Can YOU see something wrong here??

[evilivan 0]

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....average BASE jumper.....

HEY! Not happy with that comment

"If you can keep your head when all around you have lost theirs, then you probably haven't understood the seriousness of the situation."
David Brent

[tfelber 0]

Here are some things I posted on Blinc regarding this issue:

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Re: Pc In Tow Video

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I have to agree with Johnny on this. If you take a PC and drop it in still air without a force being applied to the bridle it will probably not inflate. Rather it will tend to fly in the direction of least resistance or most mass. This would tend to cause the PC to fall to one side.

When you apply force to the bridle however, the PC acts as a scoop and catches air. This scooping action deccelerates the PC until it has obtained an optimal volume. At this point the PC flies as a solid object and because the PC is symmetrical and the force is centralized about the apex the PC acts as a brake.

IMO, the very minuscule vacuum created by a PC traveling through air could hardly counteract the force of gravity on the material, much less pull a canopy, weighing many times it's own weight, from a container. Maybe I misunderstood your original post???

It seems to me that the primary function of a round canopy or PC is to present an even, controllable surface in the direction of movement to resist the force of gravity. The air inside the parachute is still and the addition of vents is to increase the stability of the surface presented to the direction of motion or to create the ability to steer the canopy.


After further research I found this:

"Because of the system's rapid descent, air rushes in through the tube's opening and accumulates at the apex of the canopy to create a high pressure air "bubble". Steady inflow continues to build up internal pressure, thus allowing the bubble's volume to expand horizontally as well as vertically....Like for any blunt objects moving through air, wake turbulence generated on the downwind side of the parachute causes the external pressure to be lower than the internal pressure near the apex....On the other hand, rapid expansion generates a large external pressure which squeezes the bubble on its upwind side, thus slowing down the expansion." - 'Parachute Inflation' By Dr. Jean Potvin

More can be found[URL]http://www.pcprg.com/inflate.htm]here[/URL]

This seems to support my theories regarding inflation, but I'm still wrestling with the dynamics of a round canopy once it has inflated...

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Re: Pc In Tow Video

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It's not contradictory but it lends credibility to some of what you mentioned in your post. From the quote I included in my previous post, the external pressure being less than the internal pressure aids in filling the canopy, but the bridle pulling the mouth of the canopy through the air captures the greatest volume of air and therfore seems to be the primary contributor to the inflation. If the mouth was covered with F111 fabric, as you previously mentioned, rather than the mesh we see on most PC's the inflation time would seem to be much greater, probably proportional to the diameter of the hole in the fabric.

For instance, look at pulling a trash bag through a swimming pool. There is very lttle resistance to the motion of the trash bag until it is full of water and the action of pulling the bag through water does little to induce a change in pressure at the apex of the bag until the bag is full. Once the bag is full the resistance to movement of the bag through the water is proportional to the diameter of the mouth of the bag and has little to do with any "vacuum" created on the downwind side of the bag.

In air, the system is more dynamic in that air is compressible. Now the difference in external vs. internal pressure may cause "surging" as an equillibrium pressure is found. But, once the PC is inflated, I would think these small changes in pressure can be ignored and the PC would experience forces similar to the trash bag in water.

These opinions are based on my observations and a very little amount of research (two articles from one Google search). I am in no way saying this is how it works. I'm just expressing my thoughts. As you said, this is an interesting and eye-opening discussion and our lives depend on this phenomenon.
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[JohnnyUtah 0]

I dont have a PhD, and I cant rattle off a bunch of scientific terminology or theories, but I try to maintain a simple understanding of this stuff. This conversation has really developed, so here are some more comments of the way I understand it all. I could be wrong or I could be right, or maybe a little of both.
Bottom line, get a parachute open before hard deck.

Some basics:

Square canopy
A square canopy is an airfoil, a wing, a glider. The pressurized wing generates lift because it is an airfoil shape. In this thread, round canopies seem to be the topic, so…

Round canopy
A round canopy is a drag device. The canopy captures air as it is pulled in a direction. This capturing of air pressurizes the canopy with static pressure holding the round canopy in its inflated shape. The shape of the round canopy creates drag, which is the opposing force to whatever is pulling the payload in the said direction (i.e. gravity). It is this drag that allows a person to land with a safe rate of decent under a round parachute.

Inflation of a round
For a round parachute to be considered inflated, it needs to be pressurized with air. For air to get into the round parachute, it goes in through the opening at the skirt, not the apex. Air exits through any hole at the apex.
During a round deployment with some airspeed, the canopy starts out in a stretched out streamer type of shape. The skirt of the canopy is gathered together in the center and therefore the opening at the bottom is very small if any. Because there is some airspeed, the airflow across the canopys outer surface creates a low pressure. Since there is a lower pressure on the outer surface of the canopy than on the inside of the canopy, the streamer shape begins to expand. The opening at the skirt begins to expand as well.
As this happens, air does go in there.
The air keeps collecting inside there, and it is at the apex of the canopy where it collects for the most part. Once the opening at the skirt becomes large enough to let in enough air to completely expand open and pressurize the canopy, then that is what happens.

The low air pressure on the outside of the canopy during initial inflation, helps the canopy change from a streamer shape to an expanded, more open shape.

IMHO- If a round parachute were deployed with its skirt completely opened up (full diameter), then it would inflate (pressurize) immediately and the need for an external low-pressure to help expand the streamer and open the skirt up (as during initiating inflation), would not be needed. I agree that the static pressure created by the canopy capturing air is greater than the external pressure and therefore the canopy stays open.

Pilot Chutes
I have no doubt in my mind that when doing a high airspeed deployment, there is a low pressure created on the outside of the PC which helps it to initially expand. It seems to work very well and that is why I use a regular mushroom for those types of jumps.

In this thread we were talking about a very low airspeed PC hesitation. My point of view is that if you can get the mesh/rip-stop seam of the PC opened quicker (and reliably) on its own, then air will go right in there and pressurize the PC that much quicker. That is why I was mentioning the Super Mushroom. Thats its purpose.

If you pack the PC so that it is dependent on airspeed to begin to open, then you need sufficient airspeed at deployment time or enough altitude to get away with a hesitation (like we see in the video).

With the regular mushroom, sometimes you can do a throw and go and the PC opens immediately with nil to very little airspeed. I believe this works because enough air successfully flows through the opening of the seam and successfully inflates it. However, sometimes the airflow is not so successful to get in and inflate the PC at low airspeeds with the regular mushroom. The result is an occasional hesitation. In the video, it probably hesitated until an external low pressure did happen and helped get the PC to open. Not ideal in my opinion.

Have Fun, Don't Die!
Johnny Utah
My Website
email:[email protected]

[460 0]

hey no problem. i think an intuitive understanding is generally the best understanding. i'll research the issue if anything dramatically differs from these common sense aerodynamics.

Looks like a death sandwich without the bread - Steve Deadman Morrell, BASE 174

[wwarped 0]

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I am confused. My understanding is essentially that the principle is a simle expression relating pressure of a fluid to its velocity, and roughly explains certain aspects of wings, etc. Is there a reference you can point me to confirm your thoughts or mine? My background is not in air flow theory but in atomic theory.

I pulled out the most basic book on aerodynamics on my shelf and created a response. Bernoulli's equation is fairly easily derived from Newton.

too bad I did not see an e-mail address listed in your profile, otherwise I'd PM you.

don't think the forum wants to see THAT much detail...

btw, my background is in Aeronautical/Astronautical engineering and I too have worked for NASA.

DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

[TomAiello 25]

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don't think the forum wants to see THAT much detail...

I'd like to see it. Either here, or via email or PM.

Thanks!

-- Tom Aiello

[email protected]
SnakeRiverBASE.com

[wwarped 0]

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I'd like to see it. Either here, or via email or PM.

it may be a bit off topic, but at the moderator's request...

I pulled out my first textbook regarding aerodynamics.
(Introduction to Flight by John D. Anderson, Jr. 1978, McGraw-Hill)

section 4.3 on pages 69-73 discuss momentum along a streamline. (a stream line is the path along which an infinitesimally volume of air flows.)

mass:
the volume of the piece of air is dx dy dz.
therefore it's mass = ro(dx dy dz) (ro is the density of air)

acceleration:
a = dV/dt = DV/dx dx/dt = DV/dx V (V is velocity, t is time)

Newton:
F=ma

-dp/dx (dx dy dz) = ro(dx dy dz)V dV/dx (p is pressure)
(pressure x area = F = ma)

dp = -ro V dv (Euler's equation)

or

dp + ro V dV = 0

integrating it along a streamline yields

p + 1/2 ro V^2 = constant (Bernoulli's equation)

(static pressure + dynamic pressure = total pressure = constant)

static pressure is the pressure as measured perpendicular to the airstream.

dynamic pressure is the pressure resulting from the motion of the air.

assumptions:
-gravity is insignificant
-frictionless air
-the formula is for any point on the same streamline
-incompressible flow (i.e. ro is a constant)

"if all the streamlines have the same value of p and V far upstream, then the constant in Bernoulli's equation is the same for all streamlines."

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air cannot flow into a solid object.
at best it will flow parallel to the surface.
thus the pressures affecting a wing, and creating lift, are STATIC pressures.

freefalling bodies will feel TOTAL pressures.

otherwise the following statements conflict:
“the faster air over the top of the wing results in a lower pressure.”
“the faster you fall the ‘harder’ the air becomes.”

we would be better served if we prefaced “pressure” properly…

(side note: aircraft use pitot tubes to determine airspeed. these devices compare the pressure parallel to the airstream with that perpendicular to the airstream.)

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as far as pilot chutes go...
it is a very complicated situation.
at the risk of committing over-simplification,

obviously, the airflow within the pilot chute is less than the flow outside.
Bernoulli thus states that the static pressure within the pilot chute is greater than that without.
but, an uninflated pilot chute can not support a load. any hint of pressure differences will cause the fabric to move.

therefore, the phrases “the air inside an inflating pilot chute pushes the fabric out” or “the air flowing by the inflating pilot chute pulls the pilot chute out” are equally valid.

DON'T PANIC
The lies in learning how to throw yourself at the ground and miss.
sloppy habits -> sloppy jumps -> injury or worse

[pope 0]

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mass:
the volume of the piece of air is dx dy dz.
therefore it's mass = ro(dx dy dz) (ro is the density of air)

acceleration:
a = dV/dt = DV/dx dx/dt = DV/dx V (V is velocity, t is time)

Newton:
F=ma

-dp/dx (dx dy dz) = ro(dx dy dz)V dV/dx (p is pressure)
(pressure x area = F = ma)

dp = -ro V dv (Euler's equation)

or

dp + ro V dV = 0

integrating it along a streamline yields

p + 1/2 ro V^2 = constant (Bernoulli's equation)

way to kill a perfectly good thread.

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assumptions:
-gravity is insignificant

Man--I wish you would have told me this 10 years ago! I NEVER would have taken up Skydiving and BASE!!!
Oh well...

pope