Lim Jyue (lim_jyue@pacific.net.sg)
Tue, 31 Oct 2000 01:11:17 +0800


At 11:29 10/29/2000 -0800, Mark Simmons wrote:
>and since most publishers just get all their specs from
>the MS Encyclopedia nowadays, as well as echoing all its numerous errors,
>they've also lost this useful data.

        That's a major headache. Worse, if the new animators take the values
from this source, the UC MS they produced are likely to be way above or
below MSes from the same period.

> From a specs perspective, you can divide the full weight by the mass
>ratio to find the weight without propellant.

        That's provided you have the mass ratio to begin with. I did a bit
of checking: My MSZ-006 C1 Zeta Plus kit manual had it (1.83), as well as
the AMX-009 Dreissen (1.5), but that's it. From CCA and 0080 onwards, they
didn't seemed to have included this data.

>the results are consistent enough that it's easy to extrapolate to other
>series.

        Again, while it is possible to extrapolate, I think we have no way
to account for weapon-weight reduction over the intervening years, so the
extrapolation may be a little off.

> No, not necessary - the numbers you have are _already_ expressed in
>terms of Gs. To convert Gs to meters/second^2, _then_ you'd multiply by 9.8.

        Wait a minute.. Since I'm a bit rusty with my physics, I'll like to
run through the process with you to make sure I got it right.

        To determine acceleration, I'll just use the simple F=ma formula.
Now, F needs to be in Newton (S.I, unit), m in kg, and a in ms^-2.

        In this case, T is our force, but it's in kg instead of N. Hence, we
need to multiply by G, approximately 9.8 (by your assumption in another
mail). This furfills the F unit requirement (to be in Newtons). The mass is
already in kg, so we don't worry about that. So, basically:

        T[kg]*G[ms^-2] = m[kg]*a[ms^-2] -- which is unit consistent on both
sides.

        Manipulating both sides, we get: T/m [no unit] = a/G [no unit].

        Hence, the T/W (more accurately, T/m) ratio I derived is essentially
the straight line acceleration [in G] of the MS? Is this correct?

-------------
Lim Jyue
ICQ: 24737555

I am careful not to confuse excellence with perfection.
Excellence I can reach for; perfection is God's business.

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