Re: [FT] Armour (long)(was Re:Armor)

Brian Bell wrote:

>>>Hmm... could use the same trick they use for screens - set a
>>>minimum MASS - much harder to justify though!
>>
>>Not very. Assuming a roughly similar shape for all ships regardless
of
>>size, the Mass of the armour will realistically be proportional to
>>(ship's volume)^(2/3). [snip]
> 
> Oerjan,
> Where did you get the (ship's volume)^(2/3) figure?

You seem to have interpreted my post as saying that the surface area is
EQUIVALENT to volume^(2/3). If you read a little more carefully you'll
find that I said that it is PROPORTIONAL to volume^(2/3). The
difference is that when things are only proportional to a function
instead of equivalent, there's an undetermined but constant factor in
front of the function; when you have equivalence that constant factor
has also been determined.

Assumption: "roughly similar shape for all ship classes" = the ratios
between the ship's length, width and height remain more or less the
same for all ship classes.

This means that you can use any of the three (W, D or L) as a generic
"size measurement" - it doesn't matter much which you use, since
converting from one to the other simply involves multiplying with a
fixed factor. Below I use L as the "size measurement".

With the assumption above, the ship's volume scales as L^3 - or,
conversely, L scales as volume(1/3). 

It's surface area scales as L^2, but L was proportional to volume^(1/3)
which means that the surface area is proportional to volume^(2/3).

The Mass of the armour is directly proportional to the area it covers,
so for any given thickness of armour the Mass of the armour is also
proportional to volume^(2/3), which is what I wrote above.

>A sphere has the least surface area per volume of any shape. 

The relation between the sphere's surface area A and volume V is

A = 4*Pi*(3/4Pi)^(2/3) * V^(2/3) (roughly 4.84 * V^(2/3) ), ie. A is
*proportional* to V^(2/3). It isn't *equivalent* to V^(2/3).

For a cube, the relation is A = 6 * V^(2/3), ie A is *proportional* but
not *equivalent* to V^(2/3).

For a box with L = 2X, W = H = X the relation is  
A = 10 * (1/2)^(2/3) * V^(2/3) (roughly 6.3 * V^2/3), ie. again A is
*proportional* but not *equivalent* to V^(2/3).

For a needle with L = 100X, W = H = X the relation is

A = 402 * (1/100)^(2/3) * V^(2/3) (~18.7 * V^(2/3) ). *Still*
proportional but not equivalent to V^(2/3), provided you keep the same
L:W:H ratios for all values of the "generic size measurement" X.

No matter what basic shape you use, if you just scale the size up and
down while keeping the general shape intact you will *always* get an
area:volume relationship of A = [constant] * V^(2/3). The actual
constant will depend on the specific shape (ie., the L:W:H ratios), but
it doesn't depend on the *size* of the body.

The value of your constant should be determined WRT game balance and
not WRT the actual shape of the ships. Remember that since you're
interested in the armour *mass*, not its *area*, the constant also
includes a rather arbitrary armour density and thickness.

>I would suggest that the surface area of the average ship if FTFB has
>2-3 times the surface area of a cube made with the same volume. So >a
formula of (ship volume)^(4/3) or (ship volume)^2 would be more
>accurate.

Brian, I don't know how much maths you've read and I do hope that this
is redundant info for you, but squaring a number is only the same as
multiplying it by 2 or 3 is a couple of very specific cases (2 and 3
respectively, to be exact) - the further away you get from those
numbers, the more wildly inaccurate the formula becomes. 

So, if you say that 2 times the surface area of a cube gives the
formula area = volume^(4/3), you're effectively stating the exact
volume of the ship. That formula is correct for exactly 1 Mass rating
for each basic hull shape (OK, mathematically there are 2, but the
other mass rating is less than 0 :-/ ).

Regards,

Oerjan Ohlson
oerjan.ohlson@telia.com

"Life is like a sewer.
  What you get out of it, depends on what you put into it."
- Hen3ry