| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1441.1 | | CSC32::MACGREGOR | Colorado: the TRUE mid-west | Tue May 27 1997 11:17 | 6 |
|
I don't know if it is posted in here anywhere, but I remember
discussing this once before verbally. I'd love to show the work, but
I'm missing the radius of the earth.
Marc
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| 1441.2 | | BUSY::SLAB | Audiophiles do it 'til it hertz! | Tue May 27 1997 11:55 | 4 |
|
Doesn't really matter, but round it off to a 25K-mile circumference,
or a 3979-mile radius.
|
| 1441.3 | | BUSY::SLAB | Audiophiles do it 'til it hertz! | Tue May 27 1997 11:57 | 3 |
|
BTW, note 82 sort of touches on this problem.
|
| 1441.4 | How about 2" | 39702::BERGART | Jeff-the-ref | Tue May 27 1997 12:10 | 24 |
| IMHO, you do not need to know the radius of the earth.
(Using C for circum. of the earth, and D for its Diameter)
C = Pie * D
Now solve for the new diameter:
C + 12" = Pie * (D + delta). Using the distributed law;
C + 12" = (Pie * D) + (Pie * delta)
The C and (Pie * D) terms can be crossed out since they are equal!
leaving:
12" = Pie * delta
Using 22/7 for Pie, one gets:
(12"/22)*7 = delta or
delta = approx. 3.8" therefore, the change to the radius is half that
or approx. 1.9"
|