Hello, Robert.  I overheard you talking with Matt about - -

-=> ROBERT SAYRE wrote to MATT MC_CARTHY <=-


 MM> During liftoff, as the pictues show, a "piece of
something" broke loose fr
 MM> the fuel tank and hit the shuttle.  This doesn't bother me too much, as th
 MM> "whatever it was" was travelling at the same speed as the
shuttle at that
 MM> instant in time, and would not have been the likely cause of extensive or
 MM> serious damage.  In any case, that subject is a whole train of consequence
 MM> itself.

 RS>  As I understand it, there is a very high rate of
 RS> accelleration during liftoff and it causes many
 RS> G's of force upon the crew.

 RS>  Wouldn't those same forces be on a piece that was
 RS> no longer attached to the accellerating craft?

For simplicity, assuming the shuttle is launched vertically - -

For the main vehicle,  a = F/m - - acceleration equals force divided by
mass of the object.  Forces active are the upward thrust of the shuttle
main engines, plus upward thrust of the solid rocket boosters, plus
downward force of gravity and downward force of air friction.

Say instantaneous velocity when a piece falls off is Vo, and this
is clearly the same for main vehicle and the piece.

Since V = Vo + a * t, (new velocity = original velocity plus
acceleration multiplied by time) the main vehicle continues accelerating
upward, and its upward velocity continues to increase.

As soon as it separates from the main vehicle, the only forces active on
the falling-off piece are gravity and air friction - - both acting
downward. Its new acceleration is a = F/m,  the forces due to gravity
and air friction, divided by the mass of the piece.  So the piece starts
slowing down according to the same formula - - V = Vo + a * t.

Making some off-the wall assumptions here - -  At supersonic velocities,
air friction gets to be large, so assuming 1G force from that, 1G on the
piece, and lets say a net of 4G in opposite direction for the main
vehicle, relative acceleration difference of 6G total, with a
gravitation constant of g = 32 ft /sec *sec, and an object falling off
say 64 ft above the shuttle wing - -

One of the other basic formulas - - V * V = 2 * a * s

V * V = 2 * 6 * 32 * 64 = 24,576  ft sq /sec sq

V = square root of 24,576 = 157 ft /sec, an impact velocity of about
107 miles per hour.

So, at that speed, a good-sized block of ice could definitely
deliver quite an impact.

Hoping I didn't make a mistake in the math.

- - -  JimH.

... 2 + 2 = 5 for extremely large values of 2.
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