quote:
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.
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
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