Re: [asa] Rudwick does it again (back to Adam)

If I understand your assertion correctly, a trip of 30 mi. over a perfect sphere
would only result in a "drop" of 1.5 inches. This is incorrect (if it is what
you meant). Actually over a perfect sphere with radius = about 6400km, one
would need to be 2.0 meters tall to have a tangent line of sight with a horizon
5060 m (a mere 3 miles) away (simple use of Pythagorean theorem). Of course
waves and hills would cause some scatter around that average. This is the
reason (I think) that some people get the impression that Crater lake (in
Washington) has a "mysterious" bulge in the middle. Actually, since that lake
is more than 5 miles across, you would have to be more than 2 meters above your
shore line to have an unobstructed view of the far shore line over a perfectly
calm lake. --All just from normal earth curvature; no mystery about it!

--Merv

Quoting Don Winterstein <dfwinterstein@msn.com>:

> This has little--or nothing, in most cases--to do with Earth's surface
> curvature. Trigonometry and a knowledge of Earth's radius show that the
> distance an object 30 miles away from an observer would "sink" below a line
> tangent to the surface at the observer's location is less than 1.5 inches
> when both the observer and the object are at sea level. Unaided human eyes
> can't resolve towers or much else besides mountains at a distance of 30
> miles. In short, there's no way unaided human eyes can detect Earth's
> curvature by observing how objects "sink" as they move away at sea. The rate
> of increase of "sinking" with distance is far too slow.
>
> Don
>

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Received on Thu Aug 21 12:02:07 2008