Quote from: chopsticks on 12/11/2022 05:02 pmWhat is the cross range of SS? It's a lot more than a capsule, but not as good as an airplane.You want to look at down range capacity as well, i.e. the total landing ellipse (or whatever shape it is)I assume this is in the context of an abort from orbit, since this is the only case where unplanned cross range is needed.* In that case you also need to look at down range options. You are moving at 8km/s. Even if you need to get on the ground NOW several hundred km of down range requires nothing but timing. There are situations where ms count, but there are many more where a few minutes don't matter.* An abort during launch should be preplanned, if necessary you arrange for suitable diversion landing points. If it's non-flat land send in the bulldozers months ahead of time. If its non-land send in the drone ship. An abort during a planned landing is going to end up pretty much where you planned. You might divert from LZ-1 to LZ-2 or the skid strip if someone parks a car on your destination (or some such idiocy), but this does not require much cross range and only minimal planning.
What is the cross range of SS? It's a lot more than a capsule, but not as good as an airplane.
Unless I did the math wrong (and I might have - someone check me as this doesn't seem right), the drag on a hovering SS at sea level in 25m/s winds is about 5MN, or about 3 Raptors at full throttle pushing straight sideways.
Quote from: Lee Jay on 12/11/2022 08:56 pmUnless I did the math wrong (and I might have - someone check me as this doesn't seem right), the drag on a hovering SS at sea level in 25m/s winds is about 5MN, or about 3 Raptors at full throttle pushing straight sideways.FD = ρ * v2 * CD * A / 2Approximating Starship as a cylinder, and plugging in: ρ = 1.2 kg/m3 v = 25 m/s CD = 1.17 A = 50 m * 9 m = 450 m2I get FD ≈ 200 kN, or about 20 tonne force, which is a factor 25 lower than you got. That sounds more reasonable.(Edit: Considering the factor 25 difference, did you perhaps cube the air velocity instead of squaring it?)
I assume this is in the context of an abort from orbit, since this is the only case where unplanned cross range is needed.
Quote from: tbellman on 12/11/2022 10:56 pmQuote from: Lee Jay on 12/11/2022 08:56 pmUnless I did the math wrong (and I might have - someone check me as this doesn't seem right), the drag on a hovering SS at sea level in 25m/s winds is about 5MN, or about 3 Raptors at full throttle pushing straight sideways.FD = ρ * v2 * CD * A / 2Approximating Starship as a cylinder, and plugging in: ρ = 1.2 kg/m3 v = 25 m/s CD = 1.17 A = 50 m * 9 m = 450 m2I get FD ≈ 200 kN, or about 20 tonne force, which is a factor 25 lower than you got. That sounds more reasonable.(Edit: Considering the factor 25 difference, did you perhaps cube the air velocity instead of squaring it?)Probably! I do wind energy, and the formula for that is 1/2 * rho * A * V^3 * Cp. I probably hit "cube" out of habit!Thanks for checking. That's why I said it didn't seem right!
Can you eyeball an entry corridor without necessarily knowing the related landing ellipse?
Quote from: TheRadicalModerate on 12/12/2022 01:36 amCan you eyeball an entry corridor without necessarily knowing the related landing ellipse?Yes. This was tested on the Apollo missions, to see if a successful landing could be made even if all communications were lost. (I believe the entry corridor corrections were sent from the ground, not computed on board).Nowadays, at least for Earth, if you are close enough to re-enter you've likely got GPS (it works at least as high as GSO) and plenty of onboard compute power.
Quote from: Lee Jay on 12/12/2022 12:02 amQuote from: tbellman on 12/11/2022 10:56 pmQuote from: Lee Jay on 12/11/2022 08:56 pmUnless I did the math wrong (and I might have - someone check me as this doesn't seem right), the drag on a hovering SS at sea level in 25m/s winds is about 5MN, or about 3 Raptors at full throttle pushing straight sideways.FD = ρ * v2 * CD * A / 2Approximating Starship as a cylinder, and plugging in: ρ = 1.2 kg/m3 v = 25 m/s CD = 1.17 A = 50 m * 9 m = 450 m2I get FD ≈ 200 kN, or about 20 tonne force, which is a factor 25 lower than you got. That sounds more reasonable.(Edit: Considering the factor 25 difference, did you perhaps cube the air velocity instead of squaring it?)Probably! I do wind energy, and the formula for that is 1/2 * rho * A * V^3 * Cp. I probably hit "cube" out of habit!Thanks for checking. That's why I said it didn't seem right!Side note, but what do you think about wind electricity to make hydrogen on-site?
Quote from: Barley on 12/11/2022 02:57 amQuote from: Lee Jay on 12/11/2022 01:12 amAnd "reasonably flat land" isn't as common as you think.and runways are all over the place.Reasonably flat land is at least as common as runwaysRunways are reasonably flat, if you can land a spaceplane you can land vertically. That assumes you can steer to reasonably flat land. SS has very little cross range. A space plane can steer to a runway because it has enormous cross range - in the 1000km range.
Quote from: Lee Jay on 12/11/2022 01:12 amAnd "reasonably flat land" isn't as common as you think.and runways are all over the place.Reasonably flat land is at least as common as runwaysRunways are reasonably flat, if you can land a spaceplane you can land vertically.
And "reasonably flat land" isn't as common as you think.and runways are all over the place.
Quote from: Barley on 12/11/2022 07:18 pmI assume this is in the context of an abort from orbit, since this is the only case where unplanned cross range is needed.Does anybody have a handle on how big a nav error one can have from either a translunar or interplanetary direct entry? If this is a problem, you can obviate it simply by not doing direct entries, relying instead on aerocaptures, post-capture orbit computation/refinement, and then doing an entry. But that assumes a non-emergency situation. If there's an emergency (an imminent ECLSS failure, a medical emergency, a dodgy structural or guidance issue that's more likely to succeed with one entry instead of two, etc.), direct entry is still a problem that would need to be bounded for how far off-target you're going.
Quote from: sebk on 12/11/2022 12:01 amWrong. It could emergency land on any flat patch of land. Contrary to space planes which are runway or bust. But the whole reasoning is fallacious in the first place anyway.Without legs that could be a problem as the current version of starship is using a catch system
Wrong. It could emergency land on any flat patch of land. Contrary to space planes which are runway or bust. But the whole reasoning is fallacious in the first place anyway.
Quote from: Lee Jay on 12/11/2022 02:54 pmQuote from: Barley on 12/11/2022 02:57 amQuote from: Lee Jay on 12/11/2022 01:12 amAnd "reasonably flat land" isn't as common as you think.and runways are all over the place.Reasonably flat land is at least as common as runwaysRunways are reasonably flat, if you can land a spaceplane you can land vertically. That assumes you can steer to reasonably flat land. SS has very little cross range. A space plane can steer to a runway because it has enormous cross range - in the 1000km range.SS has about half of cross range of Shuttle, i.e. right with the other space planes (Shuttle had an extreme cross range even for a spaceplane because AOA for polar orbits requirement; its cross range was about 2500km).I've no idea where did't you come with the notion that SS entry is ballistic (or anywhere close to that).
SpaceX WILL have variants of Starship that land (after reentry) on legs. Point to point cargo for the military, Mars Starship. Lunar HLS, though it won’t have heatshield tiles, it will definitely have legs.SpaceX has, of course, tested various Starship prototypes with legs on multiple occasions successfully.“SpaceX may sometimes try to catch Starship with the catching arms” is not the same thing as “it’s not feasible for SpaceX’s Starship to use legs if required for some reason.” And I’m not sure why people have difficulty with this logic.
Survivable re-entry emergencies would be stuff like when one of the fins seized and controls are now very limited and surviving hypersonic portion takes priority and let's worry about landing spot once the vehicle is transsonic.