EDUCATOR’S PAGE


Figure 3. Diagram of the Earth, where “c” is the center of the planet (exaggerated). Here, “h” is the height of a person looking out toward the horizon, on a smooth


Earth, where d1-d2 is the dis- tance seen in the direction of the horizon.


Figure 2. The differences between Earth’s actual topographic surface (solid line), ellipsoid surface (dashed line), and geoid surface (dotted line). Excesses and deficits of mass are shown as pluses and minuses in relation to the geoid surface. All surfaces are approximations for the sole sake of illustration.


surface of the planet, instruments (e.g., gravimeters) can be used to very accurately determine the gravitational accelera- tion everywhere, and consequently, determine the location of the geoid everywhere, for purposes of calculating elevation and establishing a vertical datum. Currently, the vertical and horizontal datums, or sets of data used for referencing point locations in geodetic surveying work, are the NAVD 88 and NAD 83 respectively (National Geodetic Survey, 2015).


Returning, again, to the topic of Earth’s general shape, the planet exhibits curvature and its surface consists of irregular topography. Both are important considerations when measur- ing distances and elevations, such as in practical everyday situations and in formal mapping.


How Far to the Edge of the World?


A practical application of Earth’s curvature relates to finding the horizon. People occasionally ask “how far away is the horizon?” Or, alternatively, “how far can I see (into the distance) from here?” The answers to these questions vary depending on several things: chiefly, flatness of the terrain and height above the surface that the observation is made. Being on a hill in elevated terrain or on the shore of a beach next to the ocean would change the determination of how far one’s sight could reach; also of concern is the height above the surface of any distant object one may see. Approximating distance to the horizon involves relatively routine mathemat- ics, after making a couple of assumptions: imagine Earth to be completely smooth and perfectly spherical (and neglect atmospheric refraction, for now).


Figure 3 illustrates how a right triangle can be used to show the relationship of the height at which observations are made to the farthest distance seen (a point tangent to the


surface of the planet, in this case, d2). Because the shape in the figure is that of a right triangle, the Pythagorean Theorem can be invoked to provide a determination of the length of line


segment d1-d2, or simply “d” for distance. The length of rE (radius of Earth), from point c to point d2, is the radius of the Earth (6371 km); the length of segment rE plus h is a known


value – whatever the height (h)


of observation is plus the radius of Earth. For a standing per- son, the line of sight to the distant horizon is from the eyes, so for simplicity, consider an example where the ground-to-eye height (h) is about 1.5 m (or, almost 5 feet).


The mathematical process is as follows (Plait, 2009; Stern, 2004):


Pythagorean Theorem: a2 + b2 = c2 (where each letter denotes the length of one side of the right triangle, with c2 being the longest side or hypotenuse). In this example, the


value of the hypotenuse is equal to (rE + h). Therefore, with substitution (where d = length of d1-d2) and rearrangement: (rE)2+ (d)2= (rE+h)2 (d)2= (rE+h)2 – (rE)2


Adding in our values for the variables, the math can be worked out and an answer obtained (additional decimal place values are retained):


(d)2 = (6371 km + 0.0015 km)2 – (6371 km)2 (d)2 = (6371.0015 km)2 – (6371 km)2 (d)2 = (40589660.11 km2) – (40589641 km2) dkm2) d = 4.371 km


So, in answering the question – what is the visual distance to the horizon2 for an observer with eyes set 1.5 meters above the ground over level terrain? – a value of about 4.4 kilometers is found. For an observer with eyes set at a height of about 5 feet, the greatest distance seen is 2.7 miles3 over level terrain.


Deriving a formula for the procedure above is similarly straightforward. Instead of working out an example, we leave the variables in the equation and simplify:


(rE)2 + (d)2 = (rE + h)2


h2 = rE2 + 2rE (h) + h2 This leads to:


We know: (rE + h)2 = (rE + h)(rE + h)= rE2 + rE (h) + rE (h)+


rE2 + (d)2 = rE2 + 2rE (h) + h2 rE2 – rE2 + (d)2 = 2rE (h) + h2 d2 = 2rE (h) + h2 d2 = h(2rE + h) dh(2rE + h)


Or, as an approximation4: dh(2rE ) 2. This distance to the horizon neglects refraction of light, which increases the actual distance seen.


3. Each meter is about 39.5 inches, so a height of 1.5 meters corresponds to 39.5 inches plus 19.75 inches, which is 59 inches or about 5 feet; and, since 1 mile is equal to about 1.61 km, 4.4 km divided by 1.61 gives 2.7 miles.


4. The height of a person will not likely exceed 2 meters, and so a quick calculation shows that the ratio of h:rE will not typically exceed 0.0025 km:6371 km, or 3.9 x 10-7. Therefore, the final estimate of distance is not affected until the 6th or 7th decimal place when “2rE + h” is used instead of “2rE.”


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