That "droppy" feeling

[labrys 0]

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The 'two bullets' thought experiment doesn't help us here because it ignores the powerful effects of aerodynamic lift and drag.

The "two bullets" example is not a thought experiment and it was used to point out that vertical and horizontal acceleration are independent of each other. It's a perfectly valid example of that fact. Lift and drag have nothing to do with it unless the bullets are different shapes or sizes.

Owned by Remi #?

[JohnMitchell 14]

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When a skydiver leaves the aircraft in When a skydiver leaves an aircraft with horizontal velocity he decelerates horizontally while accelerating vertically. These forces of acceleration and deceleration cancel one another and your body does not feel the "zero G" sensation.

Sorry to be so late getting into this thread. It's been a busy night at the radar scopes.

It's not that the forces "cancel each other", They are quite independent.

When you jump out of a plane, you immediately start to slow down along your horizontal path because of wind resistance, correct? This decelleration causes your body, including your inner ear, to undergo a positive G, not towards the Earth, but towards the forward horizon. For the first few moments out the door, "straight down" is towards the front of the plane, not just from a relative wind point of view, but from your gravity vector as well. As you continue to bleed off your forward speed and pick up your vertical speed, your gravity vector moves down towards the planet. It's not the lack of gravity, but the moving about of the vector that can be disorienting on exit.

If you exit at an airspeed equal to your terminal velocity, you feel exactly 1 G of wind force accelerating you backwards. Faster or slower exit speeds give you more or less G load. Zero airspeed, such as a balloon or hovering helicopter, give you zero G loading, therefore making your stomach "feel funny."

[michaelmullins 77]

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When a skydiver leaves the aircraft in When a skydiver leaves an aircraft with horizontal velocity he decelerates horizontally while accelerating vertically. These forces of acceleration and deceleration cancel one another and your body does not feel the "zero G" sensation.

Sorry to be so late getting into this thread. It's been a busy night at the radar scopes.

It's not that the forces "cancel each other", They are quite independent.

When you jump out of a plane, you immediately start to slow down along your horizontal path because of wind resistance, correct? This decelleration causes your body, including your inner ear, to undergo a positive G, not towards the Earth, but towards the forward horizon. For the first few moments out the door, "straight down" is towards the front of the plane, not just from a relative wind point of view, but from your gravity vector as well. As you continue to bleed off your forward speed and pick up your vertical speed, your gravity vector moves down towards the planet. It's not the lack of gravity, but the moving about of the vector that can be disorienting on exit.

If you exit at an airspeed equal to your terminal velocity, you feel exactly 1 G of wind force accelerating you backwards. Faster or slower exit speeds give you more or less G load. Zero airspeed, such as a balloon or hovering helicopter, give you zero G loading, therefore making your stomach "feel funny."

Indeed I had stated in my previous post that:
"Horizontal acceleration/deceleration and vertical acceleration/deceleration are totally independent events and one does not affect the other". However, it is a fact that the net result of the G of the two forces can cancel one another as far as the sensation your body feels (or make the sensation more positive or more zero G).

If you push a an aircraft over in flight you can get zero or negative G. If you bank the aircraft while pushing you can get the positive G of the bank to negate the less than 1 G of the push and what you feel is 1G and this is a common technique in high performance aircraft to avoid zero or negative G when you need to level off quickly.

What I said is the slower the airspeed of the exit the less positive G you feel (that "droopy" feeling when you are subjected to less than 1G) which is the same as you have stated. At faster airspeeds you will feel over 1 G.

Mike

[JohnMitchell 14]

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Now you hit the relative wind. The drag starts accelerating you backwards, trying to slow you down to zero speed with respect to the wind. That's an acceleration of about 10 feet per second, and you feel it acting towards the tail. That means you are feeling 1/3 of the gravitational force you usually feel and it is in an unusual direction. This is perceived by many as a feeling of falling.

Bill, I say that leaving the plane flying at your terminal velocity, let's say 120 mph, you'll feel not one third G, but 1 G of acceleration, since that's the wind force of a 120 mph wind on your body. That's, of course, the amount of wind it takes to balance the force of gravity in regular freefall, correct? So 120 mph wind force = 1 G.

As you well know, wind resistance is equal to the square of the speed, so a 168 mph exit would give you 2 G's, a 60 mph exit would give you a quarter G. I certainly agree with you about those high speed exits. You can feel the "oooph" when you hit the relative wind.

[JohnMitchell 14]

Sorry to be so late getting into this thread. It's been a busy night at the radar scopes.

Indeed I had stated in my previous post that:
"Horizontal acceleration/deceleration and vertical acceleration/deceleration are totally independent events and one does not affect the other".

Sorry Mike, you sure had. I had glimpsed the start of the thread early in my shift, then posted late last night w/o a good chance to read the whole thread. I'd buy beer, but it's not my first time.

I still find it interesting that our gravity vector moves so quickly when we jump out. We're so used to working "on the hill" we take it for granted. I try to explain to students that when you first exit, straight down is actually towards the front of the plane. We understand the relative wind principle. I guess I'd just like to add that gravity is relative to that wind too.

[bochen280 0]

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I'm re-posting this here to keep from taking off-topic the other thread in which it appeared.

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"There is little or no "droppy feeling" associated with skydiving from an airplane. That feeling comes from acceleration. Going from a slow speed to fast, etc. When you jump from an airplane going 80-90 mph, you're already going that fast when you exit. Speeding up to 120 mph over the next 10 seconds doesn't create much of a "droppy feeling"

Is that realy correct?

I think a lot of skydivers say they don't experience the droppy feeling because they have just gotten used to it over time, and it doesn't take very long. But initially, I think novice jumpers DO feel it for the first several jumps.

And I don't think it has anything to do with the horizontal speed of the aircraft or the wind when you exit. It's entirely a function of VERTICAL acceleration only. If you were moving horizontally at 120 mph on a roller coaster and suddenly hit a steep drop, I think you ARE going to feel the droppy feeling, simply because of the vertical drop alone. Likewise, if you jumped from a jet fighter flying horizontally at 600 mph, you would still feel the droppy feeling as you accelerated downward vertically, despite your high horizontal speed.

The droppy feeling comes from negative G-forces, i.e. being lighter than the pull of gravity.

G-forces are defined as x, y and z, where x is the forward direction, -x is backwards, y is to the left, -y is to the right, z is upward, and -z is downward. Or something like that. Positive z forces produce the "heavy" feeling as your weight is increased by more than normal acceleration. Negative z forces produce the "droppy" feeling as your weight is decreased by less than normal acceleration. That's the way I see it.

I think I agree that it IS about acceleration, but disagree that the DIRECTION of acceleration is irrelevant. And there is probably some human physiology involved too...

So, which view is correct?

I hope this was not partially making reference to something I posted at another post/thread earlier. If I was mistaken then I apologize but if this was indeed referring to what I think it may be.... then I should probably elaborate a little bit more on what exactly I meant by that.

I only have one first-hand empirical observation of one tandem skydiving, so that is one subjective data point is hardly conclusive evidence. Prior to skydiving/tandem I had in the past on several occasions read anecdotal accounts that although the "free fall" at terminal velocity segment of skydiving does not feel like falling or dropping or the zero-g sensation, it has been mentioned by some that the first couple of seconds of skydiving (from the moment one jumps out of the aircraft) has the same empty stomach droppy feeling that one gets when a roller coaster suddenly plummets to the ground, or akin to the first fraction of a second when a fast elevator first starts "going down", or a aircraft flying an inverse parabolic curve to match the rate of free fall in vacuum at 9.8 meters per second squared to attempt to simulate zero-gravity, etc....

That feeling is one that I find unpleasant and want to avoid. I think astronauts who spend extended time in space eventually become desensitized to that feeling, adapt and adjust and completely forget that the sensation is even there. In fact they have reported that it is only when they come back down to earth that they feel like they weigh like a ton of bricks.... since the zero-gravity has become the new "norm"...

But back to the topic. From a physics standpoint that ""droppy feeling"" you make mention to does not come from "acceleration" per se. It actually comes from a lack of acceleration forces being enacted on said individual, which ironically in a constant gravitational field (ie near earth) means yielding and physically accelerating towards the center of the gravitational object without anything like air resistance getting i the way. (when you step on the gas of your car you feel the increased "acceleration" forces but that doesn't give you the ""droppy feeling"", and imagine being in a powered elevator in a very tall vacuum chamber that "fell to the ground" at exactly 2g (19.6m/s squared) you would FEEL that up was down and down was up and that you were back on the ground again but the ground would actually be pointing toward the sky!... thus it is not acceleration that gives us the "droppy feeling" but a total LACK of any acceleration or vector forces) That is, after all, what "zero-g" is all about. When the summed up composite of all vector forces being enacted on a body cancels out and/or approximates to zero, that is when the "zero g" sensation is highest. It is actually the resistance against falling to earth that creates the sensation of acceleration forces. It may seem counter-intuitive but in essence we are (because we are in the presence of earth, if we were in deep space far away from any huge gravitational object that would be a different story) experiencing acceleration forces when we are stationary and NOT physically "accelerating" and in fact we experience "no acceleration forces" when we are in free fall (in perfect vacuum) and physically accelerate at exactly 9.8m/s2 downwards toward center of earth. When you are stationary and standing on the ground of the earth or sitting in the confines of an airplane (etc) there are platforms holding you up and not allowing you to "fall" at the say if you were in the vacuum of space around earth unconnected and totally suspended in "mid air/space" without being tethered to anything else. So there is a constant 9.8m/s2 force of (gravitational) acceleration on your body. Our body and mind mentally adapts and adjusts to that natural value.... so whenever that force "lessens" we feel that droppy uncomfortable sensation... Think about when you drive your car very fast down a hill... sometimes for a moment you feel the car almost beginning to lose its grip from the ground or suspend in air... and that temporary "lessened gravity" experience gives you like lightheaded, floating, "droppy feeling"...

This is why the first few seconds of skydiving theoretically SHOULD feel like "falling" or the "droppy feeling"... if we were on the moon or earth lacked an atmosphere then this "droppy feeling" would never go away until we impacted with the ground (and of course parachutes would not work to slow terminal velocity down from 120mph to 20mph!) Since force of gravity (acc) is constant at 9.8m/s squared, this means velocity increases at a constant rate of g (~9.82 meters per second) and position increases (or altitude decreases in the case of skydiving) at an exponential rate (all this is in ideal conditions of perfect vacuum without atmosphere) .... but because earth does have air and atmosphere what effectively happens is after a few seconds of "falling" (or rather approximate "free falling") the velocity increases to a certain amount whereby the friction caused by the air molecules rubbing along the surface area of the skydiver starts to impede his otherwise constant rate of increasing downward velocity and exponential rate of altitude loss.... and the frictional forces continues to get stronger until it completely counteracts and balances out and cancels out the effects of gravity acc forces on his body. (in essence he becomes a parachute, only difference is the parachute he carries on his back offers a LOT more surface area and thus slows his terminal velocity down from suicidal 120mph to a survivable 20mph) So terminal velocity is a function of surface area of his body (and whatever is attached and connected to it), his total mass, (basically density) and the thickness of the air molecules surrounding him plus the instantaneously gravitational forces (which for the purposes of skydiving, doesn't really change that much) .... In fact it is a misnomer... we tend to think of "terminal velocity" stage of skydiving as "Free falling"... when in fact it is the other way around... the physics term of "free falling" would be more apt to be applied to the first few seconds of jumping out of an airplane and the rest of the way down at "terminal velocity" with not be a true "free fall" because the body is once again suspended on a platform (this time by the friction of this "air cushion" rather than say on an airplane or in an elevator, or on the ground or in a skyscraper, etc) and thus that is why during the skydiving definition of "free fall" (terminal velocity) the "droppy feeling" is gone.... not because there is no more increase in velocity (which is true, pretty much levels out at around 120mph when in arch position) but because like standing on the surface of the ground on earth, the skydiver at terminal velocity in the air also once again feels a constant vertical gravitational acceleration of -9.8m/s2. It is only the first few moment when he first jumps out of an airplane when his downward velocity has not yet had the time to encounter enough air resistance that theoretical he SHOULD feel lthe "droppy feeling" caused by a DECREASE in gravitational forces on his body DUE to his "giving in to gravity" by "free-falling" towards the earth.

Jumping out from a skyscraper (or base jumping) or falling from a very tall ladder WOULD probably give one the feeling of that "droppy feeling" for the first couple of seconds... because unlike jumping out from a moving aircraft there would be no vertical movement or temporary lateral forces caused by the relative wind / frictional forces / etc.

In fact as far as the body is concerned, definitions of "up" and "down" and "vertical" or "lateral" are meaningless. Visual acuity notwithstanding, the human body itself cannot distinguish between "natural force of gravitational pull" and other artificial means of acceleration forces. The "droppy" zero-g sensation only occurs /happens when there is exactly that, near ZERO (significantly less than 1g aka 9.8m/s2) NET acceleration forces being applied on the human body...

When exiting from a 90mph moving aircraft, the first couple of seconds those lateral wind forces DENY the skydiving the brief experience of "zero g" sensation because there are artificial acceleration forces enacted upon him (well actually he is decelerating laterally from 90mph to zero mph the forward velocity of the aircraft due to again air resistance) and thus momentarily when he left the plane his BODY might feel as if the earth rotated 90 degrees and he was still subjected to "gravity"... and then a short couple seconds later when his horizontal spend decreases to near zero (unlike the airplane, he wasn't moving much "forward" anymore) by that time his downward vertical spend would have increased to nearly 120mph to offer enough air resistance to give him back the feeling of 1 g (one gravitational force) just like being back on the ground surface of earth. Therefore, in this configuration, at no time during the fall was he ever in even an approximate zero-gravity situation in which he felt no acceleration forces being acted upon him, and the total vector of said acceleration forces is a combination of lateral forces (wind resistance from the forward velocity of the aircraft) that switched over to vertical forces (wind resistance from reaching terminal velocity)

[bochen280 0]

Also, I forgot to add and wanted to point out... there is nothing "special" about the earth's gravitational pull... in reality there is no distinction between "vertical" forces and "lateral" forces.... forces are all forces. And there is no real "up" or "down"... and even altitude is a matter of attitude. It is all a matter of relative frame of reference.


EDIT:

Also I'd like to make mention of the following.... Although 90mph relative winds is not enough to make up for the sudden loss of 1g when instantaneously jumping out of the aircraft, the difference in this form of skydiving from moving aircraft scenario **vs** the roller-coaster plunge, sudden elevator drop, falling from tall building, etc is that in the former case the ~0.8(approx) g is felt almost immediately upon exit of the aircraft (infact the "acceleration" forces peaks at its highest/strongest/most-intense the moment one exits the plane and then quickly tapers down to zero) whereas in the latter case the zero g is felt almost immediately...

It is the difference between going from 1 g to ~0.8 g (a loss of only 0.2g) to going from 1 g to 0 g (a loss of entire 1.0g)...

The lateral "acceleration" (one could call this "artificial gravity") forces is MOST strong the moment one first exits the moving plane and then slowly reduces to zero but during that time it has given the "vertical" acceleration forces the opportunity to play "catch up" and supplement the loss of lateral forces. In essence it is gradual and gentle throughout...

Whereas in typical freefall from non-moving platforms the zero-g sensation is HIGHEST precisely at the moment of freefall and then reduces after a few seconds... It is this stark juxtaposition that gives us the more immediate intensity of this droppy feeling. It is a "sudden" sensation that catches us by suprise.

It is this contrast that makes all the difference. Effectively this is why one feels the "droppy feeling" much more intensely in a roller coaster ride than say skydiving from moving airplane.

[bochen280 0]

Sorry for being so long winded... here is more concise reply.

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I'm re-posting this here to keep from taking off-topic the other thread in which it appeared.

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"There is little or no "droppy feeling" associated with skydiving from an airplane. That feeling comes from acceleration. Going from a slow speed to fast, etc. When you jump from an airplane going 80-90 mph, you're already going that fast when you exit. Speeding up to 120 mph over the next 10 seconds doesn't create much of a "droppy feeling"

Is that realy correct?

The "droppy feeling" comes from a lack of (net) acceleration forces being enacted on a body. It (that zero g feeling) is "only" present when the total 3-dimensional sum composite of all the vector acceleration forces being acted on any given individual comes out to be near zero. Which in the case of being on earth usually means physically accelerating towards the planet's gravity center at a rate that cancels out the relative affects of said gravitational planet, which again is complicated by the fact that as far as traditional skydiving goes, there are other temporarily acc forces at play that also acts on the body in addition to the earth's forces (ie atmosphere, relative winds caused by forward moving aircraft, etc)

I think a lot of skydivers say they don't experience the droppy feeling because they have just gotten used to it over time, and it doesn't take very long. But initially, I think novice jumpers DO feel it for the first several jumps.

Desensitization plays a part of it. But there is more to it than just that.

And I don't think it has anything to do with the horizontal speed of the aircraft or the wind when you exit. It's entirely a function of VERTICAL acceleration only. If you were moving horizontally at 120 mph on a roller coaster and suddenly hit a steep drop, I think you ARE going to feel the droppy feeling, simply because of the vertical drop alone. Likewise, if you jumped from a jet fighter flying horizontally at 600 mph, you would still feel the droppy feeling as you accelerated downward vertically, despite your high horizontal speed.

Actually, he would not feel the "droppy feeling" at all. If anything, to his body the earth would have suddenly oriented 90 degrees around and gravity would have gotten much stronger than 9.8m/s squared. If you think about it, "vertical" and "horizontal" is just a matter of body position.

The droppy feeling comes from negative G-forces, i.e. being lighter than the pull of gravity.

There is actually no such thing as "negative G-forces", strictly speaking. We can have micro-gravity or almost near zero-gravity but not "negative G-forces". Negative G-Forces is really just acceleration in the opposite direction. "Deceleration" is simply inverse directional acceleration. Probably what you mean by "negative G-forces" is a reduction of g-forces towards zero....(but never quite reaching EXACTLY zero) because TOO much "negative" g forces simply turns out to be positive G forces in the opposite direction. And it is not so much about being lighter than the pull of gravity (?) as it is about "giving in to the pull of gravity"

G-forces are defined as x, y and z, where x is the forward direction, -x is backwards, y is to the left, -y is to the right, z is upward, and -z is downward. Or something like that. Positive z forces produce the "heavy" feeling as your weight is increased by more than normal acceleration. Negative z forces produce the "droppy" feeling as your weight is decreased by less than normal acceleration. That's the way I see it.

There is no absolute x, y, z. It is all relative to your own frame of reference.


I think I agree that it IS about acceleration, but disagree that the DIRECTION of acceleration is irrelevant. And there is probably some human physiology involved too...

It is not so much about acceleration as it is about the lack of net acceleration forces being applied to a body (which usually on or near earth it means accelerating towards the gravity center of the most predominant force of gravity, basically the planet earth) ... And "directional" is largely meaningless and irrelevant in terms of how you are applying it here... to the human body it makes no difference if the force is coming from earth or a centrifuge. Forces are forces, direction is relative to the individual only.

So, which view is correct?

[bochen280 0]

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Now you hit the relative wind. The drag starts accelerating you backwards, trying to slow you down to zero speed with respect to the wind. That's an acceleration of about 10 feet per second, and you feel it acting towards the tail. That means you are feeling 1/3 of the gravitational force you usually feel and it is in an unusual direction. This is perceived by many as a feeling of falling.

Bill, I say that leaving the plane flying at your terminal velocity, let's say 120 mph, you'll feel not one third G, but 1 G of acceleration, since that's the wind force of a 120 mph wind on your body. That's, of course, the amount of wind it takes to balance the force of gravity in regular freefall, correct? So 120 mph wind force = 1 G.

As you well know, wind resistance is equal to the square of the speed, so a 168 mph exit would give you 2 G's, a 60 mph exit would give you a quarter G. I certainly agree with you about those high speed exits. You can feel the "oooph" when you hit the relative wind.

It would still probably be less than one full g because you are not exiting in arch position standing up parallel to and facing the front of the aircraft. Given the same initial relative wind velocity and less surface area contact (since the wind is hitting you sideways) it would probably be less than a full earth gravity unit of force. Conversely, if you were skydiving and "free-falling" on your sides, your new terminal velocity would be a lot faster than 120mph.

[billvon 2,379]

>It would still probably be less than one full g because you are not exiting
>in arch position standing up parallel to and facing the front of the aircraft.

Doing any kind of normal exit at your terminal velocity will give you one G. For 4-way, for example, the goal is to present the formation to the wind so that they are "belly to wind." Thus their drag is similar, and at terminal they always feel an acceleration of 32 fps^2.

However you could do "tricks" to get around this, like launch a head-down at belly terminal. This would indeed result in less than one G because you're not at _your_ terminal speed.

[JohnMitchell 14]

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It would still probably be less than one full g because you are not exiting in arch position standing up parallel to and facing the front of the aircraft.

Yes I am. Or diving out, belly into the relative wind. Same thing, same flat presentation.

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Given the same initial relative wind velocity and less surface area contact (since the wind is hitting you sideways) it would probably be less than a full earth gravity unit of force.

That's what we call a "poor exit".

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Conversely, if you were skydiving and "free-falling" on your sides, your new terminal velocity would be a lot faster than 120mph.

True. And head down and sit flying people experience even less "wind G load" because of their higher terminal velocities than belly fliers. Notice that in all my explanations my exit speed equaled my terminal velocity.

[MariusM 0]

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You are right, acceleration is a vector. However the inner ear/vestibular system does not change that. Think about RW... you are clearly moving, in freefall, and have absolutely no sense of that. You are flying with other people.... no sense of falling. Even when tracking, you have changed direction, still no sense of falling despite adding horizontal vectors.

There is no or very small vertical accelaration in freefall (assuming speed is already at terminal), so your vestibular system doesn't feel anything. But you do feel accelaration if you turn 360 degrees. At least if you do it fast enough. Same as being in the car and going through the corner. If you pass the corner slowly, you notice nothing. If you'd pass that corner fast enough (even if speed remains the same all the time), you get pushed in the outer direction and you definately feel it.

[DaVincisEnvy 0]

A lot of confusion in this thread seems to come from a lack of definition of the frames of reference involved.

From the frame of reference of the earth, a skydiver who exits an aircraft experiences an immediate vertical acceleration (due to the force of gravity) and an immediate horizontal deceleration (due to the drag force). Vertical acceleration continues until terminal velocity is reached and horizontal deceleration continues until the skydiver is moving with the same velocity as the horizontal wind.

However, the human body does not experience sensation from the frame of reference of the earth. Rather, it experiences sensation (e.g. acceleration) relative to its own orientation. From the frame of reference of the human body exiting a turbine aircraft flying at 100mph (in a belly-to-relative-wind orientation) there is very little acceleration.

Assume that, in the body's frame of reference, "down" is defined by an arrow extending out of the belly button and, if you remain stable, is always pointing into the relative wind.

When you exit an aircraft and present to the relative wind, you have a "downward" (away from the belly) speed equal to the aircraft's horizontal speed.

While on the hill, your body rotates relative to the reference frame of the earth, but not relative to the reference frame of the body. If you remain stable, "down" (defined by that arrow out of the belly button) remains in the direction of the relative wind.

After a few seconds of stable freefall, your velocity is approximately 120mph "down".

The net acceleration experienced in the frame of reference of the human body in a stable belly-to-relative-wind orientation is:

[(terminal speed) - (horizontal speed of the aircraft)]/(time needed to accelerate to terminal speed)

If you exit from a fast-flying aircraft, the difference in speed is minimal and, therefore, so is the acceleration in the body's reference frame.

If you exit from a helo or balloon, the difference in speed is significant and, therefore, so is the acceleration.

[billvon 2,379]

>a skydiver who exits an aircraft experiences an immediate vertical
>acceleration (due to the force of gravity) and an immediate horizontal
>deceleration (due to the drag force).

He experiences no vertical acceleration; at the instant he exits is in a true "freefall" at least in the vertical axis. He does feel a forward acceleration from the drag. This vector rotates during his exit as he slows down horizontally (= less acceleration from drag) and increases his speed vertically (= more acceleration from drag.) Eventually it is back at 1G pointing upwards.

Thus someone who exits a plane at 120mph experiences 1G and not 1.4G's.

>The net acceleration experienced in the frame of reference of the human
>body in a stable belly-to-relative-wind orientation is:

>[(terminal speed) - (horizontal speed of the aircraft)]/(time needed to
>accelerate to terminal speed)

A few problems here. First, that's a scalar, and acceleration is a vector. For example, if you exited from an airplane, deployed a powered wing (think Yves Rossy) then flew back into the plane, your net scalar acceleration (in any direction) would be zero. But you'd still experience various accelerations along the way,

Secondly, we all experience acceleration every day of our lives; it's the 1G that gravity provides, and that acceleration is transmitted through the floor or our chairs as the normal force opposing gravity. It is the _absence_ of that acceleration we notice during low speed exits.

[robinheid 0]

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...there is probably some human physiology involved too...

Bingo.

Unless I missed it, everyone so far in this thread has forgotten to consider Occam and his razor in their dissertations.

You don't get "rollercoaster stomach" when you jump from a plane or a building or a balloon because you're operating at 1 G.

You do get "rollercoaster stomach" on a rollercoaster because as you go from 1 G to zero G, the urine in your bladder stays where it is in space while your body moves downward through that same space -- causing the top of your bladder to collide with the floating urine, thus causing that tingly "droppy" feeling routinely mis-identified as being in your stomach.

44

SCR-6933 / SCS-3463 / D-5533 / BASE 44 / CCS-37 / 82d Airborne (Ret.)

"The beginning of wisdom is to first call things by their right names."

[DaVincisEnvy 0]

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He experiences no vertical acceleration; at the instant he exits is in a true "freefall" at least in the vertical axis.

Freefall is not the absence of acceleration, gravitational or otherwise. Even astronauts in "freefall" orbit around the earth are experiencing an acceleration toward the earth. In fact, the force of gravity provides the centripetal acceleration necessary to keep the astronauts in orbit rather than flying off on a tangential path.

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A few problems here. First, that's a scalar, and acceleration is a vector.

True and already accounted for in the definition of the body's reference frame. Since the body's direction of travel (the vector portion of velocity) in its own reference frame is always "down" (as defined by the arrow out of the belly button) during a stable exit-to-freefall transition, there is no change in direction within the body's reference frame. Therefore, only changes in the speed (magnitude portion of velocity) contribute to acceleration.

[JohnMitchell 14]

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You don't get "rollercoaster stomach" when you jump from a plane or a building or a balloon because you're operating at 1 G.

Robin, sorry but you're wrong. Jumping at zero airspeed (balloon, hovering helo or BASE) will definitely have you experiencing almost zero G until you build up airspeed. It's simple physics.

[JohnMitchell 14]

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Freefall is not the absence of acceleration, gravitational or otherwise. Even astronauts in "freefall" orbit around the earth are experiencing an acceleration toward the earth. In fact, the force of gravity provides the centripetal acceleration necessary to keep the astronauts in orbit rather than flying off on a tangential path.

It's simpler if you just remember that gravity is a curvature of space-time that we perceive as a force. The astronauts in orbit are traveling in a straight path thru curved space.

[oldwomanc6 38]

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You don't get "rollercoaster stomach" when you jump from a plane or a building or a balloon because you're operating at 1 G.

Robin, sorry but you're wrong. Jumping at zero airspeed (balloon, hovering helo or BASE) will definitely have you experiencing almost zero G until you build up airspeed. It's simple physics.

I agree that jumping from a slow or non moving object will indeed result in the "elevator drop" sensation. Helo jumps for me, and I'm surprised that no one has brought up a cut-away. Definitely had the big droppy feeling when I did one.

But, Robin's pee explanation was enlightening and entertaining.

lisa
WSCR 594
FB 1023
CBDB 9

[EOCS 0]

The only time i ever felt a droppy feeling was on my first, and hopefully last, cutaway. Was a line over on a large stable canopy so was more or less not moving downward. after the cut there was like 2 sec of falling feeling them BAM canopy overhead.

Other then that i dont remember anything like that on my first jumps, too much wind already i think

[billvon 2,379]

>Freefall is not the absence of acceleration, gravitational or otherwise.

Actually it is. Gravitational acceleration means that you feel 1G of acceleration when you are standing on the ground. Remove that and you're in freefall.

(Note that freefall from a physics perspective is what we experience for a few seconds on a balloon jump; terminal velocity is not true freefall because you have normal gravitational acceleration, provided by drag.)

>Since the body's direction of travel (the vector portion of velocity) in its
>own reference frame is always "down" (as defined by the arrow out of the
>belly button) during a stable exit-to-freefall transition, there is no change
>in direction within the body's reference frame.

Agreed. But there are rapid orientation changes and we do sense that. That's part of people's disorientation on exit; the rapid change in that acceleration vector.

[DaVincisEnvy 0]

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Actually it is. Gravitational acceleration means that you feel 1G of acceleration when you are standing on the ground. Remove that and you're in freefall.

You are still experiencing 9.81m/s^2 (or "1G") of acceleration in freefall. That acceleration is responsible for your increase in vertical speed. If you experienced no vertical acceleration on a skydive, you would literally hover at altitude (i.e. your velocity would remain constant).

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terminal velocity is not true freefall because you have normal gravitational acceleration, provided by drag.)

Drag does not cause gravitational acceleration. The mass of the planet earth interacting with the mass of the skydiver through the force of gravity causes gravitational acceleration.

Drag can only cause deceleration because it acts always to oppose the motion of an object. At terminal velocity, a skydiver experiences no net acceleration because the downward force of gravity is precisely balanced by the upward force of drag.

You're absolutely right that falling at terminal velocity is not true free fall, although for a different reason than you stated earlier. The skydiving term "freefall" is actually different from the physics definition of "free fall". In pure free fall, there is no wind resistance (no drag), and the only force acting on an object is the force of gravity. The object therefore accelerates with a magnitude of F/m (where F is the force of gravity and m is the mass of the object) in the direction of the force of gravity. Near Earth's surface, the acceleration due to gravity is fairly constant and given by 9.81 m/s^2.

Quoting Physics for Scientists and Engineers, Volume 1, 4th Ed., by Paul A Tipler, pg. 87:

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If we drop an object near the earth's surface, it accelerates toward the earth. If we neglect air resistance, all objects have the same acceleration, called the acceleration due to gravity, at any given point in space. The force causing this acceleration is the force of gravity on the object, called its weight. If weight is the only force action on an object, it is said to be in free fall.

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Agreed. But there are rapid orientation changes and we do sense that. That's part of people's disorientation on exit

Very true, but I specifically emphasized in my analysis that it applies to a stable exit-to-freefall transition during which the skydiver maintains a belly-to-the-relative-wind (and therefore belly-in-the-direction-of-motion) orientation the entire time. We do not feel disorientation during such exits precisely because our orientation to our direction of travel does not change.

However, during a funnel or other "fun" exit, we certainly do experience changes in orientation and certainly do feel it (or at least I do).

[robinheid 0]

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You don't get "rollercoaster stomach" when you jump from a plane or a building or a balloon because you're operating at 1 G.

Robin, sorry but you're wrong. Jumping at zero airspeed (balloon, hovering helo or BASE) will definitely have you experiencing almost zero G until you build up airspeed. It's simple physics.

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You don't get "rollercoaster stomach" when you jump from a plane or a building or a balloon because you're operating at 1 G.

Robin, sorry but you're wrong. Jumping at zero airspeed (balloon, hovering helo or BASE) will definitely have you experiencing almost zero G until you build up airspeed. It's simple physics.

C'mon, John. I love ya, but it is simple physics and you're making it complicated.

I just responded to your comment about the role of physiology in that "droppy feeling" and now you're conflating airspeed and gravity.

Simple physics, John:

1. Zero airspeed does not equal zero gravity.

2. Zero airspeed equals zero airspeed.

3. Zero gravity is achieved only when the centrifugal force on a mass offsets the gravitational attraction to it.

So let me rephrase what I said above:

You do get "rollercoaster stomach" on a rollercoaster because as the centrifugal force of the rollercoaster going over the top of its hill offsets gravity, the urine in your bladder stays where it is in space while your body moves downward through that same space -- causing the top of your bladder to collide with the floating urine, thus causing that tingly "droppy" feeling routinely mis-identified as being in your stomach.

That's the only way you get to "almost zero G" -- "jumping at zero airspeed" has exactly zero to do with it.

It's simple physics.

44

SCR-6933 / SCS-3463 / D-5533 / BASE 44 / CCS-37 / 82d Airborne (Ret.)

"The beginning of wisdom is to first call things by their right names."

[JohnMitchell 14]

You feel no gravity when you're falling, Robin. You only feel the G force when you resist it by falling thru air at terminal velocity, standing on solid ground, sitting in chair, floating on (or in) a body of water.

Are the astronaut trainees not weightless during the parabolic "zero G" rides in the Vomit Comet? They're are certainly not at orbital velocities. It's just that their aircraft is in a ballistic trajectory. It's wings are not creating lift, the plane is not resisting gravity. BTW, the weightlessness for them starts during the pushover from the initial zoom climb, while they are still traveling upwards. Wrap your head around that.

Imagine being in an elevator when someone cuts the cable. Will you not feel weightless inside the cabin as you plunge to the basement?

A quote from you "Zero airspeed equals zero airspeed. " Yes, and as long as you stay on the top of the building or in the balloon, you'll feel 1 G. But as soon as you leave you will feel weightess, zero G, until you're falling fast enough that the wind resists gravity for you. I know you have way many more base jumps than I do, but I've got my share of cutaways and helo jumps. I feel weightless on them.

Once again, gravity is a tricky thing to think about. We're under its influence throughout the universe, but we only feel it when we resist it. Remember, too, that any constant acceleration is indistinguishable from the acceleration of gravity. Once again, it's our perception of gravity as a force that causes our confusion.

[rss_v 0]

The theory has been discussed further than I can usefully add to, but if I may offer my practical experience for anyone scanning the thread and wondering what they might feel:

As a novice with only ten skydives to my name, I've never felt "that droppy feeling" when leaving the aircraft. The only time I have during the jump is when releasing a strong flare under canopy (high up, not for landing) and when you then surge forward it feels like that.

I haven't done any BASE jumps but I have jumped from high bridges into deep water, and I most certainly felt it then.