Venator Mass

[McC]

Most people will find this post utterly pointless and totally academic. Sue me, it's 0430

I was sitting in bed trying to work out the accelerations for various starfighters based on ICS datapoints. I got sidetracked, though, when I started thinking about the 40,000 tonnes/second figure for the Venator. I started playing with all manner of formulae before the obvious finally clicked: if ion drives are accelerating their exhaust mass at c, and if a Venator devotes its full reactor load to thrust, then 40,000 tonnes/second not only gives us a power rating for the ship, but also a Force rating for its engines (since it's telling us a known mass quantity, at a known velocity, over a known unit of time). Working that out:

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P = mv
P = (4E7 kg)(c)
P = 1.2E16 kg•m/s

Ft = mv
Ft = P
F = P/t
T = 1 second
F = 1.2E16 kg•m/s^2

From here, we know that Venators are definitely 'faster' than Trade Federation coreships (300G) and probably slower than Acclamators (3500 G).

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 300G < a < 3500G
F = ma
a = F/m
F = 1.2E16 kg•m/s^2
300G < (1.2E16 kg•m/s^2)/m < 3500G

Doing the math out, you end up with figures ranging from 3.5E11 kg to 4.1E12 kg, or between 350 million and 4.1 billion metric tonnes. With a precise acceleration figure, we could nail it down.

EDIT: And I'm retarded. We do have an exact acceleration figure: 3000G. With this number, it comes out to 410 million metric tonnes.

Obviously, this assumes the accelerations and momentums calculated represent the ship's absolute maximums, since it's never going to divert all of its power to its engines, but it's at least a good ballpark...I hope.

Thoughts?

[Firefox]

Would you be interested in measurements of the nozzles on the Revell model? They should be close enough to the real thing to derive accurate thrust figures.

[phongn]

How is the exhast mass being accelerated at c, though?

[McC]

How is the exhast mass being accelerated at c, though?

Isn't the exhaust mass exiting the thruster at almost-c a basic assumption of ion-based rockets?

[phongn]

Shouldn't you be calculating relativistic momentum then?

[McC]

Shouldn't you be calculating relativistic momentum then?

...yes

In my fervor last night, I forgot about that. Apologies.

Relativistic momentum is defined (for those who don't know) as mv/sqrt(1-v^2/c^2), with the 1/sqrt(1-v^2/c^2) quantity being defined as gamma, y. This equation should set off immediate warning bells, since we can't just plug c into it (else we'll be dividing by 0). Also, the problem the relativistic momentum introduces is that the total momentum imparted can change enormously depending on what fraction of c we use (obviously).

A few examples:

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m0 = 4E7 kg
v = .99c
y = 7.08881205
p = 8.4E16 kg•m/s

v = .999c
y = 22.3662720
p = 2.7E17 kg•m/s

v = .9999c
y = 70.7124460
p = 8.5E17 kg•m/s

Old Method:
p = 1.2E16 kg•m/s

Pretty large disparity. We'd need a definite figure for the mean velocity of the exhaust mass in order to truly lock in this mass. But, using the ranges above...

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p = 8.4E16 kg•m/s
F = 8.4E16 kg•m/s^2
F = m(3000G)
m = 2.9E12 kg

p = 2.7E17 kg•m/s
m = 9.2E12 kg

p = 8.5E17 kg•m/s
m = 2.9E13 kg

So, values ranging from 2.9 billion to 29 billion metric tonnes. A fair site heavier than 410 million.

*sigh* Thanks, phongn, go and ruin my bubble of satisfaction and achievement (Though to be fair, I'd rather be wrong and accurate than "right" and clueless)

[Ender]

Shouldn't you be calculating relativistic momentum then?

Ideally? Yes, we'd have to find an exhaust velocity, and work back from there. I've been doing that for some time to get figures.

Realistically? C is perfect of rhte back of the envelope calcs that you do to get the numbers needed here for mass, power, accel, etc.

McC: My latest putzing around with this is such - Peak powr of DS1, divided by accel (100 Gs = 980 m/s^2) times C gets you mass. Mass divided by volume gives you density. Find the volume of other ships and multiply by the density to get their mass. Mass times c times accel gets you peak power. The returned values are about what you get for the given values, and very close to what you get from scaling the reactor down (confirming the method as valid). In fact, closer then what you get from scaling down, indicating that this might be how some of them were derived.

BTW spartan, my playing around like this is why I haven't sent you the calcs yet.

[Illuminatus Primus]

Why are Acclamators so fast?

[Ender]

Why are Acclamators so fast?

Probably a combination of the two methods. Scale down to the reactor, adjusting for the logirithmic scale, divide by the estimated mass and C.