Using the site suggested by Chris Clark, and assuming the density of air to be about 80% of the sea level value, I get a terminal velocity of 32.43 m/sec. This will require much less altitude. I suspect that the velocity of a meteor will be up to twice that amount because of its irregular shape and the fact that it is decelerating. Brent __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

Just a (crazy) thought, but if forward velocity (imparted by the aircraft) is a concern, I could always bail out and release whatever we decide to drop from under canopy... Patrick Brent Watson wrote:

BTW, Why not drop rocks that are similar to meteors instead of a bowling ball? They could be painted bright red to help identify them.

Brent

I personally have two bowling balls to donate to the cause....just let me make sure there aren't any distinguishing marks on them that could lead the authorities back to me ;-) dlb On Saturday, January 4, 2003, at 09:44 AM, Chris Clark wrote:

But to heck with all the numbers: all I wanna know is how many bowling balls we can safely load into a 182. :-)

Patrick: when have you scheduled the official bowling-ball "drop day"? :-)

Chris

Chris, go to the web site and plug the numbers into the calculator. I agree that the number seems a bit small, but look at the examples of other objects there. A baseball is only 74 mph. I may have the density of the atmosphere wrong but you can see the results yourself. I used 33 kg for the mass, and 22 cm for the diameter. The number I used for the density of air is way too large. The actual density is 1.2929kg/m^3 at sea level. At 4000 feet, the pressure of air is aboput 80% that of sea level, and so the density should follow that. This would give a value of about 1.0343 kg/m^3. At 10,000 feet, the pressure is about 66% that of sea level. If the density follows this, then the density would be .8404 kg/m^3. These numbers are a factor of 100 smaller than what I used. Your site also talks about the numerical coefficient of drag ,"C", and that it can be up to four times higher than that of a sphere. This would mean a terminal velocity one half that of a shpere. This is why I suggested using rocks that approximate the shape of a meteor better. It seems to me that a way to help track the balls as they are falling is to place a streamer on one and drop it at the same time as one without a streamer. The one would be visible, and would fall near where the other one falls. I still think it would be better to drop rocks that approximate the size and density of a meteor. I have some lead I'll donate to the cause to up the density if necessary. As far as the number of bowling balls, I think the upper limit is around 30 for my aircraft. I don't know where you would put more than that. --- Chris Clark <cpclark@xmission.com> wrote:

On Fri, 3 Jan 2003, Brent Watson wrote:

...

Using the site suggested by Chris Clark, and assuming the density of air to be about 80% of the sea level value, I get a terminal velocity of 32.43 m/sec.

So you're saying that a bowling ball would freefall at "only" 72 miles per hour? Something intuitively doesn't compute there, because if anything I'd expect a bowling ball to have a higher terminal velocity than a skydiver freefalling in a flat "belly to earth" position (at a t.v. of about 110-120 mph (~ 50 m/s)). What values did you use for ball radius and density to reach the 32 m/s conclusion?

...

I suspect that the velocity of a meteor will be up to twice that amount because of its irregular shape and the fact that it is decelerating.

Several ostensibly-authoritative web sites claim that most meteors (those below a certain mass, that is) lose all of their "cosmic velocity" high in the earth's atmosphere and then begin "normal" acceleration under gravity until the object's atmospheric terminal velocity is reached. In other words, meteors have long since stopped decelerating by the time of impact. A good exposition can be found on this page:

http://www.amsmeteors.org/fireball/faqf.html#12

But to heck with all the numbers: all I wanna know is how many bowling balls we can safely load into a 182. :-)

Patrick: when have you scheduled the official bowling-ball "drop day"? :-)

Chris

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