RoundEarth

How We Know Earth Is NOT Flat Earth Dimensions The earth diameter is an The Spherical Astronomy method recognizes this important element of maps and geography. The formula as S=(ΔS)(360)/(A2-A1), with A1 being zero spherical diameter is used to scale the land area in Eratosthenes’ original case, & ΔS the length of a between grid points. Also this discussion is given line along same meridian, with A2 & A1 being angles to rebuff “flat earth” proponents, if they will of sun’s shadow at identical times. Statistical listen to logic. In 1687, Isaac Newton published Variants of this method were used in 18th & 19th the Principia in which he included a proof that a century to improve diameter estimates. rotating self-gravitating fluid body in equilibrium Posidonius (135-51BC), calculated the Earth's takes the form of an oblate ellipsoid of revolution circumference by reference to (spheroid). The WGS84 GPS map uses an the position of the star inverse flatness ratio of 298.26. In general terms, Canopus. As explained by for a sphere, the ratio of Circumference to Cleomedes, Posidonius Diameter is the value of Pi, π, 3.14159, diameter observed Canopus on but never being twice the radius. above the horizon at Rhodes, while at Alexandria he Eratosthenes' (275-194BC), method to calculate saw it ascend as far as 7½ degrees above the horizon. Earth Diameter is described in simplified version. Since he thought Rhodes was 5,000 stadia due north Where-in was given a distance from Alexandria to of Alexandria, and the difference in the star's Aswan (ancient Syene) of 5000 Stadia (~500 statute elevation indicated the distance between the two miles) & the angle of a stick shadow at high noon locales was 1/48 of the circle, he multiplied 5,000 by upon Summer Solstice in Alexandria was 1/50 of a 48 to arrive at a figure of 240,000 stadia circle and at Aswan there was no shadow. He (24,000miles) for the circumference of the Earth. considered these 2 points to fall upon the same Ptolemy, (100-170AD), used Strabo’s, (63BC-24AD) meridian. So the circumference of earth was 50*500 lower revision of 180,000 stadia (about 33% too low) or 25,000 miles. Today’s accepted distance is for the Earth's circumference in his Geography. Ptolemy’s number was then used by Christopher 24,902miles. Eratosthenes’ value differed by only Columbus (1492AD) to underestimate the distance to 0.39% of the modern circumference. India as 70,000 stadia. It is generally thought that the stadia used was 1/10 of a modern statute mile. Whilst the assumptions of these early writers were not precisely correct, it was sufficient until the era of modern physics. Great Triangle Methods (Line-of-Sight, LoS) Lastly are the Line-of-Sight (LoS) methods. With this change in methods comes a change in variable. From earth circumference to earth radius, Re =24,902/2/3.14=3963miles. Interesting aside, 25/ (2*4)=3-1/8 or 3.13 very nearly π. In the modern world radar is routinely used as a navigation aid. The LoS equation for radar also has a factor for refraction by atmosphere. The equations are shown in the following diagram: The formula for calculating LoS is below with Re the radius of Earth and Ho, height above sea level. The term Ho2 can be dropped because (2ReHo >>Ho2) December2020 OP Armstrong 1 of 3 

How We Know Earth Is NOT Flat & LoS2=(2ReHo+Ho2)+(Re2-Re2) LoS2=(2ReHo+Ho2), light-rays skim (tangent to) the sea. Let LoS be this (LoS2-Ho2)/2Ho=Re as D=2R observer’s distance from the Pharos, and Re his (LoS2 /Ho)-Ho=D, or simply as LoS=√(2ReHo) distance from the Earth’s center, while the Pharos’ The formula is LoS=√(2ReHo) = √(1.5Ho), with Ho in flame is Re+Ho from that center Ho being the Pharos’ feet, LoS in miles, 1.5 is from Re= 3963 miles so height and Re the ideally-spherical Earth’s radius. At 2(3963){X.ft/(5280ft/mile)=1.501~1.5. Re=LoS2/2Ho the observer’s position, the angle between a Atmospheric refraction effect for radar adds an extra skimming-light-ray vector and the Earth-radius 15.4% to the LoS, which changes the 1.5 under the vector is a right angle. Assume an airless Earth bracket to 2.0, LoR=√(2Ho). Where LoR is Line-of- (which permits straight-line light-rays), and use Radar. For visible light the effect has been estimated Pythagoras’ Theorem. between 10% and 17% depending upon the But to find the Real Lighthouse Equation (based on atmospherical conditions. Earth-with-atmosphere) at sea level, one must Some examples: the radar horizon for 1-mile (5280’) account for horizontal atmospheric refraction, which altitude will be 102.76-mile and at 75 feet height will stretches artificially by the square root of 6/5 since be 12.25-mile. At 14,000’ LoS=145mi, LoR=167mi. horizontal light is bent with curvature equal to 1/6 of At 200’H, LoS=17.3mi & LoR=20miles. the Earth’s curvature (S.Newcomb 1906 “Josephus J.War 4.613 says, “the flame of the Pharos pp.198-203). so LoS2 is augmented by factor 6/5, Light House in Alexandria was visible to ships for producing an Earth-radius high by +20%. (Curvature 300 stades,” which would make it the world’s then- is the inverse of radius) To return the problem to the tallest building (exceeding the Great Pyramid); yet it straight-ray Pythagorean math requires undoing the was never so described. Solution to Josephus’ datum: effect of the ray’s curvature. Ancients may have the crow’s-nests of tall ancient ships were roughly suspected atmospheric refraction, but no evidence for 1/4 of the Pharos’ height, meaning that about 1/3 of quantitative corrections exist until Tycho (c.1600 Josephus’ 300 stadia was due to ship-height; so Ho ≈ AD). 200 stadia is an adequate rough estimate for the Biruni’s Method used the same Great Triangle but Pharos’ visibility-distance at sea level. looked upon the arrangement slightly different, with At whatever distant point the Pharos’ flame starts use of the angle to the horizon, α. (due to Earth-curvature) becoming invisible to a But the angle to horizon is also receding observer on the sea, is where the Pharos’ affected by refraction. Since his result was not corrected for refraction, then either the results or inputs must remain suspect. Also as the angle to horizon is very small, it would demand a hard to determine correction for refraction. Re=HoCosα/(1-Cosα), for small angles the Taylor series of Cosα, α must be in radians, Cosα=1-α2/2, so Re=2Ho/α2. Rads=2πdeg/360. For a 10000 foot hill 1o46’ & for 1055 foot hill 0o34’, average Radius 4002miles. For such small angles refraction corrections should be used. This online calculator for un-refracted curvature may be helpful. December2020 OP Armstrong 2 of 3

How We Know Earth Is NOT Flat Appendix on Triangle and changes to degree. In some routines this may Solution Methods not be required if function “ACOS” is set to degree Law of Sines’ when Given rather than radians. 2 Angles and 1 Side find other Side L’s. Let capitol “A” be the referenced angle and let small “a” refer to the length of side “a” opposite angle “A”, likewise B,b & C,c. Given c=600, A=57.6O & B=64.85O, find a & b. 600.33=SIN(57.6)÷SIN(180-57.6-64.85)×600=b & 643.61=SIN(64.85)÷SIN(180-57.6-64.85)×600=a. The angle functions must accept degrees else use rads as 2π rads per 360degrees of angle. Law of Tangents with Law of Cosines given: 2 sides and one (1) known angle L3 = (L1^2+L2^2−2×L1×L2×COS(θ))^0.5 Given L1=345; L2=174.07: θ=37.333degree nd L3 to be 232.001 & Α =115.6deg, B=27.065deg Given AO, BO, & d of d=76’, find h Law of Tan&Cosin 1<) & 2L =76’×TAN(27)×TAN(52)÷{TAN(52)−TAN(27)}=64.3’ L1 18.739 464.700 367.000 345.000 Given 3 sides nd Angles by Law of L2 7.642 289.300 498.000 174.070 Cosine For a=345, <)12 45.308 87.063 37.730 37.333 b=174.07, c=232.001, & s/ L3 14.426 534.660 305.929 232.001 2=Σ=375.54 with s/2 being just <)a 112.600 60.230 47.230 115.602 (a+b+c)/2 we nd angles as <)b 22.120 32.710 95.040 27.065 A-B= 2×(ATAN((L1−L2)÷(L1+L2)÷TAN(θ/2)))=λ A+B=(180-θ)=φ & A=(λ+φ)/2 & B=φ-Α Example Data Sets Included. A=2×DEGREES(ACOS((Σ×(Σ−a)/b/c)^0.5))=115.6 B=2×DEGREES(ACOS((Σ×(Σ−b)/a/c)^0.5))=27.07 C=2×DEGREES(ACOS((Σ×(Σ−c)/b/a)^0.5))=37.33, The 3rd angle could be determined by 180 less 1st less 2nd angles but best to prove solution validity by summing all angles. This sum should be 180. The function “DEGREES” takes the result in radians December2020 OP Armstrong 3 of 3 ififif