Climber & Rappeller UtilityYes, a climber would need something other than chemical energy. Even HVDC power lines are inefficient and unsuitable over MSE distances. It may be that areosync PV + a superconducting tether power line will be needed, to power an MSE climber system with performance superior to Mars launch.A cargo descent rappeller is more efficient than a climber: it uses 0 kW. And it does so while saving the tremendous energy that would otherwise be devoted to propellant manufacture on Mars, for the cargo ship's surface launch. There's utility in that, yes?
I think a more likely use for a Rapeller could be with a 1400 km tether to drop cargo to LMO in order to build infrastructure there.
The projectiles are guided by the magnetic rings, in your drawing they would be the sections connected to the ground-cables. Between, the projectiles are free-flying. The meteors achieve nothing. See below.
Okay so we've eliminated the enclosing ring continuously accelerating the projectilesThen the projectiles flying at faster than orbital velocity would not be flying a nice circular orbit from one platform to another.You would have to aim the ricochet to hit the next platform.
It wouldn't work if the particles were just in a natural circular orbit.
But explaining how it works is off-topic for this thread, especially zombied from back in August. If you want to dig around, there might be an existing thread in Advanced Concepts that would be more appropriately re-animated.
No TankersNotably, one of the benefits of the Omaha Trail proposal is that cargo flights can be launched without dedicated tanker ships. No tankers at all. Just returning cargo ships.This could be especially beneficial to the construction phase of SpaceX's Mars City, which could take decades. The work might go 10x faster at Omaha Crater, but still, it's a benefit.Cargo flight staging. Deimos propellant. Mars Lift space elevator in gold.
By establishing a propellant depot in the asteroid belt or one of the moons of Jupiter, you can make flights from Mars to Jupiter, no problem.
...produce propellant [from water in the lunar soil]. You’d put that in a propellant depot in lunar orbit to allow humans to go anywhere in the solar system.
That slide didn't garner comment, but it shows in-orbit depot use, an ambition at SpaceX and elsewhere.
Here the depot is at Deimos, with propellant shuttled to LEO. Typical #s, allowing notional 55 t propellant reserves:(1) Burn 397 t of 1008 t propellant after DRL, for transit and LEO circularization.
(2) Launch cargo to LEO.(3) Transfer 540 t in LEO.
(4) Burn 16 t for EDL.
Option: To double payload, (1) is performed with 1066 t propellant load, twice, filling tanks in (3). A second Earth launch transfers the second cargo. With full tanks (5) delivers 300 t to Arestation.
All is done without an Earth tanker fleet, tanker booster fleet, and their propellant. Cost savings would follow efficiency.
False. My previous post was directly discussing that slide.Your architecture expects you to then fully fuel that other ship for it to transfer to Marscompare with SpaceX's architecture apples to apples, 50 tons return to Earth
Not sure why you are limiting yourself to 1008 t
to just do the 1.55 km/s burn to get on an Earth trajectory it would take 399 t with your claimed propellant load and the 50 t cargo needed for this architecture to be compared to SpaceX's.This does not include losses in transit or LEO circularization.
I'm curious where you got this number, [Burn 16 t for EDL.]
I don't think anyone outside SpaceX has a good idea how much fuel is needed forth this on BFR
No, you didn't mention the slide, and your text is not consistent with that flight plan.
There's no need to "fully fuel" cargo ships in LEO. That's required only for crewed missions, where delta-v must be maximized. Different flight plan.
Also there's no cargo on the return flight.
These cargo ships never land on Mars. So unless you assume a booming market for Deimos cinder block at Home Depot, return cargo is naturally 0 t.
Quote from: meberbs on 01/05/2018 02:49 pmNot sure why you are limiting yourself to 1008 t Because that's the minimum needed. We add more for options, such as doubled payload.
Quote from: meberbs on 01/05/2018 02:49 pm to just do the 1.55 km/s burn to get on an Earth trajectory it would take 399 t with your claimed propellant load and the 50 t cargo needed for this architecture to be compared to SpaceX's.This does not include losses in transit or LEO circularization. As we noted previously, perihelion requires ~1.55 km/s, aphelion requires ~0.8 km/s. "Typical" #s are intermediate, not the extremes. For example, 1.16 km/s.
Then we tacked on 0.5 km/s for circularization. (Your 0.94 km/s reasoning isn't right, and typical #s are smaller.)
As for losses, long-duration tankage scenarios expect some ZBOT method. Any further loss would get some of the 92 t left at Deimos.
Quote from: meberbs on 01/05/2018 02:49 pmI'm curious where you got this number, [Burn 16 t for EDL.]ITS drops 99%+ of KE aerodynamically, leaving only a landing burn. We used a 40 s hover.
Redo the cargo flight plan to see how the numbers improve.
Quote from: LMT on 01/06/2018 09:01 pmNo, you didn't mention the slide, and your text is not consistent with that flight plan. I shouldn't have to explicitly mention the slide, it should be obvious that is what I was discussing, and it follows your mission plan exactly, other than pointing out that you have entirely insufficient fuel.
Quote from: LMT on 01/06/2018 09:01 pmThen we tacked on 0.5 km/s for circularization. (Your 0.94 km/s reasoning isn't right, and typical #s are smaller.)The 0.94 number has nothing to do with the LEO circularization, and everything to do with the fact that you used 0.94 km/s to get from Deimos to an elliptical orbit with a periapsis close to Mars. In the best case scenario, that is also the orbit you would end up in after an aerocapture maneuver (actually you would need a lower periapsis to be deep enough in the atmosphere, but that just makes things worse for you). Circularizing that elliptical orbit to match Deimos takes exactly the same delta-v as entering that orbit from Deimos took in the other direction.
Not at all. The given, specific cargo flight plan has no requirements or assumptions for full LEO tanks or 50 t return cargo -- and of course in this flight plan there's no Mars landing for a 50-t cargo pickup anyway.
So our mass and delta-v numbers work as given, with corresponding efficiency improvements over SpaceX baseline. The numbers shouldn't be very surprising, especially given the familiarity of the route and spacecraft.
Quote from: meberbs on 01/06/2018 11:32 pmQuote from: LMT on 01/06/2018 09:01 pmThen we tacked on 0.5 km/s for circularization. (Your 0.94 km/s reasoning isn't right, and typical #s are smaller.)The 0.94 number has nothing to do with the LEO circularization, and everything to do with the fact that you used 0.94 km/s to get from Deimos to an elliptical orbit with a periapsis close to Mars. In the best case scenario, that is also the orbit you would end up in after an aerocapture maneuver (actually you would need a lower periapsis to be deep enough in the atmosphere, but that just makes things worse for you). Circularizing that elliptical orbit to match Deimos takes exactly the same delta-v as entering that orbit from Deimos took in the other direction.That reasoning is quite wrong. The delta-v coming off the DRL by no means sets the delta-v for aerocapture circularization. For one thing, the DRL launch vector is certainly not the most efficient for lowering periapsis; so it's not the delta-v that one must "take". That's easy to see with the example GMAT script: tweak the DRL VNB to reach the same periapsis using much less than 0.94 km/s delta-v....Second, the mission literature shows you what's actually needed for circularization. In one Lockheed Martin flight plan the spacecraft changes plane and circularizes to Deimos with a pair of burns totaling 607 m/s delta-v.
But the rocket equation and orbit changes are, by themselves, OT. Focus should be on possibilities for the Mars Space Elevator and related tether systems.
(1) Burn 397 t of 1008 t propellant after DRL, for transit and LEO circularization.
I just used the [DRL] number you provided for simplicity, assuming that you had bothered to orient the vector in close to optimal direction.
Quote from: LMT on 01/10/2018 06:41 amThe given, specific cargo flight plan has no requirements or assumptions for full LEO tanks or 50 t return cargo -- and of course in this flight plan there's no Mars landing for a 50-t cargo pickup anyway. 50 tons return cargo is part of the SpaceX baseline you claim to be comparing to, you can't just throw out capability and claim your system is equivalent but more efficient.Full LEO tanks isn't an assumption, it is a fact of what SpaceX has presented as their plan.
The given, specific cargo flight plan has no requirements or assumptions for full LEO tanks or 50 t return cargo -- and of course in this flight plan there's no Mars landing for a 50-t cargo pickup anyway.
"Mars is really begging for a space elevator," said Manning. "I think it has great potential. That would solve a lot of problems, and Mars would be an excellent platform to try it."
Quote from: meberbs on 01/06/2018 11:32 pmI just used the [DRL] number you provided for simplicity, assuming that you had bothered to orient the vector in close to optimal direction.? The DRL vector isn't remotely "close to optimal direction", obviously. You don't seem to understand VNB impulse. You might "look it up," as you put it. OT here.
Quote from: meberbs on 01/10/2018 08:41 amQuote from: LMT on 01/10/2018 06:41 amThe given, specific cargo flight plan has no requirements or assumptions for full LEO tanks or 50 t return cargo -- and of course in this flight plan there's no Mars landing for a 50-t cargo pickup anyway. 50 tons return cargo is part of the SpaceX baseline you claim to be comparing to, you can't just throw out capability and claim your system is equivalent but more efficient.Full LEO tanks isn't an assumption, it is a fact of what SpaceX has presented as their plan.Oh, it's fair comparison because equivalent and properly calculated. Same ship, same cargo, out and back, with and without Omaha Trail infrastructure. "150 out / 0 back" is naturally the baseline ITS requirement.
SpaceX hasn't proposed an Omaha Trail themselves. But if tomorrow SpaceX proposed it under their own logo, I'm pretty sure no one would say, "But this isn't the plan! They can't claim this is more efficient!" More likely folks would brainstorm ways to deliver tunnel-boring machines and Tesla Roadsters by elevator.
Quote from: stefan r on 07/18/2017 10:53 pmSome of the containers get loaded in Shanghai and travel through the Panama canal before getting unloaded from container ships in New Orleans. Some of the containers have a thousand kilometer trips before and after shipping. Some of the components in the products already made ocean trips before assembly. Of course some containers could be Huston to New Orleans. So you're not just talking about the cranes at the New Orleans harbor. You are also talking about all the ships that move between New Orleans and points throughout the planet. You're also talking about the Panama Canal which is rather massive.Also a ship traveling on the Atlantic has different fuel requirements than a tug moving stuff to different orbits. Delta V to move a mass from Phobos to Low Mars Orbit is about 1.2 km/s. By my arithmetic it would take about 3.1e15 kilograms of hydrogen/oxygen bipropellent. The world's annual production of oil was about 77,500,000 barrels per years as of 2014. That comes to about 4.4e12 kg of oil.So the hydrogen/oxygen bipropellent to move Phobos would need to mass about 700 times as much as the world's annual oil production.What is the mass of the infrastructure you imagine?This is an important factor that people seem to want to ignore.
Some of the containers get loaded in Shanghai and travel through the Panama canal before getting unloaded from container ships in New Orleans. Some of the containers have a thousand kilometer trips before and after shipping. Some of the components in the products already made ocean trips before assembly. Of course some containers could be Huston to New Orleans.
The real issue that you are failing to address is that you don't have the available delta V
Quote from: meberbs on 01/12/2018 06:45 amThe real issue that you are failing to address is that you don't have the available delta VNo, our very first thread presentation gave the novel propellant and delta-v numbers for the Omaha Trail. Anyone could plug those numbers into the rocket equation, to compare against famous Earth/Mars baseline, and verify. That much is easy.You, Paul451 and Lar imagined otherwise, and ignored the presentation numbers.
An EM DRL delta-v of 1.9 km/s could, if vectored optimally, execute Earth-return without rocket propulsion. Getting that vector is the first trick.
This could only support 1 launch every 2.5 days, so a maximum of about 25 launches assuming a 2 month window...By the time there is resources at Mars for a project on that scale, we will want more than 30 BFS per year of combined cargo and crew.
Quote from: meberbs on 03/06/2018 02:20 pmThis could only support 1 launch every 2.5 days, so a maximum of about 25 launches assuming a 2 month window...By the time there is resources at Mars for a project on that scale, we will want more than 30 BFS per year of combined cargo and crew. ? The DRL is always shown with paired launches, for balance; it's twice the launch rate you imagined.
That's the rate for one pendular DRL -- and there's no cosmic limit of one.i.e., roll out another.
The envisioned pendular DRL delivers full tanks to Earth aerobrake.
As cargo ships unload only at Arestation (5), they have no return cargo.
I haven't seen you state with words that it is always dual launches before now.
Each pair of ITS tankers launching on the pendular DRL aerobrakes at Earth...
We have plugged the numbers in for you in this thread. They do not add up
Can an ITS cargo ship load enough water to do the job, and achieve perfect delivery efficiency? We plug numbers into the rocket equation and find...