This is the third lab
week for the year 2 project and our progress is quite slow. This week we were mainly focused on the calculations. All members of the group where trying to understand the complex algorithms that needed to be implemented into the code which was easier said than done, as it seemed each equation led to another unknown!
A complete calculation was finally achieved by Dominyka! Her entire working is shown below. This demonstrates how complex the algorithm is as all of this working merely shows the position at one specific time on a certain date.
Calculating
the position of the Moon for 05/03/2017 at 11:32. (first quarter) [1]
1) Conversion
of time and date to Julian days
Julian
day = 864000s
Julian
year = 365.25d
Julian
century = 36525d
JD=365.25*year’+30.6001*(month’+1)-15+1720996.5+day+(hour+minute/60+second/3600)/24
If month<=2, then month’=month+1,
year’=year-1,
if month>2, then month’=month,
year’=year.
Number of Julien centuries: T=(JD-JD0)/36525
JD0 is the Julian day for
01/01/2000 at 12:00UT
JD0=365.25*1999+30.6001*14-15+1720996.5+1+12/24=2451546.151
JD is the Julian day for 05/03/2017
at 11:32UT
JD=365.25*2017+30.60001*4-15+1720996.5+5+(11+32/60)/24=2457818.631
T=6272.479596/36525=0.171731132
2) Astronomical
algorithms
·
Ecliptic latitude B and longitude L of the
Moon [2]
B=-5.171º
L=75.062º
·
Convert B and L to right ascension RA and
declination delta
eps=23+26/60+21.448/3600-(46.8150*T+0.00059*T2-0.001813*T3)/3600
X=cosB*cosL
Y=cos(eps)*cosB*sinL-sin(eps)*sinB
Z=sin(eps)*cosB*sinL+cos(eps)*sinB
R=
delta=180/π*arctg(Z/R) //π=180º
RA=180/π*arctg(sinL*cos(eps)-(tgB*sin(eps))/cosL)
eps=23+26/60+21.448/3600-(46.8150*0.171731132+0.1717311322*0.00059-0.001813*0.1717311323)/3600=23.43705789º≈23.437º
X=cos(-5.171º)*cos(75.062º)=0.2567
Y=cos(23.437º)*cos(-5.171º)*sin(75.062º)-sin(23.437º)*sin(-5.171º)=0.9187
Z=sin(23.437º)*cos(-5.171º)*sin(75.062º)-sin(23.437º)*sin(-5.171º)=0.3
R=0.953926
delta=1*arctg(0.3/0.953926)=17.4578º
RA=arctg(sin(75.062º)*cos(23.437º)-
(tg(-5.171º)*sin(23.437º))/cos(75.062º))=45.73877694º≈45.74º
·
Sidereal time at Greenwich [3]
theta0=280.46061837+360.98564736629*(JD-JD0)
Local time: theta=theta0+longitude(E+,W-)
Hour angle: tau=theta-RA
theta0=280.46061837+360.98564736629*6272.479596=2264555.568
2264555.568/360=6290.432134
6290.432134-6290=0.4321338119
theta0=0.4321338119*360=155.5681723º
theta=155.5681723º-2.963827º=152.6043453º
tau=152.6043453º-45.74º=406.8643453º
3) Final
results
·
Convert tau and delta to horizon
coordinates h(altitude) and az(azimuth) of the observer (53.406773º,
-2.965723º)
sinh=sin(beta)*sin(delta)+cos(beta)*cos(delta)*cos(tau) //beta-latitude
tg(az)=-sin(tau)/(cos(beta)*tg(delta)-sin(beta)*cos(tau))
h=sin-1(sin(53.406773º)*sin(17.4578º)+
cos(53.406773º)*cos(17.4578º)*cos(106.8643453º))=4.353º
az=arctg(-sin(106.8643453º)/(cos(53.406773º)*tg(17.4578º)-sin(53.406773º)*cos(106.8643453º)))=66.285º
·
Compute the parallax in altitude
horParal=r/a
(rad) //r-radius of the
Earth(6378km)
//a-distance
to the Moon(384400km)
Converting
to degrees: horParal=horParal*57.2957795
altParal=sin-1(cosh*sin(horParal))
Apparent
altitude: H=h-altParal
horParal=0.0165921rad=0.9506568201º
altParal= sin-1(cos(4.353º)*sin(0.9506568201º))=0.9479142571º
H=4.353º-0.9479142571º=3.41º
Notes for week 4:
- Finish implementing the code and begin testing.
- Design the poster and discuss what should be included.
[3] - http://www2.arnes.si/~gljsentvid10/sidereal.htm