I have a hard time figuring out why a longitudinally unstable aircraft has a lower trim drag. This seems to be a thing especially in fighter aircraft, so does it have to do anything with supersonic flight as well?

enter image description here

  • $\begingroup$ There seems to be a lot of confusion on this issue. Perhaps if you share your sources of information I might be able to give you a better answer. $\endgroup$ – DLH Jun 7 at 20:22
  • $\begingroup$ Well that's the thing, I don't have any sources. I actually stumbled upon it during my exam preparations, it's an old exam question but no answer are provided. I added an image from the lecture materials though. $\endgroup$ – Wouterr G Jun 7 at 20:39
  • 1
    $\begingroup$ The aircraft is only unstable in subsonic flight. In supersonic flight it is stable again, but much less stable than a subsonically stable version. Since stability costs trim drag, supersonic trim drag is greatly reduced. As you correctly note, there is no drag reduction in subsonic flight. However, an unstable layout allows to reduce wing and tail area, so there is some drag reduction indeed. $\endgroup$ – Peter Kämpf Jun 7 at 21:40

It might be helpful to look at this article from Boldmethod.

The basic idea is that the farther away the center of gravity is from the center of pressure (or center of lift depending on your terminology) the more lift the horizontal stabilizer will have to generate. The more lift an airfoil has to generate will result in more induced drag. Trim drag is specifically the induced drag created by the horizontal stabilizer. The horizontally stabilizer can actually produce both positive (upward) and negative (downward) lift but the negative lift has the even worse affect of requiring the wings to create more lift to compensate for the downward lift as demonstrated in the Boldmethod article.

Having the center of gravity at the center of pressure also allows the aircraft to be more maneuverable since less force is required from the horizontal stabilizer to initiate a maneuver. This is why fighter jets tend to be longitudinally unstable and use fly-by-wire to compensate for it.

  • $\begingroup$ This is why fighter jets tend to be longitudinally unstable - Really? If it were as you say, they should be indifferent. Instability will again cause trim drag, just as the answer supposes. $\endgroup$ – Peter Kämpf Jun 8 at 8:16
  • $\begingroup$ I though the longitudinal instability of fighters was to be more reactive in pitch. $\endgroup$ – Manu H Jul 10 at 8:54

Longitudinal trim is achieved when the total pitching moment on the aircraft is zero. Except for some special cases, the tail (or elevons in the case of tailless aircraft) will generate some lift to trim out the pitching moment from the wingbody. Therefore, if the tail generates negative lift (thereby providing a nose up moment), the wingbody needs to work harder (i.e. higher AOA) to generate the aircraft level $C_L$ required for level flight.

At this point, let's define what trim drag actually is. Trim drag is the ensemble of: additional induced drag from tail incidence (or elevators or elevons), plus additional induced drag from the wingbody due to a higher AOA required to achieve the total $C_L$, plus additional interference/viscous drag due to control surface deflections. The first and third components are actually relatively small compared to the second component. Unless you have a big control surface deflection, the majority of the trim drag actually comes from the loss of lift!

When the aircraft is pitch stable and trimmed, there is a reduction of $C_{L_{\alpha}}$ compared to the untrimmed case. As the static margin is reduced, amount of negative lift from the tail is reduced and the lift slope improves. This also improves the trim drag.

The following graphs illustrate the effect of static margin with trimmed $C_{L}$, amount of h-stab needed to trim and trim drag, generated with typical aircraft geometries and aerodynamics:

CL vs AOA Stab to trim Trim drag

  • $\begingroup$ Excellent graphs. Folks should remember with negative static stability there will be an increase in drag due to pitch oscillations (which can be computer controlled) and an increased danger of stalling the aircraft. Static stability (speed stable) is a valuable safety factor to pilot under high workload, as the aircraft can be trimmed to hold speed and AOA. A staticly unstable aircraft is much more dangerous in slow flight. $\endgroup$ – Robert DiGiovanni Jul 10 at 23:59

It's really simple. To hold the aircraft up in the air, the TOTAL Sum of Up Lift and Down lift (plus or minus) must be equal to, and oppose, the weight of the aircraft. If the lift from the tail is pulling the aircraft down, towards the earth, (as in a positively static stable aircraft), then the lift from the wing must be higher (by twice the tail amount pulling down) to counteract it. The total lift up minus the total lift down must be equal to the aircraft weight. So since the total lift Up and Down are both producing drag, then there must be more drag.

  Say the aircraft weighs 1000 pounds and say that drag is 10% of total lift
          Wing Lift      tail Lift     Result       Total lift    Drag
 Stable    1200 lbs      -200 lbs      1000 lbs      1400 lbs    140 lbs    
 Unstable  800 lbs        200 lbs      1000 lbs      1000 lbs    100 lbs     
  • $\begingroup$ You may wish to express tail force in torque as -200 ft lbs (as it would have a longer lever arm than the wing). $\endgroup$ – Robert DiGiovanni Jul 10 at 23:49
  • $\begingroup$ True, but this issue is not related to the moment (torque) it is simply related to the force the aerodynamic surface is exerting on the airframe. The tail is pulling the aircraft down in an aircraft with positive static stability, and it is pulling the airframe up in one with negative static stability. True, these forces generate moments, but it is simply the force being exerted that explains why the trim drag is higher in one case than in the other. $\endgroup$ – Charles Bretana Jul 11 at 2:39
  • $\begingroup$ "The tail is pulling the aircraft down in an aircraft with positive static stability, and it is pulling the airframe up in one with negative static stability. " -- I suspect this is not true. A frequent topic of discussion on this site. I don't think it's always true that a download on tail is required for positive static stability. $\endgroup$ – quiet flyer Jul 11 at 12:53
  • $\begingroup$ See for example some of the comments under here- -aviation.stackexchange.com/questions/66311/… $\endgroup$ – quiet flyer Jul 11 at 12:57
  • $\begingroup$ @quietflye, yes you are right, it is not ALWAYS the case. Canard aircraft, for example, violate that premise. They can have positive stability with both the canard and the main wing (aft) generating positive lift. But generally, in a conventional aircraft, with a aft mounted tail, if the CG is forward of the AC, BY DEFINITION, the sum of all aerodynamic forces (which BY DEFINITION can be thought of as acting through the AC, will cause a nose down pitching moment. Obviously, to counteract this, a tail mounted elevator is AFT of the CG, and must exert downwards lift to generate a nose up moment. $\endgroup$ – Charles Bretana Jul 12 at 3:14

Different question, same answer. Active stability allows for the $c.g.$ to be behind the centre of lift, thereby compensating for the associated aerodynamic instability.

enter image description here

For passive longitudinal static stability (by aeroforces), the total centre of lift $C_N$ must be behind the centre of gravity. At all angles of attack and all velocities, with stalled wings etc. The only passive solution that is always safe in all circumstances, is the aerodynamic centre $a.c._w$ behind the $c.g.$, always creating a nose down moment which must then be compensated by an aerodynamic nose up moment from the tailplane: negative lift. So we need to compensate this by more lift from the main wing, with associated induced drag.

That's the passive, aerodynamic solution. If we allow $n.p._{fixed}$ to be in front of the $c.g.$, the tailplane will always help in creating lift, not destroy it. At cruise, we can trim the aeroplane for neutral pitching moment, but if there is a disturbance in angle of attack (like a vertical gust) the main wing will create more lift than the tailplane (it makes sense to make the main wing the most efficient one.) But that means that any disturbance in AoA will create a sudden, unstabilised nose up reaction: static instability.

The only solution to be able to use the $n.p._{fixed}$ before $c.g.$ situation, is by using active stability. Any disturbance in pitching moment is immediately counteracted by an automatic elevator deflection, like balancing a stick vertically on an open palm, or riding a unicycle.

This principle goes for both subsonic and supersonic flight. But going supersonic means that the Centre of Pressure shifts rearwards: Mach tuck. The aeroplane could be:

  • passive statically stable in supersonic flight, unstable in subsonic flight.
  • passive statically unstable in supersonic flight, and much more unstable in subsonic flight.
  • $\begingroup$ These answer seems to be at odds w/ the fact that many old-fashioned free-flight model airplanes clearly had lifting tails. $\endgroup$ – quiet flyer Jul 13 at 12:29
  • $\begingroup$ @quietflyer the drawing shows a statically stable configuration with a lifting tail. $\endgroup$ – Koyovis Jul 13 at 22:17
  • $\begingroup$ You can be staticly stable with a lifting tail but still get into trouble if the rear wing stalls first. A solution is to use a straight front/delta rear. This was a very popular combination for fighter biplanes, but totally unnecessary for an airliner. But it is understandable that these behemoths would be extremely sluggish if TOO stable. But too unstable is unsafe. $\endgroup$ – Robert DiGiovanni Jul 14 at 8:59

You need a picture of a 3rd airplane with the center of gravity right underneath the wing lift arrow. Notice the elevons would be neither up or down. This is the lowest drag configuration, as up or down elevons adds drag. Up elevons (staticly stable) are slightly more draggy (because they force the wing to work harder against their downforce) than down elevons (staticly unstable), but both are more draggy than no elevon deflection.


It is the horizontal stabilizers job to set optimal AOA of the single wing (since the 1920s) while the Hstab is at 0 (lowest drag) angle of attack in flight. The difference in incidence is called DECALAGE of the horizontal stabilizer (See B-52). The center of gravity optimally belongs DIRECTLY UNDER the center of lift. The horizontal stabilizer must have ADEQUATE VOLUME to hold that wing in place.

Next one decides how much static stability (speed stability) one wants in the plane for safety. This is dependent on weight placement of fuel and payload, as well as potential shifts in CP due to change in AOA and ENGINE THRUST torque factors. Without active stability control (computer control) this is generally set up to be positive.

Setting up for static stability and trimming aerodynamicly is usually accomplished with a small trim tab. Sadly, this seems to have lead to a belief among modern designers that a tiny horizontal tail volume is OK, and computers will solve everything.

We even have graphs showing us that creating a staticly unstable tandem (biplane) will save fuel.

Starting every morning by looking at a Piper Cub may be helpful.


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.