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Forum index » Science and Technology » Math » Recreational
elevation
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gcotterl@co.riverside.ca.
science forum beginner


Joined: 13 Sep 2005
Posts: 17

PostPosted: Mon Jun 19, 2006 3:42 am    Post subject: elevation Reply with quote

Considering the curvature of the earth, what is the "observed"
elevation and the elevation angle of a mountain peak (5,710' above sea
level) seen from 851' above sea level, 47 miles away?
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marc.t.davies@gmail.com
science forum addict


Joined: 31 May 2006
Posts: 52

PostPosted: Mon Jun 19, 2006 10:50 am    Post subject: Re: elevation Reply with quote

Taking O as our point of observation, P as the peak of the mountain, G
as the central base point of the mountain, R as the radius of the
Earth, and S as the segment OG, we can calculate the angle alpha
between OR and OG from first principles (Circumference = 2*pi*|R|).
(We'll assume |OR| = |GR| = 3960 miles for simplicity)

Added picture for clarity:
http://www.geocities.com/anarchistic_progressivist/thing.GIF

Knowing alpha, we can now draw where the tangent line t at O strikes
RP, denoting that point M. Clearly, angle beta (between OM and OP)
denotes the effective elevation angle and |PM| the effective height.

To find beta, we should note that the angle between OG and OM is alpha,
hence between OG and OP is (alpha+beta). But we can easily find |OG|
and |GP|, hence (alpha+beta); from there on, finding the various
lengths involved from the ones we already have is a matter of plugging
in the numbers.
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Barry Schwarz
science forum beginner


Joined: 30 Apr 2005
Posts: 31

PostPosted: Tue Jun 20, 2006 2:53 am    Post subject: Re: elevation Reply with quote

On 18 Jun 2006 20:42:00 -0700, gcotterl@co.riverside.ca.us wrote:

Quote:
Considering the curvature of the earth, what is the "observed"
elevation and the elevation angle of a mountain peak (5,710' above sea
level) seen from 851' above sea level, 47 miles away?

It depends on how you measure the 47 miles. Let R be the radius of
the earth in feet:

If the distance is the straight line from the observer to the
peak, we have a triangle with a vertex at the center of the earth and
sides of R+851, R+5710, and 47*5280. The observer's angle can be
computed from the law of cosines and then subtract 90 to convert from
vertical to horizontal. We also have a right triangle formed by the
tangent from the observer to the ground, the radius at the tangent
position, and the extended radius up to the observer. Trig gives us
the angle the tangent makes with the extended radius. The tangent
line can be extended until it meets the extended radius (at P1) to the
mountain peak. The law of cosines yields the angle between the two
extended radii and then we also know the third angle. The law of
sines gives us the distance from the center of the earth to P1 and
then subtraction tells how much of the extended radius is above P1.

If the 47 miles is the arc distance along the ground between
the point directly below the observer to the point directly below the
peak, we can compute the angle between the two extended radii and from
the law of cosines compute the straight line distance between the
observer and the peak. After that, the previous discussion will work.


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STUARTe
science forum beginner


Joined: 05 Feb 2006
Posts: 8

PostPosted: Tue Jun 20, 2006 3:23 am    Post subject: Re: elevation Reply with quote

using the equation (distance to horizon) = 90*(sqrt(h)) h is height
in miles

in this case use 47 miles for the distance and solve for h

subtract h from the height of the mountain

that comes out to roughly .27 miles or 3600 feet

move the whole ensemble up 871 feet yielding about 4500 feet above sea
level
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