Talk-Polywell.org

Tom Ligon wrote:Yes, Mars has an atmosphere that is largely CO2, but not much atmosphere. And using solar to convert it to methane is plagued by not having much solar, either.

it may be a better strategy to boot-strap power production first.

i mean sure, start with a basic converter if just for oxygen production.

but then i'd say work on an in-situ solar membranes/concentrators/whatever (and wires) so you can start multiplying your energy capacity.

then you build you have sustainable greenhouse, more room, robotics repair...

plastics...

seems like it'd be a while until rocket fuel production would be the next thing to scale up

then again, seems like a while before you'd be sending 100 people at a time...

maybe creating a solar furnace, or multiple, might be a good early step for bootstrapping. since i imagine most of the energy use in materials production would be for heating.

better conversion efficiency if you don't convert it at all.

and much easier to make mirrors (very thin layer of aluminum or silicon-carbide) than photo-voltaics.

and lets be clear - there is no wood on mars.

happyjack27 wrote: 3. fuel cost per deltav is not linear - fuel has mass. so your fuel requirement is going to grow a lot faster than your deltav requirement. a deltav saving of say 3 to 1 is a fuel saving of much more than that. exponentially. https://en.wikipedia.org/wiki/Tsiolkovs ... t_equation

You are using this incorrectly. You cannot "save" deltav, as it is a fixed value for any specific flight path. Earth -> LEO -> TLI -> LLO -> Lunar Landing is always the same number, no matter what vehicle you send on that path. You are correct that the fuel needed changes, but that is directly related to vehicle mass/fuel mass combined with the Isp of the engines. No matter what combination of engine/fuel/vehicle-size you pick, the deltav value is always the same. For a return trip (Luna -> LLO -> TEI -> LEO -> Earth Landing) it is the exact same deltav total, however, you can use the Earth's atmosphere instead of shipboard fuel for some of the required change in velocity. You are still not 'saving deltav', though -- you are saving fuel (and thus mass).

edit: Rereading stuff, if you are referring to the deltav capabilities of the vehicle then you are not actually 'wrong'. If that is the case, then yes, you can 'save deltav' in that you can do things in how you design the flight path such that a change in velocity without using fuel happens (aerocapture/-brake, gravity assist, etc), saving fuel which can then be used for additional changes in velocity.

krenshala wrote:

happyjack27 wrote: 3. fuel cost per deltav is not linear - fuel has mass. so your fuel requirement is going to grow a lot faster than your deltav requirement. a deltav saving of say 3 to 1 is a fuel saving of much more than that. exponentially. https://en.wikipedia.org/wiki/Tsiolkovs ... t_equation

You are using this incorrectly. You cannot "save" deltav, as it is a fixed value for any specific flight path. Earth -> LEO -> TLI -> LLO -> Lunar Landing is always the same number, no matter what vehicle you send on that path. You are correct that the fuel needed changes, but that is directly related to vehicle mass/fuel mass combined with the Isp of the engines. No matter what combination of engine/fuel/vehicle-size you pick, the deltav value is always the same. For a return trip (Luna -> LLO -> TEI -> LEO -> Earth Landing) it is the exact same deltav total, however, you can use the Earth's atmosphere instead of shipboard fuel for some of the required change in velocity. You are still not 'saving deltav', though -- you are saving fuel (and thus mass).

edit: Rereading stuff, if you are referring to the deltav capabilities of the vehicle then you are not actually 'wrong'. If that is the case, then yes, you can 'save deltav' in that you can do things in how you design the flight path such that a change in velocity without using fuel happens (aerocapture/-brake, gravity assist, etc), saving fuel which can then be used for additional changes in velocity.

i was talking about the deltav of the path. i.e. Earth -> LEO is one path, landed at moon -> GTO is a different path. I know deltav is constant for a given path. you're going form one inertia frame to another. deltav is essentially the "distance" between these frames as measured by change in velocity.

I was analyzing the economics of refueling from the moon, rather than from earth.

If the craft you're refueling is in LEO, it's cheaper to refuel from earth, because you can aerobrake. If it's in GTO (and you can rendezvous in GTO), it's cheaper from the moon, though you need to reserve 3.2 delta-v for a return trip (whereas refueling from earth you don't).

Due to the rocket equation, this linear savings in delta-v translates to an exponential fuel savings and/or a linear dry mass capability increase.

At about 2/3 of the delta-v to refuel from earth to LEO, to deliver the same amount of fuel from moon to GTO would take 2/3rd as many trips. (since each trip you could carry 3/2 as much)

I goofed that - for moon to GTO refueling, you only have to carry the payload for half the trip.

So presuming the payload is the bulk of the dry mass, it's more like 3x the deliverable mass per trip. So you'd only need 1 trip to refuel a SpaceX interplanetary.

That's assuming you can somehow get a booster stage on there, (which doesn't need nearly as much thrust, and should e optimized vacu rather than sea level.)

Seeing as though you needn't contend with atmosphere, you might as well make your moon-based refueler looked like Armadillo aerospace's "pixel" - four spherical tanks and a rocket. Thus minimizing it's dry mass.

Still could probably benefit from two stages - with the fuel mass and thrust ratios choosen intelligently.

And regarding the rocket equation and Elon's claim that multiple launches are more efficient:

Imagine two IPS's, stapled together. Now their fuel cost would be double and their payload delivery would be double.

Now let's say instead of stapling them together you launch them separately. This saves a substantial amount on zinc and iron. But apart from that, it's about the same.

So I wonder how this reasoning fits in with the rocket equation.

Well the rocket equation is really based on the ratio of total mass to dry mass.

You double both, and you keep that ratio the same. And thus you keep the delta-v the same.

Paging Delta:

"As part of the investigation, SpaceX officials had come across something suspicious they wanted to check out, according to three industry officials with knowledge of the episode. SpaceX had still images from video that appeared to show an odd shadow, then a white spot on the roof of a nearby building belonging to ULA, a joint venture between Lockheed Martin and Boeing."
Implication of sabotage adds intrigue to SpaceX investigation

http://waitbutwhy.com/2016/09/spacexs-b ... story.html

I see the grid fins but where are the landing legs?

Aero wrote:I see the grid fins but where are the landing legs?

If we estimate that one Raptor Engine is between 1 and 1.3 Tons of weight it means that in the bottom of the rocket you have a ballast mass of anything between 40 and 50 Tons. Add to that also the low height/width ratio and I would not be surprised if the center of gravity of the Booster is low enough to get rid of the landing legs all together.
Landing could than be safely done in a conic hole like the video showed.

using numbers from wikipedia (right side): https://en.wikipedia.org/wiki/ITS_launc ... ing_launch

i calculated deltav for various configurations: https://docs.google.com/spreadsheets/d/ ... sp=sharing

using 9.4 as the deltav needed to reach LEO, a few discoveries:

* all 3 ships - booster, tanker, and transport (with no cargo), can reach LEO by themselves. (assuming no payload)
* a transport with no payload, launched as a second stage from the booster, will have 340 tons of fuel remaining after reaching leo.
* a tanker with no payload, launched as a second stage from the booster, will have 710 tons of fuel remaining after reaching leo.
* a fully loaded ship - lets say 450 tons of cargo - with the help of a booster, reaches leo with 0.15 dv to spare (this is assuming the booster saves no fuel for reentry)
** from there, after full refueling, it can travel an additional 5.4 deltav
*** now from leo to mars landing without aerobraking or aerocapture is 10.2 deltav. So a fully loaded ship would have to do 10.2-5.4 = 4.8 dv worth of aerobraking/capture to land on mars at 0 m/s.

* the fuel capacity of the transport is 1950 tons, so that means it would take no more than 3 refuelings, if we assume no fuel is needed for reentry.

I wonder how big/how much a Polyell will be? Say, 100 MW.

Mass is the big issue there, with size/volume only really being an issue while in atmosphere ... depending on configuration.

...updated my spreadsheet:

a "heavy" configuration (2 additional boosters) would weight about twice as much and deliver a little over twice the fuel in one launch.

(why not 3x? because you still only have 1 tanker)

so half as many launches, and half as many uses of the tanker (though 3/2 as many uses of the boosters)

the middle booster won't reach leo, so won't need to do a de-orbit burn. will be going a little over twice as fast than the side boosters were when it decouples.

presumably the middle booster engine configuration could be optimized (less thrust needed and spends more time in high altitude) to reduce mass and increase isp a little.


the tanker has a better full mass / empty mass ratio than the booster, AND has better isp. (the former presumably on the account of not needing so much thrust, and the latter on account of near vacuum.) consequently, by increasing the booster-to-tanker ratio, you're slightly increasing how much fuel you need to lift each ton of fuel. (from 12 tons per ton to 14 tons per ton)