Thursday, August 23, 2018

How to raise a tower to space? The ancient Stonehenge builders may have had the right idea. And applications.

Copyright 2018 Robert Clark
(patents pending)

 Stonehenge has been a mystery for a long time. How did these pre-technology peoples, in the time range of 3,000 to 2,000 BC, raise these huge stones weighing several tons?

  A recent theory gaining credence is that they let gravity do the work. The stones actually extend down into the ground. The theory for how they were raised was that they dug a pit under the horizontal stones extending from about a third of the way from one edge. The stones were constructed thicker and so heavier at this edge. Then gravity would cause that end to fall into the pit. This part would then be covered up letting the remaining part be visible above ground.

 See for example these videos explaining the theory:

How Stonehenge site was built.

AMAZING VIDEO: Man Lifts 20 Ton Block By Hand.

 So rather than the many concerns about building the high altitude towers vertically, instead they would be built horizontally, and we would then follow the Stonehenge builders by allowing gravity to raise them vertically.

  For a space tower that might be up to 100 km high however, we would need a deep pit for the one end to fall into, likely kilometers deep. For the purpose, we'll use the deep depths of the ocean. The ocean basins can be in the range of 5 km deep.

 We will then put a heavy weight on one end to cause the rest of the tower to rise vertically. 

 How much will the tower weigh and much will we have to use as a weight on the bottom to raise it? This article gives the formulas for the weight and taper ratio of a tower based on height and materials used. 

Optimal Solid Space Tower.
Alexander Bolonkin
(Submitted on 8 Jan 2007)

 In section 3 are given formulas for the taper of the tower and for the tower mass. The formulas are complicated for the general case where you have to consider the variation of gravity with altitude and centrifugal forces when the tower or elevator may extend thousands of kilometers into space. However, for the shorter case of less than 100 km, it reduces to being exponential in the ratio of the height to the characteristic length.

 The characteristic length is the maximal length of a straight, untapered, column of the material that can support its own weight. It is given by Lc = σ/(ρ*g), σ, the compressive strength of the material, ρ is the density and g, the gravitational acceleration at the surface. Usually, you want the tower to be tapered to minimize mass even if it is less than the characteristic length. If it is larger than the characteristic length then it will necessarily have to be tapered..

 The formula for the taper is, A = A0*exp[-L/Lc], where A is the area at the top of the tower, A0 is the area of the base, and L the length of the tower. And the formula for the tower mass is, M = M0*(exp[L/Lc] - 1).

 Various grades of steel have varying strengths. An especially strong, and expensive, grade is 350 Maraging Steel at a 2,400 MPa compressive strength, and 8,200 kg/m3 density, for a characteristic length of 29.8 km.

 There are some composite materials with better strength-to-weight ratios. See for example here:

Mechanical Properties of Carbon Fibre Composite Materials

Mechanical Properties of Carbon Fibre Composite Materials, Fibre / Epoxy resin (120°C Cure)

Fibres @ 0° (UD), 0/90° (fabric) to loading axis, Dry, Room Temperature, Vf = 60% (UD), 50% (fabric)
SymbolUnitsStd CF
E glass
Std CF
E glass
Tit. dtd
Young’s Modulus 0°E1GPa70852530135175300407520020772110
Young’s Modulus 90°E2GPa7085253010812861520772110
In-plane Shear ModulusG12GPa55455554258025
Major Poisson’s Ratiov120.
Ult. Tensile Strength 0°XtMPa600350440480150010001600100013001400990460
Ult. Comp. Strength 0°XcMPa570150425190120085013006002802800
Ult. Tensile Strength 90°YtMPa600350440480504050303090
Ult. Comp. Strength 90°YcMPa570150425190250200250110140280
Ult. In-plane Shear Stren.SMPa903540507060754060140
Ult. Tensile Strain 0°ext%0.850.401.751.601.050.552.501.700.70
Ult. Comp. Strain 0°exc%0.800.151.700.600.850.451.500.351.40
Ult. Tensile Strain 90°eyt%0.850.401.751.600.500.500.350.500.60
Ult. Comp. Strain 90°eyc%0.800.151.700.602.502.501.352.301.85
Ult. In-plane shear straines%1.800.701.001.001.401.
Thermal Exp. Co-ef. 0°Alpha1Strain/K2.101.1011.607.40-0.30-0.30-0.306.004.0018.00
Thermal Exp. Co-ef. 90°Alpha2Strain/K2.101.1011.607.4028.0025.0028.0035.0040.0040.00
Moisture Exp. Co-ef 0°Beta1Strain/K0.
Moisture Exp. Co-ef 90°Beta2Strain/K0.
** Calculated figures

Fibres @ +/-45 Deg. to loading axis, Dry, Room Temperature, Vf = 60% (UD), 50% (fabric)
SymbolUnitsStd. CFHM CFE GlassStd. CF fabricE Glass fabricSteelAl
Longitudinal ModulusE1GPa171712.319.112.220772
Transverse ModulusE2GPa171712.319.112.220772
In Plane Shear ModulusG12GPa3347113088025
Poisson’s Ratiov12.
Tensile StrengthXtMPa11011090120120990460
Compressive StrengthXcMPa11011090120120990460
In Plane Shear StrengthSMPa260210100310150
Thermal Expansion Co-efAlpha1Strain/K2.15 E-60.9 E-612 E-64.9 E-610 E-611 E-623 E-6
Moisture Co-efBeta1Strain/K3.22 E-42.49 E-46.9 E-4
** Calculated figures

These tables are for reference / information only and are NOT a guarantee of performance
1 GPa = 1000 MPa = 1000 N/mm² = 145,000 PSI

These tables relate to only 2 of the many fibre orientations possible. Most components are made using combinations of the above materials and with the fibre orientations being dictated by the performance requirements of the product. Performance Composites Ltd. can assist with the design of components where appropriate.

 The standard carbon fiber fabric has a compressive strength of 570 MPa at a density of 1.60 g/cc, 1,600 kg/m3, for a characteristic length of 36.3 km. This is the type of fabric that has the fibers aligned in multiple directions.

 Unidirectional composites (UD) however have greater strength in the direction of the fibers, and markedly reduced strength outside of that direction. The carbon fiber unidirectional composite (CF UD) has a compressive strength of 1,200 MPa at a density of 1,600 kg/m3, for a characteristic length of 76.5 km, but only in the direction of the fibers.

 Another unidirectional composite attains even greater strength using boron fibers. It's compressive strength is listed as 2,800 MPa at a density of 2,000 kg/m3, for a characteristic length of 142.7 km, but again this is only in the direction of the fibers.

 The question is whether the unidirectional composites will suffice if the fibers are oriented in the direction of the greatest compressive stress.

 Another composite, high-strength structure is the isotruss:

 It gains its strength from its unique geometry:

Isotruss Tower
280' tower installed in Spanish Fork, Utah

(II.)Tall Launch Towers.
Geoffrey Landis has calculated that a tower at a 25 km height could make a single stage to orbit(SSTO) financially feasible, increasing the payload by 122%:

High Altitude Launch for a Practical SSTO.

 Landis considers using it for a rather large launch vehicle at 2,000 metric tons gross mass. However, I'll consider it for smaller vehicle in the range of 100 metric tons, such as for example the Falcon 9 upper stage, which does have the mass ratio to be SSTO.

 We'll calculate the mass of the 25 km tower bearing a 100 ton weight at the top using the various materials. As commonly done for the space tower or space elevator calculations, we'll use a safety factor of 2. That is, in calculating the characteristic length we'll use a compressive strength half of its actual value, giving a characteristic length half of its actual value.

 So for the 350 grade maraging steel the exponential term in the mass formula, (exp[L/Lc] - 1), amounts to 4.35. So the mass of the tower would be 4.35 times the 100 ton weight at the top, or 435 tons. Note as a point of comparison, the structural mass of the Eiffel tower at only 300 meters tall is 7,300 metric tons.

 For the standard carbon fiber fabric, the exponential term would be 2.96, for a tower mass of 296 tons.

 If the unidirectional composites can be used, then the UD carbon fiber gives an exponential term of 0.922, for a tower mass of 92.2 tons, and the UD boron fiber results in a tower mass of 0.42 times 100 tons or 42 tons.

 (III.)Launch tracks to orbit.
 Some recent proposals for achieving orbit even want to beyond the low costs of reusable SSTO's. They propose creating high altitude tracks on which the space vehicles would be accelerated electromagnetically all the way to orbital velocity. One such proposal is the Lofstrom Launch Loop:

 This proposal though would have the tracks be kept aloft dynamically via electrodynamic methods, which would entail the risk of catastrophic failure if the dynamic forces holding it aloft failed. Then it may be preferable to use a static support method.

 If the tracks weighed, say, 100 tons between towers supporting the tracks and using the best composite material in the boron UD, the mass of a 100 km tall tower would be 3.06 times 100 tons, or 306 tons.

 The towers as 100 km tall pylon supports would be erected using the Stonehenge method one at a time along a line over the oceans. Their mass of, say, 306 tons, would not be an impediment to their being towed out to sea since off-shore drilling platforms weighing hundreds of thousands of tons have been towed out to sea towards their operating locations.

 There would be the question of getting the tracks up to the tops of the pylons once they are all erected. The pylons would have a small height at first then gradually having a taller heights until they reach the altitude of the acceleration track, The pylons would then proceed horizontally at common heights. Then each section of the track would be drawn up slowly until they reached their desired positions.

 The track sections would need to be able to span the gaps between the pylons though. An automated way of doing it might be this automatic bridge building machine:

(IV.)Jet stream powered wind mills.
 The winds in the jet stream can be in the range of 100 mph, 160 km/hr, 5 to 10 times the common wind speeds at the ground.

The polar jet stream can travel at speeds greater than 100 miles per hour (160 km/h). Here, the fastest winds are coloured red; slower winds are blue.

  Since the power of the turbine varies as the cube power of the speed, this means a wind turbine in the jet stream could provide a hundred to a thousand times more power than one on the ground. A single wind turbine might supply 1 gigawatt of power, enough for an entire city of a million people.

 This led to several proposals for harnessing the power of the jet stream:

Scientists look high in the sky for power / Jet stream could fill global energy needs, researchers say
Keay Davidson, Chronicle Science Writer Published 4:00 am PDT, Monday, May 7, 2007

 And Bill Gates intends to invest billions in renewable energy such as wind power.

 To get to the winds of the jet stream we could use the standard wind mill form of a tall tower with a turbine attached, except we now have the capability to construct those towers to the altitude of the jet stream, ca. 10 km.

 As a point of comparison, the Nordex N100/2500 wind turbine generates ca. 2.5 MW using a 100 m wide rotor. The rotor made of carbon composite weighs less than 10 tons. We'll calculate the mass of the tower to 10 km to support that 10 ton weight, using a safety factor of 2 in each case.

 For the 350 grade maraging steel, the exponential term in the mass formula is 0.96, so the tower mass to support the 10 ton rotor would be 9.6 tons. For the standard carbon fiber factor, the exponential term would 0.73, for a tower mass of 7.3 tons. And for the boron fiber UD, the exponential term is 0.15, for a tower mass of 1.5 tons.

 That 1.5 ton weight using the boron fiber UD  is quite remarkable for a tower to reach 10 km high,  the altitude jet airliners cruise at. This illuminates how important it is to determine if unidirectional composites can be used for construction of tapered towers.

  Bob Clark

Friday, August 17, 2018

Pumping pressurized fluids to high altitude for the space tower and for fighting forest fires, Page 2: high volume, high head, single pump solution.

Copyright 2018 Robert Clark
(patents pending)

 In the blog post "Pumping pressurized fluids to high altitude for the space tower and for fighting forest fires," I mentioned high volume pumps such as used on fireboats to send kilometers-long water streams from lakes or rivers to fight wildfires.

 However, the water turrets on fireboats would not have the "head", i.e., altitude capability, to send the water streams kilometers in the air on their own. In fact, I didn't think such pumps existed. So instead I suggested adding an additional pump that would trade flow rate for altitude. A pump that could do it would be the famous ram or hydraulic ram pump.

 I was surprised though after doing a web search that there do exist pumps that do have both high volume and high head. These are so-called "mud pumps" that are used for dredging the sea floor and for pumping out mud and debris deep within mines. In addition to the high volume for dredging and mining purposes, they need the high head since they will also be pumping up mud and rocks, which are denser, so heavier, than just water.

 One is the 14-P-220 mud pump. It is capable of 7,500 psi at 375 gpm.

  Another thing I didn't realize is such pumps themselves have the capability to vary their psi vs. flow rate values, i.e., they are already able to trade flow rate for head. See for example the table here:

   But there is a limit to which a pump's pressure values can be varied. Going beyond that limits the pumps lifetime or can damage it. A psi value of 7,500 psi is 500 bar. This corresponds to a vertical head of 5,000 m. Using the analogy of the ball thrown at a 45 degree angle, the horizontal distance possible would be twice that or 10,000 m, 10 km. It's likely the pump's head can be increased beyond the rated value somewhat at the cost of reduced lifetime, but in any case we can still use the ram pump or other methods to further trade flow rate for distance.

 Note, actually for a ball or a water stream, because of air drag or frictional drag in a pipe, the angle for maximum horizontal distance will actually be less than 45 degrees and the distance will be less than twice the vertical height.

 In a correspondence, Giottis Motsanos suggested I look into laminar flow to maximize distance. This is when the water flow is in a uniform, parallel stream. I had been assuming the flow had to be within a pipe to maximize distance. But some remarkable instances of laminar flow raise the possibility we can achieve these long distances without using a pipe:

Making a Laminar Flow Nozzle.

AMAZING Under $300.00 Home Made Laminar Water Jet.

 And this video specifically mentions laminar flow allows longer water streams for firefighting:

Digitally Controlled laminar Fountain in Burj Al Arab Building

 If so, this would eliminate the problem of how to support the weight of the pipe over kilometers-long distances.

  Bob Clark

Tuesday, August 7, 2018

Pumping pressurized fluids to high altitude for the space tower and for fighting forest fires.

Copyright 2018 Robert Clark
(patents pending)

  Attached below were some discussions I had on news forums about producing a self-standing tower that can reach to space. It is known current materials do not have sufficient strength-to-weight to accomplish this. So the idea would be to do it dynamically by pressurized fluids pumped upwards. In the ensuing discussion it occurred to me we could also use it to fight forest fires, a currently serious problem here in the U.S. 

(I.) Calculations suggest we currently have pumps at sufficient power to carry water long distances through piping supported by pressurized water or air. The formula for the power required of a pump to carry a liquid vertically is proportional to its density, flow rate, and altitude, or head, as indicated here:

 For water, 10 meters, 33 feet, of altitude corresponds to one additional bar of pressure, or 14.7 psi. So the formula for the power of a pump can also be written in terms of pressure.

 From proportionality, for a pump of a certain rated power, we can trade flow rate for  altitude, and vice versa. That is, in order to increase the altitude the water is pumped to by a factor of 10 with the same pumping power, we can decrease the flow rate by a factor of 10.

 Now consider the pumping power of the fireboat the Warner Lawrence:

 It can throw as much as 38,000 gallons per minute 400 feet upwards in the air, using its twin 1,575 horsepower motors and 10 water turrets. Then it could send 3,800 gallons per minute 4,000 feet in the air. 

 Then the idea would be to transport these type of pumping motors, if the ship itself is too difficult to transport, to a lake nearby to a wildfire or even the ocean, which can be several miles away from the fire.

 These motors would suffice to transport 3,800 GPM to 4,000 foot altitude. The idea would be to then turn the pipes horizontally and angled downward to allow the water to flow horizontally and down to reach the wildfire.

 The question would then be how is the pipe supported vertically and horizontally? The plan behind this is illustrated by the diagram in post #2 below. Exhaust ports at periodic intervals would direct a (small) proportion of the high pressure water downwards to create an upwards force.

 How much force? For the vertical portion, the water supports itself by the water pressure. But how about the water pipe? From the 400 foot head of the pumps, this corresponds to about 12 bar pressure above ambient.

 Now the thrust produced would be analogous to that of a water rocket

 That is, twice the product of the water pressure above ambient times the cross-sectional area. From the appearance of the size of the water turrets, I'll take their diameter as 8", among the largest sizes for hoses used by firefighters.. 

 Each of the 10 turrets at 8" diameter and 12 bar pressure will be converted to an 8" hose at 120 bar pressure. This is 120*100,000 Pascals = 12,000,000 Pa. The diameter of 8", is 0.20 m. Then the total thrust possible would be 2*(12,000,000 Pa)*(π*0.10^2) = 750,000 N, or 76,000 kilogram-force.

 It turns out this is the maximum that could be lifted by using the periodically placed exhaust ports along the length of the pipe. The problem is that the horizontal section would be too heavy to be supported by this thrust because for this part you would have to support both the pipe and the water, at least for this pipe size.

 You could further trade water flow rate for pressure thereby giving greater pressure and increasing thrust and at the same time reducing the water mass that needed to be supported but rather than doing that I'll work with the possibility suggested in post #5 below of sending the water in a parabolic arc.

 Note this is the path the water would take anyway when sent at the appropriate angle so in this case we should not need to support the water mass.

 For the pipe mass, let's estimate it first for aluminum. The ratio of the wall thickness of a cylindrical pressure vessel to its radius is in the same proportion as the contained pressure to the vessel material's tensile strength
Stress in thin-walled pressure vessels
Stress in a shallow-walled pressure vessel in the shape of a sphere is
where  is hoop stress, or stress in the circumferential direction,  is stress in the longitudinal direction, p is internal gauge pressure, r is the inner radius of the sphere, and t is thickness of the sphere wall. A vessel can be considered "shallow-walled" if the diameter is at least 10 times (sometimes cited as 20 times) greater than the wall depth.[14]

Stress in the cylinder body of a pressure vessel.
Stress in a shallow-walled pressure vessel in the shape of a cylinder is

  •  is hoop stress, or stress in the circumferential direction
  •  is stress in the longitudinal direction
  • p is internal gauge pressure
  • r is the inner radius of the cylinder
  • t is thickness of the cylinder wall.

 The tensile strength of standard aluminum is 45,000 psi. The pressure of water at 120 bar would be 1,800 psi, which corresponds to a ratio of 25 to 1. Then the ratio of the wall thickness to the pipe radius would also be 25 to 1, giving a thickness of 0.1 m/25 = 0.004 m.

 Say the length of the pipe were 10 km, 10,000 m. Then the volume of the pipe wall would be π*0.20*0.004*10,000 = 25 m^3. At an aluminum density of 2,700 kg/m^3 this would weigh, 68,000 kg. This is uncomfortably close to the 76,000 kilo-force that could be supplied by the water flow. Then all but about a tenth of the water would be shooting out of the exhaust ports to support the pipe mass rather than being applied to fight the fire.

 The situation is made worse when you take into account the frictional losses, especially over long distances.

 So instead of standard aluminum we'll use other materials that have higher strength-to-weight ratio than standard aluminum. For instance, carbon fiber composites can cut the weight in half for pressure vessels. This would give the pipe a weight in the range of 34,000 kg, and about half the water would be available to fight the fire. 

 This would mean in the range of 1,900 GPM could be applied to the fire. Note that a common 5" diameter fire hose might only supply 500 GPM so this would be like four 5" size fire hoses being applied simultaneously and continuously to the fire.

 The weight for the pipe might be actually smaller by an additional factor of 2. The reason is as the water gets higher the pressure reduces so the required thickness of the pipe can also be reduced. If the pipe were straight it would be exactly half, but in curved parabolic shape the degree of reduction would need to be calculated.

 Also, as noted in post #5 below it may be the weight of the pipe will be supported by the flow of the water anyway and we might only need the exhaust ports for stability. This happens for example with those "air dancers":

 We still need to calculate the frictional losses however. For each of the ten 8" pipes the initial inflow would 1/10th of the total 3,800 GPM, or 380 GPM. 

 This page gives a frictional loss calculator:

Hazen-Williams Equation - calculating Head Loss in Water Pipes. 

 For an 8" pipe, of 10,000 m, 32,500 ft, length, and 380 GPM flow rate, the loss by this calculator is only 38 psi out of the 1,800 psi.

 The pumps for the Warner Lawrence were only used as an example of what could be done. It could also be done by other large size pumps, such as this: 

 This can pump 28,000 GPM with a head of 96 feet. It is meant to be portable despite its 27,000 pound weight.

 I have said that water flow rate can be traded for altitude for a given pump power. But I didn't actually say how to do that. Certainly you can make the pump to begin with that can reach such high altitudes and also have high flow rates as needed in this application of fighting wildfires. But pumps don't commonly come with this kind of combination of ratings. 

 For water pumps with pressure ratings in the thousands of psi these are usually just used for pressure washers or for waterjet cutting systems. In such cases the water flow rate is small. 

 But to use this for fighting the current wildfires, rather than waiting for pumps of this special type to be designed, developed, tested and manufactured, which might take years, we need another method that can be implemented quickly.

 There is a rather well known method for converting water flow rate for altitude: the ram pump, or hydraulic ram. This is a rather simple type of pump that does not even need a motor. It works by using water flowing downhill to provide the energy to drive a smaller amount of water uphill to an even greater height than the initial water source, i.e., it trades water flow rate for altitude:

Hydraulic Ram.
"A hydraulic ram, or hydram, is a cyclic water pump powered by hydropower. It takes in water at one "hydraulic head" (pressure) and flow rate, and outputs water at a higher hydraulic head and lower flow rate. The device uses the water hammer effect to develop pressure that allows a portion of the input water that powers the pump to be lifted to a point higher than where the water originally started. The hydraulic ram is sometimes used in remote areas, where there is both a source of low-head hydropower and a need for pumping water to a destination higher in elevation than the source. In this situation, the ram is often useful, since it requires no outside source of power other than the kinetic energy of flowing water."

 Ram pumps are of such simple construction that they are often made by amateurs either for their water pump needs or simply for experiment. See for example here:

How to build a RAM PUMP.

 They can also be rather easily made of large size:

Largest Ram Pump in the World.

 There is at least one other company investigating pumping water up to high altitudes to fight fires, though in this case it's only going up vertically and only to 900 feet. The water hose in this case is only being supported by a drone: 

  As mentioned there these methods would allow fighting fires in high rises at heights above which normal fire trucks can reach with their hoses. They would allow the hoses to even be sent inside the buildings at high floors to direct the water to the specific locations inside the buildings of the fire. They also could be use for rescue for heights beyond which firetruck ladders can reach. They could have prevented tragedies such as the Grenell Tower fire in London.

 Another application would be drought relief when a fresh water source is miles away, and also for natural disasters such as hurricanes and earthquakes when normal water service is interrupted.

 And it's not just areas of unexpected drought or natural disasters. Many areas in the developing nations suffer from fresh water scarcity. In these areas a significant part of the day is devoted simply to walking miles to reach a source of fresh water and carrying it back, just as a way of life. The people of these communities could accomplish much more without having to spend hours just transporting water.

 Note for this application, unlike the fire fighting application, most of the energy for transporting it could be recovered. For you could attach turbines at the bottom of the pipe to gain energy from the falling water. Some of the energy would be lost, due to frictional losses. But considering that much of the energy could be recovered, it's likely the water transport could be powered by solar power alone. 

(II.) The original impetus of this work was to produce a self-standing tower to reach the altitude of space, 100 km or above. The payload that could be launched would be increased significantly in that case. However, the pressures required for a single pump to accomplish this would for a large amount of water flow would be impractical. 

 The suggestion instead was to do the pumping in stages, as illustrated in post #2. You would need lightweight pumps that would be supported by the pressurized flow exiting the exhaust ports.

  I'll discuss a small pump to serve as a proof-of-principle to reach, say, 10 km. A low cost example of the individual pumps that might work is the battery operated Worx 325 psi Hydroshot, at 0.5 GPM, 0.5 in hose diameter, running at 20 volts, 2 amps.

 At the 325 psi = 2,210,000 Pa pressure and a 0.5 in diameter hose, this could generate a thrust of 2*(2,210,000 Pa)*π*(0.25/39.37 m)^2 = 560 N, 57 kilogram-force. The pump only weighs 3.7 pounds, 1.7 kg. 

 As before, we'll suppose the water itself is self-supported by the pressurized, directed upwards, water stream so we need to determine the weight of the pipe. Again using aluminum as a baseline, its 45,000 psi tensile strength is about 140 times the 325 psi internal pressure of the pipe. So the thickness of the pipe wall would be the size of the pipe radius 0.25 divided by 140: 0.25/140 = 0.001785 in. And the wall volume of a 10 km = 10,000 m pipe, with the diameter and thickness converted to meters, would be 
π*(0.5/39.37)*(0.001785/39.37)*10,000 m = 0.018 m^3. At a 2,700 kg/m^3 density of aluminum this would weigh 49 kg. Likely, we'll want to reduce the weight by half by using carbon fiber composite instead, so 25 kg.

 How many of the pumps would we need? The 22 bar pressure of the pump could send a water stream 21 bar*10 = 210 meters in the air. Say we put a pump every 200 meters, then to go the full 10,000 m length we would need 50 pumps. 

 Given the 25 kg weight for the full 10,000 m length of the carbon fiber pipe, each of the fifty 200 m long segments would only weigh 0.5 kilo. 

  The water thrust sent out the exhaust ports would need to support the 1.7 kg weight of the Worx, the 0.5 kg weight of the 200 m long pipe segment, and the battery. Say, we use a 0.3 kg battery, for a total weight of 2.5 kg. Then this is only a fraction of 2.5/57 of the full weight that can be supported. But since this is being exhausted out the ports, this same fraction 2.5/57 of the 0.5 gallon per minute of the flow would be lost. 

 Note also though this same fraction would be lost for each of the 50 stages. So the amount remaining at the end would be (1 - (2.5/57) )^50 = 0.1 of the original 0.5 GPM, or 0.05 GPM.

 As for the run time for the battery, commonly non-rechargeable lithium batteries can get 500 watt-hours per kilogram, and rechargeable lithiums about 250 wh per kilo. So a 0.3 non-rechargeable battery would have 150 wh. The 400 watt Worx can then run for 0.375 hours, 22 minutes. For the rechargeable battery it would be half that.

 This would be for a proof-of-principle demonstration to show you can reach thousands of kilometers at a relatively low cost. There are pressure washers though that put out thousands of psi, not just 325 psi, and put out 2 or more gallons per minute, not just 0.5 GPM. These could be used to reach the 100 km altitude for space.

 They can also be combined at hundreds of pumps at each stage to produce a water stream of sufficient flow to fight forest fires. 

 Another possibility is offered by a new development called the Pyrolance. It uses ultra high pressure water to cut through brick and concrete for the purposes of firefighting. It operates at up to 20 GPM and 1,450 PSI. Ten of these combined just on the ground, i.e., without staging, could be used to generate a kilometers long stream to fight forest fires.

There is also another pressure washer that operates at even high pressures, at 40,000 psi. This puts out 11 GPM, the VORTEX™ 36-9950-15A:

 Still, for higher altitudes and for longer distances it would better to use staging. The problem is significant amounts of water can be lost using the exhaust ports method to keep the pipe aloft. 

 I mentioned before the other company that proposes using a drone to keep the water hose aloft. You would need extra engines though for this. But if you're giving the water all this energy to reach high altitude why not use it as well to keep the pipe aloft? One way that would not lose water through the exhaust ports would have a turbine inside the pipe that would turn with the water flow. This turbine could then be connected through gears to propellers on the outside to support the pipe in the air.

 Another way would be to use a turbine or paddle wheel on the inside to connect to a piston on the outside. The piston would compress air, then release the air when a sufficient level of compression was reached to act like a compressed air rocket.

  Bob Clark

From: "Robert Clark" <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: "Rockets not carrying fuel" and the space tower.
Date: 28 Mar 2005 12:52:00 -0800

 I copied below a proposal for space access where the propulsion is
provided by a highly pressurized fluid piped up from the ground [you
may need to use a fixed-width font such as Courier New to properly view
the diagram.]

  A key aspect of the proposal as desccribed below is that the material
forming the pipeline does not have to be especially strong as for
example to support its own weight. The force for supporting each
portion of the pipeline is provided by the thrust produced by
pressurized fluid vented along the entire length of the pipeline.

 Note that this also would provide a means of producing a space tower
or space elevator (to low Earth orbit).  You really wouldn't need to
attach a rocket to the end of the rocket itself. You would use the
pipeline to *slowly* raise the payload to the required altitude for
LEO. Then you could use a rocket attached to the payload only to give
the payload the required tangential orbital velocity. Note that the
fuel and rocket that would need to be attached to the payload would be
significantly less since this fuel would not be used for getting it up
to altitude.

 You could have this "space fountain" raised only when you wanted to
launch a payload, or you could have it permanently raised in the air.
This would work if you located the fountain next to a large permanently
flowing source of water such as a river. Then for example a ram pump,
which requires no moving parts, could be used to raise the water in the

Contents for the pulser pump section of Gaiatech.

Designing a Hydraulic Ram Pump.

   Bob Clark

Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
From: (Robert Clark)
Date: 20 Nov 2004 17:04:01 -0800
Local: Sat,  Nov 20 2004  5:04 pm
Subject: Re: "Rockets not carrying fuel" for orbital transfer.

"George Dishman" <> wrote in message
> "Robert Clark" <> wrote in message
> >...

> > I came up with two other ideas for reducing the weight of the fluid
> > that had to be supported by the rocket as the tube trails behind the
> > rocket.
> > Firstly, I wanted to investigate both the possibilities of using
> > gaseous hydrogen or liquid hydrogen for the fluid carried by the tube.

> That drops the density so you need much higher
> speeds for the same flow rate so makes everything
> more difficult.

> > However, the liquid hydrogen scenario just gave too much weight. But
> > suppose the rocket didn't have to provide the propulsion for the fluid
> > in the tube? This is what I envision:

> <Snip pictures>

> All you have done is use a compound engine. The
> same mass is being accelerated to the same speed
> so will need the same fuel. You are forgetting
> the engines not only lift the craft but also the
> fuel needed to lift themselves. In fact with more
> engines, you have greatly increased the mass and
> the fuel needed, and all these schemes create a
> huge drag with air friction on the tube which also
> needs more fuel.
> Instead, imagine using a nearly rigid pipe as the
> arm of a trebuchet to pump fuel only over the first
> few seconds. That might be practical though the
> risks during disconnection are significant.

> George

 I'm also investigating the possibility of using a rigid structure to
reach into LEO. However, I think the efficiency of the tube method is
better than you suggest.
 Let's go back to the case of launch from Earth to LEO. I'm still
considering here that you're not using engines to combust fuel but are
only conducting a high pressure fluid up the tube to provide
propulsion. So the weight of the exhaust ports is quite small, not
that of a full blown engine.
 Let's estimate the the size of these exhaust ports.

Towards the rocket.

       |            |
       |            |<----Fluid carrying tube.
       |            |
       |            |
       |            |
       |___      ___|
       /__ |    | __\
      //  ||    ||  \\
     //|            |\\
    // |            | \\
       |            |
       |            |
       |            |
       |            |
       |            |
       |            |
       |___      ___|
       /__ |    | __\
      //  ||    ||  \\
     //|            |\\<---Exhaust ports directed aft.
    // |            | \\
       |            |
       |            |
       |            |
       |            |
       |            |
       |            |
       |___      ___|
       /__ |    | __\
      //  ||    ||  \\
     //|            |\\
    // |            | \\
       |            |
       |            |
       |            |

 Let's say you put a pair of these ports every 100 meters. Then each
pair of ports would only have to provide the thrust to support the
weight of 100 meters of the tube and fluid. Let's use liquid hydrogen
now. Its density is 71 kg/m^3. The volume of a 100 m tube, .3m wide is
Pi*(.15)^2*100 = 7.07 m^3. So the mass is 71 kg/m^3 times this or
about 502 kg, 1104 lbs.

 We're still using the presumption that we can communicate, say, a
pressure like the 6400 psi pressure produced by the shuttle liquid
hydrogen turbopumps up the tube. (Whatever type of pumps we use would
be located on the ground not the rocket so can be quite large.) Now we
want two exhaust ports to support 1104 lbs., or 552 lbs. each. So 552
lbs = (pressure)*(square area of ports) = 6400 * Pi * (1/4)*(diameter
of ports)^2 . We get a diameter of .33 in or 8 millimeters. Actually
they might even be smaller than this by using convergent-divergent
type nozzles used with rockets.

 Now remember the entire tubes weight is supported by these exhaust
ports so the great majority of the fluid that reaches the rocket will
be driving only  the payload and rocket. For a .3m = 12in wide tube
this could be a thrust of 6400 * Pi * 6^2 = 723,824 lbs. that is
solely used to loft the payload and (engineless) rocket, and again we
can probably do better than this using the nozzles normally used on

 Note that we can get even more thrust from the exhaust ports by
making them wider or by using more than 2 at each level. This is
important since we can also solve the hypersonic drag problem. These
exhaust ports are not engines but it would be a simple (and light
weight matter) to give them directional ability. Then you could have
them automatically direct their thrust to counteract the drag caused
by each portion of the tube.

    Bob Clark

Newsgroups: sci.astro, sci.physics, sci.mech.fluids, sci.engr.mech,
From: "Robert Clark" <>
Date: 28 Mar 2005 12:52:00 -0800
Subject: "Rockets not carrying fuel" and the space tower.

 I suggested in the posts on that thread of using pressurized fluid, liquid or gas, supplied from the ground to provide propulsion to a rocket and/or to support a tower to high altitude. The *principle* behind this of using pressurized fluid supplied from the ground to provide the thrust has now been demonstrated, if not to high altitude:

Fluid Motion: JetLev-Flyer H2O-Propelled Jet Pack By Chuck Squatriglia February 02, 2009 | 5:02:03 PM

James Bond-style jetpack powered by high-pressure water invented. 
A German entrepreneur, Hermann Ramke, has invented a James Bond-like jetpack powered by high-pressure water, called the JetLev-Flyer. 
Last Updated: 10:35AM GMT 17 Feb 2009 

 The patent on the device is described here:

Water-Jet Pack Patented: The Jet Ski of 2020? By Rob Beschizza August 27, 2007 | 7:40:28 PM

   Hmmm, I wonder where they got the idea for this from ...

       Bob Clark

From: Robert Clark <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: Re: Proven principle: "Rockets not carrying fuel" and the space tower.
Date: Thu, 19 Feb 2009 00:42:42 -0800 (PST)
Xref: sci.astro:18935 sci.physics:123293 sci.mech.fluids:108 sci.engr.mech:1710

On Feb 18, 8:29 pm, Brian Whatcott <> wrote:
> Robert Clark wrote:
> >...
> > Fluid Motion: JetLev-Flyer H2O-Propelled Jet Pack
> > By Chuck Squatriglia
> > February 02, 2009 | 5:02:03 PM
> >
> > ...
> >    Hmmm, I wonder where they got the idea for this from ...
> >        Bob Clark
> Reminds me of the Mythbusters episode which set out to replicate
>   a uTube exploit: raising a car using fire truck hoses.
>   The fearless MB gang had to take the engine out, but they did it.
> 25 ft up or so, and reasonably stable
> Brian W

 Thanks for that info:

* Hovering car *.

Myth Busters-Full size firehose car.

 Actually, the fire hoses reminded me of a possible application - as a
rescue platform to the high upper floors of skyscrapers. It could also
be used to deliver water to fight the fires at the high upper floors.

 However, using water would be dropping quite large amounts of water
down below and at high pressure, possibly dangerously to those below.
This might be solved just be using pumps on the ground that used
compressed air instead of water. To get the high pressures and volume
of air required would require quite large and heavy pumps so these
would, likely, still have to stay on the ground, though air is
ubiquitous unlike the water case.

 If we did use water, could the water be recycled? What I'm thinking
of is encasing the hose(s) in a wider shroud so that the water is not
lost, and can be recycled. But what I'm puzzling about is whether
capturing this water will make you lose the upward lifting force. For
the water coming down will be hitting on the shroud with a downwards
force component which will tend to want to bring the structure
downwards. Perhaps just making the shroud wide enough so that
relatively little of the water hits the sides would work.

 Another possible solution might be to have the hose(s) rising in the
form of an arch. Is it possible to make the arch stay aloft if no
water is actually exiting higher up as with the jet packs? If you have
a horizontal water pipe with an elbow bend you can certainly provide a
force pushing radially outward from the elbow against a wall for
example. If you directed instead the water in the pipe or hose upwards
at an angle could you with sufficiently high pressure get the pipe or
hose to stay in the vertical arch orientation? I'm thinking it would
be unstable because as soon as it tended to lean over, there be
nothing to prevent if from further toppling over, unlike when there
are directional jets at the top of the structure. But imagine the
material were rigid as with a pipe. The pressure of the water is
provided at the bottom and the direction it is squirted into the pipe
can also be altered. If the pipe were short then squirting the water
in a direction opposite to the one it is falling since it is rigid
would tend to counteract the fall. But we are imaging a very long pipe
(hose). There would be long lever arm for the part that is falling and
a quite short one at the bottom which is provided for a force to
counteract the fall. The required force of the water to counteract the
fall might then be impractically large.

 If we made the pipe be flexibly jointed all along its length, could
we have directional nozzles inside the pipe at each joint to direct a
counteracting force of water, still only inside the pipe, that only
had to do this for the section of the pipe directly above it?

     Bob Clark

From: Robert Clark <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: Re: Proven principle: "Rockets not carrying fuel" and the space tower.
Date: Fri, 20 Feb 2009 10:33:10 -0800 (PST)

 You know that new outdoor advertising method that has an air blown
dancing puppet outside a store? They are called "air dancers":

Sky Dancers Torero Inflatables.

 Here, air that is constrained to lie within a tube supports the tube
vertically by being blown vertically at high speed. If the air blower
is tipped at an angle the "dancer" will even be supported at an angle.

 Now, imagine your ordinary garden hose. If you point it upwards but
at an angle the water flow will come out in a parabolic arc. Then if
you had a very lightweight, flexible material wrapped in a tube shape
attached to the end of the hose the water should still be able to
support the weight of the lightweight tube which should then also
follow the parabolic arc. I'm thinking of some material like
polyethylene, which is used to make plastic bags. The reason why you
don't see this normally with your garden hose is because it is too
heavy to be supported by the water flow for any appreciable distance.
But at a couple of inches so, it will also.

 To test this you could cut a rectangle from a plastic bag, then tape
or glue an edge to make a cylinder of the same diameter as  your
garden hose, then attach it tightly to the end of the hose. See if you
can hold the hose at an angle to make the lightweight tube follow the
same parabolic arc as the water flow normally does all the way from
the end of the hose, upwards, and back down to the ground.

   Bob Clark

From: Robert Clark <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: Re: Proven principle: "Rockets not carrying fuel" and the space tower.
Date: Sat, 21 Feb 2009 06:35:21 -0800 (PST)

On Feb 19, 5:30 am, tadchem <> wrote:
> ...
> You could also use it as a drought-buster.  You would only need to get
> up a few miles.
> ;-)
> Tom Davidson
> Richmond, VA

 I like that idea. You could also use it fight forest fires. A big
problem though is that the entire horizontal distance, which could
range into many miles to reach a water source, would have to be a no
fly zone. But if I remember correctly for people who fly model rockets
for example you don't have to get special exemptions from the FAA as
long as your rockets don't travel over say a few hundred to a thousand
feet high. So perhaps we could keep the pipeline within that altitude

 You also have the problem I mentioned before of supporting the tube
horizontally. It's possible it could work to be supported in a
parabolic arc by the water flow itself. However, as you see in the
case of the "air dancers" high winds can sometimes blow it all the way
to the ground. We wouldn't want that. Possibly a jointed pipeline with
directional, internal nozzles at each joint would also work.

 Another possibility is suggested by how bottle rockets actually work.
It's not just the water that supplies the thrust but also pressurized
air. Then we could use both pressurized water and air and have
external nozzles along its length to provide jet thrust to support the
weight and to counteract winds. Then we could use molecular sieves,
which can be tailored just to pass water or to pass only air, in front
of the nozzles so that only air exits out of these nozzles providing
the thrust along the horizontal length.

 Instead of using molecular sieves, we might also just have the
pressurized air flowing in an outer layer within the pipeline to go to
the external nozzles, while the pressurized water only flows in an
inner tube.

  Bob Clark

From: Robert Clark <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: Re: Proven principle: "Rockets not carrying fuel" and the space tower.
Date: Fri, 27 Feb 2009 20:43:31 -0800 (PST)

On Feb 23, 2:03=A0pm, wrote:
> Clark, why don't you do some analysis before posting these physically
> impossible ideas.
> Just quoting sources doesn't prover your point
> Also while you are at it, get a clue. Do a little research on
> hydrostatic pressure

Hydrogen storage in capillary arrays.
N.K. Zhevago and V.I. Glebova
"We have developed the technology of hydrogen storage in capillary
arrays, and we present the corresponding theoretical background. The
technology can be effectively used for safe transportation and storage
of highly pressurized hydrogen in mobile systems, ranging from
domestic electronic devices to ground and sea vehicles."
Energy Conversion and Management, Volume 48, Issue 5, May 2007, Pages

A New Technology for Hydrogen Storage in Capillary Arrays.
Dan Eliezer, Martin Beckmann-Kluge, Merek Gebauer, Kai Holtappels.
"The major obstacles of hydrogen use, in replacing fossil oil, in cars
and trucks is the volume and mass of the tank and safety issues of
transferring and releasing the hydrogen.
C.En Ltd. has developed an innovative technology based on capillary
arrays for a safe infusing, storage and controlled release of
hydrogen. The company=92s technology enables the storage of a
significantly greater amount of hydrogen than other technologies. It
will allow cars =96 equipped with a 60-liter tank weighing no more that
50 kg to travel more than 500 km - more than any
alternative technologies using hydrogen.
Experiments and testing of C.En's patents-pending system for storage
in and release of hydrogen from capillary arrays is underway at BAM,
Germany's highly-respected Federal Institute for Materials Research
and Testing."

 Sounds like an interesting idea right? Using hollow glass fibers for
hydrogen storage. Afterall, high strength glass fibers such as S-glass
are among the strongest of fibers:

Tensile strength.

From: "Robert Clark" <>
Date: 22 Apr 2005 12:27:13 -0700
Subject: High strength fibers for high pressure tubes.
Newsgroups: sci.astro, sci.physics, sci.materials, sci.engr.mech,

    Bob Clark

From: Robert Clark <>
Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,
Subject: Re: Proven principle: "Rockets not carrying fuel" and the space tower.
Date: Thu, 19 Mar 2009 10:45:23 -0700 (PDT)

On Feb 21, 10:35am, Robert Clark <> wrote:
> On Feb 19, 5:30 am, tadchem <> wrote:
> > You could also use it as a drought-buster. You would only need to get
> > up a few miles.
> > ;-)
> > Tom Davidson
> > Richmond, VA
> I like that idea. You could also use it fight forest fires. A big
> problem though is that the entire horizontal distance, which could
> range into many miles to reach a water source, would have to be a no
> fly zone. But if I remember correctly for people who fly modelrockets
> for example you don't have to get special exemptions from the FAA as
> long as yourrocketsdon't travel over say a few hundred to a thousand
> feet high. So perhaps we could keep the pipeline within that altitude
> range.
>  You also have the problem I mentioned before of supporting the tube
> horizontally. It's possible it could work to be supported in a
> parabolic arc by the water flow itself. However, as you see in the
> case of the "air dancers" high winds can sometimes blow it all the way
> to the ground. We wouldn't want that. Possibly a jointed pipeline with
> directional, internal nozzles at each joint would also work.
> Another possibility is suggested by how bottle rockets actually work.
> It's not just the water that supplies the thrust but also pressurized
> air. Then we could use both pressurized water and air and have
> external nozzles along its length to provide jet thrust to support the
> weight and to counteract winds. Then we could use molecular sieves,
> which can be tailored just to pass water or to pass only air, in front
> of the nozzles so that only air exits out of these nozzles providing
> the thrust along the horizontal length.
> Instead of using molecular sieves, we might also just have the
> pressurized air flowing in an outer layer within the pipeline to go to
> the external nozzles, while the pressurized water only flows in an
> inner tube.
> Bob Clark

 If this is feasible we might use it also to limit the destructiveness
of hurricanes. A single hurricane can cause thousands of deaths and
billions of dollars in damage. Interestingly it is known that a
reduction of only 5 degrees Fahrenheit in the ocean surface
temperature can cause a hurricane to dissipate. Because of that I
suggested putting cooling chemicals or ice in the path of a hurricane
to reduce its strength:

Newsgroups: sci.physics, sci.chem, sci.geo.meteorology
Date: 24 Sep 2005 16:51:45 -0700
Subject: Could we use endothermic(heat absorbing) reactions to reduce
hurricane strength?

 The problem would be getting the required massive amounts to the area
in time. The raised pipeline might provide a means to accomplish this.
There are pumps that can put out hundreds of cubic meters or hundreds
of metric tons of water a second:

Parched Saurashtra now has "world's largest" pump.
Posted: Mar 22, 2007 at 0054 hrs IST

 If you pumped the water mixed with the right freezing point lowering
salt, the water could be kept liquid at temperatures of dozens of
degrees below freezing and you might only need in the range of tens of
thousands of metric tons to accomplish the temperature drop, which
could be pumped with these largest pumps in a matter of minutes.

     Bob Clark


A route to aircraft-like reusability for rocket engines.

  Copyright 2024 Robert Clark   A general fact about aircraft jet engines may offer a route to achieve aircraft-like reusability for rockets...