# Will a heavier aircraft be more efficient at high speeds(cruise)? [closed]

Since heavier aircraft have more mass and inertia, they will have less drag because it will be easier for them to push through the air so does that mean that we can achieve the same cruise speed with less engine power=more efficiency?

• Wrong. Drag has nothing to do with mass or inertia. Drag is an aerodynamic force, caused by the force created by air molecules impacting and bouncing from the surface of the aircraft. The mass of the air molecules matters, but aircraft mass or inertia has nothing to do with it. Sep 11, 2022 at 6:09
• I think he wants to say that the aircraft will have more energy so it will be "easier to push through the air" and there will be less energy wasted in opposing drag Sep 11, 2022 at 6:34
• All the rest being equal, more mass implies more power needed to accelerate or decelerate the aircraft and more lift and drag to keep if flying Sep 11, 2022 at 14:13
• This works for projectiles like bullets which "store" kinetic energy and don't need lift. If you need lift and can store the energy as height (glider) it roughly cancels. For any other energy source weight is bad, although a plane so flimsy it breaks up (Helios) or planes that can be blown away while parked aren't exactly good either. Sep 14, 2022 at 6:46

Inertia is defined by Oxford languages as

a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.

The key here is "the external force".

The equation for drag force is

$$F_d=\frac12ρv^2c_dA$$

While you are correct in that the drag force is linked into the interia as it is an external force acting on an object, you can see that drag equation does not have the mass of the (flying) object in it.

To maintain straight and level flight, the drag has to be matched by engine thrust, inertia has no effect here because the plane is not accelerating. In that sense the mass of the aircraft is insignificant.

However, as we know the aircraft needs lift to stay in the air, and the more mass it has, the more lift is needed. To put it simply, that lift is ultimately generated by pushing air downwards by wings, rotors etc, and for that we ultimately need engine power. So the heavier the plane, the more power is needed for cruise.

It might be more efficient per unit mass to fly a heavier plane up to a certain point, but it is not because of inertia.

Will a heavier aircraft be more efficient at high speeds(cruise)?

If the aircraft is a glider, then yes. Gliders very often carry water ballast to increase their weight, and thus reduce their sink rate and increase their efficiency (expressed in terms of glide ratio, or approximately, L/D ratio1), at a given, high, target airspeed that is well above the unballasted, still-air best-glide-angle speed.

Footnotes:

1. Only in constant-airspeed flight in still air does glide ratio exactly equal L/D ratio. But regardless of the details, the point remains that increasing the mass of a glider allows flight at a higher angle-of-attack, and also often a lower sink rate, and thus a higher glide ratio relative to the ground, for flight at a given high airspeed.
• Interesting point of view Sep 14, 2022 at 6:08

Inertia is a property related to acceleration, as in Newton’s first law $$F = m \cdot a$$. If we want to change the current velocity of the aeroplane, we need more force to do so when inertia is higher.

Aerodynamic drag is a force proportional to velocity. At any constant velocity the drag remains constant, no matter what the inertia of the plane is.

..does that mean that we can achieve the same cruise speed with less engine power..

Heavy aeroplanes require more thrust and power:

• To get to cruise speed, it has taken more force to accelerate the higher inertia. The plane's kinetic energy content $$E_{kin} = ½mV^2$$ is higher at a higher mass m.
• At the same cruise speed, a heavy aeroplane requires more thrust than the same light aeroplane in the same cruise conditions. It requires more lift, which increases drag D = $$C_D \cdot ½ \rho V^2 \cdot S$$, with $$C_D = C_{D0} + k{C_L}^2$$ (k = a constant based on aeroplane geometry)
• "a heavy aeroplane requires the same thrust as the same light aeroplane". It would actually require more lift, more drag and therefore more thrust Sep 13, 2022 at 6:09
• @sophit Indeed, have amended the answer, thanks for that. Not a good idea to post an answer after a transcontinental flight. Sep 14, 2022 at 4:16
• Eh eh eh, no :-) Sep 14, 2022 at 4:31