I couldn't find anything about the time it takes for a ship to move through the void. Does anybody know of any rules, or of any fluff that gives an idea of how many days it takes for a ship to travel from planet to planet or from warp point to planet?

Not trying to be a smart ass but there's a very obvious indicator of how fast a ship can move through the void. Its speed.

Fluff is inconsistent on this point; some pieces indicate that it takes weeks to get to the goldilocks zone of a system from the outer fringes, while in others it's bare hours. I tend to hedge it at about a week at full burn.

As a sidenote, if you assume a Speed of 8 VU, one VU being 10,000km, it would take approximately 4.21 years to get from Earth to Pluto.

I completely overlooked that, but you're very right

However if I set the speed of an average ship (about 6) and the length of a void unit (10,000km) against the distance in our own solar system the voyage gets to be rather long. A warp translation point is, I think, always at the edge of a solar system, so (according to wikipedia) the farthest distance of pluto to the sun is about 50AU, or 7,500,000,000km (to get really clear of the solar system you'd probably have to travel even further). An average ship would travel 60,000km per half our, or 2,88 million km per day. To get to earth you'd have to travel at least 49AU, or 73,500,000,000km, making it a trip lasting about 70 years, and then you would have to make a return trip to travel to a warp translation point...

Errant said:

Actually I could live with an inconsistency like this. That would just mean that I as a GM could decide the amount of time they'd spent inside a system, and make it anything from a couple of hours to a couple of weeks. However having to make it anything from 4 years to 70 years for a one way trip would make the characters rather short lived

ships can make 2 warp jumps. 1 to get to the system and another to get close to ,relatively speaking, to planets. look at frozen reaches. the orks jump into the middle of your fleet. it is insanely close to the planet,within hours, but jump in a day or two away from the one you want after you have taken a look from out system. it's more dangerous but it saves a lot of time.

Yeeees, but about six of the nine vessels that attempt that are torn apart by the system's gravity well. I doubt most Rogue Traders would be terribly happy with those odds

that was my point about the distance from the planet the orks jumped. they jumped into real space within 12 hours of the planet. they lost around 1/4 of their ships as written in the adventure. jump to 2 days out from the planet. outside the planetary and moons grav well but still much closer than pluto's orbit.

I'm 95% certain the adventure specifically states that the ork role are torn apart by the STAR's gravity well, not the planet's.

Anyway, I found a paragraph that talks about void travel outside of combat. It pretty much leaves it up to the GM but says ships should move a lot faster out of combat than in combat, similar to combat speed and narrative speed for vehicles.I'm pretty sure the paragraph was in RT core but I'm not positive.

Page 183 on in the core rulebook talks about traveling through the warp

Red Bart said:

I think one could argue that in narrative time you can continue to accelerate and accelerate for hours and hours, eventually reaching speeds faster than speed x VU. I think going 1/4 the speed of light could be reasonable, making 50 AU's take only a few days. I believe the book says it takes several weeks to cross a planetary system (which I would interpret as the full width of it).

If you are trying to bring real-world physics into this, you're in for some pain.

Remember, there is no air-resistance in space, the only "drag" you have to worry about is inertia. And once overcome, a moving object will keep moving at a constant velocity without any further thrust.

At "Full burn", you're not measuring a constant "max speed", you're measuring a constant "max acceleration". As long as you're working those engines, you will keep going ever faster and faster, until you eventually approach light speed and relativity kicks in and limits your effective speed.

So for practical purposes I'd rule that the speed value for a ship (ie 6 VU p Combat turn) is an approximation of the distance a ship can cover while turning and maneuvering (and thus not benefiting from extended acceleration). But for long trips, no ship will be limited to this limit.

Even at a modest 1G acceleration, you will travel 1 AU in the first 2 days. At that point, your speed will be over 6 million kilometers pr hour (about 609 Void Units/hr), or about 1 AU / day. Keep accelerating, and you'll hit 2 AU / day after 4 days, 4 AU after 8 days, and so forth.

At your journeys half-way point, you would need to start decelerating at 1 G, in order to end up stationary upon your arrival. Whether this is necessary to perform a warp jump is not known, but let's assume so.

To travel 50 AU then, from standstill and to a complete stop, you would need to accelerate at 1 G for 10 days. This would take you 25 AU, and leave you with a top speed of some 3050 VU / hr. Then you simply hit the brakes (or more likely, turn the ship around and let the engines thrust as normal giving you the same deceleration as you had as acceleration). For the next 10 days you will cover the same distance, and end up stationary on the edge of our solar system after 20 days of travel.

If you did not need to slow down, you could be there after only 14 days at "full burn", but then you'd be moving at almost 4300 VU/h

And this is at 1 G. The level of acceleration you experience if you jump of something here on earth. With fancy-gravplating and inertial dampeners and other sci-fi tropes you could easily increase this, drastically reducing travel times. At 2G you'd just need 7 days before reaching your mid-point. At 3G, you're down to 6.

Of course, keeping this "full thrust" thing going would require a lot of fuel, which would require storage. Or some fancy extract-from-space-as-you-go thing. Or a lot of handwavium.

Or instead of constant acceleration, just accelerate until you're satisfied and turn off your engines and then coast to your warp transition point. Would save a bunch of fuel. Take a little longer, but cost less fuel.

As a GM, I go for approximately 2 weeks at "full burn" to get to safe warp tansit range for a normal star system (ie, like our own Sol star system). Modify that time by the max G accel of the ship class (roughly), a 2g ship will take 2 weeks, a 1g ship 4 weeks etc. A formula for that would be time in days = 28/g. Seems to work in my campaigns, and makes faster ships (in strategic terms) have a bit of an edge with travel times.

As an example, take the Ambition and Dictator class cruisers, both have a combat speed of 5VU, but the ambition has 3g sustainable acceleration vs the dictators 2.5. The Ambition can make a safe warp-point in 9 days and 8 hours while the Dictator can get there in 11 days and 5 hours.

Darth Smeg, any chance you could work up a light formula like the one Hygric uses? Going on your 1g = 20 days to clear the Sol system, something like

Small System: 10/G Days

Medium System: 20/G Days (Sol size)

Large System: 30/G Days

Would that be anywhere near accurate?

Problem is the equation is not linear.

However, the formula is something like this: Time taken (in days) = Square root (408,9 / G)

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.