A kerosene-fueled X-33 as a single stage to orbit vehicle.

Table of Contents. I.)Introduction. II.)Lightweight propellant tanks. III.)Kerosene fuel and engines for the X-33/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, single-stage-to-orbit 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 1-3, 1996 http://www.osti.gov/bridge/servlets/purl/379977-2LwFyZ/webviewable/379977.pdf
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 hydrogen-fueled 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 X-33/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 L-M X-33's weight growth was the "multi- lobed" propellant tanks growing considerably heavier than promised. Neither Rockwell nor McDonnell-Douglas bid these; both used proven circular-section tanks. X-33's graphite-epoxy "multi-lobed" 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 circular-section tanks - yet an X-33 tank still split open in test last fall. Going over to aluminum will make the problem worse; X- 33's aluminum multi-lobed liquid oxygen tank is nearly four times as heavy relative to the weight of propellant carried as Shuttle's aluminum circular-section equivalent." http://www.space-access.org/updates/sau91.html
The X-33'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 X-33 Program in the Midst of Final Testing and Validation of Key Components. http://www.xs4all.nl/~carlkop/x33.html
The weight of the propellant carried by the X-33 was supposed to be 210,000 lb. So the propellant to tank mass ratio for the X-33 was only about 14 to 1(!). This would be a severe problem for the full-scale VentureStar. Its gross lift off weight was supposed to be 2,186,000 lbs with a fuel weight of 1,929,000 lbs:
X-33 Advanced Technology Demonstrator. http://teacherlink.ed.usu.edu/tlnasa/OtherPRINT/Lithographs/X33.Advanced.Technology.Demonstrator.PDF
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 X-33, 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 X-33 tanks on this forum:
http://www.bautforum.com/space-exploration/86728-passenger-market-suborbital-hypersonic-transports-5.html#post1495726
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 X-33 would be 210,000 lbs/35 = 6,000 lbs, saving 9,200 lbs off the vehicle dry weight. This would allow the hydrogen-fueled X-33 to achieve its original Mach 15 maximum velocity. The same idea applied to the full-scale hydrogen-fueled 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 = 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 X-33'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 X-33 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. http://www.efunda.com/formulae/solid_mechanics/plates/calculators/SSSS_PUniform.cfm
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 X-33 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. http://en.wikipedia.org/wiki/Poisson%27s_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 non-standard honeycombed structures that can even have negative Poisson ratios:
Chiral honeycomb. http://silver.neep.wisc.edu/~lakes/PoissonChiral.html
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 non-compressible 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 hydrogen-fueled engines of the X-33/ VentureStar with kerosene ones would allow the X-33 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 X-33 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 = 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 Flight-Weight Tank for the X-33 Reusable Launch Vehicle. http://www.grc.nasa.gov/WWW/RT/RT2001/5000/5870tomsik.html
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 X-33 LH/LOX tanks weighed 15,200 lbs, or 6,900 kg. So the empty weight of the X-33 is reduced from 63,000 lbs, 28,600 kg, to 28,600kg - 6,900kg + 3070kg = 24,800 kg. How about the engines? The X-33 is to be reusable so you want to use reusable kerosene engines. The RS-84 might be ideal when it is completed for the full-scale VentureStar, but it turns out it's a bit too heavy for the X-33. It would have a weight of about 15,000 lbs, 6,800 kg:
RS-84. http://www.astronautix.com/engines/rs84.htm
about the weight of the two aerospike engines currently on the X-33:
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." http://www.memagazine.org/backissues/membersonly/october98/features/launch/launch.html
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 RS-84 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 delta-v 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 X-33 I'll look at the cases of the lighter for its thrust NK-33, 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 NK-33 probably could make 5 to 10 flights before needing to be replaced.The NK-33 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:
NK-33. http://www.astronautix.com/engines/nk33.htm
I'll take the average Isp as 315 s. With three NK-33 engines the mass of the X-33 becomes 21,700 kg, and the mass ratio becomes 15.15. Then with an average Isp of 315 s, the total delta-v 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 delta-v typically given for orbit of 9,200 m/s. However, it should be noted that the extra delta-v 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 A-44L: 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 http://gravityloss.wordpress.com/2008/01/10/drag-loss-in-ascent-gain-in-descent-and-what-it-means-for-scalability /
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 X-33 with three NK-33 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 X-33 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 delta-v.
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 2000-1045 A Multidisciplinary Performance Analysis of a Lifting-Body Single- Stage-to-Orbit Vehicle. Paul V. Tartabini, Roger A. Lepsch, J. J. Korte, Kathryn E. Wurster 38th Aerospace Sciences Meeting & Exhibit. 10-13 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 angles-of-attack and resulted in greater stress on the vehicle structure. Accordingly, the nominal trajectory was constrained to keep the parameter q-_ below a 1500 psf-deg 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 non-lifting trajectory resulted in a payload penalty of over 1000 lbs compared to the nominal case." http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000031364_2000025539.pdf
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 = 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 straight-line 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) = 7,925 m/s additional delta-v to reach orbit. Then the total delta-v to orbit might only be in the 8,100 m/s range. Note this is significantly less than the 9,200 m/s delta-v 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 center-line 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 re-directed 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 straight-line. 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 straight-line 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 straight-line 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 X-33/ VentureStar is rather poor, only around 1.2, barely better than the space shuttle:
AIAA-99-4162 X-33 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 http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20040087108_2004091447.pdf
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. http://www.aerospaceweb.org/design/waverider/waverider.shtml
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 kerosene-fueled X-33's thrust and weight values. Here's the calculation for constant L/D at a constant angle θ (theta). I'll regard the straight-line path as my X-axis and the perpendicular to this as the Y-axis. 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 straight-line path, our X-axis, is Fx = T - gMsin(θ) - D and the force component along the Y-axis is Fy = L - gMcos(θ). We'll set L = gMcos(θ). Then the force along the straight- line is Fx = T - gMsin(θ) - gMcos(θ)/(L/D). As with the calculation 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) = Ve*ln(M0/Mf) -g*tsin(θ) - g*tcos(θ)/(L/D), with M0 the initial mass, and Mf, the mass at time t, a la the rocket equation. If you make the angle θ (theta) be shallow, the g*tsin(θ) 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 straight-line path is used for a time when the altitude is high enough to use the vacuum Ve of 331s*9.8 m/s^2 = 3244 m/s. According to the Astronautix page, 3 NK-33'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) = 1,515 kg/sec. I'll use the formula: V(t) = Ve*ln(M0/Mf) - g*tsin(θ) - g*tcos(θ)/(L/D) , to calculate the velocity along the inclined straight-line 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 straight-line path as 100 sec. Then Mf = 328,700kg -100s*(1,515 kg/s) = 177,200 kg. After trying some examples an angle of 30º provides a good savings over just using the usual non- lifting trajectory. Then V(t) = 3244*ln(328,700/177,200) - 9.8*100(sin (30º) + cos(30º)/5) = 1,345 m/s. Then the vertical component of this velocity is Vy = 1,135*sin(30º) = 672.3 m/s and the horizontal, Vx = 1,135*cos(30º) = 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) = 3244*ln(M0/Mf) - 9.8*t - 100 m/s, where M0 =328,700 kg and Mf = 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 = 85 sec, V(85) = 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) = 3244*ln(M0/Mf). Note in this case M0 = 328,700 - 85*1,515 = 199,925 kg, which is the total mass left after you burned off the propellant during the vertical portion, and so Mf = 199,925 - t*1,515. Trying different values of t gives for t = 40, V(40) = 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 = 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 non-lift 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 = 1.14 larger dry mass could be lofted, which could be used for greater payload. For a reconfigured X-33 dry mass of 21,700 kg, this means 3,038 kg extra payload. Remember though this is for our imagined new X-33 lifting shape that is able to keep a high L/D ratio of 5 at hypersonic speed, not for the current X-33 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 X-33'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 = 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 RS-84 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 RS-84 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 = 3,031,735 kg, and the ending dry mass would be 207,335 kg, for a mass ratio of 14.6. Then the total delta-v would be 3200ln(14.6) = 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
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Why not stick with H2O2 and propargyl alcohol, or H2O2 and cyclopropane?
~ BG
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I had a debate with you a few months ago about which should be the propellant to use for a single stage to orbit vehicle. I argued it should be hydrogen/LOX because that gave the highest Isp and therefore the lowest propellant mass. I also argued that both the DC-X demonstrator and the proposed VentureStar vehicles used hydrogen fuel. However, I now realize Isp is not the only key variable. There are other variables that can overwhelm the Isp advantage of hydrogen/LOX. Then propellant combinations with dense fuels and/or oxidizers, including H2O2, may indeed be better suited for a SSTO vehicle. Here's one report that discusses using H2O2 as the oxidizer for a SSTO:
A Single Stage to Orbit Rocket with Non-Cryogenic Propellants. Abstract "Different propellant combinations for single-stage-to-orbit-rocket applications were compared to oxygen/hydrogen, including nitrogen tetroxide/hydrazine, oxygen/methane, oxygen/propane, oxygen/RP-1, solid core nuclear/hydrogen, and hydrogen peroxide/JP-5. Results show that hydrogen peroxide and JP-5, which have a specific impulse of 328 s in vacuum and a density of 1,330 kg/cu m. This high-density jet fuel offers 1.79 times the payload specific energy of oxygen and hydrogen. By catalytically decomposing the hydrogen peroxide to steam and oxygen before injection into the thrust chamber, the JP-5 can be injected as a liquid into a high-temperature gas flow. This would yield superior combustion stability and permit easy throttling of the engine by adjusting the amount of JP-5 in the mixture. It is concluded that development of modern hydrogen peroxide/JP-5 engines, combined with modern structural technology, could lead to a simple, robust, and versatile single-stage-to-orbit capability." http://www.erps.org/docs/SSTORwNCP.pdf [full text]
Bob Clark
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We'll view the X-33 hydrogen tanks standing vertically as conical with flattened front and back. This report on page 19 by the PDF file page numbering gives the dimensions of the X-33 hydrogen tanks as 28.5 feet long, 20 feet wide and 14 feet high:
Final Report of the X-33 Liquid Hydrogen Tank Test Investigation Team. http://alpha.tamu.edu/public/jae/misc/tankreport.pdf
Call it 9 meters long, 6 meters wide, and 4.3 meters deep for this calculation. I'll simplify the calculation by approximating the shape as rectangular, i.e., uniformly 6 meters wide. See the attached image. Note that the rounded portions of the sides, top and bottom will be considered separately. I'll call the vertical length of each section x, and the bulkhead thickness h. Since the length of the tank is 9m, the number of sections is 9/x. I'm doing the calculation for kerosene/LOX propellant tanks, but approx. same size as the X-33 tanks. Typically these are pressurized in the 20-40 psi range. I'll take it as 30 psi; call it 2 bar, 2x10^5 Pa. Referring to the drawing of the tank, each bulkhead takes part in supporting the internal pressure of the two sections on either side of it. This means for each section the internal pressure is supported by one-half of each bulkhead on either side of it, which is equivalent to saying each bulkhead supports the internal pressure of one section. The force on each section is the cross-sectional area times the internal pressure, so 6m*x*(2*10^5 Pa), with x as in the diagram the vertical length of each section. The bulkhead cross-sectional area is 6m*h, with h the thickness of the bulkheads. Then the pressure the bulkheads have to withstand is 6m*x*(2*10^5 Pa)/6m*h = (2*10^5 Pa)*x/ h. The volume of each bulkhead is 6m*h*4.3m. The density of aluminum- lithium alloy is somewhat less than aluminum, call it 2,600 kg/m^3. So the mass of each bulkhead is (2,600 kg/m^3)*6m*h*4.3m = 67,080*h. Then the total mass of all the 9/x bulkheads is (9/x)*67080*h = 603,720*(h/ x). Note that additionally to the horizontal bulkheads shown there will be vertical bulkheads on the sides. These will have less than 1/10 the mass of the horizontal bulkheads because the length of each section x will be small compared to the width of 6m, and will have likewise small contribution to the support of the internal pressure. The tensile strength of some high strength aluminum-lithium alloys can reach 700 MPa, 7*10^8 Pa. Then the pressure the bulkheads are subjected to has to be less than or equal to this: (2*10^5 Pa)*x/h <= 7*10^8 Pa, so x/h <= 3,500, and h/x => 1/3,500. Therefore the total mass of the bulkheads = 603,720*(h/x) => 172.5 kg. Note we have not said yet how thick the bulkheads have to be only that their total mass is at or above 172.5 kg, for one of the twin rear tanks. It's twin would also require 172.5 kg in bulkhead mass. The third, forward, tank had about 2/3rds the volume of these twin rear tanks so I'll estimate the bulkhead mass it will require as 2/3rds of 172.5 kg, 115 kg. Then the total bulkhead mass would be 460 kg, about 15% of the 3,070 kg tank mass I calculated for the reconfigured X-33. This is for the bulkheads resisting the outwards pressure of the sections. Notice I did not calculate the pressure inside the tank on the bulkheads from the propellant on either side. This is because the pressure will be equalized on either side of the bulkheads. However, we will have to be concerned about the pressure on the rounded right and left sides of the tank, and the rounded top and bottom of the tank, where the pressure is not equalized on the outside of the tank. Before we get to that, remember the purpose of partitioning the tank was to minimize the bowing out of the front and back sides from the internal pressure. Consider this page then that calculates the deflection of a flat plat under a uniform load:
eFunda: Plate Calculator -- Clamped rectangular plate with uniformly distributed loading.
https://www.efunda.com/formulae/solid_mechanics/plates/images / CCCC_PUniform.gif "This calculator computes the maximum displacement and stress of a clamped (fixed) rectangular plate under a uniformly distributed load." http://www.efunda.com/formulae/solid_mechanics/plates/calculators/CCCC_PUniform.cfm
In the data input boxes, we'll put 200 kPa for the uniform load, 6 meters for the horizontal distance, .3 m, say, for the vertical distance x, and 6 mm for the thickness of the plate. For the vertical distance x I'm taking a value proportionally small compared to the tank width, but which won't result in an inordinate number of partitioned sections of the tank. For the thickness I'm taking a value at 1/1000th the width of the tank, which is common for cylindrical tanks. For the material specifications for aluminum-lithium we can take the Young's modulus as 90 GPa. Then the calculator gives the deflection as only 2.35mm, probably adequate. However, we still have to consider what happens to the rounded sides and the bottom and top. Look at the last figure on this page:
Thin-Walled Pressure Vessels.
https://www.efunda.com/formulae/solid_mechanics/mat_mechanics / images/PressureVesselCylindricalC.gif http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
It shows the calculation for the hoop stress of a cylindrical pressure vessel. The calculation given is 2*s*t*dx = p*2*r*dx, using s for the hoop stress. This implies, s = p*r/t, or equivalently t = p*r/s. So for a given material strength s, the thickness will depend only on the radius and internal pressure. However, what's key here is the same argument will apply in the figure if one of the sides shown is flat, instead of curved. Therefore in our scenario, the rounded sides, top and bottom, which we regard as half- cylinders, will only need the thickness corresponding to a cylinder of their same diameter, i.e., one of a diameter of 4.3m. So the rounded portions actually require a smaller thickness than what would be needed for a cylinder of diameter of the full 6m width of the tank. This means the partitioned tank requires material of somewhat less mass than a cylindrical tank of dimension the full width of the tank plus about 15% of that mass as bulkheads.
Bob Clark
tank drawing.
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The SpaceLaunchReport.com site operated by Ed Kyle provides the specifications of some launch vehicles. Here's the page for the Falcon 1:
Space Launch Report: SpaceX Falcon Data Sheet. http://www.spacelaunchreport.com/falcon.html
Quite interesting is that the total mass and dry mass values for the Falcon 1 first stage with Merlin 1C engine give a mass ratio of about 20 to 1. This is notable because a 20 to 1 mass ratio is the value usually given for a kerosene-fueled vehicle to be SSTO. However, this is for the engine having high vacuum Isp ca. 350 s. The Merlin 1C with a vacuum Isp of 304 s probably wouldn't work. However, there are some high performance Russian kerosene engines that could work. Some possibilities:
Engine Model: RD-120M. http://www.astronautix.com/engines/rd120.htm#RD-120M
RD-0124. http://www.astronautix.com/engines/rd0124.htm
Engine Model: RD-0234-HC. http://www.astronautix.com/engines/rd0234.htm
However, I don't know if this third one was actually built, being a modification of another engine that burned aerozine.
Some other possibilities can be found on the Astronautix site:
Lox/Kerosene. http://www.astronautix.com/props/loxosene.htm
And on this list of Russian rocket engines:
Russian/Ukrainian space-rocket and missile liquid-propellant engines. http://www.b14643.de/Spacerockets_1/Diverse/Russian%20engines/engines.htm
The problem is the engine has to have good Isp as well as a good T/W ratio for this SSTO application. There are some engines listed that even have a vacuum Isp above 360 s. However, these generally are the small engines used for example as reaction control thrusters in orbit and usually have poor T/W ratios. For the required delta-V I'll use the fact that a dense propellant vehicle may only require a delta-V of 8,900 m/s, compared to a hydrogen-fueled vehicle which may require in the range of 9,100 to 9,200 m/s. The reason for this is explained here:
Hydrogen delta-V. http://yarchive.net/space/rocket/fuels/hydrogen_deltav.html
Then when you add on the fact that launching near the equator gives you 462 m/s for free from the Earth's rotation, we can take the required delta-V that has to be supplied by the kerosene-fueled vehicle as 8,500 m/s. I'll focus on the RD-0124 because of its high Isp, 359 s vacuum and 331 s sea level. On the "Russian/Ukrainian space-rocket and missile liquid-propellant engines" page its sea level thrust is given as 253,200 N, 25,840 kgf. However, the Falcon 1 first stage weighs 28,553 kg. So we'll need two of them. Each weighs 480 kg, so two would be 960 kg. This is 300 kg more than the single Merlin 1C. So the dry mass of the Falcon 1 first stage is raised to 1,751 kg. There is a RD-0124M listed on the Astronautix page that only weighs 360 kg, but its sea level Isp and thrust are not given, so we'll use the RD-0124 until further info on the RD-0124M is available. Taking the midpoint value of the Isp as 345 s we get a delta-V of 345*9.8ln(1 + 27102/1751) = 9,474 m/s (!) Note also the achieved delta- V would actually be higher than this because the trajectory averaged Isp is closer to the vacuum value since the rocket spends most of the time at altitude. This calculation did not include the nose cone fairing weight of 136 kg. However, the dry mass for the first stage probably includes the interstage weight, which is not listed, since this remains behind with the first stage when the second stage fires. Note then that the interstage would be removed for the SSTO application. From looking at the images of the Falcon 1, the size of the cylindrical interstage in comparison to the conical nose cone fairing suggests the interstage should weigh more. So I'll keep the dry mass as 1,751 kg. Now considering that we only need 8,500 m/s delta-V we can add 636 kg of payload. But this is even higher than the payload capacity of the two stage Falcon 1! We saw that the thrust value of the RD-0124 is not much smaller than the gross weight of the Falcon 1 first stage. So we can get a vehicle capable of being lifted by a single RD-0124 by reducing the propellant somewhat, say by 25%. This reduces the dry weight now since one RD-0124 weighs less than a Merlin 1C and the tank mass would also be reduced 25%. Using an analogous calculation as before, the payload capacity of this SSTO would be in the range of 500 kg. We can perform a similar analysis on the Falcon 1e first stage that uses the upgraded Merlin 1C+ engine. Assuming the T/W ratio of the Merlin 1C+ is the same as that of the Merlin 1C, the mass of the two of the RD-124's would now be only 100 kg more than the Merlin 1C+. The dry mass and total mass numbers on the SpaceLaunchReport page for the Falcon 1e are estimated. But accepting these values we would be able to get a payload in the range of 1,800 kg. This is again higher than the payload capacity of the original two stage Falcon 1e. In fact it could place into orbit the 1-man Mercury capsule. The launch cost of the Falcon 1, Falcon 1e is only about $8 million - $9 million. So we could have the first stage for that amount or perhaps less since we don't need the engines which make up the bulk of the cost. How much could we buy the Russian engines for? This article says the much higher thrust RD-180 cost $10 million:
From Russia, With 1 Million Pounds of Thrust. Why the workhorse RD-180 may be the future of US rocketry. Issue 9.12 | Dec 2001 "This engine cost $10 million and produces almost 1 million pounds of thrust. You can't do that with an American-made engine." http://www.wired.com/wired/archive/9.12/rd-180.html
This report gives the price of the also much higher thrust AJ26-60, derived from the Russian NK-43, as $4 milliion:
A Study of Air Launch Methods for RLVs. Marti Sarigul-Klijn, Ph.D. and Nesrin Sarigul-Klijn, Ph.D. AIAA 2001-4619 "The main engine is currently proposed as the 3,260 lb. RP-LOX Aerojet AJ26-60, which is the former Russian NK-43 engine. Thrust to weight of 122 to 1 compares to the Space Shuttle Main Engines (SSME) 67 to 1 and specific impulse (Isp = 348.3 seconds vacuum) is 50 to 60 seconds better than the Atlas II, Delta II, or Delta III RP-LOX engines. A total of 831 engines have been tested for 194,000 seconds. These engines are available for $4 million each, which is about 10% the cost of a SSME." http://mae.ucdavis.edu/faculty/sarigul/aiaa2001-4619.pdf
Then the much lower thrust RD-0124 could quite likely be purchased for less than $4 million. So the single RD-0124 powered SSTO could be purchased for less than $12 million.
Even though the mathematics says it should be possible, and has been for decades, it is still commonly believed that SSTO performance with chemical propulsion is not possible even among experts in the space industry:
Space Tourism is a Hoax By Fredrick Engstrom and Heinz Pfeffer 11/16/09 09:02 AM ET "In 1903, the Russian scientist Konstantin Tsiolkovsky established the so-called rocket equation, which calculates the initial mass of a rocket needed to put a certain payload into orbit, given that the orbital speed is fixed at 28,000 kilometers per hour, and that the maximum speed of the gas exhausted from the rocket that propels it forward is also fixed. "You quickly find that the structure and the tanks needed to contain the fuel are so heavy that you will never be able to orbit a significant payload with a single-stage rocket. Thus, it is necessary to use several rocket stages that are dumped on the way up to get any net mass, i.e. payload, into orbit. "Let us look at the most successful rocket on the market the European Ariane 5. Its start weight is 750 tons, of which 650 tons are fuel, 80 tons are structure and around 20 tons are left for low Earth orbit payload. "You can have a different number of stages, and you can look for minor improvements, but you can never get around the fact that you need big machines that are staged to reach orbital speed. Not much has happened in propulsion in a fundamental sense since Wernher von Brauns Saturn rocket. And there is nothing on the horizon, if you discount controlling gravity or some exotic technology like that. In any case, it is not for tomorrow." http://www.spacenews.com/commentaries/091116-space-tourism-hoax.html
The Cold Equations Of Spaceflight. by Jeffrey F. Bell Honolulu HI (SPX) Sep 09, 2005 "Why isn't Mike Griffin pulling out the blueprints for X-30/NASP, DC-X/ Delta Clipper, or X-33/VentureStar? Billions of dollars were spent on these programs before they were cancelled. Why aren't we using all that research to design a cheap, reusable, Single-Stage-To-Orbit vehicle that operates just like an airplane and doesn't fall in the ocean after one flight?" "The answer to this question is: All of these vehicles were fantasy projects. They violated basic laws of physics and engineering. They were impossible with current technology, or any technology we can afford to develop on the timescale and budgets available to NASA. They were doomed attempts to avoid the Cold Equations of Spaceflight." http://www.spacedaily.com/news/oped-05zy.html
Then it is important that such a SSTO vehicle be produced even if first expendable to remove the psychological barrier that it can not be done. Once it is seen that it can be done, and in fact how easily and cheaply it can be done, then there it will be seen that in fact the production of SSTO vehicles are really no more difficult than those of multistage vehicles. Then will be opened the floodgates to reusable SSTO vehicles, and low cost passenger space access as commonplace as trans-oceanic air travel.
Bob Clark
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Another question asked in email and on different forums about this is, if the Russians already had these high performance kerosene engines that make SSTO possible why aren't they doing it? I wondered about that too. I thought their new Angara rocket should be SSTO capable since it will be using these high performance kerosene engines. But then I checked the specifications of the Angara rocket on the SpaceLaunchReport.com site:
Space Launch Report: New Launchers - Angara. http://www.spacelaunchreport.com/angara.html
I found that the Angara mass ratios are significantly worse than for the SpaceX Falcon launchers. In fact, in general the Russian launchers are not as well mass optimized as the American launchers. This probably is a big part of the reason the Russians have had this great drive to increase the performance of their kerosene engines - out of necessity. Then to get SSTO you use the best features of both the American and Russian designs combined into one.
Bob Clark

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I wanted to get some info on minimal trajectories to orbit. I found this page after a web search:
Atmospheric Flight of a Launch Vehicle. http://astro.u-strasbg.fr/~koppen/launcher/launcher.html
It has a JAVA applet for calculating the ascent to orbit for some known rockets. It requires you to tweak the ascent angles, but fills in much of the information for you. Problem is I haven't been able to get it to work right. For the given rockets, we know they were able to achieve orbit but every time I try it I get that the orbit was not achieved. Perhaps someone here can figure out how to work it.
Bob Clark

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It is known that you can reduce the fuel requirements to orbit by using a lifting trajectory. For instance it was explored for use on a reusable orbital craft in the 80's:
Fire in the sky: the Air Launched Sortie Vehicle of the early 1980s (part 1) by Dwayne Day Monday, February 22, 2010 http://www.thespacereview.com/article/1569/1
In the first post in this thread I suggested this fact be used for a lifting-body SSTO. Here is the argument I made:
------------------------------------------------------------------ "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 = 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 straight-line 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) = 7,925 m/s additional delta-v to reach orbit. Then the total delta-v to orbit might only be in the 8,100 m/s range. Note this is significantly less than the 9,200 m/s delta-v 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 center-line 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 re-directed 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." ------------------------------------------------------------------
The key fact I want to focus on is that sqrt(7800^2 + 1400^2) = 7,925 m/s number. Actually if you add on the ca. 460 m/s velocity you get for free from the Earth's rotation you might be able to reduce this to sqrt(7400^2 + 1400^2) = 7,531 m/s. So I what I want to investigate is if it is possible for a rocket that does not have wings or lifting surfaces to travel at such a straight-line trajectory at an angle from lift-off so that the achieved velocity will be in this range (but keeping in mind this might not be the same as the equivalent "delta-V" that actually has to be put out by the engines.) But I have question: if you angle the rocket launch from the start with the thrust vector along the center line with the trajectory angle such that the vertical component of the thrust equals the rocket weight could you have the rocket travel at a straight-line all the way to orbit? I'm inclined to say no because the gravity is operating at the center of gravity of the rocket not at the tail where the thrust is operating. This would certainly work if you had a point particle, but I'm not sure if it would work when your body has some linear extent. This method for traveling at a straight-line at an angle for some or all of the trip would make my calculation easier. However, I'll show in a following post there is another way to do it even if this first method doesn't work. The second method though would require some modification to the usual design of rockets and is more computationally complicated.
Bob Clark
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The question can be boiled down to this: imagine you have a long cylindrical object, could be a pencil, could be broom stick. You can give it an initial thrust at the bottom and push it away at an angle. It will then follow a parabolic trajectory with its center of mass following a parabolic arc, disregarding air drag. What I'm asking is will it work to supply a continual push at the bottom with the force maintained at the bottom at a fixed angle to the horizontal so that the vertical component of this force is the cylindrical body's weight? Will the body maintain a continual straight-line trajectory at this set fixed angle?
Bob Clark
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This is really actually a question in continuum mechanics sometimes called solid mechanics. In physics we sometimes idealize a body subject to forces as a point particle. Idealized this was, the thrust force applied to the rocket would add as a vector to the force of gravity so it would cancel gravity no matter what the angle of the trajectory. But in continuum mechanics you have to consider the physical extent of the body and where on the body the forces are applied. I imagine this is a common type of problem addressed in mechanical engineering and civil engineering. A force applied at one position on the body won't have the same effect as when it is applied to another point for instance in regards to the torque produced. Torque is measuring the turning "force" on the body. But it's defined as the cross product of the force vector times the radial vector to the center of rotation. Intuitively, what we have to worry about is rotation of the rocket with the thrust applied only at the tail. But if the rocket were to rotate it would be about the center of gravity. However, if we make it so the thrust is always along the center line, the radial vector to the cg and the force vector are parallel, resulting in a 0 torque vector. That would mean there would be no rotation around the center of gravity so we should be able to maintain our straight-line trajectory. This would be valid if the center of gravity were fixed. But the cg is accelerating as the thrust is applied. So I'm not sure if this argument applies in that case.
Bob Clark
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