65 lines
1.7 KiB
JavaScript
65 lines
1.7 KiB
JavaScript
'use strict';
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Object.defineProperty(exports, '__esModule', { value: true });
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/**
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* @copyright 2013 Sonia Keys
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* @copyright 2016 commenthol
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* @license MIT
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* @module circle
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*/
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/**
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* Smallest finds the smallest circle containing three points.
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*
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* Arguments should represent coordinates in right ascension and declination
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* or longitude and latitude. Result Δ is the diameter of the circle, typeI
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* is true if solution is of type I.
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*
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* @param {Coord} c1 - ra, dec point 1
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* @param {Coord} c2 - ra, dec point 2
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* @param {Coord} c3 - ra, dec point 3
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* @returns {Array} [Δ, typeI]
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* {Number} Δ - diameter of the circle
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* {Number} typeI - true - Two points on circle, one interior.
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* false - All three points on circle.
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*/
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function smallest (c1, c2, c3) {
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// Using haversine formula
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const cd1 = Math.cos(c1.dec);
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const cd2 = Math.cos(c2.dec);
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const cd3 = Math.cos(c3.dec);
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let a = 2 * Math.asin(Math.sqrt(hav(c2.dec - c1.dec) + cd1 * cd2 * hav(c2.ra - c1.ra)));
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let b = 2 * Math.asin(Math.sqrt(hav(c3.dec - c2.dec) + cd2 * cd3 * hav(c3.ra - c2.ra)));
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let c = 2 * Math.asin(Math.sqrt(hav(c1.dec - c3.dec) + cd3 * cd1 * hav(c1.ra - c3.ra)));
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if (b > a) {
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[a, b] = noswap(b, a);
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}
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if (c > a) {
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[a, c] = noswap(c, a);
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}
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if (a * a >= b * b + c * c) {
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return [a, true]
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}
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// (20.1) p. 128
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return [2 * a * b * c / Math.sqrt((a + b + c) * (a + b - c) * (b + c - a) * (a + c - b)), false]
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}
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const noswap = function (a, b) {
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return [a, b]
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};
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/**
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* haversine function (17.5) p. 115
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*/
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const hav = function (a) {
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return 0.5 * (1 - Math.cos(a))
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};
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var circle = {
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smallest
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};
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exports["default"] = circle;
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exports.smallest = smallest;
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