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A ballute is generally defined as a hybrid between a balloon and a parachute. To the best of my knowledge, there is so far only 2 real applications of the ballute, in reentry physics and in air to surface bombs, but that's about it. Well I propose using a ballute as the next generation of parachutes
in skydiving:
A ballute filled with helium could be used to achieve a skydive that has a much slower terminal velocity than with a regular parachute, and thereby would give a longer and much more enjoyable experience for both experienced and novice skydivers alike. In addition to a very slow terminal velocity that a ballute has to offer, it would also be very possible for a skydiver to achieve a slow vertical climb in the presence of an upward gust of wind or in a thermal.
The ballute would come with a pressurized helium tank that could be carried by the skydiver so that buoyancy could be increased, and a helium release valve would be present on the ballute in case the skydiver wants to lose buoyancy.
ballute
http://www.answers....lute?cat=technology
[quantum_flux, Jul 27 2007]
Ballute Experiments
http://www.andrews-...php?subsection=MTA0
[quantum_flux, Jul 27 2007]
Wikipedia: Terminal velocity
http://en.wikipedia...i/Terminal_velocity
Vt = sqrt( 2*m*g / rho*A*Cd ) [jutta, Jul 28 2007]
Not baloot
http://en.wikipedia.org/wiki/Ballute
[normzone, Jul 28 2007]
And definitely not balut
http://en.wikipedia.org/wiki/Balut
[imaginality, Jul 29 2007]
Nor is it anything to do with Ute Ball
http://www.spaff.com/poesy/ute_ball.html
[normzone, Jul 29 2007]
Helium Balloon Flight
http://video.google...pe=search&plindex=1
[quantum_flux, Feb 28 2008]
[link]
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Why helium? Why not fill it with air, just as you fill a parachute with air? The ballute is not large enough to provide any buoyancy. |
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I've tried it (edit: Baloot) once, but only once. |
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Quantum, you seem to be confusing the
freefall part of skydiving with the
subsequent parachute (or ballute) descent. |
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the ballute could open up like an airbag canister, heck, divers woud be able to pop open multiples of these canisters if necessary |
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fox 1, fox 2, fox 3! wahoo, it worked on fox 3! we have bouyancy, although we now have too much bouyancy so we're rising now, release some helium out of bag 2 to equalize the weight and the lift and let the parachute do the rest of the job! (note: all 3 ballutes would be under 1 common parachute in this scenario) |
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Tsk, tsk, [quantum_flux]. If you're going to delete my link to baloot, why didn't you delete my previous anno as well? |
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didn't know that it, my idea or your anno, was related to food. |
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Oh. SERIOUSLY - you meant your idea to be taken seriously. I must have missed that. |
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well, yeah, I think it is a legitimate idea. I just submit them here to be proved wrong though! |
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Oh, it's a perfectly legitimate idea. |
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I'm trying to do the math on this one, and (as is common with me and Physics) I'm running into some trouble. Specifically, what part of the terminal velocity equation over to the left changes when one adds helium? |
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I understand that there's a buoyancy that's pushing a balloon up, and I'd expect a helium ballute to reach its terminal velocity later than an air ballute - but I'm not sure they actually have different terminal velocities. The *mass* of the helium ballute is a little smaller, but not much. No? |
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Help from the physics-literate? |
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This is still screwey. The original idea,
and the way the phrase "terminal
velocity" is used, suggest that the
ballute is intended for use in freefall;
yet the other bits of the idea, and some
of the annos, suggest that the ballute is
intended to replace a parachute for use
after the freefall. So which is it? |
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[Edit - I overlapped with [bigsleep]'s
anno; but using a ballute as a drogue in
freefall would be pointless - just use a
tiny drogue chute as is done for tandem
skydives] |
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Anyway, Jutta, the terminal velocity
equation from Wikipedia assumes that
the body is dense enough that it's
weight is effectively its mass (ie, it
displaces a negligible volume of air). If
you want to make the equation precise,
then the "m" should be replaced by
"Mo-Ma", where Mo is the mass of the
object and Ma is the mass of the air it
displaces. |
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This gets screwy when Ma is greater
than Mo (eg, a helium balloon, since
you're left having to take the sqrt of a
negative term -this is because the
equation is sloppily formulated). In this
case, though, just use the absolute
value of [Mo-Ma], and remember that
the terminal velocity will be upwards
rather than downwards! |
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// a parachute for use after the freefall.
Hang on. (No pun intended.) Isn't a parachute in freefall, just with a lot more drag, hence a low terminal velocity? |
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Since it takes a *lot* of helium to float a person, I was writing on the assumption that the paraloon, er, ballute with someone hanging off it doesn't rise like a balloon would - it weighs less, but it still weighs something. |
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It feels to me like the thing we should be messing with is not m, but g - where g should be |Fg - Fb(m)|, force of gravity minus force of buoyancy (which, of course, depends on the volume of m.) Meh. It probably doesn't matter whether you have a hidden buoyancy and gravity in the mass or a hidden mass in the buoyancy... |
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Isn't the inclusion of the weight of a helium tank defeating the lift of the ballon as well? |
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Somebody let me know when we're ready to readdress the subject of baloot. |
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1) Inflated helium has a larger surface area than compressed helium, and therefore it has a higher drag coefficient which results in a smaller terminal velocity. |
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2) Inflated helium takes up more space, i.e. it displaces a greater volume of air, than compressed helium. Because of this, more lift is generated by the bouyancy force, and thereby the net weight experienced by the ballute is decreased.... therefore, the freefaller will experience a slower acceleration (negative in some cases) to get to that aforementioned terminal velocity. |
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[Jutta] Yes, you're right that a
parachute is in "freefall" (technically, the
skydiver is never in freefall, since he is
always supported to some extent by air
resistance, whether before or after
opening his 'chute). However,
conventionally the pre-parachute-
opening phase of the descent is called
"freefall", whereas the descent under
canopy is the canopy ride. But these
are semantics. My point was that, in
the original idea and annos, it was not
clear if this was intended for use during
the "freefall" phase of the jump or as an
alternative to a parachute for the
canopy ride. |
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Anyway, as regards //we should be
messing with is not m, but g// , well,
not really. If you want to be really
rigorous you'd work with the mass of
the falling object and the mass of the
air it displaces and go through all that
rigmarole (g is constant, and exerts
forces both on the falling object and on
the volume of air it's displacing). |
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However, it's simpler to assume (and
gives the same answer) that the mass of
the falling object is effectiely it's own
mass (which is always positive, even for
helium) minus the mass of the air it
displaces (which is always positive;
fairly negligible for a human body, but
substantial for a large balloon). |
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Quantum flux is correct in what he
says. Don't mix up the two ideas of
mass (which determines the force of
gravity on the body) and drag (which
determines the speed at which air
resistance balances that force, and
hence the terminal velocity of the body
- be it downwards or upwards). |
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// Isn't the inclusion of the weight of a helium tank defeating the lift of the ballon as well?//
Yes, but I presume QF will simply drop the tank, which will either plunge to earth, reaching a spectacular terminal velocity, or float down on its own parachute. |
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Certainly, by changing the shape of an airfoil, you also happen to change the drag and lift coefficients. |
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oh dear lord i just learned what balut was. my innocence is gone. so is my appetite. |
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// oh dear lord i just learned what balut was. my innocence is gone. so is my appetite. \\ |
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"Balut" doesn't sound anything like a "Skydiver's Ballute Pack" geeze! Also, how are you ever going to travel to Asia with such a weak appetite? |
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According to my helium ballon design program, it takes 5000 cubic feet of helium to lift 345 pounds (that is payload and ballon and supporting rig. Large, fully loaded high pressure helium tanks holds about 245 cubic feet of gas at normal pressure...the tank weighs in at about 85 pounds...with regulators a bit more. So, if you were just rtying to lighten the load to get a bit longer sky ride, I suspect you may want to displace about 150 pounds...making an average man and rig about 100 pounds rather than 250 or so, there are some other problems. Trying for that fall-load goal, you would need about ten tanks or 850 pounds or so of gas pressure cylinders. Also, your freefall time is quite limited...measured in seconds, I understand.......you will have to have some way of quickly discharging all the tanks to get the benefit of the filled ballute. This rapid release of highly pressurized helium is going to create a super cooling effect that is going to greatly reduce the lifting ability and volume of the released gas....might take quite a while for the gas to reach temperature equilibrim with the surrounding air which is important to achieve full lift capacity of the lifting gas. |
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This pressurized gas rig could be lightened by using low pressure but well insulated liquid helium tanks...so long as you empty them rather quickly and be careful not to spill any of the super cold stuff on you....or your friends. But, the rig is still going to be quite heavy and terribly cumbersome. And, you will still have the cyrogenic effects to overcome as the liquid helium turns to a gas. (Hmmmm)...perhaps some sort of super heater fueled by butane...like a hot air balloon burner....there goes the cumbersome effect again, though. |
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I think the idea is just a bit impractical.....since you could just use a fully equipped helium balloon to fly to altitude and then just deflate it until it was heavy enough to descend and you get the basic same effect. |
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