PB666

Jet engines (.i.e modern efficient turbofans) have bypass flow, a jet engine works in two part process. the first part is that air goes into a turbine, were it reacts with gas to create work on the blades of the turbine, this is used to force air around the turbine to the back were both are mixed, the heat from the gas mixes with the forces air creating more work (the colder the incoming air temperature the higher the expansion ratio, the more pressure at the back of the engine). The outflow of the turbine and the impelling of air around the turbine create a larger gas flow. That is less important than what the jet does. A propellar on a plane creates work, but the tips of the propeller are forces to work against Mach forces when the blade travels fast enough, propellers cannot compress air well in the direction orthogonal to motion and make the air go faster (jet) than a certain speed. A jet compresses the air inward toward the axis of forward motion, in doing this it accelerates the air without needing for impeller blades to match that pesky Mach speed. So for aircraft that fly about one-half Mach or lower a propellar (or turboprop) is an appropriate propulsion. Once the craft exceeds 170 m/s in speed than Mach but below Mach speed a jet engine is more appropriate. Note (wikipedia-United Airlines 777 N797UA LAX.jpg) that the front of this Rolls Royce Trent Engine is about 2 workman in height, the back is about is a small fraction. So that if the Air comes in a laminar flow at say 250m/s (something that may occur close to Mach speed) then if the area at the back is say 1/2th the area then its exiting at 500 m/s. Note: wikipedia - File:Turbofan3_Labelled.gif Note that as one is closer to the axis of rotation the blue arrows move faster. Inside the turbine (yellow) they move the fastest. This engine converts some of the work done on gas in turbine (slowing it down, into work done on the bypass air (speeding it up). If we remember the equation F = 2 * eff * E/t / exhaust velocity. A jet engine (modern tubofan) does not expell gas at 3500 m/s, it more like 300 meters per second, so it can use the energy to create magnitudes more force, or in the case of a jet engine versus rocket 10 times the Force for the same amount of power input. For a low velocity turbofan it can do this magnitudes more force per gram of fuel. But that defeats the purpose, because the atmosphere is 1/5th as thick at 13 km as MSL and because jet engines work better when the bypass air is closer to 0K, you can not take advantage of high altitude flight if your jet engine is at its coffin's corner at 7 km. So that jet engines are designed to fly generally at a certain altitude, forgoing some thrust by having some higher exhaust velocity so that they can travel at sub-mach speed higher, the higher the flight the more efficient the system works. So jet engines are not designed to create the most amount of Forced air per unit of fuel, but designed to have an exhaust velocity at high altitude that allows optimal system efficiency in a zone between 25,000 and 45,000 feet. (The reason they are not designed to work at a higher altitude which they could, wing loading, the craft would need a larger wing area, ozone levels increase above 45,000, the engines would be larger an bulkier to increase inlet pressure . . . .unless you are flying above Mach 1 at 45,000 feet . . . .the designs are not really beneficial).

This is why one want to place a spacecraft on the edge of the safe "zone" at 400 km or so (that way you can wave to the ISS as you pass). The problem is that at 250km your craft is traveling at 7700 m / second, you are scooping up particles which means, they are imparting all their momentum into your space craft. To compensate you must expel that gas at >7700 So for your typical spacecraft you have your vessel which is a boxy thing a couple meters in diameter (or less) and these big bulky solar panels (often which cross sectional areas 3 or 4 times the area of your craft. To make this work you would need to expell gas at more than 3 to 4 times the area you collect over (since you cannot collect what bounces off your solar panels). this means the exhaust velocity needs to be at least 32000 (ISP ~ 3300) which is on the high end of commercially available ION drives. In terms of Insertion into GTO, lol, your power would not bee the limiting step, fuel flow would be, talk about nanometers/sq.second of acceleration.

Its a bird, its a plane its . . . . . . . .no it was a bird.

You quoted the reason yourself, so why did you post this. Even if one uses ION kicks, there is generally never enough thrust in the last burn past the periapsis to take advantage of the speed within a gravity well (oberth-like effect) We have been over this sooooo many times in this forum. 'Even if you have ION drive, you might want to carry a small chemical rocket to add speed as one passes the planet one last time'. And your not paying attention, satellites going to GTO now-a-days are using ION drives, this has been a trend since an ION drive rescued a chemical propelled rocket about a decade ago allowing it to complete its mission. ION is the way to go, and manf who can't figure out a way to get the solar-electric prop system past VE belts should probably seek a different occupation. NOTE: in image above, many satelittes with solar panels are operating in the Van Allen belts. The rings indicate belts where satellties are stationed.

So what happened to the Chinese alternative to the ISS, are they abandoning having stations in space?

There are a couple of issues here, let me start with the most abstract. Evolution is really fast (and if need be it can accelerate also), in 500 million years human descendants (500 million years ago, fish did not exist, something like tunicate that had a free swimming stage), if we survive, we be nothing like people today, so that our needs today may not represent our needs tomorrow. More to the point if we started having children in space, overtime humans would evolve to be compatible with life in space (zero gravity tolerance, more tolerant to radiation). Provided an ample supply of nutrients species can adapt to anything, e.g. chernobyl. So the idea here is that in 300,000 million years the Earth and all similar yellow stars will expand and wipe out all the species living on those planets, so how do species survive . . . . .go interstellar, prevent your star from expanding . . . The problem here is efficiency. Where you are at, a yellow star, we could think of a yellow star as a nursery, this is where life is more likely to start, the goldilocks zone has a sure point in which life can arise and produce sentiency, where as white blue stars produce to much UV (although species could adapt, the problem is that the light gasses in the atmosphere cannot, they evolve into space), Red stars produce more heat than useful radiation, to be close enough to the star to get photosynthesis basically burns off alot of water. Thus the goldilocks zone is actually a zone where life might exist, if even transiently, but a much much smaller zone where sentients evolve. The next issue is preserving your star, but why? there are so many long-lived red stars, why not just move to one of those. Red stars are mostly unstable, but as they age they stabilize and they then burn for 10s of billions of years with very predictable outputs. Moving into orbit around red stars is a solution. But what are the problems. Heat, there is alot more heat than photovoltaic radiation. e = hf there is more energy to be gained in higher wavelengths and easier to obtain, current modern PVs can be stacked to absorb at three wavelengths, getting efficiency up to 42%, but this is star relative efficiency. If we moved those to a red star, efficiency would drop. The second problem is to make use you have to get so close to the star, you have a heat problem, and panel weights increases . . . . .HUman need to become grossly more efficient at managing heat before we could even make an interstellar ship (the fusion problem) or live around a red star. So basically extracting hydrogen from a yellow star turns it into a red star, thus by that time you are so good with fusion, you don't care about solar. If you are good with fusion you are good at managing heat and extracting energy, so the heat is not a problem either. But just because you can extract heat does not mean every thing goes cold, it just means you get work out of it (somewhere along the line moving the heat from one system to another), the heat is still there. If we are doing it properly we take infrared radiation and release energy CMBR temperature and red shows up where we are working probably somewhere out in the system where it dark. You could use the heat to convert say helium back into hydrogen and dueterium, or iron back into lighter elements (we don't know how to do that, particularly, but you could store energy in hydrogen). You could be using the energy you store to create photonic racetracks that use solar and heat conversion to propel craft between the stars (which increasing looks like what interstellar travel will be, slow generational ships carried on rivers of photons). Somewhere (sooner rather than later) however the energy of the star needs to be given up as heat and increase entropy or the system doesn't work. So the bottom line is that humans descendant species (if we survive) will leave Earth and venture the stars . . . . . .and to do that we need to manage (exploit, manipulate, otherwise exploit physical systems and redirect the minutia of entropy) energy in our solar system better and a dyson swarm is one solution. Another solution is you exploit only enough to leave and find a more stable star(s). Dyson swarms and star farming is something that involves fusion. Three things 1. we have to get it to work 2. it cannot be dumping huge amounts of energy into the craft but power that can be easily transferred where needed 3. And do this without a prohibitive amount of mass. Otherwise all the energy produced will be sunk into pushing the mass around. Or to put otherwise, to become a Kardashev scale being, significant evolution to that effect has occurred, this may include non-organic sentients.

This is the chinese space-station, was it meant to come down so quickly?

But it could not possibly compile a list of all 13,000 asteroid belt objects if one the one hand its mission was to detect the motion of moving objects but at the same time plot X,Y,Z positions. The data it is sending back are x,Px and if the program is at all working its got a psuedo ellipse and a time reference to some clock like J2000. To know the relative position of an object based on 'the suns time' when you cannot simultaneously see all those objects means that you have a Newtonian understanding of the motion of those objects with respect to time. (X, Y, Z, T)

ION driven spacecraft with fully deployed solar panels are moving though the Van Allen belts at liberty, currently. The only known take down of solar panels in recent history that I know of are the loss of 2 panels on Hayabusa on its way to intercept its target by a freak intercept of a solar ion storm. Van allen belts are particularly harmful to humans, not so much to modern satellites, since the overwhelming majority of satellites use solar panels. . . . . . Compare with: http://www.nasa.gov/images/content/668517main_vab-orig.jpg ION drives are the way to go, 20 years ago few space craft have ION drives, now many have ION drives. No pesky and dangerous RCS to deal with.

That's an assumption, since customers are paying 120 to 132 million per F9 launch and reusable version of the FH is 90 million. Of course this is a wish-wash answer, since if the core cost 40 million, and BE is <64 for an reusable, then the price should be 100 million, which means the customers are paying for additional goods or services that are opaque (and so I used the term several times). Some of the services that they are providing will be service vessels to the ISS (Dragon capsule). The are charging more money for expendable, how much more we cannot tell, but it looks like 40 million-ish more. I think that Musk implied the other day that if they had enough customers to GTO they could change the amount of fuel the second stage carries, offering a higher performance to GTO. They could, if they wanted to, put metholox on board. But another point is that the second stage does not have to circularize to GTO if the PL has its own ion driven PL, it can push to GTO itself. If instead of pushing 18 T to to LEO it pushes 63 T to LEO they could have any number of rocket systems that would get them to GTO. - - - - Devinating the dV required to get to equatorial GTO from a cape Canaveral LEO300(because you were to too cheap to pay for Arianspace, https://en.wikipedia.org/wiki/American_Samoa, https://en.wikipedia.org/wiki/Puerto_Rico, https://en.wikipedia.org/wiki/Guam) - - - µ = 3.986 x 1014, P = (sec/d)(d/y -1)/(d/y) = 86163, f = 1/P, w = 2πf , w = 2π/P, w2r = µ/r2 , r = (µ/w2)1/3 42.163 Mm =aGTO 6.371 Mm + 0.3Mm = 6.671 Mm aLEO δSPE = µ/6671Km - µ/42163Km = 50,297,430 j/kg δSKE = (µ/6671 km - µ/42163km)/2 = δSPE/2 The theoretical amount of energy required to reach GTO from LEO300km = δSPE/2. But as we know a Hohmann transfer perfectly done requires two impulses of dV, and because we were too cheap to buy a ride from the equator we need to correct the inclination somewhere in orbit (preferably at the point of lowest velocity). Assumming that we were intellgent enough to burn at the equator LEO, then its easy enough However it needs to be done in two steps, in the second step is where you want to start a = (6671 km + 42163km)/2 = 24417 Km, Vapogee = V42163 Km = SQRT(µ * (2/r - 1/a)) = 1607.135 m/s (<--- note: this is the point of lowest velocity), the amount of δV we need to correct the orbit is. δV = [SQRT(µ/42163 Km) = 3074] - 1607.135 = 1467.4 m/s. Next, inclination change. (1607 + 3074)/2 = 2340.5 since we launched from cape canaveral at 28.6 degrees north, we need to change inclination by 28.6 degrees. We could use the nifty equation from my other thread, but in a circular orbit the maximum extent (you launch bearing 90' = due east) is on the unit of 1 is actually the angle your orbit crosses the equator because the orbital plane crosses the center of the earth at also the same angle. So basically 2340.5 * sin 28.6 = 1120.37 meters per second (on the tangent or orthogonal). The total δV = SQRT(1055.62 + 1467.42] = 1846.2 m/s 35.73° ±N. Note that would could have burnt in two legs at apogee this would have been 769.38 + 1467.4 = 2,236.7 m/s so better to burn on the diagonals than to burn on the orthogonals. IOW you would have saved 350 dV by launching from the equator, but paid 5 times as much for the ride. Vperigee, a=24417 KM = V6671 Km = SQRT(µ * (2/r - 1/a)) = 10,157 m/s our initial velocity was from LEO along the transfer elliptical. This means we needed to add 10,157.644 - 7729.3 = 2428.3 Therefore our total dV from LEO300 to GTO having launched from Canaveral was 4274.3. We can lower this cost even more, during the equatorial transfer burn we could have burnt at an ever so slight angle, lessening the dV required for inclination/circularization at apogee. This is because the 1 - cos (some small angle) is a number close to 1. For example cos 10° = 98.4%, loss = 1.6% but Sin is 17.3%, the closer one is to Θ = 0 the higher the gain of Y per loss in X. - - - Devinating the best way to get to GTO to LEO - - - - - - - - - - - - - - - - - ION thrusters. So all GTO satellites have solar panels, thus we assume they have power, but also ION drives are cheap, and the typical ION drive has an ISP between 1500 and 3500. http://busek.com/technologies__ion.htm These little thrusters weigh grams to kilograms So the question is how much fuel do you need per kilogram of satellite weight to get to GTO. δV = ISP * 9.81 * ln(mi/mf) for 1500 ISP: 0.2883 = ln(mi/mf) , 1.33 = mi /mf > 25% of the weight of the spacecraft + 10% storage penalty = 27.5% of the satellite is fuel. If you had a 63 ton payload to LEO this means 17.3t is fuel, 45.7 t is yours to keep in GTO. So basically that basically blows any other launch system out of GTO except SLS launching to LEO and doing the same but getting 90t to GTO. (but the problem is that there is not enough panels you can put around increasingly large space craft to do this to indefinite size, so its possible, but not practical, is better to launch in pieces). for 3300 ISP: 1.141 = mi /mf = 12.36%/87.64% Again for a FH LEO of 63 thats 55.2 t to GTO. The problem of solar panels is severe, but not to bad if you are launching 4 satellites to GTO, better request better payload fairings. Note that the higher the ISP of your ION thruster, the more losses from diffuse impulse. But on the other hand you can use the ION thrusters indefinitely for station keeping, no need for RCS or other thrust systems, and the ION drives only need a tiny stainless steel tubing (like HPLC grade), to feed propellant to 6 motors each weighing less than a kg each. With a 55t space craft the small amount of weight added for 6-12 thrusters is not going to keep you awake at night. The good of ION is that if you are patient, you can lift maximal amounts to GTO, but drag is the enemy, such that if your launch provider can give you a decent push toward GTO so that you can circularize out of LEO, then this is a good choice, another choice is a variable voltage variable power ION thruster, which pushes low ISP in LEO and high ISP in GTO range. Compared to other engines, this is off the shelf stuff, order today and get . . . well at least before the other guys. Cryogenic hydrolox. We have deep space engines that range in ISP from 425 to 462 ISP, these are terrible expensive, and there appears to be a delay, but these will get payloads to GTO and they typically weigh a couple of hundred kilograms. If you need your payload to GTO tomorrow, these are your deep space engines (except it will take two + years for your RL10b-2 to be delivered, you are a shrewed planner (who just had a heart attack when you recieved the 30dCOC bill for the thruster). for 462 ISP: 2.56 = mi /mf = 61.0%/38.94% The cost is that you need >4% storage (61.0% * 0.04%) = 2.44% and the cost of one engine, like the ION drive you are probably making multiple kicks at LEO to boost a. With fuel and tank and engine factored out, the RL10b-2 can get 22.73t to GTO. IMHO, unless you can find a cheap Russian or Chinese substitute, you are better off going with an expendable which can deliver the same payload for 130 to 150 million. An alternative that would be clever is simply use the cheap whatever cryolox engine to push the 2400 dV required to push 3/4ths the way GTO and use the ION drive engines to do a sloppy circulization. This gets the PL into the 40t range and save alot of time. The weakness of ION drive engines is in attempt to burn at LEO perigee, they own the skys at apogee, so that if you place at high enough apogee where they get full sunlight, they can circularize via spiral mechanics and in a decent amount of time reach GTO. Again since you are dealing with space X you might want to consider methane over hydrogen. Metholox for 375 ISP: 3.91 = mi /mf = 68.8%/31.2% The cost is that you need >4% storage (61.0% * 0.04%) = 2.75% and the cost of one Raptor engine, like the ION drive you are probably making multiple kicks at LEO to boost a. WIth fuel and tank and engine? factored out, the raptor can get 16.92t to GTO.

The Isaac Arthur video depicts the use of fusion energy and a dyson swarm to create a magnetic field such that electrically charged particles end up at the poles, where somehow passing ships siphon off the gas and store it, according to the logic if you siphon off enough gas, young enough in a stars life, then it becomes a brown dwarf that lives forever, therefore prolonging the life of your civilization. A couple of flaws in the theory 1. Current fusion power does not create such a power or have a level of efficiency that it might be used in space. 2. the creation of the magnetic field requires ships to be fairly close to the star, an operating environment hostile to any of our known materials. 3. although alpha particles are charged, the solar gases tend to be protonic in nature, the first electron of a proton is rather more easily removed and because they accelerate more quickly, this would cause an older start basically want to go the direction of a white dwarf. Over time the amount of helium, carbon, oxygen would build up in the core. The author of the video, when interviewed elsewhere appears to have a great deal of skepticism whether type II and III civilizations as star farming would be notable signs of a sentients presence, even in relatively closeby galaxies. No such evidence of dyson swarms have been observed. In particular if you are converting the stars sunlight to energy as close range, your swarm is going to have a huge infrared signature, but not show much in the form of light.

He's right though, Block 5 is the first full production for reusability, It will be interesting to see what their actual reusability or expendability prices will be say in a year or two from now.

I just mod the stock stuff with RL stats. The exception is the RL10b variants (cause the game version looks nothing like an RL10b. What kills however is that tank masses in game are rediculously high. The STS ET had a 4% tank mass to fuel ratio and that was a liquid hydrogen tank, but on the other hand I just launched 55 kt of rocket tonight and put 3300 kt in LEO shooting 800 kt of metal into transmartian orbit. I bet I can make a crater on any upstart colony with that, lol.

Hint: don't submit job application to chinese space agency.

Well if you had a model for the second stage, you could actually use FH for every rocket and have enough fuel in the second stage to re land once it dumped its payload, I wouldn't count on it. Elon Musk still considers the second stage an expendable part of the rocket, estimates that it cost in the 10 million dollar range, if we argue that a core and two boosters cost 58 million dollars you cannot justify the added launch cost to bring back the second stage. What they will probably is make the second stage cheaper (less costly to build). SFRBs are not the way to go, they make a rocket more complex, as F9 did pair your side boosters the same diameter as the main rocket, if you are not going to use fuel feeds from the booster to the core, then what you do as you approach 250 m/s then action/toggle some of the cores engines off (assuming you have a multi-engine core), after passing MaxQ tag them back or leave them off until booster separation. This keeps fuel in the core. Asparagas may not be used but it is handy in one particlar situtation, if you have a PL that is extremely bulky and you want to delay MaxQ until the static pressure is hideously low, then just go strait up and drop boosters, the FH shows how cheap boosters can be (10-20 million or so per each), over there in the corner ULA is charging 400 millon dollars, so dropping 4 boosters at 56 million on a set.

When I was trying to look up the price of the FH heavy expendable, I found a range of prices, on the high end it was 150 million on the low end it was 132 million and just expending the core was much lower even still. Even at that 150 minus 90 is 60 million, much of which is in the core and core upgrades, this means a booster could be in the 14 million dollar range, if that was the case then the value to the company in the core could be in the 32 million dollar range. But he claims that 75 percent of the cost of the rocket is in the core, so a unmodified core would be cheaper (28-30M maybe) thus the cost of a F9 would not be 80 million, but 40 million. This is what I am saying, there are alot of contradictory statements, you will have to do some statistics on the flights secured from now once block 5 is up and running seeing how much are expendable or not. But at between 120 million and 132 million per contract launch he is either offering great opaque perks, they are making alot of money or what they put out on their flyer in meaningless. On of the potential purchasers mentioned they would be the first to buy a ride on an expendable for 30 million, I don't think SpaceX took them up on that, but that may be reflective of what the actual cost of an expendable is. My guess is that the final price on a FH is in the 120 to 150 range, on the expendable is 150 to 200 (again, not sure what is being paid for, maybe lavish vacations at some Seaside resort in the Sechelles for the purchaser) this is based on the average contract price for the F9 of 120 to 132 million dollars.

Got a power supply? With a tiny 100,000 Ve ION drive and a small amount of Xenon we could get that into the 50ish km/second range (over 30 years). Gliese will come and will go, but it will never approach a distance where 15ish km/sec would be useful to transit.

I wanted to post this in an F9 thread but there wasn't one, SpaceX thread is a dirty laundry list as it is, and the BFR thread is already polluted by off-topic discussion of cost. Here is a link to the cost of the Falcon 9. http://spacenews.com/spacexs-reusable-falcon-9-what-are-the-real-cost-savings-for-customers/ Its from two years ago and prices have gone up. What we can surmise from a variety of sources is that Falcon9 expendable runs 62 million dollars, this is not the cost but what SpaceX charges. There have been varying reports some reporting that spaceX make 2 to 5 million on the launch, some report they make 25 million, I would more predict they are making in the low millions, but other cost of doing business are not included, for example Autonomous drone ships, Boca Chica development, product testing. A falcon heavy is 90 million about 28 million dollars different, be we know this is not just the cost of the boosters, it also includes the cost of reinforcement. There have been estimates on the internet concerning the profit SpaceX makes per launch, these range from 24.8 million (see above) to losses in the 10s of millions, what is known is that besides Musk investors are pooring money in and spaceX is contracted to launch 70 public sector satellites worth 10 billion, which it can now do at the rate of about 3 to 5 per month. Assuming these are all F9 thats 142 million dollars per satellite. Another company offers that SpaceX has 12 billion dollars worth of contracts for 100 contracted launches, 2 which we know of are falcon heavy, so from this we can argue that they are gathering somewhat less than 120,000,000 per F9 launch. Some of these include the cost of the dragon vehicle, which is either currently recyclable or will be recyclable in the future. https://www.investopedia.com/articles/small-business/011717/space-exploration-technologies-corp.asp http://www.vault.com/company-profiles/aerospace-and-defense/space-exploration-technologies-corporation/company-overview Thus is appears that 62M is the base sticker price, that price dealers put on the car that always seems to at the back of the repair lot when they ask the salesman to see it. If we work between the two extremes, 37 - 67 m then we have a confidence range for the core, based on the statement of Elon Musk that 75% of the cost is the core-booster, then we can guess at the cost of the booster as 27 million to 46.5 million.So bring the the cost of the boosters in-line with some reasonable fraction of the core the cores being each 10 million or so each then the core cost should not exceed 40 million. We can also add to this the 4 to 6 million dollar cost of the fairing, which means the reusable costs in the vehicle range from 31 to 46 million dollars. It is nearly impossible to judge the cost of the vehicle with profit estimates based on offered price and sale prices so widely divergent. We have to basically assume that the cost of the F9 is 62 million and thus the cost of the core is <46.5 and the cost of the fairings 5 million dollars. The cost of the boosters for the FH should be less than <14 million dollars. There are two basic models, the first model is Elon the sly businessman model, in this model he is charging the customers cost for expendible but gives no discount for reusing a rocket, the installation cost of the core is 5 million and the recycling cost is 5 million. While recycling the fairing changes the shape of the curve (higher saturation limit), because he is so overcharging the customers on the reuse, there is little relative to be gained after 10 launches and he would be better off vending it as an expendible. If he discounts the reusable flight then the gain in profit will be more slowly Model - - - - cost/l - - -profit/l Maiden 62 0 1 reuse 48.8 13.25 2 reuses 44.7 17.7 3 reuses 42.1 19.8 5 reuses 39.9 22.1 10 reuses 37.9 24.1 20 reuses 36.9 25.2 IN this same model, if he offers reusable rocket for 50 million, then we would see a -12 millon dollar average profit for the first launch 11.5 million/l by the 10th reuse and 13.2 mil/l by the 20th, and 13.6 by the 30th reuse, after which the marginal gain for continued reuse versus expending it at 62 million dollars (for a customer who wants higher dV). Lets say they could reuse the core every two months, the means that for the additional 10 more flights earning just .5M more for flight the would have to forgo the interest of there immediate gain of 12 million dollars. So it is really marginal to keep reusing the rocket, if there is an immediate interest in selling the whole rocket for more money. Not to mention the insurance underwriter for the rocket is probably back there saying they will get some discount if they expend the rocket and start over with a new one. There will also be improvements over time. In the next model far less money because the core is a smaller part and there is more labor on installing the booster and recycling the booster than the above model, in this model recycling the fairing makes a considerable difference, but not to be included in order to focus how many flights to reach a point of marginal gain. Model - - - - cost/l - - -profit/l Maiden 62 0 1 reuse 58.5 3.5 2 reuses 57.3 4.7 3 reuses 56.7 5.3 5 reuses 56.2 5.83 10 reuses 55.6 6.4 20 reuses 55.4 6.65 30 reuses 55.8 6.8 In this model there is very little room to discount, so that if the offer is made to expend the rocket, there is little profit margin. But still after 20 to 30 flights, there is probably some utility and sending the rocket on an expendible mission there is just no profit in doing it. One of the advantages of selling expendibles is to provide a basis for rocket manufacturing, this model assumes that low price will stimulate demand which will then cause increase usage, if not the manufacturing part of the business will suffer, and eventually will not be profitable, whereas a robust manufacturing process can support more capital. Of course as we see the contract prices are way higher than 62 million dollars per launch, which means they are providing other services that may be the cash cows of the operation, marginal profit of the core is trivial, what they really want to achieve is more customers (like themselves in a satellite telecommunications business). The other thing that appears to be in this rather opaque business model is that customers who use a reused rocket will get a discount, but how much is not clear because they are paying more for other services, and so the discounts may come on the other services. Again there is some information, but the information on the cost and pricing is not complete. We see that F9 is bing offered for 62 million, but the customers are contracting for much higher prices for other services.

Their payroll per flight is too high, this is due in part to launch delays working between a couple of sites, once the boca chica site is operational, if two sites are delayed they can still launch from the third. Southwest does not build their own 737s, SpaceX build their own rockets, if they can build more efficiently, their cost will go down, if they can launch more frequently and more predictably, then the cost of launch crew per launch goes down. BTW SW relative competitiveness is not asymptotic, but saturation kinetics.

They generally borrow money to buy the fleet, the increased cost of interest on the loans is leveraged against increased sales. So that gets factored in. BTW this is always the case, in carrying a capital inventory the alternative use of the money has to be factored in, even if you don't borrow. The good thing is the cost goes down as the fleet depreciates in value (value goes off the balance sheet and onto the P/L sheet), if you consign an aircraft to a depreciable life after say 15 years and thereafter has only salvage value, and you use your fleet for a long period of time then that cost will drop to a trivial level. So what determines the vehicles lifespan, why doesn't an airline buy a new airframe after 15 years and consign the old air-frame to the desert or sell it to Middle of Nowhere Airlines? The issue is competition, if the competition is fierce (i.e. the mean net profit per ticket is low), then to keep cost down you use the plane as long as you can, as long as your passengers will accept it, as long as the FAA will allow, and as long as your repair crew is not charging more to repair an older vehicle. IOW, the cost of your capital inventory is low and expensed cost is also low. SW defines competition in the medium distance commercial airline market, so they use a single type of plane, lowering cost and optimizing maintenance and use it repeatedly and force ticket prices down. In the case of SpaceX, the longevity is more market driven, alternative prices per launch are several fold higher than what they are offering, the are trying to attract new customers (including their own sat. tele. business) which they see that in many aspects of space they can outcompete by reusing the same cores, boosters and fairings. But the price driver for their satellite communication will be in getting the cost of launch down and since they are the customer they can set their own QC (Hughes is certainly not looking amorously at their move). For the other market, as long as their competitors are selling at 3 to 10 times their price the same service, then its probably good for them to use that differential to replace their cores from time to time. Essentially they are building infrastructure that they need to keep going. They are not in exactly the same situation as SW, but it a nice selling point to offer '1000' reuses. We have to factor out one thing. If you have an old core, and you need to place a payload that is in the F9 expendable range, you can either offer the customer a ride on a F9, recover the core and boosters, or offer a used core to be expended. My guess that factoring in the rockets known depreciation they will probably offer and F9 expendable more cheaply than an FH reusable if the core is of a certain number of reuses. Again, part of your business is building cores, then you want to keep that operation going (otherwise your workers may be working for your competitors as their business dies). In terms of the BFR, I am not sure how their product testing will play out. In order to stay on topic I just want to mention, its hard to compare something in the engineering/early construction phase, with the operating model for a 737 that has been in flying since 1968 or the F9 which has been recycling for the last couple of years. These handwaving arguments are often points where threads derail into off-topic never-land. I caution about making economic conclusions at this point, and I would remind everyone that SpaceX has yet to turn a profit in any of their their operations, they are just throwing more skin into the game, and frequently (as in the oil boom of 2014) this does not have a pretty outcome.

Largest to go to GTO, iirc. I have no idea if this was accurate, it was late and I was watching another TV program at the time.

When ß is non-zero, α is essentially (or can be transformed easily to) the longditude of the ascending node, ß is the inclination, Γ is the argument of the perisapsis all amp to the reference frame. If you know Γ and you know your crafts position in R3 then you also know your position in E3 . The difference is that in E3 you map to a spherical coordinate system, and in R3 you map to a polar coordinate system as long as you know the E3 coordiantes of your perisapsis. Once you have ß fixed, then α globally references all possible ellipses with inclination ß and so set of all Γ can map there semi-major axes comprehensively. Imagine it like this ß is a rotation about an Xαe1, Yαe2. Γ0 is Xe1, Ye2 in the sense that X'r1 = 1 and Y'r2 = 0, for all coordinate in the plane defined by Z'r3 = 0. If you know where Γ0 is you can know where any point in the plane is, including the argument of periapsis (say Γ1) and the true anomoly (say Γ2) and all past periapsis in the same plane (say Γ3 - Γn) as long as you can map back to Γ0. Since these are all unit vector then radius is simply a position vector in the 2-D plane orthogonal to r3. Orbital parameters are all preserved. So why use this, the reason is accuracy, on the computer if you are constantly using the keplarian laws of motion, then every process is a function of sin, cosine followed by inverse sin and inverse cosine, this is highly inaccurate if use for each moment of acceleration. It is better to use X, Y, Z (newtonian) coordinate system to prevent drift of the orbit, i've tested this and the drift is something like 10-100 times faster. The problem is when you are working with piddly ION drives where the moment of thrust is tiny and the burns can last years the relative variance is problematic. IOW, once you are in a distinct keplarian orbit your are fine, and if you are perform short bursts of thrust you probably don't have a problem. But with ION driven spacecraft, you do not want to be altering your periapsis continuously using a keplarian equations, it will get filthy. So that once we have placed the craft in a Z = 0 plane, the math greatly simplifies and as long as we stay in that plane we can continue mapping the coordinates of our periapsis Γ relative to Xαe1, Yαe2. Quaternions are an artifact that turn out to be useful in computer graphics, the problem is the goal here is not to create computer graphics or simulate physics on the computer in 3D. The specific goal is to use a set of coordinates that are most easily tested using the La grangian (the least action required) in this case to achieve desired orbit with regard to energy spent and time. Because of this we want drift to be minimal per small unit of time. Let me give an example. A goal of burning would be to project a vector in space in a given direction. In the case we can set an exit position in E3 (say a Mars orbit intercepting Hohmann transfer ellipse that projects off of Earth's SOI (or hill sphere). There are many small referencing problems here. But the first problem is that this exit position in E3 maps to a coordinate position defined at the solar system. We can define the points along the Hohmann transfer in the heliocentric coordinate system let's call it H3. Xh1, Y'h2, Zh3 And like the motion of the Earth. We can also project onto the sphere of Earths SOI where the Hohmann ellipse will intersect that sphere. So these are relatively easy maniupulations and can be done anyway you like, but then you need to convert to Earth's coordinate system, again this can be done by any number of techniques, it's not important. The reason these operation are 'trivial' is that you only do them once, if you were to repeat them 100 times a second for 10 years, the method you choose would not be trivial. But then there is a single ellipse within that system (E3 is infinite but SOI is not) that projects out of the SOI. But the ellipse that leaves, given that impulse is not ever instantaneous in the real world, a set of evolving ellipses with time, to be accurate many different ellipses in Z'=0 all with very small intervals of time. So given any small range of acceleration (i.e. a =10-4 to 10-2) a large amount of dV. The ISP, dV and amount of time burning can change. The question is when is best to burn, along what angles, and at what ISP (N = KW * eff./exhaust velocity) to achieve the fastest exit times with the least amount of fuel burnt. So for any given point along an ellipse we need to momentarily optimize parameters to achieve the highest benefit for the lowest cost. These transformations put me where I want to be, this is what I am going to use, I just thought I would present them here. Take it or leave it.

lets not go there, matrix multiplication and minor determinants are bad enough.

First off, let me define what the need is. When you launch from a planet you are basically using a spherical cartesian system. Longitude and latitude are replaced by a non-newtonian coordinate system. Once you get into space your orbit has no fixation relative to the surface unless your are in GTO. These orbits are typically oriented to the position of the sun at the northern vernal equinox. This system can have two orientations, the one orientation is relative to earths poles, the second can be relative to the normal of a surface pointing in the direction of the sun (essentially consider a sphere of radius r, its the path along the surface that intercepts a line between the Earths' center and the sun). Within the EM system there can also be other coordinate systems. There could be a lunar-centric/earth-referenced system, lunar-centric/solar-referenced, L1centric, L2-centric. Then there are the myriad of planetary and satellite coordinate systems. Going between points in the solar system can be cumbersome if you use earth rotational coordinate system marked on the vernal equinox because everything (including Earth's orbital inclination, by ~ 23'). There is a preferential plane of motion (marked largely by the motion of Jupiter) that is used to define the plane of the system). And some day a spacecraft might voyager might use a galactic coordinate system based on an orientation to some far off galactic center close to the plane of the galaxy. Each coordinate system has a basis. The earth's X,Y, Z has a line that goes through Greenwhich and some point in the Atlantic near West Africa marks the X=r point Z = r close to the geographic north pole. If you read the wiki page on Change of Basis you are given the following: c = cosine, sin = sine. α , ß, gamma are the three angles by which any vector can be defined by a change of x -> x' (rotation around z), z->z' (rotation around x'), and x'->x'' (rotation around z') all counterclockwise. Source wikipedia - change of basis. While this is correct, this basis, quoting wikipedia. "R be a new basis given by its Euler angles. The matrix of the basis will have as columns the components of each vector. Therefore, this matrix will be [above matrix]". But in fact this is not very useful, because by the fact that you have the angles that define the rotations, you pretty much already know where R is with respect E3. This matrix tells you what you already know. By saying its not very useful, its useful on the 'return trip back to the previous (home) coordinate system, but not so useful on the leg of a journey that creates the need for the return home conversion. But what you really, really badly want to know is how to map a vector already mapped in E3 in your new basis R and here is what the wiki article says. " Again, any vector of the space can be changed to this new basis by left-multiplying its components by the inverse of this matrix." This single quotations underlies a fairly complex problem that can be simplified with certain information that wikipedia article does not readily provide. The first bit of information one needs to know the determinant of the Matrix above. To know that you have to how that matrix was derived. The matrix was derived from three matrices which are principally defined here;https://en.wikipedia.org/wiki/Rotation_matrix#Basic_rotations and converted into symbolism, defined above, below. https://i.imgur.com/bwHo4O9.png One law of determinants is that if all the determinants are 1 then the multiplication of the Matrices is also 1. https://en.wikipedia.org/wiki/Determinant With this rather innocuous fact one can derived the inverse matrix. https://i.imgur.com/8ftGcsI.png This is of course not what the math actually looks like. The only way to know for certain if the proposed R-1 is to multiply R-1R = I (the identity matrix, I3 below) This is the first stage after multiplying everything out and getting rid of extra parenthesis. Irow , column or Ii,j i=row# and j= column# I1,1 = c21c23 + s21c22s23 - 2cs1c2cs3 + c21s23+s'1c22c23 + 2cs1c2cs3 + s21s22 I1,2 = c21c2cs3-cs1c22s23 - s21c2cs3+cs1c23 + cs1s23 + s21c2cs3 - c21c2cs3 - cs1c2'2c23 - cs1s22 I1,3 = c1s2cs3 - s1cs2s23 - c1s2cs3 - s1cs2c23 + s1cs2 I2,1 = cs1c23 + c21c2cs3 - cs1c22s23 - s21c2cs3 + s21c2cs3+cs1s23 - c21c2cs3-cs1c22c23 - s22cs1 I2,2 = s21c23 + c21c22s23 + 2cs1c2cs3 + c21c22c23 + s21s23 - 2cs1c2cs3 + c21s22 I2,3 = c1cs2s23+s1s2cs3-s1s2cs3+c1cs2c23-c1cs2 I3,1 = c1s2cs3 - s1cs2s23 - c1s2cs3 - s1cs2c23 + s1cs2 I3,2 = c1cs2s23 + s1s2cs3 - s1s2cs3 + c1cs2c23 - c1cs2 I3,3 = s22s23 + s22c'3 + c22 https://en.wikipedia.org/wiki/File:Matrix.svg Source: Wikipedia - Identity Matrix An Example (the easiest). I3,3 = s22s23 + s22c23 + c22 (converting the identity subscripts to angles, since excel does not handle text markup well) I3,3 = s2ßs2Γ + s2ßc2Γ + c2ß (rearranging) I3,3 = s2ßc2Γ + s2ßs2Γ + c2ß (factoring) I3,3 = s2ß(c2Γ + s2Γ)+ c2ß (Replacing, c2Θ+s2Θ = 1) I3,3 = s2ß(1)+ c2ß (Reducing) I3,3 = s2ß+ c2ß (Replacing, c2Θ+s2Θ = 1) I3,3 = 1 Pretty much terms disappear in the diagonal finally as replacing a final c2Θ+s2Θ as 1, and off diagonal by a final subtraction of x + -x = 0 resulting in the identity matrix. An Example (the tie for hardest, witchcraft no less) I1,2 = c21c2cs3-cs1c22s23 - s21c2cs3+cs1c23 + cs1s23 + s21c2cs3 - c21c2cs3 - cs1c2'2c23 - cs1s22 (getting rid of ambiguous symbolism and markups, c = cos, s = sin) I1,2 = c2αcßcΓsΓ - cαsαc2ßs2Γ - s2αcßcΓsΓ + cαsαc2Γ + cαsαs2Γ + s2αcßcΓsΓ - c2αcßcΓsΓ - cαsαc2ßc2Γ - cαsαs2ß (rearranging) I1,2 = c2αcßcΓsΓ + s2αcßcΓsΓ + cαsαc2Γ + cαsαs2Γ - c2αcßcΓsΓ - s2αcßcΓsΓ - cαsαc2ßc2Γ- cαsαc2ßs2Γ - cαsαs2ß (factoring) I1,2 = (c2α + s2α)cßcΓsΓ + cαsα(c2Γ + s2Γ) - (c2α+ s2α)cßcΓsΓ - cαsαc2ß(c2Γ + s2Γ) - cαsαs2ß (Replacing, c2Θ+s2Θ = 1) I1,2 = (1)cßcΓsΓ + cαsα(1) - (1)cßcΓsΓ - cαsαc2ß(1) - cαsαs2ß (Reducing) I1,2 = cßcΓsΓ + cαsα - cßcΓsΓ - cαsαc2ß - cαsαs2ß (rearranging) I1,2 = (cßcΓsΓ - cßcΓsΓ)+ cαsα - cαsα(c2ß + s2ß ) (rearranging & factoring) I1,2 = (cßcΓsΓ - cßcΓsΓ)+ cαsα - cαsα(c2ß + s2ß ) (replacing, c2Θ+s2Θ = 1; x + -x = 0 ) I1,2 = (0)+ cαsα - cαsα(1) (reducing) I1,2 = cαsα - cαsα (replacing; x + -x = 0 ) I1,2 = 0 [If you were having trouble sleeping the above 12 lines of linear algebra is much more effective than any sleep medication, I discovered this while watching lectures on quantum entanglement; if that still doesn't work try solving the unsolved elements for 1 or 0] So how does one use the inverted Basis Matrix (R-1), wait, why would someone, anyone want to use this. Chances are you already are. When you reach the edge of an SOI in KSP, you are going to move to a Kerbols SOI. Lets give an example, you are heading radially outward along a prograde AoA from Kerbin, north is above your head, and Kerbol is to your right. You reach the SOI, now instead of heading radially at Vkerbin you are now heading you are traveling a different velocity, circum Kerbol at a Velocity in the opposite direction (Kerbins orbital velocity - the last radial velocity inside kerbin upon exit). You also might notice you are accelerating toward kerbol, this is because your new velocity creates a centripedal acceleration that is below the current force of gravity from Kerbol. (this would be a useful problem except KSP likes to map the Kerbols coordinate system versus the Kerbins, so . . .we need a real system) There's two sets of problems here though, acceleration is a moment problem that we do not have to worry about at the change of basis, the velocity 'blip' can be converted using the inverse Matrix and the position vector can be determined by determining the current position of planet from its star, the distance of the ship from the planet, the path the planet travels. and the current angle of the vessel from the path at the paths exit. Basic trigonometry. Since velocity in a 3-D coordinate system can be a vector independent of the coordinate system, we can superimpose the stars coordinate system on planets center, ascertain the change of basis angles, the apply the inverse matrix to the velocity vector to get its new velocity vector in the stars coordinate system (the magnitude of the velocity is still referenced to the planet) so the planets velocity vector relative to the star needs to be added to the new velocity vector to deduce the ships velocity relative to the star while still inside the planets SOI. Since it is easier on an escape vector to calculate both the planets position (and therefore its gravity vectors) the stars position (and therefore its gravity vectors) one can simply change coordinates once an escape trajectory is realized. One way to do that is to change basis as the ship crosses the line from the planet to the star of from the star through the planet and out the other side (outer system transfers). Anyway ev = Xe1, Ye2, Ze3 and rv = X'r1, Y'r2, Z'r3 You start with your planets X, Y , Z coordinates in the local system they would looke like this https://www.khanacademy.org/math/precalculus/precalc-matrices/matrices-as-transformations/e/multiplying_a_matrix_by_a_vectorThis is a process otherwise known as a transformation, but in the new coordinate system we have a new set of axis (so in the video you would draw another set of axis and plot your transform in that new set of axis) X' = XcαcΓ - XsαcßsΓ - YsαcΓ + YcαcßsΓ + ZsßsΓ Y' = -XcαsΓ - XsαcßcΓ - YsαsΓ + YcαcßcΓ + ZsßcΓ Z' = Xsαsß - Ycαsß + Zcß And written finally out in a fashion that even a college graduate might understand. Everyone wants a summary. X' = X * cos α *cos Γ - X * sin α * cos ß * sin Γ - Y * sinα * cosΓ + Y * cos α * cos ß * sinΓ + Z * sinß * sinΓ Y' = -X * cos α * sinΓ - X * sinα * cos ß * cosΓ - Y * sinα * sinΓ + Y * cos α * cos ß * cosΓ + Z * sinß * cosΓ Z' = X * sin α * sin ß - Y * cos α * sin ß + Z * cos ß Suppose that you only rotated around one axis, say the Z axis. The above matrix is still good: not that element R-133 = cos ß if no rotation along X' axis. If then ß = 0 and the cosine of zero is 1 while the sin 0 is zero. So whatever Z value you have, Z' = X * 0 + Y * 0 + Z * 1, in actuality, if you never make a rotation ß around the X' then basically you apply the A = matrix above in which the α = (your α + Γ). Does this happen? if your thrust changes the position of your periapsis but not the inclination relative to your external reference frame, then before the thrust your elliptical reference frame is at angle α relative to the basis (E3) and the movement of the periapsis move Γ relative to the old elliptical, to place the crafts position and velocity vectors in the new elliptical reference frame you could use the equation above applying α and Γ. Summary, this is as hairy as the math gets, but using it you can follow position and velocity in two coordinate systems (or even three as the above shows) at one time, moving flawlessly between different systems of plotting. I think its pretty obvious how to got to a R coordinate system back to the E basis system (the equations are very close).

Since both fuels are cryogenic the moment of acceleration is simply to move the fuels to the transfer outlet. The pressure of the cryogenics will move the fuel, but also push the craft away from each other. Its not a robust fuel transfer system. To achieve an efficient transfer there needs to a condesation system in the recipient tank, otherwise a vapor lock will occur and you will just that, push the craft away from each other. The recipient tanks need to be precooled and toward the end of the process transferred under pressure (as fuel flows from the first adiabatic cooling occurs and the second tank will warm, meaning fuel wants to flow backwards). A better way, to connect the tanks in such a way that there is centrifugal force operating. The force is pushing liquids down the tube in which the rate of flow is determined by the spin, the heat can be radiated from the recipient tank or the degassing liquid can be transferred back to the top of the source tank.