JRS: In article , dated Thu, 18
May 2006 20:32:38 remote, seen in news:sci.space.station, Jorge R. Frank
posted :
Dr John Stockton wrote in news:$n9DF9AxAObEFwW9
:

What does the drag amount to in actual Newtons and pounds-of-force, for
a typical or mean attitude? It can hardly be 166 kgf/m^2.

Right, the ballistic number is not pressure (force per unit area) but
rather mass per unit area. So 166 kg/m^2 means the station has 166 kg of
mass for every square meter of projected area.

Drag force is F = 0.5*rho*v^2*Cd*A, drag deceleration is a = F/m. The
ballistic number is BN = m/(Cd*A) so drag simplifies to a = 0.5*rho*v^2/BN
or F = 0.5*rho*v^2*m/BN. m is around 185,000 kg, v is around 7700 m/s. I
don't have my atmosphere models handy so plug in your favorite value for
rho and chug away.

That's all very well, no doubt; but neither do I.

The Heavens-Above site indicates that, between boosts, ISS drops by
about 3 kilometres/month or 1 millimetre/second.

Now ISTM that dR/dt = A T / pi where T is the period (Does anyone
know if that is right?) which puts the deceleration A at around 0.05
micro-gee and the force therefore as about one centi-newton (1 gram).
Someone please check that!

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