> #7   Super dream ......   I want flat packs strapped to the chest and
back of the passengers that activate in free fall or manually to
create a "fall ball" around the user that slows their descent to the
ground and then provides adequate cushioning to stop them safely.
Maybe a new sport.
/>

You can want ... :-).


NB E&OE !!!!!!!!!!!!

See my prior formula for terminal velocity of parachutes skydivers, bowling
balls, field mice and raindrops. Correct assumptions have to be made, but it
works quite well enough to get a  feel for things..

1. Parachute descent speed at sea level V ~~~~~~= sqrt(2.m.g/A) in mksA
units.  ie A = chute projected area in m^2. g ~= 9.8 m/s/s V = m/s
/>>

Note that "to survive" 40 mph impact yuou need to reduce g forces to a level
which is acceptable.
Take 10g as an OK shock for a [properly oriented and padded person. The
padding simply spreads the forces.
40 mph  ~= 18 metre/second.
2. Distance = V^2/(2a) = 18^2//2/(9.8*10) = 1.65 metre.
ie a linear 1.65 metre deceleration will slow a body from 40 mph to rest at
10g.

ie you need about a 6 foot 'airbag' with linear deceleration to make you
impact survivable.
If you can tolerate 20g (getting 'rather unpleasant' you can halve that
distance. Getting the deceleration linear and spreading the forces may be
'difficult'

To get your 'fall ball' large enough to achieve that sped as a terminal
velocity is another matter.
using formula 1. above

V ~~~~~~= sqrt(2.m.g/A)
or A ~~= 2.m.g/V^2
Here V = 18, m = say 100 kg, g =~ 10
A =~ 6 m^2 per 100 kg to achieve 18 m/s or 40 mph.
Reduce that to a more survivable 10 mph and area increases by a factor of 16
to around 100 m^2.
= 5.64 m radius or 11.3m dia if circular.

Then, there's the bungee parachute ! :-)
http://www.4p8.com/eric.brasseur/bungee_parachute.html

This could be useful.
Haven't checked to see if it agrees with me :-)
http://www.psc473.org/howto/MathematicsParachutes.doc


E&OE - after all, it's Pi a.m.

          Russell
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