Re: Orientation of TiPS tether

Jim Varney (jvarney@mail2.quiknet.com)
Sat, 13 Jul 1996 10:13:59 -0700

Robert McNaught wrote:

>Comment has been made on the orientation of the TiPS tether relative to the
>direction of motion.  IF the tether always points towards the center of the
>Earth, 

Does TiPS hang vertically?

Let's take a look at the TiPS system:

           o  10.3 kg (Norton)
           |
           |     
           |    4 km tether, mass 5.5 kg, 2 mm dia
           |
           o  37.7 kg (Ralph)

Notice the unequal distribution of mass.  I calculate the center of mass to
be located 0.98 km above Ralph.

Now consider atmospheric drag.  A cylinder has the highest coefficient of drag
of any regular shape, so the tether will contribute most of the drag in this
system.  The frontal area of the tether is (4000 m)(.002 mm) = 8 sq meters.
If the two end masses are on the order of 0.5 m in diameter, they add another
0.4 sq m to the system.  The drag force will be distributed equally from top
to bottom.

So we have 8.4 sq m of frontal area for a system that has a total mass of 
53.5 kg.  The ballistic coefficient, which can be thought of as an indicator
of susceptibility to drag, is (0.5)(C-sub-d)(A/m) = (0.5)(4)(8.4/53.5) =
0.31 sq m/kg.  By comparison an Agena r/b is 10.7 sq m and has a dry mass
of about 600 kg, for a ballistic coefficient of about 0.036 sq m/kg.  TiPS
will be affected by drag more than a typical rocket body by a factor of
ten.

At 1000 km height, the MSIS atmosphere model gives an atmospheric density of
5.5 x 10 -16 kg/m3.  Using the drag equation

             2
 f= C  A p  V
     D      -     v = velocity = sqrt (u/r) = 7.4 km/sec
            2
                                                 2
  = (3)(8.4 sq m)(5.5 x 10 -16 kg/m3)(7.4 km/sec) /2  = 3.8 x 10 -7  Newtons.

A very small force.

Conveniently ignoring the tether, Ralph being 4 km closer to the geocenter
than Norton means that Ralph feels a higher force of gravity.  I figure this
gravity gradient to be about .00081 g's, so Ralph exerts a tension force
of about 0.3 Newtons on the cable.  This is a pretty small force, which is
why they can get away with using a cable of only 2mm in diameter.

But it can be seen that the drag force is on the order of 10x-6 times
smaller than the cable tension, so the system should be very close to
vertical.

At lower heights drag will become much more significant.  Rho at 250 km is
about 10x 6 times more dense than it is at 1000 km.  At this height the drag
force will be roughly equal to the gravity gradient force, and a significant
deflection of the cable from vertical will occur.  Of course the deflection
will develop gradually as TiPS descends. 

I could calculate the deflection, but I haven't used calculus in over 10 
years and it would be too painful for me :)  

Anyone know the tensile strength of the cable?  It would be interesting to
model if the cable is strong enough for Ralph to pull Norton along at
decay height, say 110 km.  Would the cable snap due to drag?  Or will
Ralph pull Norton along down to the very end? 

Good Passes,

Jim





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Jim Varney      |  121^ 23' 54" W,  38^ 27' 28" N   |           Sacramento, CA
Member, SeeSat-L|            Elev. 31 ft.           |jvarney@mail2.quiknet.com
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