[StratComm]

Scientifically, yes.  Unfortunately the way MOI is used in FS, size does have a pretty big determining factor.  Shape counts, but how big the shape is also matters.

[Goober5000]

Size always matters in MOI, because the MOI of a point depends on its radius.

You ought to be able to get a good estimate of the MOI of a ship by finding the MOI of a bar with the same mass and length.

[Nuke]

well none the les can you come up with a means to calculate moi based on an objects geometry.?

[Raven2001]

Originally posted by Bobboau
well holy crap, defineing the normal arbitraraly actualy made it infinitely simpler

(((1/(60 norm.z))((((pt3.x (((-pt1.y) + pt2.y)) + pt2.x ((pt1.y - pt3.y)) + pt1.x (((-pt2.y) + pt3.y)))) ((norm.y ((4 pt3.x^2 pt1.y + 2 pt1.y^3 - pt3.x^2pt2.y + 2 pt1.y^2 pt2.y + 2 pt1.y pt2.y^2 - 3 pt2.y^3 + pt1.x ((pt3.x ((3 pt1.y - pt2.y - 2 pt3.y)) + pt2.x ((3 pt1.y - 2 pt2.y - pt3.y)))) + pt2.x^2 ((4 pt1.y - 3 pt2.y - pt3.y)) + pt1.x^2 ((2 pt1.y - pt2.y - pt3.y)) + 2 pt2.x pt3.x ((2 pt1.y - pt2.y - pt3.y)) - 3 pt3.x^2 pt3.y + 2 pt1.y^2 pt3.y + 2 pt1.y pt2.y pt3.y - 3 pt2.y^2 pt3.y + 2 pt1.y pt3.y^2 - 3 pt2.y pt3.y^2 - 3pt3.y^3)) + norm.x ((8 pt1.x^3 + 3 pt2.x^3 + 3 pt2.x^2 pt3.x + 8 pt1.x^2 ((pt2.x + pt3.x)) + pt2.x ((3 pt3.x^2 + pt1.y^2 + 2 pt1.y pt2.y + 3 pt2.y^2 + pt1.y pt3.y + 2 pt2.y pt3.y + pt3.y^2)) + pt3.x ((3 pt3.x^2 + pt1.y^2 + pt1.y pt2.y + pt2.y^2 + 2 pt1.y pt3.y + 2 pt2.y  pt3.y + 3 pt3.y^2)) + pt1.x ((8 pt2.x^2 + 8 pt2.x pt3.x + 8 pt3.x^2 + 8 pt1.y^2 + 7 pt1.y pt2.y + 6 pt2.y^2 + 7 pt1.y pt3.y + 6 pt2.y pt3.y + 6 pt3.y^2)))) + 5 norm.z ((pt1.x^2 + pt2.x^2 + pt2.x pt3.x + pt3.x^2 + pt1.x ((pt2.x + pt3.x)) + pt1.y^2 + pt1.y pt2.y + pt2.y^2 +pt1.y pt3.y + pt2.y pt3.y + pt3.y^2)) pt1.z))))))

[edit]made into almost-C

Ah, yes much simpler, now any of us common mortals can understand it, don't we??...

[Fieari]

It's simple enough.  It's an equation... y'know, like the pythagorean theorem or the quadratic equation.  You plug in the numbers and out pops the value.  I don't personally understand WHY the equation would work, but then, I flunked college physics I (PHYS151 IIRC) twice in a row, so the reasons why are beyond me.  I find it interesting that only three points are referenced though... huh.  And I don't know which three points to use there, but... meh.  It's an equation.  Plug in the values.

[Sticks]

Actually this is really really overcomplicated, not to mention not really very applicable.

Newtonion inertia is most often estimated in space using an infintely small point of finite mass. So, for example, you place this point at the object CoG, and give it the mass of the entire object, and you have inertia. In fact, this estimation is generally so close that NASA most often uses it in mission planning.

Shape has no real effect outside of fluid dynamics, the only things you need are mass and CoG. Density however, does have an effect, but for the perposes here you can assume the entire object has the average density of the whole.

Check out http://www.saburchill.com/physics/chapters/0024.html for a simple MoI of a point mass.

[Bobboau]

so by your estimation rotateing the colosus around it's z axis would be nearly identical to rotateing it around it's y axis in terms of energy requiered.

[Sticks]

Yes, because you need to rotate the entire mass anyhow. I suppose if there were huge density differences about the ship, it would make a difference, but for simplicity's sake, yes.

[FireCrack]

So you're saying that if you pick up a big long stick of wood (contstant density) it takes the same energy so spin it along it's axis as it does to spin it perpendicular to it's axis.

[Sesquipedalian]

That's not quite what he was saying.  Follow the link he gave.

[FireCrack]

Ah, ok i think i get what he's saying now, use just the inertia of the objects cog and distance from point of rotation to find moi? what happens when cog is center of rotation?

[Bobboau]

I can tell you there is huge diference in MoIs for diferent shaped objects and it makes a very noticeable diference, if you don't beleve me take a pencil, hold it by one end and spin it along it's length, then grab it from it's center and spin it perpendicular to it's length.
in the game we don't try to rotate a mass along a ridgid length (like a weight taped to a ruler) we are rotateing ridged bodies a herc's MoI is going to be radicaly diferent to a Valc's. when you rotate an object you are aplying a torqe, and the object is applying back, it comes down to this the sum of the mass of all the atom's of the object * there distance from the axis of rotation squared. if we were to represent all the objects in game as point masses, then there MoIs' would all be 0, because it's mass*r^2 were r is the distance of the point from the CoM wich because it's just a single point is 0 wich would mean the slightest feather's touch anywere on the object would instantly cause the object to start rotateing that way. from the site you yourself pointed to
http://www.saburchill.com/physics/chapters/0024.html
"and the moment of inertia of any body can be found by adding together the moments of inertia of all its component particles."
I am integrateing the MoI's across a polyhreia volume.
"M. of I. of a uniform rod of length l*     I = (ml²)/3
M. of I. of a disc or cylinder**    I = (mr²)/2
M. of I. of a hollow cylinder or ring    I = mr²
M. of I. of a sphere**    I = 2(mr²)/5"
notice how extreamly diferent all of these are.

and this is just for finding MoI for a single axis of rotation, what we need is a MoI tensor that will allow us MoI for an arbitrary axis of rotation.

now after all of this there is a simple way to aproximate MoI, you make a 3d grid and determine if the points in that grid are inside or outside the object, you add all the MoI calculations up from the points inside, divide them by the number of points inside, and bam! done.

if your looking for something to do, I'll give you detailed explaination on what all 9 elements of the MoI tensor are about (even though FS's implementation I'm still not completely sure about)

[Sticks]

In one sense you're right and in another sense I am. I wish I could devote more time to the subject, but I'm on the road and my laptop battery is running low.

If you want a phenomenal source of info on the subject as relates to games, try http://www.d6.com/users/checker/dynamics.htm

It looks to be mostly integrating several point masses to get a reasonable answer. A perhaps simpler solution would be using the MoI of a cylinder, since most ship fall vaguely into that shape range. The other issue to consider is where you are applying your force, i.e. where are the manuvering thrusters located about the ship, etc.