[meteorite-list] A "Strike" with a spare ball

Mail actions: [ respond to this message ] [ mail a new topic ]
Contemporary messages sorted: [ by date ] [ by thread ] [ by subject ] [ by author ]

From: MexicoDoug_at_aol.com <MexicoDoug_at_meteoritecentral.com>
Date: Thu Apr 22 10:32:45 2004
Message-ID: <123.2b856990.2d768afb_at_aol.com>

--part1_123.2b856990.2d768afb_boundary
Content-Type: text/plain; charset="US-ASCII"
Content-Transfer-Encoding: 7bit

Another way to make this more "intuitive" is to imagine sticking a bowling
ball out the window of a car traveling 155 mph. Do you think that 14 pounds of
air resistance would be on it? Imagine at over 300 mph (the 140 m/s you
mentioned).

Saludos
Doug

En un mensaje con fecha 03/02/2004 6:57:27 PM Mexico Standard Time,
MexicoDoug escribe:

> Hi Rob,
>
> Thanks for the note. I was thinking about the bowling ball problem, too
> especially after seeing your post. The 70 meters per second you mention (as a
> maximum in a friction free fall from 250 meters altitude), as you eluded to,
> is not what the velocity the ball reaches. This is due to the drag
> coefficient. At a typical drag coefficient of 0.5, the distance will be greatly
> multiplied to reach that speed. I would think that their 1,000 meter height is
> about right to reach the 70 m/s (155 mph), without trying to figure that one
> out.
>
> When you say "a bit less" due to the friction, I'd argue "substantially
> less" to the point that it is what is most important.
>
> As to reaching 140 m/s (312 mph), as you suggest from 2,000 meters height,
> you can forget that for a bowling ball. I calculate the terminal velocity of
> a bowling ball to be about the same: 70 m/s (155 mph) (see below). This may
> not be intuitive, but then again intuition is a gift in physics to precious
> few as are experienced skydivers as well. The reason is because a bowling
> ball is basically made of plastic or polymer, perhaps with a heavy small core,
> around as light (say 1.015 g/cc bulk) as water. It is just not that dense and
> is not a good candidate for simulating meteorite densities (3 - 8 g/cc). It
> hurts when it falls on your foot due to the overall mass and relative
> incompressibility, not the density. To make it intuitive, imaging a bowling ball
> made of fine octahedrite. It would probably go through the floor when you let
> it loose, if you could swing it, and it would weigh well over 100 pounds.
> BTW, did you know some premium balls have "oriented" cores of to get special
> "flow" direction?
>
> So the bowling ball actually free falls at a lower speed than an experienced
> skydiver who weighs 10 - 12 times as much but torpedo dives with a cross
> sectional area not way different (maybe a factor of only 2), and reaches twice
> that terminal velocity. The ball will fall at bicycling speed faster than a
> skydiver. That's probably why the guy in the article wants to drop it and let
> it go. That would be fun, as he could find the mass to watch it fall, say
> at his speed and play catch up - fall back with it.
>
> Calculation assumptions:
> Drag coefficient = 0.5 (typical for a sphere)
> Bowling ball diameter = 9 inches (typical)
> Bowling ball weight = 14 pounds (typical most heavy ball)
> Bulk density = 1.01525
>
> I hope I haven't screwed up the numbers, but I still have a headache.
>
> Saludos
> Doug Dawn
> Mexico
>
> En un mensaje con fecha 03/02/2004 5:29:00 PM Mexico Standard Time,
> ROBERT.D.MATSON_at_saic.com escribe:
>
>
> >> On the subject of dropping objects from high places:
>>
>> >On Feb. 13 a single-engine Cessna flew low over the Utah desert toward the
>>
>> >Bonneville Seabase at 80 knots. Pilot Patrick Wiggins checked his
>> altimeter.
>> >As planned, he was just 820 feet (250 meters) above the surface.
>>
>> At this low altitude, the bowling ball would only have reached a velocity
>> of
>> 70 m/sec in a vacuum -- a bit less going through air. But isn't typical
>> terminal
>> velocity for a good-sized meteorite more like 200 m/sec? If so, then the
>> drop
>> altitude would need to be over 2000 meters (over 6500 feet) to reach
>> terminal
>> velocity.
>>
>> >Wiggins hopes to make the next drop at more than a half-mile up (1,000
>> meters),
>> >from where an object should achieve "terminal velocity," or top vertical
>> speed,
>> >zooming straight down and having lost all its horizontal movement induced
>> >by the airplane ride.
>>
>> At 1000-meter altitude, the bowling ball only has time to get up to 140
>> m/sec,
>> which is probably fine for a smaller meteorite, but larger ones will be
>> faster.
>>
>> --Rob
>>
>>
>>
>>
>>
>>
>>
>>
>>
>
>

--part1_123.2b856990.2d768afb_boundary
Content-Type: text/html; charset="US-ASCII"
Content-Transfer-Encoding: quoted-printable

<HTML><HTML>Another way to make this more "intu=
itive" is to imagine sticking a bowling ball out the window of a car traveli=
ng 155 mph.  Do you think that 14 pounds of air resistance would be on=20=
it?  Imagine at over 300 mph (the 140 m/s you mentioned). 
 
Saludos 
Doug 
 
En un mensaje con fecha 03/02/2004 6:57:27 PM Mexico Standard Time, MexicoDo=
ug escribe: 
 
<BLOCKQUOTE TYPE=3DCITE style=3D"BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT=
: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">Hi Rob, 
 
Thanks for the note.  I was thinking about the bowling ball problem, to=
o especially after seeing your post.  The 70 meters per second you ment=
ion (as a maximum in a friction free fall from 250 meters altitude), as you=20=
eluded to, is not what the velocity the ball reaches.  This is due to t=
he drag coefficient.  At a typical drag coefficient of 0.5, the distanc=
e will be greatly multiplied to reach that speed.  I would think that t=
heir 1,000 meter height is about right to reach the 70 m/s (155 mph), withou=
t trying to figure that one out. 
 
When you say "a bit less" due to the friction, I'd argue "substantially less=
" to the point that it is what is most important. 
 
As to reaching 140 m/s (312 mph), as you suggest from 2,000 meters height, y=
ou can forget that for a bowling ball.  I calculate the terminal veloci=
ty of a bowling ball to be about the same: 70 m/s (155 mph) (see below).&nbs=
p; This may not be intuitive, but then again intuition is a gift in physics=20=
to precious few as are experienced skydivers as well.  The reason is be=
cause a bowling ball is basically made of plastic or polymer, perhaps with a=
heavy small core, around as light (say 1.015 g/cc bulk) as water.  It=20=
is just not that dense and is not a good candidate for simulating meteorite=20=
densities (3 - 8 g/cc).  It hurts when it falls on your foot due to the=
overall mass and relative incompressibility, not the density.  To make=
it intuitive, imaging a bowling ball made of fine octahedrite.  It wou=
ld probably go through the floor when you let it loose, if you could swing i=
t, and it would weigh well over 100 pounds.  BTW, did you know some pre=
mium balls have "oriented" cores of to get special "flow" direction? 
 
So the bowling ball actually free falls at a lower speed than an experienced=
skydiver who weighs 10 - 12 times as much but torpedo dives with a cross se=
ctional area not way different (maybe a factor of only 2), and reaches twice=
that terminal velocity.  The ball will fall at bicycling speed faster=20=
than a skydiver.  That's probably why the guy in the article wants to d=
rop it and let it go.  That would be fun, as he could find the mass to=20=
watch it fall, say at his speed and play catch up - fall back with it. 
 
Calculation assumptions: 
Drag coefficient =3D 0.5 (typical for a sphere) 
Bowling ball diameter =3D 9 inches (typical) 
Bowling ball weight =3D 14 pounds (typical most heavy ball) 
Bulk density =3D 1.01525 
 
I hope I haven't screwed up the numbers, but I still have a headache. 
 
Saludos 
Doug Dawn 
Mexico 
 
En un mensaje con fecha 03/02/2004 5:29:00 PM Mexico Standard Time, ROBERT.D=
.MATSON_at_saic.com escribe: 
 
 
<BLOCKQUOTE TYPE=3DCITE style=3D"BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT=
: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">On the subject of dropping obj=
ects from high places: 
 
>On Feb. 13 a single-engine Cessna flew low over the Utah desert towar=
d the 
>Bonneville Seabase at 80 knots. Pilot Patrick Wiggins checked his alt=
imeter. 
>As planned, he was just 820 feet (250 meters) above the surface.</FON=
T> </F=
ONT><B=
R>
 
At this low altitude, the bowling ball would only have reached a velocity o=
f 
70 m/sec in a vacuum -- a bit less going through air.  But isn't typic=
al terminal 
velocity for a good-sized meteorite more like 200 m/sec?  If so, then=20=
the drop 
altitude would need to be over 2000 meters (over 6500 feet) to reach termin=
al 
velocity. 
 
>Wiggins hopes to make the next drop at more than a half-mile up (1,00=
0 meters), 
>from where an object should achieve "terminal velocity," or top verti=
cal speed, 
>zooming straight down and having lost all its horizontal movement ind=
uced 
>by the airplane ride. 
 
At 1000-meter altitude, the bowling ball only has time to get up to 140 m/s=
ec, 
which is probably fine for a smaller meteorite, but larger ones will b=
e faster. 
 
--Rob 
 
 
 
 
 
 
 
 
</BLOCKQUOTE> 
 
</BLOCKQUOTE> 
 
</HTML>
--part1_123.2b856990.2d768afb_boundary--
Received on Tue 02 Mar 2004 08:12:27 PM PST

This message: [ Message body ]
Next message: Matson, Robert: "[meteorite-list] A "Strike" with a spare ball"
Previous message: Steve Arnold, Chicago!!!: "[meteorite-list] WHAT IS THIS????? ACCUSED OF RIPPING PEOPLE OFF!!"

Mail actions: [ respond to this message ] [ mail a new topic ]
Contemporary messages sorted: [ by date ] [ by thread ] [ by subject ] [ by author ]

StumbleUpon

del.icio.us

Yahoo MyWeb