Travel times, acceleration and power plants

[Allensh]

I built a 600-ton ship with 4-G of acceleration. I estimated it can accelerate constantly for 11.25 space combat turns (6 minutes each, roughly 67.5 hours) before the batteries run dry. This puts it outside the distance given for a Close Gas Giant (600,000,000 km) but not to a Far Gas Giant. Math is most definitley not my strong suit but the text says that you have to be 100 diameters out from any object (assuming they mean planet, star, asteroid..anything with a signficant enough gravitational field). This means 140,000,000 km for the sun, 34,760 km for the moon and 12, 756, 320 km for Terra. Of course, actual orbital positions will change this (if for example, the Earth is between you and the sun and the moon is between you and the Earth this may add to the jump distance limit) but this may not be signficant enough to count.

Now...it is apparent to me that my ship that comes out of jump space at the Far Gas Giant (app. 900,000,000 km from the..star? main world?) cannot reach the main world (if that world is in roughly the same location as Terra) by using constant acceleration. it would have to stop and recharge to at least some degree before continuing. This throws the travel time tables into complete disarray.

I see two possible solutions:

1. Allow the purchasing of additional batteries separate from the power plant. This would allow increasing the storage capacity so as to allow longer maneuvering (or increased weapons fire or whatever).

2. Alter the power plant output to that it can power the maneuver drive of the same letter continually without relying on battery power. Alter the power requirements of the jump drive so they empty the storage capacity of a power plant of the same letter. (This would be for the highest jump number of course, maneuver drives and jump drives of lesser letters would use less power).

As it stand right now, the power plant system seems to break the concept of constant acceleration and impacts system travel rather severely.

Allen

 

[EDG]

I built a 600-ton ship with 4-G of acceleration. I estimated it can accelerate constantly for 11.25 space combat turns (6 minutes each, roughly 67.5 hours) before the batteries run dry. This puts it outside the distance given for a Close Gas Giant (600,000,000 km) but not to a Far Gas Giant.

11.25 * 6 minutes = 67.5 minutes (4050 seconds), not 67.5 hours!

The equation of motion to figure out distance is s = ut +0.5(at^2). So accelerating at 4G from rest (ut=0) should get you out to a distance of (0.5*40*(4050^2) =) 328,050,000 metres from your starting position.

So accelerating at a constant 4G from rest at a constant rate of 4G should get you out to 328,050 km. (If you did that for 67.5 hours, you'd get out to 1.18098 billion km).

The formula on page 136 of the playtest document divides that distance by 2 again (since it's accelerating to halfway, then decelerating from there to rest, and that does look correct), so the total distance travelled according to that equation after 67.5 minutes would be 164,025 km. And for 67.5 hours it would be about 590 million km (which is probably what you were thinking of). That assumes that the ship is constantly accelerating at 4G to halfway and then constantly decelerating at the same rate.

So that seems to exacerbate the problem somewhat , since it can't even travel a distance equivalent to about half the earth-moon distance in 67.5 minutes using that method.

This means 140,000,000 km for the sun, 34,760 km for the moon

(You missed a zero for the moon, it should be 347,600 km.)

Now...it is apparent to me that my ship that comes out of jump space at the Far Gas Giant (app. 900,000,000 km from the..star? main world?) cannot reach the main world (if that world is in roughly the same location as Terra) by using constant acceleration. it would have to stop and recharge to at least some degree before continuing. This throws the travel time tables into complete disarray.

Or the ship doesn't do the constant acceleration, turnaround and constant deceleration - if you picture that on a graph with velocity on the y-axis and time on the x-axis, the graph line would look like an upward slope followed by a downward slope (a triangle, with constant accel then constant decel).

However, it could instead accelerate for a fixed period, then coast at a constant velocity, then decelerate for a fixed period. On that graph, the velocity curve would appear as an upward slope (acceleration), then a flat plateau (constant velocity), then a downward slope (deceleration). Only problem with that is that it'd significantly increase travel times.


Hm. That's got me curious now... I'm just going ramble off on a tangent here to see how long it would take with a coast phase. Let's say that it accelerates constantly at 4G for 10 minutes, coasts for 47.5 minutes, then decelerates constantly at 4G for 10 minutes to get back to rest at its destination. How far will it have travelled?

Acceleration phase:
acceleration time = 10 mins = 600 seconds.
distance travelled = 0.5*at^2 = 0.5*40*(600^2) = 7,200 km.
velocity at end of acceleration = at = 40*600 = 24 km/s.

Coast phase:
velocity = 24 km/s
coast time = 47.5 minutes = 2850 seconds
distance travelled = vt = 24*2850 = 68,400 km.

Deceleration phase:
deceleration time = 10 minutes = 600 seconds.
distance travelled = 0.5*at^2 = 0.5*40*(600^2) = 7,200 km
velocity at end of deceleration = at = -40*600 = 24-24 km/s = 0 km/s.

So the total distance travelled with a coast phase = AT^2 + at^2 + v(Tc)

where A = acceleration, T = acceleration time, a = deceleration, t = deceleration time, v = coast velocity, (Tc) = coast time.

So at the end of all that it's travelled 7200 + 7200 + 68,400 km = 82,800 km. That's about half the distance it would have travelled under constant accleration/turnaround/constant deceleration. But it's got about 7 combat turns worth of fuel left after that, as opposed to no fuel at all.

Or put another way - to travel 164,025 km with the same 10 minute accel and decel, it'd have to coast for 149,625 km at 24 km/s, which translates to 6234 seconds. Therefore it'd be accelerating for 10 minutes, coasting for about 104 minutes and then decelerating for 10 minutes, for a total of 124 minutes. So to travel the same distance as it would have travelled in 67.5 minutes under constant accel/turnaround/constant decel, it'd actually take 124 minutes - just over twice as long, but it's still got 7 combat turns of fuel left.

That would however allow it to travel 590 million km as well though - it can coast for as long as it likes and still have the same amount of fuel left, since it's only burning it during accel and decel. But it'd take about 284.6 days to travel that far if it did the 10 minute 4G accel/decel and coasted the rest.

 

[Allensh]

oh geez....

Partially I was going by the travel table which said to travel 600,000.000 km took 60.9 hours at 4 G, but obviously I really messed up.

I hate math...

Allen

 

[EDG]

Well, you're not really going to escape maths in an SF game . Though in this case, you muddled up your units (hours vs minutes) so it's not really the maths at fault .

I think the trick with math is to (a) be sure to check your results and (b) check the result against what you'd expect (if you already have any understanding of the situation). e.g. if you get a result in million of km and you're expecting a result in thousands of km, then you're probably doing something wrong so it's worth checking again.

 

[EDG]

That said, where does it tell you how to calculate how many combat turns of acceleration that the fuel can last for? I can't find that in the 3.1 playtest document, all I can find is that it says you need fuel for the powerplant and the jump drive, but nothing about how that relates to how long you can accelerate using manoeuvre drive.

 

[captainjack23]

That said, where does it tell you how to calculate how many combat turns of acceleration that the fuel can last for? I can't find that in the 3.1 playtest document, all I can find is that it says you need fuel for the powerplant and the jump drive, but nothing about how that relates to how long you can accelerate using manoeuvre drive.

Traditionally, the reactionless M drive didn't use fuel per se. It ran on juice provided by the Power plant, which could run it at fulll load for (depending on which design system) about 14-28 days on a basic fuel load (for the power plant).

CT may or may not have been reactionless, but still used the power plant to run the M drive - at (IIRC) a general factor of ten greater than HG, and was stated to give 288 turns of acceleration (1000 secs) on its Powerplant fuel load.

 

[EDG]

Well it looks like this needs to be clarified, since being able to accelerate for only a matter of 10 minutes or so is going to somewhat slow down the travel times .

But there's nothing in the playtest doc that shows how to calculated how long you can burn the M-drives for, right?

 

[captainjack23]

Well it looks like this needs to be clarified, since being able to accelerate for only a matter of 10 minutes or so is going to somewhat slow down the travel times .

But there's nothing in the playtest doc that shows how to calculated how long you can burn the M-drives for, right?

I haven't read it with a fine toothed comb for that fact, but as far as I can see, no.

That said, Gar did post that the powerplants will equal the drive requirements at a minumum, so its being fixed. I just doubt if we're going to see it, given his deadline.

 

[Allensh]

That said, where does it tell you how to calculate how many combat turns of acceleration that the fuel can last for? I can't find that in the 3.1 playtest document, all I can find is that it says you need fuel for the powerplant and the jump drive, but nothing about how that relates to how long you can accelerate using manoeuvre drive.

I looked at how much power the maneuver drive takes, and how much of that would have to come from the battery since the power plant output doesn't cover it all. I then divided the energy listed for storage by the difference between what the plant puts out what the maneuver drive needs. I assumed these were 6 minute combat turns (which is what the book says the space combat turns are). I came up with 11.25 turns times 6, which is 67.5, So, you can accelerate for an hour and 17.5 minutes before the batteries will be exhausted and you'l have to recharge in order to accelerate more. My mistake was in getting crossed up by the table on page 11 of the Spacecraft Operations chapter, and that was just some weird brain thing, I guess.

That table says at 4 G's it takes 53 minutes to go 100,000 km. That isn't even the 400,000 km to a Orbiting Satellite on the Common Distances table right below it.

So we have a travel time table that assumes constant acceleration and deceleration, and a ship that cannot do that with the power plant and maneuver drive it is supposed to have, more or less.

I'm sure this is being addressed in the upcoming document but at this point I am almost ready to simply ignore the whole power thing altogether when I run the game.

Allen