It is not a bug, it is becouse algorithm wwork with flat sphere. And I need to calculate scattering on the mountain.
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Turbochist
Member Since 26 Oct 2013Offline Last Active May 27 2016 05:59 AM
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In Topic: O'Neil atmosphere mountains bug.
27 May 2016  01:09 AM
In Topic: GLSL optimization. What is faster?
17 April 2016  03:57 PM
Yes! Thank you very much WiredCat and Alex! I think I understand. Thank you!
In Topic: How to calculate sun and planets positions around the Earth by real date and...
03 April 2016  03:45 AM
Actually I've used this calculations you can watch results on http://www.openglobus.org:
It's not best like Simon's but good enough for me now and comparatively low calculations.
og.math.TWO_PI = 2.0 * Math.PI; og.math.PI_TWO = Math.PI / 2.0; og.math.RADIANS = Math.PI / 180.0; og.math.DEGREES = 180.0 / Math.PI; // Angle between J2000 mean equator and the ecliptic plane. // 23 deg 26' 21".448 (Seidelmann, _Explanatory Supplement to the // Astronomical Almanac_ (1992), eqn 3.2221. og.astro.J2000_OBLIQUITY = 23.4392911; og.math.rev = function (x) { return x  Math.floor(x / 360.0) * 360.0; }; og.math.Quaternion.yRotation = function (a) { a *= 0.5; return new og.math.Quaternion(0.0, Math.sin(a), 0.0, Math.cos(a)); }; /** * http://stjarnhimlen.se/comp/tutorial.html * a Mean distance, or semimajor axis * e Eccentricity * T Time at perihelion * * q Perihelion distance = a * (1  e) * Q Aphelion distance = a * (1 + e) * * i Inclination, i.e. the "tilt" of the orbit relative to the * ecliptic. The inclination varies from 0 to 180 degrees. If * the inclination is larger than 90 degrees, the planet is in * a retrogade orbit, i.e. it moves "backwards". The most * wellknown celestial body with retrogade motion is Comet Halley. * * N (usually written as "Capital Omega") Longitude of Ascending * Node. This is the angle, along the ecliptic, from the Vernal * Point to the Ascending Node, which is the intersection between * the orbit and the ecliptic, where the planet moves from south * of to north of the ecliptic, i.e. from negative to positive * latitudes. * * w (usually written as "small Omega") The angle from the Ascending * node to the Perihelion, along the orbit. * * P Orbital period = 365.256898326 * a**1.5/sqrt(1+m) days, * where m = the mass of the planet in solar masses (0 for * comets and asteroids). sqrt() is the square root function. * * n Daily motion = 360_deg / P degrees/day * * t Some epoch as a day count, e.g. Julian Day Number. The Time * at Perihelion, T, should then be expressed as the same day count. * * t  T Time since Perihelion, usually in days * * M Mean Anomaly = n * (t  T) = (t  T) * 360_deg / P * Mean Anomaly is 0 at perihelion and 180 degrees at aphelion * * L Mean Longitude = M + w + N * * E Eccentric anomaly, defined by Kepler's equation: M = E  e * sin(E) * An auxiliary angle to compute the position in an elliptic orbit * * v True anomaly: the angle from perihelion to the planet, as seen * from the Sun * * r Heliocentric distance: the planet's distance from the Sun. * * x,y,z Rectangular coordinates. Used e.g. when a heliocentric * position (seen from the Sun) should be converted to a * corresponding geocentric position (seen from the Earth). */ og.astro.earth.getSunPosition = function (jd) { var d = jd  og.jd.J2000; var w = 282.9404 + 4.70935E5 * d; //longitude of perihelion var a = 1.000000; //mean distance, a.u. var e = 0.016709  1.151E9 * d; //eccentricity var M = og.math.rev(356.0470 + 0.9856002585 * d); //mean anomaly var oblecl = og.astro.J2000_OBLIQUITY  3.563E7 * d; //obliquity of the ecliptic var L = og.math.rev(w + M); //Sun's mean longitude var E = M + og.math.DEGREES * e * Math.sin(M * og.math.RADIANS) * (1 + e * Math.cos(M * og.math.RADIANS)); //eccentric anomaly //Sun rectangular coordiantes, where the X axis points towards the perihelion var x = Math.cos(E * og.math.RADIANS)  e; var y = Math.sin(E * og.math.RADIANS) * Math.sqrt(1  e * e); var r = Math.sqrt(x * x + y * y); // distance var v = Math.atan2(y, x) * og.math.DEGREES; // true anomaly var lon = og.math.rev(v + w); //longitude of the Sun //the Sun's ecliptic rectangular coordinates x = r * Math.cos(lon * og.math.RADIANS); y = r * Math.sin(lon * og.math.RADIANS); //We use oblecl, and rotate these coordinates var xequat = x; var yequat = y * Math.cos(oblecl * og.math.RADIANS); var zequat = y * Math.sin(oblecl * og.math.RADIANS); var theta = og.math.TWO_PI * (d * 24.0 / 23.9344694  259.853 / 360.0); // Siderial spin time return og.math.Quaternion.yRotation(theta).mulVec3(new og.math.Vector3(yequat * og.astro.AU_TO_METERS, zequat * og.astro.AU_TO_METERS, xequat * og.astro.AU_TO_METERS)); //Convert to RA and Decl //var RA = Math.atan2(yequat, xequat) * og.math.DEGREES; //var Decl = Math.atan2(zequat, Math.sqrt(xequat * xequat + yequat * yequat)) * og.math.DEGREES; };
Thanks guys!
In Topic: How to calculate sun and planets positions around the Earth by real date and...
29 March 2016  12:40 AM
I he found Bretagnon and Simon method for the Sun position calculation. Seems what I m looking.
In Topic: How to calculate sun and planets positions around the Earth by real date and...
28 March 2016  03:58 AM
Im not real clear to understand how to convert and what convert to my 3d space cartesian. I mean have I to convert ecliptical Earth coordinates to cartesian and how?
For example I'm sure it is not righ but what I think:
1. By date and time(juliat tai) I get Earth ecliptical coordinates and distance to the Sun.
2. Convert some how ecliptical to cartesian.
3. By time(seconds of a day) and day by calculated nutation and precession etc. I'll get the rotation of the Earth in geocentric coordiantes. and rotate it.
4. Place the Sun to the cartesian position respect to the rotated Earth.
Could you help me understand is it right way?