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To: Albertafriend

Try this calculation. Take the area of the oval (U used Google Earth) approx .21 miles across by .16 miles sidewalk to sidewalk. That's an area of .0336 square miles or 936,714 square feet.

If you really packed the oval wall to wall using approx 9.36 sf per person (that's pretty snug in a big crowd), you would get 100,000 people. Clearly looking at the photo there is only one small dense area that approaches or exceeds 9sf per person. This area is 1/8th of the oval's area. The rest is empty or sparsely populated.

Now of course some kooks are already marching, but the estimate of 100,000 total is way off, half that at best for everyone. There are maybe 10,000 in the oval.

48 posted on 09/26/2005 2:19:54 PM PDT by finnman69 (cum puella incedit minore medio corpore sub quo manifestu s globus, inflammare animos)

To: finnman69; Albertafriend
OK.

Oval is accounted for - pun intended. Maybe 10,000 there by your calc's and comparisons.

Devil's advocate here speaking:

There is at least that many (who were in the 1/5 of the oval that was populated) again in the rectangular area in the street : How can we (reasonably) account for the street, the "sides" of the crowd on the street not in the picture, and those hidden in the trees? (No tree-hugging monkey jokes, please.)
72 posted on 09/26/2005 9:11:56 PM PDT by Robert A Cook PE (-I contribute to FR monthly, but ABBCNNBCBS supports Hillary's Secular Sexual Socialism every day.)

To: finnman69; Albertafriend

http://mathforum.org/library/drmath/view/55402.html

Area of an oval is (w/2)*(h/2)*3.14

In our case, (.21/2)*(.16/2)*3.15 or 0.0263 sq miles.

73 posted on 09/26/2005 9:21:07 PM PDT by Robert A Cook PE (-I contribute to FR monthly, but ABBCNNBCBS supports Hillary's Secular Sexual Socialism every day.)

To: finnman69; Albertafriend

http://mathforum.org/library/drmath/view/55402.html