formulas for working out travel time

[Xerxeskingofking]

I'm posting this here because its about ships, and this is, as far as i can tell the "gearheads" section of the forum.

anyway, i would like to know what the formula is for working out how long a object travelling at a constant acceleration/deceleration takes to travel a given distance.(ie a traveller ship under M-drive to reach a jump point, or other arbitary distance.).

are their any physic students out thier that can help me?

 

[aramis]

The three continuous steady-burn acceleration formulae:

T=√(2D/A)
D=0.5AT²
A=2D/T²
D= Distance in Meters
A= Acceleration in m/s² (Real G's are 9.8m/s², Traveller G's are 10m/s²)
T= Time in seconds​

All of this is in book 2 of CT.
Top of the three is "Given Distance D and Acceleration A, how long does it take"
Second is "Given Time T and Acceleration A, how far can/will it have gone"
Third is "Given distance D and needing to be there in time T, how much acceleration must it use."

A quick google gets 1.49598E11m per AU, but 1.5E11 is close enough.
An hour is 3600s
A day is 8.64E4s (86400s)
A week is 6.048E5s

 

[BillDowns]

The usual form for Newtonian movement is:
[FONT=&quot]s = ½ at² + (v₀)t + s₀[/FONT]
[FONT=&quot]where s = final displacement[/FONT]
[FONT=&quot]a = acceleration[/FONT]
[FONT=&quot]t = elapsed time[/FONT]
[FONT=&quot]v₀ = initial velocity[/FONT]
[FONT=&quot]s₀ = initial displacement[/FONT]

For 3 dimensional space, everything except time would be a vector. Units of measure must correspond. The equation can be solved for any 1 of the 5 variables by fixing the other 4 as shown in Aramis' examples where initial velocity and initial displacement are assumed to be 0.

For a real-world rocket, it gets far more complicated very quickly as the fuel burns, reducing mass which increases acceleration, and as the engines heat up, changing their thrust performance. Then, one has to take into effect the gravitational pull of the planet/moon/body you are departing.

Even with Traveller's M-drives, this would not be a straight line, but curved as the gravity pulls the ship around. And even on a surface, initial velocity and displacement would not be zero; the body would be rotating thus imparting motion, and, of course, the ship would not be in the center of the body.

So, how close is close enough for your game?

BTW, I apologize for the formula - I could not figure out how to get superscripts and subscripts.

 

[Andrew Boulton]

The three continuous steady-burn acceleration formulae:

T=√(2D/A)​

​

In this case, of course, you'll be going pretty fast when you get there. If you want to be stationary when you arrive, you use

T=2√(D/A)

 

[Xerxeskingofking]

those are more than statisfactory for my purposes. I'm not looking for a second perfect answer. it's just all i had to go on before was the rather unsatifactory table in MgT.

 

[aramis]

[/INDENT]In this case, of course, you'll be going pretty fast when you get there. If you want to be stationary when you arrive, you use

T=2√(D/A)

Oops, yeah, I gave full speed at end...

The ones for stop at end, as given in TTB, p54, are
T=2√(D/A)
D=0.25AT²
A=4D/T²

 

[Sturn]

I made this a while back for my own use:

Reaction Drive Trip Times


It isn't meant for steady burn though, for reaction drives. Only a slight tweaking would make it for standard steady burn MD's.

 

[Icosahedron]

those are more than statisfactory for my purposes. I'm not looking for a second perfect answer. it's just all i had to go on before was the rather unsatifactory table in MgT.

<checks Mongoose core>
Yeah, I see what you mean about unsatisfactory. I hate tables for that very reason - they give a man a fish, where a formula teaches a man to fish...

In case you are unsure about the formulas given above, one relates to constant acceleration over the whole journey that will have you whizzing past your target at immense speed, and the other refers to turning around at mid-point and decelerating the second half of the journey so you reach the destination at zero speed.

Note that these formulae don't take relativity into account. If your acceleration continues long enough to bring you to an appreciable proportion of lightspeed (eg accelerating between stars) the formula stops working. Within a solar system, it's fine.

 

[Xerxeskingofking]

sorry to drag this up after a few years, but i've been trying to use those sums recently and it appears my grasp of algebra is not as good as i thought.

i'm plugging some numbers into what i understand D=0.25AT² to be, and get an an answer, seemingly so far so good.

However, when I double check the answer by using that out D as the input for T=2√(D/A), I get a value of T that is exactly half the value of T used in the first sum, when it should be the same.

as i understand it, D=0.25AT² means D=(0.25*A)*(T*T). but this must be wrong, so what am i doing wrong?



if anyone cares, I was trying to work out how far a ship could go at 6g in one week, i.e. the maximum range a sublight ship would be quicker than a jump-ship.

 

[Carlobrand]

CT Book 2 and The Traveller Book both offered a set of typical travel distances for those who didn't want to be bothered with calculations. The Traveller Book went farther and pointed out which distances were useful for what - for example, the typical travel distance to jump point of a size X world, or the typical travel time to a far gas giant. Your distance exceeds that by a good bit, but it's a useful resource

Lessee, assuming you want to stop at the end, I get about 5.5 billion kilometers. I do assume you want to stop, or else the jump ship will get there with whatever its velocity was and you'll be racing past at, umm, "ludicrous" speed.

 

[Xerxeskingofking]

odd, cos thats the answer i'm getting as well.

I say odd, becuase that would mean i'm gettting the other sum wrong, which was the one i thought i understood.

hmm..... I don't see how else you could expand T=2√(D/A), apart form to T= square root of (D divided by A), but a D of 5.5 billion K and an A of 60 M/S gives me a T of 302,400 seconds, or half a week, whereas the T i put into D=0.25AT² was 604,800 seconds.

surely, if i'm doing the sum in reverse, the output should be the same?

 

[Carlobrand]

odd, cos thats the answer i'm getting as well.

I say odd, becuase that would mean i'm gettting the other sum wrong, which was the one i thought i understood.

hmm..... I don't see how else you could expand T=2√(D/A), apart form to T= square root of (D divided by A), but a D of 5.5 billion K and an A of 60 M/S gives me a T of 302,400 seconds, or half a week, whereas the T i put into D=0.25AT² was 604,800 seconds.

surely, if i'm doing the sum in reverse, the output should be the same?

a=60 m/s
t=7 days=604,800 seconds
d=0.25at²
t²=365,783,040,000
at²=21,946,982,400,000
0.25at²=5,486,745,600,000 meters, or about 5.5 billion klicks.

t=2√(d/a)
d=5,486,745,600,000
a=60
d/a=91,445,760,000
√(d/a)=302,400
2√(d/a)=604,800 seconds

I'm not sure what you're missing, but it seems to work.

 

[Xerxeskingofking]

i've just clicked. i was reading 2√ as "square root", not "twice the square root", hence why my answer was half the correct size.

 

[leo knight]

After being ridiculed by a player for taking the time to calculate in-system travel time, I came up with a shortcut for myself. At 1g acceleration, find the distance in AUs, take the square root, and multiply by 68. That's how many hours the trip takes. Higher accelerations divide trip time by their square root, e.g. 1.4 at 2g, 1.7 at 3g, 2 at 4g, etc.

 

[shaunhilburn]

Ii would like to know what the formula is for working out how long a object travelling at a constant acceleration/deceleration takes to travel a given distance.(ie a traveller ship under M-drive to reach a jump point, or other arbitary distance

For a "Burn & Turn" voyage, with continual acceleration to midpoint with turnaround & braking, and no coasting: Voyage Hours = sqrt(AU/G)×68.6167

For a "Burn No Turn" mission, Hours = sqrt(AU/G)×48.5193.


A voyage at 2G takes 0.707 times as long as a 1G voyage, sqrt(1/2).
A 3G voyage takes 0.816 times as long as a 2G voyage, sqrt(2/3).

 

[Carlobrand]

After being ridiculed by a player for taking the time to calculate in-system travel time, ...

Well, there is the point that the planets move, therefore those distances are always changing. Me, I just look it up in my handy Traveller Book (which, unlike Book 2, is nice enough to give examples) and then add a bit according to my whim. Who's to know? :devil:

 

[leo knight]

Well, there is the point that the planets move, therefore those distances are always changing. Me, I just look it up in my handy Traveller Book (which, unlike Book 2, is nice enough to give examples) and then add a bit according to my whim. Who's to know? :devil:

I had been reading a lot at Atomic Rockets, and I was in an unusually realistic mood. I had already calculated radio times, in case they sent a distress signal, and after the session figured out how long rescue/ relief might come, at different accelerations. But yeah, especially with my group, I should have realized any answer would work.

 

[Tobias]

Small necro: Does anybody have handy formulas for travel times involving non-constant acceleration? In other words, for X burn-hours at Y Gs?

 

[aramis]

Ignoring difference in vectors...
T_t=((D-AT_a²)/AT_a)+2T_aT_t = Travel Time in seconds
T_a = Burn Time in seconds (assumed to be symmetrical)
D = Distance in m
A = Acceleration in m/s²

 

[Tobias]

Thanks!