Quickie calc. - Re: 5.0 in Indiana?-3.5 in New Mexico
Posted by 2cents on June 18, 2002 at 22:50:53:

Well, as you know:

Using NEIC locations-
30.79S 70.96W 53.0 6.6 A CHILE-ARGENTINA BORDER
12.56S 166.35E 33.0 6.7 A SANTA CRUZ ISLANDS
38.07N 87.68W 5.0 5.0 A SOUTHERN INDIANA

Distance Calculation Results (from link):

-Distance between 30.79S 70.96W and 38.07N 87.68W is 7859.6995 km

-Distance between 12.56S 166.35E and 38.07N 87.68W is 12268.5843 km

( This calculation assumes the earth is a perfect sphere with a radius of 6378.0 km )

Circumference = 2 * PI * rad. = 40074.16 km

So for Chile/Indiana the angle is:
(7859.6995 /40074.16 )* 360.0 = 70.6 degrees.

So for Santa Cruz/Indiana the angle is:
(12268.5843 / 40074.16 )* 360.0 = 110.21 degrees.

The first one might be near an FFA angle and the 2nd is sorta maybe near the 107....

So for Indiana location it looks like a 50/50 proposition (as far as being actually on an FFA angle).

.02



Follow Ups:
     ● Re: Quickie calc. - Re: 5.0 in Indiana?-3.5 in New Mexico - 2cents  03:18:37 - 6/19/2002  (16072)  (1)
        ● 3.7 in Texas panhandle - chris in suburbia  08:49:44 - 6/19/2002  (16073)  (1)
           ● Re: 3.7 in Texas panhandle - lowell  00:42:46 - 6/20/2002  (16079)  (0)