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Amk service between NYC and Boston restored, airlines badly fouled
On 1/13/2011 10:56 PM, John Levine wrote:
I'm not going to fly to Asia from Newark.- Hide quoted text - Because you don't need to go to Asia, or because you have some objection to Newark? Because several of you in this newsgroup cannot read a compass. When I look at my compass (assisted by the handy Great Circle Mapper site), I see that EWR-SIN goes 2.7 deg N, and LAX-SIN at 302.8 NW, that is, north, not west. From LAX it's 8770 miles, from EWR it's 9535 miles, so EWR is about 800 miles farther. But ORD-EWR is 719 mi, and ORD-LAX is 1744 mi, so the total distance is less via Newark. Dunno if there are flights from SFO, but if so, it's still farther than via Newark. Great Circle URL: http://gc.kls2.com/ And, the orthodromic calculations back that up: Great-circle distance The great-circle distance or orthodromic distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere's interior). Because spherical geometry is rather different from ordinary Euclidean geometry, the equations for distance take on a different form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere). Between any two different points on a sphere which are not directly opposite each other, there is a unique great circle. The two points separate the great circle into two arcs. The length of the shorter arc is the great-circle distance between the points. A great circle endowed with such a distance is the Riemannian circle. Between two points which are directly opposite each other, called antipodal points, there are infinitely many great circles, but all great circle arcs between antipodal points have the same length, i.e. half the circumference of the circle, or πr, where r is the radius of the sphere. Because the Earth is approximately spherical (see Earth radius), the equations for great-circle distance are important for finding the shortest distance between points on the surface of the Earth (as the crow flies), and so have important applications in navigation. Formulae Let \phi_s,\lambda_s;\ \phi_f,\lambda_f\;\! be the geographical latitude and longitude of two points (a base "standpoint" and the destination "forepoint"), respectively, and \Delta\phi,\Delta\lambda\;\! their differences and \Delta\widehat{\sigma}\;\! the (spherical) angular difference/distance, or central angle, which can be constituted from the spherical law of cosines: {\color{white}\Big|}\Delta\widehat{\sigma}=\arccos \big(\sin\phi_s\sin\phi_f+\cos\phi_s\cos\phi_f\cos \Delta\lambda\big).\;\! A useful way to remember this formula is cos(central angle)= cos(longitude difference CTM ) , where CTM could be taken to mean 'Only the cos terms in longitude angle difference cosine expansion to be multiplied with cos(latitude difference)'. The central angle is alternately expressed in terms of latitude and longitude differences dlat,dlong, using only cosines, as: arccos( cos(dlat) - cos(lat1)*cos(lat2)*(1 - cos(dlong) ) . The distance d, i.e. the arc length, for a sphere of radius r and \Delta \widehat{\sigma}\! given in radians, is then: d = r \, \Delta\widehat{\sigma}.\,\! This arccosine formula above can have large rounding errors for the common case where the distance is small, so it is not normally used for manual calculations. Instead, an equation known historically as the haversine formula was preferred, which is much more numerically stable for small distances:[1] {\color{white}\frac{\bigg|}{|}}\Delta\widehat{\sig ma} =2\arcsin\left(\sqrt{\sin^2\left(\frac{\Delta\phi} {2}\right)+\cos{\phi_s}\cos{\phi_f}\sin^2\left(\fr ac{\Delta\lambda}{2}\right)}\right).\;\! Historically, the use of this formula was simplified by the availability of tables for the haversine function: hav(θ) = sin2 (θ/2). Although this formula is accurate for most distances, it too suffers from rounding errors for the special (and somewhat unusual) case of antipodal points (on opposite ends of the sphere). A more complicated formula that is accurate for all distances is the following special case (a sphere, which is an ellipsoid with equal major and minor axes) of the Vincenty formula (which more generally is a method to compute distances on ellipsoids):[2] {\color{white}\frac{\bigg|}{|}|}\Delta\widehat{\si gma}=\arctan\left(\frac{\sqrt{\left(\cos\phi_f\sin \Delta\lambda\right)^2+\left(\cos\phi_s\sin\phi_f-\sin\phi_s\cos\phi_f\cos\Delta\lambda\right)^2}}{\ sin\phi_s\sin\phi_f+\cos\phi_s\cos\phi_f\cos\Delta \lambda}\right).\;\! When programming a computer, one should use the atan2() function rather than the ordinary arctangent function (atan()), in order to simplify handling of the case where the denominator is zero, and to compute \Delta\widehat{\sigma}\;\! unambiguously in all quadrants. If r is the great-circle radius of the sphere, then the great-circle distance is r\,\Delta\widehat{\sigma}\;\!. . . . The shape of the Earth closely resembles a flattened sphere (a spheroid) with equatorial radius a of 6,378.137 km; distance b from the center of the spheroid to each pole is 6356.752 km. When calculating the length of a short north-south line at the equator, the sphere that best approximates that part of the spheroid has a radius of b2 / a, or 6,335.439 km, while the spheroid at the poles is best approximated by a sphere of radius a2 / b, or 6,399.594 km, a 1% difference. So as long as we're assuming a spherical Earth, any single formula for distance on the Earth is only guaranteed correct within 0.5% (though we can do better if our formula is only intended to apply to a limited area). The average radius for a spherical approximation of the figure of the Earth with respect to surface is approximately 6371.01 km, while the quadratic mean or root mean square approximation of the average great-circle circumference derives a radius of about 6372.8 km.[5] [edit] Worked example For an example of the formula in practice, take the latitude and longitude of two airports: * Nashville International Airport (BNA) in Nashville, TN, USA: N 36°7.2', W 86°40.2' * Los Angeles International Airport (LAX) in Los Angeles, CA, USA: N 33°56.4', W 118°24.0' First convert the co-ordinates to decimal degrees, keeping in mind, of course, these values are presumed converted to radians for the actual trigonometric calulation: \scriptstyle V^\circ=\text{Sign}\times\left(\text{Deg}+\frac{\t ext{Min}}{60}+\frac{\text{Sec}}{3600}\right)\to V=V^\circ\frac{\pi}{180};\quad\text{E.g.,}\,\sin(V )=\sin\left(V^\circ\frac{\pi}{180}\right);\! * BNA: \;\phi_s=36.12^\circ\tfrac{\pi}{180}, \quad\lambda_s=-86.67^\circ\tfrac{\pi}{180}\;\! * LAX: \;\phi_f=33.94^\circ\tfrac{\pi}{180}, \quad\lambda_f=-118.40^\circ\tfrac{\pi}{180};\;\! Plug these values into the law of cosines \cos\Delta\widehat{\sigma}=\sin\phi_s\sin\phi_f+\c os\phi_s\cos\phi_f\cos\Delta\lambda\,\! \Delta\widehat{\sigma} comes out to be 25.958 degrees, or 0.45306 radians, and the great-circle distance is the assumed radius times that angle: r\,\Delta\widehat{\sigma}\approx 6372.8 \times 0.45306 \approx 2887.26\mbox{ km}.\;\! Thus the distance between LAX and BNA is about 2887 km or 1794 statute miles (× 0.621371) or 1559 nautical miles (× 0.539957). Actual distance between the given coordinates on the GRS 80/WGS 84 spheroid is about 2892.8 km. |
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