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Space Forum / Astronomy / April 2005About "Rockets not carrying fuel": how far does the shuttle deviate from its planned trajectory?
In the thread below I mention two options for implementing the idea of pumping fuel or reaction mass up a long pipeline for space access: Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech, sci.space.policy From: "Robert Clark" <rgregorycl...@yahoo.com> Date: 28 Mar 2005 12:52:00 -0800 Subject: "Rockets not carrying fuel" and the space tower. http://groups-beta.google.com/group/sci.astro/browse_frm/thread/ab0c8a53330a521a One is to have the pipeline be accelerated along with the rocket as it travels to orbit, the other to have the pipeline or tower in place and the spacecraft (slowly) climb up it. In the first method I have some qualms about a hundred kilometer pipeline carrying thousands of tons of mass attached to the spacecraft, especially for a man-rated craft. Any unexpected aerodynamic deviation along the hundred kilometers could be catastrophic for the spacecraft considering the pipelines great momentum. But in the second method it would seem wasteful to have all that fuel or mass within the pipeline or tower and not use it to propel the payload to the required orbital velocity, using a rocket instead after detaching from the pipeline/tower. So you would prefer to have the spacecraft accelerate to orbital velocity as it moves up the tower. But then I'm not very comfortable with the idea of the spacecraft moving at up to 17,000 mph right next to a fixed structure, especially considering the number of train derailments that occur at speeds less than only 100 mph. So another possibility I'm considering is to have the tower be fixed in place but a much shorter pipeline be attached, say 10 km long to the tower. I believe the required diameter of this shorter tube would also be less. So this would have a much lesser effect on the momentum on the rocket. But, still at the highest speed to reach orbit around 17,000 mph, it is moving around 10 km/sec so if the spacecraft deviated from the intended trajectory for only 1 second and if it was in the direction of the space tower you could have a collision. So my question: how great a distance does the space shuttle normally deviate from its intended trajectory? Can it be by more than 10 km at any point? Bob Clark > One is to have the pipeline be accelerated along with the rocket as > it travels to orbit, the other to have the pipeline or tower in place [quoted text clipped - 4 lines] > along the hundred kilometers could be catastrophic for the spacecraft > considering the pipelines great momentum. how about the vehicle rides up on top of a nested hydaulic cylinder? Use steering motors, and aerodynamic surfaces under computer control, to stabilize the rising central cylinder (and thus the whole launch tower). I was visualizing more like 25km, rather than 100 km tall. Each segment 2.5 meters long, only 2 meters would extend. Half a meter would stiffen the top of the next lower cylinder. The question is 'can we achieve escape velocity at the top of the tower, if each section is expanding at a much lower rate. I think so. How could we make sure that all cylinders expand at the same time? How about a computer controlled 'ratchet arm' and geared track that rate limits each cylinder's expansion. I estimated this would take less than a cubic mile of hydraulic fluid (water?).  Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap "bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message > I was visualizing more like 25km, rather than 100 km tall. Static pressure at base = 10200 bar =150000 PSI. ! > I estimated this would take less than a cubic mile of hydraulic fluid > (water?). 1 cubic mile of water weighs 4160000000 tonnes. to pump it upwards at 1 mph with a 100 % eficient pump(s) and zero loss, the pump(s) would need motors delivering 27000000000000 Horse power. This bozo must be posting from sci.astro... perish the thought he ( or she ) has anything to do with engineering. SignatureJonathan Barnes's theorem; for every foolproof device there is a fool greater than the proof. To reply remove AT > "bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message >> I was visualizing more like 25km, rather than 100 km tall. >> > Static pressure at base = 10200 bar =150000 PSI. ! 25000 ft of water head = 10846.8 PSI (neglecting the weight of the tower itself). >> I estimated this would take less than a cubic mile of hydraulic fluid >> (water?). I said less than a cubic mile. I actually got 0.459 mi^3 which would weigh 443 Mega ton, considerably less than 4 billion tons. > 1 cubic mile of water weighs 4,160,000,000 tonnes. > > to pump it upwards at 1 mph with a 100 % eficient pump(s) and zero loss, > the pump(s) would need motors delivering 27000000000000 Horse power. The water doesn't all have to be pumped to the top of the tower. In fact, the bulk of the volume would be close to the earth. Some could even be gravity fed from a ring of 'water towers' around base of the tower. Of course, pumps would be needed to raise the tower over about 50 feet. >> This bozo must be posting from sci.astro... perish the thought he ( or >> she ) has anything to do with engineering. sci.physics. How about you? 1) I didn't make the original suggestion. I did find it interesting. 2) I have made some suggestions that might make it MORE possible. 3) I make no claim that it is practical. 4) I don't call you names or make fun of you, even when you make mistakes. Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap >>"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message >> [quoted text clipped - 4 lines] > 25000 ft of water head = 10846.8 PSI > (neglecting the weight of the tower itself). Why are almost all the numbers calculated wrongly in this thread? Try doing everything by two different methods. 25 km = 25,000 m, NOT 25,000 ft, but = 82,021 ft Density of water = 1000 kg/m^3 = 1.938 slug/ft^3 g = 9.81 m/s^2 = 32.2 ft/sec^2 Pressure = Rho*g*h = 1000*9.81*25000 = 245,250 kPa = 35,570 psi or = 1.938*32.2*82021 = 5,118,000 lb/ft^2 = 35,540 psi Note that the correct answer is neither the 150,000 psi from JB nor the 10,846 from BZ! >>>I estimated this would take less than a cubic mile of hydraulic fluid >>>(water?). [quoted text clipped - 5 lines] > >>1 cubic mile of water weighs 4,160,000,000 tonnes. 1 mile = 1,609 m = 5,280 ft 1 mile^3 = 4.165*10^9 m^3 = 1.472*10^11 ft^3 Mass = Rho*Volume = 1000*4.165^10^9 = 4.165*10^12 kg 9.18*10^12 lb = 4.099*10^9 ton = 4.099 billion tons or = 1.938*1.472*10^11 = 2.853*10^11 slugs weighing 9.185*10^12 lbs = 4.10 billion tons So it looks as though JB was right on this one. Even if the volume was 0.459 miles^3 the weight would still be 1.88 billion tons, NOT the 443 Mtons given by BZ. >>to pump it upwards at 1 mph with a 100 % eficient pump(s) and zero loss, >>the pump(s) would need motors delivering 27000000000000 Horse power. 1 mph = 0.447 m/s = 1.467 ft/sec 0.459 mile^3 = 1.912*10^9 m^3 = 6.756*10^10 ft^3 For height = 25,000 m = 82,021 ft this implies a crossectional area of 1.912*10^9/25000 = 76,480 m^2 = 823,000 ft^2 or 6.756*10^10/ 82,021 = 824,000 ft^2 This implies a pipe diameter of 1,024 ft. Where did this come from? RC's pipes were about 0.2 m diam the last I read. So all the volumes, powers etc. are way out by many orders of magnitude. This is typical of this thread which has absolutely no connection to real engineering. > The water doesn't all have to be pumped to the top of the tower. In fact, > the bulk of the volume would be close to the earth. Some could even be [quoted text clipped - 10 lines] > 3) I make no claim that it is practical. > 4) I don't call you names or make fun of you, even when you make mistakes. David Wilkinson <david@wilkinson6337.freeserve.co.uk> wrote in news:d3nnpn $jo3$1@newsg1.svr.pol.co.uk: >>>"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message >>> [quoted text clipped - 9 lines] > > 25 km = 25,000 m, NOT 25,000 ft, but = 82,021 ft I said km but meant ft. Sorry. You are correct > Density of water = 1000 kg/m^3 = 1.938 slug/ft^3 > [quoted text clipped - 3 lines] > > = 1000*9.81*25000 = 245,250 kPa = 35,570 psi correct for km > or > [quoted text clipped - 16 lines] > > 1 mile^3 = 4.165*10^9 m^3 = 1.472*10^11 ft^3 correct. > Mass = Rho*Volume > > = 1000*4.165^10^9 = 4.165*10^12 kg 9.18*10^12 lb = 4.099*10^9 ton = > 4.099 billion tons mi^3 * 997.1 kg/m3 * 2.399e10 m^3/mi^3 = 4.58e9 ton close enough > Even if the volume was 0.459 miles^3 the weight would still be 1.88 > billion tons, NOT the 443 Mtons given by BZ. ahhhh. I used (.459*mi)^3. It was a stupid mistake. ... Comes from not double checking everything. Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap > >>"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message > >> [quoted text clipped - 24 lines] > Note that the correct answer is neither the 150,000 psi from JB nor the > 10,846 from BZ! J.B.'s reply.. possibly working in several unit systems is unhelpful. to recalculate, working in metric system. P = Rho g h P = 1000 ( kg/m^3) x 9.81 ( m/s^2) x 25000 ( m ) = 245000000 (Pa (N/m^2)) 1 bar = 1x10e5 Pa, so pressure is 2450 bar, 1 bar = 14.7 psi so pressure is 36000 psi my origonal post is nearer to being correct for the 100 Km tower ( salty water :-) ) > >>>I estimated this would take less than a cubic mile of hydraulic fluid > >>>(water?). [quoted text clipped - 47 lines] > magnitude. This is typical of this thread which has absolutely no > connection to real engineering. J.B.'s reply.. I was using Power = foce x speed, working on the assuption that force would be the weight of fluid, speed was an arbitary one for the whole mass being raised, ( Working metric 4160,000,000,000 kg x 0.477 m/s = ?wats, 745 w = 1 HP )I did not make any alowance for the hydralc cylinders themselves, and was working on the assumtion that the base ring bearing the weight was being pumped up. as to the supply pipes I made no claim as to there size and number, or the speed of flow down them.. ( I think the calculation is the size of the base ram presurised at the required static pressure to suport the mass of fluid but I could be wrong.) > > The water doesn't all have to be pumped to the top of the tower. In fact, > > the bulk of the volume would be close to the earth. Some could even be [quoted text clipped - 7 lines] > > > > 1) I didn't make the original suggestion. I did find it interesting. J.B.'s reply... It's so ludicrus it realy does not bear consideration... but then I'm wasting my time as well :-) > > 2) I have made some suggestions that might make it MORE possible. J,B. says: Now that IS a waste of time > > 3) I make no claim that it is practical. J.B. O.K. just as well > > 4) I don't call you names or make fun of you, even when you make mistakes. J.B. says: I stick my hand up for getting the pressure wrong... but by less than an order of magnitude, a trifle in this debate :-) I'm sorrey about the name calling, I am just fed up with hevily cross posted threads about totaly impossible devices that seem to run forever... I just lashed out. I post from sci.engr.mech SignatureJonathan Barnes's theorem; for every foolproof device there is a fool greater than the proof. To reply remove AT "Jonathan Barnes" <jbarnes6@btinternet.com> wrote in news:d3o6k0$rg6$1 @titan.btinternet.com: > I am just fed up with hevily cross posted > threads about totaly impossible devices that seem to run forever. I second that emotion. I vow to be more careful in my calculations. With mathcad, I usually don't worry, as long at the units in the answer make sense because it does conversion from system to system just fine. When I do something stupid like (.4 mi^3) rather than .4 mi^3, it is quite happy for me to continue on in ignorance. [twack twack twack {sound of my hand hitting the side of my head}] :) Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap >>>>to pump it upwards at 1 mph with a 100 % eficient pump(s) and zero loss, >>>>the pump(s) would need motors delivering 27000000000000 Horse power. [quoted text clipped - 30 lines] > ram presurised at the required static pressure to suport the mass of fluid > but I could be wrong.) The point was that the power used in pumping is the volume flow rate in m^3/s times the pressure rise in N/m^2 giving power in Nm/s. As I only had the speed of 1 mph I needed the cross-sectional area as well to find the flow rate. This should have come from the supposed total volume in the column of 0.459 mile^3 and the height of 25 km. However, as this gave a diameter for the column of 1,024 ft, above, compared to numbers like 0.2 m mentioned in earlier postings something seemed to be seriously wrong. Looking at it another way, in consistent units, a height of 25 km = 15.5 miles so a volume of 0.459 miles^3 implies a cross-section = 0.459/15.5 = 0.0296 mile^2. Since for a diameter D, Pi/4*D^2 = 0.0296, this implies D = 0.194 miles = 1,025 ft, as above. While I could now work out the flow rate and power it seems pointless because the volume of 0.459 mile^3 is probably another wild miscalculation by many orders of magnitude. There really is no sense in any of this. The concepts are daft, all the sums are wrong and there is no possibility of it ever working in the real world. Have you not got anything better to do than float crackpot schemes that belong in science fiction only? > > One is to have the pipeline be accelerated along with the rocket as > > it travels to orbit, the other to have the pipeline or tower in place [quoted text clipped - 22 lines] > > I estimated this would take less than a cubic mile of hydraulic fluid > (water?). > > ... If the sections are nested you might have the same problem that Wilkinson mentioned that if would be too wide at that bottom even if tapered. I haven't done the calculation. See how wide it would be if tapered with a cross-section area taper ratio of A = A0*e^(L/Lc) when you take into account the thickness of each section and each section has to fit into the next. How much volume would it take up? Bob Clark >> > One is to have the pipeline be accelerated along with the rocket > as [quoted text clipped - 16 lines] >> stabilize the rising central cylinder (and thus the whole launch > tower). .... > If the sections are nested you might have the same problem that > Wilkinson mentioned that if would be too wide at that bottom even if [quoted text clipped - 4 lines] > into the next. > How much volume would it take up? I figured each section was .2 m in diameter larger than the smaller one and each is 2 meters in extended length. We can approximate volume as the volume of a cone, 1/2 base area times height, neglecting the volume of the pipe and the unextended portion thereof. I made a "small error" in my previous calculations. I was using 25,000 ft, rather than 25,000 km. 25,000 km would be impractical as we would have 1.25e7 sections and a base that would be 2500 km diameter, volume 1.4e10 mi^3, pressure 1e7 psi. For a 100 km high tower, 50,000 sections and a 10.01 km base, you get a volume of 944 mi^3, with 43k psi at the base. Not practical. For a 10 km high tower, 5000 sections and a 1.01 km base, a volume of 0.962 mi^3 with 4339 psi. Maybe possible. Put it on top of a high mountain range near the equator. For 25k ft, 3810 sections, 0.772 km base, volume 0.428 mi^3, pressure 3306 psi. You can make the walls thinner and sections longer as you go higher, this would reduce the volume. Do I think it is practical? Unfortunately, probably not. Signaturebz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap Dear bz: ... >> If the sections are nested you might have the same >> problem that Wilkinson mentioned that if would be [quoted text clipped - 40 lines] > Do I think it is practical? Unfortunately, > probably not. The earth underneath this behemoth will agree. I suspect resulting change in plate tectonics will make short work of the tower, and its foundation. David A. Smith >> One is to have the pipeline be accelerated along with the rocket as >> it travels to orbit, the other to have the pipeline or tower in place [quoted text clipped - 10 lines] > > I was visualizing more like 25km, rather than 100 km tall. .... > The question is 'can we achieve escape velocity at the top of the tower, Escape velocity is 11km/s. To reach that speed in 25km, the top of the cylinder must accelerate at 2420m/s^2 or 247g. The time taken will be 4.55s Using Bob's figure of 0.2m diameter, the volume is 3142m^3. > if > each section is expanding at a much lower rate. Although each tube section may be sliding slowly, the fluid is incompressible so the whole column of fluid inside the sections is moving at the same rate. Using water, you have to accelerate a mass of 3142 tonnes from rest to 24,600 mph in 4.55 seconds. And that assumes you can prevent a cylinder 25km tall and 0.2m in diameter from buckling. Oh and we are neglecting any friction between the fluid and pipe sections when the fluid is moving at something like Mach 32 but the base pipe sections are barely moving. > I think so. I think not. George And don't forget Coriolis forces - to accelerate something vertically upwards from a rotating earth will impose a relatively high lateral force as well. >>>One is to have the pipeline be accelerated along with the rocket as >>>it travels to orbit, the other to have the pipeline or tower in place [quoted text clipped - 44 lines] > > George >Although each tube section may be sliding slowly, >the fluid is incompressible so the whole column >of fluid inside the sections is moving at the same >rate. Using water, you have to accelerate a mass of >3142 tonnes from rest to 24,600 mph in 4.55 seconds. With a column of water 25km high, you'll be taking hydraulic pressures of something like 245 million pascals at the base. Water is quite compressable at that kind of pressure! (About 11% if I did the math right...) >I think not. Me neither :D =Smidge= > I estimated this would take less than a cubic mile of hydraulic fluid > (water?). The more you think and type about this whole scheme, the worse it gets. A cubic mile of fluid, at something around 69 lbs per cubic foot, would weigh over 10 Trillion pounds. Unless it had a footprint the size of Ohio, it'd sink into the earth right up to its hilt. Hmmmm! Come to think of it, this really COULD work. Fill the tube with water, keep it vertical and straight with guy wires and winches, and let it sink 100km deep into the Earth. Then, the Earth's interior would super-heat all the water at the bottom, and turn it to ultra high pressure steam! When you let go of the winches and cables, the whole damned thing would come blasting out of the ground like a bullet from a rifle barrel. I bet you could get escape velocity easy! I don't know how/when/where you'd attach the passengers or payload; but those are prolly just details to be worked out by the engineers. Worst case, you'd have a really big tube of hot water zipping through space, and astronauts roaming around in conventional spacecraft could drop by for a nice hot shower whenever they needed one. Oh. You might want to put a few dollars in your budget plan for dealing with that really deep hole in the middle of Ohio where there didn't used to be a volcanic vent. And for compensating the poor folks who lived within the two-hundred-mile radius of farms, houses, and cities blown flat by steam exhaust before they were covered in lava. Just a thought. KG > In the thread below I mention two options for implementing the idea of > pumping fuel or reaction mass up a long pipeline for space access: [snip horrible crap] Oops - no remaining post. Learn some hydraulics, git - like the concept of "head." SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > > In the thread below I mention two options for implementing the idea of > > pumping fuel or reaction mass up a long pipeline for space access: [quoted text clipped - 8 lines] > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf Enlighten us and explain why the idea is "impossible". And afterwards make reference to this page of Bill Beatty's which he says originated with you: THE OFFICIAL TRUTH. http://www.amasci.com/freenrg/laughed.html Bob Clark think they need to fix the problem of fuel, maybe an alternative method of moving the ship forward. -- Cowly the Game player Please click on my links <a href=http://www.gamestotal.com>http://www.gamestotal.com</a> <a href=http://www.gamestotal.org>http://www.gamestotal.org</a> <a href=http://www.spacefederation.net>http://www.spacefederation.net</a> <a href=http://uc.gamestotal.com>http://uc.gamestotal.com</a> <a href=http://aw.gamestotal.com>http://aw.gamestotal.com</a> <a href=http://gc.gamestotal.com>http://gc.gamestotal.com</a> <a href=http://mmorpg.gamestotal.com>http://mmorpg.gamestotal.com</a> <a href=http://3700ad.gamestotal.com>http://3700ad.gamestotal.com</a> <a href=http://ballmonster.gamestotal.com>http://ballmonster.gamestotal.com</a> <a href=http://www.gamestotal.com/strategygames/>http://www.gamestotal.com/strategygames/</a> <a href=http://spacefed.com/news/>http://spacefed.com/news/</a> <a href=http://spacefed.com>http://spacefed.com</a> -- |
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