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This question includes the quote:

A C42 doesn't weigh much, so to pull the 5+ G required to exceed the design specifications of it's wing spar and fixings would require extreme speeds.

I'm not an aircraft design expert, but it seems to me that the weight of a plane doesn't affect its ability to pull G (in a maneuver, not sustained). With sufficiently large control surfaces and a good amount of power a light aircraft would find it easier to pull G than a heavy one. Am I wrong?

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  • $\begingroup$ You may get a different answer if you're considering nearly instantaneous loading, or the ability to sustain (like, in level flight) a particular G loading. May be worth clarifying that. $\endgroup$
    – Ralph J
    Commented Apr 29 at 19:09
  • $\begingroup$ It depends more on the wing loading $W/S$. A sustained turn might be a good way to illustrate it, let's see if someone feels like elaborating a proper answer. $\endgroup$
    – sophit
    Commented Apr 29 at 20:27
  • $\begingroup$ Well, lighter and smaller tolerates more G's. Radio-controlled pylon racers and dynamic-soaring sailplanes weighing 20 to 40 pounds commonly pull over 40 G's. $\endgroup$ Commented Apr 29 at 20:31
  • $\begingroup$ @CamilleGoudeseune Is that because it doesn't really matter if they break up in flight? And there isn't crew to die a horrible squishy death? $\endgroup$ Commented Apr 29 at 20:38
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    $\begingroup$ The reason RC aircraft can pull such loads is the square-cube law. The weight cost of components that can stand the stresses is quite low. The bigger the plane the greater the percent of it's total mass is devoted to keeping it from breaking. The RC also must be built stronger because of the handling loads. Try picking that jetliner up by it's tail and see what happens to it. $\endgroup$ Commented May 1 at 4:14

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Statistics give a different picture: Smaller airplanes usually have lower wing loadings and higher G limits than heavier, larger ones. But the short answer is: You are right and the reasoning in the quote is wrong.

Now let me first get some things clarified:

  1. Weight is mass times acceleration, mostly gravitational but also centrifugal acceleration. Weight is therefore a force. From here on, I will interpret your question as "does wing loading of an aircraft affect its ability to pull G?", wing loading being mass per reference area.
  2. You can either fly multiple gs by maneuvering or by flying into gusts. Both need to be looked at separately.

Now to the first way of pulling Gs: Tight circles. If we don't care about height loss (instationary turns), the g limit is a design factor and independent of wing loading. Lighter airplanes require less speed to reach their G limit, so the quote is wrong. Less speed is required.

The stick force for pulling Gs goes up with the stability margin and dynamic pressure. Higher wing loading, therefore, will require more sick force, all other things being equal. The large control surface helps to reach the G limit with maximum forward CG. Installed power will limit the height loss in instationary turns, so both help to reach the G limit.

If you want a light stick force to pull maximum Gs, fly with a rear CG location and light wing loading. If we consider stick forces to be a factor, a low wing loading helps to reach high Gs.

Now for the gust case: Here, gust strength and flight speed determine the G load. Given again two identical planes which differ only in wing loading, the heavier one needs to fly faster, so the same gust will result in a lower angle of attack change from flight into a vertical gust and a lower gust load factor compared to the lighter plane. Make no mistake: The structural stresses will be higher for the heavier airplane, but its higher mass requires proportionally more lift for the same G loading. With the speed ratio between flight speed and gust speed lower for the heavier airplane, it will experience less Gs.

Ergo: In the gust case a lighter wing loading will result in higher Gs. Which also explains why lighter planes have higher G limits. Another explanation are scaling laws: Larger airplanes will have higher wing loading.


P.S. Someone felt that the g needs to be a capital G. Normally, the gravitational acceleration is denoted with a lowercase g, but maybe using the capital G improves readability. Also, I normally don't put an apostrophe between a word and a plural s, but even SE seems to succumb to Internet spelling, so I leave it at that.

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    $\begingroup$ Re: "The g limit is a design factor and independent of wing loading". It is not always the case. G-limit usually decreases as gross weight is increased above a certain threshold in the same aircraft. When the aircraft is light enough, with minimal fuel or stores, then the g-limit can stay relatively the same. This decrease in g-limit with increasing weight is seen on aircraft such as F-15, F-16, F-18 and Su-27. $\endgroup$
    – LJQCN101
    Commented Apr 30 at 5:51
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    $\begingroup$ @LJQCN101 My focus is on smaller airplanes. Of course, if you add supersonic planes and types which dump 15% of their take-off mass in one instance, things become more complicated. Remember, this started with the Ikarus C42 ultralight. $\endgroup$ Commented Apr 30 at 10:04
  • $\begingroup$ Sure but not only supersonic planes, as there’re plenty of manuals out there, and one thing in common is that both g-limit for maneuvering and gust takes weight into account. The maneuvering g-limit can be constant for a ranges of weight, as I said above. I’ll be adding other aircraft if people need it. $\endgroup$
    – LJQCN101
    Commented Apr 30 at 11:25
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    $\begingroup$ @LJQCN101 It's not so simple. Much depends where the weight is. Fuel in tip tanks has a different influence than bombs in a central bomb bay. Please don't assume I don't know this, but I wanted to keep the answer on point. $\endgroup$ Commented Apr 30 at 15:38
  • $\begingroup$ @LJQCN101 You will see this with reversible control when a combination of forward cg and high gross weight means too high stick forces to pull out of a dive. Not all airplanes use hydraulic control systems! $\endgroup$ Commented May 3 at 7:50
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Short answer: A heavier weight, by loading more fuel or stores, decreases the ability of the same aircraft to pull more instantaneous Gs (available load factor) aerodynamically.

The load factor in the direction of Lift can be calculated as:

  • n = (Total Lift + Thrust * sinα) / Weight

So with a heavier weight, you'll need a higher lift (e.g. by increasing AOA), and/or a greater thrust, to achieve the same load factor.

For an example, you can refer to the F-16C/D SUPPLEMENTAL FLIGHT MANUAL T.O. GR1F-16CJ-1-1, linked https://publicintelligence.net/hellenic-air-force-f-16cd-flight-manuals/

In the 'Available Load Factor' chart at Figure A1-8 (block52) or Figure B1-8 (block50), where there's clearly an entry for GW (gross weight). A heavier weight leads to a lower attainable G for the same condition.

Block52 available Gs

Regarding the pilot observed / designed / allowable G-limits, it also takes gross weight into consideration, as the wing is designed to withstand a certain amount of force. For a simplified force analysis, you can assume the force is in the body Z axis, so the wing withstands a force of Nz*weight, where Nz is the normal load factor.

Just to be sure, load factor in the lift direction (stability Z axis) is not to be confused with normal load factor (Nz), which is in the body Z axis:

  • Nz = Total Normal Force / Weight = (Total Lift * cosα  + Total Drag * sinα) / Weight

So here, Nz takes into account the lift and drag components in the body axes, while load factor takes into account the thrust component in the stability axes.

F16 Axes system

The G-limit of F-16C, versus gross weight:

F16 G-limits

The maneuver and gust G-limits of B-52H, versus gross weight: B52 G-limits2

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  • $\begingroup$ Actual load on the wing spar will be G load × weight, not just G factor. $\endgroup$ Commented Apr 30 at 9:42
  • $\begingroup$ @RobertDiGiovanni To be specific, the actual force that is perpendicular to the wing is the normal force in the body Z axis, which equals to Nz*weight, or (Total Lift * cosα + Total Drag * sinα). Nz is normal load factor, not any other factors. $\endgroup$
    – LJQCN101
    Commented Apr 30 at 10:20
  • $\begingroup$ That's right. That's how hard the wing is tugging on the fuselage. I think this is why some folks switched to Newtons. In this case, G and mass change, resulting in the same (instantaneous) load on the spar! (Even though the pilot will experience higher G if the plane is lighter. $\endgroup$ Commented Apr 30 at 17:11
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The crux of the analysis seems to be the potential change in angle of attack from a straight and level flying condition (though other conditions are possible).

All aircraft have a limit in their flight envelope where an abrupt change in angle of attack, from maneuvering or change in relative wind from a gust, can cause either a stall or G forces exceeding the aircraft limit.

Below this maneuvering speed, AoA can increase until the aircraft stalls without exceeding G limits.

Because the AoA required for adequate lift is lower in lighter aircraft, there is more "room" to increase AoA before stall. Within the range of AoA before stall, the relationship of lift to AoA is more or less linear.

So, a lighter aircraft producing 1 G at 3 degrees AoA can produce 5 G at 15 degrees AoA before stalling (if the stall AoA of that particular airfoil is 15 degrees).

The same aircraft with more weight might need 6 degrees AoA at the same speed or 3 degrees at a higher airspeed.

Working at the same speed, we see that the heavier aircraft can only produce 2.5 G before reaching its stall limit.

Crucial to understanding all of this is that an extremely light aircraft, like a paper airplane, can change directions more rapidly than it can increase AoA. Not so with heavier aircraft.

So, now it boils down to rate of pitch vs rate of change in direction.

The lighter airplane, with the same control input, at the same speed, has less inertia to overcome. It can pitch faster$^1$ than the heavier one!

Asking the question about ability to pull G "in a maneuver, not sustained" put us on the right track, as this, by virtue of the change in AoA, is where the greatest G forces are found.

More reading here.

$^1$ Peter Kampf also pointed out moving CG back has the same effect: " lighter stick forces "

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    $\begingroup$ Realizing the heavier aircraft weights 2x more, therefor the load on the wing would be the same with equal pitch rate! $\endgroup$ Commented Apr 29 at 22:20
  • $\begingroup$ A heavier aircraft does not immediately equal to a lower pitch rate. As told in NASA TP-1538, the component of aircraft pitch acceleration can be calculated as: pitching moment coefficient Cm * dynamic pressure * mean air chord * reference wing area / moment of inertia Iy. As you can see, increasing Cm by e.g. reducing static margin, can make up for the increased Iy. But the ability to pull G is more related to CL than Cm, due to how load factor is calculated. But Cm decides whether an AOA can be reachable. $\endgroup$
    – LJQCN101
    Commented Apr 30 at 7:35
  • $\begingroup$ @LJQCN101 this is why G forces felt by the pilot (of constant weight) may be different from load forces (from varying wright). GA aircraft specs "get away with" G load limits because aircraft weight does not vary much. Really waiting for a 747, 777, C5 Galaxy pilot to "weigh in" on this subject. $\endgroup$ Commented Apr 30 at 9:40

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