calculate the time when the sun is X degrees below/above the Horizon
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O

1

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I want to know what is the time when the sun is X degrees below/above the Horizon.

For example, I want to find the time when the sun is 19.75 degrees below the horizon. I think it has something to do with the zenith in the function date_sunrise/date_sunset but I'm not sure.

Thanks in advance!

Oddball answered 18/3, 2014 at 23:12 Comment(3)
You'll find most of the algorithms here but you also need date and position on the earth's surface as well as altitudeWeller
@MarkBaker Hey! I found many equations on the webpage, but couldn't find the equation where I give it degrees below/above the horizon. Could you help me here?Oddball
btw you need visual, physical or geometric Sun's position? (see atmospheric refraction,time of light travel is 8 min so you see position from last 8 min) what precision you need (if big then you need also include aberations)Kidron
K
2
  1. collect Sun ephemerids data for day you need

take 1 hour steps and get Suns position in azimuthal coordinates for the geo-location you need. Either use equations you found or use some WEB service like:

  • JPL Horizons don't like this one as they use some weird output reference frames that does not correspond to my measurements, but it is more likely I transform something wrong along the way ...

  • Presov observatory this is mine favorite (but it is in Slovak) output is easily copyable to mine engines and the output is corresponding with mine observations, computations, estimations and measurements. Just fill in the:

    • geo-location (Miesto pozorovania)
    • date,time (Dátum a čas pozorovania)
    • in the bottom from left: interval[days],interval step[days]
    • click on the button for Sun(Slnko), Moon(Mesiac),Planets(Planety)

there are many such pages out there just look but always check if they outputting correct data. I use Kepler's laws/equations form planetary motions (lower precision but for Earth-Sun should be OK). Nowadays engines use gravity model instead (but that is unstable with longer times from epoch)

  1. handle the data as set of 3D points along polyline (azimut,height,time)

  2. now just find in the data 2 points

one below desired angle and the next above desired angle. Booth points must be neighboring. If any point is on the desired angle then you already have the solution so stop

  1. interpolate the height angle crossing time

enter image description here

so if desired height angle is b and wanted time t then:

  • a0,a1 are azimutal angles
  • b0,b1 are height angles
  • t0,t1 are times

then just solve this linear system:

    b=b0+(b1-b0)*u
    t=t0+(t1-t0)*u

so if I did not make some silly mistake:

    t=t0+((t1-t0)*(b-b0)/(b1-b0))

[Notes]

if you do not need too high precision (and usage above 100 years) and geo-location is fixed then you can table whole year and use this data periodically. This way you will not need step 1 on runtime.

[Edit1] Kepler's law

if you want to go this way look here. You will need orbital and rotation parameters of Earth. These are extracted from mine ephemeris engine *.ini for solar system:

[Earth]
txr_map=Earth_Map.jpg
txr_nor=Earth_Normal.jpg
txr_clouds=Earth_Cloud.jpg
txr_lights=Earth_Light.jpg
txr_ring_map=
txr_ring_alpha=
is_star=0
mother=Sun
re=6378141.2
rp=6356754.79506139
r0=-1
r1=-1
ha=60000
vd=250000
B0r=0.1981
B0g=0.4656
B0b=0.8625
B0a=0.75
t0=-0.0833333333333333 ; this means 1.1.2000 00:00:00 UT
a=149597896927.617
da=-0.122872993839836
e=0.01673163
de=-1.00232717316906E-9
i=-9.48516635288838E-6
di=-6.38963964003634E-9
O=-0.004695
dO=-1.15274665428334E-7
o=1.79646842620403
do=1.51932094052745E-7
M  =1.7464
dM =0.0172021242603194
ddM=0
rota0 =3.0707963267949
rotda =6.30038738085328
prea0 =1.5707963267949
preda =-6.68704522111755E-7
prei  =0.409124584728753
predi =0
nuta  =0
nutda =0
nutia =0
nutdia=0
nutii =0
nutdii=0

and here are the explanations:

[Name]      [string id] object ID name
txr_map     [filename] surface texture
txr_nor     [filename] surface normal/bump texture 
txr_clouds  [filename] cloud blend texture (white cloud, black clear sky)
txr_lights  [filename] night surface texture
txr_ring_map    [filename] rings color texture 
txr_ring_alpha  [filename] rings alpha texture (alpha0 transparent, alpha1 solid)
is_star     [0/1] is star ?
mother      [string] "" or owner object name
re      [m] equator radius
rp      [m] polar radius
r0      [m] -1 or rings inner radius
r1      [m] -1 or rings outer radius
ha      [m]  0 or atmosphere thickness
vd      [m] -1 or atmosphere view depth
B0r     <0,1> star R light or atmosphere color
B0g     <0,1> star G light or atmosphere color
B0b     <0,1> star B light or atmosphere color
B0a     <0,1> overglow of star below horizont
t0      [day]     t0 time the parameters are taken after 1.1.2000 00:00:00
a       [m]       a main semiaxis
da      [m/day]   a change in time
e       [-]       e eccentricity
de      [-/day]   e change in time
i       [rad]     i inclination
di      [rad/day] i change in time
O       [rad]     O (node n) position of inclination axis
dO      [rad/day] O node shift (pi2/T)
o       [rad]     o perihelium (no change in inclination position)
do      [rad/day] o perihelium shift (pi2/T)
M       [rad]     M rotation around owner position in t0
dM      [rad/day] dM orbital rotation (pi2/draconic month)
ddM0        [rad/day^2] dM change in time
rota0       [rad]     rota0 rotation around self axis position in t0
rotda       [rad/day] rotda mean rotation around self axis
prea0       [rad]     prea rotation axis position in t0
preda       [rad/day] preda precession rotation (pi2/Platonic year)
prei        [rad]     prei equator inclination to ecliptic
predi       [rad/day] prei change in time
nuta        [rad]     nuta angle position on nutation ellipse
nutda       [rad/day] nutation rotation (pi2/T)
nutia       [rad]     nutia nutation (of rotation axis) ellipse semiaxis  axis in ecliptic plane
nutdia      [rad/day] nutia change in time
nutii       [rad]     nutii nutation (of rotation axis) ellipse semiaxis  axis in rotation axis direction
nutdii      [rad/day] nutii change in time

Ignore the is_star, textures, rings and atmosphere parameters. So:

  1. get Sun to position (0,0,0) in cartesian coordinates
  2. compute Earth position (x,y,z) from Kepler's law

Sun is then (-x,-y,-z) in geocentric coordinates

  1. rotate back by dayly rotation,precession,nutation (-x,-y,-z) -> (x',y',z')

  2. compute NEH frame for your geolocation (North,East,High(Up))

  3. convert (x',y',z') to NEH local coordinates (xx,yy,zz)

  4. compute:

     azimut=atanxy(-xx,-yy)
     height=atanxy(sqrt((xx*xx)+(yy*yy)),-zz)
    

and that is it

here is mine Heliocentric astro body position computation:

void astro_body::compute(double t)
    {
      // t is time in days after 1.1.2000 00:00:00
      // double pos[3] is output heliocentric position [m]
      // reper rep is output heliocentric position [m] and orientation transform matrix (mine class) where Z is rotation axis (North pole) and X is long=0,lat=0

    rot_a.compute(t); // compute actual value for orbital parameters changing in time
    pre_a.compute(t); // the actual parameter is in XXX.a you can ignore this part
    pre_i.compute(t);
    nut_a.compute(t);
    nut_ia.compute(t);
    nut_ii.compute(t);

//  pre_a=pre_a0+(pre_da.a*dt)+(nut_ia*cos(nut_a)); // some old legacy dead code
//  pre_i=pre_i0+(pre_di.a*sin(pre_e))+(nut_ii*sin(nut_a));

    rep.reset(); // rep is the transform matrix representing body coordinate system (orientation and position)
    rep.lrotz(pre_a.a); // local rotation around reps Z axis by pre_a.a [rad] angle
    rep.lroty(pre_i.a);
    rep.lrotx(nut_ia.a*cos(nut_a.a));
    rep.lroty(nut_ii.a*sin(nut_a.a));
    rep.lrotz(rot_a.a);

    a.compute(t); // the same as above can ignore this part
    e.compute(t);
    i.compute(t);
    O.compute(t);
    o.compute(t);
    M.compute(t);
    M.compute(t);

    double  c0,c1,c2,sO,si,cO,ci,b;       // trajectory constants
    double  x,y;
    int     q;

    if (e.a>=1.0) e.a=0;
    c0=sqrt((1.0-e.a)/(1.0+e.a));       // some helper constants computation
    c1=sqrt((1.0+e.a)/(1.0-e.a));
    c2=a.a*(1-e.a*e.a);
    sO=sin(O.a);
    cO=cos(O.a);
    si=sin(-i.a);
    ci=cos(-i.a);
    b=a.a*sqrt(1.0-e.a);

    M.a-=o.a;                           // correction
    M.a=M.a-pi2*floor(M.a/pi2);
    E=M.a;
    for (q=0;q<20;q++) E=M.a+e.a*sin(E); // Kepler's equation
    V=2.0*atan(c1*tan(E/2.0));
    r=c2/(1.0+e.a*cos(V));
    pos[0]=r*cos(V+o.a-O.a);  // now just compute heliocentric position along ecliptic ellipse
    pos[1]=r*sin(V+o.a-O.a);  // and then rotate by inclination
    pos[2]=-pos[1]*si;
    pos[1]=+pos[1]*ci;
    x=pos[0]; y=pos[1];
    pos[0]=x*cO-y*sO;
    pos[1]=x*sO+y*cO;

    if ((mother>=0)&&(tab!=NULL)) vector_add(pos,pos,tab[mother].pos); // if satelite like Moon add owners position
    rep.gpos_set(pos); // set the global position to transform matrix also
    }
//---------------------------------------------------------------------------

reper class is quite complicated (something like GLM) the only thing you need from it are local rotations (all other stuff is basic). this is how lrotx works:

double c=cos(ang),s=sin(ang);
double rot[16],inv[16]; // rot is the rotation around x transform matrix
rot=(1, 0, 0, 0,
      0, c,-s, 0,
      0, s, c, 0,
      0, 0, 0, 1);
inv=inverse(rep); // inverse is inverse matrix 4x4
inv=inv*rot
rep=inverse(inv);
  • rep is the input and output matrix
  • ang is the rotation angle [rad]

now you can use rep to convert to/from Earth local coordinate system

  • LCS to GCS (l2g) ... (gx,gy,gz)=rep*(lx,ly,lz)
  • GCS to LCS (g2l) ... (lx,ly,lz)=inverse(rep)*(gx,gy,gz)

local is Earth's coordinate system and global Sun's coordinate system

Kidron answered 8/9, 2014 at 10:37 Comment(2)
i think the presov URL changed to astropresov.sk/astro-info/v-cislach/efemeridyOsteology
@Osteology thx for the update ... I do not use online effemerids ... I got my own C++ one using this approach (working for 20+ years now) with just minor error which is still not a big deal for me as I use this to look when is a good time to get my telescope to action and where ...I updated the link thx againKidron

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