I.)Introduction.
II.)Lightweight propellant tanks.
III.)Kerosene fuel and engines for the X33/Venture star.
IVa.)Aerodynamic lift applied to ascent to orbit.
b.)Estimation of fuel saving using lift.
V.)Kerosene fueled VentureStar payload to orbit.
I.) A debate among those questing for the Holy Grail of a reusable,
singlestagetoorbit vehicle is whether it should be powered by
hydrogen or a dense hydrocarbon such as kerosene. Most concepts for
such a vehicle centered on hydrogen, since a hydrogen/LOX combination
provides a higher Isp. However, some have argued that dense fuels
should be used since they take up less volume (equivalently more fuel
mass can be carried in the same sized tank) so they incur less air
drag and also since the largest hydrocarbon engines produce greater
thrust they can get to the desired altitude more quickly so they also
incur lower gravity drag loss.
Another key fact is that for dense fuels the ratio of propellant mass
to tank mass is higher, i.e., you need less tank mass for the same
mass of propellant. This fact is explored in this report:
Single Stage To Orbit Mass Budgets Derived From Propellant Density and
Specific Impulse.
John C. Whitehead
32nd AIAA/ASME/SAE/ASEE Joint Propulsion ConferenceLake Buena Vista,
FLJuly 13, 1996
Whitehead notes that the propellant mass to tank mass ratio for
kerosene/LOX is typically around 100 to 1, while for liquid hydrogen/
LOX it's about 35 to 1, which would result in a significantly greater
dry mass for the hydrogenfueled case just in tank weight alone. Based
on calculations such as these Whitehead concludes the best option for
a SSTO would be to use kerosene/LOX.
The case for the X33/VentureStar is even worse because the unusual
shape of the tanks requires them to use more tank mass than a
comparably sized cylindrical tank. This is discussed here:
Space Access Update #91 2/7/00.
The Last Five Years: NASA Gets Handed The Ball, And Drops It.
"...part of LM X33's weight growth was the "multi
lobed" propellant tanks growing considerably heavier than promised.
Neither Rockwell nor McDonnellDouglas bid these; both used proven
circularsection tanks. X33's graphiteepoxy "multilobed" liquid
hydrogen tanks have ended up over twice as heavy relative to the
weight of propellant carried as the Shuttle's 70's vintage aluminum
circularsection tanks  yet an X33 tank still split open in test
last fall. Going over to aluminum will make the problem worse; X
33's aluminum multilobed liquid oxygen tank is nearly four times as
heavy relative to the weight of propellant carried as Shuttle's
aluminum circularsection equivalent."
The X33's twin liquid hydrogen tanks had a weight of 4,600 pounds
each, and the liquid oxygen tank a weight of 6,000 pounds, for total
of 15,200 pounds for the tanks:
Marshall Space Flight Center
Lockheed Martin Skunk Works
Sept. 28, 1999
X33 Program in the Midst of Final Testing and Validation of Key
Components.
The weight of the propellant carried by the X33 was supposed to be
210,000 lb. So the propellant to tank mass ratio for the X33 was only
about 14 to 1(!). This would be a severe problem for the fullscale
VentureStar. Its gross lift off weight was supposed to be 2,186,000
lbs with a fuel weight of 1,929,000 lbs:
X33 Advanced Technology Demonstrator.
So the VentureStar would have a dry mass of 257,000 lbs. Since the
same design would be used for the VentureStar tanks as those of the
X33, the propellant mass to tank mass ratio would also be 14 to 1, so
the tank mass would be 138,000 lbs. But this means the empty tank mass
alone would be over half of the vehicle's dry weight (!)
It would have been extremely difficult for the VentureStar to have
made orbit with such a large weight penalty from the start. From all
accounts the weight problem with the tanks drove other problems such
as the need to add larger wings, increasing the weight problem
further. NASA wound up canceling the program when Lockheed couldn't
deliver the working liquid hydrogen tanks even at this excessive
weight. However, rather than canceling the program I believe the
better course would have been to open up competition for coming up
with alternative, creative solutions for reducing the weight of the
tanks. This would also have resolved some of the problems with the
vehicles weight growth.
II.) I have proposed one possibility for lightweighting the X33 tanks
on this forum:
The idea would be to achieve the same lightweight tanks as
cylindrical ones by using multiple, small diameter, aluminum
cylindrical tanks. You could get the same volume by using varying
lengths and diameters of the multiple cylinders to fill up the volume
taken up by the tanks. The cylinders would not have to be especially
small. In fact they could be at centimeter to millimeter diameters, so
would be of commonly used sizes for aluminum tubes and pipes.
The weight of the tanks could be brought down to the usual 35 to 1
ratio for propellant to tank mass. Then the mass of the tanks on the
X33 would be 210,000 lbs/35 =3D 6,000 lbs, saving 9,200 lbs off the
vehicle dry weight. This would allow the hydrogenfueled X33 to
achieve its original Mach 15 maximum velocity.
The same idea applied to the fullscale hydrogenfueled VentureStar
would allow it to significantly increase its payload carrying
capacity. At a 35 to 1 ratio of propellant mass to tank mass, the
1,929,000 lbs propellant mass would require a mass of 1,929,000/35 =3D
55,000 lbs for the tanks, a saving of 83,000 lbs off the original tank
mass. This could go to extra payload, so from 45,000 lbs max payload
to 128,000 lbs max payload.
An analogous possibility might be to use a honeycombed structure for
the entire internal makeup of the tank. The X33's carbon composite
tank was to have a honeycombed structure for the skin alone. Using a
honeycomb structure throughout the interior might result in a lighter
tank in the same way as does multiple cylinders throughout the
interior.
Still another method might be to model the tanks standing vertically
as conical but with a flat front and back, and rounded sides. Then the
problem with the front and back naturally trying to balloon out to a
circular cross section might be solved by having supporting flat
panels at regular intervals within the interior. The X33 composite
tanks did have support arches to help prevent the tanks from
ballooning but these only went partially the way through into the
interior. You might get stronger a result by having these panels go
all the way through to the other side.
These would partition the tanks into portions. This could still work
if you had separate fuel lines, pressurizing gas lines, etc. for each
of these partitions and each got used in turn sequentially. A
preliminary calculation based on the deflection of flat plates under
pressure shows with the tank made of aluminum alloy and allowing
deflection of the flat front and back to be only of millimeters that
the support panels might add only 10% to 20% to the weight of the
tanks, while getting similar propellant mass to tank mass ratio as
cylindrical tank. See this page for an online calculator of the
deflection of flat plates:
eFunda: Plate Calculator  Simply supported rectangular plate with
uniformly distributed loading.
Note you might not need to have a partitioned tank, with separate
fuel lines, etc., if the panels had openings to allow the fuel to pass
through. These would look analogous to the wing spars in aircraft
wings that allow fuel to pass through. You might have the panels be in
a honeycomb form for high strength at lightweight that still allowed
the fuel to flow through the tank. Or you might have separate beams
with a spaces between them instead of solid panels that allowed the
fuel to pass through between the beams.
Another method is also related to the current design of having a
honeycombed skin for the composite hydrogen tanks. Supposed we filled
these honeycombed cells with fluid. It is known that pressurized tanks
can provide great compressive strength. This is in fact used to
provide some of the structural strength for the X33 that would
otherwise have to be provided by heavy strengthening members. This
idea would be to apply fluid filled honycombed cells. However, what we
need for our pressurized propellant tanks is *tensile strength*.
A possible way tensile strength could be provided would be to use the
Poisson's ratio of the honeycombed cells:
Poisson's ratio.
Poisson's ratio refers to the tendency of a material stretched in one
direction to shrink in length in an orthogonal direction. Most
isotropic solid materials have Poisson's ratio of about .3. However,
the usual hexagonal honeycombed structure, not being isotropic, can
have Poisson's ratios in the range of +1. This is mentioned in this
article about nonstandard honeycombed structures that can even have
negative Poisson ratios:
Chiral honeycomb.
However, note that from the formula for the volumetric change in the
Wikipedia Poisson's ratio page, a stretching of a material with a +1
Poisson's ratio implies a *decrease* in volume; actually this is true
for any case where the Poisson's ratio is greater than +.5. Then fluid
filled honeycombed cells would resist the stretching of tensile strain
by the resistance to volume compression. This would be present with
both gases and liquids. Gases are lighter. However, they are highly
compressible and it might take too large an internal pressure in the
cells to provide sufficient resistance, and so also too thick cell
walls to hold this pressure. Liquids are heavier but they are highly
noncompressible so could provide strong resistance to the volume
compression and thereby to the tensile strain.
Then for liquid hydrogen tanks we might use liquid hydrogen filled
cells within the skin of the tanks. Hydrogen is rather light compared
to other liquids at a density of only about 72 kg/m^3. This then could
provide high tensile strength at a much lower weight than typical
solid wall tanks.
Kerosene and liquid oxygen would be used in the honeyombed cells for
their corresponding tanks, to keep the storage temperatures
comparable. These are heavier liquids than liquid hydrogen,
approximately in the density range of liquid water. Still these liquid
filled honeycombed cells would provide much lighter tanks than
comparable solid wall tanks.
III.) Any of these methods might allow you to reduce the weight of the
tanks to be similar to that of cylindrical tanks and thus raise the
payload to over 100,000 lbs. This would be for keeping the original
hydrogen/LOX propellant. However, in keeping with the analyses that
show dense propellants would be more appropriate for a SSTO vehicle
I'll show that replacing the hydrogenfueled engines of the X33/
VentureStar with kerosene ones would allow the X33 to actually now
become an *orbital* craft instead of just suborbital, and the payload
capacity of the VentureStar would increase to be comparable to that
proposed for Ares V.
The volume of the X33 liquid hydrogen tanks was 29,000 gallons each
and the liquid oxygen tank, 20,000 gallons, for a total of 78,000
gallons volume for propellant. This is 78,000gal*3.8 L/gal =3D 296,000
liters, 296 cubic meters. How much mass of kerosene/LOX could we fit
here if we used these as our propellants? Typically the oxidizer to
fuel ratio for kerosene/LOX engines is in the range of 2.5 to 2.7 to
1. I'll take the O/F ratio as 2.7 to 1. The density of kerosene is
about 806 kg/m^3 and we can take the density of liquid oxygen to be
1160 kg/m^3 when densified by subcooling:
Liquid Oxygen Propellant Densification Unit Ground Tested With a Large
Scale FlightWeight Tank for the X33 Reusable Launch Vehicle.
These requirements of the propellants' total volume and densities,
result in a total propellant mass of 307,000 kg, with 83,000 kg in
kerosene and 224,000 kg in LOX. Kerosene/LOX tanks weigh typically
1/100th the propellant mass, so the tank mass would be 3,070 kg. The
current X33 LH/LOX tanks weighed 15,200 lbs, or 6,900 kg. So the
empty weight of the X33 is reduced from 63,000 lbs, 28,600 kg, to
28,600kg  6,900kg + 3070kg =3D 24,800 kg.
How about the engines? The X33 is to be reusable so you want to use
reusable kerosene engines. The RS84 might be ideal when it is
completed for the fullscale VentureStar, but it turns out it's a bit
too heavy for the X33. It would have a weight of about 15,000 lbs,
6,800 kg:
RS84.
about the weight of the two aerospike engines currently on the X33:
Bringing launch costs down to earth.
"Three federally funded projects are underway to develop new rocket
engines that can make it more affordable to send payloads into orbit."
With 307,000 kg kerosene/LOX fuel and 24,800 kg dry weight, the mass
ratio would be 13.4. According to the Astronautix page, the sea level
Isp of the RS84 would be 301 s, and the vacuum 335 s. Take the
average Isp as 320s. The total Isp for a rocket to orbit including
gravity and air drag losses is usually taken to be about 9,200 m/s.
Then an average exhaust velocity of 3200 m/s and mass ratio of 13.4
would give a total deltav of 8,300 m/s. Even if you add on the 462 m/
s additional velocity you can get for free by launching at the equator
this would not be enough for orbit.
So for the X33 I'll look at the cases of the lighter for its thrust
NK33, used as a trio. Note that though not designed to be a reusable
engine to make, say, 100 flights, all liquid fuel rocket engines
undergo extensive static firings during testing so the NK33 probably
could make 5 to 10 flights before needing to be replaced.The NK33 is
almost legendary for its thrust to weight ratio of 136. According to
the Astronautix page its weight is 1,222 kg , with a sea level Isp of
297 sec and a vacuum Isp of 331:
NK33.
I'll take the average Isp as 315 s. With three NK33 engines the mass
of the X33 becomes 21,700 kg, and the mass ratio becomes 15.15. Then
with an average Isp of 315 s, the total deltav would be 8561 m/s and
if you add on the 462 m/s additional equatorial velocity it's 9,023 m/
s. Still slightly below the deltav typically given for orbit of 9,200
m/s.
However, it should be noted that the extra deltav required beyond the
7,800 m/s orbital velocity is highly dependent on the vehicle and
trajectory. Here's a page that gives the gravity loss and air drag
loss for some orbital rockets:
Drag: Loss in Ascent, Gain in Descent, and What It Means for
Scalability.
Thursday 2008.01.10 by gravityloss
* Ariane A44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
* Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
* Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
* Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
* Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
* Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s
Note that the gravity loss for the Delta 7925 is particularly small.
As a general principle the gravity loss can be minimized if you have a
high thrust vehicle that rapidly develops high vertical velocity
sufficient to reach the altitude for orbit. For then it can more
quickly apply the horizontal thrust required to achieve the 7,800 m/s
orbital, tangential, velocity, there being no gravity loss over the
horizontal thrust portion. Note then the liftoff thrust to liftoff
weight ratio for the Delta 7925 is the relatively large 1.4; in
comparison, for the Saturn V it was only 1.14. And note now also that
the X33 with three NK33 engines has total mass of 328,700 kg and
total thrust of 4,530,600 N giving a liftoff thrust to liftoff weight
ratio of 1.4. Then this reconfigured X33 would likely have comparable
gravity drag loss as the Delta 7925. When you take into account the
high thrust means it would rapidly reach high altitude, implying the
Isp would quickly get close to the vacuum Isp, the average Isp over
the trajectory is most likely closer to the 331 s vacuum Isp than just
315 s giving an actually higher achieved deltav.
IV.a) It's capability of reaching orbit and possibly even with a small
payload could be increased with another additional factor. Among the
questors for a SSTO vehicle, the idea to use wings for a horizontal
landing has been derided because of the view they were just dead
weight on ascent and you need to save as much weight off the empty
weight of the vehicle as possible to achieve orbit. However, the key
fact is that wings or a lifting body shape can reduce the total delta
v for orbit by using aerodynamic lift to supply the force to raise the
vehicle to a large portion of the altitude of orbit rather than this
force being entirely supplied by the thrust of the engines. This fact
is discussed on page 4 of this report:
AIAA 20001045
A Multidisciplinary Performance Analysis of a LiftingBody Single
StagetoOrbit Vehicle.
Paul V. Tartabini, Roger A. Lepsch, J. J. Korte, Kathryn E. Wurster
38th Aerospace Sciences Meeting & Exhibit.
1013 January 2000 / Reno, NV
"One feature of the VentureStar design that could
be exploited during ascent was its lifting body shape.
By flying a lilting trajectory, it was possible to significantly
decrease the amount of gravity losses, thereby
improving vehicle performance and payload capability.
Yet increasing the amount of lift during ascent generally
required flight at higher anglesofattack and resulted
in greater stress on the vehicle structure. Accordingly,
the nominal trajectory was constrained to keep the parameter
q_ below a 1500 psfdeg structural design limit
to ensure that the aerodynamic loads did not exceed
the structural capability of the vehicle. The effect of this
trajectory constraint on vehicle performance is shown
in Fig. 3. There was a substantial benefit associated with
using lift during ascent since flying a nonlifting trajectory
resulted in a payload penalty of over 1000 lbs compared
to the nominal case."
Then the gravity losses could be further reduced by flying a lifting
trajectory, which would also increase the payload capability by a
small percentage.
The trajectory I'll use to illustrate this will first be straight
line at an angle up to some high altitude that still allows
aerodynamic lift to operate. At the end of this portion the vehicle
will have some horizontal and vertical component to its velocity.
We'll have the vertical component be sufficient to allow the vehicle
to reach 100 km, altitude. The usual way to estimate this vertical
velocity is by using the relation between kinetic energy and potential
energy. It gives the speed of v =3D sqrt(2gh) to reach an altitude of h
meters. At 100,000 m, v is 1,400 m/s.
Now to have orbital velocity you need 7,800 m/s tangential, i.e.,
horizontal velocity. If you were able to fly at a straightline at a
constant angle to reach 7,800 m/s horizontal velocity and 1,400 m/s
vertical velocity and such that the air drag was kept at the usual low
100 to 150 m/s then you would only need sqrt(7800^2 + 1400^2) =3D 7,925
m/s additional deltav to reach orbit. Then the total deltav to orbit
might only be in the 8,100 m/s range. Note this is significantly less
than the 9,200 m/s deltav typically needed for orbit, including
gravity and air drag.
The problem is with usual rocket propulsion to orbit not using lift
the thrust vector has to be more or less along the centerline of the
rocket otherwise the rocket would tumble. You can gimbal the engines
only for a short time to change the rocket's attitude but the engines
have to be then redirected along the center line. However, the center
line has to be more or less pointing into the airstream, i.e.,
pointing in the same direction as the velocity vector, to reduce
aerodynamic stress and drag on the vehicle. But the rocket thrust
having to counter act gravity means a large portion of the thrust has
to be in the vertical component which means the thrust vector has to
be nearly vertical at least for the early part of the trip when the
gross mass is high. Then the thrust vector couldn't be along the
center line of a nearly horizontally traveling rocket at least during
the early part of the trip.
However, using lift you are able to get this large upwards vertical
component for the force on the rocket to allow it to travel along this
straightline. A problem now though is that at an altitude short of
that of space, the air density will not be enough for aerodynamic
lift. Therefore we will use lift for the first portion of the
trajectory, traveling in a straightline at an angle. Then after that,
with sufficient vertical velocity component attained to coast to 100
km altitude, we will supply only horizontal thrust during the second
portion to reach the 7,800 m/s horizontal velocity component required
for orbital velocity.
IV.b) How much fuel could we save using a lifting straightline
portion of the trajectory? I'll give an example calculation that
illustrates the fuel savings from using aerodynamic lift during
ascent. First note that just as for aircraft fuel savings are best at
a high L/D ratio. However, the hypersonic lift /drag ratio of the X33/
VentureStar is rather poor, only around 1.2, barely better than the
space shuttle:
AIAA994162
X33 Hypersonic Aerodynamic Characteristics.
Kelly J, Murphy, Robert J, Nowak, Richard A, Thompson, Brian R, Hollis
NASA Langley Research Center
Ramadas K. Prabhu
Lockheed Martin Engineering &Sciences Company
This explains the low increase in payload, about 1,000 lbs., less
than .5% of the vehicle dry weight, by using a lifting trajectory for
the VentureStar. However, some lifting body designs can have a lift/
drag ratio of from 6 to 8 at hypersonic speeds:
Waverider Design.
The L/D is usually optimized for a specific speed range but we can
imagine "morphing" wings that allow a good L/D ratio over a wide speed
range. For instance note on the "Waverider Design" web page the
vehicles optimized for the highest hypersonic speeds have a long,
slender shape, compared to those for the slower hypersonic speeds.
Then for an orbital craft we could have telescoping sides of the
vehicle that would be extended when full of fuel at the slower speeds,
and retracted, producing a slimmer vehicle, when most of the fuel is
burned off and the vehicle is flying faster. Note that a good L/D
ratio at the highest hypersonic speeds also means the vehicle will
experience less aerothermal heating on return.
Then we can imagine a second generation lifting trajectory vehicle
having this high L/D ratio over a wide speed range. So in the example
I'll take the supersonic/hypersonic L/D ratio as 5, and for lack of a
another vehicle I'll use the reconfigured kerosenefueled X33's
thrust and weight values.
Here's the calculation for constant L/D at a constant angle =CE=B8
(theta). I'll regard the straightline path as my Xaxis and the
perpendicular to this as the Yaxis. Note this means my axes look like
they are at an angle to the usual horizontal and vertical axes, but it
makes the calculation easier. Call the thrust T, and the mass, M. Then
the force component along the straightline path, our Xaxis, is Fx =3D
T  gMsin(=CE=B8)  D and the force component along the Yaxis is Fy =3D L =

gMcos(=CE=B8). We'll set L =3D gMcos(=CE=B8). Then the force along the stra=
ight
line is Fx =3D T  gMsin(=CE=B8)  gMcos(=CE=B8)/(L/D). As with the calcula=
tion
for the usual rocket equation, divide this by M to get the
acceleration along this line, and integrate to get the velocity. The
result is V(t) =3D Ve*ln(M0/Mf) g*tsin(=CE=B8)  g*tcos(=CE=B8)/(L/D), wit=
h M0
the initial mass, and Mf, the mass at time t, a la the rocket
equation. If you make the angle =CE=B8 (theta) be shallow, the g*tsin(=CE=
=B8)
term will be smaller than the usual gravity drag loss of g*t and the
(L/D) divisor will make the cosine term smaller as well.
I'll assume the straightline path is used for a time when the
altitude is high enough to use the vacuum Ve of 331s*9.8 m/s^2 =3D 3244
m/s. According to the Astronautix page, 3 NK33's would have a total
vacuum thrust of 4,914,000 N and for an Isp of 331s, the propellant
flow rate would be 4,914,000/(331x9.8) =3D 1,515 kg/sec. I'll use the
formula: V(t) =3D Ve*ln(M0/Mf)  g*tsin(=CE=B8)  g*tcos(=CE=B8)/(L/D) , to
calculate the velocity along the inclined straightline path. There
are a couple of key facts in this formula. First note that it includes
*both* the gravity and air drag. Secondly, note that though using
aerodynamic lift generates additional, large, induced drag, this is
covered by the fact that the L/D ratio includes this induced drag,
since it involves the *total* drag.
I'll take the time along the straightline path as 100 sec. Then Mf =3D
328,700kg 100s*(1,515 kg/s) =3D 177,200 kg. After trying some examples
an angle of 30=C2=BA provides a good savings over just using the usual non
lifting trajectory. Then V(t) =3D 3244*ln(328,700/177,200)  9.8*100(sin
(30=C2=BA) + cos(30=C2=BA)/5) =3D 1,345 m/s. Then the vertical component of=
this
velocity is Vy =3D 1,135*sin(30=C2=BA) =3D 672.3 m/s and the horizontal, Vx=
=3D
1,135*cos(30=C2=BA) =3D 1,164.5 m/s.
To compare this to a usual rocket trajectory I'll calculate how much
fuel would be needed to first make a vertical trip to reach a vertical
speed of 672.3 m/s subject to gravity and air drag, and then to apply
horizontal thrust to reach a 1,164.5 m/s horizontal speed.
The air drag for a usual rocket is in the range of 100 m/s to 200 m/s.
I'll take the air drag loss as 100 m/s for this vertical portion. Then
the equation for the velocity along this vertical part including the
gravity loss and the air drag loss would be V(t) =3D 3244*ln(M0/Mf) 
9.8*t  100 m/s, where M0 =3D328,700 kg and Mf =3D 328,700  t(1,515). You
want to find the t so that this velocity matches the vertical
component in the inclined case of 672.3 m/s. Plugging in different
values of t, gives for t =3D 85 sec, V(85) =3D 680 m/s.
Now to find the horizontal velocity burn. Since this is horizontal
there is no gravity loss, and I'll assume this part is at very high
altitude so has negligible air drag loss. Then the velocity fomula is V
(t) =3D 3244*ln(M0/Mf). Note in this case M0 =3D 328,700  85*1,515 =3D
199,925 kg, which is the total mass left after you burned off the
propellant during the vertical portion, and so Mf =3D 199,925  t*1,515.
Trying different values of t gives for t =3D 40, V(40) =3D 1,171.5 m/s.
Then doing it this way results in a total of 125 sec of fuel burn, 25
percent higher than in the aerodynamic lift case, specifically
25s*1,515 kg/s =3D 37,875 kg more. Or viewed the other way, the
aerodynamic lift case requires 20% less fuel over this portion of the
trip than the usual nonlift trajectory. With a 307,000 kg total fuel
load, thus corresponds to a 12.3% reduction in the total fuel that
would actually be needed. Or keeping the same fuel load, a factor
1/.877 =3D 1.14 larger dry mass could be lofted, which could be used for
greater payload. For a reconfigured X33 dry mass of 21,700 kg, this
means 3,038 kg extra payload. Remember though this is for our imagined
new X33 lifting shape that is able to keep a high L/D ratio of 5 at
hypersonic speed, not for the current X33 shape which only has a
hypersonic L/D of 1.2.
With the possibility of using morphing lifting body or wings with
high hypersonic L/D ratio allowing a large reduction in fuel
requirements to orbit, this may be something that could be tested by
amateurs or by the "New Space" launch companies.
V.) Now for the calculation of the payload the VentureStar could carry
using kerosene/LOX engines. The propellant mass of the VentureStar was
1,929,000 lbs. compared to the X33's 210,000 lbs., i.e., 9.2 times
more. Then its propellant tank volume would also be 9.2 times higher,
and the kerosene/LOX they could contain would also be 9.2 times
higher, or to 9.2*307,000 =3D 2,824,400 kg.
We saw the VentureStar dry mass was 257,000 lbs, 116,818 kg, with half
of this as just the mass of the LH2/LOX tanks, at 138,000 lbs, 62,727
kg. However, going to kerosene/LOX propellant means the tanks would
only have to be 1/100th the mass of the propellant so only 28,244 kg.
Then the dry mass would be reduced to 82,335 kg. We need kerosene/LOX
engines now. I suggest the RS84 be completed and used for the
purpose. You would need seven of them to lift the heavier propellant
load. They weigh about the same as the aerospike engines on the
current version of the VentureStar so you wouldn't gain any weight
savings here.
To calculate how much we could lift to orbit I'll take the average
Isp of the RS84 as 320. Then if we took the payload as 125,000 kg the
total liftoff mass would be 2,824,400 + 82,335 + 125,000 =3D 3,031,735
kg, and the ending dry mass would be 207,335 kg, for a mass ratio of
14.6. Then the total deltav would be 3200ln(14.6) =3D 8,580 m/s. Adding
on the 462 m/s equatorial speed brings this to 9042 m/s. With the
reduction in gravity drag using a lifting trajectory this would
suffice for orbit.
Bob Clark