Hi, 2009/2/8 Jos Kint <jos.k...@skynet.be>
> Hello sundialists around the world! > > Who can help me? I am looking for the algebraic relation between the > Equation of time and the eccentricity of the earth orbital. It will help me > for calculating this eccentricity with my own sun dial. > Chris Lusby Tailor gave me a nice hint with the Ptolemaios model, which was > rather effective ( I found e= 0,0172 instead of 0,0167) , but I think that > there must be also some algebraic solution to the problem. Who is aware of > such a solution? > Find below my solution to the problem. If you know the equation of time, you can calculate the local mean solar time when the sun crosses the meridian. From this you can calculate the local sideral time, which is also sun's right ascension R.A. at that moment. Knowing the obliquity i (23.439) of the ecliptic sun's ecliptic longitude L results then from the equation (1) tan(L)=tan(R.A.)/cos(i). Next we must know or guess the time, when the sun is at the perigee. We can safely assume Jan 3 and improve the date later when our analysis progress. As before, we can now calculate sun's ecliptic longitude Lp at the perigee at which its true anomaly v is (2) v=L-Lp. Next we calculate P, the number of days from the perigee to the date of the observation to find sun's mean anomaly M (in degrees) (3) M=P*360/365.2422 If e is the eccentricity of the orbit and E sun's eccentric anomaly then according to Kepler's equation (4) E = M + e sin(E). Further (5) tan(v/2) = q tan(E/2), where q= square root [(1+e)/(1-e)] or differently (6) sin(E) = r sin(v)/[1+e cos(v)], where r = square root(1-e*e). To solve equations (4) and (6) the method of iteration may be used. Change all angles to radians and take E=(M+v)/2 as a starting value. Count then e = (E-M)/sin(E) sin(Ek)= r sin(v)/(1+e cos(v)) (note that approximately r=1) Ek=arc sin (sin(Ek)) (take same quadrant where M is) E=(E+Ek)/2 and start counting again as long as E is changing. If you have many observations, you may try to seek the date of the perigee, too. If you guess wrong, the values of the iterations are totally wrong and different from each others. In my tests I have gained an accuracy of +- one day. As can be seen, there is much to calculate. So a good calculator is demanded. I have used my own PcCalculator. You can find its home page here http://pc-calculator.sourceforge.net/ Aimo Niemi > > Jos Kint > Email: jos.k...@skynet.be > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > >
--------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial
