Coriolis Station Physics

[Cosmo Bozo]

It is definitely more than 2km when I flew yesterday. Have to actually get in game to measure that of course Well, maybe in 3.1

Impeccable is 2km long isn't it?
It definitely seems bigger than that. Feels to me like you could fit impeccable in the landing area if it would fit through the slot.

Of course I cant get in game to check these things now.

 

[Iskariot]

I just listened to the relevant part of the Lave broadcast again and Mr Braben says that the 'outside' of the station is supposed to have approximately 1 G and the docking area 0.1 G.

 

[drew]

We've had hints that some of the bigger stations are going to be 10km + in size. I think a Coriolis at 4km feels about right.

My in book descriptions of the interior require this sort of size.

Cheers,

Drew.

 

[Dirtmuncher]

In case of emergency rectangular doors are easier to close up then a round Iris. And your playing elite space jockeys so 1 cm clearance should be enough to get you inside otherwise put it on autopilot, like they do with docking with the ISS.

 

[Slythe]

The door doesn't need to be bigger, it just needs to be rounder

If you take the existing door width and convert it into a circular door of the same width you end up with a larger area to block with shields. If the existing door has a 5:1 width/height ratio, then the door area would be almost 4 times larger if circular.

 

[Rog]

If the station is 3km in diameter, and the rotation takes about 80 seconds, then the force at the circumference is about 9.3ms^2 which is about right?

We need a big tape measure.

 

[Cmdr Eagleboy]

Impeccable is 2km long isn't it?
It definitely seems bigger than that. Feels to me like you could fit impeccable in the landing area if it would fit through the slot.

Of course I cant get in game to check these things now.

It is way bigger than Impeccable, so 4km sounds very very true.

 

[drew]

If the station is 3km in diameter, and the rotation takes about 80 seconds, then the force at the circumference is about 9.3ms^2 which is about right?

We need a big tape measure.

Which is what I said in my opening post.

Cheers,

Drew.

 

[Slythe]

If the station is 3km in diameter, and the rotation takes about 80 seconds, then the force at the circumference is about 9.3ms^2 which is about right?

We need a big tape measure.

You need someone to park their 21m wide Sidewinder in the hatch while another player takes a screenshot. Then we can use that to accurately measure the size of the hatch and after that the entire station, based on the hatch... simples!

 

[Cmdr Eagleboy]

If you take the existing door width and convert it into a circular door of the same width you end up with a larger area to block with shields. If the existing door has a 5:1 width/height ratio, then the door area would be almost 4 times larger if circular.

This. Current exit not only makes sense from legacy POV, it is also feels right from security and support POV.

 

[Rog]

Which is what I said in my opening post.

Cheers,

Drew.

I plead hangover-induced brane failure.

 

[horncastle]

This is why I love Elite D, actual detailed physics being modelled and the players having the same eye for detail.

I have visions of the SC team .....lets make this station huge man like America hhhhhuuuu haarr high fives all round.

 

[j6wbs]

Which is what I said in my opening post.

Cheers,

Drew.

We do have a tape measure, if the navigation beacon is in a fixed position relative to the station.

I have landed on the far side of the station, as high as I can, and I have landed on the docking side of the station, both times taking the distance reading from targeting the navigation beacon. I get 3,621m and 753m respectively.
This gives an approximate distance of 2,878m. (give or take a little for error)

This is the distance of one of the diagonals of the large squares on the face of the cuboctohedron. Using Pythagoras' theorem, this tells us the side of one of the squares is approx 2,035m.

And if the station was held snugly within a globe, it would have the diameter of 4,070m (give or take a few metres for error) (i.e. sqrt(2,878^2 + 2,878^2))

Does that seem right?

Jez

 

[Slythe]

We do have a tape measure, if the navigation beacon is in a fixed position relative to the station.

I have landed on the far side of the station, as high as I can, and I have landed on the docking side of the station, both times taking the distance reading from targeting the navigation beacon. I get 3,621m and 753m respectively.
This gives an approximate distance of 2,878m. (give or take a little for error)

This is the distance of one of the diagonals of the large squares on the face of the cuboctohedron. Using Pythagoras' theorem, this tells us the side of one of the squares is approx 2,035m.

And if the station was held snugly within a globe, it would have the diameter of 4,070m (give or take a few metres for error) (i.e. sqrt(2,878^2 + 2,878^2))

Does that seem right?

Jez

All calculations checked and correct!

To add to that, the main hatch is roughly 1/8th of the width of the station (measured using my finger on the screen! ) which makes it about 360m wide, based on your measurements. And the height of the hatch is roughly 1/4th of its' width; 90m.

It should be possible to double-check it using the Sidewinder if someone can get a screenshot of one in the hatchway.

 

[j6wbs]

All calculations checked and correct!

To add to that I would that the main hatch is roughly 1/8th of the width of the station (measured using my finger on the screen! ) which makes it about 360m wide, based on your measurements. And the height of the hatch is roughly 1/4th of its width; 90m.

It should be possible to double-check it using the Sidewinder if someone can get a screenshot of one in the hatchway.

Thanks for the extra measurements Slythe, so far I think it is something like this then...

Jez.

 

[Maxeren]

Thanks for the extra measurements Slythe, so far I think it is something like this then...

Jez.

So the station is "only" 2.8km wide?

 

[Slythe]

So the station is "only" 2.8km wide?

Yes, or about 4km diagonally.

 

[CaptainKremmen]

I don't know why, but I just had a vision of a cardigan-wearing Federation supervisor carrying a clipboard, stopwatch and 100g weight... standing at the outermost level of the space station, performing drop-tests... and tut-tutting to himself as he crosses off his checklist items.

K.

 

[drew]

Excellent work. Now we just need someone who can get into the Alpha to give us an accurate rotation speed!

Can someone also work out the diameter of the docking area inside the station? We can then check the simulated gravity on the pads.

Cheers,

Drew.

 

[CMDR Texas Stu]

No idea WHY but the rectangular 'slot' opening allows for two ships to pass easily in both directions. We'll just have to decide which is 'in' and which side is 'out'. In deference to you Brits I've been entering on the left (unless I'm upside down then I've been entering on the right)!