56.6: Vincenty’s Formulæ- Inverse Problem
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In the inverse problem, we're given two points on the Earth's surface (ϕ1,L1) and (ϕ2,L2) and want to calculate the distance s between them, as well as the direction from one to the other. We'll use the constants defining the Earth's ellipsoid as before:
- Earth ellipsoid semi-major axis (i.e. equatorial radius): a=6378137.0 meters.
- Earth flattening factor f=1/298.257223563. This is defined as the difference between semi-major and semi-minor axes, divided by the semi-major axis: f=(a−b)/a.
In performing the following calculations, work with all angles in radians. We begin by calculating
U1=tan−1[(1−f)tanϕ1]U2=tan−1[(1−f)tanϕ2]L=L2−L1b=(1−f)a
Now set an initial value λ=L. Then iterate on Eqs. (53.28) through (53.35) until λ converges:
sinσ=√(cosU2sinλ)2+(cosU1sinU2−sinU1cosU2cosλ)2cosσ=sinU1sinU2+cosU1cosU2cosλσ=arctansinσcosσsinα=cosU1cosU2sinλsinσcos2α=1−sin2αcos2(2σm)=cosσ−2sinU1sinU2cos2αC=f16cos2α[4+f(4−3cos2α)]λ=L+(1−C)fsinα{σ+Csinσ[cos(2σm)+Ccosσ(−1+2cos2(2σm))]}
When λ has converged to the desired degree of accuracy, continue calculating:
u2=(cos2α)(a2−b2b2)A=1+u216384{4096+u2[−768+u2(320−175u2)]}B=u21024{256+u2[−128+u2(74−47u2)]}Δσ=Bsinσ{cos(2σm)+14B[cosσ(−1+2cos2(2σm))−16Bcos(2σm)(−3+4sin2σ)(−3+4cos2(2σm))]}s=bA(σ−Δσ)α1=arctan(cosU2sinλcosU1sinU2−sinU1cosU2cosλ)α2=arctan(cosU1sinλ−sinU1cosU2+cosU1sinU2cosλ)
Then s is the distance between the two points.
Find the distance between the sounding rocket in Chesapeake Hall at Prince George's Community College ( 38∘53′16.87′′N,76∘49′23.14′′W ) and the top (apex) of the Great Pyramid of Giza in Egypt (29∘58′45.03′′N,31∘08′03.69′′E).
Solution
The given parameters are the coordinates ϕ1=38.888019∘,L1=−76.823094∘,ϕ2= 29.979175∘,L2=+31.134358∘. Employing Vincenty’s formulæ (inverse method), we find:
U1=38.794230∘U2=29.895958∘L=339.15856744∘b=6356752.3 meters λ=108.139490∘u2=0.00393162979A=1.00098218405082B=9.809796134747123×10−4Δσ=0.054160886∘s=9351378.858 meters α1=55.910048∘α2=131.801775∘
So the distance s=9351.378858 km (5280 miles, 3576 feet, 10 inches), in the direction 55.910048∘ (10.91 south of northeast).