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..

True enough. I was influenced by Diaspora, where you make a similar journey to the slipgates outside the system.

I can't remember the rationale here, but at least in this game you do in fact start decelerating at half-way, in order to achieve complete stop before "jumping".

But with the warp, this may not be a requirement.

Unless there is any canon source on this, I would say it isn't needed. I can see no obvious story advantages either. Although in 'Fire Upon the Deep' by Vernon Vinge, a group used asteroids impelled to high differential velocity and jumped them on collision course with a planet… Not possible in wh40k.

The original fluff (Space Fleet vintage, if not the ORIGINAL Rogue Trader game) put a (soft) limit of 0.01c* on travel out to the Warp Zone. Since, even then, ships were capable of sustained acceleration that would bring them to that speed far sooner than the weeks/months it posited taking to reach the Warp Zone and entering the Immaterium, this suggests that entering the warp must occur at or below that velocity relative to the star (Imperial ships aren't equipped with forward facing main drive tubes, and for some reason making turnover to slow down quicker doesn't appear to be a common maneuver).

*The phrasing is "may reach speeds of up to 1% the speed of light".

Well, as discussed before, a ship's velocity wrt the local star seems a very odd limitation. It strains for a explanation to anyone with a modest physics background. Obviously the physics of warp translation are up to GW to define if they care to, but why make 'speed' a factor? What 'explanation' for it could be offered that would have verisimilitude?

I can't see any good reason from a story perspective. Unless the author wishes to plagiarise the naval engagements in 'Hornblower' et al… Perhaps this is the heart of it. Rather than develop a coherent vision of how spacecraft would manoeuvre and engage each other, you limit their relative velocity and narrate their motion as you would sea vessels. I hope this isn't the case. It hasn't been in any BL books I have read.

P.S. Not letting ships turn 180 degrees is getting silly…

At a guess, it's because warp conditions could be different further on, and putting a cap of .01c on your velocity at translation ensures that pre-warp calculations are likely to be still valid (ie, you won't have moved too far on). I personally doubt that there's an actual speed limit (or at least, not one that low- obviously particle densities are going to be a problem once you get fast enough); if memory serves there's an example in the Horus Heresy of a ship entering combat at .6c.

And you're right, there's no reason for ships to be unable to make 180 turns for slow-down, it just doesn't seem to be how the fluff plays it.

Warp conditions different further on? Further on what? Even 0.5c is crawling like a snail wrt interstellar travel.

The first step in understanding the theory of relativity is to abandon the notion that speed exists as an independent variable.

Q: What is the velocity of any object? A: It's relative motion with respect to the reference frame you choose to measure it against.

The choice of reference frame is entirely up to you. The universe don't care. Is the ship moving at 0.01c with respect to the nearest star or is the star moving at 0.01c with respect to the ship? Or are they both moving with respect to the galactic centre? How fast is the ship moving relative to a galaxy nearly red-shifted beyond detection? What about some completely arbitrary reference frame? All measures are equally valid.

What is this business about particles in deep space? Such particles would often hit you at a high fraction of c whatever speed you choose to say you are travelling…

The 40K universe has always tried to mimic The Age of Sail when it comes to the nature of ships traveling the oceans between the stars. So I think this is probably the reason for the various limitations.

As for the warp, well it's another dimension. Perhaps you must be stationary (against whatever. I doubt the designers thought of the constantly moving space) in order to project precisely into this other dimension. Perhaps your vector going in is reflected when going out, and you need slow (0?) speeds to be able to safely emerge into the unknown void at your destination.

The old descriptions actually made the trip to the edge of the system the major part of the trip, rather than the warp jump itself (excepting particularly long jumps). The actual warp jumps seemed to most commonly last a few hours or days from the ships perspective (from a few days to a couple of weeks realside), while it was the trip to the edge of system would take at least a week (and then probably another week at the other end to get to the destination).

More recent stuff (including the RT book) seems to suggest much longer (time wise) warp jumps, and play down the real side travel. Simpler that way, but leads to certain logical issues (surprise attacks on planets should be much easier).

I think I decided for my purposes for simplicity's sake it should be about week each end for intra-system travel.

Something which isn't yet mentioned is the fact that your target planet, may be on the other side of the star you've just translated in on. This could double your in-system travel time.

I've read a handful of sci-fi novels based on less fantasical fiction. One of things I liked about them is the "micro-jump". I do think it is possible to bring this into 40k, maybe it is the energy required for a full translation that causes the issues, a "micro-jump" uses less energy. I don't know, I just like the idea of the micro-jump, but you would still have issue's if you attempted, more than 1 or 2 micro-jumps.

I see this topic has progressed on. Im a bit late with my reply on the last but still willing to give my 0.02ct .

The one thing that has been mentioned but not really explored is the warp itself. Basically its outside our laws of physics hence anything goes. E.g. when exiting the warp:

1. your speed is always 0 in respect the closest gravitational system (CGS) so in effect relative to the speed of its star
2. your speed is equal to your relative speed from the CGS you left to the closest CGS you enter
3. anything else: half, double, triple, arbitrary, die roll etc. etc.

lets pick 2, which is the most interesting, the least 'arbitrary', and seems to be on most peeps mind (that most who actually think about this stuff).

Then there are 2 very interesting options:

1. direction (you can exit the warp in the direction of the system, but all other angles, including AWAY from the system, are also possible)
2. position (you can exit the warp at a point where THE planet is between you and its start, but its also possible that the star is between you and THE planet., and any other combination)

both are a matter of navigation where time tracking is vital (exactly as it did in the old days and with currently with our modern GPS systems). And interestingly enough, the time estimation is one of the navigators rolls!!

What is also interesting is HOW does entry and exit of the Warp work?

Here's a thought: say that two ships decide to travel from A to B. They enter the warp at A with the same speed and in the same direction (basically they fly next to eachother). Somehow they manage to keep in track of eachother in the warp. Now they exit at point B. Is it possible (by choice) that ship 1 exits the warp with a relative speed (RS) TOWARDS the CGS while ship 2 exits the warp with a RS AWAY from the system??

If it can, the travel in the void can be minimized by good navigation cause you always want to exit the warp with a RS TOWARDS the CGS AND with THE planet between you and its star. However you do need to know the time (to know where the planet is in respect to your entry point and its star)

If it can't then when you exit with a RS facing the CGS, the planet could be on the other side of the sun, which would add roughly add 2,5 AU's to the travel distance. (Radius of our system is 40 AU, distance Start-Planet is 1 AU, so 39 ideal, 41 if its on the other side and you don't want to go THROUGH the star so you need a 0,5 AU detour). Come to think of it, its of course also possible to exit the warp 'above' or 'below' the system, as a solar system is flat! But again this would have little impact on the 'extra' distance to travel. It boils down to (worst case) of only 6% of the ideal travel time, so actually no issue at all. @shockwave: it will certainly NOT *double* the time.

So TO CONCLUDE:

If I were to houserule:

1. Every system is 40 AU (our solar system which is fairly average)
2. THE planet (of your interest) is always at a distance of 1 AU from its star*
3. Your relative speed (RS) on entry to the closest gravitational system (CGS) is the same as your RS to the CGS you left.
4. That means to LEAVE a system you only need to ACCELERATE and to ENTER a system you only need to DECELERATE

*keep in mind that if its NOT that you still need to decelerate the same distance as from the system you just came from!!

That means that there is no half way point and that means that the used formula is [days] = SQRT(4.1[AU]/[g]) with 40AU this becomes (a bit rounded) [days] = SQRT(165/[g]) , for e.g. a merchant class raider travelling at 5g this becomes 5.7 days to exit or enter a system. If we take the 'the star can be between you and your destination detour' into account than this is +6% MAX, which equals 8 hours. Basically I would ignore this. If a navigator REALLY screws up his her rolls then they're 'somewhere else' anyway .

And to make it really easy here a table:

G (DAYS, both for exit and entry)

g (Days)
1 (13)
1,5 (10,5)
2 (9)
2,5 (8)
3 (7,5)
4 (6,5)
5 (5,5)
6 (5)

You could assume that your Realspace velocity and vector is maintained. Otherwise you are not conserving momentum…

You could also assume that navigators can sense where a nearby star 'is' while within the warp. So once he has 'found' the star-system he is aiming for (he thinks) he can always position the Realspace translation point at a suitable distance. If he maintains knowledge of what his Realspace vector is, then he can position the ship so that it is heading into the system.

Fresnel said:

nicely put! Thats what my assumptions boils down to. Here you find (on the first page) an xls tool that calculates the total travel time (excluding the warp part) in the void from A to B.