Something closer to a utility truck rather than an extreme sports car with razor thin margins.
Did not Musk say somewhere that there would be approximately 10 - 100 ton cargo flights for every one 100 passenger flight? Cargo being necessary for living areas, food growing units, power stations, fuel manufacturing equipment, mining equipment, long term food storage, etc. To me it would be safer and easier just to send 10 people with each MCT. All vehicles would be the same. All would have cargo and people to work, but not be overloaded with large expensive human habitat area on the vehicles. At least until a fairly large viable colony is built for massive transfers of people.
Quote from: Lars-J on 11/03/2015 04:26 amSomething closer to a utility truck rather than an extreme sports car with razor thin margins.Maybe that's the problem. You're saying a utility truck while Musk, the person who's actually going to decide how the MCT is built, is saying a bus type vehicle (and maybe even a cruise ship vehicle later) that has access to landing pads. You're just on the wrong thread.
Quote from: stoker5432 on 11/03/2015 03:18 pmQuote from: Lars-J on 11/03/2015 04:26 amSomething closer to a utility truck rather than an extreme sports car with razor thin margins.Maybe that's the problem. You're saying a utility truck while Musk, the person who's actually going to decide how the MCT is built, is saying a bus type vehicle (and maybe even a cruise ship vehicle later) that has access to landing pads. You're just on the wrong thread.Ah, the lost art of intentionally mis-reading a post.
Quote from: dror on 11/01/2015 03:58 pmQuote from: Robotbeat on 10/26/2015 10:42 pmMusk is borrowing somewhat from Mars Direct (or is it Semi-Direct?) where an already-fueled ascent vehicle is fueled up on the surface. I don't see a good reason not to have a fueled up vehicle ready when they arrive.I see no reason why to land the fuel production equipment at all. It can do the same job on orbit.you have said it yourself: http://forum.nasaspaceflight.com/index.php?topic=17984.msg620589#msg620589CO/O2 is really neat, but orbital propellant collection is very difficult, and although I certainly support its development, I have absolutely no reason to think that's what SpaceX is planning. In fact, we can be reasonably certain that Raptor will use methane (and certainly it will use some sort of hydrocarbon), and that's what MCT will use.
Quote from: Robotbeat on 10/26/2015 10:42 pmMusk is borrowing somewhat from Mars Direct (or is it Semi-Direct?) where an already-fueled ascent vehicle is fueled up on the surface. I don't see a good reason not to have a fueled up vehicle ready when they arrive.I see no reason why to land the fuel production equipment at all. It can do the same job on orbit.you have said it yourself: http://forum.nasaspaceflight.com/index.php?topic=17984.msg620589#msg620589
Musk is borrowing somewhat from Mars Direct (or is it Semi-Direct?) where an already-fueled ascent vehicle is fueled up on the surface. I don't see a good reason not to have a fueled up vehicle ready when they arrive.
Where is the 12 flight requirement coming from?And suppose there is some requirement. You're implying customer flights, and revenue-generating ones at that, are cheaper than SpaceX flight? For a reusable rocket, test flights are the cheapest option. Quote "A rocket is typically tested only once with a dummy payload with subsequent launches being with paid customers who have satellites." But the whole point was that this is a reusable rocket. Your "typically" applies to expendable rockets that operate in a field where there are worthy second-rate payloads that can both generate revenue and be of lesser value if lost.None of that applies here.If you're building a reusable rocket, then you can test it like you test any aircraft. Send it up (you can test the first stage independently of the rest of the rocket), certify it, then start using it.You can't use "typically" across a paradigm shift.Quote from: Impaler on 11/02/2015 03:53 amQuote from: meekGee on 11/02/2015 02:41 amBut we know that SpaceX, even in the fast build-up scenario, needs to send multiple unmanned payloads to Mars first.BFR is an Earth-to-orbit vehicle, and so you're not risking a two-year delay when you first launch it. Yes, the first launch may not carry an MCT, but by the same token, it won't carry any other one-of-a-kind payload. Being reusable, it's a complete non-brainer to fly a dummy payload. If it works, you you relaunch. It it doesn't, good thing you didn't.Actually, once you're shifted to a reusable rocket, test flights are cheap enough that you could do multiple dummy payloads before you put an MCT on top - and it's not like you wasted multiple rockets doing it.So I think the "build up flight history" argument doesn't apply.It will be cheaper per kg to LEO, but it is by no means going to be cheap enough to build an adequate flight history (~12 launches) with nothing but dummy payloads all at SpaceX's own expense. A rocket is typically tested only once with a dummy payload with subsequent launches being with paid customers who have satellites. I expect prices of ~200 million per launch even with recovery (about $1000 per kg, Musk's optimistic goal) so it would cost Billions to do these launches without customers.I expect that 1st stage recovered will be attempted on every single BFR flight with very likely full success from the start. If first stage recovery works on the dummy flight and is declared a solved problem then the price point will likely be set such that SpaceX is fully covering the cost of the 2nd stage in case it is lost as I expect 2nd stage recover to require a long campaign of attempts with lots of failures and redesigns as we have seen with F9, the customer will not care any more about the success or failure of these recovery attempts any more then they care about 1st stage recovery attempts now. By the time you have your 12 flight history your close to getting the 2nd stage to recover reliably and can drop the price to perhaps 100 million and try to get more volume.
"A rocket is typically tested only once with a dummy payload with subsequent launches being with paid customers who have satellites."
Quote from: meekGee on 11/02/2015 02:41 amBut we know that SpaceX, even in the fast build-up scenario, needs to send multiple unmanned payloads to Mars first.BFR is an Earth-to-orbit vehicle, and so you're not risking a two-year delay when you first launch it. Yes, the first launch may not carry an MCT, but by the same token, it won't carry any other one-of-a-kind payload. Being reusable, it's a complete non-brainer to fly a dummy payload. If it works, you you relaunch. It it doesn't, good thing you didn't.Actually, once you're shifted to a reusable rocket, test flights are cheap enough that you could do multiple dummy payloads before you put an MCT on top - and it's not like you wasted multiple rockets doing it.So I think the "build up flight history" argument doesn't apply.It will be cheaper per kg to LEO, but it is by no means going to be cheap enough to build an adequate flight history (~12 launches) with nothing but dummy payloads all at SpaceX's own expense. A rocket is typically tested only once with a dummy payload with subsequent launches being with paid customers who have satellites. I expect prices of ~200 million per launch even with recovery (about $1000 per kg, Musk's optimistic goal) so it would cost Billions to do these launches without customers.I expect that 1st stage recovered will be attempted on every single BFR flight with very likely full success from the start. If first stage recovery works on the dummy flight and is declared a solved problem then the price point will likely be set such that SpaceX is fully covering the cost of the 2nd stage in case it is lost as I expect 2nd stage recover to require a long campaign of attempts with lots of failures and redesigns as we have seen with F9, the customer will not care any more about the success or failure of these recovery attempts any more then they care about 1st stage recovery attempts now. By the time you have your 12 flight history your close to getting the 2nd stage to recover reliably and can drop the price to perhaps 100 million and try to get more volume.
But we know that SpaceX, even in the fast build-up scenario, needs to send multiple unmanned payloads to Mars first.BFR is an Earth-to-orbit vehicle, and so you're not risking a two-year delay when you first launch it. Yes, the first launch may not carry an MCT, but by the same token, it won't carry any other one-of-a-kind payload. Being reusable, it's a complete non-brainer to fly a dummy payload. If it works, you you relaunch. It it doesn't, good thing you didn't.Actually, once you're shifted to a reusable rocket, test flights are cheap enough that you could do multiple dummy payloads before you put an MCT on top - and it's not like you wasted multiple rockets doing it.So I think the "build up flight history" argument doesn't apply.
You're implying customer flights, and revenue-generating ones at that, are cheaper than SpaceX flight? For a reusable rocket, test flights are the cheapest option.
Methane and O2 can be produced on orbit provided a supply of H2 from the surface.that way you get 11 times the fuel on orbit then what you launch.
Quote from: dror on 11/03/2015 06:11 pmMethane and O2 can be produced on orbit provided a supply of H2 from the surface.that way you get 11 times the fuel on orbit then what you launch.Errr, where does the carbon and oxygen come from?
1. Fuel will be needed to maintain a low orbit, constant adjustments. It would be easier and quicker just to land, refuel, then launch back. 2. Mars' atmosphere isn't that thick. An orbit low enough to scoop enough atmosphere is probably going to be too low, like to stay in orbit maybe as low as 20 or 30 miles. Might as well land and relaunch.
You should not be so easy to dismiss an idea. check out that thread and see that a lot of work was put into atmospheric scooping starting in the 60's or so. Quote from: spacenut on 11/03/2015 07:09 pm1. Fuel will be needed to maintain a low orbit, constant adjustments. It would be easier and quicker just to land, refuel, then launch back. 2. Mars' atmosphere isn't that thick. An orbit low enough to scoop enough atmosphere is probably going to be too low, like to stay in orbit maybe as low as 20 or 30 miles. Might as well land and relaunch.1. fuel will be there. that's the point. some of it will be used to maintain altitude and some will be gained.2. than have a bigger scoop or an eliptical orbit.
Can we leave the crackpot ideas off this forum thread? The idea that you can scoop enough enough atmosphere to generate enough propellant to overcome the friction and *also* generate extra propellant is just two degrees short of a perpetual motion machine. It has been studied many times and also immediately rejected as many times.
Quote from: Lars-J on 11/03/2015 09:14 pmCan we leave the crackpot ideas off this forum thread? The idea that you can scoop enough enough atmosphere to generate enough propellant to overcome the friction and *also* generate extra propellant is just two degrees short of a perpetual motion machine. It has been studied many times and also immediately rejected as many times.Using SEP, assuming highest performance panels and highest possible thrust drive, the raw numbers just barely work. Although we're talking a scoop in the hundred of metres length range. TRL is maybe 2, if we're generous. It doesn't work with chemical engines, therefore you can't use the scooped fuel to power the SEP. So in practice, TRL is zero or negative. So yeah, it doesn't belong here.
-snip-From Hop_David's spreadsheet, I offer five cases:Straight Hohmann with propulsive capture - 259 daysA straight-up Hohmann burn from a {1AUx1AU heliocentric, 300km x 300km altitude geocentric} orbit to a {1.524AUx1.524AU heliocentric, 3697x23459km semimajor axis aereocentric} coplanar orbit using a {1AUx1.5240001AU heliocentric} transfer costs 510m/s plus change on Earth departure, over and above the 3220m/s escape velocity burn. Then to propulsively capture into a highly elliptical Mars orbit it costs 1021m/s. Conservative aerobraking can reduce that ellipse down to LMO for another 668m/s that we don't need to pay, where standard EDL penalties apply (I'll assume them to be 2km/s for purposes of this discussion).Total: 3220 + 510 + 1021 + 2000 = 6751m/sWet to drymass ratio at 380s Isp: 6.12 to 1The spreadsheet author's suggested non-Hohmann trajectory with propulsive capture - 102 daysThe author would reduce perihelion in order to balance the perigee and periaerion burn into something sensible, assuming that propulsive capture is necessary. He uses non-prograde burns or a suboptimal burn time in Earth orbit, which is highly inefficient, to balance out the Earth and Mars sides for minimal total dV. From a {1AUx1AU heliocentric, 300km x 300km altitude geocentric} orbit to a {1.524AUx1.524AU heliocentric, 3697x23459km semimajor axis aereocentric} coplanar orbit using a {0.87AUx1.8AU heliocentric} transfer orbit costs 3146m/s plus change on Earth departure, over and above the 3220m/s escape velocity burn. Then to propulsively capture into a highly elliptical Mars orbit it costs 4668m/s. Conservative aerobraking can reduce that ellipse down to LMO for another 668m/s that we don't need to pay, where standard EDL penalties apply (I'll assume them to be 2km/s for purposes of this discussion).Total: 3220 + 3146 + 4668 + 2000 = 13034m/sWet to drymass ratio at 380s Isp: 33.1 to 1A 100 day semi-Hohmann transit given perfect free aerocapture - 100 daysFrom a {1AUx1AU heliocentric, 300km x 300km altitude geocentric} orbit to a {1.524AUx1.524AU heliocentric, 3697x23459km semimajor axis aereocentric} coplanar orbit using a {1AUx3.31AU heliocentric} transfer orbit costs 2241m/s plus change on Earth departure, over and above the 3220m/s escape velocity burn. Then to propulsively capture into a highly elliptical Mars orbit it costs 9276m/s, but we're not going to do that: Instead, magnetoshell or some other aerocapture technology is going to do that all in one go; Then it's going to, in the same step, further reduce the elliptical orbit down to LMO for another 668m/s that we don't need to pay, then go directly into EDL, where standard EDL penalties apply (I'll assume them to be 2km/s for purposes of this discussion).Total: 3220 + 2241 + 2000 = 7461m/sWet to drymass ratio at 380s Isp: 7.41 to 1A 100 day semi-Hohmann transit given propulsive capture - 100 daysExamining the previous proposition without the non-prograde burns that the spreadsheet author makes. From a {1AUx1AU heliocentric, 300km x 300km altitude geocentric} orbit to a {1.524AUx1.524AU heliocentric, 3697x23459km semimajor axis aereocentric} coplanar orbit using a {1AUx3.31AU heliocentric} transfer orbit costs 2241m/s plus change on Earth departure, over and above the 3220m/s escape velocity burn. Then to propulsively capture into a highly elliptical Mars orbit it costs 9276m/s. Conservative aerobraking can reduce that ellipse down to LMO for another 668m/s that we don't need to pay, where standard EDL penalties apply (I'll assume them to be 2km/s for purposes of this discussion).Total: 3220 + 2241 + 9276 + 2000 = 16737m/sWet to drymass ratio at 380s Isp: 89.3 to 1A reasonable 180 day near-Hohmann transfer with mild aerocapture - 180 daysFrom a {1AUx1AU heliocentric, 300km x 300km altitude geocentric} orbit to a {1.524AUx1.524AU heliocentric, 3697x23459km semimajor axis aereocentric} coplanar orbit using a {1AUx1.652AU heliocentric} transfer orbit costs 655m/s plus change on Earth departure, over and above the 3220m/s escape velocity burn. Then to propulsively capture into a highly elliptical Mars orbit it costs 2443m/s, but we're not going to do that: Instead, magnetoshell or some other aerocapture technology is going to do that. Conservative aerobraking can reduce that ellipse down to LMO for another 668m/s that we don't need to pay, where standard EDL penalties apply (I'll assume them to be 2km/s for purposes of this discussion).Total: 3220 + 655 + 2000 = 5875m/sWet to drymass ratio at 380s Isp: 4.84 to 1
Quote from: Robotbeat on 10/15/2015 10:37 pmQuote from: Impaler on 10/15/2015 02:17 amQuote from: Burninate on 10/14/2015 11:14 pmWould anyone like to do the work of reversing that spreadsheet so we could look at more refined estimates of return delta V? I might try it at some point, but not today.Orbits are time reversible so don't we just need to look at the 'Vinf at mars' and calculate the DeltaV needed to archive that escape velocity from mars surface, then we would (if we pointed ourselves in the right direction) be headed back down the equivalent half of the outbound orbit and we should reach Earth in the specified time and with the specified Vinf so we know what we need to do to aerocapture at Earth as well.It would be nice to have this done for us on the spreadsheet though.They're reversible, but you're forgetting the Oberth effect: because on the way from Earth to Mars, you can dump your exhaust in a deeper gravity well than Mars to Earth, it takes less delta-v.I was just referring to the heliocentric portion of the transit being reversible, we would obviously need to take mars's smaller gravity well into account, that's why I said the Vinf at mars is needs to be looked at. Here let me take some of Burnate's example and reverse them into mars-Earth transits and show how I would do it.Straight Hohmann with propulsive capture - 259 daysVinf at mars is 2.652818827 km/s that is our velocity at this point in the heliocentric orbit before we enter the mars gravity well, naturally we accelerate when falling into that gravity well but we would just come out again with the same speed aka were hyperbolic. To get back to Earth we need that same heliocentric speed (pointed back in now) when leaving mars to return to Earth in 259 days. To determine this we just take the escape velocity at mars surface 5 km/s, square it along with the needed Vinf add them together and get the square root which is 5.660 km/s. Likewise we can use the previously 'escape' burn from Earth to know how much deceleration we need to shed to go into that high orbit at Earth after which we will aerobrake down to LEO, I'll assume that we can be refueled by tanker at this point in order to do our Earth landing as that will avoid having to blast off from mars with propellant we wont need until now in LEO where we know their is a continuous refueling process in place.Earth return reversalTotal: 5660 + 510 = 6170m/sWet to drymass ratio at 380s Isp: 5.24 to 1The spreadsheet author's suggested non-Hohmann trajectory with propulsive capture - 102 daysEarth return reversalTotal 9241 + 3145 = 12386 m/sWet to drymass ratio at 380s Isp: 27.82 to 1A 100 day semi-Hohmann transit given perfect free aerocapture - 100 daysEarth return reversalTotal 13816 m/sWet to drymass ratio at 380s Isp: 40.85 to 1This can be improved over simply reversing the outbound orbit which was {1AUx3.31AU heliocentric}, I've found {.76AUx1.67AU heliocentric} with the same travel time which can decrease the mars escape burn substantially but at the cost of raising Earth side capture needs.Earth return reversalTotal 8792 m/sWet to drymass ratio at 380s Isp: 10.6 to 1A 100 day semi-Hohmann transit given propulsive capture - 100 daysEarth return reversalTotal 13816 + 2241 = 16057 m/s Wet to drymass ratio at 380s Isp: 74.57 to 1Using my alternative trajectoryTotal 8792 + 4597 m/s = 13389Wet to drymass ratio at 380s Isp: 36.42 to 1A reasonable 180 day near-Hohmann transfer with mild aerocapture - 180 daysEarth return reversalTotal 7047 m/sWet to drymass ratio at 380s Isp: 6.63 to 1Conclusions: There seems to be no viable fast (100 day) returns from mars, unlike on the outbound even with perfect free aerocapture doesn't allow it at any kind of propellant fraction and DeltaV which would be believable. The problem is that were starting from mars surface unlike LEO in the the outbound assumptions, that's putting us around 3600 m/s behind where we would be. If we had propellant depots in mars orbit fast return would be a reasonable number if we also had free aerocapture as the wet/dry ratio would be 4:1, which would require something around 300 mT of propellant in mars orbit a huge amount.
Quote from: Impaler on 10/15/2015 02:17 amQuote from: Burninate on 10/14/2015 11:14 pmWould anyone like to do the work of reversing that spreadsheet so we could look at more refined estimates of return delta V? I might try it at some point, but not today.Orbits are time reversible so don't we just need to look at the 'Vinf at mars' and calculate the DeltaV needed to archive that escape velocity from mars surface, then we would (if we pointed ourselves in the right direction) be headed back down the equivalent half of the outbound orbit and we should reach Earth in the specified time and with the specified Vinf so we know what we need to do to aerocapture at Earth as well.It would be nice to have this done for us on the spreadsheet though.They're reversible, but you're forgetting the Oberth effect: because on the way from Earth to Mars, you can dump your exhaust in a deeper gravity well than Mars to Earth, it takes less delta-v.
Quote from: Burninate on 10/14/2015 11:14 pmWould anyone like to do the work of reversing that spreadsheet so we could look at more refined estimates of return delta V? I might try it at some point, but not today.Orbits are time reversible so don't we just need to look at the 'Vinf at mars' and calculate the DeltaV needed to archive that escape velocity from mars surface, then we would (if we pointed ourselves in the right direction) be headed back down the equivalent half of the outbound orbit and we should reach Earth in the specified time and with the specified Vinf so we know what we need to do to aerocapture at Earth as well.It would be nice to have this done for us on the spreadsheet though.
Would anyone like to do the work of reversing that spreadsheet so we could look at more refined estimates of return delta V? I might try it at some point, but not today.