What does 408.9 represent?

Time taken for the journey (in days) squared, then multiplied by the number of Gs in the acceleration.

This is just re-arranging the law of uniform acceleration, in order to solve for time (distance = 1/2 * Acceleration * time^2)

The value 408,9 is a bit arbitrary, but results mainly from your choice of units. Ie, we're talking Days and G's here. You might want to talk about Hours and other units of acceleration, in which case this factor would be different. It is also dependent on the distance traveled. In this case, 50AU.

Re-working to factor in distance as a variable, you divide 408,9 by 50, and then multiply with distance you want (in AUs).

So the new and improved (but still complicated) formula is then:

Time taken (in days) = Square root ((Distance * 8,2) / G)
 

I think that for practical game purposes, I would go for a simpler formula ala 28/g, or make a little table with "common" values.

The complex formula is nice since due to time constraints and life, I can only play PbP games on RPOL.com.

After doing some light googling, I've learned that our solar system has a 60 AU diameter (out to the Kuiper belt only, not the oort cloud which is 190,000+ AU), plus a fraction. Half that would be 30. The hospitable zone of a solar system is (rounding) 1 to 3 AUs from the star. So with a Sol sized solar system, you're looking at 29 AU which translates into 7.5 days to reach warp safety for a Dauntless Light Cruiser and 9.75 days in a Lunar Class Cruiser (two common hull types players use)

I think all that's right!

Now for the big question, one that none of my googling turned up. How big, by comparison, is the Sol system? Are we big? Small? Average? Friggin huge?

Hm, just realized a flaw. A solar system's size has nothing to do with it. The only factor is the star's mass and consequently the size and strength of its gravity well. The larger the star, the further out you need to be before you can safely enter/exit the warp. I quote The Frozen Reaches as my example:

"....but this should serve as a graphic remind of the dangers of using a warp drive too close to a star."
     The Frozen Reaches, page 44, third paragraph from the bottom on the left side

What I get from that is that there is either a specific distance away from the star that you must be to enter the warp safely no matter how big/massive the star is or you have to be further out the more massive a star and thus the larger and stronger its gravity well is. If the latter is the case, then we really have nothing at all to go on unless there's some line in one of the hundreds of WH40K novels that an author states exactly how many AU away from Sol that the ship was before it entered the warp. Or at least how many days it took to reach that point as long as the ship's class type was identified so that we can do the math and find the distance ourselves.

Yes, I'm greatly overcomplicating this. But I love learning! And theory.

WhiteLycan said:

The beauty of geekdom! I completely agree.

I did see some interesting pictures showing our planets relative to our sun, then our sun relative to other known stars. Sol is tiny. But whether those large stars have habitable planets around them I have no idea.

WhiteLycan said:

Well, look at it this way:  The bigger the star, the bigger the star's gravity well, and therefore the more room you have to fit planets around it.  So, you can estimate the system size simply by looking at the size of the star.  (There can be other complications, but I'll get into those later.)   Sol is a G2 main sequence star.  In brief, the main sequence stars fit into one of the following classes:  O, B, A, F, G, K, M.  That's in order from largest to smallest.   Each class is further subdivided from 0 to 9.  Again, that's in order from largest to smallest.  That means Sol is fairly average, around the middle of the main sequence.  [If you want a much more detailed explanation, go to Wikipedia, type 'main sequence star' into the search box, and be prepared to spend a good half hour wiki-walking.]

Now, about those complications...

• The O and B class stars are friggin' huge and very bright, but they are also very rare.  Big stars are less likely to form, and they burn their fuel proportionately faster than smaller stars do.  The biggest stars burn their fuel very fast and go nova after just a few million years.  The smaller the star, the longer its lifespan - Sol is 4.5 billion years old or so.  The combination of lifespan and formation rate means smaller stars are much more common than bigger ones.  More than half the stars in the galaxy are the small, dim, red M class.
• Binary star systems are very common.  More than half the stars in the galaxy are in binary pairs, where two (occasionally more) stars orbit each other.  The distance between the two stars can range from very great to very little.   Two massive objects orbiting each other like that is going to have ... interesting ... complications for the system's gravity well.  The gravity well won't be symmetric, it'll be a spinning dumbbell or peanut shape, and that means the strength and angle of the system's pull will vary both with time and the direction of approach.
• Red giant stars don't fit the above classification scheme.  These are bloated, many times bigger than their mass would suggest, and are likely to have consumed or incinerated any planets they might once have had.

Right, since we've gone this far, might as well discuss habitable planets (aka profit-making opportunities).  Yes, Lord-Captain, I thought that would get your attention.

1. Stars are colour coded for your convenience.  O stars are blue, B stars are blue-white, A stars are white, F stars are yellow-white, G stars are yellow, K stars are orange, and M stars are red.  The star's colour provides a rough idea of its size.  (Chemical composition, metal content, magnetic field strength, and spin rate can all influence the colour somewhat - like I said, its a rough estimate.)
2. Remember, the larger the star, the shorter the lifespan.  Larger stars will be young, guaranteed.  Smaller stars can be anywhere from very young to very old.  Younger stars are less likely to have life, but more likely to have valuable metals. 
3. Larger stars have potentially more planets than smaller ones, but that potential doesn't always pay off.  There are many, many things that can disrupt planetary formation or strip planets away.  Radiation pressure from a nearby huge star during planetary formation, a close encounter with another star with its gravity pulling away planets, or simply the wrong chemical composition can mean few or no planets regardless of a star's size.
4. Binary systems are a poor choice.  They are very likely going to be dangerous to navigate due to the complicated gravity well.  Worse, the gravity issues will also disrupt or eject planets from the system.  A binary system will only have planets if A) the two stars are relatively far apart and the planets are tucked in close, or B) the two stars are very close and the planets orbit a common center of mass.  They are more likely to have several twisted asteroid belts.  (Which make a great spot for a hidden pirate base, by the way.)
5. The bigger the star, the wider the habitable zone. So, a larger star is more likely to have a planet (or planets) orbiting at a habitable temperature. But, if the star is too big, the planets will be too young for life to form.  Gas giant planets emit some heat, and can warm their moons, so a habitable moon is possible even outside the "normal" biozone.
6. Given all that, the best candidates for a habitable planet is a solitary F, G, or K class star.  The larger stars are too young and spew too much radiation to be a likely life candidate (but may well have metal rich planets).  The very common M class stars have a wafer thin habitable zone and are likely to be metal poor.  But, if they do have life, it may well be ancient as such stars can be very old...  Most of the Halo Stars (as in 'Halo Devices') are described as dim and ancient.  That'd be a small red one, then. 

Just my 2 thrones worth,

- V.

I love you Vandergraffe!! All that information gave me a mind boner. I'm eventually going to record all this for easy access.

One thing I literally just thought about a few minutes ago. Since we're on the subject of entering/exiting things, what about a planet's atmosphere? How long would it take for a gun-cutter or a halo barge to make it from the earth's surface to orbit? I ask because does it take half an hour? 3 hours? 10? It's an important thing to know, especially in The Frozen Reaches.

Solar System Enter/Exit Times: In order to find out how long it will take for a ship to enter or exit a solar system, one must first determine both the size of the solar system (which is related to the size of the star; see below) and the location of the ship within the solar system at the point in space in which they set a course for their warp translation coordinates.

To establish the size of a solar system that (like most) doesn't have a listed size, there are a few things you need to determine. First off, what type of star is it? The three star types most likely to have a large enough habitable zone are F (Large), G (Medium), and K (Small) class stars (the sizes are simply for reference). F-class stars are yellow-white, G-class stars are yellow, and K-class stars are orange. For comparison, our own star is a larger-than-average G-class star (a G2, to be precise).

Following the letter in a star's classification is a number from 0-9. The higher the number, the smaller the star. To randomly determine a star's sub-classification, simply roll a 1d10 then invert that number (2 becomes 8, 7 becomes 3, 6 becomes 4, etc.) and multiply it by 2. Then add that number to the AU in the section below.

F-class stars are typically 80 AU across, G-class stars are typically 60 AU across, and K-class stars are typically 40 AU across. A star's Habitable zone is a ring that is, on average, 1 to 3 AU away from the star. If the players are indeed on a habitable planet, then the process to find their distance from the edge of their solar system is simple: Divide the AU diameter of the solar system by 2 (thereby finding the system's radius), then subtract the distance that the planet is away from the star.
Finally, the formula to figure out how fast a specific ship can make it from one point in a solar system to the outer edge of the solar system is this: The Square Root of ((8.2 x AU Distance)/G). G is the acceleration of the ship which can be found in all ships' entries. 8.2 represents... don't worry about it, it's complicated.

EXAMPLE: Aspyce is in a solar system with a K7 star. That means the solar system is about 46 AU across (40 from K-class, 6 from sub-classification) which means the radius is 23. The planet she is in orbit around is 2 AU from the star. This means she is approximately 21 AU from the edge of the solar system. She multiplies 21 (radius of the solar system minus the distance of the planet from its star) by 8.2 to get 172.2. Then she divides that by her ship's acceleration, which is 4.3, to get 40.04 and a bunch of decimals. Lastly, she determines the square root of that number. The result is 6.3, the approximate amount of time that it will take, in days, for the ship to reach the edge of the solar system.

Okay. I'm happy with all of that except the whole sub-classification number. There's got to be a better way of doing that. It works for me, but it might confuse some other people.

WhiteLycan said:

Uhhhhh, you're welcome... I think.

Right, getting to orbit.  The real answer is "however long the GM wants" and it should also depend on both the planet and the flyer. 

In real life, it doesn't take that long.  Rockets, and the recently retired space shuttle, blast off relatively quickly.  The space shuttle's main engines shut down eight minutes after launch, and by that point, the shuttle's already out of the atmosphere.  However, it still needs to use the OMS [Orbital Maneuvering System] engines to lift it up to orbit.  This is mostly coasting (there's no air drag to slow it down), punctuated by a few modest engine burns.  The OMS engines have a total burn time of 1250 seconds, and that includes the fuel needed to deorbit the shuttle and bring it back to Earth.

Then again, that's a rocket that blasts its way out of the atmosphere, straight up.  A flyer has a different ascent profile.  If 40K engines are much more fuel efficient than today's rockets (very likely cos rockets have a fuel economy numbers of CRAP), then a flyer probably takes a slower, lower angled climb.  Then again, they don't have to coast to conserve fuel once they're out of the atmosphere.  Since you seem to like formulas, here's one I just made up:

G * 200 minutes / S = T

G = Planet's gravity, in G's.  (Earth = 1)

S = Flyer's air speed in aeronautical units (AU)

T = Time to low orbit, in minutes.

That way, an Arvus lighter, speed 10, can make low orbit from an Earthlike planet in 20 minutes.  Twenty to thirty minutes sounds about right.  As for reentry, there is a need to not burn up due to excessive speed, so most flyers probably take at least 20 minutes there regardless of how fast a vessel is.  The only exception I'd make is for Drop Pods, which explicitly have a very fast reentry and land hard... but are designed to do that to avoid anti-flyer fire from the planet.

Cheers,

- V.

WARNING HIGH NERD CONTENTS

Ok the exercise has already been done twice in this topic, but no one actually made a proper breakdown as where the number 8.2 comes from and how the calculation is made. Since I'm the one bitching about it, let me also be the one to give the breakdown:
Stuff needed:
- 1 g (the acceleration you experience on earth due to its gravitational field): 9.807 m/s^2 (meters per square second), in free space this means that every second our speed increases with roughly 10 m/s. So after 1 second we travel 10 meters/second, after 2 seconds we travel 20 meters/second, then 30, 40, etc.
- 1 AU (astronomical unit, the distance between the sun and the earth): 149,597,870.7 km (kilometres)
- Radius of our solar system (based on the average distance between Pluto and our sun): 39.5 AU
- So distance edge of our solar system and the earth: 39.5-1 = 38.5 AU
- As explained before, you'll need to accelerate half the distance and decelerate half the distance, so you calculate the acceleration over 38.5/2 = 19.25 AU (the half way point) and multiply the time that takes by 2 as you will be decelerating the same amount of time.

Ok now we bring in the big formula for the law of uniform acceleration which has been mentioned before: distance = 1/2 acceleration * time^2. Acceleration in this case is g so this becomes:
distance = 1/2 * g * time^2. We need time so so rewriting the equation: time = square root(2*distance/g)
In terms of units this now is: [seconds] = SQRT(2*[meters]/[meters/seconds^2]) 

However we like to talk about [days], [AU's] and [g's] so we need some conversion factored in. To do this lets first just do the calculation with what we've got. For this we need to at least convert 1 AU into meters which is:
1 (AU) * 149,597,870.7 (km) * 1000 (m) = 149,597,870,700 meters.
So the travelling time to the half way point is: SQRT(2 * 19.25 * 149,597,870,700 / 9,807) = 766346 seconds = 9 days (1 day = 24*60*60 = 86,400 seconds)
As the decelleration takes the same amount of time it will take 18 days (and 1 hour) to reach Earth from the edge of the solar system.

Now to obtain THE formula
To get the answer we used 3 conversion factors: 149,597,870,700 (AU in meters), 86,400 (seconds in a day) and 9.807 (g in m/s^2). So putting that into the equation:
[seconds] = SQRT(2*[meters]/[meters/seconds^2])
changing acceleration from m/s^2 to g:
[seconds] = SQRT(2*[meters]/(9.807*[g]))
changing answer from seconds to days:
[days] = SQRT(2*[meters]/(9.807*[g])) / 86,400
changing distance from meters to atronomical units (AU):
[days] = SQRT(2*149,597,870,700[AU]/(9.807*[g])) / 86,400
dividing the total distance by 2 (to the halfway point) and multiplying the time need by 2 (from edge to halfway and from halfway to destination):
[days] = 2 * SQRT(2 * 1/2 * 149,597,870,700[AU] * 1/9.807 * 1/[g]) / 86,400
cleaning up the formula by moving all the numbers together and calculating the result
[days] = SQRT(15,254,192,994.80[AU] / [g]) / 43,200
[days] = SQRT((15,254,192,994.80[AU] * 1/(43,200^2) * 1/[g])
[days] = SQRT(8.17[AU]/[g])

round the number and you get the illustrious 8.2 factor mentioned before:
Time [days] = SQRT(8.2 * distance [AU] / acceleration [g])

the simpler version as mentioned before would be to simply use 50 AU as default for any solar trip so inside the sqrt you get 8.17 * 50[AU] = 408.5 :
Time [days] = SQRT(408.5 / acceleration [g])

Note:

if its possible to make a warp jump WITHOUT the need to have speed 0 then the halfway point needs to be factored out and you can then just keep accererating to the edge of the solar system. Similarly, if you EXIT the warp space you could also assume that you have the same speed as when you entered it meaning that you would only need to decelarate to get to the correct point. Of course if you accelerated from a 50AU solar made a jump to the edge of a 30AU solar system, then you'll be missing target by 20AU!!. Stuff can get a bit complex this way. Anyway, the corresponding formula assuming NO halfway point (just take out the 2* and 1/2 from the above steps):
[days] = SQRT(4.1[AU]/[g])

This might be an interesting read as well

wolph42 said:

Well done. Thanks for saving me all of the work.

 There is no such thing as zero speed in space. So forget having to 'slow down' to enter warp IMHO.

If you emerge at the same relative speed and direction as you enter then the navigator needs to try to emerge so the ship is heading towards the planet of interest. 

 Well, there is stationary relative to the local star. But I cannot recall ever having read that this was an issue, so it really is up to each GM to decide whether this is required or not.

It might also be very dangerous to come out of the warp at very high speeds; you may be on a collision course with something big. Then there is the fact that if you enter a smaller system, and have a shorter distance to travel than the one from which you originate, then you will be traveling too fast to slow down in time.

The idea that warp jumps need to be done in deep space dates back the Asimov - earliest I am aware of. The concept was that the calculations to safely enter/exit could only be done in regions where space was 'flat' enough.

For relative speed wrt the local star to be a factor then the dC/dT (where C is the local Curvature of time-space) would have to be a factor. Given the tiny dC/dT values we would be dealing with here, I suspect background gravity waves would be greater in magnetude, preventing all warp jumps. So IMO this cannot be the case.

Speeds in space are astronomic. If you emerged on collision course with anything you are in trouble regardless whether you call your speed zero. In any case deep space is vastly empty. The chance of emerging near anything is negligible.

Wrt overshooting, it would be part of the job of the navigator to emerge at a suitable point to make an efficient system entry. Obviously he should build in margin so the ship doesn't overshoot, but this isn't a big deal anyway. Slightly embarrassing perhaps..