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Lunar Siyuz flight
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One possibility to be on the safe side is to simply assume that three BlockDMs are required. This would increase the hardwareweight by 4600. The amount of propellant these three Block DMs could contain then would be 46,500 kg. If still 15,500 kg would be sufficient for the translunar injection then there would be no further question because I suppose that the same amount of propellant is sufficient to do the orbital insertion burn around Earth. At the Moon there is no orbital insertion planned for the lunar Soyuzflight  so I don't consider it regarding the reusable replacements. The third Block DM I am assuming at the moment to be required for nonorbital insertions at the Moon. I am assuming it at the moment  this doesn't mean that I keep the assumption. Since 15,500 kg of propellant will be than required the remainder, the surplus perhaps will fit into other requirements I suppose to be faced to possibly.
The question comes up to me if the CXV might be equipped so that it can do those burns itself  not requiring the third Block DM. Before I go on another remark is required. By Block DM not only its replacement(s) are meant here  it might be a replacement that is one unit as large as three replacements/Block DMs. The complete weight of the CXVlike vehicle together with the increased booster would be 3,600 + 6,900 kg = 10,500 kg without the propellant. This is still less than the sum of the weight of Apollo CM and Apollo SM  by 19,832 kg. Making use of this number and applying the former method the new resulting required amount of propellant would be (10,500 kg/30,332 kg) * 25,300 kg = 8,758.08 kg. This is far less than the 46,500 kg of propellant three Block DMs would contain all together. But there is the weight of the propellant required to reenter the orbit after returning from the Moon and to be carried to the Moon around it and back to Earth. This weight also will be 8758.08 kg. Consequently the weight to be launched out of the orbit for the Moon will be 19,258.08. (19,258.08 kg/30,332 kg) * 25,500 kg = 16,190.20 kg of propellant is required if no propellant is needed to get araound the Moon without entering a lunar orbit  still 30,301.80 less than the proepllant contained by three Block DMs if the capacity of their tanks is completely filled. To think over the propellant required at the Moon now it seems to be clear that it will be less than 8758.08 kg  for the following reasons: 1. The lunar escape velocity is far less than the earthian one  2.5 km/s if I remember correct. Ass far as I remember the earthian escape velocity is 11 km/s. This means that the lunar escape velocity is less by the factor of 4.4. 2. A vehicle orbiting the Moon is moving by a share of the lunar escape velocity already. 3. The vehicle considered here approaches the Moon at more than the lunar escape velocity if I am right (I will check it perhaps) and wouldn't be decelerated to enter a lunar orbit. To try to have a safety margin I add an amount of propellant of 4,400 kg here. This means that 23,658.08 kg would have to launched out of the earthian orbit and the required amount of propellant I get is (23,658.08 kg / 30,332 kg) * 25,500 kg = 19,889.26 kg would be required here. This weight is got by taking into account also that the CXV is significantly lighter than the Soyuzcapsule and that so the CXVlike vehicle considered here can be assumed to be lighter also. But since I was looking at Apollo in calculating the recent numbers I still have to look for the consequences of the different propellenats Apollo and the Block DM use. According to Wikipedia Apollo uses 50% hydrazine, 50% ADMH and dinitrogentetroxide. To adjust the number calculated to cerosene/LOX I simply take half of the Isp got by hydrazine  2845/2 = 1422.5  plus half the Isp got by ADMH  2804/2 = 1402  resulting in 2824.5 and divide it by the Isp got by the Isp of cerosene. 2824.5/2700 = 1,05  this seems to be the factor by which I have to multiply the number(s) calculated to get the required amount of cerosene 20,883.72 (all numbers got from www.berndleitenberger.de) Based on all this I now calculate the costs of the reusabilityversion of the former modifications of the concepts for lunar Soyuzflights. What I said here seems to deserve at least one check but I at the moment I want to proceed to according cost numbers. The replacement of the Block DM is heavier now and since more than 4600 kg  2 Block DMs is more than the weight of one CXV I don't want to consider a launch of that booster via VLA/QuickReach no more  a Falcon is required. Obviously it will have to be a Falcon 9 and so the launch will cost $ 27 mio. But this is an investment to be depreciated over the number of flights since the booster will stay in space for ever. The new amount of propellant will have to be launched by a Falcon 9 S9 at transportation costs of $ 78 mio. If the capacity of the Falcon 9 S9 is completely used than the share the propellant will have of the transportation costs is 20,883.72/24,750 = 0.84. 0.84 * $ 78 mio = $ 65,520,000. Taking rounded numbers the flight seems to cost $ 66 mio in its reusabilityversion. Then the lostinspacecosts will be $ 66 mio also. The price per customer will be $ 70 mio then  $ 4 mio to get into orbit together with the second customer and three orbital customers plus $ 66 mio flight costs and coverage of lostinspacecosts. Obviously now again two flights will be required to cover all the lostinspacecosts if only $ 55 mio will be taken to cover them. This would be valid if the price would be $ 4 mio + $ 33 mio + $ 27.5 mio = $ 64.5 mio. When all lostinspace costs are covered the price could be dropped by $ 27.5 mio down to $ 37 mio. This would be $ 16.5 mio to $ 12 mio more than in the original version. I will think about this and perhaps put it into a better system of thinking etc. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Up to now I didn't find any problems with what I did in the previous post  which doesn't mean yet that there might not be found problems. So I will continue to look at it all.
In between I also thought about the costs of transportation via the VLA and QuickReach. The transportation costs for propellant the would be 20,883.72/3600 = 5,8 flights. The results of the CXVthread allow for the possibility that one launch costs a bit more than $ 10 mio. Then the transporation costs viy VLA/QuickReach would be $ 58 mio if 20% of the capacity of the last launch would be proepallnt for anothher flight than the lunar one or for the next flight or any other cargo. This would mean that another $ 8 mio are saved. Both the flight costs and the lostinspace costs would be dropped to $ 58 mio  the initial price could be $ 4 mio orbital + $ 29 mio lunar propellant + $ 27.5 mio lostinspace costs = $ 60.5 mio. The second flight would have to cover a remainder of $ 3 mio of lostinspacecosts then resulting in a price of $ 36 mio or the initial price could be $ 62 mio. The final price could be $ 33 mio then. It seems that the price could be closer to the price I got in the case that the CXVlike vehicle is its last flight. But the above is valid only if really the costs of a launch via VLA/QuickReach eren't higher than $ 11,379,310.34. On the other hand the CEVconcept of t/Space included tankers to deliver sufficient propellant to the CEV when it is at the Moon. May be that this propellant was meant for the landing vehicles but the point is that such delivery would not cost too much. Interesting: For a special reason at this point a safety margin could be included. Prof. Collins (refer to the Collinsthread please) in his answer to me supposed that a lunar flight would have a price of $ 200,000 in 2030  24 years from now. He estimates an orbital flight $ 20,000 will have to be paid for that time. Obviously he estimates the price of a lunar flight to be ten times the price of an orbital flight then. Applying this factor in this thread the possible price of $ 4 mio per person for an orbital flight via the CXV would result in a price of $ 40 mio for an orbital flight via the vehicle which is financially CXVlike but flies arond the Moon. This safety margin would add $ 3 mio to the price got in the previous post or $ 7 mio to the price got in this post. Another interesting point: The Prarom the Russians are thinking about for the orbital flights of Clipper reminds me to the reusable lunar booster replacing the Block DM in this thread. I don't know if the Prarom is fueled or simply will be used to steer the Clipper to the ISS or anwhere else which I consider to be more likely. But in principle the Prarom is similar to the reusable booster to be parked in orbit: 1. The Prarom will be parked in orbit 2. The Prarom is reusable 3. It might be possible to develop a Prarom that can be fueled 4. The Prarom can do manoevers. 5. The Clipper explicitly is intended to be used for tourism flights. 6. The Clipper will dock to the Prarom after arriving in orbit. I will go on in the next post but I am still no way content reagrding checking the previous post for problems, mistakes, errors etc. It seems to be best to try if I get similar results if I use informations about another vehicle than Apollo  I simply didn't find one I would consider to be proper. So I go on to look for such a vehicle. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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I am not sure if I remeber right that one of the two articles reported that after the visit to the ISS two cosmonauts will reenter the Soyuz and that including these two four persons will be in the Soyuz going travelling to the Moon altogether. The Soyuz can carry four persons but it will be very narrow for them then. So I am doubting if I remember correct.
But the CXV really can carry six persons according to t/Space and so I consider this capacity now. Up to now I considered three persons travelling around the Moon by a financially CXVlike vehicle â€“ one pilot and two tourists. This number is equal to the number of astronauts having gone to the Moon by Apollo. Since one person weights around 80 kg three persons weigh 240 kg and six persons 480 kg â€“ so this weight seems will not have that much impact on my results. Thre persons are considered already  meaning that the weight will be increased by 240 kg only. Food, potable water per day and clothes are light also and at least partially included also. So it seems that the only major equipment left is the amount of oxygen for breathing and its tank. Apollo had an oxygentank of 144 liters volume. According to www.berndleitenberger.de oxygen has a density of 1.27 kg per liter. So 144 liters of oxygen weigh 182.88 kg. So three persons more will mean additional 182.88 kg of oxygen â€“ thatâ€™s not too much and Apollo needed oxygen for more than six days begiining with Apollo 11. If I remebr right the minimum time was 8 days and so an amount of 75% of 2* 182.88 kg seems to be sufficient â€“ this would be 274.32 kg. So only a difference of 91.44 kg of oxygen need to be added to the amount of oxygen for breathing Apollo carried. Now some of this was required for breathing in the earthian orbit â€“ this requirement holds for the CXV also and it will be covered already. Result: The additioanl amount of oxygen required for six persons doing atrip around the Moon can be neglected regaridng costs. The tank for this oxygen will not contribute much weight as far as I know â€“ there necessaryly will be a tank included in the CXV already and so it is included into the financially CXVlike vehicle already. This means that it is already part of the weight or mass to be launched out of orbit already. The only question left seems to be if sufficient space is left for six persons. 144 liters are much less than one cubemeter â€“ 0.144 m^3 only. The additional amount of oxygen of 91.44 kg will require a volume of 48.38 liters only â€“ 0.004838 m^3. So the total volume required is 0.148838 m^3. So I finally suppose that it is no problem to apply the costs already calculated to six persons travelling around the Moon. Five of the six persons only will be tourists. So the flight costs of $ 58 mio as well as the lostinapce costs could be covered by five tourists instead of only two. This means $ 11.6 mio flight costs per person plus initially additional $ 11.6 mio lostinspace costs per person which fall apart later. The initial price per person could be $ 4 mio into the orbit + $ 11.6 mio around the Moon + $ 11.6 lostinspace = $ 27.2 mio initially and $ 15.6 mio when all lostinspace costs are covered. This valid only if five persons plus a pilot are carried around the Moon. The lostinspace costs would be covered by one flight only in this fivepersonscase. The price is only four times the orbital price now â€“ and less than the orbital price via Soyuz! And there is a particular aspect to be considered in the next post. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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At the end of this post there is an enhanced table of the steps calculated previously.
Up to now I allways left away the depreciations of the booster waiting in the orbit, the depreciations of the financially CXVlike vehicle also waiting in the orbit or returning to the surface. And I left away the total price of the propellant itself. I did so because I assume that the numbers for those depreciations and for the propellant are small in comparison to the flight costs of the CXV according to t/Space. Like I already calculated the amount of propellant of the Block DM cost only $ 9.000 to $ 10,000 if it were cerosene completely  but it is LOX partially which is much cheaper than cerosene. $ 10,000 are really small compared to $ 20,000,000  a twothousands only. But what about the depreciations? In general there will be a difference between the depreciations for an investment to be used between surface and orbit and another investment of identical monetary size and scale to be used in space only. The reason is that the investment to be used in space only normaly doesn't experience renetry stresses and so doesn't require the refurbishments, repairs, shields etc. needed by the CXV for example. A large number of such investments is orbiting Earth since long  telecommunications satellites and the like. And as is known theses satellites have lifetimes of several years during which they orbit Earth around 5,840 times per year. Their lifetimes in principle end at destruction by any debris. Of course running short of fuel sets an end also but this isn't valid here because the calculations include refuelling before each lunar flight  or even after each lunar flight. Perhaps more differences between the satellites and the investments considered here can be found ... So it can be concluded that the lifetime of the CXVlike vehicle and its booster considered here are significantly longer in doubt than the lifetime of the CXV  or more precisely if their lifetimes are equal the lifetime of the CXVlike vehicle and its booster both waiting in orbit is cheaper. In the Accumulationsthread I one time used a lifetime of 800 flights. Since the investment into a CXV will be $ 400 mio the depreciation per flight is $ 125,000. This means that the depreciation of the financially CXVlike vehicle to fly around the Moon would cost depreciations per flight of $ 125,000 if it requires an investment of $ 400 mio. This would be $ 62,500 per passenger in the twopersonscase or $ 25,000 in the fivepersonscase. What's left is the booster. Let's take the investment into the Block DM and apply it three times  I remember having got a bit less than $ 10 mio for the Block DM, round that up to $ 10 mio and so get $ 30 mio for the booster. At 800 flights this would mean $ 37,500 per flight. Another $ 40 mio are to be included as investment costs since both the vehciel and the booster are of use only in orbit in this thread. Three VLA/QuickReachflights are required to install a booster based on three segments and another VLA/QuickReachflight I include to install the vehicle in orbit. The depreciation per flight at 800 flights then is $ 50,000. Then all depreciations plus propellant per flight in total would be $ 222,500 using the numbers listed in this post up to now  since the amount of propellant used in the previous post is around 25% more I added another $ 2,500 and get $ 225,000 per flight. This is between $ 112,500 and $ 45,000 per passenger to be added to the millions of transportation costs for delivering propellant into the orbit. So at 800 flights the depreciations for hardware and the propellant costs are that small that the prices and costs remain close to the numbers already calculated if they are included. If it is assumed that the passengers "feel" these additional cost if they would be at $ 500,000 then the number of flights could be reduced down to between 180 and 80  which would mean between 900 and 160 passengers and so less than the number of potential customers Space Adventures have identified. There is a safety margin included in these numbers of flights. The question at this point is how much will have to be invetsed into the financially CXVlike vehicle  $ 400 mio? Less than $ 400 mio? or more than that? It would reuqire the watercooling for reentry for example ... The financially CXVlike vehicle might be not a complete vehicle but a vehicle a CXV could be incorporated into temporarily. The CXV could dock to it providing access to a larger volume to be inhabited during flight and so on. Having said this I can go on later. Dipl.Volkswirt (bdvb) Augustin (Political Economist) Table about the steps resulting in a largely reduced price  variable costs only rightmost column "price  lostinspace" corrected. That column lists the price after all lostinspacecosts are covered. Explanation of the last row: The price includes the orbital price per person via the CXV and the numbers listed at the end of the previous post. Code: step revenue price costs lostin cov. flts price  space lstin cov lost costs spcco inspace orig. 200 mio 100 mio 145 mio 145 mio 55 mio 3 72.5 mio concept CXV in 155 mio 77.5 mio 100 mio 100 mio 55 mio 2 50 mio stead Soyuz reusbl. 135 mio 67.5 mio 80 mio 80 mio 55 mio 2 40 mio booster 129 mio 64.5 mio 74 mio 74 mio 37 mio kept in orbit less 117 mio 58.5 mio 62 mio 62 mio 55 mio 2 31 mio prop.by 108 mio 54 mio 53 mio 53 mio  26.5 mio less weight 3 pass. 105 mio 52.5 mio 50 mio 50 mio 55 mio 1 25 mio ISS etc  96 mio 48 mio 41 mio 41 mio 20.5 mio + pass lunar more 129 mio 64.5 mio 66 mio 66 mio 55 mio 2 37 mio prop.for Moon + reenter erthian orbit deliv. 121 mio 60.5 mio 58 mio 58 mio 55 mio 2 33 mio by VLA+ Quick Reach 5 136 mio 27.2 mio 58 mio 58 mio 55 mio 2 15.6 mio toorists 
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Since I am using a financially CXVlike vehicle here I now refer to the post(s) about the CXV in the Accumulationthread and in the Collinsthread. Both in these threads I described which way the price into orbit could be dropped down to $ 25,000.
Before I go on here I have to do a correction of what I said about it in the Accumulationthread. There I said that the price could be dropped down to $ 25,000 if 5 passengers are flown and the lifetime of the CXV is 41,000 flights. The number of 41,000 flights is incorrect  41,000 passengers is correct. This number is got by dividing the sum of the $ 400 mio for the CXV and the $ 10 mio for a reusable booster is divided by the difference between the price of $ 25,000 and the propellantprice per passenger of $ 15,000 which is $ 10,000  $ 410 mio/$ 10,000 depreciation per passenger = 41,000 passengers. The number of $ 10,000 depreciations per passenger is based on 5 passengers per flight and so the required lifetime of the CXV would be 41,000 passengers/5 = 8,200 flights. At that lifetime the ticketprice including propellant could be $ 25,000. This would mean $ 125,000 to be paid for one flight of five passengers. Then it is possible to apply this price to the lunar flight also  it would be the portion of the total lunar price that is to be paid from surface into orbit. But the real interesting aspect is that this price for the complete orbital flight is to be applied for the delivery of propellant into orbit also  the propellant to be used for the lunar flight. As I calculated earlier in this thread 5.8 flights are required to deliver the complete amount of propellant  consequently the transportation costs of the propellant are 5.8 * $ 125,000 = $ 725,000. A lunar flight would cost $ 725,000 then. This would be the lostinspacecosts also  and one flight would be sufficient to cover them...  and this wouldn't reuqire $ 55 mio but less than a million! Two passengers initially would have to pay $ 25,000 into orbit + $ 362,500 lunar + lostinspacecosts = $ 387,500 + lostinspacecosts. If  different to the previous posts  the lostinsapcecosts are assumed to be $ 725,000 then the two passengers would have to pay $ 750,000 initially  $ 387,500 after the first flight. At five passengers per flight  which seems to be realistic according to some results in the previous posts  the initial price would have to be $ 25,000 orbital + $ 145,000 lunar + $ 145,000 lostinspacecosts = $ 315,000. After the first flight the price could drop down to $ 170,000...! Before I go on let's have a look at the 8,200 flights. One orbital flight per week assumed theses would be 157.69 years of operation. If two flights per week are possible this time drops to 78.85 years. This appears to be unrealistically long. But it might be possible that the number of passengers per flight is increased. Better yet is a way to do price differentiation. As said in the Accumulationthread there are several ways to apply price differentiation. To keep the price calculated here this differentiation would have to increase the orbital price for particular orbital flights. These particular flights could be flights urgently required by space agencies, companies and the like. I avoid here the possibility to use the lunar flight itself for price differentiation since this would wipe out the segmentation of the market by the kind of flight  the difference between orbital flight and lunar flight would disappear then. The point of view would be changed which requires a new thread. To go on  please compare the price of $ 170,000 per passenger in the fivepassengercase to the depreciations of between $ 112,500 to $ 45,000 per passenger  the depreciations are no longer small compared to the other costs and so have to be added. The price has to be betwwen $ 452,500 and $ 385,000 initially and between $ 282,500 and $ 215,000 after the first flight. Only but at least the lower price is close to $ 200,000. The higher price is $ 35,000 above ten times the applied orbital price of $ 25,000. But this requires 800 lunar flights. Please note once again here  all the prices I calculated here don't include no profits. Also keep in mind that these numbers are based on the financial properties of the CXV one of which is that $ 400 mio are estimated to be reuired by t/Space. SpaceDev estimates $ 100 mio for an orbital version of their suborbital vehicle  a fourth or 25% of the CXVinvestment. Consequently SpaceDev's orbital vehicle could drop the depreciations down to 25%  between $ 28,125 and $ 11,250. Then both the prices after the first flight would be less than $ 200,000 then. The orbital price would be reduced down to $ 17,500 at five persons and 8,200 flights or price differentiation  this would drop the price per passenger once more. I didn't calculate the result(s) but the price would be significantly below $ 170,000 then  including the depreciations. And the solution for the unrealistice large number of flights per lifetime of the CXV as well as the 800 lunar flights could be vehicles of larger passenger capacity. Since one passenger weighs 80 kg in average only ten passengers will not be that large a problem  in particular for the lunar flight. two times the number of passengers applied here doesn't mean two times the propellant and two engines to be required. The reason is that the amount of material required to build the vehicle is less than doubled in that case  far less. ... The lunar flights considered up to now in this thread are all going once around the far side of the Moon  they didn't enter a lunar orbit and they didn't include a landing. In the next post I will extend the flights. Dipl.Volkswirt (bdvb) Augustin (Political Economist) Table about the steps resulting in a largely reduced price  variable costs only Code: step revenue price costs lostin cov. flts price  space lstin cov lost costs spcco inspace orig. 200 mio 100 mio 145 mio 145 mio 55 mio 3 72.5 mio concept CXV in 155 mio 77.5 mio 100 mio 100 mio 55 mio 2 50 mio stead Soyuz reusbl. 135 mio 67.5 mio 80 mio 80 mio 55 mio 2 40 mio booster 129 mio 64.5 mio 74 mio 74 mio 37 mio kept in orbit less 117 mio 58.5 mio 62 mio 62 mio 55 mio 2 31 mio prop.by 108 mio 54 mio 53 mio 53 mio  26.5 mio less weight 3 pass. 105 mio 52.5 mio 50 mio 50 mio 55 mio 1 25 mio ISS etc  96 mio 48 mio 41 mio 41 mio 20.5 mio + pass lunar more 129 mio 64.5 mio 66 mio 66 mio 55 mio 2 37 mio prop.for Moon + reenter erthian orbit deliv. 121 mio 60.5 mio 58 mio 58 mio 55 mio 2 33 mio by VLA+ Quick Reach 5 136 mio 27.2 mio 58 mio 58 mio 55 mio 2 15.6 mio tourists 2 1.5 mio 750000 725000 725000 725000 1 387500 tourists $ 25,000 orbital 5 1.575 mio 315000 725000 725000 725000 1 170000 tourists $ 25000 orbital For the last two rows the depreciations have to be included yet which has been done in the text only. 
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The concept of thinking about the landing is that the amount of propellant to launch from the lunar surface is much less than that required to launch from the earthian surface. This amount I use as a standard also consumed to land on the Moon. Since the lunar escape velocity is much less than the velocity the vehicle is approaching the Moon by from Earth and also less than the  in principle equal  velocity it will return by to Earth The booster for leaving the earthian orbit and to return to Earth will be left in orbit. So also propellant to enter a lunar orbit is to be calculated here.
Next the calculation of the costs of landing on and launching from the lunar surface can't be based on the data about the CXV since the lack of a lunar atmosphere doesn't allow for airplanebased launches. It would be optimal to apply data about Armadillo Aerospace's vehicle(s) but no applicable data are available  no price, no flight costs etc. I was thinking about Starchaser's vehicle since there is a price  but that price is suborbital and far from being proper for the purpose here. The only vehicle/rocket left is SpaceX's Falcon  Falcon V, Falcon IX, Falcon IX S5 or/and Falcon IX S9. Only the stages of these will be considered here  not the launch equipment etc. The rocket(s) will be used only to carry the weight of one CXV down onto the lunar surface and to launch it later again  but they will never land on or launch from Earth except for one launch to install them in orbit where they will or can be docked or connected or incorpoarted into the reusable booster already considered or to the financially CXVlike vehicle waiting there. For this reason I continue to consider the variable costs only and include the investment costs of the rocket(s) later like done up to now. Also I will start where the flight around the far side of the Moon without entering any orbit cost $ 15.6 mio. Let's have a limited look to the data about the Falcon V now (a detailed look like that on CXV and Soyuz for the comparison deserves and requires another thread): Length: 47 m Width: 3.7 m Mass: 154,500 kg according to SpaceX. So the cylindrical volume of the Falcon V is 173.9 m^3. This is the maximum volume of LOX/cerosene it could contain. The real amount will be less but I will apply this amount here. It weighs between 129.903300 tons and 146.076000 tons if I apply the density according to wikipedia. Like I already mentioned in another post of this thread the lunar escape velocity is less than the earthian one by a faytor of 4.4. Applying the principle to linearily calculate lower amounts etc. from higher ones the amount of cerosene for a lunar Falcon V would be less than between 29.52 tons and 33.20 tons. One remark seems to be required here: I am calculating as if only cerosene would be required but no oxygen  I am doing so because cerosene costs so much more than oxygen. The weight got is by between 9.52 and 13.20 tons higher than the weight required to leave the erathian orbit, fly around the far side of the Moon and reenter the earthian orbit after return including a safety amount. It might be that the real Falcon V requires the amount calculated only if the payload capacity is completely used which isn't the case here. So the numbers might be less in reality if all would be done like apllied for these calculations. The thoughts will go on in the next post. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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PLEASY NOTE URGENTLY: FALCON AND FALCONBASED NUMBER CORRECTED ==> PLEASE READ THE COMPLETE POST FROM THE BEGINNING
CORRECETD NUMBERS PAST MY SIGNATURE The essential result of the last post was that a Falcon V used to land on and launch from the lunar surface would reuqire an amount of cerosene of between 29.52 and 33.20 tons. Let's round these numbers to 30 tons and 34 tons. The numbers had been compared to the numbers got for the flight around the Moon without orbiting or landing. Using the rounded numbers the differemce is 10 tons to 14 tons now. In between I had a look to the Lunar Module of the ApolloMissions. The informations at Wikipedia say that the Ascent Stage of the LM contained 2.353 tons of propellant while the Descent stage contained 8.165 tons of propellant  10.518 tons in total. Obviously there is a difference to the numbers got from the Falcon V of more than 19 or 23 tons. Looking into the reasons of this difference there are the following differences to the Falcon V: 1. The Ascent Stage of the LM weighed 4.670 tons including fuel, the Descent Stage 10.334 tons which are summing up to 15.004 tons while the difference between the weight of the real Falcon V of 154.500 tons and the weight of the propellant claculated in the previous post is between 24.500 tons (rounded) and 8 tons (rounded). Since I simply used the Falcon V's cylindrical volume it can be suspected that the real and actual amount of propellant required is less which menas that the Falcon is to suspected to be heavier. Also I am not sure if the weight listed by SpaceX is including or excluding propellant. 2. Only the light Ascent Stage has been launched back into lunar orbit to doc again with the CSM  not only the weight of the Ascent Stage INCLUDING propellant being 4.670 tons is lighter than the earthian Falcon V is  more importantly the Ascent Stage WITHOUT propellant weighed 2.317 tons only while the earthian real and actual Falcon V is going to weigh more than thrice that weight. 3. The combined weight of the Ascent Stage and the Descent Stage of the LM was 2.317 tons + 2.169 tons = 4.486 tons which also is significantly less than the real and actual Falcon V as it is designed. 4. The lunar vehicle used here will return to the orbit unchanged comapred to the landing while the LM was changed at return to orbit since the Descent Stage has been left on the lunar surface. I referred to the earthian real and actual Falcon V all the time here because I don't apply any numbers about the a lunar Falcon Vlike vehicle up to now. I don't have an idea about how to approximate it best up to now. It certainly will be lighter  but not by linear large factors. I think so because the reduction of the amounts of propellant required result in the reduction of the VOLUME of the tanks mainly while the reduction of the surface of the tanks around the empty space within them necessaryly is reduced subproportionally to the volume. Next tanks will be light compared to engines and the weight of the engines will not be changed only because of reduced required propellant amounts. Obviuosly the complete weight of the Ascent Stage, the Descent Stage and the propellant of the Ascent Stage  6.839 tons  required 8.165 tons of propellant. If I reduce the weight to be landed by the weight of the propellant of the Ascent Stage the total weight to be landed would be 4.486 tons only. This is 0.55 times the original weight landed. Calculating linearily down from the originally required amount of propellant 4.4091 tons are got. This is 1.88 times the original required amount of propellant the Ascent Stage alone needed  the number is rounded up and includes a small safety margin I tended to apply in previous posts also. Then 8.8951 tons of propellant seem to be required to return both the Ascent Stage and the Descent Stage of the LM to the lunar orbit. This could be rounded up to 9 to get a small safety margin and to 10 to get a better safety margin. This also will cover the increase of the required amount of propellant for the descent not calculated here (caused by the increased amount for the ascent). 10 tons of propellant to land on the Moon and in total return the LM to the orbit still is by 20 tons to 24 tons less than calculated from the earthian Falcon V as it is designed. The difference might be caused by the following reasons: 1. I applied the escape velocities of Earth and Moon to get the numbers for the Falcon V adjusted to the Moon. The Apollo 8 Press Kit, page 6, says that Quote: A lunar orbit insertion burn with the spacecraft service propulsion engine will place the spaceecraft into a 60 x 170 nm (69 x 196 sm. 111 x 314.8 km) elliptical lunar orbit which later will be circularized at 60 nm (69 sm, 111 km). Quote: Except for about 45 minutes of every twohour lunar orbit, Apollo 8 will be "in view" of at least one of three 85foot deepspace tracking antennas at Canberra, Australia, Madrid, Spain, and Goldstone, California. 2. The earthian Falcon V as it is designed up to now is staged and each of the stages requires at least one engine. The first stage will have five engines. All these engines have a weight and this weight will be significant if compared to the empty tanks. A lunar Falcon might or will be singlestaged because of the very much reduced requirments of propellant even for escape velocity as calculated in the previous post. 3. I choosed the Falcon V because it is reusability. This reusability might add weight the LM didn't include 4. I calculated the reduction of the required amount of propellant linearily which overestimates the amounts. In particular here a difference in velocity is concerned. There is one additional interesting point here  the combined weight of the Ascent Stage and the Descent Stage of the LM without all the fuel fits completely into the payload capacity of the Falcon V. But today an LMequivalent vehicle can be expected to weigh less than that since even the CXV according to the informations from and about t(Space weighs less than that. At this point the results are: 1. The numbers got by the consideration(s) of the Falcon V might include an amount of propellant required only to get into an orbit significantly higher than the Apolloorbit(s). 2. The numbers got for or by the LM of the Apollomissions are valid for a lower orbit than applied for the consideration(s) of the Falcon V. Now it is required to have look back to what has been said about the propellant required to fly around the Moon without entering a lunar orbit: Quote: One possibility to be on the safe side is to simply assume that three BlockDMs are required. This would increase the hardwareweight by 4600. The amount of propellant these three Block DMs could contain then would be 46,500 kg. If still 15,500 kg would be sufficient for the translunar injection then there would be no further question because I suppose that the same amount of propellant is sufficient to do the orbital insertion burn around Earth. At the Moon there is no orbital insertion planned for the lunar Soyuzflight  so I don't consider it regarding the reusable replacements. The third Block DM I am assuming at the moment to be required for nonorbital insertions at the Moon. I am assuming it at the moment  this doesn't mean that I keep the assumption. Since 15,500 kg of propellant will be than required the remainder, the surplus perhaps will fit into other requirements I suppose to be faced to possibly. The question comes up to me if the CXV might be equipped so that it can do those burns itself  not requiring the third Block DM. Ass far as I could check it I applied that throuout all further posts and so there is an amount of propellant of 46,500 kg in total  15,500 kg for translunar injection  15,500 kg for reentering earthian orbit = 15,500 left which are assumed to be reuqired for the nonorbital manoevers etc at the Moon. 15,500 kg of propellant for nonorbital lunar manoevers etc. seems to be a bit much to me since the mass of the Moon is 1/83.1 the mass of Earth. Additionally an orbit of certain altitude around the Moon is much shorter than an orbit of identical altitude around Earth. And there is more propellant included and incorporated in that post: Quote: To try to have a safety margin I add an amount of propellant of 4,400 kg here. So that post provided 29,900 kg of propellant  more than the lower boundary of the amounts calculated in the previous post and only 4,100 kg less than the rounded upper boundary used in this post. Giving up the former assumptions mentioned in the first quote that amount is sufficient to enter a lunar orbit at least. But because of the much shorter length of the lunar orbit and the much smaller mass of the Moon I suppose this amount to include the requirments of the TransEarthInjection also. Next the calculations of the post I am quoting above apply the lunar escape velocity. This means that they are compatible with the upper boundaries of the required amounts of propellant for the lunar Falcon V. So I very simply can calculate two first but still to be corrected numbers of the propellant costs for a toruistic lunar trip like that reported about but including a landing: Quote: This weight is got by taking into account also that the CXV is significantly lighter than the Soyuzcapsule and that so the CXVlike vehicle considered here can be assumed to be lighter also. But since I was looking at Apollo in calculating the recent numbers I still have to look for the consequences of the different propellants Apollo and the Block DM use. According to Wikipedia Apollo uses 50% hydrazine, 50% ADMH and dinitrogentetroxide. To adjust the number calculated to cerosene/LOX I simply take half of the Isp got by hydrazine  2845/2 = 1422.5  plus half the Isp got by ADMH  2804/2 = 1402  resulting in 2824.5 and divide it by the Isp got by the Isp of cerosene. 2824.5/2700 = 1,05  this seems to be the factor by which I have to multiply the number(s) calculated to get the required amount of cerosene 20,883.72 (all numbers got from www.berndleitenberger.de) So 21,000 kg of propellant to arrive at the Moon, to return to Earth and to enter and leave a high lunar orbit where the orbital velocity is close to escape velocity + 30,000 kg to 34,000 kg of propellant to land on the lunar surface and later launch from it to that very high lunar orbit again via a kind of lunar Falcon V are required. This results in a required total amount of propellant for a touristic lunar flight including a landing of 51,000 kg to 55,000 kg. This amount has to be transported into an earthian orbit before the lunar trip can start. A Falcon 9 S9 can lift 24,750 kg and its launch costs $ 78 mio  two of these will transport 49,500 kg at costs of $ 156 mio. Then 1,500 kg to 5,500 kg are left  the first would costs another $ 6.6 mio via a Falcon V if the remainder of the capacity is completely used to carry other payloads while the latter will cost $ 16 mio if a Falcon 9 is used and the remainder of the capacity is completely used for other payloads. Then in total the transportation costs for the new and higher required amount of propellant are between $ 162.6 mio and $ 172 mio. Per person these would mean between $ 81.3 mio and $ 86 mio in the twopersoncase and between $ 32.52 mio and $ 36.4 mio. Using the QuickReach and applying flight costs of $ 10 mio like done earlier the total transportation costs would be between $ 141.666... mio and $ 152.777... mio  let's say $ 142 mio to $ 153 mio. Per person these would be $ 71 mio to $ 76.5 mio at two persons and $ 28.4 mio to $ 30.6 mio. Applying the very low flights costs related to the very high number of flights during th long lifetime of a reusable QuickReach the transportation costs would be 1,770,833.33 to 1,909,722.22. Then per person the propellant costs are between 885,416.67 and 954,861.11 in the twopersonscase and between $ 354,166.67 and $ 381,944.44 in the fivepersoncase. Remembering that the most recent price got was below or close to $ 300,000 the landing seems to be increasing the price by a significant amount  but it still is in tha range below $ 1 mio. Please note urgently: I simply did as if the Falcon repeatedly lands on and launches from the lunar surface while it isn't carried there during the trip carrying th passengers to the Moon. I am using a concept here where there is one vehcile stationed at the Moon as a taxi between lunar surface and lunar orbit only. This could be done really if the Moon would have cerosene ressources and if the lunar Falcon would be built at the Moon using lunar arch mined and lunar production facilities. If the lunar Falcon V would be produced on Earth and transported to the Moon to be stationed there then the costs of that would be investment costs which  like earlier  are to be considered later and separately. But nothing of that can be applied yet  the cerosene for the Falcon as well as the lunar Falcon itself has to be carried from Earth each flight and  except the cerosene completely consumed  carried back into the earthian orbit each flight. This requires extra propellant not considered here yet  the numbers aren't valid at present. For this reason I don't include them into the table here. For the same reason I don't calculate here the costs for the case the Apolloorbit is used  it would be wasted time. I'll go on in the next post. Dipl.Volkswirt (bdvb) Augustin (Political Economist) EDIT: CORRECTED NUMBERS INCLUDING CONSEQUENCES Just some moments after posting this post I have become aware that there seems to be something wrong with the cylindrical volume of the Falcon I calculated. I tried to find out which way I calculated it but didn't get the number again no way. So I can't find out what has gone wrong. The correct cylindrical volume of the earthian Falcon is 452.2 m^3 which is 2.60034502587694077055779183438758times the wrong value of 173.9 m^3. So all numbers based on that former and wrong value have to be corrected by that factor  here are the corrected numbers: weight of that voulme of cerosene: 337.7934 kg to 379.848 kg lunar adjustion by the factor of 4.4: 76.78 kg to 86.33 kg difference to nonorbital roundtrip: 56.78 kg to 66.33 kg new rounded lunar weight: 77 kg to 87 kg new rounded difference: 57 kg to 67 kg new difference of LM to lunar Falcon: 66 kg to 76 kg In the first row of the first list of numbered rows the subtraction of the weight of the propellant from the weight of the Falcon seems to have to be removed and the complete 154,500 kg of the Falcon have to be applied in that comparison. This way the huge difference between the required amounts of propellant seems to be plausible by far. new difference modified LMamount of: 67 kg to 77 kg propellant new result of adjustion to : 49.56 kg to 55.72 kg Apollo 8orbit new difference after adjustion: 39.56 kg to 45.72 kg difference new to old: 0.59 to 0.59 So my remark that the propellant calculated for the nonorbital roundtrip already includes the propellant required for the lunar Falcon is NO LONGER VALID!!!!! Since that was the essential reason for all further calculations those caÃ¶culations are NO LONGER VALID BUT WRONG!!! I WILL NOT APPLY THEM ANY LONGER  it is required what I nonetheless will do: modify the numbers got for the nonorbital lunar roundtrip. 
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In the Ã¼revious post I tried to analyze the difference of required amount of propellant for a lunar Falcon V and a ApolloLM.
But the numbers for that lunar Falcon V arenâ€™t correct yet. The propellant requirement caÃ¶culated is sufficient for one takeoff only â€“ but the vehicle has to land again. Or â€“ more precisely it has to land and then to launch again. This I remark because it will be reuired to reverse the view applied up to now later. To do so I checked the factor by which the lunar Falcon has to be adjusted again. The correct factor is larger yet â€“ according to wikipedia the earthian escape velocity is 11,186 km/s while the lunar is 2,38 km/s giving a factor of 4,7. So the required amount of propellant will have a weight of between 71.87 tons and 80.82 tons. To land again another time this weight will be required because no parachuted landing is possible on the Moon â€“ 143.74 tons to 161.64 tons in total. Because of the weight of the propellant required to land again the previously calculated amounts have to be increased. Since I took the weight of the Falcon V as listed by SpaceX as the empty weight I use it for an approximation here. That weight is only slightly less than the weight calculated above and the amount of propellant required to launch it from the lunar surface has been calculated already: 71.87 tons and 80.82 tons. To launch half the weight of that I get 35.935 tons to 40.41 tons calculating down linearly from the larger number(s). Neyt the weight applied is around 7 tons more than the earthian Falcon V weighs â€“ so I apply weights that are increased by 5 tons and that are rounded â€“ 41 tons to 46 tons. So for the total required amount for the lunar Falcon V I get 185 tons to 208 tons. Thatâ€˜s more than 9 to 10 times the proeppalnt calculated for the lunar roundtrip. I didnâ€™t find numbers about the altitude of the lunar roundtrip above the lunar surface yet nor I know the altitude of the 2.38 km/sorbit yet. For these reasons I suppose that the numbers includea very high safety margin â€“ and I suspect that the ApolloLM is very much closer to the correct result. The amount of 1o tons for the modified and reusable ApolloLM was 10 tons â€“ and this amount now really fits into the numbers got in an earlier post for the lunar roundtrip: Quote: ... I am assuming it at the moment  this doesn't mean that I keep the assumption. Since 15,500 kg of propellant will be than required the remainder, the surplus perhaps will fit into other requirements I suppose to be faced to possibly. The 10 tons are much better based than the 185 tons to 28 tons because a real landing has been done. Because of this I now go on to calculate the lower boundary. First of all the 10 tons are ADMH and dinitrogentetroxide. According to the post already quoted I now apply the weight to calculate the according weight of cerosene â€“ Quote: ... To adjust the number calculated to cerosene/LOX I simply take half of the Isp got by hydrazine  2845/2 = 1422.5  plus half the Isp got by ADMH  2804/2 = 1402  resulting in 2824.5 and divide it by the Isp got by the Isp of cerosene. 2824.5/2700 = 1,05  this seems to be the factor by which I have to multiply the number(s) calculated (all numbers got from www.berndleitenberger.de) Of 15.5 tons 5 tons then are left. The complete weight of hardware used for the lunar roundtrip was 3,600 kg CXVlike vehicle + 6,900 kg reusable booster = 10,500 kg in total. For the complete lunar roundtrip a required amount of propellant of 20,883.72 had been got. This was much less than the booster could contain â€“ 46.5 tons. The total weight to be launched out of the earthian orbit was 23,658.08 kg â€“ including a safety margin of 4,400 kg of propellant. The required amount of propellant to launch for the Moon was 20,833.72 kg. Before launch that weight was orbiting Earth by around 7.8 to 7.9 km/s. Since the escape velocity is 11,186 km/s the difference is 3.286 to 3.386 km/s which the 20,833.72 kg are consumed for. So to enter the lunar Apollorobit at least this amount seems to be required again. Next the velocity of that lunar orbit is 1.613.km/s â€“ 6.187 to 6.3 less than the initial earthian orbit. So another 20,833.72 kg are required. Whatâ€™s left are 2.801 km/s to 3.014 km/s are left which linearly calculated require another 2.801/3.386 * 20,833.72 kg = 17,234.2 kg to 3.014/3.286 * 20,833.72 kg = 19,109.20 kg. This all sums up to between 58,901.64 kg and 60,776.64 kg would be required at the Moon to decelerate the weight of 23,658.08 kg into the Apolloorbit. Obviously the booster used for the lunar roundtrip isnâ€™t sufficient nomore if a landing is included. And again â€“ the lunar orbit has to be left yet which requires another 58,901.64 kg to 60.776.64 kg at this point of calculations. 117,803.28 kg to 121,553.28 kg in total â€“ the number seem to be converging to the numbers for the lunar Falcon. But the comparison isnâ€™t valid â€“ the lunar Falcon is launched to the highest lunar orbit from the lunar surface while here a vehicle is decelerated into a low lunar orbit coming from Earth. However I am doubting these numbers because they are that high â€“ multiple times the amounts Apollo required. Because of that I will look for additional data and informations I didnâ€™t need yet. Iâ€™ll go on later. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Because of my doubts expressed at the end of the previous post and caused by comparisons to Apollo I had a look to Wikipedia, www.berndleitenberger.de etc. to find out details about Lunar Orbital Insertion not needed yet.
First of all I looked to the J2 engine. Obviuosly the maximum weight of propellent it consumes is 106,600 kg. The least burn time listed is 475 seconds â€“ so the consumption per second is around 224.42 kg.. According to the data I got the burn time for Translunar Insertion is six minutes = 360 seconds. So 360 * around 224.42 kg = 80,791,58 kg are consumed. This multiplied by weight of the CXV including the booster divided by the weight of Apollo 10,500 kg/30,332 kg = 0.35 is giving 27,967.55 kg. This agian multiplied by the Isp of LOX/LH2 divided by the Isp of LOX/cerosene 3,830/2,945 = 1.30. results in 36,357.82 kg. This number can be used for comparisons later if required which I am suspecting... Next the source www.christa.org/lunar.htm says that the earthian gravitation slows down Apollo to 2,000 mph which is 3,200 km/h. Since this is far less than the earthian escape velocity applied up to now in the recent calculations this explains to a large degree the very large propellant requirements got. It also might indicate how large the impact of the linearity applied during the calculations is. I remember that I read in the book â€žApollo 13â€œ of a velocity of 5000 km/h but this may be a number told of regarding the way back to Earth. On the other hand the Moon accelerates the vehicle again. I didnâ€™t find informations about the velocity when Lunar Orbit Injection occurs and so I need to switch over to another way this moment. The Apollo Lunar Mission Tutorial chapter Lunar Orbit Insertion says that Quote: ...the Service Propulsion engine was fired for 370.8 seconds... Next the propellant for Trans Earth Injection needs to be taken into account also â€“ about that the Apollo Lunar Mission Tutorial chapter Trans Earth Injection says that Quote: The maneuver required that this engine  the one ticket home for three astronauts  start, burn consistently and accurately for (typically) 2.5 minutes... Quote: During the Apollo 14 mission, the TransEarth Injection maneuver was initiated ... following a 149 second SPS engine burn ... Then in total 88% of the propellant are required for orbital insertion and entering a return path to Earth â€“ the remainder of 12 % is available for corrections. Knowing these percentages isnâ€™t sufficient here and so I looked for more data. Doing so I detected that it looks as if I applied too large a number for the propellant requirements earlier in this thread â€“ which means a good safety margin. I used 25.5 tons of propellant because of informations from Wikipedia. Astronautix is talking about 18,488 kg. So orbital insertion plus return path injection consume 16,269.44 kg propellant â€“ applying the 88% to the formerly used amount of 25,500 kg of propellant the resulting amount for Lunar Orbital Insertion and Trans Earth Injection is 22,440 kg. Since I am nonetheless looking to the formerly applied numbers I think it to be useful to continue that now â€“ it seems to be nothing else than a revision of former remarks under the aspect of the additional data applied here. First of all obviuosly 16,269.44 kg of the correct amount of propellant Apollo carried or 22,440 kg of the amount of propellant I applied earlier are required to enter and to leave a lunar orbit later only and only 1,730.56 kg or 3,060 kg of propellant are required for corrections. To these amounts the factor of 1,05 already applied earlier to transit from ADMH/Dinitrogentetroxide to cerosene. The values then are 17082,91 kg or 23,562 kg of cerosene for Lunar Orbital Insertion and Trans Earth Injection but only 1,817.09 kg or 3,213 kg of cerosene for corrections. For the lunar roundtrip I added a safety margin of 4,400 kg.. â€“ which completely covers the amount got just above for corrections. And before adding that safety margin I calculated 8758.08 kg of propellant to be required at the Moon â€“ both number still not multiplied by 1.05. In total these are 13,158.08 kg and multiplied by 1,05 they mean 13,815.99 kg of cerosene. Of these 10,602 kg seem to be available for Lunar Orbital Insertion and Trans Earth Injection. That seems to be insufficient yet. Next the correction for the mass difference between Apollo and CXV has to be done.But I donâ€™t apply the factor of 0.35 used above because that factor included the mass of the propellant not to be consumed that moment yet. I am going to apply the weight of the empty Apollo. This weight according to the new sources applied here is 6.6 to 7 tons ( www.projektapollo.com , german source). CXV/Apollo then is 3,600/6,600 = 0.55. This correction results in 9,317.96 kg or 12,852 kg of cerosene required for Lunar Orbital Insertion and Trans Earth Injection of the CXV without its booster. The former of these is the based on th correct amount of Apollopropellant informed about by the new sources while the latter is based on the to high number got earlier. The number based on the new informations fits well in the available amount already got. But the requirements for corrections need to be adjusted to the CXV also â€“ they are 991,14 kg or 1,752.55 kg. These amounts required for corrections together with the insertion and injectionamounts are 10,309.10 kg or 14,604.55 kg. The difference is very small â€“ but there is the booster and the propellant it carries to enter an earthian orbit after return. I was thinking about applying the number got above for the Trans Lunar Injection but this would be invalid for the simple reason that already the numbers for the Block DM have been used and there are no Block DM data yet for Trans Lunar Injection. Also the Block DM like the replacement of it used here consumes cerosene and no adjustments are to be calculated. Additionaly the propellant capacity of the Block DM is less than the number got at the beginning of this post. The requirements onviuosly excedd the available propellant by 788.56 kg. Next the lower boundary of the propellant requirements of the lander are 10 tons = 10,000 kg which multiplied by 1.05 mena 10500 kg cerosene. To that 4,486 kg for the mass of the empty lander have to be added. Then the total weight to be added to the weight calculated already for the roundtrip which turned out to include a surplus because of safety margins... â€“ the weight to added to that is 788.56 kg + 10,500 kg + 4,486 kg = 15,774.56 kg more to be launched out of the earthian orbit. To include another small safety margin I round that up to 16,000 kg. Whatâ€™s known about the propellant requirements without the required calculations is that to the former 20,883.72 788.56 and 10,500 kg have to be added â€“ 32,172.28 kg are to be transported now from the earthian surface into the orbit. So the costs of transportation of the propellant are increased by more than 50%  which seems to be the first hint that the lower boundary isnâ€™t that far off the boundaries calculated for the roundtrip and that the number will be close to what rpspeck has said very recently in the General Microspace Forum. The amount of 20,883.72 kg of propellant I got for 23,658.08 kg to be launched. I now try to go an easy way and apply linearity from below instead of from above as done else to be at the safe side â€“ I next add a new safety margin because of the loss of safety. The factor is 16,000 kg/23,658.08 kg = 0,676. Then the additional amount of propellant is 14,117.40 kg. Together with the above already calculated amount of 32,172.28 the total amount is 46,289.68 kg completely fitting into the capacity of the three boosters of 46,500 kg. The safety margin I have been speaking of could be 3,710.32 kg to get 50 tons or 4110.32 kg to get a number that can be divided by the mass of the CXV. I apply the larger number because of the safety as well as because of the round integer number the CXVmass has to be mulitplied by to get the new value â€“ since that number is the number of launches required to deliver the propellant into the orbit via QuickReach. So the amount of propellant got is 50,400 kg now. This amount requires 2.04 Falcon 9 S9 launches or 14 QuickReachlaunches â€“ resulting in transportation costs of $ 159.12 mio or $ 140 mio in the present situation and would drop down to $ 1,750,000 if or when there is the situation of the reusable QuickReach applied earlier already. All these numbers are valid also as lostinspace costs. The $ 159.12 mio of the Falcon 9 S9s require 3 flights to cover the lostinspace costs, the $ 140 mio for the present QuickReach also requires three flights to cover them while the thirs situation requires only one flight fÃ¼r coverage and the lostin.space costs are far below the $ 55 mio used as standard here. Then the initial costs are $ 214.12 mio $ 195 mio and $ 3,500,000. Added the orbital price the ticket prices are as follows Code: Lunar Costs 2 passengers 5 passengers $ 214.12 mio $ 111.06 mio $ 46.824 mio $ 159.12 mio $ 83.56 mio $ 35.824 mio $ 195 mio $ 101.5 mio $ 43 mio $ 140 mio $ 74 mio $ 18 mio $ 3.5 mio $ 1.775 mio $ 725,000 $ 1.75 mio $ 900,000 $ 375,000 Obviously the lower boundary of the price remains at the levles already got for the roundtrip â€“ and even for the data valid at present a price for a touristic lunar landing of less than the touristic orbital Soyuzprice can be calculated. These numbers I will add to the table given earlier later since first I have to proceed with the upper boundary yet where I use a lunar Falcon. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Contents of this post
Additional Aspects Application of the Aspects â€“ a few problems Weight of the Lunar Falcon Amount of Propellant for Lunar Falcon recalculated Portion of Total Propellant already known Additional Requirement except Lunar Orbital Insertion and Trans Earth Injection Weights to be inserted or injected Propellant for Insertion and Injection Total Propellant and Transportation Costs How could the astronomical numbers got rid of? Appendix Sources used in this post and the previous post Table of Costs and Prices (enhanced) One remark about what I am doing here. The calculations and consideartions are only meant to get economical insights  bacuase of the proiperties of the subject(s) of Economics it is not possible to keep the precision and correctness of physical, technical and technological calculations and methods. I am not out on getting physical etc. correct results but only to get boundaries of prices and costs which all include components that are outside of Physics and Engineering. Weights, propellants, engines etc. are simply tools to get ideas about the costs and prices. The Falcon is used here as a tool â€“ and this turns out to be a test of that tool. Regarding if itâ€™s fitting into the task. To get the upper boundary of the costs and price of a touristic luanr landing proceeding the way gone since initiating this thread I now have to turn to the lunar Falcon again. I already got 185 tons to 208 tons being 9 to 10 times the propellant calculated for the lunar roundtrip which was 20 tons to 21 tons. Looking back into my calculations I have applied the factor by which the lunar escape velocity fits into the earthian escape velocity. Additional Aspects In between I have become aware of a circumstance not considered yet and I have found a partcular new information used in the previous post. The first of these is the high difference between the earthian and the lunar gravitation due to the difference between their masses while the second is the deceleration of the vehicle by the Earth and its acceleration by the Moon at a distance of 344000 km from the Moon. I also neglected that Apollo didnâ€™t go by esacep velocity but by a velocity slightly less â€“ the difference is around 1 km/s. Now the correct velocities to be applied would increase the required amount of propellant I got â€“ to apply them first would mean to mix earthian conditions with lunar conditions. The comparison up to now was that between an earthian Falcon launching from Earth and a lunar Falcon launching from the Moon â€“ and I simply applied the diffrerence of the escape velocities. As long as the vehicles didnâ€™t reach an orbit I should consider them to be taking off from the surface here. Takeoff requires thrust to get rid of the gravitational pull of the planet. This thrust requires propellant. Since the mass of the Moon is 1 / 81.3 the earthian mass I am wrong to reduce the propellant simply by the factor earthian escape velocity/lunar escape velocity â€“ regardless of 2.38 km/s or 11.186 km/s for all velocities a launch from the Moon will require less propellant than a launch from Earth. Application of the Aspects â€“ a few problems But it is obvious that it would be wrong to simply divide the amount got by 81.3 â€“ the result would be less than the amount got for the completely reusable LM by a factor of around 4. Now the information about the velocity applied in the previous post is of help. At a distance of 344,000 km from the Moon the velocity of Apollo was 3,200 km/h (the distance I didnâ€™t list but it is listed in the sources I used). These are 0,89 km/s which is much less than the lunar escape velocity. So the fact that Apollo had to apply a Lunar Orbital Insertion burn means that the Moon accelerated Apollo from 0.89 km/s to more than the orbital velocity. The orbital velocity Apollo had to have was slightly more than 1.6 km/s and Apollo required a Lunar Orbital Insertion burn to be decelerated down to that velocity. So it will have been more than 1.6 km/s â€“ and 1.613 / 0.889 is 1,82. This factor still is too low to counter the factor of four by which the result got by the gravitational difference would be wrong. And it easyly can be seen that the lunar escape velocity also would have results that can be applied. Weight of the Lunar Falcon But there is another fact â€“ the weight of the empty completely reusable LM would be 4.486 tons while I didnâ€™t calculate a weight of an empty lunar Falcon yet. The amount of propellant the lunar Falcon would require has been got by linearly calculating down â€“ and both the original value and the resulting value fill avolume inside the tanks. This mean that the weight of the tanks in turn has been reduced sublinearly by a high degree â€“ and so the complete lunar Falcon ha sbecome light far less than proportionally to the reduction of the required amount of propellant. The upper boundary of the amount of propellant the lunar Falcon might need is 208 tons while the amount got for Earth was 379.848 tons â€“ a factor of 0,55. SpaceX says that the earthian Falcon weighs 154.500 tons. Applying the factor would result in a weight of 84.602 tons which will be far too low I think because of the considerations before the calculation. But there is a way out  all the informations I already used in the Financial Barriers section seem to include that engines contribute a large share of the weight. If I remember correct then I alreday said that the lunar Falcon can be assumed to be singlestaged while the earthian Falcon consists of two stages. This difference in stages means that the luanr Falcon has at least one set of engines less than the earthian one. This might justify to assume that an empty lunar Falcon weighs 85 tons. Amount of Propellant for Lunar Falcon recalculated These 85 tons then mean that I have a third factor. I can apply the 81.3, the 1.82 and I can apply 85 / 4.486 = 18,95. This thirs factor I round up to 20. Then I divide the weight of the lunar Falcon got earlier by 81.3 mulitply the result by 1.82 and then another time by 20 â€“ this is a resulting factor of 0,45 the required amount of propellant is to be multiplied by. Then I get 82.83 tons to 93.13 tons for both launch and landing. These then I round up to 83 tons and 94 tons and then decide to apply 100 tons. This is only five times yet the amount got for the lunar roundtrip and may be a reasonable upper boundary. The way I got the boundary is marked by several lacks of informations, by uncertainty, helplessness and it is voluntary to some degree â€“ but the amount got is 10 times that got for the totally reusable lander of the lower boundary. For this reason I suppose it to be acceptable. Portion of Total Propellant already known To the former 20,883 kg of propellant of the lunar roundtrip now at least 100,000 kg of propellant have to be added. But the lunar Falcon is assumed to weigh 85 tons which have to be added to the mass to be launched out of the arthian orbit and to be reentered later again.. So the amount of propellant known without additional calculations is 120,883 kg of cerosene and the weight of hardware already known is one CXVlike vehicle of 3,600 kg, a booster like three Block DMs weighing 6,900 kg in total and a lunar Falcon assumed to weigh 85,000 kg here â€“ total hardware then is 95,500 kg. But while the increase of hardware weight for the lower boundary can be neglected because it is the lower case this is by far impossible regarding this upper boundary â€“ the difference is too high and it is an upper boundary which in contrary to the lower one should be kept as unexcessable as possible. Additional Requirement except Lunar Orbital Insertion and Trans Earth Injection And the Lunar Orbital Insertion isnâ€™t added yet â€“ but letâ€™s start with what is listed already. To the roundtrip up to now 100,000 kg of propellant and 85,000 kg of hardware are added â€“ 185,000 kg in total. Calculating up linearly which is unsafe results in a factor of 185,000 / 10,500 = 17.62. The propellant for the roundtrip mulitplied by this factor reasults in 367,938.57 kg of propellant which is 23.74 times the capacity of one Block DM. So 24 Block DMlike boosters are to be assumed. Three of them already have been included into the lunar roundtrip. Then 21 of their replacements are to be added â€“ weighing 48,300 kg. These require another 96,061.8 kg of propellant â€“ between 6 and seven more Block DMreplacements. Assuming seven is more than sufficient because they could contain 108,500 kg of propellant while six would be insufficient 93,000 kg. The seven replacements would weigh 16,100 kg which is less than twice the original hardware weight and another 42,000 kg will be more than sufficient to cover that. â€“ meaning three more replacements weighing 6,900 kg. So I at this point end up with one more booster and one additional amount of 15,500 kg of propellant. So the amount of propellant got now in total is the number 120,883 kg above plus 367,938.57 kg plus 108,500 kg plus 42,000 kg plus 15,500 kg. This sums up to 654,821.57 kg in total which include 12,000 kg to 20,000 kg safety margin at least. The safety margin is larger than those 12 to 20 tons since the 20.883 kg of the lunar roundtrip included the propellant for reentering the earthian orbit except the propellant used for Trans Lunar Injection â€“ may be also the propellant consmued at the Moon for corrections was left away. So I am doing here as if the 100 tons of propellant to land on the Moon and to launch from it later again would be carried back into the earthian orbit. So I could remove the effect of this amount of propellant to some degree. But the huge surplus contains a margin required for Lunar Orbital Insertion and Trans Earth Injection. Weights to be inserted or injected At Lunar Orbital Insertion the mass is reduced down to 100 tons of propellant plus 85 tons for the Falcon plus 3.6 tons of the CXV plus 73.6 tons of 32 replacements for the Block DM(s). So in total 262.2 tons would have to be inserted into the highest lunar orbit. Only 162.2 tons would have to do the Trans Earth Injection because the propellant for landing on and launching from the Moon are to be assumed to be consumed totally. These numbers donâ€™t include yet the propellant required to leave the Moon for Earth and to reneter an earthian orbit. Since the 654,821.57 kg would be for a lunar roundtrip and include a safety margin I now take the risk to apply to low numbers â€“ I subtract the lower safety margin and divide the remainder by 2. This results in 321,410.785 kg. This now I add to the weight or mass to be inserted into lunar orbit and to leave for Earth later again. So I get 583,610.785 for Lunar Orbital Insertion and 483,610.785 for Trans Earth Injection. The mass of the Apollo CSM going to be applied now is 25.5 tons as correct value or 50 tons as the formerly applied value if I remember correct. To get the amount of propellant required here I now first divide the total weight to be inserted by the lower value because this will end up in the higher value which is the safer one. The quotient is 22.89 â€“ this the amount Apollo required would be to be divided by but Apollo has gone into a much lower orbit than I calculated the lunar Falcon for. So I also apply the quotient of the velocity of the Apolloorbit divided by the lunar escapevelocity  0,68. These both factors multiplied with each other result in 15.57. Propellant for Insertion and Injection By this factor I now multiply the amount of propellant required for Lunar Orbital Insertion. Apollo consumed 16,269.44 kg of propellant for both the manoevers which was 88% of total. 63 % only were consumed at the first manoever â€“ which were 16,269.44 kg * ( 63 / 88 ) = 11,647.44 kg. This number next has to be multiplied by 1.05 for adjustment to cerosene â€“ 12,229.812 kg. This multiplied by 15.57 results in 190,418.17284 kg. The total weight after the return of the lunar Falcon from the lunar surface is 483,610.785 kg. This divided by 25,500 results in the factor 18.97. Multiplication by 0.68 is giving 12.90. 25% of Apolloâ€™s propellant have been consumed at Trans Earth Injection â€“ 16,269.44 kg * ( 25 / 88 ) = 4,622 kg. These are 4,853.1 kg adjusted to cerosene. Multiplication by 12.90 results in 62,604.99 kg. In total 253,023.16284 kg are required at the Moon. For these would be another 17 Block DMreplacements weighing 39,100 kg. These require another 84,000 kg meaning additional six replacements which then weigh 13,800 kg. Another 15,500 kg of propellant and another booster seem to be sufficient since I also have the safety margin. Obviously in total 253,023.16284 kg + 84,000 kg + 15,500 kg = 352,523.16284 kg have to be added now. But there is additional weight to be inserted and to injected â€“ 55,200 kg of hardware. But the division by 25,500 and the multiplication by 0.68 resulkts in a value that I think that an increase of the propellant by 45,000 kg is sufficient â€“ these are three more replacments weighing 6,900 kg. Total Propellant and Transportation Costs So in total I get a required amount of propellant of 654,821.57 kg + 352,523.16284 kg + 45,000 kg = 1,052,344.73284 kg which I round up to 1,052,345 kg. These require 292.32 flights for transportation if the QuickReach is used or 42.52 flights if the Falcon 9 S9 is used. The transportation costs via the Falcon would be $ 3,316.5 mio. $ 2,923.2 mio via the expendable QuickReach and $ 36.54 mio via the reusable QuickReach.. These are also the lostinspace costs â€“ to cover them by portions of $ 55 mio 61 flights, 54 flights and 1 flight are required The tickets would have a price of $ 1,662.25 mio, $ 1,465.6 mio and $ 22.27 mio at two persons and of $ 663.3 mio, $ 584.64 mio and $ 7.308 mio at five persons after the lostinspacecosts are covered. â€“ before that the proces are $ 1,687.75 mio, $ 1,483.1 mio, $ 49.77 mio, $ 674.3 mio, $ 595.64 mio and $ 18.308 mio. These numbers are astronomical â€“ in principle I mustnâ€™t use them any longer but I didnâ€™t find a way for a better upper boundary yet. But the Falcon obviously is a bad How could the astronomical numbers got rid of? But an entrepreneur would look if he can get rid of these numbers. This would include the idea to leave the lunar Falcon in the lunar orbit once carried there instead of carrying it back to Earth each trip. That low lunar orbits are instable due to the gravitational anomaly of the Moon wouldnâ€™t be no problem â€“ the lunar Falcon would be left in the highest orbit and it would be used often. The reusable replacment of Apolloâ€™s LM could be left in a selenostationary orbit or in an orbit high enough to let it not crash into the Moon before the next touristic trip. This way the 85 tons of the lunar lander would be saved as well as the lesss than 5 tons of the LMreplacement. The huge amount of propellant for the lunar Falcon also have to be got rid of. An entrepreneur might decide to carry the propellant via the roundtrips until a sufficient amount has been gathered in the lunar orbit. He also would design a delivery plan. t/Space say that they would deliver propellant to the Moon via a trajectory that takes three months to the Moon to be able to carry more propellant without increasing the weight of the vehicle and without needing more propellant for the tanker itself. Next the entrepreneur will offer roundtrips as well as orbital visits as well as landings and use the first two to crossfinance the last. Because of all this I will leave the inacceptable upper boundary as it is and proceed later. I wll replace it if and when I find a better upper boundary. These all are results I am very far from being content with but the next steps will remove some of the problems at least because now new possibilities are available. Dipl.Volkswirt (bdvb) Augustin (Political Economist5) Appendix Sources used in this post and the previous post www.christa.org/lunar.htm www.berndleitenberger.de Wikipedia Astronautix Apollo Lunar Mission Tutorial www.projektapollo.com Table of Costs and Prices (enhanced) Table about the steps resulting in a largely reduced price  variable costs only Code: step revenue price costs lostin cov. flts price  space lstin cov lost costs spcco inspace orig. 200 mio 100 mio 145 mio 145 mio 55 mio 3 72.5 mio concept CXV in 155 mio 77.5 mio 100 mio 100 mio 55 mio 2 50 mio stead Soyuz reusbl. 135 mio 67.5 mio 80 mio 80 mio 55 mio 2 40 mio booster 129 mio 64.5 mio 74 mio 74 mio 37 mio kept in orbit less 117 mio 58.5 mio 62 mio 62 mio 55 mio 2 31 mio prop.by 108 mio 54 mio 53 mio 53 mio  26.5 mio less weight 3 pass. 105 mio 52.5 mio 50 mio 50 mio 55 mio 1 25 mio ISS etc  96 mio 48 mio 41 mio 41 mio 20.5 mio + pass lunar more 129 mio 64.5 mio 66 mio 66 mio 55 mio 2 37 mio prop.for Moon + reenter erthian orbit deliv. 121 mio 60.5 mio 58 mio 58 mio 55 mio 2 33 mio by VLA+ Quick Reach 5 136 mio 27.2 mio 58 mio 58 mio 55 mio 2 15.6 mio tourists 2 1.5 mio 750000 725000 725000 725000 1 387500 tourists $ 25,000 orbital 5 1.575 mio 315000 725000 725000 725000 1 170000 tourists $ 25000 orbital For the last two rows the depreciations have to be included yet which has been done in an earlier post.  Lunar Trip including Landing 2 215 mio 112 mio 215 mio 215 mio 55 mio 4 83.6 mio tourists 3.4 bio 1.7 bio 3.4 bio 3.4 bio 61  1.7 bio Falcon 5 215 mio 84 mio 215 mio 215 mio 55 mio 4 35.9 mio tourists 3.4 bio 675 mio 3.4 bio 3.4 bio 61 664 mio Falcon 2 195 mio 102 mio 195 mio 195 mio 55 mio 4 74 mio tourists 3.0 bio 1.5 bio 3.0 bio 3.0 bio 54  1.5 bio Quick Reach 5 195 mio 43 mio 195 mio 195 mio 55 mio 4 18 mio tourists 3.0 bio 596 mio 3.0 bio 3.0 bio 54 585 mio Quick Reach 2 3.5 mio 1.8 mio 3.5 mio 3.5 mio 3.5 mio 1 725,000 tourists  37 mio  50 mio  37 mio  37 mio  37 mio  23 mio 25,000 orbital 5 1.8 mio 900,000 1.8 mio 1.8 mio 1.8 mio 1 375,000 tourists  37 mio  19 mio  37 mio  37 mio  37 mio 7.4 mio 25,000 orbital 
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Getting rid of the astronomical numbers Using the now available new possibilities Numbers based on the new possibilities Difference to previous posts Getting rid of the astronomical numbers To get rid of my problems etc. experienced during the calculation of the upper boundary I can act like the entrepreneur â€“ but at some point(s) this will differ from what that entreprenuer would do really: 1. An empty â€“ nonfueled â€“ vehicle to land on and launch from the Moon from and into a lunar orbit could be carried to the Moon. This would save the propellant required because of the 100 tons of propellant the lunar Falcon needs. The factor of 17.92 is reduced to 85,000/10,500 = 5.6 then and the additonal propellant is 115,353.8 kg yet. This saves 7 Block DMreplacements and a hardware weight of 16,100 kg and 32,020.6 kg of propellant to carry them â€“ so another two replacements can be saved at least. The amount for Lunar Orbital Insertion was 190,418.17284 kg. These were needed for 100 tons of cerosen + 85 tons of the lunar Falcon + the weight of the CXVlike vehicle. 50 % of that would be the weight of the nonfueled lunar Falcon + a large fraction of the CXVlike vehicle and all it needs and so I take 60 % for the moment  114250,903704. Next the lunar Falcon will be left in lunar orbit and not carried back to Earth â€“ which saves a large protion of whatâ€™s required for Trans Earth Injection â€“ letâ€™s assume two thirds this moment and round up the result: 23,000 are left. It seems that this first step reduces the requirements by more than 50 %. But still the problems would be reduced only and thus left which I donâ€™t want to base something at any longer. 2. An empty lunar Falcon is carried to the Moon using a Block DMreplacement as large as 47 boosters. These then contain 728,500 kg of propellant which is remarkably less than the original amount but seem to be too lot to me. This will be due to the circumstance that I didnâ€™t calculate the impact of the saved weight of boosters and other things on the propellant requirements. If I would have done so I would have got less boosters and less proepallent capacity. But the essential point here is that there is the tank of the nonfueled lunar Falcon. The lunar Falcon could be fueled which would save six more Block DMreplacements. Originally there wer a Block DMreplacemnet of the capacity of 62 Block DMs â€“ now there are left only 41 of them and this amount still is too large because I didnâ€™t iterate the claculations as done in the previous post. Whatâ€™s essential here is the way to get rid of the numbers â€“ not the numbers themselves. Thatâ€™s the reason why I calculate them over the thumb far more at present. 3. The lunar Falcon might be carried without the CXVlike vehicle and thus automatically. This saves the weight of four more boosters. Then I am at 37 â€“ and still the impacts of the saved weight arenâ€™t included. 24 boosters weigh 55,200 tons. Thinking about â€“ still by the thumbb I end up at less than 20 replacements. It seems that more reductions might be achieved by carrying the lunar Falcon in pieces into the lunar orbit and mounting them together there. This way a remaining portion of the problems of the previous post also is turned into investment â€“ this I have to consider later. But the lunar Falcon is unfueled when it finally arrives in the lunar orbit â€“ so still the touristic flights would have to carry 100 tons of cerosene to the Moon each trip. Obviously the problems arenâ€™t got rid of yet. At this point I am going to to do something different than an entreprenuer might do.perhaps. In the previous post I already said that the propellant could be carried to the Moon a way t/Space has in mind â€“ tankers flying along long term trajectories to be able to carry larger amounts of propellant. To calculate such deliveries plus the different times plus the larger amounts etc. is another consideration to be done in parallel. This is an additional expense of time which can be avoided. 4. Without calculating anything the lunar Falcon should land on the Moon â€“ and then it is nonfueled. This would increase the weight to be carried and the capacity of the required Block DMreplacement. But what is won this way? The lunar Falcon still would have to be fueled and the only way still seems to be the tankers going the long term trajectories and now they would have to land also. Using the now available new possibilities At this point the possibilities available now should be used â€“ and this will remove the problems. Of course â€“ at present it is specualtive and more data have to be waited for. But they will be gathered. The lunar dust conatines oxygen which will be won by technologies NASA set a Centennial Challenges Prize for. And â€“ much more important and interesting here â€“ it is supposed that there might be hydrogen in polar craters which can be won also. May be also that there is water in those craters. If all this hydrogen or water is sufficient to allow for letâ€™s say 100 years launches and living on the Moon then it wouldnâ€™t cause too much problems to use that hydrogen or water for touristic flights also. This would free each operator from propellant deliveries from Earth to the Moon. So the lunar Falcon will land and be refueled at the lunar surface â€“ and this method will be applied each touristic trip to the lunar surface. Result: The complete flight from Earth to Moon is freed from the requirement to carry propellant for the lunar Falcon. And this applies for the reusable LM of the lower boundary also! So the real touristic trips require only one modification of the lunar roundtrip â€“ a Lunar Orbital Insertion and a Trans Earth Injection. The lander regardless of that being the reusable LM or the lunar Falcon now can be considered independently. Numbers based on the new possibilities That the complete flight from Earth to Moon is freed from the propellant for launching from and ladning on the Moon as well as from the weight of the lander means that I can take the number(s) I got when I looked if the safety margins of the lunar roundtrip are sufficient to cover the requirements of Lunar Orbital Insertion plus Trans Earth Injection. I found that only around 1,000 kg were missing This menas that for the journey from the earthian orbit into the lunar orbit only 20,883 kg + 23,658.08 kg to be launched out of the earthian orbit â€“ 10,500 weight of CXV + booster + 1,000 kg = 36,041.08 kg are required. Then the additional weight to launch the additional 1,000 kg is neglected â€“ to try to get rid of that and to provide a safety margin I round up to 40,000 kg. These 40,000 kg require 1.62 Falcon 9 S9 launches and thus cost $ 126.36 mio â€“ regardless if the lower or the upper boundary is calculated. If the QuickReach of the CXV is used then 11.11 flights are required which cost $ 111,111,111.11 mio for the expendale QuickReach and 1,388,889 for the reusable QuickReach These numbers are interesting in so far as the entrepreneur could offer vacations in the lunar orbit â€“ this would be a trip between the roundtrip and the landing. It would allow for price differentiation â€“ the luanr orbital customers would pay another price than the landing tourists. This would allow for more passengers into the orbit and might help to avoid high landing prices. Per passenger the prices would be $ 67.18 mio at two passngers and the Falcon, $ 59,555,555.55 at two passengers and the expendable QuickReach, $ 719,445 at two passengers and the reusable QuickReach, $ 29.272 mio at 5 passengers and the Falcon, $ 26,222,222.22 at 5 passengers and the expendable QuickReach and $ 301,778 at 5 passengers and the reusable QuickReach. These are the prices after the lostinspace costs are covered â€“ I will include them when I post the enhanced table about the prices and costs. Still the costs for landing and launching are left. In the previous posts I calculated the amounts of cerosene which were 10,500 kg fr the resuable ApolloLM and 100,000 kg for the lunar Falcon. These amounts of cerosene in this post are replaced by amounts of hydrogen. I now again apply the qutient of the Isp of LOX/cerosene and the Isp of LOX/LH2. Applying the numbers I find under www.berndleitenberger.de the ratio is 2945/3830 = 0,77. In reality this ratio might be better perhaps since in vacuum the Isps are higher â€“ but I am not sure if the ratio is changed then. But it menas that less proepllant will be required and there is another safety margin here. Applying the ratio 8,085 kg or 77,000 kg of hydrogen are required. The only informations about the price of hydrogen I found are german informations. For cars the price is said to be 0.56 Euro per liter because of subsidies. According to www.berndletienberger.de the average denisty of LOX/LH2 is 0.28. This menas that the lower amount is 28,875 liters and the upper boundary is 275,000 liters. These then cost 16,170 Euros or 154,000 Euros. In another thread I applied the exchange rate of 1 Euro = 0,8287 Dollar valid valid at the 18th of July 2005. This results in $ 13,400.08 as lower boundary or $ 127,619.80 as upper boundary. Since these numbers are subsidized prices higher numbers should be applied. The problem is that I donâ€™t know the share of the subsideis. So lets assume that the upper boundary is $ 15 mio â€“ a hundred times the upper boundary calculated but still less than the flight costs of the CXV from the earthian surface into the earthian orbit of the ISS.. Obviously the lower boundary of $ 6,700 at two persons and $ 2,680 at five persons has nearly no impact on the price of the trip while the upper boundary of 7.5 mio at two perons and $ 3 mio at five persons has an impact but not too much if the cersone is transported into the earthian orbit via the Falcon 9 S9 or the expendable QuickReach. But the impact of the upper boundary is significant if the reusable QuickReach is used. Because of this I am thinking about using the subsidized upper boundary also â€“ which is $ 63,810 at two persons or $ 25,524 at five persons. The prices including landing at the lower boundary then are $ 67.1867 mio at two passngers and the Falcon, $ 59,562,255.55 at two passengers and the expendable QuickReach, $ 726,145 at two passengers and the reusable QuickReach, $ 29.27468 mio at 5 passengers and the Falcon, $ 26,224,702.22 at 5 passengers and the expendable QuickReach and $ 304,458 at 5 passengers and the reusable QuickReach. At the upper boundary the prices including landing are $ 74.68 mio at two passngers and the Falcon, $ 67,055,555.55 at two passengers and the expendable QuickReach, $ 8,219,445 at two passengers and the reusable QuickReach, $ 32.272 mio at 5 passengers and the Falcon, $ 29,222,222.22 at 5 passengers and the expendable QuickReach and $ 3,301,778 at 5 passengers and the reusable QuickReach The medium â€žboundaryâ€œ I am thinking about would be at $ 67.243810 mio at two passngers and the Falcon, $ 59,619,365.55 at two passengers and the expendable QuickReach, $ 783,255 at two passengers and the reusable QuickReach, $ 29.297524 mio at 5 passengers and the Falcon, $ 26,247,746.22 at 5 passengers and the expendable QuickReach and $ 327,302 at 5 passengers and the reusable QuickReach. Difference to previous posts Now I have included the price of the propellant itself here while I didnâ€™t do that regarding the cerosene yet. This I have done because no propellant is carried into the lunar orbit in this post. If I wouldnâ€™t have inlcuded the price of the hydrogen it would have looked to me as if I forgot the costs od the propellant. To do as I have done also shows me how low the numbers are now. I wonâ€™t incorporate the numbers including hte hydrogen price into the tabel of steps but leave away it and add it later when I include all propellant costs. As can be seen I got rid of the problems of the previous post(s) â€“ something like this an entrepreneur will prefer. Still more is possible yet. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Propellant reuired to return to Earth The lunar ressources Removing costs Lunar costs Resualting costs Propellant reuired to return to Earth The CXVlike vehicle including its booster has to carry the propellant for the Trans Earth Injection and to reneter the earthian orbit. Several posts earlier I calculated the propellant to renter plus the propellant required at the Moon in the roundtripcase to be 8,758.08 kg. In the previous posts I said that 1,000 kg have to be added to cover the requirements of Lunar Orbital Insertion and Trans Earth Injection. The first number has to be adjusted by the factor of 1.05 but I levae it away here because I will round up to get a safety margin. For both the Insertion and the Injection I used the number of 23,562 kg ealier â€“ this is 88% of the complete propellant I used for the Moon that time while additional 12 % were available for corrections. Of the 88% are share of 25/88 were required for Trans Earth Injection. These are (25 * 23,562 kg)/88 =6,693,75 kg. Rounding up the first number to 9,000, adding 1,000 kg that were required additionally and then adding another 7,000 kg required for Trans Earth Injection I get an amount of propellant of 17,000 kg required to return to Earth from the lunar orbit. The lunar ressources This is interesting for one simple reason. The lunar lander uses lunar ressources in between. So these ressoruces may be carried into the lunar orbit also to refuel the booster of the CXVlike vehicle â€“ and this now would free that booster from carrying from Earth fuel for the return to Earth if I assume for a while that that booster and its engines can contain and consume LOX/LH2 as well as LOX/cerosene without any changes, modifications and so on. Of course this assumption is quite incorrect technically but I am looking for the costs and the financial barriers here only and this is only one further step but not the last one. Later this assumption will be removed again. Removing costs The possibility to free the booster from 17,000 kg of propellant means that at least one Block DMcapacity of 15,500 kg of propellant can be removed â€“ the weight of one booster of 2,300 kg is saved. Then a remainder of 1,500 kg of propellant is left and I keep it as a safety margin. This means that 15,500 kg of propellant + 2,300 kg of one BlockDM = 17,800 kg less are to be launched out of the earthian orbit. These are 1.69 CXVboostercombinations. Regarding the luanr roundtrip I calculated 8,758.08 kg to be reuqired to launch one such combination. Again the factor of 1.05 would have to be applied but I am going to use the nonadjusted number here because I will subtract the savings from the numbers of the previous post â€“ this way another safety margin will be included. This also justifies to apply the first post in which I calculated the propellant and the boostersize for the lunar roundtrip. And another small safety margin is already included implictly because to save one BlockDMcapacity also saves a bit propellant. So the 40,000 kg used in the previous post can be reduced by 15,500 kg down to 24,500 kg. I already was iterating the reduction but recognized that this would include an error I will think about â€“ I obviously was neglecting essential earlier thoughts because of the way I am calculating here. The 22,500 kg require only 22,500/24,750 = 0,99 Falcon 9 S9 launches yet which would mean costs of $ 77.22 mio. Using QuickReach 6.81 launches would be required meaning $ 68.1 mio in the expendable case or $ 851,250 in the reusable case. Lunar costs To these the costs of transportation of LOX/LH2 from the lunar surface into the lunar orbit have to be added. In the previous post I got propellant costs for launching the reusable Apollo Lm or the lunar Falcon of $ 13,400.08 as lower boundary and $ 127,619.80 as upper boundary. I also have been introducing another upper boundary of $ 15 mio for sefaty margin reasons. The lower boundary is the reusable Apollo LM I am not sure if it weighes 4 ,700 kg, 4,400 kg or 4,000 kg â€“ but I will apply the number of 3,600 kg here. First this shows that I in principle tend to think about a CXVlike lander regarding weight which would save propellant required per launch. Next this would simplify calculations because it would be astandrad for calculations but thrid this increases the number of launches to deliver propellant into the lunar orbit â€“ which is another safety margin regarding costs. I also will apply the 3,600 kg in the case of the lunar Falcon. This includes a safety margin of costs also since the the Falcon V â€“ earthian â€“ can launch much more than 3,6ÃŸÃŸ kg. The saved and now to be delivered weight of propellant was LOX/cerosene taken as completely cerosene here. This amount has to modified into hydrogen. The saved amount has to be multiplied by 0.77 giving 11,935 kg. To launch that weight into the lunar orbit 3.32 launches are required Each launch including retrun to the lunar surface costs have to be added. This results in Moomlunarorbit transportation costs of $ 42,880.26, $ 408,383.36 and $ 48 mio. These numbers donâ€™t include depreciations yet â€“ the only depreciations included in the calculations up to now are those of the CXV throughout the thread and may be that i in one case included depreciations of the CXVlike vehicle already â€“ there was a post where I compared the depreciations to the costs to show that they have a very small sharre in percent now but will have a significant share in the case of the reusable booster. I showed the number of launches, trips required to keep them low per flight and pointed to alternative ways to kepp them low. But I leave them away here and add them when I complete all the steps by propellant costs and depreciations. Resualting costs So I get the follwoing total costs of the trip including klanding now: Falcon + lower boundary $ 77.22 mio + $ 43,000 = $ 77.263 mio Falcon + upper boundary $ 77.22 mio + $ 409,000 = $ 77.629 mio Falcon + safe boundary $ 77.22 mio + $ 48 mio = $ 125.22 mio QuickReach expendable + lower $ 68.1 mio + $ 43,000 = $ 68.143 mio expendable + upper $ 68.1 mio + $ 409,000 = $ 68.509 mio expendable + safe $ 68.1 mio + $ 48 mio = $ 116.1 mio reusable + lower $ 851,250 + $ 43,000 = $ 894,250 reusable + upper $ 851,250 + $ 409,000 = $ 1.26025 mio reusable + safe $ 851,250 + $ 48 mio = $ 48.851250 mio Including the flight from the earthian surface into the earthian orbit I get: Per passenger these are: 2 pass. Falcon lower $ 38.6315 mio + $ 4 mio = $ 42.6315 mio ïƒ¨ around $ 43 mio 2 pass. Falcon upper $ 38.8145 mio + $ 4 mio = $ 42.8145 mio ïƒ¨ around $ 43 mio 2 pass. Falcon safe $ 62.61 mio + $ 4 mio = $ 68 mio 5 pass. Falcon lower $ 15.4526 mio + $ 4 mio = $ 19.4526 mio ïƒ¨ around $ 20 mio 5 pass. Falcon upper $ 15.5258 mio + $ 4 mio = $ 19.5258 mio ïƒ¨ around $ 20 mio 5 pass. Falcon safe $ 25.44 mio + $ 4 mio = 29.44 mio ïƒ¨ around $ 30 mio 2 pass. QR exp. lower $ 34.0715 mio + $ 4 mio = $ 38.0715 mio ïƒ¨ around $ 38 mio 2 pass. QR exp. upper $ 34.2545 mio + $ 4 mio = $ 38.2245 mio ïƒ¨ around $ 39 mio 2 pass. QR exp. safe $ 58.05 mio + $ 4 mio = $ 62.05 mio ïƒ¨ around $ 62 miu 5 pass. QR exp. lower $ 13.6286 mio + $ 4 mio = $ 17.6286 mio ïƒ¨ around $ 18 mio 5 pass. QR exp. upper $ 13.7018 mio + $ 4 mio = $ 17.7018 mio ïƒ¨ around $ 18 mio 5 pass. QR exp. safe $ 23.22 mio + $ 4 mio = $ 23.22 mio ïƒ¨ aound $ 24 mio 2 pass. QR reu. lower $ 447,125 + $ 25,000 = $ 472,125 ïƒ¨ around $ 500,000 2 pass. QR reu. upper $ 630,125 + $ 25,000 = $ 655,125 ïƒ¨ around $ 700,000 2 pass. QR reu. safe $ 24.425625 mio + $ 25,000 = $ 24.450625 mio ïƒ¨ around $ 25 mio 5 pass. QR reu. lower $ 178,850 + $ 25,000 = $ 203,850 ïƒ¨ around $ 210,000 5 pass. QR reu. upper $ 252,050 + $ 25,000 = $ 277,050 ïƒ¨ around $ 280,000 5 pass. QR reu. safe $ 9.77025 mio + $ 25,000 = $ 9.79524 mio ïƒ¨ around $ 10 mio In the case of the reusable QuickReach the safe upper boundary may be not proper any longer since it seems to express the expendable earthian case more than anything else. So please look at the results for the lower and the upper boundary in particular â€“ at present the numbers are very close to the roundtripcase. This also holds for the expendable case. I am delaying the updated table f steps again because I donâ€™t want to include all these numbers there and want to think about the way to include them until the next post. To a certain degree this will determine how to do the next step also. What I have done in this post can be considered as a germ of an infrastructure to some degree. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Post includes the enhanced table of steps in the APPENDIX
The step I am going to do now toa high degree is a concept of t/Space. In their document Transformational Space Projects Lunar Exploration and Development.htm to be found on their website they say: Quote: The t/Space CEV, because it lands on the lunar surface, can be directly refueled with Moon propellants. (In Apollostyle architectures, the main vehicles remain in orbit and require the creation of specialized surfacetoorbit tankers.) In addition, CEVsized tankers can lift off from the Moon with propellant to take back to LEO to fuel Marsbound ships. (Launching from the Moon back to LEO takes only 5% of the energy required to launch from the Earth up to LEO â€“so it's more efficient for spacebased vehicles to get their propellant from the Moon than from Earth.) Prof. Red Whittaker and his associate Matthew DiGioia at the Robotics Institute of Carnegie Mellon University have developed for t/Space a system of automated devices to mine polar ice and turn it into propellant. Their research indicates that the cost and mass of this system is rapidly repaid by the cost savings of no longer hauling propellant up the steep gravity well from Earth's surface. By the way â€“ this quote also menas that what I am doing here might not mean that large a consumption of limited lunar hydrogen or water depots. For the application here I now modify the concept to be able to do caculations based on what I already did in the previous post(s): 1. The tanker will be landersized and will carry 3,600 kg of propellant. 2. The tanker will go like the CXVlike vehicle goes but not the threemonthstrajectory 3. The propellant delivered to LEO will be used for lunar trips of course â€“ NOT fÃ¼r trips to Mars The first trips until the first touristic trip leaves LEO are left out of consideration here because these first trips are investment here â€“ I still have to think about the question how much of them are lostinspace costs if any but there are no testflights for touristice flights included. So whatâ€™s to be taken into account here are the costs of the launch of the tanker off the lunar surface into orbit, the Trans Earth Injection costs, the Earthian Orbital Insertion costs, the Translunar Injection costs, the Lunar Orbital Insertion costs and the costs to land on the Moon again â€“ these all together are the transportation costs of the propellant now. Since the weight applied here is the same weight used for the CXVlike vehicle the changes of the number can be expected to be caused by three factors only: 1. change of the propellant 2. costs of launches and landings off the lunar surface 3. no costs of propellant transportation from the earthian surface into LEO At the Moon the booster has to be refueled, next the tanker has to be launched â€“ but the propellant requirements to be met at the Moon are increased now because the tanker has to return to the Moon from the Earthian orbit and then to land again. Regarding the propellant of the tanker itself there seems to be no change â€“ it has to go up first and to go down again but simply later in this case. Also the weight of the vehicle isnâ€™t increased because it can be assumed that the passenger capsule is replaces by the tanker simply. Furthermore the tanker will weigh significantly less when it returns to the Moon because the propellant is fueled into another vehicle or into a proepplant depot then. Because of all the above it can be expected that the numbers of previous posts cann be applied to a large degree here. First the transporation costs at the Moon now. According to the previous post they are $ 13,400.08 as lower boundary and $ 127,619.80 as upper boundary. $ 15 mio for sefaty margin reasons. The weight to be delivered into the lunar orbit according to the previous post was 11,935 kg. This amount of propellant is completely consumed when the passenger vehicle has entered the earthian orbit again. Next the propellant to return to the Moon has to be considered. It was 24,500 kg of LOX/cerosene â€“ which has to be modified into LOX/LH2 and then will be taken as if it were LH2 completely. Again I apply the factor of 0.77 and get 18,865 kg â€“ this amount has to be carried from the Moon to avoid transportation from the earthian surface which would be more expensive. The only alternative I can think about this moment is to keep empty tankers waiting in LEO until via permament deliveries the required amount has been stored to return one tanker â€“ this then has to do with logistics and Economics of Storing (correct translation? The german term is â€žLagerhaltung). It would require to apply more complex formulars of Enterprise Economic including interest rate, tiem until exhaustion and more â€“ which I want to avoid at present. But it would show how interesting it will be to establish such a lunarerathian infrastructure for space trips, space fairing and even scintific missions like those of ESA, NASA, ROSKOSMOS, JAXA and others. The additional 18,865 kg cause an increase of the required propellant to do Trans Earth Injection. If I remember correct the weight returned back to Earth consuming the 11,935 kg refueled at the Moon was 10,500 kg â€“ CXV + its booster. Applying the not that safe way of calculating uo linearly this would mean that additional 1.80 * 11.935 kg are required which are 21483 kg. These require 1.4 additional Block DMreplacements weighing 3,220 kg. Linearly these would require another 4,000 kg of propellant if it were a third of the CXV+booster but it is less. Together with the 21,483 and rounded up this would reuslt in 26,000 kg. Since I have gone the unsafe linear way I now assume 40,000 kg which I think to be safe enough. So the delivery of 3,600 kg of propellant into the earthian orbit would require 11.935 kg + 18.865 kg + 40,000 kg = 70,800 kg. To this the amount of 3,600 to be delivered into LEO have to be added â€“ 73,600 kg are to be launched of the Moon by several launches. The number of launches required is 20.44. The the numbers are $ 273,897.64 as lower boundary and $ 2,608,548.72 as upper boundary $ 306,600,000 for safety margin reasons These are the transporations costs for 3,600 kg of propellant. For comparison â€“ the costs to deliver these 3,600 kg from the earthian surface were $ 10 mio via the expendable QuckReach and $ 78 mio * 3,600 kg/24,750 kg = $ 11.35 mio via the Falcon 9 S9 â€“ what a huge saving. To get the transportation costs for one touristical lunar trip including landing 5.8 such delivery are required which results in transportation costs per trip of $ 1,588,606.32 as lower boundary and $ 15,129,582.58 as upper boundary $ 1,778,280,000 for safety margin reasons Obviously the safety margin boundary is astronomical again. And there is no doubt that it is above the transportation costs from the earthian surface got in the previous posts. But that#S no problem in this step â€“ the entrepreneur or the company simply would prefer transportation from the earthian surface if the delivery from the Moon isnâ€™t significantly below the delievery from Earth. So I in this step reduce the safety margin boundary down to $ 65 mio for the total of trnasportation costs for the complete touristic flight â€“ above that the transportation from the earthian surface would be preferred. Per delivery flight this would be a safety margin boundary of $ 65 mio / 5.8 = $ 11,206,896.56. These are the costs to deliver all the propellant into the lunar orbit â€“ the proepllant required to deliver as well as the amount to be delivered. These were 20.44 launches. By this number the$ 11 mio have to be dived again to get the safety margin boundary to be applied instead of the fromer $ 15 mio. The result is $ 548,282.62. This boundary might added to the previous previous step also but I personally think that the $ 15 mio would be accepted in that case because the price still would be significantly below the price Space Adventures have published for the actual and present circumstances. The numbers now are as follows: Costs per lunar flight Persons carried into orbit using expendable QuickReach Lower 2personscase $ 8 mio orbital + $ 1.6 mio to the Moon + $ 43,000 landing = $ 9.643 mio Upper 2personscase $ 8 mio orbital + $ 16 mio to the Moon + $ 409,000 landing = $ 24.409 mio Safety 2personscase $ 8 mio orbital + $ 65 mio to the Moon + $ 550,000 landing = $ 73.55 mio Lower 5personscase $ 20 mio orbital + $ 1.6 mio to the Moon + $ 43,000 landing = $ 21.643 mio Upper 5personscase $ 20 mio orbital + $ 16 mio to the Moon + $ 409,000 landing = $ 36.409 mio Safety 5personscase $ 20 mio orbital + $ 65 mio to the Moon + $ 550,000 landing = $ 85.55 mio Persons carried into orbit using reusable QuickReach Lower 2personscase $ 50,000 orbital + $ 1.6 mio to the Moon + $ 43,000 landing = $ 1.693 mio Upper 2personscase $ 50,000 orbital + $ 16 mio to the Moon + $ 409,000 landing = $ 16.459 mio Safety 2personscase $ 50,000 orbital + $ 65 mio to the Moon + $ 550,000 landing = $ 65.6 mio Lower 5personscase $ 125,000 orbital + $ 1.6 mio to the Moon + $ 43,000 landing = $ 1.768 mio Upper 5personscase $ 125,000 orbital + $ 16 mio to the Moon + $ 409,000 landing = $ 16.534 mio Safety 5personscase $ 125,000 orbital + $ 65 mio to the Moon + $ 550,000 landing = $ 65.675 mio Costs per passenger after lostinspace costs Persons carried into orbit using expendable QuickReach Lower 2personscase $ 4.8215 mio Upper 2personscase $ 12.2045 mio Safety 2personscase $ 32.775 mio Lower 5personscase $ 4.3286 mio Upper 5personscase $ 7.2818 mio Safety 5personscase $ 17.11 mio Persons carried into orbit using reusable QuickReach Lower 2personscase $ 846,500 Upper 2personscase $ 8.2295 mio Safety 2personscase $ 32.8 mio Lower 5personscase $ 353,600 Upper 5personscase $ 3.3068 mio Safety 5personscase $ 13.135 mio For any reason the least number is above the least number of the previous post. This reason I will look for later when I systematically check all numbers etc and put all steps into one system to provide transparency. Interesting also is that the amount got here for Trans Lunar Injection is 18,865 kg while it was 80,791,58 kg for Apollo and was adjusted to CXVlike vehicle + booster at 27,967.55 kg. I will have an eye on that later. Next the costs of the cerosene or the hydrogen itself have to calculated and added for each step and I have to look where the depreciations have to be added also yet. Dipl.Volkswirt (bdvb) Augustin (Political Economist) APPENDIX Table about the steps resulting in a largely reduced price  variable costs only Some of the numbers have changed in comparison to the post(s) because of correction of detected errors and roundings because of limited space Code: step revenue price costs lostin cov. flts price  space lstin cov lost costs spcco inspace orig. 200 mio 100 mio 145 mio 145 mio 55 mio 3 72.5 mio concept CXV in 155 mio 77.5 mio 100 mio 100 mio 55 mio 2 50 mio stead Soyuz reusbl. 135 mio 67.5 mio 80 mio 80 mio 55 mio 2 40 mio booster 129 mio 64.5 mio 74 mio 74 mio 37 mio kept in orbit less 117 mio 58.5 mio 62 mio 62 mio 55 mio 2 31 mio prop.by 108 mio 54 mio 53 mio 53 mio  26.5 mio less weight 3 pass. 105 mio 52.5 mio 50 mio 50 mio 55 mio 1 25 mio ISS etc  96 mio 48 mio 41 mio 41 mio 20.5 mio + pass lunar more 129 mio 64.5 mio 66 mio 66 mio 55 mio 2 37 mio prop.for Moon + reenter erthian orbit deliv. 121 mio 60.5 mio 58 mio 58 mio 55 mio 2 33 mio by VLA+ Quick Reach 5 136 mio 27.2 mio 58 mio 58 mio 55 mio 2 15.6 mio tourists 2 1.5 mio 750000 725000 725000 725000 1 387500 tourists $ 25,000 orbital 5 1.575 mio 315000 725000 725000 725000 1 170000 tourists $ 25000 orbital For the last two rows the depreciations have to be included yet which has been done in an earlier post.  Lunar Trip including Landing 2 215 mio 112 mio 215 mio 215 mio 55 mio 4 83.6 mio tourists 3.4 bio 1.7 bio 3.4 bio 3.4 bio 61  1.7 bio Falcon 5 215 mio 84 mio 215 mio 215 mio 55 mio 4 35.9 mio tourists 3.4 bio 675 mio 3.4 bio 3.4 bio 61 664 mio Falcon 2 195 mio 102 mio 195 mio 195 mio 55 mio 4 74 mio tourists 3.0 bio 1.5 bio 3.0 bio 3.0 bio 54  1.5 bio Quick Reach 5 195 mio 43 mio 195 mio 195 mio 55 mio 4 18 mio tourists 3.0 bio 596 mio 3.0 bio 3.0 bio 54 585 mio Quick Reach 2 3.5 mio 1.8 mio 3.5 mio 3.5 mio 3.5 mio 1 725,000 tourists  37 mio  50 mio  37 mio  37 mio  37 mio  23 mio 25,000 orbital 5 1.8 mio 900,000 1.8 mio 1.8 mio 1.8 mio 1 375,000 tourists  37 mio  19 mio  37 mio  37 mio  37 mio 7.4 mio 25,000 orbital  Lunar Orbital Trip 2 pass. 190 mio 95 mio 127 mio 127 mio 55 mio 3 67.5 mio Falcons 2 pass. 175 mio 87.5 mio 112 mio 112 mio 55 mio 3 60 mio exp. QR 2 pass. 3 mio 1.5 mio 1.4 mio 1.4 mio 1.4 mio 1 800,000 reu. QR 5 pass. 202 mio 40.4 mio 127 mio 127 mio 55 mio 3 29.4 mio Falcons 5 pass. 187 mio 37.5 mio 112 mio 112 mio 55 mio 3 26.5 mio exp. QR 5 pass. 3 mio 600,000 1.4 mio 1.4 mio 1.4 mio 1 320,000 reu. QR  Lunar Trip including Landing  Using lunar propellant reservoirs Numbers except the costs of the propellant itself and except depreciations  lunar propellant ressources used for the lander only 2 pass. 190 mio 95 mio 127 mio 127 mio 55 mio 3 67.5 mio Falcons 2 pass. 175 mio 87.5 mio 112 mio 112 mio 55 mio 3 60 mio exp. QR 2 pass. 3 mio 1.5 mio 1.4 mio 1.4 mio 1.4 mio 1 800,000 reu. QR 5 pass. 202 mio 40.4 mio 127 mio 127 mio 55 mio 3 29.4 mio Falcons 5 pass. 187 mio 37.5 mio 112 mio 112 mio 55 mio 3 26.5 mio exp. QR 5 pass. 3 mio 600,000 1.4 mio 1.4 mio 1.4 mio 1 320,000 reu. QR  lunar propellant ressources used for the lander and return flight into the earthian orbit 2 pass. 141 mio 70.5 mio 86 mio 86 mio 55 mio 2 43 mio Falcons 189 mio 94.5 mio 134 mio 134 mio 3 67 mio 2 pass. 132 mio 66 mio 77 mio 77 mio 55 mio 2 30.5 mio exp. QR 180 mio 90 mio 125 mio 125 mio 3  63 mio 2 pass. 1.9 mio 950,000 944,250 944,250 944,250 1 477,750 reu. QR 2.8 mio 1.4 mio 1.4mio 1.4mio 1.4mio 700,000 104 mio 52 mio 49 mio 49 mio 55 mio 24.5 mio 5 pass. 153 mio 30.6 mio 98 mio 98 mio 55 mio 2 20 mio Falcons 201 mio 40.2 mio 146 mio 146 mio 3  30 mio 5 pass. 144 mio 28.8 mio 89 mio 89 mio 55 mio 2 18 mio exp. QR 192 mio 38.4 mio 137 mio 137 mio 3  28 mio 5 pass. 2.4 mio 480,000 1.2 mio 1.2 mio 1.2 mio 1 240,000 reu. QR  3 mio 600,000 1.5mio 1.5mio 1.5mio 300,000 104 mio 20.8 mio 49.1mio 49.1mio 55 mio 9.8 mio  lunar propellant ressources used for all together 2 pass. 19.4 mio 9.7 mio 9.7mio 9.7mio 9.7mio 1 4.85 mio exp. QR 49 mio 24.5 mio 24.5mio 24.5mio 24.5mio 12.3mio 147.2 mio 73.6 mio 73.6mio 73.6mio 55 mio 2 38.8 mio 5 pass. 43.4 mio 8.7 mio 21.7mio 21.7mio 21.7mio 1 4.34 mio exp. QR 73 mio 14.6mio 36.5mio 36.5mio 36.5mio 7.3 mio 171.2 mio 34.3 mio 85.6mio 85.6mio 55 mio 2 17.2 mio 2 pass. 3.4 mio 1.7 mio 1.7mio 1.7mio 1.7mio 1 850,000 reu. QR 33 mio 16.5 mio 16.5mio 16.5mio 16.5mio 8.25mio 131.2 mio 65.6 mio 65.6mio 65.6mio 55 mio 2 32.8 mio 5 pass. 3.6 mio 720,000 1.8 mio 1.8 mio 1.8 mio 1 360,000 reu. QR 33.2 mio 6.7 mio 16.6mio 16.6mio 16.6mio 3.4 mio 131.4 mio 26.3 mio 65.7mio 65.7mio 55 mio 2 13.14mio END OF APPENDIX 
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Remark Propellant costs Depreciations Hello, Stefan Sigwarth Thank You Very Much for the document you sent me. I didnâ€™t find time yet to read it but will look to make use of it in this thread later. Remark While looking through all posts for the required varable costs and depreciations to be added yet I detected that I have omitted the costs to get into orbit via the CXV in two steps regarding the roundtrip â€“ this I will correct in parallel with checking the calculations for errors and mistakes. In the next post I will list all ways I applied to get too high lower and upper boundaries by. I delay that because it might be that I detect errors and mistakes causing to high numbers also â€“ these I should add to the list with the remark that an error or mistake is behind them. I will have found them then because I will check my doings for errors and mistakes â€“ and I already said in a previous post that I wondered about something and so the check is required and may modify the result(s) up or down. But it may require more time. Because of this the numbers at present should be taken and used as they are. Also thi numbers for propellant itself and the depreciations have to be included into the check and so should be available first. The errored or wrong information from Wikipedia I will handle as error or mistake also â€“ may also be that I misunderstood Wikipedia simply. May be that I also change the way of calculation to keep all within one system â€“ this provides trnasparency or establishes it. May be too that some numbers become smaller while others become smaller â€“ but I already detetced a detail that will cause a reduction. Furthermore I may detect interesting details which result in modifications Propellant costs Using the numbers calculated and used in previous posts I get the follwoing tabel of propellant costs: Propellant costs per flight Code: serv trip boun amount propel price total var. ice dary lant costs round lower 9,300 kg cerosene $ 0.65 $ 6,045 trip round upper 10,230 kg cerosene $ 0.65 $ 6,649.50 trip round 20,883.72 kg cerosene $ 0.65 $ 14,874.418 trip landing lower 50,400 kg cerosene $ 0.65 $ 32,760 landing upper 1,052,345 kg cerosene $ 0.65 $ 684,024.25 orbital 40,000 kg cerosene $ 0.65 $ 26,000 no landing lower total $ 39,400.08 refuel no landing upper total $ 153,619.80 refuel lunar landing lower total $ 47,806.149 refuel lunar landing upper total $ 162,025.869 refuel lunar landing lower total $ 165,350.512 deliv. lunar landing upper total $ 279,570.232 deliv. The impacts of these numbers on the total variable costs per flight are marginal â€“ in principle it can be seen at costs below the $ 2 mioboundary only. Propellant costs per passenger Code: serv trip boun 2 passengers 5 passengers Ice dary round lower $ 3,022.50 $ 1,209 trip round upper $ 3,324.75 $ 1,329.90 trip round $ 7,437.209 $ 2,974.8836 trip landing lower $ 16,380 $ 6,552 landing upper $ 342,012.125 $ 136,804.85 orbital $ 13,000 $ 5,200 no landing lower $ 19,700.04 $ 7,880.016 refuel no landing upper $ 76,809.90 $ 30,723.96 refuel lunar landing lower $ 23,903.0745 $ 9,561.2298 refuel lunar landing upper $ 81,012.9345 $ 32,405.1738 refuel lunar landing lower $ 82,675.256 $ 33,070.1024 deliv. lunar landing upper $ 139,785.116 $ 55,914.0464 deliv. These numbers seem to mean that at the prices below the $ 1 miomark the prices are increased now by more than 10% of what I got in the previous posts.at the upper boundary while the lower boundary has nearly no impact Since the lower boundary is the reusable AopolloLM which might be CXVlike also the pricelevel got in the previously still is valid. So obviously the result is relatively stable after including the propellant costs because they are that small compared to the transportation costs of the propellant. Depreciations I already considered depreciations in another post but ignore that here completely. Since there are a lot of alternative numbers of flights or passengers to depreciate by the selction of one or several particular numbers would be very volunatry. Because of this I select a maximum depreciation from which then a ceratin number of flights and/or passengers results. This way provides additional useful conclusions â€“ those flight numbers and passenger numbers are critical for keeping the costs near the level of variable costs as they are calculated up to now. Thisn way a criterion is got to evaluate later if the world, a company or a vehicle fits into the numbers calculated simply by looking at the number of flights or passegners during the lifetime of the vehicle. The maximum depreciation I am going to apply here is 10% of the total variable cost listed above. The numbers of passengers and flights are to be considered to be independent of each other. Into the investments also have to be included the iinstallations of the hardware â€“ CXVlike vehcile, booster(s) in LEO and lander and tanker at the Moon. The CXVlike vehicle and the booster(s) require only an orbital launch while the lander and the tanker require at least one entire lunar trip. Regarding the lunar installation the following considerations are helping â€“ although they are very raw. The lowerboundarylander weighs between 4,400 kg and 5,000 kg and needs 10,500 kg propellant â€“ usually. At installation less than 10,500 kg will be required because the lander has to land only then. Since I will do a check and then invlove details I donâ€™t look for the required amount here. For descent only significant less propellant is required than for descent plus ascent because the follwoing ascent would require additional propellant which can be left at installation. So I make use of 5,250 kg of propellant here. I add on 5,000 kg and get 10,250 kg at installation â€“ this is less than three times the weight of a CXV. Thatâ€™s the weight I allways was using here. It is the easiest way to use three times a lunar trip including a landing here. The result is too high and so includes a very large safety margin â€“ in particular becasue the weight of the CXV isnâ€™t carried back at installation. The same is valid for the tanker. In other words â€“ I use the booster(s) to be used for carrying tourists for installation also because this menas to apply the reusable hardware that is depreciated by a lot of flghts or passengers. Here are the investments to be depreciated: [code] investm. invested number number passengers flights CXVlike 400 mio 40 80  200 CXVlike 420 mio 53  106  +booster 103 515 lower CXVlike 426 mio 54  108  +booster 104 520 upper CXVlike 460 mio 63  126  +3boost. 70 350 lower CXVlike 478 mio 65  130  +3boost 6052 12104 upper CXVlike 1640 mio 5  10  +62 bst. 8949 17898 lower CXVlike 2012 mio 6  12  +62 bst. 10978 21956 upper CXVlike 800 mio 63  126  + <20bst 5611 11222 lower CXVlike 920 mio 73  146  + <20bst 6452 12904 upper CXVlike 802 mio 63  126  < 20 bst 5558 11116 install. lower CXVlike 1060 mio 73  146  < 20 bst 6392 12784 install upper CXVlike 762 mio 53  106  < 18 bst 7661 15322 install refuel lower CXVlike 1029 mio 60  120  < 18 bst 8750 17500 install refuel upper CXVlike 1604 mio 94  188 < 20 bst 4289 8578 install lower CXVlike 2075 mio 108  216 < 20 bst 4933 9866 install upper [/code The highest numbers allways are required in the case of the lowest variable costs. The numbers fro the installations are got by first dividing the total variable costs by ten and then multiplying them by eleven again. To this the investment into the lander and the CXVlike vechicle are added then. The CXVlike vehicle is taken as $ 400 mio which I consider to be justified because the article â€žEXCLUSIVE: Bigelow Orbital Module Launchedâ€œ ( www.space.com/missionlaunches/060712_ge ... aunch.html ) says: Quote: Bankrolling the expandable space module conceptâ€”now roughly a $75 million investmentâ€”is businessman, Robert Bigelow, owner of the Budget Suites of America Hotel Chain among other enterprises, and head of Bigelow Aerospace. The Genesis Pathfinder 1 has a volume of between 4 m^3 and 5 m^3 so that doesnâ€™t need to be much larger to have the volume the CXV has. Whatâ€™s left are the engines and tanks. May be that this is sufficient. The booster is taken at $ 3 mio to $ 9 mio plus $ 17 mio to install it in space. This moment I recognize that I left away the costs to install the CXVlike vehicle but these costs I only can take as $ 10 mio here â€“ it doesnâ€™t have that much an impact and so I included them while doing the check. The tanker has been included by taking the landerinstalltion two times and taking the tanker as $ 400 mio. Obviuosly the maximum number of potential customers Space Adventures have listed â€“ 1000 â€“ fits into the numbers above. Of course the upper numbers are a multiple of that but those number have their meaning regarding the future. I may have forgotten something I wanted to say here and I also delay a particular remarkk at present. The checks have already been started and it will be included. By the way â€“ there will be a little universe of possibilities to reduce the costs further. About a few I already have an idea. I no way have in mind to calculate them all but I will list them as I have done in the Accumulationsthread. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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