PB666

What is your proof that this is the reason, what if they didn't want Mongolian picking up their 'accidents' and handing them over to the another group?

The solid fuel, oh no, not much more expensive. Solid fuel has aluminum in it. I mean Space X could get the Mexicans to make the housing and float it across the riogrande, lol. Here are some KSP dynamics, Prelaunch a = 300,360 I built a 9.95 payload rocket and on a 40.3 ton rocket with a SSME on it, no booster it made a = 646.8 km I added a very small attachment point added 4 sepratons to added 0.5t (5 second and separation) it made a = 672.5 km Replaced the seps with 2 Flea - aking total mass = 46.8t (8 seconds, with an aero fuel on top giving about 15 seconds of burn time to engine) a = 1270 km Increased main fuel tank added 4 hammer mass = 70.9 ton (side tanks add another 30 seconds of burn time to main engine) a = N/A escape velocity + 3700m/s at 72km. Replaced boosters with 2,2,2 asparagas mass = 81.3 to (side tanks add 70 seconds of burn time to main tank) escape velocity 70km alt v = 4700 m/s (requires a very aerodynamic payload to withstand maximum dynamic pressure) Payload size remain constant, in the last three only the booster system changed. The point- Accelerate quickly to the spike - Disposable boosters primary use is to accelerate the rocket as fast as possible to maximum dynamic pressure. Handle the pressure spike - They push the rocket against the greatest resistance to forward motion during the flight Dispose of the mass tax created by the boosters after the spike- After which they are disposed of, allowing the main engines to efficiently take the craft into space. Implicit in these three is they add critical kinetic energy, momentum, and altitude all of which act to make the main engine more efficiently and thus add more energy to the payload. 1. fast acceleration means less gravitational losses while 'hoovering' over the launch pad. The differential between gravity and omega^2*r plotted against time acculates as waste, faster acceleration along the prograde to circularlize reduces this waste. 2. Increased altitude adds potential energy and decrease gravitational pull (an artifact of KSP is that gravity drops about 10 time more quickly than it does on earth for each unit of altitude gained). Reaching these more quickly lowers the gravity costs even more. 3. Increased velocity means dV added by the main engines substantially increases kinetic energy, which can be converted to potential energy by Apo affording a lower cost burn to circularize 4. the cost to get to initial apo in terms of main engine dV is lower, which it has more dV left at apo to circularize and do other stuff (like say carry more payload at the expense of fuel). Even the smallest booster at launch can have the most profound effect if the main engines a scaled properly for high altitude climb. As we can clearly see, if separation of an orbit from the surface is the goal, boosters get the rocket to extreme heights without the need of stacked stage. Stacked stages need to increase by a factor of 5 for each lower stage added. There is a bit of a declining utility in boosters if they aren't aerodynamic, because for the first 20 seconds they are accelerating, the force of drag is less than gravity but once the force of drag = gravity the moment of total force goes up with velocity rapidly, there is some advantage to asparagas because you can control speed and prevent escalating drag cost, but those cost are most apparent for a very short period of time.

Where do you folks come up with these ideas 1. Elevator cons No material No means of elevaing the mass needed to make it, especially given no sufficiently strong material The wire would pass through GSO with a counter mass outside to keep it GS, this would interfere with station keep at GSO. No means of powering the elevator (conductor length exceeds limitation of powering, elevator would need to be self-powered, solar power would elevate very slowly, passengers would suffocate) No means of stabilizing the wire against resonance created by surface winds. 2. Space fountain. Completely and totally rediculous. 1. build a mountain about 20,000 feet taller than mount everest 2. build a mass accelerator in the mountain 3. at a 45 degree angle you have 15,000 meters x 1.414 = 21200 meters 4. You have a limitation on human acceleration of 8 g = 78 m/s 21200 = 1/2*78* t^2 t = 23 seconds (78-9.8)*23 = 1600 meters per second - exact same problem with space planes. you still have another 6500 m/s to go and this only affords you 64,000 km, you still need about 25% more verticel kinetic energy to reach a burn point. 5. No current materials capable of building such a structure. Thestructure would have to be strong enough to contain a partial vacuum.

Or another way to think of it. When you travel about the sun you have potential and kinetic energy, imagine if you could borrow temporarily some potential energy and turn it into kinetic energy, and then give it back, you could not do this because to give potential means you lower altitude. Crossing a planets SOI does this, you give up some of your potential energy momentarily and take kinetic energy (not to the star but to the planet), but when you fire your engines you gain alot more kinetic energy because you are going faster (work = force * distance traveled), then when you leave the SOI you give back the kinetic energy loan and take back the potential energy and get to keep all the extra kinetic energy.

If the fuel is 80% of the cost why bother.

Seriously, look at the map, the BCLP is east of north of the Mexican coast line, a line due south does not really reach habitable land for about 500 miles. And most polar orbits are not directly over the poles so they could go SSE a bit to give even more clearance, the bay to the south of Rio grand is a scarcely populated area.

How much fuel (in tons) is needed for Falcon 9 to land back near landing pad? How much does landing legs weight? How much weights landing gear? The solution is to design a side tank with a cheap engine, cheap fuel, and barely suitable wall.....otherwise known as a SRB.

really? I would say the on paper only credibility rate is quite low.

Right you burn a different higher pe which if you are leaving that system is unneccesary also, e doesn't really matter that much if you are leaving, its all over one, but why would you invest in a higher pe, if once you leave the system you are not going use that pe. Its sauce for the goose, because the most efficient place to burn out is always at periapse, the place were speed is fastest. BUt the most important thing is that the reason Apo increases to Escape point followed by increases of post-Escape velocities is that burning at pE increases kinetic energy, part of which is converted to potential energy as it leaves, and the rest remains kinetic energy relative to the system central mass. The reason that Transfer energetics is not ideal in the 2 step process between orbits that start small and then increase 5 or six fold is because the second burn is velocity wise, far from the first, if it is close to the first, its efficient, and if it close to escape burn its efficient (because at escape velocity the second burn is basically 0 relative to the planet, its a limit that can never be reached in a comoving space-time system) a second burn is not needed. Its all about the energetics, work is the addition of force over distance, if you travel into a low point around a celestial and then do work, you are moving faster relative to the celestial. Your craft has an absolute limit of thrust to weight a measure acceleration (increase in velocity) and of force/time, it does not matter if it is stationary to the celestial or moving 1/10th the speed of light. Since work is force * distance and since velocity is distance * time the force/time * distance (=velocity * time) therefore force/time * velocity* time Kinetic Energy= integral (t=0 to t=end thrust) thrust*velocity. since the moment of thrust can be converted to dV as dV = f(thrust. time) the two velocity aspects multiply by each other. This aspect of rocket behavior can be used in the atmosphere also, conservation of motion during a launch is extremely important, for example the atmosphere aspect decreases allows increased velocity and ground referenced acceleration, loss of acceleratoin in stage changes cost efficiency for several reasons, thus its a good idea to minimize the stage losses. The more one is pushing against gravity to stay aloft the more important these are. One strategy to avoid this is to shed boosters from the side, in which the main engines continue thrusting, even if the booster engines have energy left. Once a craft has turned and has significant omega^2 r centripetal acceleration, it can relieve its stack stages without much cost. Even though you are adding the same dV you have produced alot more energy. That celestial only took u/r potential energy coming in and will only take u/r going out, so the large ramp up of kinetic energy at the pe, is all for the space craft to keep once its leaves the SOI, if it wait halfway up the exit to add more dV it will be going slower relative to. I think the way we can look at this is via enthalpy available in warped space-time complexes, there because space-time is warped by energy. Bodies that enter these systems cannot gain or loose total energy and random applications of force will result in a loss of energy from the bodies (such as frictional forces causing the collapse of a nebula protosystem disk). Nature assumes the application of force is random and so enthralpy will always tend from energetic to less energetic state, application of force intelligently can favor the improved energetics of one object at the expense of something else. In this case the exhaust of gases at pE in a retrograde direction almost certain results in those gases falling into the celestial. The system assumes a continuous motion toward disorder but ordering certain aspects can be done, even randomly objects are tossed from systems into space, others are tossed into the star, and the order in a planet is sometimes smashed into collisions.

That would be a major faux pas. Gee we need precious metal x- to build rockets to go search for precious metal x after we already run low, when if we had a space program in a time of plenty that had already found them, we would not be spending 50 years looking in space, while we have already run out. Some people just don't get science. You don't do science today for tomorrows jollies, you do science today for the benefit of decades and generations in the future. What ever kind process gave a brain to think abstractly gave the ability to rationalize future needs. I think by now we are figuring out that the earth is in a perilous way, if not in its due course, then by anthropomorphic acceleration, if you wait until the suns corona has melted the bottle in which your corona once was, it will be too late to find a place to sip your next corona.

e = (rapo - rpe) / 2a v = sqrt(u)*sqrt(2/r - 1/a) vis visa equation. e = 0.11728 a = 759018 rap= 848036.2967 solved using the 2525.4 = SQRT(3.53E12)*SQRT(2/670000-1/a) 2525.4 = 1879255.172 * SQRT(1/335000 - 1/a) ::Insert here because GUI screwed up on cut and paste::::: 1.80588E-06 = 2.98507E-06 - 1/a ::::::::::: 1/a = 1.17919E-06

I can put another. Try this form a low kerbol orbit, say a million km, burn to reach Jool orbit, record starting and final velocity. Next starting at the same orbit burn about half the dv of first, wait until you have lost about have of the velocity and then burn to make jool orbit. The some of the two dVs requiredis more than that required by a single burn. this is because when you a very close to a celestial,myou are going very fast, you have a square function of energy, when you add dV you add even more energy, howver there is a finite amount of exit energy required, determined by u and the periapsis radius, once you have superceded that your residual is a funciotion of 1/2v2 - energy required , the residual energy converts to velocity squar root (2 times residual energy). Remembering that you are traveling along an orbit of the celestial, so that once you are out of the SOI your residual velocity is added on top. Essentially you are exploiting the curvature of space time, which likes to hold onto to lower energetic trajectories longer, and release higher energetic trajectories, as the object is receding faster from a celestial mass it is experiencing gravity for a shorter oeriod of time, it keeps more of its velocity. If you are given two positions in a orbit to add velocity and you want to add dV to go away from that body, its always better to add velocity when the object is going the fastest. The closer to that fastest point you add dV the further and faster you will go away. Another example of this is kicking (repeated burns at pe for a high isp, low thrust rocket) to maximize apo or escape velocity). If the celestial is orbiting a star, then to gain orbit about the star, burn at a pe that is a sunset of the orbit (angle to prograde is greater than or equal to 180, depending on how far away you want to be) you might also consider adding normal or antinormsl intonthe burn, its cheap and it might lessen a inclination burn later.. To loose a around the celestial at or just before angle tonprograde reaches zero, dawn. If you are going to moho and for instances you can time the position correctly with regrad to the 6 degress of inclination change you need to make, you can save alot of dV)

Don't believe everything you read on the internet, particularly things written by NASA bashers.

That plan is lunacy.

Lets talk about the equipiment bay, litium ion batteries and Ethiopian and Japan Airlines.

Musk has the power to create interest in start up projects that Boeing cannot do. It takes Boeing 15 years to redesign/modernize a wide body aircraft. Does anyone know who the CEO of boeing is? Seen him on TV, even cares what his aspirations are. Musk has the power of personality, the public often feeds off delusions (look at the current body politic). His selling point is talking about going to Mars and waving his hands with plans. He can create a cult of personality and I noticed we have a few followers in this group. Boeing, a stodgy old corporation that has bought up most of its competition and its only remaining competition is Airbus [cough, gasp], which is owned by a bunch of European countries who are at the present trying to undermine each others economics, foreign policies and internal defenses (or simply deciding whether to leave the union) and absolutely paralyzed in dealing with the Syrian crisis.

More than that. The ISS has a balance, if they chain supply vessels it will change the center of gravity and cost more in dV required to station keep. There is an KSP addon called recyling bin as part of the EL launchpad pakage, I simply shove my supply drones into the recycling bin. Out comes metal and I can build a satellite with them.

Whether you trust the government or not, I doubt SpaceX is going to build a significant scientific foundation for other government or industries to work off of. So even if they make it would be a bit of a wasted effort. The point about governments doing it, even if slow is to develope a resource base (knowledge) that other explorers can use.

I have updated this with formula for the alterantive, It assumes the user will manage the dV. There it is only for space craft design. Its convinient because it might shorten the list of post-launch-minimal orbit dV needed u of course is the celestials gravitational constant. Earth is 3.986E14 (in standard metric units). Eccentricity (e) is (rapo - rpe) / (rapo + rpe) Happy square root day http://www.illinoisscience.org/2016/04/happy-square-root-day/ e Correction factor (k) e Correction factor (k) e Correction factor (k) 0.001 1.001 0.26 1.2176 0.65 1.4019 0.002 1.002 0.27 1.2245 0.66 1.4038 0.004 1.004 0.28 1.2313 0.67 1.4056 0.006 1.006 0.29 1.2380 0.68 1.4071 0.008 1.008 0.30 1.2445 0.69 1.4084 0.010 1.010 0.31 1.2510 0.70 1.4094 0.015 1.015 0.32 1.2574 0.71 1.4103 0.020 1.020 0.33 1.2636 0.72 1.4108 0.025 1.025 0.34 1.2698 0.73 1.4112 0.030 1.029 0.35 1.2758 0.74 1.4112 0.035 1.034 0.36 1.2818 0.75 1.4110 0.040 1.039 0.37 1.2876 0.76 1.4104 0.045 1.044 0.38 1.2934 0.77 1.4096 0.050 1.048 0.39 1.2990 0.78 1.4083 0.055 1.053 0.40 1.3045 0.79 1.4068 0.060 1.057 0.41 1.3099 0.80 1.4048 0.065 1.062 0.42 1.3152 0.81 1.4024 0.070 1.067 0.43 1.3204 0.82 1.3995 0.075 1.071 0.44 1.3255 0.83 1.3961 0.080 1.076 0.45 1.3304 0.84 1.3922 0.085 1.080 0.46 1.3353 0.85 1.3877 0.090 1.084 0.47 1.3400 0.86 1.3825 0.095 1.089 0.48 1.3446 0.87 1.3766 0.100 1.093 0.49 1.3491 0.88 1.3699 0.110 1.102 0.50 1.3535 0.89 1.3622 0.120 1.110 0.51 1.3577 0.90 1.3535 0.130 1.119 0.52 1.3618 0.91 1.3435 0.140 1.127 0.53 1.3658 0.92 1.3321 0.150 1.135 0.54 1.3696 0.93 1.3189 0.160 1.143 0.55 1.3733 0.94 1.3035 0.170 1.151 0.56 1.3769 0.95 1.2854 0.180 1.159 0.57 1.3803 0.96 1.2637 0.190 1.167 0.58 1.3836 0.97 1.2367 0.200 1.174 0.59 1.3867 0.98 1.2016 0.210 1.182 0.60 1.3896 0.99 1.1504 0.220 1.189 0.61 1.3924 0.995 1.1104 0.230 1.196 0.62 1.3951 0.9975 1.0802 0.240 1.204 0.63 1.3975 0.99875 1.0577 0.250 1.211 0.64 1.3998 0.9999 1.0169

Tell that to the astronomers that came up with the fix for the Hubble. You don't have to know optical physics to be a 3rd Millennium astronomer, but you would not be a very good one. Modern day astronomy, even ground based, is about spectrophotometer (strait out of chemistry and physics), atmospheric absorption and scattering curves, anomaly correction software, light integrator, etc. The instruments are doing most of the observing, the computer is doing the processing and the product comes out in bit image graphics, not the good ole silver-embedded negatives. Its pretty much the same in many areas, in the 70s we used to take plots and cut peaks and weigh them to integrate, from the 80s on they use electronic integration with multiple overlapping peak dissection. This stuff is physics, statistics, mathematics, computer science. You want to execute science beyond the cookbook level prepakaged machine science, you also need to be technically proficient in what you are doing. If I need to due composition analysis, I better know how to deal with diamond dust, clean a photospectrometer cell, calibrate a integrator, replumb an HPLC. do I need to do this everyday, no, but I better know enough to train someone, and I should know enough how to troubleshoot anomalies in the data stream or my data is not going to be worth ____. And so you see astromers looking at a Hubble image presentation they have to know enough about optics to deduce why in the hell there image is 50 time more fuzzy than it was supposed to be. Think about what they are talking about with the James Webb Space telescope. Designing a telescope to see in the low infrared range requires a knowledge of Behavior of reflective lens and mirrors in space (don't forget hubble 1.0) Behavior of the electronics in space (including sensors) Cooling critical parts of the telescope to below 7'K. Star tracking, system protection (as in avoidance of solar radiation on critical instruments), Scan windows. Station keeping. Infrared reflector composition. This stuff is not mail-order astronomy, you get a pH.D. in astronomy you better know something about other sciences. To question is the nature of science. Question? Need to characterize objects very far away, so far away they are red-shifted 10x in wavelength, how does this differ from light telescopy? What will it require? What are the likely pitfalls? What type of backup systems might be needed? How will material physics interfere (with very large objects there are vibrations and resonances).?

Do you want to be a stone age astrologer, iron age astrologer, early industrial stage astronomer, or a 3rd Millennium astrophysicist? I chose the last, your choice is up to you.

I didn't say it didn't but try looking to the earliest stars in the Universe without Hubble. One other point - lenses used for ground based telescopes - mathematics, chemistry, physics Application - atmospheric science, meterology, astronomy Making of lenses for space telescopes - material science (chemistry and physics).

What good is theoretical physics if you can't apply it? Note: The modern systems - Hubble space telescope, Kepler observatory, gamma-ray observatory, LISA, Mariner-1,2 Voyager 1,2-LISA.

Hmm, Astro means you get into space, before you can do that you need to get through the atmosphere. First issue is propulsion systems, SRBs are pretty much alchemy to advanced chemistry, physical systems of lH2 and l02 tanks, physical systems and engineering for the turbochargers and other systems in the rocket engine Second issue it flight control, which means that you probably need a reliable gyroscope, gyrostat, which was invented by Lord Kelvin who was a mathematician and pysicist. Forth is in flight steering, this is electronics and aerodynamics, for example if you have steering vanes on your rocket. V2 has a problem, it ran into Mach Compression on the way up.So I would say materials required for the pre-Astro phase are sensors and fast radio transmittors. For this one would need electronics. In addition to this you have system aerodynamics for dealing with Mach effects which is advanced aerodynamics. These are needed to get into space. Once you are in space you need Gyroscopes and reaction wheels. Computer systems that can keep track of orbit relative orientation. Chemistry and advanced propulsion systems for the RCS Chemistry, electronics, relativistic physics for the ION drive, also need photovoltaics or nuclear physics Astronomy and advanced computer science for star tracking. Advanced Thermodynamics for the maintenance of space craft temperature. Global positioning and relativistic physics for determining inspace position. Materials chemistry for design of space resistent parts (i.e. what happened to the Cannae drive when they tried to operate it in a vacuum). Mathematics, Chemistry, (requires physics and Mathematics) Physics (Requires Mathematics to test physics you need chemistry) Electronics (Chemistry, physics and mathematics) Astronomy (itself, physics and mathematics) THese are the basis.

Last week there was a thread created that discussed the basic requirements of deltaV required to get into various positions of the moon. Other than the launch variables the statement was made or asked if deltaV tables was the best way to handle this. I looked at the from an energy perspective, first off I need to add that the classic formula for calculating delta-V between two circular orbits is - SQRT(u/r0) for the first burn (r is r0 in this case in the wiki image, ignore the v = ) r can either be an apoapse or periapsis and SQRT(u/r1) - (r is r1 in this case in the wiki image) for the second burn. r can either be an periapsis or apoapse The perfect energy requirement equal to the is close to this at in the case of the lowest and highest eccentricities (e = 1) but in the middle ranges it is considerably different. The basic problem is that elevation of a circular orbit neccesarily requires two burns. During small burns the change of velocity is small and as a consequence little momentum is lost. In changing to very eccentric orbit much momentum is lost, but the dV required to establish the second orbit is small fractional to the energy required to create the transfer orbit. At minimum escape velocity its zero. In eccentricities (e) of transfer orbits around 0.7 (e.g a geosynchronoous from LEO transfer) have substantial inefficiency because considerable momentum is lost as the satellite slows to its apoapse at which it needs to burn. So for example a station keeping burn is perfectly efficient, and also a escape orbit (minimal) is perfectly efficient (but because of N-body problems more or less a theoretical exercise) The energy requirement works within tolerances if the correction factor for eccentricity is provided dV (total)/((1-e)+LN(1+e^1.9)), up to about e=0.75 but becomes inaccurate after this. Its not perfect. I tested this with a number of orbits, a is irrelevant the error is a function of e. This means without using a table one has a minimum requirement for a single step energy plot of knowing e as well as initial radius and final radius. Its not hard to calculate e but in creates also a two step operation. Ergo the OP is correct, the two step dV plots are as simple as any other means of plotting the dV requirements of an orbital change.