Saludos
Doug
En un mensaje con fecha 03/02/2004 6:57:27 PM Mexico Standard Time, MexicoDoug escribe:
Hi Rob,On the subject of dropping objects from high places:
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, [EMAIL PROTECTED] escribe:
>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
