8
$\begingroup$

Does cruising speed of aircraft depend upon mass of aircraft?

I am asking this because as lift is generated to overcome the gravitational forces ($mg$). And lift is proportional to square of the aircraft velocity.

$\endgroup$
  • $\begingroup$ and the area of the wing $\endgroup$ – user3528438 Jan 10 '18 at 5:06
  • 3
    $\begingroup$ Cruise speed is usually a design decision, along with payload and range. The airframe, in particular the wing, and engines are then designed and selected to deliver that performance. $\endgroup$ – user27769 Jan 10 '18 at 5:20
10
$\begingroup$

The speed of the aircraft in cruise is determined by the forces in the horizontal axis: thrust and drag. Weight and lift are forces in the vertical axis; they do pop up in the horizontal axis, but are relatively tiny in cruise.

enter image description hereImage source

At low airspeeds, induced drag is the dominant factor: this is the drag that is proportional to lift and weight. At cruise speeds, induced drag drops off and parasite drag rises quadratically with airspeed. It is drag associated with the aerodynamic shape of the aircraft. At any weight, the most beneficial shape is the one that aligns the fuselage best with the free air stream: tilt the nose up or down from the Angle of Attack for minimum parasite drag and drag will increase, at any speed and weight. Wind tunnel tests and Computed Fluid Dynamics are the tools used for establishing minimal form drag.

However, for straight and level flight the vertical component of lift must indeed match the weight, so one of the parameters of the flight state must be tuned in order to get zero climb rate. Changing cruise speed is indeed one of the degrees of freedom to change lift, but not the best one. In the aircraft aerodynamic axes, lift L is given as $$L = C_L \cdot \frac{1}{2} \cdot \rho \cdot V^2 \cdot S$$

S is wing area, usually a constant that cannot be changed during cruise. So the degrees of freedom to change lift are:

  • Change lift coefficient $C_L$: this is done by changing Angle of Attack, by pitching up or down. Doing so creates extra drag and fuel burn, the least amount of drag is generally with the fuselage aligned with the free stream.
  • Change air density $\rho$: fly higher or lower. This is what is done when the aircraft gets lighter due to fuel burned: at higher altitude there is less air resistance and less thrust needs to be delivered by the engine, improving fuel burn without slowing down.
  • Change V. Flying slower as fuel is burned is beneficial for fuel burn per hour - but it takes longer for the aircraft to arrive, so fuel is burned for more hours.

So for varying weight, it is best to vary lift by varying altitude. Long range cruise aircraft would have the least fuel burn if they could gradually climb as fuel weight diminishes, however there are air traffic corridors which the aircraft needs to stay in. Cruise climb is done in steps instead.

enter image description here

From this lecture, which also provides equations for cruising at constant altitude (varying velocity) and at constant velocity (varying altitude). The latter wins.

$\endgroup$
  • 3
    $\begingroup$ Wing area is usually constant. *8') $\endgroup$ – Mark Booth Jan 10 '18 at 11:49
  • 1
    $\begingroup$ I think there should be a caveat that this is a significant simplification and that the true way of determining such things is of course finite element analysis, etc as well as wind tunnel tests. Nobody really uses that formula in practice. $\endgroup$ – user1997744 Jan 10 '18 at 12:12
  • $\begingroup$ @Koyovis I hope it didn’t come across as a negative comment. To clarify I think there should just be a caveat to the reader to take it with a pinch of salt, so to speak. $\endgroup$ – user1997744 Jan 10 '18 at 12:16
  • $\begingroup$ @user1997744 Plz check page 24 of the linked Princeton lecture, where they use the equation. $\endgroup$ – Koyovis Jan 10 '18 at 13:13
  • 1
    $\begingroup$ @CptReynolds Oh, absolutely, I wouldn't expect an aircraft operator to do that. What I meant was the manufacturer would be using such methods to determine suitable cruising speeds. $\endgroup$ – user1997744 Jan 10 '18 at 17:01
4
$\begingroup$

The most efficient cruise speed of a given aircraft depends, among other parameters, on the aircraft‘s mass in the way you hint at. The operationally flown cruise speed may differ and may thus be, within a certain speed range, selected independently of mass, e.g. for air traffic control reasons. That safe speed range, however, again depends on mass (among other parameters).

$\endgroup$
  • $\begingroup$ It means heavily loaded aircraft reach to it's destination earlier as compared with less loaded aircraft..? $\endgroup$ – Abhishek Acharya Jan 10 '18 at 5:38
  • $\begingroup$ You wrote "The operationally flown cruise speed may differ and may thus be, within a certain speed range, selected independently of mass" I didn't understand that part :) $\endgroup$ – Abhishek Acharya Jan 10 '18 at 5:42
  • $\begingroup$ @AbhishekAcharya Yes, if the same aircraft is loaded heavier, its best cruise speed will be higher than if it is lighter, all else being equal. What I meant with the other sentence is that there is a mass-dependent minimum speed (heavier mass, higher speed) and a fix maximum speed for any aircraft, and the cruise speed must be between the two. $\endgroup$ – Cpt Reynolds Jan 10 '18 at 16:45
  • $\begingroup$ if aircraft is cruising with a speed which is not equal to best cruising speed than what does pilot do to balance lift .. ? $\endgroup$ – Abhishek Acharya Jan 11 '18 at 11:11
  • $\begingroup$ @AbhishekAcharya I am not sure I understand the question. Lift force is dependent on the air density, the airspeed, the angle of attack (angle between direction of travel through air and longitudinal aircraft axis) and aircraft geometry including the wing surface. Assuming fixed geometry and air density, the aircraft can still be operated at different speeds, as long as the associated angle of attack can be flown. Simply speaking, the slower the aircraft flies, the higher the nose (even in level flight). Did that help? $\endgroup$ – Cpt Reynolds Jan 11 '18 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.