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How big is the Expanse


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#1 Kraken

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Posted 26 March 2013 - 02:09 PM

Is there any mention in the sourcebooks about the size of the Koronus Expanse?

For that matter, anyone know how large the Calixis Sector is? The Rogue Trader Core book says that sectors vary in size depending on the area, density of the stars, etc. It then mentions that a "standard" sector (if any such exists) is a square (probably more like a flattened cube) around 200 light years across.

Is the Expanse (as mapped currently and shown in the sourcebooks) the same size as the Calixis Sector, larger or smaller?

Any input, thoughts or insane specualtion welcome.



#2 Iku Rex

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Posted 26 March 2013 - 02:42 PM

 

A 200 LY cube seems as good a guess as any.



Note that given the same stellar density as in our part of the galaxy, that means that (unless I've made a mistake somewhere) the Expanse contains around 0.004*(200*200*200) = 32 000 stars. Presumably most are either uninteresting or unaccessible, but there should still be quite a few left to work with. :)



#3 Kraken

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Posted 26 March 2013 - 04:27 PM

My assumption is that stellar density would "thin out" as one travels away from the centre of the galaxy.

The Expanse is located much further out than Holy Tera, on the galactic fringes and probably has a lower density of stars, but still, even if you half it, that's a hell of a lot of stars. I don't think I'll bother try to map them all.



#4 wolph42

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Posted 27 March 2013 - 02:04 AM

Iku Rex said:

 

 

A 200 LY cube seems as good a guess as any.



Note that given the same stellar density as in our part of the galaxy, that means that (unless I've made a mistake somewhere) the Expanse contains around 0.004*(200*200*200) = 32 000 stars. Presumably most are either uninteresting or unaccessible, but there should still be quite a few left to work with. :)

 

 

 

although quite debatable. The w40k "game universe" is flat or at leat flat-ish. Obviously if you leave out the 3rd dim. entirely then it cannot contain anything, but if you set the 'thickness' to 1 LY then you get: 0.004*(200*200*1) = 160 stars!

edit: actually choosing a thickness of 1 is rather arbitrary. The 'space' per star is 6.3^3 LY. (That means that on average a piece of space with the dimensions of 6.3 x 6.3 x 6.3 LY contains 1 star)

Thus a far less arbitrary number would be 6.3 LY as that is the 'thickness' of the 'space/star' dimension. Using that you get to a total of: 0.004*(200*200*6.3) = 1008 stars! in a 'flat-ish' W40K-game universe.



#5 Iku Rex

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Posted 27 March 2013 - 02:58 AM

wolph42 said:

although quite debatable. The w40k "game universe" is flat or at leat flat-ish. 

What makes you think that?

 

Anyway, I found the relevant passage from the Core Rulebook (page 315): "Each Segmentum is divided into sectors varying in size according to local demands and stellar density. A typical sector might encompass seven million cubic light years, equivalent to a cube with sides almost 200 light-years long." 

So the default is actually around  7 000 000 * 0.004 = 28 000 stars. Stellar density influences the size of a sector, so sectors further out, like the Calixis Sector or the Koronus Expanse, are probably larger in volume.



#6 wolph42

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Posted 01 April 2013 - 08:32 PM

Iku Rex said:

wolph42 said:

 

although quite debatable. The w40k "game universe" is flat or at leat flat-ish. 

 

 

What makes you think that?

 

quite simple: all the maps they use for the games are 2-dimensional. The 'fluff' obivously is 3D but the 'game' is 2D. Another example (if not convinced already) is space combat in RT its definately NOT 3D, but 2D. For obvious reasons, but still 2D. 






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