# Island in the sky - helium launch platform

1. Jan 22, 2004

### jjalexand

This is a new launch plaform idea to assist Nasa in meeting it's extra-planetary goals.

It seems that a large helium-supported platform in the upper atmosphere could save a lot of launch energy and pollution.

The platform might be quite large, perhaps 0.5 to 2 km in diameter, and could be used for launching much smaller rockets than a Saturn 5, which would require far less fuel for a similar payload when launched from this height.

The platform could be covered in photovoltaic cells for energy, or the underside could receive microwave radiation from a power station on the ground.

Launch vehicles, fuel and supplies could be raised to this height using large helium dirgibles for minimal energy cost and pollution. The dirgible gas could be compressed to return to ground, and platform waste used for additional ballast.

Possibly an appropriate hydrogen/helium mixture would provide better boyancy without fire risk.

Solar power might be used to produce hydrogen and oxygen electrolytically on the platform from ambient water vapour, if enough is available.

The platform would include a small base for the launch and operations crew. The platform would rarely if ever return to the ground, instead forming a semi-permanent 'island in the sky'.

A larger plaform could be assembled from smaller standardized (hexagonal?) units. Units could be produced and added when required, to smooth production costs and provide flexibility and reduced risks.

Multiple units would provide redundancy and safety.

More than one platform could be assembled from such units. Different countries could supply units according to their GDP's for an international effort.

A multi-layer plaform could provide protected pressurized living and engineering workspaces.

The platform might also allow easier reentry, with the start and end of an astronaut's journey being a leisurely balloon trip between the plaform and the ground.

This option seems to have many benefits and yet it may be quite possible using existing technology taken to an extreme. The cost savings and might be dramatic over several years of operation, and the additional flexibility and reduction in pollution could provide other benefits. Upper atmosphere winds may prove a problem.

The platform(s) would probably be best sited over the oceans to avoid shadowing and visual pollution of populated areas.

Later, space tourism could start with visits to the plaform to assist in funding operations.

2. Jan 22, 2004

### enigma

Staff Emeritus
You know, one of my teammates for this project mentioned the balloon idea, but I just can't see how it would save you enough to be cost effective.

You'd save maybe 1km/sec from drag losses because you're out of the atmosphere. That still leaves over 7km/sec you'd need to spend to reach orbit.

If you've got a platform up there, how would you safely transfer the rocket from the transfer vehicle to the platform prior to launch.

Re-entry would be worse. You'd need to slow down to zero velocity at a much higher altitude than current craft do. You wouldn't be able to use much of the atmosphere for aerobraking either.

It's a cool idea, but it definately needs some kinks worked out.

[/engineer]

3. Jan 22, 2004

### jjalexand

The cost and pollution savings are from avoiding say the use of a first stage of a saturn rocket.

I'm not too sure at what height a second stage would separate (some hand waving even flailing here :), but I know most of the fuel is used just to lift the rest of the fuel.

I would not be surprised if 90% of the total liftoff fuel of the whole assembly is used by the time it gets to say 20km.

Thus a platform at such a height might save 90% of the fuel and might allow light reusable shuttle-like craft to enter space and reach orbital velocity without any booster.

Lets face it, the current brutal method of using huge amounts of fuel just to lift the rest of the fuel to a reasonable height has got serious warts on it.

I hope some engineer could clarify the above and possibly give the the approximate fuel utilization ratio from ground to different heights, assuming some reasonable staging and vehicle/fuel mass ratio assumptions.

Since fuel can be lifted to the platform by dirgible, or produced in situ photovoltaically, we would not be wasting most of it lifting itself 20km upwards.

So the whole process could be much more efficient than it is now.

The wind problem might be difficult to overcome. Does anyone know if there are any feasible heights without much wind? Or I suppose the platform could be allowed to move with the wind. It might be technically feasible to distribute smaller platforms vertically and attach them by tether, so wind at one level would have a reduced effect.

The main benefit is for launch. Rentry landing on the platform would be a bonus. Re-entry rendezvous with the platform may not be feasible for risk reasons such as accidents damaging the platform, or might use MORE fuel than current techniques, but the convenience and overall fuel savings might make up for it.

Also this idea might just be used for a commercial floating hotel, or even a small floating village, but you might pay a lot to join the 10-mile high club :)

4. Jan 22, 2004

### enigma

Staff Emeritus
It's not the up that makes the orbit. It's the fast. The up is only necessary to get out of the atmospheric drag regime which would bring the ship back down. Every spacecraft in orbit is going over 7 km/sec. There isn't any way around it.

5. Jan 22, 2004

### jjalexand

The total energy used falls into the following categories:

1. Accelerate the payload to orbital velocity (small fraction)
2. Lift the payload to orbital height (small fraction)

3. Accelerate the mass of the fuel (large with respect to mass of payload) to some average sub-orbital velocity (large fraction)
4. Lift the mass of the fuel to an average sub-orbital height(large fraction)

5. Accelerate the redundant stages to sub-orbital velocity (medium fraction)
6. Lift the redundant stages to sub-orbital height (medium fraction)

7. Heat up the atmosphere (mostly in the bottom 20km and mostly due to the first stage). (unknown probably large fraction)

Use of the platform attacks items 3,4,5,6 and 7, most of which have larger fractions of energy (fuel) use.

Feel free to post calculations to prove me wrong :)

6. Jan 23, 2004

### jjalexand

combine floating platform with rail gun

Possible the 'Island in the Sky' could carry the 'Rail-Gun in the Sky' or similar propulsive device (possibly powered by accumulated energy from microwaves beamed from the ground) to launch (probably unmanned) objects into orbit, (such as fuel brought up by dirgible, parts, vehicles, etc).

This is much more feasible than it would be from the ground as there would be approx 1/10th the pressure and approx 1/10th the distance to vacuum, leading to approx (? 1/10th x 1/10th =) 1/100th? the integrated frictional energy loss compared with a ground launch from a rail-gun.

7. Jan 23, 2004

### enigma

Staff Emeritus
Vis Viva

$$V=\sqrt{\mu*(\frac{2}{r}-\frac{1}{a})}$$

V is necessary velocity

$\mu$ is (Gravitational Constant)*(Mass of the Earth) or 398600 km^3/sec^2 for Earth

r is the distance from the center of the Earth

a is the semi-major axis of the orbital ellipse

For a circular orbit at an altitude of ~250km, r=a=6378+250km=6628km

V=sqrt(398600/6628) = 7.75 km/sec

Regardless of staging.

*******************

Going up to whatever altitude in a balloon (no more than 15-20km or so)doesn't get you going that speed. It just gets you out of the atmosphere, which bleeds off between 1 and 2 km/sec during liftoff.

8. Jan 23, 2004

### jjalexand

Isle in sky - reply to enigma calc 1

Sorry Enigma, thanks for the formulae, you are probably an excellent physicist, but all those calculations show is how to derive orbital speed, they do little to support any argument about the ratio of (a) fuel/energy used to get a Saturn 5 to 20 kms against (b) fuel/energy used to get perhaps the last stage of a Saturn 5 vehicle to orbital height and orbital velocity from a 20km standing start.

What height did a Saturn 5 drop it's first stage? What was the ratio of fuel in that first stage? What height does any reasonably sized current launch vehicle drop it's first stage? What is the ratio of horizontal velocity to orbital velocity at that point? These questions should be almost public knowledge, and would suggest an answer the question a lot more definitely.

The energy calculations are probably not that hard to do either. Energy required to accelerate single stage vehicle and payload from standing start to orbital speed vs integrated energy required to lift all fuel and stages to their respective heights from the ground and to accelerate them to their respective heights and integrated enery to overcome friction on total vehicle from ground to orbit.

Just becuase an idea is apparently new does not mean there HAS to be something wrong with it (although there often is :)

It would be great if you can find any of the info you mentioned, I think Nasa needs some help right now, and I don't think we are ever going to really get off this planet without some new ideas.

Regards

9. Jan 23, 2004

### jjalexand

Island in the Sky - Bifield-brown effect

The 'Island in the Sky' might be able to use the Bifield-Brown effect to steer, for stability, for partial motion control in wind, and to raise itself up and down, as it would be a very low density object with a relatively large surface area. Maybe also to provide some pressurisation via venturi's, etc, perhaps in one integrated mechanism.

Energy could be from photovoltaics or beamed microwave or laser energy. One advantage of the B-B effect is that it can utilise existing surfaces (with insulated segments) as a major component of it's physical mechanism, and the remaining components might be quite light.

Might be better not to use any Hydrogen in this environment :)

10. Jan 23, 2004

### enigma

Staff Emeritus
Re: Isle in sky - reply to enigma calc 1

Engineer

The reason I didn't go into more details is it isn't an easy problem to solve. There are huge tradeoffs all over the place. All the numbers depend on inert mass fractions, actual trajectories, etc. which are different for each vehicle and each launch.

Ideal rocket equation:

$$\Delta V = -Isp*g_0*ln(\frac{M_F}{M_F+M_P})$$

Isp is the specific impulse of the engine. 450 or so for a shuttle engine
g_0 is sea level acceleration: 9.8m/s^2
M_F is the final mass
M_P is the mass of the fuel expended

Now, the only way we're going to get anywhere without digging deep or running simulations is to make some approximations.

We'll assume that all atmospheric drag and non-tangential velocity can be summed up as a 2km/sec impulse. Doing the actual calculations would take a small team of engineers a few weeks, I'd imagine.

From here, we'll compare two http://liftoff.msfc.nasa.gov/academy/rocket_sci/satellites/hohmann.html. One from Earth surface at 0 velocity and one at 20km at 0 velocity. Both are going to a 250km circular orbit. We'll assume no inclination changes or Earth rotation boosts.

The final velocity at the end needs to be 7.75km/sec measured earlier.

250km transfer ellipse:

a=(6378+6628)/2 = 6503km

Velocity needed at the ground (perigee)

V=sqrt(mu*(2/6378-1/6503))=7.98km/sec

Velocity of the transfer ellipse at apogee

V=sqrt(mu*(2/6628-1/6503))=7.68km/sec

So from the ground, you need a delta V of 7.98km/sec at the ground, and .07km/sec to stabilize the orbit at apogee. Then you add the losses in. Total is about 10km/sec

For the other scenario, a = (6403+6628)/2=6515km

perigee
V=sqrt(mu*(2/6403-1/6515)) = 7.95 km/sec

apogee
V=sqrt(mu*(2/6628-1/6515)) = 7.68 km/sec

Giving a total of 7.95 km/sec at the station and .07 to stabilize the orbit. Your total is the same +- a fraction, just without the majority of the drag losses, like I stated above.

Both of those are approximations. You can't have a single impulse burn, particularly a large one like the initial firing at liftoff, but the equations aren't closed, so you need to do many calculations to come up with an approximation. Basically, all the expenditure is done in the first few minutes, you coast up to your orbital altitude, and then you do a small burn there.

When you release each stage, how you break your fuel up, etc. depends entirely on how much mass you need to put into orbit and how many stages you decide on. There are tradeoff formulas which determine how many stages is best from there. When you release each stage depends entirely on the fuel/mass breakups of the structure. They don't make a difference on the actual energy required.

The ideal rocket equation is used to determine fuel needed and gives you inert mass to fuel ratio. Fuel is cheap. Structure is not, so you try to minimize the structural requirements. Again, more tradeoffs to determine the best route (isn't engineering great?).

So now the only question which needs to be asked is this:

Is the 2km/sec or so you're gaining worth the expense and risk of using a launch platform where you'd need to have constant and relatively expensive food/oxygen/maintenance shipments, the "ground" crew would need to work in bulky pressure suits, the entire thing would need protection from UV radiation/monatomic oxygen/etc., you'd have no real abort sequence in case of minor failure (launching from the platform poses the risk of bursting the balloon, so you'd need to drop the rocket prior to ignition), it's a REALLY long way down in case of fire/emergency/accidental fall, etc.

That's why I said it's an interresting idea, but the engineering details would need to be fleshed out significantly before it's feasible.

Last edited by a moderator: Apr 20, 2017
11. Jan 23, 2004

### jjalexand

Sky Isle - partial response to Enigma calc 2

I still don't think you are really tackling the problem I thought we were discussing. When the whole argument more or less rests on things like how much energy is used to lift the average qty of fuel required over the time is is used, the assumption of a single impulse burn completely ignores the problem. Of course, if you could use all the fuel at once on the ground, you would't have to lift any fuel at all, and you might use 1/10 the the qty, yet a large part of the argument is precisely about that.

You're not really addressing the issue. We are talking about the difference in energy between getting a payload to orbital velocity and getting some other much heavier stuff like fuel and structure to some velocity roughly half that, as well as lifting a much heavier thing to a height of 20 km or so. You can't just hide all that inside the wrappings of a 2km/sec delta V with a single impulse burn, it's a joke. What delta V and energy would you get if you dropped a Saturn 5 half full of fuel from 20 km! Quite a lot I think! This is about the same energy it would take to get it up there.

Clearly there is a very big difference in the energy required in the two circumstances. Let's find out what height a Saturn 5 dropped it's first stage, rather than making up our mind's it's not going to work, then trying to prove it with accurate calculations based on possibly irrelevant or flawed models and assumptions. I bet you an Aussie dollar against your US dollar that it lost its first stage (and over 50% of it's total fuel) below 20 kms!

Also, structure may cost more than fuel economically, but what about the environment. I believe there is significant pollution even from Hydrogen burning. And all that clean Hydrogen could be used elsewhere as replacement for fossil fuels. And it probably caused a lot of pollution in it's production.

Regards

12. Jan 23, 2004

### enigma

Staff Emeritus
Re: Sky Isle - partial response to Enigma calc 2

That's where the year of orbital dynamics comes into play, I suppose. The single and only measurement of energy in space is delta V. That's it. You don't care how much fuel you have, because it depends on the rocket. You don't care what type of rocket you have because it depends on the fuel. All that matters is the delta V you obtain from the ideal rocket equation (which takes both fuel use and rocket type into account...).

The single burn is the lowest energy solution. It's an approximation, sure, but it is a decent approximation. The firing, even during liftoff of the Saturn V, is done in the first 5% of the transfer ellipse.

The fuel used doesn't matter when calculating the velocities needed. You find how fast you need to be going first and foremost. Once that is done, like I said, you plug it into the ideal rocket equation to find out how much fuel you need.

You'd still need to accelerate all the fuel in the imaginary fraction of an instant you're assuming the burn takes. It doesn't matter if you've got a 100,000,000N force for 1 second or a 100,000N force for 1000 seconds. You've still got the same impulse at the end.

If your burn takes only a second, then after half a second you've got half the fuel left which is getting accelerated along with the inert mass. The time it took is irrelevant except for course correction and aerodynamic (etc.) losses.

Instead of a constant acceleration and a smooth increase in velocity, you've got a short, powerful acceleration and a rapid increase in velocity. In either case, you need to be going at the same speed at the end. In either case, you need to accelerate all the fuel at the start of the burn, half the fuel at halfway through the burn, etc.

It is exactly what is done, joke or not. It is simply making an integration for the total negative impulse imparted during the ascent, and saying: "OK. Total loss was this."

Everything you gained in kinetic energy, you'd lose in gravitational potential energy. They're all tied into the Vis-Viva equation above.

No bet. It did. Probably even in the first 5km.

The thing you're missing is that it was going at about 2-3km/sec when it did dump that first stage. You can't just hack off a stage, start at 0 velocity and say it's no different. You saw the calculations up above. You still need to get yourself up to the same speed, over 7km/sec, even if you're at 20km altitude to start with. There was no significant difference in the required velocities. If you hack off a first stage the only effect is that the remaining stages need to be bigger.

2 H2 + O2 = 2 H20 only

Fossil fuels pollute because they have all those nasty C's and S's along with the H.

If it caused pollution in its production it would still cause pollution in its production if you're lifting it to 20km.

If you use RP-1 (which is more or less kerosene) at 20km, the byproducts will still filter down (and are less polluting than equal amounts of automobile or airplane fuel).

Again, Hohmann is an approximation, yes. Unless you're using an ion engine which operates for days, weeks, or months constantly, Hohmann is the approximation used up until you're actually planning the specific trajectory and course - very late stage planning.

Last edited: Jan 23, 2004
13. Jan 23, 2004

### enigma

Staff Emeritus
I'm looking over your posts again.

The main problem you're having with visualizing the problem is the position vs. velocity issue.

Simply being high doesn't make you in orbit.

If you're at 250km with no velocity, you'll fall down to Earth. If you're at 250km with a velocity of 6km/sec, you'll fall to Earth somewhere on the other side of the planet. Either case, you'll crash.

If you're on the ground going tangentially with a speed of 10km/sec, your momentum will pull you into a highly elliptical orbit (ignoring drag). You can even find out how high you'll get before coming back down to your starting point using the Vis-Viva equation provided in this thread.

If you're on the ground going roughly 7.8km/sec you'll orbit the Earth at an altitude of 0 km (assuming no drag or pesky mountains of course ).

Where the rocket is when it dumps its first stage doesn't matter. What matters is how fast it's going when it does it.

Last edited: Jan 23, 2004
14. Jan 24, 2004

### jjalexand

Sky Isle - response to enigma

I've done some research on this, and discovered that the Saturn 5 lost it's first stage at 61km (not 20 km or 5km), so my guesswork was wrong on this, as was Enigmas, who I now owe 1 A\$ to!

However, I have also calculated by very simple means from the stats below (assuming a maximal linear burn rate for the first stage) that it had used about 50% of the total vehicle propellant by the time it reached 20km, which I think is a significant fraction!

I'm not quite sure if Enigma ever really addressed this percentage, which was one of the critical issues in my mind at least, not delta V.

For most of the journey up to 20 kms, the Saturn 5's vector was probably more or less vertical (presumably Enigma's orbital mechanics year taught him that it is more energy efficient to travel more or less vertically upwards during the slowest part of the journey until losing most of the atmosphere to reduce non-linear air-resistance losses before trying to achieve a more rapid 'horizontal' orbital vector. (From memory, wind-resistance forces are roughly proportional to the cube of the air speed). That means go up first until more or less out of the atmosphere, then go sideways.

Also, as this was the early part of the journey, and it was still accelerating a very heavy mass (not much fuel used yet) it's acceleration and it's speed were still relatively low at a height 20km (much less than 1/3 of its separation speed at 61km).

Thus by my mental model, it had made very little progress towards establishing any 'horizontal' orbital velocity vector at 20kms. Thus the approximatly 50% of propellant used to get to 20km mostly wasted, and could have been avoided by a 20km launch, with some additional savings by using an inherently lighter and cheaper structure.

I applied the same initial rough calculation to the space shuttle launch and got similar figures, although the calculation is slightly more complicated. The space shuttle fuel mass and stage separation height figures are also held at the site below.

The Saturn 5 stats are pasted here as CSV info. You can cut and paste to a text file then load into Excel or Open Office Spreadsheet for example. These figures are compiled by me from info at http://users.commkey.net/Braeunig/space/specs.htm [Broken] by Robert A. Braeunig, with thanks.

Saturn 5,,,,,,,,,,,,,,,
,Designation,Diameter,# Engines,Eng Manuf,Eng Designation,Dry mass,Oxididant,Fuel,Propellant Mass kg,Max Burn Rate,Total Thrust,Burn #,Burn time,Altitude,Speed at Burnout
,,M,,,,Kg,,,Kg,kg/s,KN,,S,Km,km/hr
Stage 1,Saturn 1C,10.05,5,RocketDyne,F-1,137440,LOX,RP-1,2077000,12885,35140,1,150,61,9800
Stage 2,Saturn II,10.05,5,RocketDyne,J-2,45000,LOX,L.Hydrogen,427300,1205.5,4450,1,395,184,24700
Stage 3,Saturn IVB,6.6,1,RocketDyne,J-2,15240,LOX,L.Hydrogen,105200,241.1,1023,1,165,185,28060

The logic of my model is this:

Look at the propellant used to get to 20 km (unknown) as against total vehicle propellant (known).

We do know the following:

(a) Height of burnout of first stage (61 km)
(b) Speed of first stage at burnout (9,800 km/h)
(c) Time to get to first stage burnout (150 sec)
(d) Propellant used by each stage (2,077,000 427,300 105,200)
(e) Total thrust of stage 1 (12,885,000 kg.m/sec^2)
(f) Total dry mass (13,440 + 45,000 + 15,240) = 197,680 kg
(g) Stage 1 engine maximum burn propellant rate (5 x F-1 engines = 12,855 kg/sec)
(h) Dry weight of vehicle (197,680 kg)
(i) Total mass of propellant (2,609,500 kg)
(j) Mass of propellant in first stage (2,077,000 kg)
(k) Total launch mass (2,609,500 + 197,680) = 2,807,180 kg

We can deduce that propellant utilization of first stage is more or less constant until burnout because rate of use of propellant is close to maximum burn rate of 5 F-1 engines.

Thus _thrust_ to 61 km is more or less constant (but not vehicle speed or acceleration).

Initially let's assume for expository reasons that the speed of the rocket is constant during the first 61kms. The vehicle would then burn roughly 1/3 of its propellant in each of the 3 x 20 km height segments (3 x 20kms = 61kms approx :) of the first stage burn trip.

Note that under this initial (false but instructive model), the ratios of fuel used in each of the three segments would be 1:1:1

This would amount to 1/3 of the first stage 1 propellant, i,e 33% of 2,077,000 = 27% of the total propellant for all stages. (A)

However, (A) is quite wrong, because the speed of the rocket is _not_ constant during the period, because it is constantly accelerating.

The vehicle spends _a_lot_ more time in the bottom 20 kms than it does in the top 20 kms of the 60km, because it is is going a lot _slower_ in the bottom 20 km height segment.

Since burn rate is constant, propellant burned in a given time interval is proportionate to the number of seconds in that interval.

The time spent in each interval is inversely proportional to the average speed in each interval.

Let's assume for the moment that the acceleration of the vehicle constant during first stage burn.

Under this assumption, speed will increase linearly from 0 meters per second.

This would make the average speed in each of the three intervals bear a ratio to one another of 1:2:3.

Elapsed time spent in each interval would be inversely proportional to this, i.e. ratios of 3:2:1, instead of 1:1:1 in the constant speed scenario (A)

This 3:2:1 ratio would mean we would use a ratio of 3:(2+1) = 1:1 of the total stage 1 propellant in the first 20 km and the rest in the next 40 km.

This means we use 50% of the stage 1 propellant in the first 20 km segment. This amounts to 40% of the total propellant for all stages (B)

But wait, there's more. Even (B) is wrong, because of another major factor we have so far ignored.

The mass of the vehicle is changing rapidly as it burns propellant.

That means the assumption of uniform acceleration use to derive (B) is false.

We can now correct scenario (B) with a first approximation of the change in acceleration over the period from launch to first stage burnout.

by

f=ma (implies a=f/m)

if

T is the total stage 1 thrust (constant)
m1 and m2 are the masses of the vehicle at launch and just prior to stage 1 burnout respectively
a1 and a2 are the corresponding accelerations at these two points

then (since the acceleration increases linearly over the period until stage 1 burnout, as propellant is burned linearly and total vehicle mass decreases linearly)

a1 ~= (T-m1)/m1
a2 ~= (T-m2)/m2

where ~= means 'is approximately equal to' (i.e. ignoring changes in graviational field with height, non-linear air resistance, etc)

If we draw a straight line between a1 and a2, dividide it into 3 equal segments, and then mark the midpoint of each segment as follows:

a1 !-----'-----!-----'-----!-----'-----! a2,

we can see that the average acceleration at each midpoint is (5a1+a2)/6, (a1+a2)/2, (a1+5a2)/6,

and by multiplying each term by 6 we can see that the average accelerations during each of the three 20 km height segments are in a ratio of (5a1+a2):(3a1+3a2):(a1+5a2)

In our case, the initial acceleration a1 is (12,885, 000 - 2,807,180)/2,807,180 = 3.59 m/sec.

(remember how slowly the large Saturn 5 (or any very large rocket) seemed to lift off the ground?

That would have been about 3.5 meters a second, yes?)

Just pre stage 1 burnout the new total vehicle mass is approximately (2,807,180 - 2,077,000) kg = 730,180 kg,

thus a2 is (12,885,000-730,180)/730,180 = 16.65 m/sec (about 4 times a1)

Thus the ratios of the three 20 km segment midpoint accelerations is (5 x 3.65 + 16.65):(3 x 3.65 + 3 x 16.65):(3.65 + 5 x 16.65)

which is 34.9:60.9:86.9 or approx 35:61:87

This means the speed is increasing a lot more slowly in segment 1 than in 2 and 3.

We can approximate the effect of the different accelerations on the times in the three intervals by multiplying the 1:2:3 speed ratios (not the inverse 3:2:1 time ratios) derived in (B) under the assumption of constant acceleration by the new acceleration ratios of 35:61:87 (I think this is right, but I'm not quite sure about this step, perhaps someone could enlighten me (Enigma?), I'm getting a bit tired here, this might not be the correct transformation here but an additional effect is there nonetheless)

This gives new ratios of 1x35:2:61:3x87 giving 35:122:261. The time spent in each interval is inversely proportional to the speed, giving time ratios of 261:122:35.

Since propellant use is constant over time, we can use the same ratios for propellant use (261:122:35) in the three 20km segments.

This means the ratio of propellant used in the first segment is 261:(122+35) = 261:157

This means propellant used in the first 20 km segment approximately is 261/(261+157) of the total stage 1 propellant = 62% of the stage 1 propellant, or 49%, or slightly more approximately 50% of the total propellant for the launch.

Thus a similar mission launched from a stationary platform at 20 kms could use significantly less fuel, and lighter structures, although the savings may not be as large as 50% since the conventional launch vehicle does have some (presumably vertical) velocity vector of 9800 km/h at a height of 20 km which would not apply to a plaform launched mission, unless it was launched from a plaform-based rail gun or similar device.

Perhaps Enigma would like to work out this propellant usage to 20km more accurately with the conventional methods from the stats above.

One more thing for Enigma, simple chemistry is not enough when it comes to estimating pollution from burning hydrogen in air, I believe oxides of nitrogen are formed as a side effect due to the heat, and that these have a negative impact on the environment, although this is not generally realized. I think this was published in the last 6 months in the New Scientist or the Scientific American.

Last edited by a moderator: May 1, 2017
15. Jan 24, 2004

### jjalexand

Correction a1 and a2 units should be in m/sec^2

Correction: The units for a1 and a2 should be shown in m/sec^2, not m/sec.

16. Jan 24, 2004

### jjalexand

In response to Enigam last post,

I'm glad you're starting to explicitly separate the horizontal and vertical components.

Notice that the only reason to travel upwards is to get above the atmosphere.

Not travelling the vertically up the first 20 km by rocket power but by floating up in a dirgible saves a lot of fuel without any effect on the energy required to achieve to achieve an orthogonal _horizontal_ orbital velocity.

Neglecting for a moment the effects of the atmosphere, whether we start on the ground, at 20 km or at 100 kms makes little difference to the energy required to get into orbit.

I think the energy I'm saving is the potential energy of all that mass at a height of 20 kms, or alternatively the kinetic energy that would be converted from that potential energy if you dropped the Saturn 5 half full of propellant from a height of 20 kms.

17. Jan 24, 2004

### enigma

Staff Emeritus
Re: In response to Enigam last post,

Yes. All this is correct. You are gaining ~200000 J/kg plus any energy bled off from drag. The orbital kinetic energy is ~30000000 J/kg. Around 150 times more.

The point I've been trying to make throughout the thread is that the extra costs for fuel and structure of a larger rocket is much less expensive and much, much safer than the construction of a massive floating launch complex 20km up.

At the start of the thread, you were insinuating that from 20km up, the size of the rockets would be miniscule compared to the rockets used today. As has been shown, it simply isn't the case.

If you want to launch a rocket with payload and rocket inert mass weighing 20,000kg to low Earth orbit from the platform, the fuel you'd need would be over 100,000kg. Launching from the ground, it would be more like 170,000kg. Still, all a rocket is is a giant fuel tank with an engine on the bottom. Once all the guidance, control, etc. is designed, any increase in the size of the rocket doesn't add a tremendous amount to the development costs. The tanks are the cheapest part.

Lets look at this seriously. The balloon would need to be enormous on a scale we'd hardly be able to comprehend. "See it from the ground" type enormous. The crew wouldn't be able to work without pressure suits. You'd need to ship all life support supplies up regularly. You'd need to have a way to transfer any rocket flown up to it safely. You would need to drop the rocket prior to ignition, eliminating any aborts in case of minor problems in the startup sequence (electrical failure in a subsystem, for example). The list goes on and on.

The new expendable launch vehicles are saving costs by increasing their payload to space by using modular components. The Delta IV Heavy is just a Delta IV with two additional center booster cores strapped to the sides. The Delta IV Superheavy has even more. By doing it this way, they are reducing the development costs which are the primary cash drain in rocket design. They are also reducing construction costs due to 'learning curve' effects of making the same piece over and over again.

Launching from the ground is simply much cheaper and safer than building a floating city.

It is an interresting idea, I'll grant you that, but I'll go double or nothing that it won't be feasible any time soon.

Last edited: Jan 24, 2004
18. Jan 24, 2004

### Staff: Mentor

Re: Sky Isle - response to enigma

The error in your reasoning (as enigma explained more quantitatively) is that you overestimate the time and fuel it takes to get to 20km vs what's left. Getting to 20km is a relatively insignificant fraction of the distance and time the vehicle has to travel. The shuttle begins pitching something like 50 seconds into its 9 minute flight. By the time the srbs separate after 2 min its something like 160 miles downrange.

An interesting site on the shuttle ascent profile: http://www.stemnet.nf.ca/CITE/sts_ascent.htm [Broken]

Last edited by a moderator: May 1, 2017
19. Jan 24, 2004

### jjalexand

Re: Re: Sky Isle - response to enigma

I believe Enigma has made an unsubstantiated statement comparing Joules per kilogram in two different situations. I do not think this directly relates to my estimates of propellant utilization of a Saturn 5 at 20 kms.

If you see a flaw in my propellant utilization estimates, please state it. For example, do you object to the 3:2:1 ratio for time spent in each of the 20 km segment of the 60km first stage burn under a constant acceleration scenario in (B)?

I agree that many new ideas are either rubbish or require very serious review, and that Enigma and yourself are acting in that useful role. However, I do think all such objections should be fairly and sensibly made, by logical argument addressing the issues raised. Just ganging up by posting effectively unsubstantiated support for another Mentor does not really seem to fit into that pattern. New ideas are like babies that may need some nurturing to bring them to the flower of adulthood. It is accepted that the best environment for the development of new ideas is a non-critical encouraging one rather than an initially harsh & critical one.

I will respond to Enigma's last reply when I have more time to analyse the energy calculations and estimate the air resistance.

In the meantime, I would like to point out that while such a plaform would of course be very large (that's why I suggested placing it over an ocean) and of course would require pressurised living and working quarters, it is feasible with current technology and would have many other uses, for example a large but light telescope could be located on the platform. This might use unconventional lens technology with a helium gas lens in a concave plastic shell, using advanced dynamic image correction. Thus the telescope might support it's own weight. Also many functions now performed by satellites might be performed much more cheaply by floating up instrumentation and attaching it to a 'maintanance rail' around the edge of the platform. It would be nice to get rid of all those cellphone towers for example. (That would obviously require many smaller platforms.) It could also play a role in missile defence technology. Enigma is still very much thinking in terms of rocket technology, and I think the rail gun idea may have some merit at this height for launching unmanned items, where it might not be feasible on the ground.

20. Jan 25, 2004

### enigma

Staff Emeritus
Re: Re: Re: Sky Isle - response to enigma

Specific gravitational energy in J/kg is just g*h [m/s^2 and m].
Specific kinetic energy in J/kg is V^2/2 [m/s].

I used 20,000m and a constant 9.8 m/s^2 for SGPE and 7,750 m/s for SKE.

I'm going about it by a less calculation intensive route. Just look at the relative energies, and you'll see that lifting the object to 20km without imparting any tangential velocity is a small fraction of the total energy required. Don't worry about the specifics of the Saturn V. The numbers cited apply equally to all rockets, even ones which are currently in production (and weren't designed to go to the moon).

The problems I see are that the model does not match reality. Engines are throttled forward and backward several times during the ascent to keep the acceleration in check. Also, like Russ pointed out, the rockets turn fairly early on in the ascent.

Unless we know the specific throttle profile, estimating fuel use after a certain altitude would merely be guesswork.

Both Russ and I have taken at least a few aerospace classes, and the energies in question don't justify the potential expense for the proposed launch platform. Please don't think that we're trying to gang up on you.

If you want to discuss other potential uses like telescopes or communications stations, those are other topics. Considering some science experiments are currently done by balloon, I'd wager that the economics would be much more likely to work out for such a project.