Timeline for Travel times in The Expanse novels
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 19 at 5:27 | comment | added | fez | Close voters - you will note that the “real world science” close reason explicitly states “rooted clearly within a cited work of fiction. […] asking for an answer within the context of a fictional universe, even if that question requires real-world science information, is on-topic”. | |
| Apr 19 at 3:03 | review | Close votes | |||
| Apr 19 at 6:59 | |||||
| Apr 15 at 15:39 | comment | added | Matthias Ludewig | It just seems to me that in the books, the travel time is calculated as if it was linear in the distance, while really, it is proportional to a square root of the distance. | |
| Apr 15 at 15:38 | comment | added | Matthias Ludewig | @Ethan: My formula was 2*A*(t/2)^2, which is precisely what you wrote after pulling out the 1/2. In any event, there is a discrepancy of a factor way above 10. I don't think that anything regarding curved trajectories will take care of such a discrepancy. | |
| Apr 15 at 12:48 | comment | added | BradV | @PM2Ring agreed! It would be very interesting to see what "powered" trajectories would look like. I would think there would be somewhat sharply curving sections at startup and ending as velocity changes rapidly and gravity is a major factor again. The major portion of a powered trajectory would be 'straighter' than normal coasting. I was wondering this exact thing as I was reading the books. | |
| Apr 15 at 7:26 | comment | added | PM 2Ring | @BradV Those continuous burn trajectories are very different to our beloved Hohmann and bi-elliptic transfers. The acceleration dwarfs the Sun's field strength. Eg, at 1 AU, the Sun's gravity is ~ 0.00593 m/s^2 = 0.0006047 g. So if you're pulling 1/3 g you can virtually ignore the Sun's gravity. | |
| Apr 14 at 21:50 | comment | added | BradV | the equation you cite/use has the specific stipulation that it is only good for straight line motion. There are virtually NO 'straight line' travels between orbital bodies. The overly simplified flight paths commonly used to show things in movies and TV do not do justice to the actual curved path traveled. ALSO... I'd ask you be be very certain that you are using the proper units of measure for all factors. A single misapplied unit really bungles everything! Visit the space exploration section to seek understanding of transfers. | |
| Apr 14 at 21:14 | answer | added | Michael Stachowsky | timeline score: 2 | |
| Apr 14 at 21:07 | comment | added | Gouvernathor | @Ethan having read the books, it does mean a high-g burn. And having read the following books in the series, I didn't make a deep analysis of it but in particular in book 9 there seems to be big inconsistencies in travel times through systems. | |
| Apr 14 at 20:41 | comment | added | Ethan | Also note that IRL "high burn" refers to fuel consumption rather than acceleration per se. So maybe in the Expanse a high burn flight plan refers to one that uses a lot of fuel compared to some hypothetical alternative flight plan. This would not necessarily imply that the mean acceleration during travel is > 1G. | |
| Apr 14 at 18:44 | comment | added | Ethan | d = 1/2 A t^2 So you're off by at least a factor of two to start with. Beyond that I suppose you have to guess at what "high burn" means to people who live in low gravity. (I have not read the books). | |
| Apr 14 at 15:58 | review | Close votes | |||
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| Apr 14 at 15:52 | history | edited | LogicDictates | CC BY-SA 4.0 |
added 241 characters in body; edited title
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| S Apr 14 at 15:28 | review | First questions | |||
| Apr 14 at 15:52 | |||||
| S Apr 14 at 15:28 | history | asked | Matthias Ludewig | CC BY-SA 4.0 |