Deckspace in Spheres: Math Calculation Question

[Andrew Boulton]

"I too was thinking 3m decks were standard, but then you would have to use deck squares 2.16m to a side"

Standard deckplans use two 1.5m squares per dton. Decks are 3m high.

2 x (1.5 x 1.5 x 3) = 13.5

 

[RainOfSteel]

I guess I'm just trying to figure out the diameter of all my decks.

The diameter of the largest decks, #125 and #126, is of course, 1000m. 1000m / 1.5m = 666.66 squares.

But what about Deck #67? (I'm still looking at the formula previously posted, but the author of them expressed doubt in them while writing it . . .)

Figuring out how to make some piece of software let me draw decks that big comes second. Figuring out how to keep track of 250 really-big decks in the same document comes third.

 

[RainOfSteel]

Oh, and because I'm making my decks bigger, each pair of squares will displace 18,000 liters, instead of 13,500 liters. However, most of this is in extra ceiling space (or is in the 1m floor-space), and so little additional useable space is recovered. Maybe I'll have to factor that into a standard increase in the volume of some components (like apartments, offices, etc.), but not others (like armor, or the Aquarium and Garden).

 

[far-trader]

What you want is to calculate the chord (length across) of the circle for different heights. What you have are two pieces of data, the radius and the height. Lucky for you all you need for circle math are any two pieces of data and you can solve for all the rest.

Even luckier I found a website that explains it better than I could and it includes a link to an excell sheet to do all the math, talk about easy. If the link works. I'll leave that to you to check out. Here's the site link:

Math Forum - Circle faq

What you want is case 12 solve for c (chord) knowing h (height) and r (radius):

EDIT: oops and d (the distance from the center of the circle to the middle of the chord) which is r-h:

theta = 2 arccos(d/r)

c = 2r sin(theta/2)

What you need to do is work from the top (or bottom) and use the total height to the deck you want to know the diameter (equal to chord) of.

You will have to figure out the chord for both the top and bottom of your deck to know where the cieling (or floor) starts to curve, and decide then if you'll use the curved part as a feature or lost space.

I think that should be all you need. Have fun

What? Math can be fun, as long as the use is fun, right?

 

[RainOfSteel]

far-trader,

Thank you for letting me know. I'll be looking as quick as I can manage.

Right now, it's off to rescue Pharoh from the clutches of the minions of Set, somewhere in the Fifth Hour of the Night (getting out is going to be a pain, because you can't go past the Fifth Hour without being judged and dead, and there is no going back).

Then, tomorrow morning, early, it's work all day.

 

[far-trader]

You're welcome RoS.

I had some time to get on the PC and fired up the excell sheet in Open Office and it seems to work fine. Just plug in the knowns and find the results in the case of interest (case 12 in this instance).

So a quick look at the "second" deck as an example, being the one between 4m and 8m from the hull. The chord at 4m is about 126m and the chord at 8m is about 178m. So you have full deck height (4m) on deck "two" with a diameter of 126m and a curved cieling for another 26m around the perimeter. At about 12m from the hull the cieling is down to 2m. This is too easy, all those high school trig classes for what Well I can still do some of it without aid so that's something, and yes the teachers didn't lie, it's stuff you can use and at least I knew the terms to do the search to find the spreadsheet. Still it feels like cheating, like I've betrayed the old ways.

Anyway have fun when you get to it and please share an outline of the thing when you can or write it up for submission.

 

[aramis]

a Reasonable asssumption for drawing the plans is to use the median width of the deck, rather than averaging top and bottom

this puts the middle decks based upon h=2, rather than h=0, and the final decks at h=498, not 496-500.

(this actually will short a small amount of volume.)

 

[GypsyComet]

Except for the top and bottom 5% or so, the wall slope will be steep enough that that can still be useful space, even if it isn't *walking* space.

 

[JAFARR]

Originally posted by Andrew Boulton:
"I too was thinking 3m decks were standard, but then you would have to use deck squares 2.16m to a side"

Standard deckplans use two 1.5m squares per dton. Decks are 3m high.

2 x (1.5 x 1.5 x 3) = 13.5

Yes, I finally found it during my lunch break today. I had forgotten the "use 2 deck squares" part of it. And while CT uses "about 14 cubic meters" MT uses 13.5 based on the formula above.
Let me see....as I stated in an earlier post enclosed 18 wheeler box trailers are just under 3m in hight... there are 10.76 square feet in a square meter.... 4.5 square meters in a dton if deck hight is 3m...= 48.42 square feet per ton... trailer boxes are 8 feet wide....a 40 foot long trailer box is about 6.6 dtons. That makes a 200 tn Free Trader about the size of 30.3 - 8 x 40 sea/land trailer boxes.

 

[JAFARR]

As this is not a ship and does not need large masses of fuel, this alternate method would not apply here, but if it were a ship you could compute the non-fuel volume and design a cylinder of the required volume and place it within the sphere. Then the space between the cylinder and the sphere becomes fuel tanks. You might even have a cylinder inside a cylinder arrangement. 1 longer and skinny with a shorter, fatter one outside the first and both being centered inside the sphere with their centers at the "equator" of your sphere. Sort of like a "ringworld around a cylinder with the sides of the ringworld connecting to the cylinder and the whole structure inside a "Dyson sphere". All your decks would be the same diameter(s).

 

[RandyT0001]

Using 130.8 million square meters as the total floor space that would be 130.8 square kilometers or 48.7 square miles. The closest major US city (not including adjacent suburban communities) would be Boston, MA with an area of 48.4 miles^2, followed by Moreno Valley, CA with 49.1 miles^2, then Tacoma, WA with an area of 48.0 miles^2. Or put another way it would cover about 78% of Liechtenstein.

 

[RainOfSteel]

Originally posted by Randy Tyler:
Using 130.8 million square meters as the total floor space that would be 130.8 square kilometers or 48.7 square miles. The closest major US city (not including adjacent suburban communities) would be Boston, MA with an area of 48.4 miles^2, followed by Moreno Valley, CA with 49.1 miles^2, then Tacoma, WA with an area of 48.0 miles^2. Or put another way it would cover about 78% of Liechtenstein.

Good. That means there will be plenty of office space.

It's certainly in high enough demand.

 

[BetterThanLife]

I used to know the math for this.

Let me do some number crunching and a little looking. (Was a math major in College.)

 

[RainOfSteel]

Ok, there. Work and post-work Mother's day activities are over now.

I think I'll flop into bed and look at this on Monday.

No, really!

 

[RainOfSteel]

Thank you, everyone!

I'm back at work (for the moment), on this little project.

I'm continuing this over in Deckplans: Imperial Palace (aka Pie in the Sky)

 

[Fritz_Brown]

Originally posted by RainOfSteel:
</font><blockquote>quote:</font><hr />Originally posted by Fritz88:
Easy answer (

), RoS:
In Excel (or some such), set your zoom really low to start with. Highlight the whole sheet, and make your rows and columns the exact same height/width in pixels

How do you set the height of rows and columns by pixel?

I only know Menu>Format>Row/Column, and to adjust the number in the dialog box.

When I set these both to be the same value, the grid is not square.

(Note: I have Excel 2000.) </font>[/QUOTE]Sorry to be a bit late on this answer, but I had the weekend with my family and my mother in town - Happy Mother's Day

!

After highlighting all the columns/rows you want to change, put your cursor over a line between columns (or rows) till it changes to a line with two arrows coming out from it. Click and drag the column (or row) border. You should get one of those little yellow boxes (Tooltips) that shows your measurement of choice (for columns) or "points" (for rows) and pixels in parentheses. There are a couple of spots where the pixels value will jump by two instead of one, but it is in the same spot for both rows and columns.

 

[LordVince]

Okay, RoS, now that you've got all your math lined up, it's time for me to toss a monkeywrench into things...

You're designing a spherical "station" that is 1000 meters in diameter, and then you're segmenting this sphere into 250 horizontal "decks", each 4 meters in height. Yes?

Well, here's something to consider.

One of Traveller's literary "grandfathers" -- the writer H. Beam Piper -- used the sphere shape for his hyperships. However, as the humans of Piper's stories were masters of artificial gravity, they built their ships a little different. Instead of building horizontal decks -- like a stack of coins; wider at the middle, getting smaller as you go "up" or "down" -- Piper built his ships outwards from the ships center.

Powerplant and gravity generators, as well as other large engineering equipment, would be tucked into a globe big enough for them. The inside of this globe would be "Deck 1"; and was usually kept in Zero-G. All the other "decks" were slightly larger globes, built around the central globe. Piper often had ships of 200 meters diameter. For your use, you just have to calculate the outer surface area of your 1000meter globe -- that's the outershell. Then run the same calculation for a globe of 992meters, then 984meters, etc., etc. Each time making the next smaller globe's diameter 8meters shorter than the previous deck.

I'm guessing that you'll get less than 250 decks, but the AREA of each deck will be the entire surface of a globe. You'll probably get more deckspace this way.

Of course, the stations artificial gravity will pull inwards/towards the core -- just like on a planet. Ships will dock by lowering down, into open hatches -- in fact, the outer few decks will probably be dedicated to landing pits and sensor pods -- everyone will "live" on the deeper decks. But, with lots of space to play with, as well as the highdef holographic plates that are all over Traveller (i.e., programmable "windows"), this won't be an issue.

So consider the "onion-skin" design, rather than the "stack of coins". I mean, on a station oif this size, you don't have to "draw" all the floorspace of every single deck -- you just draw a few important areas. A shopping mall, a "wharf" area, warehouses, a park, office areas, things like that. Inside a globe of this size, all that crap could fit ANYwhere!

 

[RainOfSteel]

I quite like Mr. Piper's work.

However, my 1000m "station" is the Imperial Palace on Capital/Core. It's overall design is somewhat pre-determined.

 

[TKalbfus]

The Surface area of a Sphere is 4(3.1415926535)r^2.

The outer deck would be 4(3.1415926535)(1000m)^2.
The deck inside of that would be 4(3.1415926535)(992m)^2, the one inside that would be 4(3.1415926535)(988m)^2. Has anyone ever considered a starship that consisted of nested spheres instead of cross sections of a sphere? It all depends on how one arranges his artificial gravity doesn't it. Another option is to have the ship spin for gravity.

 

[RandyT0001]

Tom, you used the diameter of the sphere (1000m) instead of the radius (500m) in the equations above.

I don't know of any way to have a sphere spin in order to simulate 1G gravity across all parts of the inner surface. The areas closest to the axis of the spin wouldn't have any gravity. You would have to use grav plates to simulate the gravity.

At some point of nesting spheres within each other the curvature of the spheres becomes very noticable and somewhat disorienting to those who occupy the sphere. At the 500m radius it might not be so bad but at 100m radius the floor will curve upward (or downward depending on which direction the grav plates are installed) in a noticable way. IMO