water resistance as opposed to air resistance. (We define “hull” as incorporating all parts of a hydrofoil that are lifted out of the water, e.g., including most or all of the struts.) Thus, probably uniquely among aircraft, for hydrofoils two values of aerodynamic-medium density ρ are pertinent: ρ drag , the density pertinent to drag, and ρ lift , the density pertinent to lift. Also, probably uniquely among aircraft other than hovercraft, for hydrofoils unless there is neither wind nor water current two values of velocity v are pertinent: v drag , the velocity pertinent to drag, and v lift , the velocity pertinent to lift. Let ρ air be the density of air, ρ water be the density of water, v air be the velocity of a hydrofoil relative to the air, v water be the velocity of a hydrofoil relative to the water, ( ρ v 2 ) air = ρ air v air 2 , and ( ρ v 2 ) water = ρ water v water 2 . Drag on a hydrofoil probably can be construed most simply via 〈 ρ v 2 〉 drag , the average with respect to drag of ρ v 2 , as follows:

〈 ρ v 2 〉 drag = A frontal,eff,hull ( ρ v 2 ) air + A frontal,eff,wing ( ρ v 2 ) water A frontal,eff,hull + A frontal,eff,wing = A frontal,eff,hull ( ρ v 2 ) air + A frontal,eff,wing ( ρ v 2 ) water A frontal,eff,total . (18)

This average value 〈 ρ v 2 〉 drag can be employed in Equation (7). Note that, in Equation (18), A frontal,eff,wing is the effective frontal cross-sectional area of a hydrofoil’s wings with respect to drag as per the first two paragraphs of Section 3.1.1, not the effective surface area of its wings with respect to lift as per the third and fourth paragraphs of Section 3.1.1. Since typical speeds of hydrofoils are ~5 times typical wind speeds,^{20} v water and v air by and large differ by ~20%. (Water currents are usually much slower than winds.) In contrast with Equation (7) for drag, Equation (9) for lift requires no reinterpretation for hydrofoils other than setting ρ = ρ water and v = v water .

Because water is ≈800 times as dense as air at sea level, the wings of a hydrofoil need only have ≈1/800 of the surface area with respect to lift as the wings of a (low-speed) airplane in order to obtain the same lift (if C L is the same for both the hydrofoil and the airplane, and if v water for the hydrofoil equals v air for the airplane).

Any surface watercraft experiences an additional form of drag that we do not consider in this paper: wave drag, the energy cost of generating waves (of course not to be confused with shock-wave drag experienced by aircraft at and in the vicinity of Mach 1).^{21} But, because only the wings and at most only the lower part of the struts of a hydrofoil, which have minimal surface area, intersect the surface of the water, this form of drag is minimal for a hydrofoil if, as we assume, it cruises at sufficient speed to lift all but its wings and at most the lower part of its struts completely out of the water.

For hydrofoils, as for other aircraft, the energy cost E A for traversing given horizontal distance X is directly proportional to mg. Hence reducing mg reduces the energy cost of aerodynamic level hydrofoil flight traversing given horizontal distance X equally and in direct proportion to the reduction in mg. Thus, for example, since g on Mars is ≈0.38 of g on Earth, if atmospheric pressure on Mars were high enough for liquid water to exist, then E A , Mars for aerodynamic level flight of a hydrofoil of given mass m traversing given horizontal distance X on Mars would be ≈0.38 of E A , Earth required on Earth. (As mentioned in the second paragraph of Section 3.3.2, atmospheric pressure on Mars was high enough for liquid water to exist in the past [^{19})

Hydrofoils excepted, the range of aerodynamic-medium (air) densities on Earth, from ≈1 kg/m^{3} at sea level to ≈10^{−3} kg/m^{3} at the approximate practical (as opposed to ultimate) high-altitude limit of aerodynamic level flight, seems to be optimum for aerodynamic level flight, indeed for any aerodynamic flight. Much denser aerodynamic media, such as water in the case of underwater airplanes discussed in Section 3.3.2, allow aerodynamic level flight with much less power, but owing to the great resistance of a very dense medium such as water the speed of aerodynamic level flight will then be relatively slow. (Even hydrofoils are much slower than most airplanes.) Much more rarefied aerodynamic media allow higher speeds but entail difficulties of large required power and consequent dissipation of this power via equally large frictional aerodynamic heating, as discussed in Section 3.2 and the last paragraph of Section 3.3.1. As discussed in Section 3.3.1, lower g such as on Mars reduces the practical low-density limit somewhat from ≈10^{−3} kg/m^{3} as obtains on Earth. But if g is too small, appreciably smaller than on Mars, then an atmosphere cannot be retained at all, thus precluding aerodynamic level flight, indeed precluding any aerodynamic flight (except in pressurized indoor facilities).

The considerations of the immediately preceding paragraph are modified in the case of hydrofoils, because they obtain lift via a dense medium (water) but the vast majority of their frontal cross-sectional area (both geometrical and effective) suffers resistance or drag only from a much more rarefied one (air). Of course, if g is too small, then an atmosphere cannot be retained, and with vanishing atmospheric pressure liquid water—indeed any liquid—cannot exist (except in pressurized indoor facilities).

All flights of hand-thrown projectiles that are unpowered except for the initial throw are obviously short flights as defined in Section 2.1. Thus it is not surprising that the record traversed horizontal distances for hand-thrown projectiles obtain for those executing aerodynamic flight, e.g., Frisbees, Aerobies, and boomerangs, as opposed to those executing flight that is at least primarily ballistic, e.g., sports balls [

From among hand-thrown projectiles, we define as hand-thrown aircraft those (e.g., Frisbees, Aerobies, and boomerangs) that are capable of aerodynamic flight—of flight with aerodynamic lift exceeding weight—at achievable throwing speeds. The record horizontal flight distances for hand-thrown aircraft as of this writing include 427.2 m = 1402 ft for boomerangs, 406.3 m = 1333 ft for Aerobies, and 338.00 m = 1108 ft 11.1 in for Frisbees. Moreover (unlike primarily ballistic hand-thrown projectiles) aerodynamic Frisbees, Aerobies, and boomerangs can—since lift exceeds weight at achievable throwing speeds—traverse longer distances against the wind than with it. Hand-thrown projectiles such as sports balls (e.g., golf balls, baseballs, etc.) [^{22}, more aerodynamic lift than javelins, though still not nearly enough to equal, let alone exceed, their weights at achievable throwing speeds. Hence the hand-thrown distance records even for sports balls, while exceeding those for javelins, still fall far short of those achieved by Frisbees, let alone by Aerobies and boomerangs. And hence also javelins and even sports balls cannot in any case traverse longer distances against the wind than with it.

We should also mention the discus [

We now describe what may be the simplest example of how a hand-thrown aircraft capable of flight with aerodynamic lift exceeding weight at achievable throwing speeds, such as a discus, Frisbee, Aerobie, or boomerang, can maintain its altitude farther if thrown horizontally against the wind than with it. Let a discus, Frisbee, Aerobie, or boomerang of mass m and weight mg be thrown horizontally in calm air ( v wind = 0 ) at altitude h A , 0 and at ground speed v ground = v air , min , the minimum airspeed required for aerodynamic lift L to equal its weight mg at the angle of attack at which it is thrown. Since the air is calm, its airspeed v air also equals v air , min at the instant of being thrown. Thus with calm air at the instant of being thrown v air = v ground = v air , min . Reiterating, at the instant of being thrown aerodynamic lift L equals its weight mg at the angle of attack at which it is thrown, i.e., L = m g at this angle of attack: thus v air , min is the minimum airspeed required for aerodynamic level flight at this angle of attack. (To avoid confusion, in this paragraph and the next we employ subscripts to distinguish between v air , v ground , and v wind .) [For maximum horizontal flight distance, the angle of attack must of course be that which maximizes C L / C D and hence L / D . With increased angle of attack (up to a stall) C L increases and hence L = m g obtains at a slower v air , min . But at these increased angles of attack C D increases faster with increasing angle of attack than C L , so C L / C D and thus L / D decreases, and hence horizontal flight distance is diminished.] Since a hand-thrown aircraft such as a discus, Frisbee, Aerobie, or boomerang is unpowered except for the initial throw, immediately after being thrown aerodynamic drag D will have reduced v air (and given calm air also v ground ) to less than v air , min and hence lift L to less than weight mg. Thus immediately after being thrown it begins to lose altitude. Thus it can maintain altitude h A = h A , 0 only at the very instant when it is thrown and hence for only infinitesimal horizontal distance X = 0 . Now let the discus, Frisbee, Aerobie, or boomerang be thrown horizontally at altitude h A , 0 and at ground speed v ground = v air , min ≥ v wind with a tail wind of speed v wind . (The restriction to the speed range v ground = v air , min ≥ v wind is for simplicity, so that we can focus on the main points. The speed range 0 ≤ v ground = v air , min < v wind is discussed in Supplementary Note 16.) Thus at the instant of being thrown its airspeed is v air = v ground − v wind = v air , min − v wind < v air , min . Thus even at the instant of being thrown L < m g . Thus it cannot maintain altitude h A = h A , 0 even at the instant of being thrown and hence not even for infinitesimal horizontal distance X = 0 . Now let the discus, Frisbee, Aerobie, or boomerang be thrown horizontally at altitude h A , 0 at and at ground speed v ground = v air , min against a head wind of speed v wind . At the instant of being thrown its airspeed is v air = v air , min + v wind > v air , min . Thus at the instant of being thrown L > m g . Thus it gains altitude until drag D reduces v air to v air , min and hence L to mg, by which time it will have ascended to its peak altitude h A , peak > h A , 0 . Thereafter L < m g and it begins to lose altitude, soon descending past h A , 0 . But it will have traversed finite horizontal distance X > 0 while maintaining altitude h A > h A , 0 and attaining peak altitude h A , peak > h A , 0 if thrown horizontally at v ground = v air , min against the wind, as opposed to infinitesimal horizontal distance X = 0 and even that barely at altitude h A = h A , 0 if thrown horizontally at v ground = v air , min with no wind, and not even infinitesimal horizontal distance X = 0 barely at altitude h A = h A , 0 if thrown horizontally at v ground = v air , min ≥ v wind with the wind. By Section 3.1.1, especially Equations (7) and (8) and the associated discussions, typically D is greater for a discus, Frisbee, Aerobie, or boomerang thrown against the wind than with it by a ratio ℝ drag ≈ [ ( v ground + v wind ) ÷ ( v ground − v wind ) ] 2 . But ℝ drag is a finite number, typically only moderately larger than unity. Moreover, by Equations (9) and (10) and the associated discussions, typically L is also greater for a discus, Frisbee, Aerobie, or boomerang thrown against the wind than with it by a comparable ratio ℝ lift ≈ [ ( v ground + v wind ) ÷ ( v ground − v wind ) ] 2 . Therefore typically L / D and hence C L / C D is at least approximately equal with the wind and against the wind (and also with no wind). Hence ℝ drag typically being moderately larger than unity does not contravene our result that it is possible for a discus, Frisbee, Aerobie, or boomerang to maintain altitude farther against the wind than with it. In our specific examples, altitude is maintained for finite horizontal distance X > 0 against the wind, as opposed to infinitesimal horizontal distance X = 0 with no wind and not even infinitesimal horizontal distance X = 0 with the wind—in all three cases the initial horizontal throw being at v ground = v air , min . Indeed, untypically [

Two auxiliary points: (i) The scenario in the immediately preceding paragraph, of a discus, Frisbee, Aerobie, or boomerang thrown at v ground = v air = v air , min , i.e., corresponding to L = m g , at a given angle of attack given calm air, was chosen for simplicity. But of course discuses, Frisbees, Aerobies, and boomerangs can easily be thrown considerably faster than this. They can easily be thrown fast enough so that L > m g in calm air, i.e., at v ground = v air > v air , min , or even so that L > m g with a light or moderate tail wind, i.e., at v ground > v wind + v air , min ⇔ v air = v ground − v wind > v air , min . (ii) If the wind speed exceeds maximum achievable throwing speeds, then a discus, Frisbee, Aerobie, or boomerang cannot maintain altitude farther against the wind than with it—at least not relative to the ground. For example, consider throwing a discus, Frisbee, Aerobie, or boomerang into a Category 5 extreme-hurricane-force or an EF5 extreme-tornado-force head wind.^{23} It will still traverse finite horizontal distance X > 0 at altitude h A > h A , 0 (peaking at altitude h A , peak > h A , 0 ) if thrown horizontally against the wind—but relative to the air, not relative to the ground. Such an extreme wind will reverse the hand-thrown aircraft’s direction of motion relative to the ground almost instantaneously, well before it traverses this finite horizontal distance X > 0 (peaking at altitude h A , peak > h A , 0 ) relative to the ground.

By contrast, hand-thrown projectiles whose flights are primarily ballistic, such as javelins [

Thus far in this Section 4, we considered hand-thrown projectiles on Earth. Let us now briefly consider them on Mars. Maximum achievable throwing speeds are the same on Mars as on Earth, g on Mars is ≈0.38 of g on Earth, and low-altitude air density on Mars is ≈1/70 of that on Earth [

It may be of interest to compare the energy efficiency of both aerodynamic level flight and ballistic flight with that of horizontal surface (land and/or water) transportation. If the frictional force opposing horizontal motion of a surface vehicle of mass m is a fraction F of its weight mg, then the energy cost of its traversing horizontal distance X is

E S = m g F X . (19)

(The subscript S denotes surface transportation.)

Thus F plays the same role in surface transportation that D / L plays in aerodynamic level flight, or, equivalently, 1 / F plays the same role in surface transportation that L / D plays in aerodynamic level flight. For one circumnavigation at Earth’s surface X = 2 π R and E S = 2 π R m g F ; for N circumnavigations at Earth’s surface X = 2 π R N and E S = 2 π R m g F N . (A single-circumnavigation journey is the longest possible one whose purpose is to reach a destination on Earth, with the destination being the starting point after traveling around the world.)

Typical values of the coefficient of surface friction C S for land vehicles range from ≈0.01 to ≈1 for sliding friction (≈0.005 for some maglev trains), and as low as ≈0.001 for rolling friction of hard wheels on hard surfaces—for example low-speed to moderate-speed traditional (not maglev) railroad transportation. For land vehicles at speeds low enough that air resistance is small compared to surface (e.g., sliding, rolling, or maglev) friction, F = C S and is at least approximately independent of speed. For land vehicles at higher speeds F = C S + D / m g , D being given by Equation (7) with ρ being the density of air. (While traditional railroad transportation is more energy-efficient at low to moderate speeds, maglev trains almost completely abolish wear on the tracks, owing to their lack of mechanical contact with the tracks.) See Supplementary Note 17.

Since even minimum-altitude circular-orbit ballistic spaceflight must be above any appreciable atmosphere, r B for even minimum-altitude circular-orbit ballistic spaceflight must exceed R. But for simplicity, as in Section 2.2, we let the orbit be a minimum-altitude circular one, for which h B = r B − R ≪ R and hence can be neglected in comparison with R. (For surface transportation at sea level h S = 0 and hence r S = R .) Identically as in the case of aerodynamic flight, in order for surface transportation to be more energy-efficient than ballistic transportation, neglecting air resistance in the latter, we require F < 1 / 2 for short journeys ( X ≪ 2 π R ) and F < 1 / 4 π for single-circumnavigation journeys ( X = 2 π R ). For journeys of intermediate length (X ranging from much smaller than R to approaching 2 π R ), the value that F cannot equal or exceed if surface transportation is to be more energy-efficient than ballistic transportation, neglecting air resistance in the latter, decreases monotonically from 1/2 towards 1 / 4 π as X increases from very small values towards 2 π R . (Since if X = 2 π R minimum-energy ballistic spaceflight is a circular orbit just above appreciable atmosphere at altitude h B , it cannot begin and end at the altitude h s of surface transportation, but h B − h s ≪ R .)

The requirement F < 1 / 2 for short ( X ≪ 2 π R ) surface journeys to be more energy-efficient than short ballistic journeys neglects air resistance in ballistic flight. But air resistance is not always negligible in short ballistic flights, especially in the lower atmosphere. If it is not neglected, then the requirement is weakened to F < 1 / 2 n with n < 1 . The requirement F < 4 π for single-circumnavigation surface journeys to be more energy-efficient than single-circumnavigation minimum-altitude circular-orbit ballistic journeys neglects air resistance in minimum-altitude circular-orbit ballistic spaceflight; if it is not neglected, then the requirement is weakened to F < 1 / 4 π n with n < 1 . But it is weakened only very slightly, because air resistance even at minimum-circular-orbit spaceflight altitude is very small. Not neglecting air resistance in ballistic flight, for journeys of intermediate length (X ranging from much smaller than R to approaching 2 π R ), the value that F cannot equal or exceed if surface transportation is to be more energy-efficient than ballistic flight decreases monotonically from 1 / 2 n ( X ) towards 1 / 4 π n ( X ) as X increases from very small values towards 2 π R : n ( X ) < 1 but increases monotonically towards very nearly 1 as X increases from very small values towards 2 π R . Air resistance in minimum-altitude circular-orbit ballistic flight is very small and hence also 1 − n ( X ) is very small, i.e., n ( X ) is very nearly 1, if X = 2 π R . (Since then minimum-energy ballistic spaceflight is a circular orbit just above appreciable atmosphere at altitude h B , it hence cannot begin and end at the altitude h S of surface transportation, but h B − h S ≪ R .)

By Equation (19) the energy cost E S of surface transportation increases linearly with increasing X and hence for N-circumnavigation journeys also with increasing N, identically as with aerodynamic flight by Equation (2) and Section 2.3. By contrast, for minimum-altitude circular-orbit ballistic spaceflight, irrespective of X and hence also of N, E B remains fixed at the value given by Equation (3) for N = 1 . For, even at minimum-circular-orbit altitude, air resistance is almost negligible, i.e., space is almost frictionless; thus the energy cost of launching a spacecraft is one-time. Thus for multi-circumnavigation ( X = 2 π R N , N ≫ 1 ) journeys, the energy efficiency of even minimum-altitude circular-orbit ballistic spaceflight surpasses that of surface transportation by an arbitrarily large margin, just as it does that of aerodynamic flight by an arbitrarily large margin, the margin being even larger if spaceflight is high-orbit and even larger yet if it exceeds escape velocity. With respect to both aerodynamic flight and surface transportation the reason is that stated in Section 2.3: Space is essentially frictionless, and increasingly frictionless with increasing altitude, thus allowing spacecraft but neither aircraft nor surface vehicles to take full advantage of Newton’s first law of motion (inertia).^{5} The energy cost of speed in spaceflight is one-time; the energy cost of speed in aerodynamic flight and in surface transportation is never-ending.^{5} Spaceflight is thus the only mode of transportation that can achieve ∞ mi / gal = ∞ km / l of fuel (or the equivalent thereof)— Spaceship Earth (whose fuel for its orbital and rotational motions was part of the solar nebula’s kinetic energy) is a good example.^{5} To save time in spaceflight continuous energy expenditure can be employed, for example employing solar, laser, or on-board nuclear energy. But in spaceflight continuous energy expenditure buys acceleration; in aerodynamic flight and in surface transportation it buys only (constant) speed.^{5}

We note that lighter-than-air craft, for example dirigibles and blimps, should be included within the category of surface transportation rather than within the category of aerodynamic flight, because their lift obtains typically at least primarily and often entirely via buoyancy rather than via aerodynamics. Similarly the lift for surface (land and/or water) transportation vehicles obtains via support of the ground for land surface vehicles and via buoyancy for ships and submarines, rather than via aerodynamics. In this regard, we construe maglev and air-cushion vehicles as being surface (land and/or water) transportation vehicles rather than aircraft, because their lift obtains via support of the surface through the intermediary of a magnetic-repulsion or an air cushion. [Aircraft very near the ground obtain some extra lift from the air-cushion “ground effect” (see Supplementary Note 18).] For a dirigible or blimp in level flight F is the ratio of air resistance to the unbuoyed weight of the dirigible or blimp, i.e., F = D / m g , D being given by Equation (7) with ρ being the density of air at flight altitude. For a fully-submerged submarine F is the ratio of water resistance to the unbuoyed weight of the submarine, i.e., F = D / m g , D being given by Equation (7) with ρ being the density of water. For a ship or surface-cruising submarine, or for a hydrofoil, F is the ratio of combined water and air resistance to the unbuoyed weight of the vehicle, i.e., F = D / m g , D being given by Equation (7) with 〈 ρ v 2 〉 drag taken as for hydrofoils as per Section 3.4 [see especially Equation (18) and the associated discussions]. (F for land vehicles was discussed in the third paragraph of this Section 5.) By contrast, as previously noted [see especially Section 3.4 but also the second paragraph of Section 1, the second paragraph following that containing Equation (2), and Section 3.5], hydrofoils should be classified as aircraft, because their lift obtains primarily if not essentially entirely aerodynamically rather than via buoyancy, even though the density ρ of their aerodynamic medium (water) is ≈800 times that of air at sea level. (By lifting the hull out of water into air, drag on the hull at any given speed is reduced ≈800 times; only the wings need suffer water resistance as opposed to air resistance.) Any surface-cruising watercraft experiences wave drag, i.e., the energy cost of generating waves (of course not to be confused with shock-wave drag experienced by aircraft at and in the vicinity of Mach 1),^{21} which we have not considered in this paper. Wave drag is minimal for a hydrofoil cruising at sufficient speed to lift its hull completely out of the water, but for ships and surface-cruising submarines it is typically the largest component of drag.^{21}

While this is obvious, perhaps it is worthwhile to note that there is a minimum flight speed for any (nonhovering) aircraft, but no minimum speed for land vehicles, dirigibles, ships, or submarines. Increased induced drag is imposed on (nonhovering) aircraft at minimum flight speed (see Supplementary Notes 6 and 9). By contrast: (a) For land vehicles, as speed is reduced to zero, F = C S remains at least approximately constant. (b) For dirigibles, for ships, and for both fully-submerged and surface-cruising submarines, as speed is reduced to zero, D and hence also F = D / m g is reduced to zero.

Of course, the remarks of the last paragraph of Section 2.1, and especially the remarks of Section 2.4, distinguishing between the energy efficiency of aerodynamic or ballistic flight per se and the energy efficiency of the engine that powers aerodynamic or ballistic flight apply equally in distinguishing between the energy efficiency of surface transportation per se and the energy efficiency of the engine that powers surface transportation. The energy that must be supplied to an engine whose efficiency is ϵ in order to facilitate surface transportation requiring energy E S as per Equation (19) is of course E S / ϵ .

Generalizing the third-to-last paragraph of Section 2.1 and the last paragraph of Section 2.2 in light of this Section 5, if air resistance in ballistic flight and in surface transportation can be neglected, then for traversal of any given horizontal distance X, short or long, all three quantities E B , E A , and E S are directly proportional to mg. Hence reducing mg reduces the energy cost of ballistic flight, aerodynamic level flight, and surface transportation traversing any given horizontal distance X equally and in direct proportion to the reduction in mg, but does not alter the ratio of energy costs between these three modes of transportation. If air resistance in ballistic flight or in surface transportation cannot be neglected, then reducing mg reduces the energy cost of ballistic flight or of surface transportation, respectively, less than in direct proportion to the reduction in mg.

Generalizing the second-to-last paragraph of Section 2.1 in light of this Section 5, most typically, mg is reduced by reducing m. But we can also consider reduction of g. Two examples: (i) Aerodynamic and ballistic flight, as well as surface transportation, on Mars is at lower g. (ii) An aircraft, watercraft, or surface vehicle of mass m a fraction f ( 0 < f < 1 ) of whose weight mg is offset by buoyancy can be construed as either being of effective mass m ( 1 − f ) in a gravitational field g or as being of mass m in a gravitational field of effective strength g ( 1 − f ) . Such partial offset of weight by buoyancy obtains, for example, for a dirigible or blimp that relies on buoyancy for only part of its lift, with the balance obtaining aerodynamically, for a hydrofoil that cruises so slowly that it must rely on buoyancy for part of its lift, or for an underwater surface vehicle that is denser than water. Because all solids are hundreds to thousands of times as dense as air at sea level, f is negligible for surface vehicles on land as it is for aerodynamic vehicles (aircraft) in air. But f is not negligible for underwater surface vehicles (and for underwater airplanes as discussed in Section 3.3.2) that are denser than water, because even the densest solids are little more than 20 times as dense as water. [An evenhanded comparison, in Section 3.3.2, between a standard (atmospheric) airplane’s aerodynamic level flight in air and an underwater airplane’s aerodynamic level flight in water was facilitated by an extra 10^{3} kg/m^{3} of density for the latter to offset the buoyancy provided by water.]

In this paper in general we have not considered the energy cost of building and maintaining vehicles. In this Section 5 in particular we also have not considered the energy cost of building and maintaining pathways for surface transportation. Concerning the latter, we have not, for example, considered the energy cost of building and maintaining roads, railroads, and canals. Of course, for transportation that does not require artificially-built pathways, such as transportation in air, on or in water (except via canal), on land and/or water via hovercraft or other surface vehicles that do not require roads, or via spaceflight, this latter energy cost is zero.

Hopefully, we have provided at least somewhat helpful insights concerning energy efficiency in aerodynamic versus ballistic flight, concerning aerodynamic lift and drag, concerning selected aspects and examples of flight, in distinguishing between the energy efficiency of flight per se and the energy efficiency of the engine that powers flight, and via considering the relation between the density of an aerodynamic medium and aerodynamic level flight. Also, hopefully, our comparison with the energy efficiency of surface transportation and our discussion of surface transportation have been helpful.

While we have focused mostly on Earth, with some consideration of Mars, our results are easily generalizable to any planet or other astronomical bodies on which aerodynamic flight and/or surface travel is possible, i.e., to any planet or other astronomical bodies with an atmosphere, and/or a solid and/or liquid surface. Also, they are valid irrespective of the values of M, R, g, m, ρ , and (except as for simplicity we assume h A < h B ≪ R ) of h A , r A = R + h A , h B , and r B = R + h B [

We should emphasize the limitations of this present work. In this paper our main goal was to elucidate more conceptually than mathematically some fundamental ideas concerning energy efficiency and a number of other aspects of aerodynamic versus ballistic flight, and to provide comparison with surface transportation. We did not attempt the mathematically complex and detailed fully-quantitative analyses based on rigorous application of fluid dynamics, e.g., computational fluid dynamics, as is required in the actual design of aircraft, or the analyses required in the actual design of spacecraft or surface vehicles. We also neglected many details required in the actual operation of vehicles: to mention just one example of many, we neglected reduction of vehicle mass m as fuel is consumed. Moreover, we focused mainly on the paramount ρ v 2 functional dependency of lift and drag, which is the first-order dependency upon ρ and upon v. Considerations of departures from the first-order ρ v 2 functional dependency [^{14} [

In closing, we note that, even given all of the advances in aerodynamics, new discoveries are still being made, e.g., see Ref. [

I thank Dr. Donald H. Kobe for very helpful discussions concerning fluid dynamics (especially aerodynamic drag), and for very insightful general scientific discussions over very many years. I am very grateful to my father Amnon Denur for giving me Refs. [

The author declares no conflicts of interest regarding the publication of this paper.

Denur, J. (2019) Aerodynamic versus Ballistic Flight. Open Journal of Fluid Dynamics, 9, 346-400. https://doi.org/10.4236/ojfd.2019.94023

The Supplementary Notes in this Appendix provide auxiliary information concerning topics discussed in the text and/or in the cited references.

Supplementary Note 1: A succinct discussion of aerodynamic lift and drag is provided in Chap. Thirteen of Ref. [

Supplementary Note 2: An example of differences of viewpoint in explaining aerodynamic lift is provided in the exchange of ideas in Refs. [

Supplementary Note 3: Reference [

Supplementary Note 4: Information concerning the Eta glider is accessible in Ref. [

Supplementary Note 5: The most energy-efficient angle of attack for aerodynamic flight, which maximizes L / D , is always assumed in this present paper unless otherwise noted. Angle of attack is defined on pp. 114-115 and 139-140 of Ref. [

Note that, as per Section XVI.1 of Ref. [

Supplementary Note 6: The most energy-efficient angle of attack, which maximizes L / D and which we always assume in this present paper unless otherwise noted, is not the optimum angle of attack for all facets of flight. As per Supplementary Note 5, maximum endurance in engine-powered level flight and minimum sinking speed in gliding flight without engine power obtain at a larger angle of attack and less-than-maximum L / D . In this Supplementary Note 6 we consider landing. In order to land at the slowest possible speed, a still larger angle of attack than that corresponding to maximum endurance in engine-powered level flight and minimum sinking speed in gliding flight without engine power is optimal, namely the maximum practicable angle of attack, as near to a stall as is safe, which maximizes L for a given airspeed v at the expense of more-than-minimum sinking speed and even-lesser-than-maximum L / D . (The angle of attack should not be too near a stall, especially close to the ground, because there is not sufficient altitude to recover if an aircraft stalls too close to the ground.) See Ref. [

Supplementary Note 7: In Ref. [

Supplementary Note 8: Drag and lift are often most fundamentally expressed as functions of Reynolds number and Mach number. See, for example, Ref. [

Supplementary Note 9: Induced drag is a penalty that must be paid for lift. By Newton’s third law of motion, a wing’s action must be to deflect air downward in order to obtain the upward reaction of lift. The horizontal component of airflow encountered by an aircraft by virtue of its level (horizontal) aerodynamic flight is added vectorially to this downward component of airflow. Hence the direction of the lift force is not exactly vertical but instead is aligned at a (typically small) angle backward. This (typically small) backward component is the induced drag. According to the Prandtl-Munk formula, the minimum induced drag that a wing of span b subject to airflow at speed v must pay to generate lift L is D induced,min = 2 L 2 / π ρ v 2 b 2 . See Chaps. II-III (especially pp. 61-67) in Ref. [

D induced , min = 2 L 2 π ρ v 2 b 2 = 2 ( 1 2 C L A wing , geom ρ v 2 ) 2 π ρ v 2 b 2 = C L 2 A wing , geom 2 ( ρ v 2 ) 2 2 π ρ v 2 b 2 = C L 2 A wing , geom 2 ρ v 2 2 π b 2 = C L 2 A wing , geom ρ v 2 2 π × A wing , geom b 2 = C L 2 A wing , geom ρ v 2 2 π × b c b 2 = C L 2 A wing , geom ρ v 2 2 π × c b = C L 2 A wing , geom ρ v 2 2 π A R = C L A wing , eff ρ v 2 2 π A R = L C L π A R = m g C L π A R ,

where c is the average chord of an elliptically-shaped wing as is required for D induced = D induced , min [see Ref. [

In regards to reducing induced drag, we should mention wingtip devices. [See Refs. [

Supplementary Note 10: Skin-friction drag D skinfric results from shear in airflow adjacent to surfaces parallel to the airflow—surfaces of an aircraft parallel to its direction of motion. Air sticks to such surfaces owing to its viscosity, so immediately at such surfaces the air is at rest, or very nearly at rest, relative to the aircraft. Beyond a boundary layer of typically small thickness T boundary , airflow relative to the aircraft approaches the airspeed v of the aircraft—thus the shear v / T boundary . In short, skin-friction drag results from the viscosity of air rubbing against surfaces parallel to the airflow. According to simplified arguments, letting μ be the coefficient of viscosity of air and A | | be an aircraft’s surface area parallel to the airflow, there obtains in the simplest cases, as one might expect intuitively, the Stokes’-law-regime^{24} equation D skinfric = μ ( v / T boundary ) A | | = μ A | | v / T boundary (see for example pp. 75-78 of Ref. [

D skinfric = 0.036 A | | ρ v 2 ( R e ) − 1 / 5 = 0.036 A | | ρ v 2 ( ρ v l μ ) − 1 / 5 = 0.036 A | | ρ v 2 ( μ ρ v l ) 1 / 5 = 0.036 A | | ρ 4 / 5 v 9 / 5 ( μ l ) 1 / 5 .

A more exact result given a turbulent boundary layer is provided as per Equation (8) on p. 106 of Ref. [^{2}. As per Equations (5) and (6) on p. 106 of Ref. [^{3/2} rather than as v^{2}. And as per Equation (7) on p. 106 of Ref. [^{9/5} rather than as v^{2}. By contrast, given the paramount ρ v 2 functional dependency of pressure drag, and also—recall Supplementary Note 9—of induced drag, they are disposed to increase with increasing v as v^{2}. Hence skin-friction drag becomes less important relative to pressure drag and to induced drag with increasing v, or more fundamentally with increasing Reynolds number Re. The coefficient of skin-friction drag is typically small, a few times 10^{−3}. Hence skin-friction drag is often only a small fraction of the total drag D as given by Equation (7) of this present paper. See pp. 87-97 (especially pp. 93-94) of Ref. [

Supplementary Note 11: The transition from laminar to turbulent flow is discussed on pp. 78-79 of Ref. [

Supplementary Note 12: A sharp rise in C D as Mach 1 is approached from below followed by a sharp dip in C D at values of v slightly above Mach 1 obtains for airplane wings. See pp. 128-129 of Ref. [

Supplementary Note 13: Discussions and data pertinent to the U. S. Standard Atmosphere are provided in Chap. II (especially p. 19) of Ref. [

Supplementary Note 14: The average surface atmospheric pressure at low altitudes on Mars is ≈ 1/120 of that on Earth. But owing to both the higher average molecular weight of Mars’ largely-CO_{2} atmosphere and its lower average temperature, the average surface atmospheric density at low altitudes on Mars is ≈ 1/70 of that on Earth [

Supplementary Note 15: The hand-thrown distance records for discuses, Frisbees, Aerobies, and boomerangs are provided in the following articles (all most recently revised in 2019) at https://www.wikipedia.org: “Discus throw”, “Frisbee”, “Aerobie”, and “Boomerang”. Concerning the hand-thrown distance record for Frisbees, see especially http://www.wfdf.org/worldrecords (left-click on “Outdoor Distance” under “Distance”). The hand-thrown distance records for baseballs and golf balls are provided in the following articles at https://www.wikipedia.org (“Glen Gorbous” and “Roald Bradstock”, respectively, both most recently revised in 2019. The hand-thrown distance record for javelins is provided in the following articles (both most recently revised in 2019) at https://www.wikipedia.org: “Uwe Hohn” and “Javelin throw”. Sports-ball aerodynamics is discussed on pp. 31-34 and 40 of Ref. [

Supplementary Note 16: For symmetrical hand-thrown aircraft such as Frisbees, Aerobies, or discusses, it is the magnitude | v air | of v air that alone determines whether or not aerodynamic level flight can be sustained, irrespective of the sign of v air . (The symmetry must be at least bilateral with respect to the forward/backward direction.) If 0 ≤ v ground = v air , min < v wind , v air = v ground − v wind = v air , min − v wind < 0 , and | v air | = v wind − v ground = v wind − v air , min > 0 . Aerodynamic level flight of a symmetrical hand-thrown aircraft such as a Frisbee, Aerobie, or discuss can then be maintained if | v air | ≥ v air , min , which can obtain if v wind ≥ 2 v air , min . The analyses for unsymmetrical hand-thrown aircraft such as hand-thrown (e.g., paper or balsa) gliders if 0 ≤ v ground = v air , min < v wind ⇔ v air < 0 is more complex, and we do not attempt it. v air < 0 corresponds to an aircraft flying backwards. For a symmetrical aircraft (hand-thrown or otherwise) forward and backward flight are aerodynamically equivalent—but of course not so for unsymmetrical aircraft (hand-thrown or otherwise).

Supplementary Note 17: See the following articles (all most recently revised in 2019) at https://www.wikipedia.org: “Friction”, “Rolling resistance”, “Energy efficiency in transport”, “Magnetic levitation”, “Maglev”, and “Inductrack”. (While traditional railroad transportation is more energy-efficient at low to moderate speeds, maglev trains almost completely abolish wear on the tracks, owing to their lack of mechanical contact with the tracks: see the last three Wikipedia articles cited in this Supplementary Note 17.) See also the following articles at The Engineering Toolbox website https://www.engineeringtoolbox.com: “Friction and Friction Coefficients” (most recently revised in 2004) at https://www.engineeringtoolbox.com/friction-coefficients-d_778.html.and “Rolling Resistance” (most recently revised in 2008) at https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html.

Supplementary Note 18: Ground effect is discussed in the following articles: “Ground Effect” (most recently revised in 2017) at the SKYbrary website https://www.skybrary.aero/index.php/Main_Page; also, “Ground effect (aerodynamics)” and “Ground-effect vehicle” (both most recently revised in 2019) at https://www.wikipedia.org. Intuitively, one might attribute the ground effect for an aircraft near the ground as being due to compression of air near the ground, thus creating a lift-augmenting cushion of air under the aircraft that helps to support the aircraft. The wings, and possibly the aircraft as a whole, being inclined slightly upwards towards the direction of flight helps in this compression of air near the ground and thus in the creation of this air cushion. (The ground is usually a land surface, but could be a smooth water surface.) But this is only partially correct. Ground effect is a more complex, multifaceted, phenomenon. A perhaps less intuitively obvious contribution to the ground effect for aircraft arises because the proximity of the ground does not allow vortices at the wing tips to fully form, thus reducing induced drag. Thus ground effect results not only in increased lift but also in decreased drag. A given lift is produced at a smaller angle of attack with ground effect than without it. This is consistent with induced drag being smaller with ground effect than without it, because it corresponds to lift being directed at a smaller angle backwards (recall Supplementary Note 9). Thus at any given angle of attack ground effect results not only in increased lift but also in decreased drag. Hence a higher L / D ratio obtains at any given angle of attack with ground effect than without it owing to both increased lift and decreased drag. Also, the stalling angle of attack (recall Supplementary Note 6) is larger with ground effect than without it.