Help please with in-system travel time.

[WarriorKnight]

I'm working on the next scenerio for my next game & need a little help.The crew has accepted to run a cargo to another planet in the same system & I need help figuring how long it will take them using 2g.I'm using the Sorel system in the Glisten sector.I worked up the whole system using the "Heven & Earth" program.It puts the main planet in Sorel 38.8au's out & the planet they are hulling cargo to at 1.6au's.According to the travel time chart in the book(which by the way is the same one from CT) 1 billion km takes 5.2 days.Sorel is 5,820,000,000km out(the CT book 3 says multiply the au's by 150 million for how many km.)& Epsilon the inner planet is 240million km out from the sun.That gives me a distance of 5,580,000,000km.Should I just multiply the 5.2 days by 5 or is there some arcane mathematical formula to it.I apprecate any help I can get.

(by the way I'm not a math guenius by any streach so please feel free to "dumb it down".)

 

[Whipsnade]

Should I just multiply the 5.2 days by 5 or is there some arcane mathematical formula to it.

WK,

Don't multiply the 5.2 days by 5. You're dealing with constant acceleration here so the trip will be shorter than that.

You'll find three formulas in LBB:2 on Page 10 for just this short of thing. Each solves the time - distance - acceleration relationship for either time, distance, or acceleration. In the time formula, we first divide the acceleration in meter per second squared into the distance in meters, take the square root of the result, multiply that by 2, and get the trip time in seconds. (The 2 is for accelerating to midpoint, turning around, and decelerating to a stop.)

Your distance of 5.580e+9 kilometers works out to be 5.580e+12 meters. Running that distance though the formula with an acceleration of 19.6 meters per second squared results in 6.175 days.

Remember, the formula gives a result in seconds so you'll need to convert that to hours and/or days.

The explanation in LBB:2 is easy enough for even me to understand so you'll have no trouble at all.


Regards,
Bill

Postscript Feb 10: I forgot to multiply the square root result be two, so my answer was wrong. My answer is also slightly different than Wil's because I used 19.6 m/s^2 for 2 gees and he used 20 m/s^2 for the same.

 

[aramis]

dividing seconds by 3600 gives hours, BTW.

 

[WarriorKnight]

Thanks for help.I've always used the chart out of LLB 2,but those formulas always gave me a headache trying to figure them out.If someone gives me an example or talks me through it it'll sink in & I'll understand it.

 

[aramis]

5.580e+12=5580000000000
A=10*Maneuver G's
T=2*√(D/A)

Plug in D (Distance in meters):
T=2*√(5580000000000/A)

Plug in A (Accel in m/s/s):
T=2*√(5580000000000/20)

Do the math
T=2*√(279000000000) _____division
T=2*528204.5058497703 ___hit square root
T=1056409.0116995405 _____multiply by 2

round: 1056409
Divide by 3600 to get hours: 1056409/3600=293.44
Realize that jump takes 168±17

Jump it, it's faster. Even given a day at either end to get to jump point, you'll save 4 days.

 

[Kilgs]

I just use this baby... the information for the system doesn't always coincide but it works well.

http://www.utzig.com/cgi-bin/iai/sy...21 Im G3 V M3 D&UID=&Sectors=Spinward_Marches

EDIT: 38.8AU for the Mainworld?

 

[Hemdian]

Don’t forget that the difference in the orbit numbers will only give you the MINIMUM distance you need to travel. This will only be when the two planets pass each other. If D1 is the distance from the star to the orbit of the first planet, and D2 is the distance from star to the orbit of the second planet, then the straight line distance could be anything from (D1 – D2) to (D1 + D2) ... we can ignore orbit eccentricity as for most planets this will be relatively trivial for our purposes. And if the two planets are on opposite sides of the main star at the time (D1 + D2) then there is a star in the way, so a straight line path won’t be possible.

This was (IIRC) addressed in SOM. I don’t have that book readily to hand but I seem to remember that in the absence of knowing where along the orbit the planets were you rolled 1d6 for each to determine the arc. If both were in the same arc then use (D1 – D2), if the arcs were off by one then use (D1 – D2) x 2, if the arcs were off by two then use (D1 – D2) x 3, and if the arcs were off by three (opposite sides) then use (D1 – D2) x 4.

Once you’ve got the actual distance THEN use the formulas mentioned above to calculate the time.

Final point: If staying in the same system for a while, you can divide each planet’s ‘year’ by 6 to determine how long it stays in a given arc before moving to the next.

 

[Whipsnade]

Jump it, it's faster. Even given a day at either end to get to jump point, you'll save 4 days.

Wil,

Boy am I a dope. I forgot to multiply the result of the square root by two.

Anyway, your suggestion about jumping is a good one if Epsilon is not inside the stellar jump limit. Sorel's star must be rather huge if the mainworld orbits at ~38 AU.

IIRC, there's a system in the Marches where the jump limit and the mainworld are weeks apart at 1 gee.

Thanks for correcting my error.


Regards,
Bill

 

[aramis]

No insulting board members, Bill!

I'm a teacher; there's a reason work gets shown, and you copy the formula first, then sub in the values, and you can more easily check your work.

I don't have the stellar diameters charts to hand, but 100 diameters is not going to exceed 15 AU IIRC.

But as a simplification house rule, I place the stellar jump limit at the midway between the last inner orbit and the habitable orbit.

 

[Whipsnade]

No insulting board members, Bill!

Wil,

Oh but this time it is entirely justified.

I don't have the stellar diameters charts to hand, but 100 diameters is not going to exceed 15 AU IIRC.

There's a single system in the Marches whose name I never remember with a monstrous star and equally monstrous stellar jump limit. The system should have no planets given either 1977's or our current knowledge regarding that subject, but LBB:6 never paid much attention to that sort of thing.

But as a simplification house rule, I place the stellar jump limit at the midway between the last inner orbit and the habitable orbit.

That's a very nice rule of thumb. Earth's orbit around Sol, for instance, is just an astronomical "whisker" beyond the stellar jump limit.


Regards,
Bill

 

[WarriorKnight]

I just use this baby... the information for the system doesn't always coincide but it works well.

http://www.utzig.com/cgi-bin/iai/sy...21 Im G3 V M3 D&UID=&Sectors=Spinward_Marches

EDIT: 38.8AU for the Mainworld?

Nice,wish I'd seen that before I used the "Heaven & Earth" to make up my version.Would have saved me a lot of time. Yea H&E put it out at 38.8au from the sun.Been using this version through 2 campagins now.

 

[WarriorKnight]

Wil,

Boy am I a dope. I forgot to multiply the result of the square root by two.

Anyway, your suggestion about jumping is a good one if Epsilon is not inside the stellar jump limit. Sorel's star must be rather huge if the mainworld orbits at ~38 AU.

IIRC, there's a system in the Marches where the jump limit and the mainworld are weeks apart at 1 gee.

Thanks for correcting my error.


Regards,
Bill

Its an M5 III.Epsilon is in orbit 4 & Sorel is in orbit 9. Epsilon is one of 4 Gas Giants in system & it has 10 moons 2 of which have colonies on them. Sorel has 2 more planets ut pass it.(both of which are Gas Giants.I don't know if they'll jump or not;the ship they just got hasn't been flown in 4 years & its j2 will only do j1 right now.(the port administrator is dangling a 20% bonus if they get it there with out delay.:smirk: I've already rolled to see if they have a miss jump or not,so I'll just see if they're willing to take the risk.))

 

[WarriorKnight]

5.580e+12=5580000000000
A=10*Maneuver G's
T=2*√(D/A)

Plug in D (Distance in meters):
T=2*√(5580000000000/A)

Plug in A (Accel in m/s/s):
T=2*√(5580000000000/20)

Do the math
T=2*√(279000000000) _____division
T=2*528204.5058497703 ___hit square root
T=1056409.0116995405 _____multiply by 2

round: 1056409
Divide by 3600 to get hours: 1056409/3600=293.44
Realize that jump takes 168±17

Jump it, it's faster. Even given a day at either end to get to jump point, you'll save 4 days.

Aramis thank you,when its layed out like that it makes sense to me.My brain freezes up when algabra or geomerty figures are put up.My friend Cena is a math wiz who keeps trying to get it through my skull that when I write up a Hero System/Champions character I'm doing algabra,but I have a had time with it unless I physically see the numbers and not the abstract figures using letters.

 

[aramis]

Just do the process that same way, and it's easy to follow. Sub in one variable on each re-write.

I teach this stuff, and when I do it, I still do it that same way, almost every time.