newtonian motion

[Puke]

quote:

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Originally posted by Quikngruvn:
I know my head's still spinning from my Last post. Give me little bit and I'll get it flipped from negative to positive spin....

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you sure its not spinning in a state of supermotion? hold on while i observe it.


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"...the green, sticky spawn of the stars"
(with apologies to H.P.L.)

[Quikngruvn]

quote:

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Originally posted by Puke:
...i just figured that the time scale was such that the time and thrust necessary to turn... would be negligable compared to the thrust necessary to move a ship one square on the tactical grid.

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It may be. Let me see if I can so some rough calculations....

Kinetic Energy = 1/2 x Mass x Velocity^2
KE is expresses in joules, the unit of energy.

So, a ship of mass M must expend J joules of energy to rotate at a given velocity V.

Now, assume you have a 150 kT escort and a 1500 kT baseship, both of which want to turn around in the same amount of time. Since the baseship is ten times more massive than the escort, it must expend ten times as much energy as the escort to execute the same turn.

Further, the velocity of the basheship must be greater than the escorts.

Circumference = pi x Diameter
Distance = Velocity x Time

Assuming a 180 degree turn, the distance traveled for each end of the ship is one-half the circumference, thus:

Distance = 1/2 x pi x Diameter = Velocity x Time
Velocity = (1/2 x pi)/Time x Diameter

Since the time is the same for both ships' turns, (1/2 x pi)/Time is a constant. Thus, the only thing affecting the velocity of the spin is the diameter of the ship: the longer the ship, the greater the velocity.

Thus, the longer a ship is, the more energy it must expend to execute a turn in a given length of time. Unless there are some funky ship designs going on, the baseship will be significantly longer than the escort, and thus must expend more energy. Oh yeah, and the difference in velocity is also squared, increasing the difference in energy even more....

N.B. These are very crude calculations, really. They do not take into account the fact that different parts of the ship will move at different speeds during the turn. The true velocity of any point on a ship during a turn will depend on the distance of that point from the axis of rotation. But, seeing that it would take calculus to compute the sum total of the kinetic energies of each point on a ship, I simplified just a smidgem. However, since much of the baseship will be spinning faster than the fastest points on the escort, this shouldn't affect the gist of my argument, just the values I'm about to pull out of the air. I'll use points midway between the axis of rotation and the tip of the ship to get an "average" value for velocity and kinetic energy.

So, if you've got a 20-meter long escort and a 100-meter long baseship, the midpoints would be 5 meters and 25 meters, respectively, and so the difference in velocity would be:

Velocity = Q x radius, where Q is a constant
Velocity(baseship) = 25 x Q
Velocity(escort) = 5 x Q

Thus, the baseship's average rotational velocity is 5 times that of the escort.

So, to get back to the energy expended, we gotta go back to the kinetic energy formula and input my seemingly arbitrary values:

KE = 1/2 x m x v^2
m(baseship) = 10 x m(escort)
v(bs) = 5 x v(e)
KE(e) = 1/2 x m(e) x v(e)^2
KE(bs) = 1/2 x m(bs) x v(bs)^2
= 1/2 x 10 x m(e) x (5 x v(e))^2
= 1/2 x 10 x m(e) x 25 x v(e)^2
= 250 x 1/2 x m(e) x v(e)^2
= 250 x KE(e)

THUS, a baseship must create approximately 250 times the thrust of an escort to perform an identical turn in a given length of time, at least using arbitrary (but I thimk plausible) values.

And then on top of that, there's the counterthrust each ship must expend to overcome momentum in the ship's original direction....

Sorry, I got carried away. Puke, I think I could have put your logic in a chokehold, but I think I bored it to death instead. Now, lemme catch my brain before it dribbles out of my head completely....

Quikngruvn

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"That which does not kill you will make you stronger." -- Nietzsche

[Puke]

quote:

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Originally posted by Quikngruvn:
Sorry, I got carried away..

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not at all. i understood your point from the begining, and it is totally valid. what you are overlooking is that while it requires exponentially more thrust for a larger ship to spin, it requires them both far more thrust to move, say, one sqare on the tactical grid. unless the square is not much greater in size than the ship its self.

lacking my old physics ledgers, I cant pull up formulae for you, but let me give it a shot extrapolating from some basic geometry (granted I slept through geometry yet paid attention through physics, but guess which one comes readily to mind 7 years later? danm that. odds are my math is going to be invalid, so feel free to point out any glaring flaws) what we need to do is get a feel for the scale of the game:

in order to get an idea of scale, we need some rough idea of what a square is on the tactical grid. while they seem to have sides with length equal to the lengh of their diagnal, we will presume for a moment that they are insted standard geometric squares and not these bizarre geometrically impossible things. we can extrapolate that each of the (roughly) 63 squares along the side of the tactical map is 1/63 of a strategic square, which is in turn 1/13th the diamater of the stellar system in question. given that pluto is roughly 5.9 billion KM from the sun at its average orbital distance, we can simplify and say that the average solar system in se4 is 10 billion km in diamater. 1/819 of this distance (which is the size of a tactical square) is roughly 12 million km.

The Nimitz class carrier displaces about 97 metric KT and is roughly 330 meters long. if we figure that our escort is about 1.5 times the mass then we can presume (okay, its a crappy presumption, but its the simplest way of doing it) that its length would increase arrithmetically to 495(or 500)meters, the baseship following at 5000 meters. (and those distances are generous condisering that space vessels could be far more dense than earth based surface-ships, since they dont need to worry about silly things like displacing water)

now, with the .5*pi*D formula to find the longest distance traveled in a rotation (or half-rotation, as it were), we come up with about 7850m for the baseship and 785m for the escort. just in case you think im nitpickingon the distances you chose, im NOT: this is just to give you as much distance as possible regarding comparative lenght of a turn (that is, it would have been far more beneficial to my point if i had simply kept your numbers). I am in complete agreement that the core of your arguement reamins the same, and remains unchallenged.

BUT, we have ships that appear to be capable of transversing a minimum (basic engines) of 6*12billion meters(escort) or 2*12billion meters in one turn, from a standing start. now the original point of the thread was that they can not accelerate beyond this, and they can accelerate in the oposite direction without overcoming their own momentum, but we have obviously lost track of that at this point.

now, while I grant you that my flippant referance to turning being 'free in space' was something akin to Cisco's flatulant claims that bandwidth is free, the gist of the argument remains the same. the ammount of thrust expended to rotate a ship, a distance of no more than 7,850 meters, is significantly less than that required to transverse the 12,000,000,000 meters in one square. therefore, turning has no right costing any movement points, and even if a baseship takes longer to do it than an escort, it will happen in a space of time that is completely miniscule compared to the space of time required for either of the two example ships to move one square.

now, to forestall any other attempts at wwf-smackdown-style acrobatics being purprotrated against my logic, let me forstall your arguements by saying that you could further deconstruct this by supposing that a turn consists of a ship accelerating for the first half of its movement and decelerating for the second half, thus removing the requirement for a ship to overcome momentum on a turn.. i dont know where that would really get us, but i can imagine someone trying to take it somewhere. furthermore, i could go on to deconstruct your turning argument by stating that due to the increased length of the base ship, you can apply thrust further from the axis of rotation and achieve more velocity per joule than if they were closer to the axis of rotation (as is the escort's). but since im not actually arguing against your point, i wont go into the proof.

I usually try to keep things light-hearted, so I have to appologize if I have inadvertently turned this discussion into a logical testosterone fest. furthermore, i have to appologize for my abhorant spelling and guestimated mathmatics. Since I have been in basic agreement with you from the start, you probably dont need to expend quite so much effort running rings around me while I am only attempting to stand still and point in a related direction.

I do want to encourage further disucssion though. I enjoy a friendly intelectual conversation, and i usually tend to have more to learn than i do to offer.

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"...the green, sticky spawn of the stars"
(with apologies to H.P.L.)

[This message has been edited by Puke (edited 13 July 2001).]

[dogscoff]

*narf*

[Spoo]

quote:

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1/819 of this distance (which is the size of a tactical square) is roughly 12 million km.

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Ah, but you forget that scale has no meaning in Se4. By your system a planet would be 24 million km across -20 times the diameter of the sun! (But then again planets and stars are shown as the same size. Go figure.)

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Assume you have a 1kg squirrel
E=mc^2
E=1kg(3x10^8m/s)^2=9x10^16J
which, if I'm not mistaken, is equivilent to roughly a 50 megaton nuclear bomb.
Fear the squirrel.

[Puke]

quote:

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Originally posted by Spoo:
Ah, but you forget that scale has no meaning in Se4. By your system a planet would be 24 million km across -20 times the diameter of the sun! (But then again planets and stars are shown as the same size. Go figure.)

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well, thats because, um... the diamater of planets are measured by the diagonal of the square, which is an abstract representation of three dimensional space, and thus you can traverse what should be a shorter distance along the x axis by, umm.. utilizing the z axis, which is on a smaller scale.. and ahh... well, heck, even if you measure a square by the size of a planet, i think my point holds water.

quote:

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Originally posted by Dogscoff:

*narf*

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point taken, i stand narffed. i shall leave this thread alone.

[This message has been edited by Puke (edited 13 July 2001).]

[capnq]

It seems pretty silly to even pretend every object on the map is drawn to scale. Consider that the length of even the smallest 150kT Escort is a significant fraction of a planet's diameter.

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Cap'n Q

The most merciful thing in the world, I think, is the inability of the
human mind to correlate all of its contents. We live on a placid
island of ignorance in the midst of black seas of infinity, and it was
not meant that we should go far. -- HP Lovecraft, "The Call of Cthulhu"