Re: [sfconsim-l] Propulsion Question

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• ... Well, your deltaV is dependant on two things and two things only, the velocity of the exhaust, and the ratio of the ship s mass full vs dry. Keep in mind
Message 1 of 1 , Aug 11, 2000
Donald McLean wrote:
> So is this idea at all feasible, and how would I go about
> determining the technical characteristics of the drive,
> such as delta V given loaded and unloaded mass of the ship?

Well, your deltaV is dependant on two things and two things
only, the velocity of the exhaust, and the ratio of the ship's
mass full vs dry.

Keep in mind though that since you are using this as
a surface to orbit propulsion system, you'll need to
figure thrust as well. While the ship is close to
the planet, it will have a "tax" imposed upon it
by the planet's gravity.

Basically, if the propulsion doesn't provide more
acceleration than the deceleration provided by the
planet, the rocket isn't going anywhere.

Anyway:

Lambda = Mf / Me
where:
Lambda = ship's "mass ratio"
Mf = ship's mass when fully loaded with reaction mass
Me = ship's mass "empty" (i.e,. with empty reaction mass tanks)
Some solid fuel rockets have a mass ratio of 20 to 60,
but that's pretty extreme. Liquid chemical rockets
generally don't go over a mass ratio of 12.
For a multi stage rocket, the mass ratio is the
product of each stage's mass ratio (which is why
NASA uses so many multistage rockets).
For a merchant ship, you wouldn't want a mass ratio
over, say, 4.

total deltaV = Ve * ln[Lambda]
where:
total deltaV = the total amount of velocity change the ship is capable of
Ve = exhaust velocity

Naturally, whatever units you use for Ve will be the same ones
used for deltaV (typically kilometers per second)

Thrust = mDot * Ve
where:
Ve = exhaust velocity in kilometers per second
mDot = propellant mass flow in kilograms per second
Thrust = Thrust in kilo Newtons

Acceleration is a pain:
Accel = Thrust / Mc
where:
Mc = Ship's mass at this particular instant in time
Accel = acceleration in kilometers per second per second

Its a pain because, of course, the ships mass changes constantly
as it expends reaction mass.

At 100% efficiency, the best you are going to get out
of fusion is 6.3 x 10^14 J/kg (joules per kilogram of fusion fuel)
For details about figuring the exhaust velocity,
refer to Mr. Robinson's post.
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