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• simple example – rocket engine: not too typical for aircraft, but simple one. Rocket consumes same amount of fuel per second in order to generate a unit of thurst regardless of its size and regardless if it is pointing upward (and is static with respect to air) or forward (and moving through the air). You need to burn proportionally more fuel per second to generate higher thrust. So, for the rocket propulsion, you will save fuel in the same ratio as the necessary thrust decreases.

• Propeller or jet engines are more complicate as their thrust and fuel consumption depends on engine movement through the air too. As David K pointed out in his answer, we can use momentum and kinetic energy of accelerated air to get power needed for unit of thrust.

  With some simplifications, thrust is mass flow rate through the engine/prop multiplied by change of flow speed it causes. $T = \dot m \cdot (v_{\rm out} - v_{\rm in}) = \dot m \Delta v$. Power needed for this is $P = \dot m \cdot {1\over2}(v_{\rm out}^2 - v_{\rm in}^2)=\dot m\Delta v\cdot(v_{\rm in} + {\Delta v\over 2})$. Thus $$ {P\over T}=v_{\rm in} + {\Delta v\over 2}\,. $$

  For stationary engine holding against gravity, higher thrust compared to flying fixed wing aircraft is needed as shown above. If we do not "cheat" by increasing mass flow rate through engine (like making it helicopter rotor or using multiple engines), $\Delta v$ has to be increased in order to achieve the necessary thrust. So you not only need more power because of increased thrust but also more power because of increased Watts per unit of thrust. Note that even "helicopter cheat" does not work too well. To match power consumption of engine generating less thrust thanks to the wing's L/D you need to improve P/T too – by decreasing $\Delta v$, thus increasing mass flow (rotor/propeller radius even more than proportionally to increased thrust).

  What about decrease of P/T due to the movement through air? Well, it depends on particular engine and its $\Delta v$. It will be typically in similar order of magnitude as airspeed (or even less), so we can not neglect $v_{\rm in}$ in watts-per-thrust equation above. There is a efficiency penalty when engine works on moving aircraft. But it should be still worth of it as the gain provided by lift is greater.

  A simplified example: We have an engine capable of producing enough thrust to lift aircraft vertically. It can be throttled by changing $\Delta v$ without any practical problems or change in its internal efficiency. And let's assume that mass flow rate through it is fixed area $S$ multiplied by air density and multiplied by arithmetical average of speeds of entering and exiting air. For hovering aircraft and stationary engine producing thrust equal to aircraft's weight, $w$ it is $$w=\dot m \Delta v_{\rm hover} = \rho S \Delta v_{\rm hover}^2 / 2\,;\quad \Delta v_{\rm hover}=\sqrt{2 w \over \rho S}$$ and thus $$P_{\rm hover}=w\cdot \Delta v_{\rm hover}/2=\sqrt{w^3\over 2\rho S}\,.$$

  The same aircraft flying on its wings does only need $w\over L/D$ of thrust. Airspeed is $v_{\rm air}$. Equation for the thrust: ${w\over L/D} = \dot m \Delta v_{\rm flight} = \rho S \cdot (v_{\rm air}+{\Delta v_{\rm flight} \over 2}) \cdot \Delta v_{\rm flight}$. Thus $$\Delta v_{\rm flight}=\sqrt{{2w\over(L/D)\rho S}+v_{\rm air}^2}-v_{\rm air}$$ and $$ P_{\rm flight}={w\over L/D}\cdot(\sqrt{{w\over 2(L/D)\rho S}+{v_{\rm air}^2\over 4}}+{v_{\rm air}\over 2})\,. $$

  Unfortunately, I do not see any way how to simplify and compare $P_{\rm hover}$ and $P_{\rm flight}$ so some concrete numbers:

  • Light aircraft, 1 ton, 100 knot, $S = 5\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 290\,\rm kW$, $P_{\rm flight} = 35\,\rm kW$.

  • Heavy aircraft, 100 ton, 200 knot, $S = 50\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 90\,\rm MW$, $P_{\rm flight} = 7\,\rm MW$.

  So, based on these simplifications flying with wings with similar kind of engine should be significantly more efficient in energy terms too. And, additionally, you are already moving forward using power $P_{\rm flight}$. For vertical engine extra power would be necessary for overcoming air drag due to the movement.

• simple example – rocket engine: not too typical for aircraft, but simple one. Rocket consumes same amount of fuel per second in order to generate a unit of thurst regardless of its size and regardless if it is pointing upward (and is static with respect to air) or forward (and moving through the air). You need to burn proportionally more fuel per second to generate higher thrust. So, for the rocket propulsion, you will save fuel in the same ratio as the necessary thrust decreases.

• Propeller or jet engines are more complicate as their thrust and fuel consumption depends on engine movement through the air too. As David K pointed out in his answer, we can use momentum and kinetic energy of accelerated air to get power needed for unit of thrust.

  With some simplifications, thrust is mass flow rate through the engine/prop multiplied by change of flow speed it causes. $T = \dot m \cdot (v_{\rm out} - v_{\rm in}) = \dot m \Delta v$. Power needed for this is $P = \dot m \cdot {1\over2}(v_{\rm out}^2 - v_{\rm in}^2)=\dot m\Delta v\cdot(v_{\rm in} + {\Delta v\over 2})$. Thus $$ {P\over T}=v_{\rm in} + {\Delta v\over 2}\,. $$

  For stationary engine holding against gravity, higher thrust compared to flying fixed wing aircraft is needed as shown above. If we do not "cheat" by increasing mass flow rate through engine (like making it helicopter rotor or using multiple engines), $\Delta v$ has to be increased in order to achieve the necessary thrust. So you not only need more power because of increased thrust but also more power because of increased Watts per unit of thrust. Note that even "helicopter cheat" does not work too well. To match power consumption of engine generating less thrust thanks to the wing's L/D you need to improve P/T too – by decreasing $\Delta v$, thus increasing mass flow (rotor/propeller radius even more than proportionally to increased thrust).

  What about decrease of P/T due to the movement through air? Well, it depends on particular engine and its $\Delta v$. It will be typically in similar order of magnitude as airspeed (or even less), so we can not neglect $v_{\rm in}$ in watts-per-thrust equation above. There is a efficiency penalty when engine works on moving aircraft. But it should be still worth of it as the gain provided by lift is greater.

  A simplified example: We have an engine capable of producing enough thrust to lift aircraft vertically. It can be throttled by changing $\Delta v$ without any practical problems or change in its internal efficiency. And let's assume that mass flow rate through it is fixed area $S$ multiplied by air density and multiplied by arithmetical average of speeds of entering and exiting air. For hovering aircraft and stationary engine producing thrust equal to aircraft's weight, $w$ it is $$w=\dot m \Delta v_{\rm hover} = \rho S \Delta v_{\rm hover}^2 / 2\,;\quad \Delta v_{\rm hover}=\sqrt{2 w \over \rho S}$$ and thus $$P_{\rm hover}=w\cdot \Delta v_{\rm hover}/2=\sqrt{w^3\over 2\rho S}\,.$$

  The same aircraft flying on its wings does only need $w\over L/D$ of thrust. Airspeed is $v_{\rm air}$. Equation for the thrust: ${w\over L/D} = \dot m \Delta v_{\rm flight} = \rho S \cdot (v_{\rm air}+{\Delta v_{\rm flight} \over 2}) \cdot \Delta v_{\rm flight}$. Thus $$\Delta v_{\rm flight}=\sqrt{{2w\over(L/D)\rho S}+v_{\rm air}^2}-v_{\rm air}$$ and $$ P_{\rm flight}={w\over L/D}\cdot(\sqrt{{w\over 2(L/D)\rho S}+{v_{\rm air}^2\over 4}}+{v_{\rm air}\over 2})\,. $$

  Unfortunately, I do not see any way how to simplify and compare $P_{\rm hover}$ and $P_{\rm flight}$ so some concrete numbers:

  • Light aircraft, 1 ton, 100 knots, $S = 5\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 290\,\rm kW$, $P_{\rm flight} = 35\,\rm kW$.

  • Heavy aircraft, 100 ton, 200 knots, $S = 50\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 90\,\rm MW$, $P_{\rm flight} = 7\,\rm MW$.

  So, based on these simplifications flying with wings with similar kind of engine should be significantly more efficient in energy terms too. And, additionally, you are already moving forward using power $P_{\rm flight}$. For vertical engine extra power would be necessary for overcoming air drag due to the movement.

• simple example – rocket engine: not too typical for aircraft, but simple one. Rocket consumes same amount of fuel per second in order to generate a unit of thurst regardless of its size and regardless if it is pointing upward (and is static with respect to air) or forward (and moving through the air). You need to burn proportionally more fuel per second to generate higher thrust. So, for the rocket propulsion, you will save fuel in the same ratio as the necessary thrust decreases.

• Propeller or jet engines are more complicate as their thrust and fuel consumption depends on engine movement through the air too. As David K pointed out in his answer, we can use momentum and kinetic energy of accelerated air to get power needed for unit of thrust.

  With some simplifications, thrust is mass flow rate through the engine/prop multiplied by change of flow speed it causes. $T = \dot m \cdot (v_{\rm out} - v_{\rm in}) = \dot m \Delta v$. Power needed for this is $P = \dot m \cdot {1\over2}(v_{\rm out}^2 - v_{\rm in}^2)=\dot m\Delta v\cdot(v_{\rm in} + {\Delta v\over 2})$. Thus $$ {P\over T}=v_{\rm in} + {\Delta v\over 2}\,. $$

  For stationary engine holding against gravity, higher thrust compared to flying fixed wing aircraft is needed as shown above. If we do not "cheat" by increasing mass flow rate through engine (like making it helicopter rotor or using multiple engines), $\Delta v$ has to be increased in order to achieve the necessary thrust. So you not only need more power because of increased thrust but also more power because of increased Watts per unit of thrust. Note that even "helicopter cheat" does not work too well. To match power consumption of engine generating less thrust thanks to the wing's L/D you need to improve P/T too – by decreasing $\Delta v$, thus increasing mass flow (rotor/propeller radius even more than proportionally to increased thrust).

  What about decrease of P/T due to the movement through air? Well, it depends on particular engine and its $\Delta v$.

  A simplified example: We have an engine capable of producing enough thrust to lift aircraft vertically. It can be throttled by changing $\Delta v$ without any practical problems or change in its internal efficiency. And let's assume that mass flow rate through it is fixed area $S$ multiplied by air density and multiplied by arithmetical average of speeds of entering and exiting air. For hovering aircraft and stationary engine producing thrust equal to aircraft's weight, $w$ it is $$w=\dot m \Delta v_{\rm hover} = \rho S \Delta v_{\rm hover}^2 / 2\,;\quad \Delta v_{\rm hover}=\sqrt{2 w \over \rho S}$$ and thus $$P_{\rm hover}=w\cdot \Delta v_{\rm hover}/2=\sqrt{w^3\over 2\rho S}\,.$$

  The same aircraft flying on its wings does only need $w\over L/D$ of thrust. Airspeed is $v_{\rm air}$. Equation for the thrust: ${w\over L/D} = \dot m \Delta v_{\rm flight} = \rho S \cdot (v_{\rm air}+{\Delta v_{\rm flight} \over 2}) \cdot \Delta v_{\rm flight}$. Thus $$\Delta v_{\rm flight}=\sqrt{{2w\over(L/D)\rho S}+v_{\rm air}^2}-v_{\rm air}$$ and $$ P_{\rm flight}={w\over L/D}\cdot(\sqrt{{w\over 2(L/D)\rho S}+{v_{\rm air}^2\over 4}}+{v_{\rm air}\over 2})\,. $$

  Unfortunately, I do not see any way how to simplify and compare $P_{\rm hover}$ and $P_{\rm flight}$ so some concrete numbers:

  • Light aircraft, 1 ton, 100 knot, $S = 5\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 290\,\rm kW$, $P_{\rm flight} = 35\,\rm kW$.

  • Heavy aircraft, 100 ton, 200 knot, $S = 50\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 90\,\rm MW$, $P_{\rm flight} = 7\,\rm MW$.

  So, based on these simplifications flying with wings with similar kind of engine should be significantly more efficient in energy terms too. And, additionally, you are already moving forward using power $P_{\rm flight}$. For vertical engine extra power would be necessary for overcoming air drag due to the movement.

• simple example – rocket engine: not too typical for aircraft, but simple one. Rocket consumes same amount of fuel per second in order to generate a unit of thurst regardless of its size and regardless if it is pointing upward (and is static with respect to air) or forward (and moving through the air). You need to burn proportionally more fuel per second to generate higher thrust. So, for the rocket propulsion, you will save fuel in the same ratio as the necessary thrust decreases.

• Propeller or jet engines are more complicate as their thrust and fuel consumption depends on engine movement through the air too. As David K pointed out in his answer, we can use momentum and kinetic energy of accelerated air to get power needed for unit of thrust.

  With some simplifications, thrust is mass flow rate through the engine/prop multiplied by change of flow speed it causes. $T = \dot m \cdot (v_{\rm out} - v_{\rm in}) = \dot m \Delta v$. Power needed for this is $P = \dot m \cdot {1\over2}(v_{\rm out}^2 - v_{\rm in}^2)=\dot m\Delta v\cdot(v_{\rm in} + {\Delta v\over 2})$. Thus $$ {P\over T}=v_{\rm in} + {\Delta v\over 2}\,. $$

  For stationary engine holding against gravity, higher thrust compared to flying fixed wing aircraft is needed as shown above. If we do not "cheat" by increasing mass flow rate through engine (like making it helicopter rotor or using multiple engines), $\Delta v$ has to be increased in order to achieve the necessary thrust. So you not only need more power because of increased thrust but also more power because of increased Watts per unit of thrust. Note that even "helicopter cheat" does not work too well. To match power consumption of engine generating less thrust thanks to the wing's L/D you need to improve P/T too – by decreasing $\Delta v$, thus increasing mass flow (rotor/propeller radius even more than proportionally to increased thrust).

  What about decrease of P/T due to the movement through air? Well, it depends on particular engine and its $\Delta v$. It will be typically in similar order of magnitude as airspeed (or even less), so we can not neglect $v_{\rm in}$ in watts-per-thrust equation above. There is a efficiency penalty when engine works on moving aircraft. But it should be still worth of it as the gain provided by lift is greater.

  A simplified example: We have an engine capable of producing enough thrust to lift aircraft vertically. It can be throttled by changing $\Delta v$ without any practical problems or change in its internal efficiency. And let's assume that mass flow rate through it is fixed area $S$ multiplied by air density and multiplied by arithmetical average of speeds of entering and exiting air. For hovering aircraft and stationary engine producing thrust equal to aircraft's weight, $w$ it is $$w=\dot m \Delta v_{\rm hover} = \rho S \Delta v_{\rm hover}^2 / 2\,;\quad \Delta v_{\rm hover}=\sqrt{2 w \over \rho S}$$ and thus $$P_{\rm hover}=w\cdot \Delta v_{\rm hover}/2=\sqrt{w^3\over 2\rho S}\,.$$

  The same aircraft flying on its wings does only need $w\over L/D$ of thrust. Airspeed is $v_{\rm air}$. Equation for the thrust: ${w\over L/D} = \dot m \Delta v_{\rm flight} = \rho S \cdot (v_{\rm air}+{\Delta v_{\rm flight} \over 2}) \cdot \Delta v_{\rm flight}$. Thus $$\Delta v_{\rm flight}=\sqrt{{2w\over(L/D)\rho S}+v_{\rm air}^2}-v_{\rm air}$$ and $$ P_{\rm flight}={w\over L/D}\cdot(\sqrt{{w\over 2(L/D)\rho S}+{v_{\rm air}^2\over 4}}+{v_{\rm air}\over 2})\,. $$

  Unfortunately, I do not see any way how to simplify and compare $P_{\rm hover}$ and $P_{\rm flight}$ so some concrete numbers:

  • Light aircraft, 1 ton, 100 knot, $S = 5\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 290\,\rm kW$, $P_{\rm flight} = 35\,\rm kW$.

  • Heavy aircraft, 100 ton, 200 knot, $S = 50\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 90\,\rm MW$, $P_{\rm flight} = 7\,\rm MW$.

  So, based on these simplifications flying with wings with similar kind of engine should be significantly more efficient in energy terms too. And, additionally, you are already moving forward using power $P_{\rm flight}$. For vertical engine extra power would be necessary for overcoming air drag due to the movement.

###Additional comment: how is it related to power and energy###

The answer above is focused on how wings decrease necessary engine thrust, but original question could be interpreted in terms of energy efficiency too. I'll try to add some comments on this part.

• simple example – rocket engine: not too typical for aircraft, but simple one. Rocket consumes same amount of fuel per second in order to generate a unit of thurst regardless of its size and regardless if it is pointing upward (and is static with respect to air) or forward (and moving through the air). You need to burn proportionally more fuel per second to generate higher thrust. So, for the rocket propulsion, you will save fuel in the same ratio as the necessary thrust decreases.

• Propeller or jet engines are more complicate as their thrust and fuel consumption depends on engine movement through the air too. As David K pointed out in his answer, we can use momentum and kinetic energy of accelerated air to get power needed for unit of thrust.

  With some simplifications, thrust is mass flow rate through the engine/prop multiplied by change of flow speed it causes. $T = \dot m \cdot (v_{\rm out} - v_{\rm in}) = \dot m \Delta v$. Power needed for this is $P = \dot m \cdot {1\over2}(v_{\rm out}^2 - v_{\rm in}^2)=\dot m\Delta v\cdot(v_{\rm in} + {\Delta v\over 2})$. Thus $$ {P\over T}=v_{\rm in} + {\Delta v\over 2}\,. $$

  For stationary engine holding against gravity, higher thrust compared to flying fixed wing aircraft is needed as shown above. If we do not "cheat" by increasing mass flow rate through engine (like making it helicopter rotor or using multiple engines), $\Delta v$ has to be increased in order to achieve the necessary thrust. So you not only need more power because of increased thrust but also more power because of increased Watts per unit of thrust. Note that even "helicopter cheat" does not work too well. To match power consumption of engine generating less thrust thanks to the wing's L/D you need to improve P/T too – by decreasing $\Delta v$, thus increasing mass flow (rotor/propeller radius even more than proportionally to increased thrust).

  What about decrease of P/T due to the movement through air? Well, it depends on particular engine and its $\Delta v$.

  A simplified example: We have an engine capable of producing enough thrust to lift aircraft vertically. It can be throttled by changing $\Delta v$ without any practical problems or change in its internal efficiency. And let's assume that mass flow rate through it is fixed area $S$ multiplied by air density and multiplied by arithmetical average of speeds of entering and exiting air. For hovering aircraft and stationary engine producing thrust equal to aircraft's weight, $w$ it is $$w=\dot m \Delta v_{\rm hover} = \rho S \Delta v_{\rm hover}^2 / 2\,;\quad \Delta v_{\rm hover}=\sqrt{2 w \over \rho S}$$ and thus $$P_{\rm hover}=w\cdot \Delta v_{\rm hover}/2=\sqrt{w^3\over 2\rho S}\,.$$

  The same aircraft flying on its wings does only need $w\over L/D$ of thrust. Airspeed is $v_{\rm air}$. Equation for the thrust: ${w\over L/D} = \dot m \Delta v_{\rm flight} = \rho S \cdot (v_{\rm air}+{\Delta v_{\rm flight} \over 2}) \cdot \Delta v_{\rm flight}$. Thus $$\Delta v_{\rm flight}=\sqrt{{2w\over(L/D)\rho S}+v_{\rm air}^2}-v_{\rm air}$$ and $$ P_{\rm flight}={w\over L/D}\cdot(\sqrt{{w\over 2(L/D)\rho S}+{v_{\rm air}^2\over 4}}+{v_{\rm air}\over 2})\,. $$

  Unfortunately, I do not see any way how to simplify and compare $P_{\rm hover}$ and $P_{\rm flight}$ so some concrete numbers:

  • Light aircraft, 1 ton, 100 knot, $S = 5\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 290\,\rm kW$, $P_{\rm flight} = 35\,\rm kW$.

  • Heavy aircraft, 100 ton, 200 knot, $S = 50\,\rm m^2$, $L/D = 15$: $P_{\rm hover} = 90\,\rm MW$, $P_{\rm flight} = 7\,\rm MW$.

  So, based on these simplifications flying with wings with similar kind of engine should be significantly more efficient in energy terms too. And, additionally, you are already moving forward using power $P_{\rm flight}$. For vertical engine extra power would be necessary for overcoming air drag due to the movement.