In Top Gun Maverick (2022), Vice Admiral Beau "Cyclone" Simpson speaks to squadrons about enemy terrain:
Vice Admiral Beau "Cyclone" Simpson: Well, Lieutenant, you have a fighting chance against enemy aircraft. What are the odds of surviving a head-on collision with a mountain? You'll be attacking the target from a higher altitude, level with the north wall. Gonna be a little harder to keep your lase on target, but you will avoid the high-G climb out.
What do you mean by "high-G climb out"?
What do you mean by "high-G climb out"?
Arguably this terminology is not really correct. The pull up, where the flight path changes from horizontal (or diving) to a climb, is where the high G-loading is imposed.
G-loading is simply lift force divided by aircraft weight, and a change in direction of the flight path is a form of acceleration, and the rate of acceleration is proportional to the net force acting on the aircraft, so to "bend" the flight path upwards, you have to apply a lift force that exceeds the downward pull of gravity. Since the centripetal force required to produce a given radius of curvature of the flight path is proportional to airspeed squared, if you are flying very fast and want to bend the flight sharply upward, you'll pull (impose) quite a high G-load.
In the linear portion of the climb, the G-load is actually less than one, because some of the aircraft weight is supported by the Thrust vector rather than the Lift vector. (For an extreme case, consider a vertical climb.) For more see this related ASE question -- Does lift equal weight in a climb? -- see various answers including my own.
But if a pilot said "high-G climb out", it would be clear enough to others that he meant a high-G pull-up into a steep climb.
A high positive G-load is experienced by the pilot as an apparent force pushing him/her down into the seat. This is really the pilot's inertial resistance to the upward push of the seat against his/her body. A high negative G-load is experienced by the pilot as an apparent force pulling him/her up against the seat belts. This is really the pilot's inertial resistance to the downward push of the seat belts against his/her body. In both cases the (upward or downward) lift force generated by the wings is the ultimate cause of the sensations experienced by the pilot. If lift goes to zero, the G-loading goes to zero and the pilot tends to "float" freely in the cockpit.
Actually there's a bit of ambiguity here-- "G-loading" as used in this answer is based on the force component that acts parallel to the lift vector from the wings, which will be generally "up" or "down" in the pilot's reference frame. The traditional mechanical G-meter on an aircraft's instrument panel measures this force component, or very nearly so. Arguably the most accurate definition of G-loading would be based on the complete three-dimensional force vector generated by the aircraft. For example, to truly experience "weightlessness", not only must Lift be brought to zero, but Thrust must be set to exactly equal Drag, so that there is no net force generated by the aircraft, and the pilot experiences no "forwards" or "aftwards" pressure on his or her body. A catapult launch from an aircraft carrier is an example of a situation where there is a high G-load in the fore-and-aft direction rather than the up-and-down direction, in the aircraft's reference frame. In a steady-state (constant airspeed) vertical climb or dive-- not something most of us encounter in flight, but possible in some aircraft-- the complete 3-dimensional G-loading is 1, but a traditional mechanical G-meter on the instrument panel will read 0, because the wings are not generating any lift. In this case the pilot feels no pressure in the "upwards" or "downwards" direction in the aircraft's reference frame, but experiences the back of the seat pushing "forward" on his or her back, or experiences the seat belts pushing "backwards" against his or her chest. The aircraft's excess Thrust or excess Drag vector is the root cause of this sensation.
But when pilots speak of a "high G" situation without adding further qualifiers, they are invariably referring to a high G-load acting in the upward direction in the aircraft's reference frame, which is always due to a strong Lift force generated by the wings.
You will note from other answers that the G-loading can also be defined in terms of acceleration. Force and acceleration are intimately related. When considering these concepts, keep in mind that the force exerted by gravity, or the acceleration component due to gravity, is not considered when computing the G-loading. That's why we don't have to leave the earth's gravitational field to experience 0-G "weightlessness" -- we simply have to stop resisting gravity, and fall freely.
The aircraft is in a dive while on attack. Once the payload is dropped, the aircraft must exit the target zone safely. The aircraft has to pitch up violently in order to avoid hitting the ground and the rising mountainous terrain in the flight path. The speed of the aircraft combined with the rapidly changing pitch angle causes a “high-g” load on the aircraft so that it can “climb-out” to a safe altitude clear of any obstacles.
If the Lieutenant was attacking the target from a low altitude he would have to pull-up abruptly and steeply to avoid the mountain. This would result in a high G-force. Since military fighter jets routinely fly maneuvers resulting in high G-forces I'm assuming Simpson was referring to a high G-force that may be difficult to manage or beyond the jet's or pilot's physical limits.
As described in this FAA Brochure (Acceleration in Aviation: G-Force):
Acceleration is described in units of the force called “Gs.” A pilot in a steep turn may experience forces of acceleration equivalent to many times the force of gravity. This is especially true in military fighter jets and high-performance, aerobatic aircraft where the acceleration forces may be as high as 9 Gs.
