The first task in solving for A is to find out how far the horizon is, in angular measure. The observer is r+h distant from the center of the earth, and his line of sight to the horizon is perpendicular to a radius of the earth, r, to that point on the horizon, thus forming a right triangle with one leg r and the hypotenuse r+h.
The angle of this triangle at the center of the earth is therefore arccos(r/(r+h)). The web site http://www.rism.com/Trig/spherical_cap.htm (from Google search on Area of spherical cap), calls this angle alpha and gives a formula
A = 2 pi (1 - cos(alpha)) rho^2
for the area of the spherical cap, where rho is the radius of the sphere. This makes the formula for the visible area
A = 2 pi (1 - r/(r+h)) rho^2
where rho = 6378 km and h must be expressed in km, giving an answer in km^2.
A table:
h h area area area/area of earth
(km) (mi) km^2 mi^2 (fraction)
100 62.14 3945554 1523387 0.007718432
200 124.27 7771145 3000456 0.015202189
300 186.41 11482164 4433288 0.022461815
400 248.55 15083681 5823842 0.029507229
500 310.69 18580472 7173960 0.036347776
600 372.82 21977039 8485382 0.042992261
700 434.96 25277632 9759748 0.049448997
800 497.10 28486260 10998606 0.055725829
900 559.23 31606715 12203421 0.061830173
1000 621.37 34642581 13375575 0.067769043
2000 1242.74 61015270 23558127 0.119360229
3000 1864.11 81763585 31569097 0.159948816
4000 2485.48 98513381 38036229 0.192715359
5000 3106.86 112318934 43366583 0.219722271
6000 3728.23 123893828 47835674 0.242365487
7000 4349.60 133738284 51636640 0.261623561
8000 4970.97 142213363 54908886 0.278202810
9000 5592.34 149586207 57755558 0.292625829
10000 6213.71 156058717 60254607 0.305287581
11000 6835.08 161786318 62466046 0.316492116
12000 7456.45 166890608 64436824 0.326477310
13000 8077.83 171468086 66204198 0.335431933
14000 8699.20 175596306 67798113 0.343507704
15000 9320.57 179338315 69242910 0.350827954
16000 9941.94 182745887 70558581 0.357493967
17000 10563.31 185861940 71761696 0.363589700
18000 11184.68 188722348 72866106 0.369185331
19000 11806.05 191357332 73883479 0.374339980
20000 12427.42 193792529 74823714 0.379103799
21000 13048.80 196049832 75695263 0.383519614
22000 13670.17 198148046 76505388 0.387624216
23000 14291.54 200103418 77260362 0.391449384
24000 14912.91 201930054 77965630 0.395022714
25000 15534.28 203640262 78625945 0.398368284
26000 16155.65 205244831 79245472 0.401507196
27000 16777.02 206753253 79827877 0.404458026
28000 17398.39 208173921 80376400 0.407237187
29000 18019.76 209514275 80893914 0.409859235
30000 18641.14 210780939 81382975 0.412337127
31000 19262.51 211979826 81845869 0.414682434
32000 19883.88 213116236 82284639 0.416905519
33000 20505.25 214194928 82701124 0.419015694
34000 21126.62 215220191 83096980 0.421021348
35000 21747.99 216195897 83473702 0.422930059
36000 22369.36 217125555 83832646 0.424748690
37000 22990.73 218012351 84175039 0.426483471
38000 23612.11 218859181 84502002 0.428140069
39000 24233.48 219668687 84814554 0.429723655
40000 24854.85 220443285 85113628 0.431238950
41000 25476.22 221185183 85400077 0.432690278
42000 26097.59 221896411 85674683 0.434081607
43000 26718.96 222578832 85938167 0.435416582
44000 27340.33 223234160 86191191 0.436698559
45000 27961.70 223863978 86434365 0.437930632
The criteria for part 2 indicate that the earth, from horizon to opposite point on horizon, should subtend, at the viewer's eye, 179 degrees or less in order for the curvature of the earth to register on that observer as apparently curved. (180 degrees by any measure would not register as curved as that great circle is the equivalent of a stright line, since there'd be no way of assigning it as being curved in one direction or the other.)
In that right triangle mentioned above, therefore, the angle at the observer would be 179/2 degrees and h would be r / sin(179/2 degrees) - r. This comes out to .249 km, or .1509 miles or 797 feet. This leads me to think that the threshhold is overly optimistic in its estimation of human perception. 797 feet is lower than the 86th floor observation deck on the Empire State Building. From there, I can't honestly say I could notice that the earth was a sphere or that the horizon was curved.
If instead of the earth's subtending 179 degrees as being sufficiently small enough to see the curvature of the horizon, other figures are used, the following table shows h, in km, miles and feet:
subtend h(km) h(miles) h(feet)
179 0.24 0.15 797
178.9 0.29 0.18 964
178.8 0.35 0.22 1147
178.7 0.41 0.26 1347
178.6 0.48 0.30 1562
178.5 0.55 0.34 1793
178.4 0.62 0.39 2040
178.3 0.70 0.44 2303
178.2 0.79 0.49 2582
178.1 0.88 0.54 2877
178.0 0.97 0.60 3187
177.9 1.07 0.67 3514
177.8 1.18 0.73 3857
177.7 1.28 0.80 4216
177.6 1.40 0.87 4590
177.5 1.52 0.94 4981
177.4 1.64 1.02 5387
177.3 1.77 1.10 5810
177.2 1.90 1.18 6248
177.1 2.04 1.27 6703
177 2.19 1.36 7173
175 6.08 3.78 19935
173 11.92 7.41 39103
171 19.72 12.25 64705
169 29.50 18.33 96781
167 41.26 25.64 135380
165 55.04 34.20 180563
163 70.83 44.01 232398
161 88.69 55.11 290967
159 108.62 67.49 356362
157 130.66 81.19 428685
155 154.85 96.22 508054
153 181.23 112.61 594594
151 209.84 130.39 688446
149 240.72 149.58 789763
147 273.93 170.21 898714
145 309.52 192.33 1015481
143 347.55 215.96 1140262
141 388.09 241.15 1273271
139 431.21 267.94 1414739
137 476.99 296.39 1564917
135 525.50 326.53 1724073
133 576.83 358.43 1892498
131 631.09 392.14 2070504
129 688.37 427.73 2258426
127 748.78 465.27 2456626
125 812.44 504.83 2665492
123 879.48 546.49 2885442
121 950.04 590.33 3116924
119 1024.26 636.44 3360421
117 1102.30 684.93 3616454
115 1184.32 735.91 3885580
113 1270.53 789.47 4168403
111 1361.11 845.75 4465571
109 1456.27 904.88 4777786
107 1556.25 967.01 5105802
105 1661.29 1032.28 5450437
103 1771.67 1100.87 5812573
101 1887.68 1172.95 6193168
99 2009.62 1248.72 6593257
97 2137.86 1328.40 7013967
95 2272.75 1412.22 7456520
93 2414.70 1500.43 7922248
91 2564.16 1593.30 8412600
89 2721.61 1691.13 8929162
(Alpha is 180 degrees minus half the subtending arc.)
Note that the use of 180 degrees makes the calculation of the angle of the two intersecting great circle arcs, tangent to the ends of the visible portion of the horizon, easier to calculate than if some other separation were used. That would involve some spherical trig; the 180-degree case involves only comparing arc lengths to the ends of the 180-degree long line (great circle).
Edited on January 29, 2006, 3:21 pm
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Posted by Charlie
on 2006-01-29 15:18:13 |
discussion
