"You can't feel gravity!"
Steve Seibel
www.aeroexperiments.org
steve at aeroexperiments.org
July 15 2005 edition
You can't feel gravity, because it simultaneously pulls on every molecule of
your body. Since gravity works "from the inside", not by pressing
against a few external surfaces, no internal stresses or strains are created.
So you can't feel gravity. If you are traveling in a vehicle, gravity
simultaneously acts on every molecule of the vehicle, and on every molecule of all
the contents of the vehicle, so gravity can never pull the contents of the
vehicle--including the driver or pilot--up or down or sideways toward the floor
or roof or side of the vehicle. More on this below. Also, you can't feel "centrifugal
force", at least as it is conventionally thought of, because it doesn't really
exist. You only can feel the genuine, external, physical forces
that are imposed upon your body by your surroundings. In an aircraft, you
feel only the real aerodynamic and thrust forces created by the
aircraft. In a car, you only feel the real, tangible forces exerted by the pavement against the tires, plus any significant aerodynamic
forces. And so on and so forth.
(Note: for convenience we're going to be a bit sloppy and use
"1-G" as a measure of force--equal to the weight in pounds of the
object of concern--as well as a measure of acceleration).
Examples:
1. Person standing on ground. Forces at play: 1-G downward force of gravity
pulling down on body, 1-G upward force of earth pushing up on feet. Net force:
0 G's. Net acceleration: 0 G's. Forces "felt" by the person: all of
the above except gravity. Net force "felt" by person: 1-G upward
force of earth pushing against their feet.
Unfortunately, common convention is to express the apparent G-load as if it
were created by some mysterious force within the body, not by the external
force of the floor pressing upward, and therefore we imagine that the usual
positive 1-G loading is somehow acting in the "downward" direction,
not the "upward" direction. Intuitively this seems to make some
sense, until we realize that this apparent G-load is caused entirely by the
floor pushing upwards on the body, and really has nothing whatsoever to do with
downward pull of gravity. Read on for more on this. But no matter, this is just
a convention. The true force "felt" by the person is actually the 1-G
push of the floor, in the upward direction.
2. Pilot in normal wings-level 1-G flight. Forces at play: 1-G downward force of gravity
pulling down on pilot and on aircraft, and 1-G upward lift force created by
wings. Net force: 0 G's. Net acceleration: 0 G's. Forces "felt" by
the pilot, which are also the forces imposing loads on the aircraft structure:
all of the above except gravity. Net force "felt" by pilot and by the
aircraft structure: 1-G upward lift force created by wings.
3. In wings-level flight, a pilot has shoved the control stick forward to "unload" his aircraft's wing to the zero-lift
angle-of-attack, and has set thrust exactly equal to drag, achieving "zero G" flight or weightlessness. Forces at play: 1-G downward force
of gravity, plus a forward thrust force, plus a rearward drag force. Net force:
1-G downward. Net acceleration: 1-G downward. Forces "felt" by the
pilot: all of the above except gravity. Net force "felt" by the
pilot, and by the aircraft structure, and by the G-meter in the pilot's
instrument panel: zero G's. (This is exactly what takes place in NASA's "vomit
comet" 0-g training aircraft. And note that the curving parabolic trajectory is the result, not the cause, of the fact that the aircraft is in a so-called "zero-G" condition. The actual cause of the so-called "zero-G" condition is the fact that the aerodynamic configuration of the aircraft--and in particular, the pilot's forward motion of the control stick--is keeping the wing at the zero-lift angle-of-attack.)
4. Pilot has rolled the aircraft to a 90-degree bank, and has shoved the control stick
forward to "unload" the wing to the zero-lift angle-of-attack. He is
using the rudder (if needed) to keep the nose pointing directly into the
airflow; i.e. the aircraft is not slipping sideways through the airmass. Again,
he has set the engine to create a thrust force that exactly equals drag. Forces at
play: 1-G earthward pull of gravity, plus a forward thrust force, plus a
rearward drag force. Net force: 1-G earthward. Net acceleration: 1-G earthward.
Forces "felt" by the pilot: all of the above except gravity. Net
force "felt" by the pilot, and by the aircraft structure and
contents: zero G's. The pilot doesn't "fall" toward the low side of
the aircraft and the slip-skid ball doesn't "fall" toward the low
side of the aircraft, despite the fact that the lift force is
"inadequate" in relation to the bank angle.
*The preceding paragraph is the key to understanding slips and skids! Let's pause for a few additional notes along these lines.
*The slip-skid ball--and the pilot's body--do not respond to a "lack of lift in relation to the bank angle" or to an "excess of centripetal force" or to the "downward pull of gravity", as is commonly believed. The slip-skid ball--and the pilot's body--only respond to the real, tangible aerodynamic sideforce created by the impact of a sideways component in the airflow or relative wind against the side of the fuselage and other components of the aircraft. If the nose of the aircraft is pointing in the same direction as the aircraft is actually moving through the airmass--i.e. if the aircraft is not slipping sideways through the airmass--then there will be no aerodynamic sideforce and the slip-skid ball and the pilot's body will not be deflected to the side. If the aircraft happens to be of the flying-wing type, with minimal cross sectional area as seen from the side, then even if a large sideways component is present in the airflow over the aircraft, this will generate only a small aerodynamic sideforce, causing only a small deflection of the slip-skid ball.
*The obvious next question is this: "If the pilot doesn't use the rudder to prevent a slip, or if the aircraft has no rudder, then will an 'inadequate pitch input' or a 'lack of lift' as the aircraft rolls into a turn or flies in a banked attitude tend to make the aircraft slide sideways through the airmass?" This question is explored in great detail elsewhere in this website; I've discovered that the answer is generally "no". See for example the article on this website entitled Notes for new hang glider and trike pilots--on sideslips, which has list of additional related links at the end.
*Continuing on...
5. Same as #4, but with a less extreme bank angle, and with a non-zero lift vector. Pilot has rolled the aircraft to a 60-degree bank angle, and has not moved the control stick aft to increase the wing's angle-of-attack and lift vector, so the wing is still generating 1 G of lift rather than the 2 G's that would be required for sustained, constant-airspeed, non-diving flight at this bank angle. Therefore the flight path is curving earthwards--the aircraft is beginning to dive and accelerate. We're taking a "snapshot" of the aircraft at a moment when there has not yet been any significant downward curvature of the flight path or increase in airspeed. Again, we'll assume that the pilot is using the rudder--if needed--to keep the nose pointing directly into the
airflow; i.e. the aircraft is not slipping sideways through the airmass. And again,
he has set the engine to create a thrust force that exactly equals drag. Forces at
play: 1G earthward pull of gravity, 1G lift vector acting at a 60-degree angle from the vertical (or 30 degrees above the horizontal), plus a forward thrust force, plus a
rearward drag force. Net force: a bit of trigonometry tells us that the 1G lift vector from the banked wing includes a .5G upward component and a .87G horizontal component. Therefore when we also include the 1G downward force of gravity, the net force vector works out to have a .5G downward component and a .87G horizontal component. This all sums up to equal a 1G force, inclined 60 degrees above the vertically downward direction, or 30 degrees below the horizontal. Net acceleration: .87G, in the direction described above. Forces "felt" by the pilot, and by the aircraft structure and contents: all of the above except gravity. Net
force "felt" by the pilot, and by the aircraft structure and
contents: since thrust and drag are equal, and gravity is not "felt" by the pilot and the aircraft structure and contents, the net force felt by the pilot and the aircraft structure and contents is the 1-G lift force from the wing. This force acts squarely "upward" in the reference frame of the aircraft and pilot. Therefore the pilot doesn't "fall" toward the low side of
the aircraft and the slip-skid ball doesn't "fall" toward the low
side of the aircraft, despite the fact that the lift force is
"inadequate" in relation to the bank angle.
*Additional notes: as in case #4, the key reason that the slip-skid ball, and the pilot's body, do not "fall" toward the low side of the aircraft is that there is no real, tangible, aerodynamic sideforce in the reference frame of the aircraft and pilot. The net aerodynamic force is acting "straight up" in the pilot's reference frame, so the pilot's body and the slip-skid ball remain centered. The pull of gravity on the pilot and the slip-skid ball is not relevant, because this pull is also acting on every molecule of the rest of the aircraft.
6. Same as #5, but now as the pilot rolls to a 60-degree bank, he applies enough back pressure on the control stick to increase the angle-of-attack and lift vector enough to create the "right" amount of lift for the bank angle (2 G's). Again, we'll assume that the pilot is making whatever rudder inputs are needed to keep the nose pointing directly into the
airflow; i.e. the aircraft is not slipping sideways through the airmass. And again,
he has set the engine to create a thrust force that exactly equals drag. Forces at
play: 1G earthward pull of gravity, 2G lift vector acting at a 60-degree angle from the vertical (or 30 degrees above the horizontal), plus a forward thrust force, plus a
rearward drag force. Net force: a bit of trigonometry tells us that the 2G lift vector from the banked wing includes a 1G upward component and a 1.73G horizontal component. Therefore when we also include the 1G downward force of gravity, the net force vector works out to be 1.73 G's, acting in the horizontal direction. Forces "felt" by the pilot, and by the aircraft structure and contents: all of the above except gravity. Net
force "felt" by the pilot, and by the aircraft structure and
contents: since thrust and drag are equal, and gravity is not "felt" by the pilot and the aircraft structure and contents, the net force felt by the pilot and the aircraft structure and contents is the 2-G lift force from the wing. This force acts squarely "upward" in the reference frame of the aircraft and pilot. Therefore the pilot doesn't "fall" toward either the low side or the high side of
the aircraft, and the slip-skid ball doesn't "fall" toward either the low
or the high side of the aircraft, despite the fact that the true net force acting on the aircraft and pilot is purely horizontal in the reference frame of the external world, and not squarely "upward" in the reference frame of the aircraft and pilot.
*Enough on slips and skids and turns...we'll now return to some simpler cases...
7. Pilot beginning a 2-G pull-up to enter a loop. Forces at play: 1-G
downward pull of gravity, 2-G upward pull of wings. Net force: 1-G upward. Net
acceleration: 1-G upward. Forces "felt" by the pilot: all of the
above except gravity. Net force "felt" by the pilot and by the
aircraft structure: 2-G upward.
8. A pilot, inverted at the top of a loop, has set the wing at the
angle-of-attack which will produce "positive" 1-G, in the aircraft's
reference frame, at that airspeed. Forces at play: 1-G earthward pull of
gravity, 1-G earthward pull of wing's lift. Net force: 2-G earthward. Net
acceleration: 2-G earthward. Forces "felt" by the pilot: all of the
above except gravity. Net force "felt" by the pilot and by the
aircraft structure: 1-G earthward, which is 1-G downward as seen by an external
observer, and 1-G "upward" in the reference frame of the aircraft and
pilot. This is the exactly the same G-loading, in the reference frame of the
aircraft and pilot, as exists in normal, "upright", unaccelerated
flight.
9. Pilot in clouds without visual references. Forces at play: whatever
aerodynamic forces are being generated by the aircraft, plus gravity. Net
force: vector sum of all of the above. Forces "felt" by the pilot:
all of the above except gravity. Net force "felt" by the pilot and
the aircraft structure: only the net aerodynamic force being generated by the
aircraft--gravity plays no role here. This is why, generally speaking, a
pendulum or bubble device is of no use whatsoever for maintaining orientation
when a pilot loses visual references, though in some particular cases the
slip-skid ball can give clues about the direction the aircraft is turning due
to some detailed relationships involving the aerodynamics of turns and
sideslip. But in general a pilot in clouds will have no way to "feel"
whether he is upright, inverted, etc. In general, during cloud-flying the best
clue that the aircraft has recently been inverted is that the wings are
departing the aircraft due to catastrophic structural failure during a high-speed
dive as the aircraft nears completion of a loop. This structural failure is
caused by the pilot pulling too hard on the control stick, so that the wing
develops too much lift (too many G's), or by aeroelastic flutter due to extreme
airspeed.
10. Astronaut near earth, but beyond earth's atmosphere; rocket motor is
producing 10 G's of thrust. Forces at play: 1-G downward pull of gravity, 10-G
thrust from motor. Net force: vector sum of thrust plus gravity; here we
haven't specified the direction of travel of the rocket so we can't quite
finish the math. Net acceleration: again this will be determined by the vector
sum of thrust and gravity. Forces "felt" by astronaut: all of the
above except gravity. Net force "felt " by the astronaut: 10-G.
11. Astronaut far from earth, at some hypothetical point in space where the
gravitational pull of all the stars, planets, etc happens to be completely
negligible; rocket motor is producing 10 G's of thrust. Forces at play: 10-G
thrust from motor. Net force: 10-G. Net acceleration: 10-G. Forces
"felt" by astronaut: all of the above except gravity. Net force
"felt " by the astronaut: 10-G, i.e. the same as when near the earth.
12. Astronaut near earth, but beyond earth's atmosphere; rocket motor is
switched off. Forces at play: 1-G downward pull of gravity. Net force: 1-G
downward pull of gravity. Net acceleration: 1-G downward. Forces
"felt" by astronaut: all of the above except gravity. Net force
"felt " by the astronaut: 0-G. If the spacecraft happens to be following an orbital trajectory, this
just means that the resulting downward acceleration happens to be precisely matched to the
craft's forward velocity in such a way that the flight path curves in such a
way that the distance from the craft to the curving surface of the earth remains exactly constant. On
the other hand if the spacecraft is initially completely stationary with respect to the
earth, then it will immediately begin plunging earthwards. In this case there
is no curvature in the flight path, so we are not tempted to
"explain" away the astronaut's 0-G perceptions by saying that "centrifugal
force" is "balancing" the pull of gravity.
13. Diver in water, ballasted to neutral buoyancy, and not moving with
respect to water. Forces at play: 1-G downward force of gravity pulling down on
body, 1-G upward force of water pushing up on body. Net force: 0 G's. Net
acceleration: 0 G's. Forces "felt" by the person: all of the above
except gravity. Net force "felt" by person: 1-G upward force of water
pushing up on body. Since this buoyant force is dispersed over the entire
surface of the body, it does not create very much stress or strain within the
body, and the resulting sensation is somewhat like 0-g weightlessness.
For more, see these related articles on the Aeroexperiments website:
"What makes an aircraft turn?"
Complete analysis of forces: fully balanced turn, turn with inadequate lift or G-load, slipping turn, non-turning slip, and skidding turn
