pacplyer wrote:

Richard Lamb wrote

Based on some very shakey assumptions...

Looking for drag per killometer of cable....

1) Using EAS for speed (already corrected for temperature and pressure
at cruise altitude)
218 K EAS = 250 MPH

2) that makes Q = 160 (.00256 * V^2 where V in MPH)

3) RN = 7657 Million (that's for 1 KM chord - in line with slipstream)
Viscosity effects predominate onthe part of thecable in trail!
But I have no clue how to calculate the drag on that part...

4) For the part of the cable that is _perpendicular_ to the
slipstream...

Dp = Cd S Q
Cd = .02 ?? for a round cross section
Cable diameter guessed at 2 inches diameter
S = 546 sq feet per Km of cable length

Dp = 1750 pounds per Km of cable exposed perpendicular to the stream.


Well?

Nice formula Richard. If you don't mind I'm going to email it to my
friend at Scaled tomorrow. Don't worry; it's just for grins. I spoke
with him briefly last night and asked him Tim's question about the
line speed when the tow plane banks and gives a 3 to 1 speed advantage
to the Orbiter. He says the line going supersonic is no problem; "it
just Zings" he told me. There are no stability problems with the line
at Mach numbers. Edwards has lots of experience with supersonic tow
lines.

Just for my education, what was your source and method for calculating
your Reynolds Number? What pressure altitude did you use? I'm
thumbing through my old "Theory of Wing Sections" by Abbott and Von
Doenhoff, but it's not making much sense to me!

So if we double your drag per km number, just for ballpark
"do-abiltiy" wag calcs, and call it 3500 lbs of drag per Km of cable
exposed perpendicular to the stream, since it'll be semi-vertical just
prior to towship 45 degree bank maneuver, that yields: 84,000lbs of
drag for 24km of line.

Let's see: four engines producing 67,000lbs of thrust at SL… [note:
we don't know yet what it is at FL 410 but, P&W has those charts and
we can guess it's pretty high since we can pull 1.71 normal climb EPR
at FL410 which is only .02 off the max value on the table.] Four
engines pulling 67,000lbs plus 5,000 lbs of thrust from the APU
chugging out the tail (no I'm not making this up) gives us 273,000
lbs of potential pounds of thrust (minus minor alt effects) to
overcome the line drag coefficient with. Now most of this is going to
be used just getting the mated vehicles to 41,000 ft. But once I get
that 150,000 lb OrbitOne off my back (I keep forgetting about the
100,000lbs of winch and cable so the OrbitOne wt limit is 150,000lbs),
I'll have 190,000 lbs of thrust avaiable for aircraft speed control
above whats required to overcome the line drag. That is, when the
line's pulled out all the way 24km out the roof (84,000lbs of drag.)
But at vehicle split the line will still be short so the drag from it
is negligible, and I can accelerate up to .92 mach, in a shallow
descent. Once I get down to FL350 (where the 747 was designed to
make money,) and pay out all that heavy cable in the process (say
75,000 lbs of Vetran) I'm lighter than **** at 445,000 lbs [380k
empty wt+40Kgas remaining+25k winch pallet wt] so will 84,000 lbs of
line drag have me struggling to hold MMO .92? Don't know, lets go try
it.

Let's see: 100million a launch, or $100,000 a launch? (10,000 per hr
for 747, plus OrbitOne prep and support costs guessed at 90K.) Look
out NASA, this SpaceSlingShot thing is worth exploring further since
on paper it's 100 times cheaper to launch payloads and people into
orbit than the shuttle. ;-)

Cheers,

pac "beautiful dreamer" plyer

stay tuned for more wild numbers

Oh great! There goes my chance at getting hired!


One thing I already know about is that the Cd increases (a lot)
in the transonic range, and again (even more) for supersonic.

Problem is, the drag bill gets paid before anything else does.

So it's entirely likely that the 'zing' is gonna 'fizzle'...

Still, it will be interesting to see what the cool guys say.
(a little kissing up might keep them from laughing so hard?)

Richard