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Space Future has been on something of a hiatus of late. With the concept of Space Tourism steadily increasing in acceptance, and the advances of commercial space, much of our purpose could be said to be achieved. But this industry is still nascent, and there's much to do. this space.
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J V H Hill, April 1999, "Getting to Low Earth Orbit", April 1999 (Adapted from an original paper published in 1996).
Also downloadable from to low earth orbit.shtml

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Getting to Low Earth Orbit
James V H Hill
In this paper we discuss the requirements of getting into space and how to satisfy those requirements. We discuss examples of known examples of known available chemical energy, structural requirements for reaching space and surviving, an example space vehicle that could achieve these requirements and an example flight path for the vehicle. We then discuss the basic structure of the small start up busness needed to suceed as a private company, and reasonable estimates for all costs and an algorithm for what the prices would be as a function of traffic into space.

Just what does it require to get into Low Earth Orbit? Please remember that this was accomplished with pre-1960's technology.

First, you need something that can push, pull, drag or carry a given mass and volume about 150 kilometers up with a net velocity of at least 7,814 m/s tangent to the curve of the Earth. The height is needed to get out of most of the drag in the air, which is noticeable at the needed speeds to stay in orbit. The velocity is what is needed to stay in orbit at that altitude.

Second, you need enough instrumentation to tell how fast it is going and where it is, as well as some to monitor what is going on from time to time. Such as engines' status, remaining propellants, nearby objects....

Third, the vehicle needs to survive in the given environment long enough to deliver the cargo above, and preferably to return intact. Both the vehicle and the contents have to stay intact for this.

To survive in space a craft has to withstand extreme (both high and low) temperatures and hard vacuum. It must not disintegrate under hard UV or ion bombardments. And if you want to use off the hardware store shelf components, it has to be able to keep in an atmosphere when surrounded by vacuum.

There are many things available today to the hobbyist and basement mechanic that do these things, much more now than 40 years ago, when this was 1st done. Fluorosilicate plastics and high purity ceramics are not all closely guarded military secrets, but the stuff of Junior high School science projects and the Handyman. Instrumentation has not used vacuum tubes with soldering for a long time and more circuits are put on a greeting card now than went into the first telemetry systems. It is just a case of getting the right design for a desired end and putting the parts together.

The primary problem and usually the most difficult is imparting over 8 KPS to the craft. Here, too, it's mostly a matter of design to get a certain performance and to withstand all the stresses caused in getting that performance. This can be complicated, but is not insurmountable, since it was done with far weaker materials almost 40 years ago.

Mostly it's a matter of harnessing the energy required. The Energy needed is dependent on just 2 things: (1) The mass needed in LEO (including the mass of the craft used) and (2) the flight-path/launch-curve used to get there. This is expressed as Delta Vee, usually in meters per second or in gee-seconds of acceleration.

Here we have our first serious problem. The variables are all interdependent and there are certain minimum requirements. For instance, the lower the acceleration used the less stress on a structure, But lower acceleration means lower efficiency (since more time is spent lifting ) so more fuel is required for a given payload. This requires a larger Fuel Mass ratio which puts greater demands on the structure. Requiring a greater structural mass. On the other hand, some things are fixed. For example the GPS receiver masses the same 400 grams, whatever the final mass of the craft.

Assumptions and compromises have to be made and here is where you get all of the differences in spacecraft design. Differing assumptions or compromises yield different conclusions and designs. Many times the either/or decisions switch sides and a new design arises that was dismissed before as unworkable.

Start with Delta V to LEO. If all accelerations are at or under 2.05 gees and most are after reaching 12 Km altitude; assuming latitude of launch site, climb angle, flat plate drag and scale height, then the needed delta Vees in various directions are:

Going West: 9.72 Kps
Polar: 9.42 Kps
East : 9.12 Kps

The figures above are deliberately higher than needed, since for all requirements you want to over estimate the necessities and in all performances under estimate the expected results. Just the opposite of advertising.

To keep out of the denser atmosphere you need to achieve an altitude of 150 kilometers (93 miles) and a velocity a little under 7,814 meters per second. At 150 Km the acceleration of gravity is 9.361m/sec^2 and the gas pressure is just 3 billionths of an atmosphere. Reaching this altitude and speed both require huge amounts of energy. But that's not all of the energy required. Figuring all of the energy losses as lost velocity on the way to LEO, you need to add losses on the way up due to air resistance. Loss depends on the average flat plate drag per kilogram mass integrated over the whole flightpath. This is craft design and path specific, but can be approximated by using the total air the craft will pass through. If the average density of the air halves for every 5.5 kilometers height:

V (m/s)loss = 500/ [2^(h/5.5)]* sin (theta).

Theta is the angle of climb. If you start at 11.5 Km ( 40,000 ft MSL ) then V(loss) = 17.4/sin(theta).

At theta = 60 degrees, V(loss) < 135.6 m/s for the loss of velocity due to air alone. All the losses have to be made up for by using a greater total delta vee.

Add to this the losses due to the drag in the flight before the climb. With a craft like the DC-X(A) which takes off straight up there is no flight before climbing. With a Horizontal Take Off and Horizontal Landing ( HTOHL) craft that takes off like an air plane the losses are the same as those of an aircraft if it were flying for the same time and at the same speed. In the case of a 747-400 flying at mach .82 above 40,000 ft MSL the total thrust needed is less than 1/3rd that needed to lift the same craft against 1 Earth gravity. It's four 58,000lb thrust engines push a 800,000 lb craft at 270.6 m/s ( mach .82 at cruising height ). That's .29 gees. The needed thrust to counter atmospheric drag is less than this maximum since the engines are not run at full thrust when cruising. Over an extended cruise they only average .12 gees figuring from the fuel expenditure alone.

Due to the atmospheric drag the thrust needed for level flight for a given mass, size and altitude is proportional to the cube of the velocity. Double the velocity will need eight times the thrust. Delta Th =3D (delta V)^3 so that if the actual velocity used is 160 m/s the thrust needed over the same path is (160/270.6)^3*.29 gees or under .06 gees. The total velocity loss is that .06 gees times t, where t=3D(distance traveled before climb)/160 m/s or about 221 seconds, using the same drag coefficient as the 747 even though a lifting body is smoother than the airplane. At the lower cruising velocity:

V(loss) due to level flight < 131 m/s.
Plus that initial 160 m/s cruising velocity

The final losses are due to gravity drag. Accelerating at 20 meters/sec^2 at a 60 degree angle instead of straight up and starting at the height of 11.5 Km the net acceleration upward is near 7.5488 m/s^2, and horizontally at 10 m/s^2, allowing for the change of gravity with altitude and the acceleration vector of the craft due to the horizontal velocity. This requires just under 146 seconds of thrust to eventually reach 150 kilometers.

V(loss) < 3,080 m/s.

The two velocity components are:

Up = 953.4 m/s after losses
and horizontal = 1620 m/s

The up velocity component will be all lost as the craft gains altitude. Since the needed dV here is 7,814 m/s and the achieved velocity above is only 1620 m/s An additional 6,194 m/s is needed to get up to keep in an orbit when at 150 Km.

So the total velocity change needed is:

2,920 + 6,194 + 160 + 131 or 9,405 m/s in a polar orbit

Going West you add Cosine Latitude times 463.32 m/s to account for the Earth's spin, going East you subtract the same amount, Using a dV of 9.41 Kps and setting latitude at 48.9 degrees yields the delta Vees:

Going West : 9.71 Kps
Polar: 9.41
East : 9.11

On several occasions people have asked me how chemical reactions could possibly have enough energy to get to LEO in a single stage, when the most powerful system now used, Cry-Lox [liquid Hydrogen and liquid Oxygen (lqH2/lq O2),] can't do that.

There are more powerful fuel mixes, but they aren't needed, Cry-Lox could make an SSTO. It isn't the higher Isp that is needed here. A higher Fuel Mass Ratio will do just as well to impart the needed energy.

FMR = e^(Delta Vee/Vex), Isp = Vex /9.807, so Vex = Isp*9,807

Using an assumed Isp of 353 seconds (the value of the lowest fuel mix we expect to use), then the in vacuum Vex is 3.461 Kps (kilometers per second) and the maximum needed d(v) is 9.71 Kps so the FMR is 16.6:1. This does not compare favorably with the Space Shuttle's 5.8:1 overall ratio, but it does to the 17.08:1 ratio of the orbiter's mass (including cargo) to the mass of the lift off stack.

The actual FMR is nearer to 18.01:1, and assuming the light aeroshell mass (below) just under 37:1 per kilogram cargo put into LEO. By comparison, the typical ratio for net cargo on the shuttle and most expendable launchers is 50:1 to 70:1 and the cry-lox tanks need to be very big due to their low density (0.29 grams/cc to, at most, 0.34 gm/cc). The mixes that we are considering all have densities of about 3X that, so the tanks and all related plumbing are a third the expected sizes.

Using this exhaust velocity and the Delta Vees above. The FMRs needed are:

FMR = e^(dv/Vex)

so that

West = e^2.806 or 16.55:1
Polar =e^2.710 or 15.17:1
East = e^2.633 or 13.92:1

With another factor of 1.088 for the propellants needed to get to the starting altitude (11.5 Km). Each of the other propellant mixes has a lower overall FMR required.

On to the related problem of building a strong, light weight, craft. Many structures can be made that beat the egg's structural strength of supporting 20 times it's own weight and some composites can be made to support over 500 times their own weight. About half as strong in fact. The shuttle has a GLOM of about 2,050 tonnes and a total structural mass near 300 tonnes. The maximum acceleration of the whole stack is 1.5 gees and it can't go over this until it first looses about 25% of its fuel and jettisons the spent SRBs. So it is only supporting 6.833 times it's mass at the 1.5 gees or 10.25 times the structure's weight whereas an eggshell can support 20 times it's weight.

Using the highest FMR above, such a composite structure can carry, in addition to the propellants needed, 13 times as much total mass (structure and cargo) as the mass of structure alone. The net cargo is over 10X the mass of the empty structure and it will still withstand 2 gees fully loaded, something that the Space Shuttle cannot do. Making the Shuttle frailer then an eggshell !. (footnote 1)

With the stronger composites, the entire craft (support structure, engines, avionics, aeroshell and all) can mass only a little more than the cargo carried, or even mass a little less then the cargo and still sustain over 2 gees while fully loaded. You can prove this to yourself for a few cents:

Get three 6 Oz foam cups and a 2 liter pop bottle. Stack the cups on a flat level surface as upright, top to top and bottom to bottom. Put a full 2 liter bottle in that top cup.

The centre and bottom cups are each supporting at least 2 kilograms against 1 gravity without being crushed. Each Cup is about 2 grams mass and is supporting 1,000 times its own mass, if only for a little while. The main difficulty is in keeping the forces even. With a thin coating of fiber composites, the cup could easily support far more.

Clearly, if it could be made to survive the environment and the forces on it were evenly distributed the material of that cup could hold 500 times it's own mass even if you include the mass in coatings needed for that survival.

Assuming a requirement of 150 kg for irreducible flight hardware, a basic structure of the craft as equal in mass to the cargo to be carried (and a 250 kg minimum capacity) going due West at 150 Km, then the payload required is only 650 kg and the Gross Lift Off Mass is 11,707 kg.

Or assuming that the structure alone has to support the whole mass at 2.05 Gees with some strength to spare it can hold 250 times it's mass. For a Glom under 12,000 kg it is conceivable that the aeroshell would mass a mere 48 kg! in fact at the demonstrated 1000:1 support from the foam cups you could get by with just 12 kg. It takes more, of course, to get the aeroshell to keep its shape in the environment it must pass through, prevent any outgassing or melting, fracturing etc. The engines are the heaviest single components and make up for half the craft's empty mass.

The only remaining "everybody knows" problem is how to survive the re-entry heat. The heat is caused by crashing back through the air for a fifth of an orbit or less. The entire craft does not heat up in this process even though it does lose about 7.3 Kcal worth of kinetic energy per gram mass of craft. This is shed through the shockwave and reradiation into the surrounding air. Using the Stefan-Boltzmann equations, temperature will depend on the total energy to be radiated, radiating area and the time. Flying back prevents most of this and can be done even with an non aerodynamic shape, as was done with the Venus orbiter. With a shape designed to have both negative and positive lift it is even easier. It can stay in the thin air until its cool enough, and slow enough, to go lower.

With a craft like this, several things are possible. In order to operate it as a small freight forwarding company all the costs of R&D, Maintenance, operations, and return on investment need to be considered. Since the ticket prices will be quite high by air freight standards the ticket agents get only 10% of it, while the insurer gets 3% [ it takes a hefty bond to cover space flight] and the gross profit is about 40% of what's left. Hence the price needs to be at most 1.91 times the actual total cost of the craft and operating it. As traffic increases this ratio can lower, but not below 1.3565 and still keep gross profits at 19% of ticket price.

To figure the cost of operating a spacecraft we use an algorithm that goes by the name FoCaL, which stands for Fuels, Overhead, Capital and labour. Each component cost is added to get the cost of operations then multiplied by the ratio above.

A space craft should not be any more complicated than an aircraft. They have some of the same features as a home-built submarine and a homebuilt jet kit plane. With the learning curve as shown in Joseph P. Martino's Reason report it can cost less. The cost per unit lowers exponentially with each doubling of the number made in a production run. Historically this rate has been .8, that is, with N units built, N=3D2^x and the cost of the 1st one is $(0):

$(N) (cost per unit with N units made) = .8^x*$(0)

For a more exact figure you can split the overall costs to include different bases for the materials and the labour components of the cost. For any given manufacturing method this follows an "S" curve approaching a minimum, but all the technologies keep changing too as part of the learning curve yielding the historic, logarithmic, decrease in the cost per unit above.

Complex short run aircraft cost under $1,200 per kg and one of a kind kit planes built from mail order composites run rather less. The 650 kg craft should be able to be built, eventually, for under $ 780K and so at car loan bank rates cost under $3,240/ week in financing charges. 2 Times that if you want to replace it after its useful life time. The launch cost per kg to LEO due just to construction cost will depend on the mass of cargo the craft can carry and the number of flights per week.

Since the turn around only requires a cleaning, refueling and flight recorder check, it can be loaded and ready for another flight in under 12 hours. In fact it can be done in under 8 hours from wheels up to wheels up on consecutive flights. 3 flights per day, six days a week. Figuring only 4 flights per week with 300 kg of cargo to pay back all of the finance charges above the cost per kg to LEO due to the finances and construction costs is less then $6.00.

Similar assumptions show a propellants' cost of under $25/kg to LEO where the same amount of Cry-Lox and SRB stuffing that the Space Shuttle uses would only cost $18-$20 if that worked as well. The Shuttle actually uses about 60 kg of propellants, about $90 worth, per kg cargo. Our proposed propellant mixes all cost less, and we need less of them (under 46 kg) to launch a kilogram into LEO.

Because of the simple systems it only needs a ground crew of 6 and the total time needed is under 12 hours or 72 labour hrs at a cost of $45/hr for a labour cost somewhat under $11/kg to LEO.

For various reasons the general overhead cost of launch ing runs about the same for a given size craft and rate of flights. The formula I use covering all expected costs and a margin is a function of the amount of cargo launched per week. In tonnes per week:

Cost/kg = [((607/cargo)+6.72)*1.125] or at 4 flights of 300 kg each it's 1.2 tonnes cargo per week with an over head cost of under $576/kg.

At this rate, with all rented equipment and off the shelf parts the cost is under $618.00/ kg, for a billed price, due to the earlier assumptions for running a business, of $1,181/kg fob LEO. This drops to under $562/kg at 12 flights per week.

Of this early investors could share $179/kg at a rate of 20% to run company as a business pay primary inventors, R&D costs licenses etc., with profits. 80% pay back to investors as:

  1. 40% primary investors putting up high risk and seed capital. Roughly a total of $1.5 million.
  2. 20% secondary investors putting up 3 times as much to build craft and test fly them. approximately $4.5M
  3. 18% ordinary stock etc for expansion into fleet of vehicles. $36M to $300M.
  4. 2% employees' profit sharing plan. Bonuses.

These are shares of over $36M/year of an estimated $190M billed/ year after one year of operations using just two small craft. For 320 tonnes per year at $600.00/kg fob LEO. The business should be self financing after step A, in the first year from advances on contracts and a higher billed price per kg cargo at first. With just one craft and 6 flights per week at $900/kg fob. LEO the profits drop to $12 M/year on $67.5M worth of business, about 75 tonnes. Four flights per week at $1,200/kg still yields a gross profit of over $11M/year on 50 tonnes cargo.

With increased flights we can lower the overhead costs too. The asymptotic limit to decreasing the costs is $150.00/kg fob LEO or under 1% of today's cost on the shuttle and on the expendable launch vehicles. With payloads of 250 kg or less the cost of launches can exceed $60,000/ kg with a dedicated flight, and is always over $20K/kg. Even the still not usable DC-X(a) [delta clipper] and X34-B launchers will cost about $3,000/kg to LEO and they may not be able to launch more often then once every 2 weeks nor halve that cost. No other launchers can reasonably predict a cost under $1,500/kg fob LEO without government subsidies.

In any given market, when the cost of something lowers the total amount spent in time or money per year does not lower, but either remains just as high as before, or stays at the same fraction of what's available. Usually it will increase in both with lowering cost/unit. Roughly as:

N(new) = N(old)/(d$)^2

The 1992 market for launch to LEO was over $5.8 B at about $16,000/kg, with a 3 year long waiting list for a flight that sometimes got canceled anyway.

Many things that can't be done at a profit with a cost of $16,000/kg can be, and will be, done once the cost lowers to $600.00/kg fob LEO. Some can make a profit, even at $20K/ kg if they could just get more flights, reliably.

A small fleet of the 300 kg capacity craft would not only finance the company, but build the larger and a little more complicated 15,000 kg capacity craft to transport not just freight, but tourists, to LEO and still operate on a daily basis -like a regular airline.

This is a plan to finish the R&D, build a prototype vehicle and test all the basic concepts. Using a new design to avoid some of the problems that are perceived in other SSTOs. For instance:

To prevent the build up of a bubble of explosive mixture, the craft doesn't use a launch gantry, it takes off horizontally, like a plane. Any problems and it can glide back to a horizontal landing, fix them and try again. It also does not use any pyrotechnics, such as the usual ignitors, SRBs, explosive bolts for separation, etc, but reusable electric circuits. Nothing to throw away each time.

Ideas are like mutations. Even when 99% of them are unsuccessful, it's that other 1% that work that are responsible for all progress. New businesses are more successful and even at a 90% failure rate, in 22 tries that rate reverses to more than a 90% chance that one will succeed. If each try is cheap enough in and of itself, then it costs less to try them all (or anyway several dozen of them) than to spend far more in an attempt to get a guarantee for a single, successful, test. And it's quicker!

  1. Reason Foundation policy study 173 "Meeting Space Launch Needs - Economically"
  2. JBIS 47-3, March 1994 'The Cost of Access to Space' by M.D. Grifin and W.R. Claybough II
  3. CET-89 simulation NASA sp 273 Lewis research centre
J V H Hill, April 1999, "Getting to Low Earth Orbit", April 1999 (Adapted from an original paper published in 1996).
Also downloadable from to low earth orbit.shtml

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