VTOL AIRCRAFT WITH A THRUST-TO-WEIGHT RATIO SMALLER THAN 0.1

This invention relates to aircraft, and more particularly, to aircraft with VTOL technology.

Currently, there are three kinds of traditional VTOL technologies which get vertical lift from high-speed air flowing over upper surface of the aircraft. In the first one such as patent No: US2012/0068020, a fan independent of a main engine increases speed of low-temp air flowing over the upper surface of wing along the direction of the chord, thereby increasing vertical lift. In this way, this fan less efficiently increases the vertical lift; in the second one such as patent No: GB792993, a high-temp bypass duct directs high-temp air by a valve from a nozzle of a jet engine to flow radially over the upper surface of the aircraft, thereby generating vertical life. In this way, the high-temp bypass duct generates the vertical lift more efficiently than the first one but the upper surface of the aircraft might be burnt up; in the third one such as patent No: GB2469612, a low-temp bypass duct directs low-temp air, with a rotatable nozzle, from a low-temp duct of a turbofan engine to flow over the upper surface of the wing along the direction of the chord, thereby generating a vertical lift. But another rotatable nozzle spouts high-temp air downwardly from a high-temp duct of a turbofan engine, thereby generating another vertical lift less efficiently. In this way, the low-temp bypass duct generates the vertical lift more efficiently than above ways but thin slot outlet of the low-temp bypass duct, set in the direction of the chord, enables a part of vertical lift to lose in the long low-temp bypass duct and the problem, how the high-temp air is directed to obtain more vertical lift from the upper surface of the aircraft, is not solved. Because of these limitations, VTOL is impossible for aircraft with thrust-to-weight ratio smaller than 0.1, and more particularly for large aircraft.

The present invention is a VTOL aircraft according to claim 1.

The dependent claims define further technical features of said aircraft.

The attached drawings illustrate the invention:

• Fig. 1 is a side view of an aircraft with this invention during vertical take-off and landing.
• Fig. 2 is a top view of an aircraft with this invention during vertical take-off and landing.
• Fig. 3 is a rear view of an aircraft with this invention during vertical take-off and landing.
• Fig. 4 is J local view for Fig. 3.
• Fig. 5 is A-A Section for Fig. 4
• Fig. 6 is B-B Section for Fig. 4.
• Fig. 7 is part Section for Fig. 4.
• Fig. 8 is K local view for Fig. 3.
• Fig. 9 is C-C Section for Fig. 8.
• Fig. 10 is a side view of an aircraft with this invention during forward flight.
• Fig. 11 is a top view of an aircraft with this invention during forward flight.
• Fig. 12 is a rear view of an aircraft with this invention during forward flight.
• Fig. 13 is M local view for Fig. 12.
• Fig. 14 is D-D Section for Fig. 13.
• Fig. 15 is E-E Section for Fig. 13.
• Fig. 16 is part Section for Fig. 13.
• Fig. 17 is N local view for Fig. 12.
• Fig. 18 is F-F Section for Fig. 17.

Referring to above drawings, an aircraft comprises: ailerons (1, 2); a turbofan engine (3) including a low-temp duct (6) with an openable and closable propelling nozzle (7) and a high-temp duct (4) with an openable and closable propelling nozzle (5); a low-temp bypass duct (15) including an openable and closable inlet (14) connected to the low-temp duct 6), an outer wall (16), an inner wall (17) and a rectangular outlet (19) with a height-to-width ratio smaller than 0.1 set on the upper surface of the wing and in the direction of the wingspan; a high-temp bypass duct (9) including an openable and closable inlet (8) connected to the high-temp duct (4), an outer wall (10) and a rectangular outlet (12) with a height-to-width ratio smaller than 0.1 set above the rectangular outlet (19) of the low-temp bypass duct (15) with a height-to-width ratio smaller than 0.1 and in the direction of the wingspan.

Referring to Fig.(1-9), during vertical take-off/landing, the propelling nozzles (5, 7) of the low-temp duct (6) and the high-temp duct (4) of the turbofan engine (3) are closed. At same time, the inlets (8, 14) of the low-temp bypass duct (15) and the high-temp bypass duct (9) are opened. Then, the turbofan engine (3) starts to supply the low-temp air (18) to the low-temp duct (6) and the high-temp air (11) to the high-temp duct (4). Now, the low-temp bypass duct (15) directs the low-temp air (18) from the low-temp duct (6) and enables it, in form of low-temp planar jet (20), to flow over the upper surface of the wing along the direction of wingspan. Because the velocity of the air flowing over the upper surface of the wing is higher than that over the lower surface of the wing, and according to Principle of Bernoulli stating an increase in the speed occurs simultaneously with a decrease in pressure, a vertical lift is generated on the wing due to the lower pressure on the upper surface of the wing than that on the lower surface of the wing and the low-temp planar jet (20) enables the ailerons to control horizontal and vertical balances of the aircraft. Similarly, the high-temp bypass duct (9) directs the high-temp air (11) from the high-temp duct (4) and enables it, in form of the high-temp planar jet (13), to flow above the low-temp planar jet (20) along the direction of wingspan, thereby generating another vertical lift and enabling the ailerons to control horizontal and vertical balances of the aircraft more efficiently. The rectangular outlet (12) of the high-temp bypass duct (9) with a height-to-width ratio smaller than 0.1 set above the rectangular outlet (19) of the low-temp bypass duct (15) with a height-to-width ratio smaller than 0.1.

Referring to Fig.(10-18), during forward flight, the propelling nozzles (5, 7) of the low-temp duct (6) and high-temp duct (4) of the turbofan engine (3) are opened. At same time, the inlets (8, 14) of the low-temp bypass duct (15) and the high-temp bypass duct (9) are closed. Then, the low-temp air (18) and the high-temp air (11) spout directly from the propelling nozzles (5, 7) of the low-temp duct (6) and the high-temp duct (4) into atmosphere to generate thrusts

This invention can be used to retrofit an existing aircraft to achieve VTOL or manufacture a VTOL aircraft with a thrust-to-weight ratio smaller than 0.1.

P

Atmospheric pressure

(Unit: Pa)

R

Ideal gas constant

(Unit:J.K^-1.mol^-1)

ρ₀

Low-temp jet density at the outlet

(Unit:kg/m³)

Pi

Atmospheric density

(Unit:kg/m³)

P

Jet density on the cross-section

(Unit: kg/m³)

ρ_m

Jet density of the shaft

(Unit: kg/m³)

T₀

Jet temp at the low-temp outlet

(Unit: K)

T₁

Atmospheric temp

(Unit: K)

T

Jet temp on the cross-section

(Unit: K)

V₀

Jet speed at the outlet of low-temp bypass duct

(Unit: m/s)

$V_0^{\prime }$

Jet speed at the outlet of high-temp bypass duct

(Unit: m/s)

V

Jet speed on the cross-section

(Unit: m/s)

V_m

Jet speed of shaft

(Unit: m/s)

h₀

Jet height of the outlet

(Unit: m)

h_m

Jet height of the cross-section

(Unit: m)

h

Jet height on the cross-section

(Unit: m)

L

Width of the outlet

(Unit: m)

C

Gas specific volume

(Unit: m³/kg)

M

Molar mass

(Unit: kg/mol)

Q

Air inflow of turbofan engine

(Unit: kg/s)

B

Bypass ratio of turbofan engine

(Dimensionless unit)

G

Maximum take-off weight

(Unit: T)

X

Distance between any point and outlet of low-temp bypass duct in jet direction

(Unit: m)

X₁

Distance between wing root and low-temp outlet in jet direction

(Unit: m)

X₂

Distance between low-temp outlet and cross over point at which jet boundary intersects with trailing edge of wing in jet direction

(Unit: m)

X₃

Distance between wingtip and low-temp outlet

in jet direction

(Unit: m)

α

Angle between jet boundary and trailing edge of wing

(Unit: °)

β

Angle between chord line of wing and long side of outlet

(Unit: °)

γ

Dihedral angle of wing

(Unit: °)

δ

Sweep forward angle of jet

(Unit: °)

Ø

Included angle between axis of low-temp jet and one of the high-temp jet

(Unit: °)

θ

Sweep back angle of jet

(Unit: °)

F

Total vertical lift

(Unit: T)

F₁

Lift between wing root and low-temp outlet

(Unit: T)

F₂

Lift between low-temp outlet and cross over point at which jet boundary intersects with trailing edge of wing

(Unit: T)

F₃

Lift between wingtip and low-temp outlet

(Unit: T)

F₄

Downward pressure generated by slant upward deflected jet

(Unit: T)

F₅

Thrust generated by jet of low-temp outlet

(Unit: T)

F₆

Thrust generated by jet of high-temp outlet

(Unit: T)

b

Constant

(Dimensionless unit)

d

Constant

(Dimensionless unit)

C

Constant

(Dimensionless unit)

n₁

S/N of turbofan engine

(Dimensionless unit)

n₂

S/N of turbofan engine used in vertical Take-off/landing

(Dimensionless unit)

n₃

S/N of outlet of low-temp bypass duct

(Dimensionless unit)

TWR

Thrust-to-weight ratio of aircraft

(Dimensionless unit)

According to thermodynamic, relative values of the enthalpy of various jet cross-sections are same each other when pressures of these jet cross-sections are equal and the value of surrounding gas is starting value.

And according to the feature of planar jet, it just spread on the flat which is perpendicular to the outlet section. $$P=\frac{{\rho }_0}{M}R{T}_0=\frac{{\rho }_1}{M}R{T}_1=\frac{\rho }{M}\mathit{RT}$$ $${\rho }_0{V}_0{h}_0\mathit{LC}\left(T_0-T_1\right)={\displaystyle \int \rho VL\mathrm{}C\left(T-T_1\right)}\mathit{dh}$$ $${\rho }_0{V}_0{h}_0\left(\frac{1}{{\rho }_0}-\frac{1}{{\rho }_1}\right)={\displaystyle \int \mathit{\rho V}}\left(\frac{1}{\rho }-\frac{1}{{\rho }_1}\right)\mathit{dh}$$

And according to the similarity of velocity and density distribution on the various jet sections, $$\begin{array}{l}\sqrt{\frac{V}{V_m}}=\frac{\rho -{\rho }_1}{{\rho }_m-{\rho }_1}=1-1-{\left(\frac{h}{h_m}\right)}^{1.5}\\ h_0{V}_0\left({\rho }_0-{\rho }_1\right)={\displaystyle \int \left(\rho -{\rho }_1\right)}\mathit{Vdh}\\ \frac{h_0{V}_0\left({\rho }_0-{\rho }_1\right)}{\left({\rho }_m-{\rho }_1\right)V_m{h}_m}={{\displaystyle \int \frac{\left(\rho -{\rho }_1\right)V}{\left({\rho }_m-{\rho }_1\right)V_m{h}_m}\mathit{dh}={\displaystyle \underset{0}{\overset{h_m}{\int }}\left[1-{\left(\frac{h}{h_m}\right)}^{1.5}\right]}}}^3.\frac{1}{h_m}\mathit{dh}\\ ={{\displaystyle \underset{0}{\overset{1}{\int }}\left[1-{\left(\frac{h}{h_m}\right)}^{1.5}\right]}}^{3}d\left(\frac{h}{h_m}\right)=0.3682\\ \left({\rho }_m-{\rho }_1\right)=\frac{h_0{V}_0\left({\rho }_0-{\rho }_1\right)}{0.3682\cdot V_m{h}_m}\end{array}$$

According to dynamic characteristic of jet, momentums of the various sections, are same each other in case of equal pressures. $$\begin{array}{l}{\rho }_{0}L{h}_0{V}_0^2={\displaystyle \int \mathit{\rho L}{V}^2\mathit{dh}}\\ {\rho }_0{h}_0{V}_0^2={\displaystyle \int \rho V^2\mathit{dh}}={\displaystyle \int \left(\rho -{\rho }_1\right)V^2\mathit{dh}+{\displaystyle \int {\rho }_1{V}^2\mathit{dh}}}\\ \frac{{\rho }_0{h}_0{V}_0^2}{\left({\rho }_m-{\rho }_1\right)V_m^2{h}_m}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(\rho -{\rho }_1\right)V^2}{\left({\rho }_m-{\rho }_1\right)V_m^2}d\left(\frac{h}{h_m}\right)+{\displaystyle {\int }_0^1\frac{{\rho }_1{V}^2}{\left({\rho }_m-{\rho }_1\right)V_m^2}}}d\left(\frac{h}{h_m}\right)\\ ={{\displaystyle \underset{0}{\overset{1}{\int }}\left[1-{\left(\frac{h}{h_m}\right)}^{1.5}\right]}}^{5}d\left(\frac{h}{h_m}\right)+\frac{{\rho }_1}{\left({\rho }_m-{\rho }_1\right)}{{\displaystyle {\int }_0^1\left[1-{\left(\frac{h}{h_m}\right)}^{1.5}\right]}}^{4}d\left(\frac{h}{h_m}\right)\\ =0.2786+\frac{0.3156{\rho }_1}{\left({\rho }_m-{\rho }_1\right)}\\ \frac{0.3682V_0{\rho }_0}{V_m\left({\rho }_0-{\rho }_1\right)}=0.2786+\frac{0.1162h_m{V}_m{\rho }_1}{h_0{V}_0\left({\rho }_0-{\rho }_1\right)}\end{array}$$ Substitute (1) into (2): $$\begin{array}{l}\frac{0.1162h_m{\rho }_1}{h_0{\rho }_0}\cdot {\left(\frac{V_m}{V_0}\right)}^2+0.2786\frac{\left({\rho }_0-{\rho }_1\right)}{{\rho }_0}\cdot \left(\frac{V_m}{V_0}\right)-0.3682=0\\ \frac{V_m}{V_0}=\frac{\frac{0.2786\left({\rho }_1-{\rho }_0\right)}{{\rho }_0}+\sqrt{{\left[\frac{0.2786\left({\rho }_0-{\rho }_1\right)}{{\rho }_0}\right]}^{{}^2}+\frac{0.1711h_m{\rho }_1}{h_0{\rho }_0}}}{\frac{0.2324h_m{\rho }_1}{h_0{\rho }_0}}\\ V_m=\frac{1+\sqrt{1+\frac{0.1711h_m{\rho }_0{\rho }_1}{{0.2786}^2{h}_0{\left({\rho }_0-{\rho }_1\right)}^2}}}{\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_0-{\rho }_1\right)}}{V}_0\end{array}$$ Substitute (3) into (1) $$\begin{array}{l}\left({\rho }_m-{\rho }_1\right)=\frac{h_0{V}_0\left({\rho }_0-{\rho }_1\right)}{0.3682\cdot V_m{h}_m}\\ {\rho }_m={\rho }_1-\frac{\frac{0.2324}{0.3682\times 0.2786}{\rho }_1}{1+\sqrt{1+\frac{0.1711h_m{\rho }_0{\rho }_1}{{0.2786}^2{h}_0{\left({\rho }_0-{\rho }_1\right)}^2}}}\end{array}$$ When X ≤ X₁ or X ≤ X₂ $$F=\left[P-\left(P-\frac{1}{2}{\displaystyle \int {\rho }_m{V}_m^2}\right)\right]\mathit{Ldx}=\frac{1}{2}{\displaystyle \int {\rho }_m{V}_m^2}\mathit{Ldx}$$ Substitute (3), (4) into (2) $$\begin{array}{l}=\frac{1}{2}{\displaystyle \underset{0}{\overset{X}{\int }}\left[{\rho }_1-\frac{\frac{0.2324}{0.3682\times 0.2786}{\rho }_1}{1+\sqrt{1+\frac{0.1711h_m{\rho }_0{\rho }_1}{{0.2786}^2{h}_0{\left({\rho }_0-{\rho }_1\right)}^2}}}\right]}\cdot \left[\frac{1+\sqrt{1+\frac{0.1711h_m{\rho }_0{\rho }_1}{{0.2786}^2{h}_0{\left({\rho }_0-{\rho }_1\right)}^2}}}{\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_0-{\rho }_1\right)}}\right]\cdot V_0^2\mathit{Ldx}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{X}{\int }}\{{\rho }_1{\left[\frac{1+\sqrt{1+\frac{0.1711h_m{\rho }_0{\rho }_1}{{0.2786}^2{h}_0{\left({\rho }_0-{\rho }_1\right)}^2}}}{\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_0-{\rho }_1\right)}}\right]}^2}\\ -\frac{0.2324}{0.3682\times 0.2786}{\rho }_1\cdot \left[\frac{1+\sqrt{1+\frac{0.1711h_m{\rho }_1{\rho }_0}{{0.2786}^2{h}_0{\left({\rho }_1-{\rho }_0\right)}^2}}}{\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_1-{\rho }_0\right)}}\right]\}\cdot V_0^2\mathit{Ldx}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{X}{\int }}{\rho }_1\{\frac{2+\frac{0.1711h_m{\rho }_1{\rho }_0}{{0.2786}^2{h}_0{\left({\rho }_1-{\rho }_0\right)}^2}+2\sqrt{1+\frac{0.1711h_m{\rho }_1{\rho }_0}{{0.2786}^2{h}_0{\left({\rho }_1-{\rho }_0\right)}^2}}}{{\left[\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_1-{\rho }_0\right)}\right]}^2}}\\ -\frac{\frac{0.2324}{0.3682\times 0.2786}+\frac{0.2324}{0.3682\times 0.2786}\sqrt{1+\frac{0.1711h_m{\rho }_1{\rho }_0}{{0.2786}^2{h}_0{\left({\rho }_1-{\rho }_0\right)}^2}}}{{\left[\frac{0.2324h_m{\rho }_1}{0.2786h_0\left({\rho }_1-{\rho }_0\right)}\right]}^2}\}\cdot V_0^2\mathit{Ldx}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{X}{\int }}\left\{\frac{0.1711{\rho }_0}{{0.2324}^2\left(\frac{h_m}{h_0}\right)}-\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2{\left(\frac{h_m}{h_0}\right)}^2}-\frac{0.2655\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot \left(\frac{h_m}{h_0}\right)}}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\cdot \left(\frac{h_m}{h_0}\right)\right]}^2}\right\}}\\ \cdot V_0^2\mathit{Ldx}\end{array}$$ Because of $$\frac{h_m}{h_0}=2.44\left(\frac{0.12x}{h_0}+0.41\right)$$ When X ≤ X₂ or X ≤ X₃ $$\begin{array}{l}{F}_1=\frac{1}{2}{\displaystyle \underset{1.0004}{\overset{2.44\left(\frac{0.12X_1}{h_0}+0.41\right)}{\int }}\{\frac{0.1711{\rho }_0}{{0.2324}^2\left(\frac{h_m}{h_0}\right)}-\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2{\left(\frac{h_m}{h_0}\right)}^2}}\\ -\frac{0.2655\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot \left(\frac{h_m}{h_0}\right)}}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\cdot \left(\frac{h_m}{h_0}\right)\right]}^2}\}\cdot \frac{h_0{V}_0^{2}L}{2.44\times 0.12}d\left(\frac{h_m}{h_0}\right)\\ =\frac{1}{2}{\displaystyle \underset{1.0004}{\overset{2.44\left(\frac{0.12X_1}{h_0}+0.41\right)}{\int }}\{\frac{0.1711{\rho }_0}{{0.2324}^{2}x}-\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2{x}^2}}\\ -\frac{0.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2}\cdot \frac{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}}{\cdot x^2}\}\cdot \frac{h_0{V}_0^{2}L}{2.44\times 0.12}\mathit{dx}\\ {\displaystyle \int \frac{\sqrt{1+\mathit{bx}}}{x^2}}\mathit{dx}=-\left(\frac{\sqrt{1+\mathit{bx}}}{x}-\frac{b}{2}\mathit{ln}\frac{\sqrt{1+\mathit{bx}}-1}{\sqrt{1+\mathit{bx}}+1}+C\right)\\ F_1=\frac{0.1711{\rho }_0}{{0.2324}^2}\ln x+\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^{2}x}+\frac{0.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2}\cdot [\frac{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}}{x}\\ {-\frac{0.1711{\rho }_1{\rho }_0}{2\times {0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\ln \frac{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}-1}{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}+1}\}}_{2.44\times 0.41}^{2.44\left(\frac{0.12X_1}{h_0}+0.41\right)}\\ \times \frac{1}{2}\times \frac{h_0{V}_0^{2}L}{2.44\times 0.12}\end{array}$$ $$\begin{array}{l}{F}_2=\frac{0.1711{\rho }_0}{{0.2324}^2}ln\mathrm{}x+\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^{2}x}+\frac{0.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2}\cdot [\frac{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}}{x}\\ {-\frac{0.1711{\rho }_1{\rho }_0}{2\times {0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\mathit{ln}\frac{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}-1}{\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot x}+1}\}}_{2.44\times 0.41}^{2.44\left(\frac{0.12X_2}{h_0}+0.41\right)}\\ \times \frac{1}{2}\times \frac{h_0{V}_0^{2}L}{2.44\times 0.12}\end{array}$$ When X₂ ≤ X ≤ X₃ $$\begin{array}{l}{F}_3=\frac{1}{2}{\displaystyle \underset{X_2}{\overset{X_3}{\int }}\{\frac{0.1711{\rho }_0}{{0.2324}^2\left(\frac{h_m}{h_0}\right)}--\frac{1.2655}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\right]}^2{\left(\frac{h_m}{h_0}\right)}^2}}\\ -\frac{0.2655\sqrt{1+\frac{0.1711{\rho }_1{\rho }_0}{{0.2786}^2{\left({\rho }_1-{\rho }_0\right)}^2}\cdot \left(\frac{h_m}{h_0}\right)}}{{\left[\frac{0.2324{\rho }_1}{0.2786\left({\rho }_1-{\rho }_0\right)}\cdot \left(\frac{h_m}{h_0}\right)\right]}^2}\}\times V_0^2\left[L-\left(x-X_2\right)\times tan\mathrm{}\alpha \right]\mathit{dx}\end{array}$$ When force is zero in body axis direction, $$\begin{array}{l}{F}_4=F_5+F_6\\ \left(F_4+F_5+F_6\right)\sin \delta =\left(F_5+F_6\right)\sin \theta \\ \delta ={\sin }^{-1}\frac{1}{2}\sin \theta \end{array}$$ $$F=n_2\left(F_1+F_2+F_3\right)\cos \beta \cos \gamma -n_2\left[F_4\cos \delta +F_6\sin \left(Ø+\gamma \right)-F_6\sin \left(Ø+\gamma \right)\right]$$ VTOL can be achieved once: F > G

ANNEX 2

• n₁ = 1-4
• n₂ = 1∼2
• n₃ = 1∼4
• M = 29 kg/mol
• Q(Air inflow of Trent 900 turbofan engine) = 1204 kg/s
• B(Bypass ratio of Trent 900turbofan engine) = 8.5:1
• G(Maximum take - off weight of A380) = 560 T
• T₀ = 383K
• ρ₁ = 1.293kg/m²

Assuming:

• θ = 36.5° $$F_4=\frac{1}{n_2}Q{V}_0=\frac{1}{2}\times 1204\times 211=127022N=13\mathrm{}T$$ $${\rho }_0=\frac{M{P}_0}{R{T}_0}=\frac{101325\times 29\times 0.001}{8.31\times 383}=0.9232\mathrm{}\mathit{kg}/m^3$$
• h₀ = 0.46m
• L = 6m $$V_0=\frac{\frac{1}{n_2}\times Q\times \frac{8.5}{8.5+1}}{{\rho }_0\times L\times h_0}=\frac{\frac{1}{2}\times 1204\times \frac{8.5}{8.5+1}}{0.9232\times 6\times 0.46}=211m/s$$
• $V_0^{\prime }=V_0=211m/s$ $$F_6=\frac{1}{n_2}Q\times \frac{1}{8.5+1}\times V_0^{\prime }=11370.7368\mathrm{}N=1.4\mathrm{}T$$
• X₁ = 11 m
• X₂ = 18 m
• X₃ = 32 m
• α = 14°
• β = 5°
• γ = 5°
• θ = 36.5°
• Ø = 2 × 16.3° $$\begin{array}{l}{F}_1=209832.582[2.9246\mathrm{}ln\mathrm{}x+\frac{0.1488}{x}+0.0312\cdot [\frac{\sqrt{1+19.2419x}}{x}\\ {-9.621\mathrm{}\mathit{ln}\frac{\sqrt{1+19.2419x}-1}{\sqrt{1+19.2419x}+1}]}_{2.44\times 0.41}^{2.44\left(\frac{0.12X_1}{h_0}+0.41\right)}\\ =209832.582\{2.9246\mathrm{}\mathit{ln}\frac{2.44\left(\frac{0.24\times 11}{0.32}+0.41\right)}{2.44\times 0.46}\\ +0.1488\left[\frac{1}{2.44\left(\frac{0.12\times 11}{0.46}+0.41\right)}-\frac{1}{2.44\times 041}\right]\\ +0.0312\cdot [\frac{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 11}{0.32}+0.41\right)}}{2.44\left(\frac{0.12\times 11}{0.46}+0.41\right)}-\frac{\sqrt{1+19.2419\times 2.44\times 0.41}}{2.44\times 0.41}\\ -9.621\mathrm{}\mathit{ln}\frac{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 11}{0.46}+0.41\right)}-1}{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 11}{0.46}+0.41\right)}+1}\\ +9.621\mathrm{}\mathit{ln}\frac{\sqrt{1+19.2419\times 2.44\times 0.41}-1}{\sqrt{1+19.2419\times 2.44\times 0.41}+1}]\}\\ =209832.582\left[6.0811-0.1301-0.0918-0.0873\right]\\ =1211132.68N\\ =124T\end{array}$$ $$\begin{array}{l}{F}_2=209832.582\{2.9246\mathrm{}\mathit{ln}\frac{2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}{2.44\times 0.41}\\ +0.1488\left[\frac{1}{2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}-\frac{1}{2.44\times 0.41}\right]\\ +0.0312\cdot [\frac{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}}{2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}\\ -\frac{\sqrt{1+19.2419\times 2.44\times 0.41}}{2.44\times 0.41}]-9.621\mathrm{}\mathit{ln}\frac{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}-1}{\sqrt{1+19.2419\times 2.44\left(\frac{0.12\times 18}{0.46}+0.41\right)}+1}\\ +9.621\mathrm{}\mathit{ln}\frac{\sqrt{1+19.2419\times 2.44\times 0.41}-1}{\sqrt{1+19.2419\times 2.44\times 0.41}+1}]\}\\ =209832.582\left[7.3757-0.1368-0.1015-0.0969\right]\\ =1477326.2936N\\ =151T\end{array}$$

  $$\begin{array}{l}\delta ={\sin }^{-1}\left(\frac{1}{2}\mathrm{}sin\mathrm{}\theta \right)\\ =\mathit{si}{n}^{-1}\left(\frac{1}{2}\mathrm{}si\mathrm{}n\mathrm{}36.5°\right)\\ =17.3°\end{array}$$

According to this sweep forward angle, low-temp planar jet enables the aircraft to keep balances during vertical take-off/landing. $$\begin{array}{l}F=n_2\left(F_1+F_2+F_3\right)\mathrm{}cos\mathrm{}\beta \mathrm{}cos\mathrm{}\gamma -n_2\left[F_4\mathrm{}cos\mathrm{}\delta +F_6\mathrm{}\mathit{sin}\left(Ø+\gamma \right)-F_6\mathrm{}\mathit{sin}\left(Ø+\gamma \right)\right]\\ =2\left(124+151+24\right)\mathrm{}cos\mathrm{}5°\mathrm{}cos\mathrm{}5°-2[13\mathrm{}cos\mathrm{}17.3°+1.4\mathrm{}\mathit{sin}\left(2\times 16.3°+5°\right)\\ -1.4\mathrm{}\mathit{sin}\left(2\times 16.3°-5°\right)\\ =568\mathrm{}T\end{array}$$ $$\begin{array}{l}F-G=568-560=8\mathrm{}T>0\\ \mathit{TWR}=\frac{n_3\left(F_5+F_6\right)}{G}=\frac{n_3{F}_4}{G}=\frac{4\times 13}{560}=0.09<0.1\end{array}$$

It is clearly demonstrated above that VTOL is achievable on Airbus A380 once remodeled as shown, and more particularly, in case of using just two turbofan engines and thrust-to-weight ratio smaller than 0.1.