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Terminal Velocity
We all know the NASA method of re-entry Big shower of sparks and that's the scientists on the ground.
However, given a vehicle of 40,000 LB or 20,000Kg travelling at say 3 miles per second at 200 miles above earth. Now descend, trailing a parachute 600 feet in diameter or 282,000 square feet in cross-sectional area. Parachute is opened while in zero gravity by using small inflated rubber tubing annular rings in the rim of the chute. Given at some point the density of atmosphere rises from zero to say 0.002 kg/m3 Given that drag coefficient is 0.7 for a parachute which eventually drags/spills out as it progressively collects molecules of air as it descends. Best result I can figure is a terminal velocity of 116 m/sec So is the math wrong or what ? If the chute fills progressively from an empty vacuum into thicker and thicker air, there should be no sudden explosive fill so no destructive force and I guess more and more drag and decelleration as it falls... see calculator at http://www.calctool.org/CALC/eng/aerospace/terminal Comments please. Last edited by Niobium : March 9th 11 at 09:01 AM. |
#2
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I just found another reference where the drag of a cup-shaped object is quoted as 2.3 at http://www.aerospaceweb.org/question...cs/q0231.shtml
so plugging in values of even higher mass 140,000 LB Drag = 2.3 cross-section area 280,000 sq ft density of atmosphere 0.002 kg/m3 gravity 1 produced terminal velocity of 334 ft/sec or 227mph This would be at high altitude with very thin air. By the time the air is even one third density of ground zero the speed drops to 21 ft/sec or 14mph My point being the re-entry of a large 140,000LB object from LEO to say the stratosphere needs only a 600 ft diameter chute to arrive at somewhere between 14mph and 227mph at which point an aerodynamic wing can be utilized, with no heating or protective tiles. So why has this never been tested even with say one of the LOX tanks used to get the NASA shuttlle aloft ? |
#3
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Terminal Velocity
On 10/03/2011 1:44 AM, Niobium wrote:
I just found another reference where the drag of a cup-shaped object is quoted as 2.3 at http://www.aerospaceweb.org/question...cs/q0231.shtml so plugging in values of even higher mass 140,000 LB Drag = 2.3 cross-section area 280,000 sq ft density of atmosphere 0.002 kg/m3 gravity 1 produced terminal velocity of 334 ft/sec or 227mph This would be at high altitude with very thin air. By the time the air is even one third density of ground zero the speed drops to 21 ft/sec or 14mph My point being the re-entry of a large 140,000LB object from LEO to say the stratosphere needs only a 600 ft diameter chute to arrive at somewhere between 14mph and 227mph at which point an aerodynamic wing can be utilized, with no heating or protective tiles. So why has this never been tested even with say one of the LOX tanks used to get the NASA shuttlle aloft ? Terminal velocity is not the only issue. The parachute would have to cope with the heating effects. Sylvia. |
#4
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Terminal Velocity
Niobium wrote:
My point being the re-entry of a large 140,000LB object from LEO to say the stratosphere needs only a 600 ft diameter chute to arrive at somewhere between 14mph and 227mph at which point an aerodynamic wing can be utilized, with no heating or protective tiles. So why has this never been tested even with say one of the LOX tanks used to get the NASA shuttlle aloft ? Two points: 1. A 600 ft diameter chute is an ungodly large parachute. 2. More importantly your scheme actually makes the reentry problem *worse*. Basically the kinetic energy of an reentering object has to dissipated. The only practical way to do that is to use that energy to perform work on the surrounding atmosphere. When work is done on the surrounding atmosphere, that atmosphere becomes hot. To minimize the heat flux back into the object from that hot atmosphere one should dcelerate as slowly as possible. To decelerate slowly requires a very high, indeed, impractically high L/D (lift to drag ratio). A parachute does the opposite, however; it essentially drives your L/D to zero meaning that heat flux will be high enough to destroy any parachute quickly. Jim Davis |
#5
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Terminal Velocity
In sci.space.tech message , Wed, 9 Mar
2011 08:55:30, Niobium posted: However, given a vehicle of 40,000 LB or 20,000Kg travelling at 17,000 mph at 200 miles above earth. Now descend, trailing a parachute 600 feet in diameter or 282,000 square feet in cross-sectional diameter. Given at some point the density of atmosphere rises from zero to say 0.002 kg/m3 Given that drag coefficient is 0.7 for a parachute which eventually fills out as it collects molecules of air as it descends. Until it melts. H'mmm - there has been concern about the effects of a solid booster burst on the parachutes of a Launch Escape System. have they investigated building a parachute out of buckytube fibres? -- (c) John Stockton, near London. Web http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, and links. Correct = 4-line sig. separator as above, a line precisely "-- " (RFC5536/7) Do not Mail News to me. Before a reply, quote with "" or " " (RFC5536/7) |
#6
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Terminal Velocity
On 11/03/2011 11:58 PM, Dr J R Stockton wrote:
In sci.space.tech , Wed, 9 Mar 2011 08:55:30, posted: However, given a vehicle of 40,000 LB or 20,000Kg travelling at 17,000 mph at 200 miles above earth. Now descend, trailing a parachute 600 feet in diameter or 282,000 square feet in cross-sectional diameter. Given at some point the density of atmosphere rises from zero to say 0.002 kg/m3 Given that drag coefficient is 0.7 for a parachute which eventually fills out as it collects molecules of air as it descends. Until it melts. H'mmm - there has been concern about the effects of a solid booster burst on the parachutes of a Launch Escape System. have they investigated building a parachute out of buckytube fibres? Don't say that - they might try it, with another $Billion down the drain. Sylvia. |
#7
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Terminal Velocity
Niobium wrote:
We all know the NASA method of re-entry Big shower of sparks and that's the scientists on the ground. However, given a vehicle of 40,000 LB or 20,000Kg travelling at 17,000 mph at 200 miles above earth. Now descend, trailing a parachute 600 feet in diameter or 282,000 square feet in cross-sectional diameter. Given at some point the density of atmosphere rises from zero to say 0.002 kg/m3 Given that drag coefficient is 0.7 for a parachute which eventually fills out as it collects molecules of air as it descends. Best result I can figure is a terminal velocity of 116 m/sec So is the math wrong or what ? No, the math is right (I assume, I haven't checked the numbers), just the expectation is wrong. What you have at this point is a reentry vehicle traveling at 8,000 m/s when the terminal velocity is 116 m/s. What happens then? It's traveling much faster then it's terminal velocity, so it decelerates. Rapidly. The force is such that the shrouds burst asunder; and then the parachute itself (which is still traveling at 8 km/s) melts as it decelerates. -- Peter Fairbrother If the chute fills progressively from an empty vacuum into thicker and thicker air, there should be no sudden explosive fill so no destructive force and I guess more and more drag and decelleration as it falls... see calculator at http://www.calctool.org/CALC/eng/aerospace/terminal Comments please. |
#8
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Terminal Velocity
On Mar 9, 2:55 pm, Niobium wrote:
We all know the NASA method of re-entry Big shower of sparks and that's the scientists on the ground. ....[snip] So is the math wrong or what ? If the chute fills progressively from a n empty vacuum into thicker and thicker air, there should be no sudden explosive fill so no destructive force and I guess more and more drag and decelleration as it falls... see calculator athttp://www.calctool.org/CALC/eng/aerospace/terminal Comments please. The expectations are wrong. Say you do deploy a large parachute that is stable at hypersonic speeds (good luck). Then you slowly decenend into very thin upper atmosphere. You don't have lift (parachute vers parafoil --parafoils are not stable at hypersonic speeds either) so the first thing that happens is you slow down a little. This means you now fall into the atmosphere quicker, which slows you down faster still and hence fall even faster. This type of ballistic reentry gives about a peak g loading of ~10g regardless on how big the parachute is. For example a bigger chute will mean peak g is higher in thinner air, while something smaller will have its peak g loading in thinker air. Note total heat loading is not the same. Larger means less total heat loading. Now if we add lift things don't change as much as you may like. First even quite bad L/D ratios (0.5 IIRC) can get you down to 3g peak, even a capsule can get that. However it tuns out at hypersonic speeds getting better L/D ratios is more or less impossible. So thats about as good as it gets. Again larger area, or larger "wings" does not change the peak g loading but does change the total heat load. HS has made a number of posts on this group and others on this topic. You should be able to easily find many of these posts on google groups. greg |
#9
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I read all of these posts with interest, thank you all for the replies. So let me reply.
1. Concept is a circular parachute inflated with a rubber ring around the hem, inflated by low pressure gas. Now you have a a large area cup with drag coeff 2.3 (only if there is anything to drag against) Thats why large solar arrays flying at 50km/s dont crumple in space. 2. OK now you have said parachute falling to earth at 11km/s in essentially a vacuum but the lower thermosphere and mesosphere of the earth, will be a progressive density from zero molecules of air up to what ?? pick a number at the stratosophere. That zone N miles thick is the depth a standard capsule falls before a re-entry capsule even starts being heated. Actually its a slope since re-entry is not straight down but almost parallel to earth because a capsule needs that long to decellerate. So for that period of time in the lower mesosphere the capsule is burning up because tha drag on a small (10m wide ) heat shield is the part doing the slowing against air at 17,000mph. I.e. the heat is concentrated there and only there. 3. However, and this is my point. In the early stages of re-entry such a parachute with much larger drag area (nearly 300,000 square feet) is collecting air molecules and being slowed much higher in the thermo/mesosphere and so its drag is working long before it can happen on a small area capsule. And yes there is heating but more progressively. Heat = work agreed but the idea here is to think outside the box. Mathematically, if you have an infinitely large area decellerating against a mass of air the slowing would be immediate. If you have a small capsule it needs much denser air and its small area is hitting that air at high speed. The middle of the road approach is to start the drag at much much higher altitudes where the progressive slowing effect in the progressive increasing density has a chance to slow before hitting the stratosphere. My simple math says a parachute with 300,000 sq ft area versus a capsule with 1000 sq ft area will work 300 times sooner and 300 times better so hopefully 300 times cooler. comments? By the way many years ago in the days of valve/vacuum tubes, electonics engineers said you could never miniaturize computers cuz where would all the heat go. They never imagined a situation without heat. Last edited by Niobium : March 14th 11 at 02:29 PM. |
#10
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I am trying to think of an analogy.
Drop a feather and a steel pellet in a vacuum and they both drop at the same accelleration. As the feather hits the mesosphere or wherever, it starts to hit air molecules, it slows while the pellet with low surface area keeps falling until it is hitting much denser air and it burns up in re-entry like a meteorite. Also the air is rotating around the same speed as the planet rotation, so falling into the earths gravity is necessarily a flat glide slope. Parachutes will glide around the earth from where they started falling. NO ? Last edited by Niobium : March 14th 11 at 01:24 PM. |
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