Heroman Posted January 27, 2005 Report Share Posted January 27, 2005 Below is a chart of what should be the optimal travel times for ships. This uses the old Traveller basis of accelerating 1/2 the distance and the decelerating the other half. This is 'straight line' physics and does not take into account having to angle for a moving planet.The formula used is SQRT( 4d/a ). I believe it is based on d = (1/2) * at^22d/a=t^2SQRT(2d/a)=tWhy then SQRT(4d/a)? There was a reason, but I cannot remember why :PI am guessing something to to with halving the '1/2' (which then becomes 4 after switching its side) but not sure. The above formula jives with a table from GURPS space though so I am happy.The initial time is in minutes, then once that gets large it is listed in hours, then once that gets large is listed in days. The table shows when it changes.Spreadsheet formula (row 'a' is Gs, column '1' is distance in km):=SQRT(4*$A2*1000/(B$1*9.86))/60Use 60 for minutes, (60*60) for hours, (60*60*24) for days, etc.Again, table is listed a wee bit down the page... Distance (km) 1 2 3 4 5 6 10 1.06 0.75 0.61 0.53 0.47 0.43 100 3.36 2.37 1.94 1.68 1.5 1.37 500 7.51 5.31 4.33 3.75 3.36 3.06 1,000 10.62 7.51 6.13 5.31 4.75 4.33 2,000 15.01 10.62 8.67 7.51 6.71 6.13 5,000 23.74 16.78 13.7 11.87 10.62 9.69 10,000 33.57 23.74 19.38 16.78 15.01 13.7 20,000 47.47 33.57 27.41 23.74 21.23 19.38 50,000 75.06 53.08 43.34 37.53 33.57 30.64 Time now hours 100,000 1.77 1.25 1.02 0.88 0.79 0.72 200,000 2.5 1.77 1.44 1.25 1.12 1.02 230,000 2.68 1.9 1.55 1.34 1.2 1.1 490,000 3.92 2.77 2.26 1.96 1.75 1.6 680,000 4.61 3.26 2.66 2.31 2.06 1.88 1,220,000 6.18 4.37 3.57 3.09 2.76 2.52 1,275,200 6.32 4.47 3.65 3.16 2.83 2.58 Time now days 150,000,000 2.86 2.02 1.65 1.43 1.28 1.17 4,990,000 0.52 0.37 0.3 0.26 0.23 0.21 5,630,000 0.55 0.39 0.32 0.28 0.25 0.23 12,030,000 0.81 0.57 0.47 0.4 0.36 0.33 14,080,000 0.87 0.62 0.51 0.44 0.39 0.36 42,000,000 1.51 1.07 0.87 0.76 0.68 0.62 49,500,000 1.64 1.16 0.95 0.82 0.73 0.67 58,500,000 1.78 1.26 1.03 0.89 0.8 0.73 58,500,000 1.78 1.26 1.03 0.89 0.8 0.73 75,000,000 2.02 1.43 1.17 1.01 0.9 0.82 108,000,000 2.42 1.71 1.4 1.21 1.08 0.99 108,000,000 2.42 1.71 1.4 1.21 1.08 0.99 150,000,000 2.86 2.02 1.65 1.43 1.28 1.17 195,000,000 3.26 2.3 1.88 1.63 1.46 1.33 225,000,000 3.5 2.47 2.02 1.75 1.56 1.43 360,000,000 4.42 3.13 2.55 2.21 1.98 1.81 420,000,000 4.78 3.38 2.76 2.39 2.14 1.95 645,000,000 5.92 4.19 3.42 2.96 2.65 2.42 780,000,000 6.51 4.6 3.76 3.26 2.91 2.66 1,350,000,000 8.57 6.06 4.95 4.28 3.83 3.5 1,425,000,000 8.8 6.22 5.08 4.4 3.94 3.59 1,425,000,000 8.8 6.22 5.08 4.4 3.94 3.59 1,650,000,000 9.47 6.7 5.47 4.73 4.23 3.87 2,850,000,000 12.45 8.8 7.19 6.22 5.57 5.08 4,500,000,000 15.64 11.06 9.03 7.82 6.99 6.38 5,850,000,000 17.83 12.61 10.29 8.92 7.97 7.28 13,650,000,000 27.24 19.26 15.72 13.62 12.18 11.12 19,500,000,000 32.55 23.02 18.79 16.28 14.56 13.29 Quote Link to comment Share on other sites More sharing options...
Curufea Posted January 27, 2005 Report Share Posted January 27, 2005 Re: 'Optimal' Travel Time This is 'straight line' physics and does not take into account having to angle for a moving planet. Presumably ships have adequate starcharts and computers to be able to compute the position of destination planets by the time they arrive. Quote Link to comment Share on other sites More sharing options...
Nyrath Posted January 28, 2005 Report Share Posted January 28, 2005 Re: 'Optimal' Travel Time Actually, if your ship can do a constant acceleration like this, chances are the trip times will be short enough that the planet will not move in its orbit enough to matter. The table will be reasonably accurate just as it is. Quote Link to comment Share on other sites More sharing options...
Heroman Posted January 28, 2005 Author Report Share Posted January 28, 2005 Re: 'Optimal' Travel Time Planet positions should be moderately trivial for a computer in an advanced age to know, IMHO. What will become critical is if you change from your plotted path and time since you could risk missing the target completely. This makes it very important to plan rest breaks ahead of time, especially if your course makes you plot a intercepting 'chase' course (again mainly for planets where you are not following it's orbit, but cutting across to intercept) Earth is 150,000,000km from the sun, mars is 225,000,000km ( so travel is between 75,000,000km and 375,000,000km (assuming you can fly thru the sun Assuming you time the planets just right, that is still a pretty large amount of time. I do not add in any rest times (I for one, would not want to be under 3 gs for a full day. More importantly, the chart does not take into account any final angular velocity you need in order to match the speed of your target (ie: mars) or at least a minimal matched veloticy so you can get pulled into it's gravity well for a stable orbit. Mars, I think, has a velocity of something like 24km/s, so you would need to make sure your final net velocity is a close match to that. That depends all on if you are chasing the planet (travelling in the same direction as the orbit), approaching the planet (travelling in the opposite direction), or require a huge path which puts you at a steeper angle. Example: earth has a velocity of 29.82km/s, mars 24.2km/s. If you were 'chasing' mars, you will have a end velocity of 4.4 km/s. Not too harse to make up. If you were 'approacing' it , you would end with a relative velocity of 54km/s!!! Time to do some more breaking...I believe 5 hours at a constant 3G (that means the players should shoot short and then accel in the direction of the mars orbit for 5 hours, allowing mars to hit them in the ass. All of this overly simplifies stuff like the angular component of the velocity and other things which make it a tad bit easier or harder, so that is why I added the disclaimer of it being straight line physics. During this contemplation, I did see a interesting site that talked about this a bit (mainly earth/mars mission paths) at : http://www.stanford.edu/~klynn/mars_paper.htm. The big thing, in the end, is to remember that the actual distance they need to travel can be tiny or huge, depending on how sadistic the GM is about where the relative planets are and what they want the players to do to match speed. -hm Quote Link to comment Share on other sites More sharing options...
Adam Kadmon Posted January 28, 2005 Report Share Posted January 28, 2005 Re: 'Optimal' Travel Time This is great. I've just started working on a SF campaign and I was wondering how to work out travel times. I translated the formula into AU. put this in cell B3: =SQRT(4*L11*150000000*1000/(M$2*9.86))/60/60/24 in row B2,C2, etc put in the Gs in column A3,A4,etc put in the AU Drag the formula to the right and then down and you get the table oh, then select the table, go to format cell, goto custom, type in d:h:m:sec, and you get the answer in days, hours, mins, secs. Adam K. (:?) Quote Link to comment Share on other sites More sharing options...
Heroman Posted January 28, 2005 Author Report Share Posted January 28, 2005 Re: 'Optimal' Travel Time Glad you like! The start of the 'days' section I believe leads with 1AU. Since I also wanted to deal with sub-AU (for items such as 10dd from a gravity well, moon trips, etc) I based all my stuff on kms. Using AUs could be cool if you ran a pure no-ftl campaign where you could afford to thrust for a very long time, though. I'll have to look in OO to see if it does that formatting; would be much easier for the time values. Quote Link to comment Share on other sites More sharing options...
Agemegos Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time The formula used is SQRT( 4d/a ). I believe it is based on d = (1/2) * at^2 2d/a=t^2 SQRT(2d/a)=t Why then SQRT(4d/a)? There was a reason, but I cannot remember why It is simple. t = SQRT (2d/a) is the formula for how long it take you to get there if you don't mind flying past your destination at high speed, and therefore accelerate constantly towards it. If you want to stop at the destination you have to accelerate half-way then turn over and decelerate the second half. Decelerating make you travel slower, and therefore you arrive later. Given that you accelerate half way and it takes half the time, you asrrive at mid-point (which is d/2) at t/2: t/2 = SQRT(2*(d/2)/a) t = 2 SQRT (d/a) If you want to complicate that to t = SQRT (4d/a) that is still correct. Quote Link to comment Share on other sites More sharing options...
atlascott Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time Call me a simpleton if you must, but do you really want or need that sort of detail in an RPG? Did Han Solo ever sit down during screen time and talk to Chewbacca about this? Or did they give their computer some time to figger it out? Quote Link to comment Share on other sites More sharing options...
shadowcat1313 Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time IIRC GURPS Space borrowed the formula from Traveller among a lot of other things, where Traveller got it I have no clue its simple and doesnt require major physics to deal with Quote Link to comment Share on other sites More sharing options...
Agemegos Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time IIRC GURPS Space borrowed the formula from Traveller among a lot of other things, where Traveller got it I have no clue its simple and doesnt require major physics to deal with I doesn't have to be 'borrowed'. Anyone with a pencil and paper and 9th-grade physics can work it out in two minutes. Quote Link to comment Share on other sites More sharing options...
Curufea Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time There is the JMS method as well - "Spaceships move at the speed of plot" However, it depends on the nature of your game. If it is tactical rather than character based - then numbers are more important. Quote Link to comment Share on other sites More sharing options...
austenandrews Posted March 7, 2005 Report Share Posted March 7, 2005 Re: 'Optimal' Travel Time It's an interesting mathematical puzzle, if you like that kind of thing, but it definitely qualifies as extraneous detail in my book. Quote Link to comment Share on other sites More sharing options...
Nyrath Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time I have to agree. Calculating the travel time is something one would do designing a wargame, not an RPG. Quote Link to comment Share on other sites More sharing options...
Agemegos Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time I have to agree. Calculating the travel time is something one would do designing a wargame' date=' not an RPG.[/quote'] That depends how you draw the line. There is a style of roleplaying in which the players have no opponents as such, but still have to make decisions between alternatives subject to fixed constraints. "Can we get a spare part sent from Earth in time?" is a valid concern in some styles of play. "Would Suspect #3 have had time to get from Here to there, commit the murder, and get back?" is, in other styles of play, a question with an answer that ought to be deducible and not depend on GM fiat. I run RPGs by establishing a situation with antagonists, capacities, constraints, and motives, and then letting the conflict develop organically. Call that a wargame if you will. Quote Link to comment Share on other sites More sharing options...
Curufea Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time Exactly - being more tactical does not make a roleplaying game into a wargame. Being 100% tactical does, but that never happens. There are also levels of realism - how much science do you put into your science fiction? Farscape is a good example, they deliberately have little if any science at all, it is predominantly character driven. Then you have Ringworld - where the numbers are vitally important to the plot. There are many circumstances to have numbers, and many to not. It's a style thing. Quote Link to comment Share on other sites More sharing options...
atlascott Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time Transit times, I agree, are a necessary part of any good game. Just a bit too technical for my game, that's all. Nothing wrong with it if you and your players enjoy it! Quote Link to comment Share on other sites More sharing options...
austenandrews Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time Exactly - being more tactical does not make a roleplaying game into a wargame. Being 100% tactical does' date=' but that never happens.[/quote'] I wouldn't say never. I've seen a few groups... On the flip side, I usually roleplay at least a little even in a wargame. There are many circumstances to have numbers, and many to not. It's a style thing. I've always been the geekiest of my various groups when it comes to crunching numbers. I can't imagine a group of mostly numbers-crunchers. Quote Link to comment Share on other sites More sharing options...
CBikle Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time I recently ran a space adventure in Champs and I didn't really put a lot of thought into travel-times(which turned out to be a little more important than I thought). A couple of times, the players asked about specifics relating to arrival time and I think that having a definitive answer would've helped for purposes of suspending disbelief. Quote Link to comment Share on other sites More sharing options...
GreyGuardian Posted March 8, 2005 Report Share Posted March 8, 2005 Re: 'Optimal' Travel Time Yes this sort of detail is useful for consistancy in a star hero game. For space opera I am already fudging lots and lots of science. It's nice to have a little physics to add a touch of reality. Quote Link to comment Share on other sites More sharing options...
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