From: hees@cubic.com (Phil Hees)
Newsgroups: alt.satanism,alt.atheism,soc.motss,sci.physics,alt.religion.emacs,talk.religion.newage,alt.pagan,alt.sex,talk.religion.misc
Subject: Re: Subject?
Date: 4 Sep 92 04:41:03 GMT
Organization: Almalgamated Flamethrowers and Factcheckers, Ibid., Et.Al.


I can't believe I'm posting a physics flame in alt.sex... :)


In article <25832@dog.ee.lbl.gov> sichase@csa1.lbl.gov writes:
>In article <1992Aug28.134921.28421@galileo.cc.rochester.edu>, mwsn_ss@uhura.cc.rochester.edu (Matthew Weasner) writes...
>>In article <25660@dog.ee.lbl.gov> sichase@csa3.lbl.gov writes:
>>>In article <34544@cbmvax.commodore.com>, kominetz@cbmvax.commodore.com (John Kominetz - CATS) writes...
>>>>Bush spends.
>>>>-- 
>>>>                                      __
>>>>        John Kominetz, III            \/          Aikido Ki No Kawa
>>>>  Commodore Apps and Tech Support           "I speak for myself, not C=."
>>>> ========================================================================
>>>>  "The higher you fly, the faster you fall."          -- Laurie Anderson
>>>
>>>    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>>>
>>>That's not true.  You have a defective .sig.
>>>
>>>-Scott
>>actually, in a way, I think it is true. You accelerate at the same rate. 
>>(9.8 m/s ?) So if you are higher, you get more time to accelerate. Therefore,
>>you are going faster when you hit the ground and go <SPLAT!>. Elementary, my
>>dear Scott. What does Aikido Ki No Kawa mean anyway?
>> 
>>my 2 measly cents.
>
>Listen up, Sherlock. :-)


OK, Scott, since you keep going out of your way to pick on people
(first John and then Matthew), I'm going to have to pick on you. :)


>When an object falls through the atmosphere (which is what happens if
>it suddenly stop flying) it achieves terminal velocity very quickly.
>The actual value of this terminal velocity varies depending upon the
>object's shape -
          ^^^^^

Actually, it depends on size, shape, surface characteristics, and mass.
A few comments:

Terminal velocity is the velocity at which the force due to drag equals
the force due to gravitational attraction.  The calculation of terminal
velocity is not an easy problem, as any student of fluid dynamics knows:
there are three different formulae for drag depending on whether the
speed is in the slow, turbulent or supersonic range.

Saying that "the actual value of this terminal velocity varies depending
upon the object's shape" is very misleading.  For example, in a fluid
medium such as the atmosphere, a lead sphere will fall faster and reach
a higher terminal velocity than a styrofoam sphere of the same shape and
size.  Also, a large styrofoam sphere will fall faster than a small
styrofoam sphere (despite what Galileo said) because drag is
proportional to the square of the radius, while mass (and therefore the
force due to gravitational attraction) is proportional to the cube of
the radius.  Surprisingly, a roughly textured object may or may not fall
faster than a smooth textured object: normally the roughness will
increase drag, but in some cases the roughness may act to reduce the
onset of turbulent flow, decreasing drag.  This is one of the reasons
that wind tunnel testing is so important.

BTW, Galileo's experiment worked because the ratio of the drag force to
the gravitational force is very small for a cannonball moving at less
than 100 m/s.


> - but for a person it is probably in the vacinity of 60 - 200 mph
>(someone else may have better information - I'm sure that parachuters
>know a better value)

An experienced skydiver can vary her body orientation (varying drag) to
speed or slow her rate of descent over a considerable range.  Your
values are approximately correct for a human body, although I think that
your lower limit is too low.  However, who said we were talking about a
person?  "The higher you fly, the faster you fall" - kind of general,
isn't it?  Different terminal velocities for airplane, helicopter, bald
eagle, Goodyear Blimp, etc. :)

Also, you misspelled vicinity.  (Usenet guidelines be damned... I _like_
spelling flames!)

>at 9.8 m/s^2 (note the units - yours were wrong) you get to terminal
>velocity in 3 seconds (for 60 mph.)

Actually 2.7 seconds, although you are correct if you were limiting it
to one significant digit.

>In that time you drop only 30 meters.

44.1 meters using your figure of 3 seconds, 36.5 using my figure of 2.7
(distance = 0.5 * a * t^2).  I expected better math from someone at
Lawrence Berkeley Laboratory. :)

But actually, you don't just accelerate at a constant 9.8 m/s^2 until
you hit terminal velocity and then suddenly stop accelerating.  At any
instant, your acceleration can be calculated as (Fg - Fd)/m, where Fg is
the force due to gravitational attraction, Fd is the drag force, and m
is the mass of the object.  Fd increases with increasing velocity but
the relationship is very nonlinear, particularly at the onset of
turbulent flow.  As a first (very rough) approximation, I get:

  after 1 second       31% of terminal velocity (using 60 MPH figure)
  after 2 ...          52%
        3              67%
        4              77%
        5              84%
  after 6.2 seconds    90% of terminal velocity
  after 8.1 seconds    95% of terminal velocity

the last two entries being commonly used endpoints.  So for your basic
falling person it will take somewhat more than your 3 seconds to get to
something which we could reasonably call "terminal velocity".

>So if you are falling from any height much greater than 30 meters
>(probable if you were flying,) then you will reach terminal velocity
>early in your fall.

Again, you never "reach" terminal velocity.  It is more meaningful to
talk about reaching a value such as 90% of terminal velocity.  Granted,
this is nitpicking... :)

>The velocity with which you splat will therefore be independent of the
>height from which you fell.
>
>-Scott

Your conclusion is essentially correct, despite my nitpicking.  But
returning to the original quote, I would like to cite two good points
from Steve Franklin's post in this thread:

	Well, if you guys really want to know the answer - the air is
	thinner at higher altitudes, and because of this, the terminal
	velocity is higher in the upper reaches of the atmosphere. Thus,
	the higher you fly, the faster you will fall, but your terminal
	velocity will decrease as you continue to fall...

In this context, the original .sig was truthful.  It would also be
truthful if the falling object were sufficiently large and/or dense
that it never reached terminal velocity in the finite falling distance
that the Earth's atmosphere represents.  (Oh, the original quote didn't
specify the Earth... no terminal velocity on the Moon, is there? :)

	However, I personally think the initial statement is
	metaphorical to the use of drugs. I don't think it has anything
	to do with parachuting or free-fall.

It could also be metaphorical to political careers, showbusiness careers
or the career of Donald Trump.  In any case, considering the source of
the quote it was surely a metaphor.  Don't listen to much Laurie
Anderson, do you?  :)

--
Phil Hees
hees@cubic.com
"The higher you fly, the faster you fall."          -- Laurie Anderson