How can I calculate the elliptical shape of the sun at sunset. Is their a formula?
104230
This shape is called the "omega shape", at least as the lower limb of the sun disappears beneath the apparent horizon (the sun has already set quite a bit farther beneath the horizon and one is only seeing so much of the sun because of refraction effects). It is called an omega because it is not strictly an oval or ellipse at that point, and actually has a pair of "feet" at the bottom like an omega. \n\n The actual shape will vary, depending on refraction effects, and those depend on the temperature profile of the atmospheric layers. The shape is typically computed in a simulation by ray-tracing, although one might be able to do some sort of approximation by an ellipse. See the attached link for more details.
320123
Depends on the time of year.\nSummer solstice = 0\nWinter solstice = 24\nSpring/fall equinox = 12\n\nAll other days, some gradual change from 0 to 24. This is true of any land within the Arctic or Antarctic circles (of course the times are reversed for the Arctic circle). Those circles are defined by the places on Earth that have at least one day of complete light and complete darkness during the year.
950528
Using the chronometer as a starting point, you figure out when noon is supposed to be.  \n\nYou use the Sextant to measure the Declination (it has heavy, heavy dark glass for this) and monitor the sun until it hits the highest point.   It actually appears to 'hover' there for a few seconds. Then you compare it to the chronometer.  The distance it is off of from Greenwich time gives you the longitude.
1056603
http://www.wsanford.com/~wsanford/exo/sundials/horizontal_sundial_setup.html\n\nhere's how\n\nA horizontal sundial consists of the dial plate, marked off in hours, and the gnomon (including the style) that sits on the meridian (noon hour line) and sticks up from the dial plate at a right angle relative to the plane of the dial face. Place the sundial on a horizontal surface (hence the name, "horizontal sundial") and align the gnomon with true north (on the dial face, 12 noon should point north). \nFinding the Direction of True North \n\nUse a compass (compensate for magnetic declination). \nAlign with Polaris, the North Star.
624444
Is there a way to quantify/measure a value for solar radiation with respect to time, it seems to me like you could come up with a calculation based on the angle of the rays. However, I think you would run into some problems with exact calculations due to the nature of our atmosphere (i.e. moisture, dust, etc.)
958764
Ok so we now the location and the percentage of sunlight that it is receiving, like the north pole is getting 100% sunlight which equals 24 hours of the day where as the equator is only receiving 50% sunlight equaling 12 hours of the day.\n\nSince we know that it is a ratio, we can say NYC receives 60% sunlight during 24 hours.  To find the number of hours we would multiply 24 hours by 60%   (24 x 0.6 = 14.4)   to get 14.4 hours.\n\nTo convert hours to minutes multiply by 60.  Like 14.4 hours times 60 minutes per hour gives us 864 minutes.\n\nHope this helps.
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
970854
Use Kepler's law, which states that the orbital period squared is proportional to the cube of the orbital radius.  See http://en.wikipedia.org/wiki/Kepler%27s_law
509207
Yes. You can use sine or cosine to determine the angle.\n\nSine x = Opposite / Hypotonuse\nSin x=5/13\nSin x=.38\nx=22.6*\n\nCosine x = Adjacent / Hypotonuse\nCos x=12/13\nCos x=.92\nx=22.6*
817446
The ecliptic is defined as the path thourgh the sky where the sun appears over the course of the year.  Thus, no matter when you see the sun it is on the ecliptic.  On the equinoxes the ecliptic and the equator of the earth line up, and on those two days you can find the sun on the equator, but the sun is always on the ecliptic.
1071951
NASA/JPL lists a detailed ephemeris for this object with celestial coordinates ==> http://neo.jpl.nasa.gov/cgi-bin/db_shm?sstr=2004%20XP14
275059
yes, search yahoo.\nHere's a good one\nhttp://www.solarviews.com/cap/sun/sundiag.htm i think...
40284
(1) Subtitute the damn values for what you have.\n\n(2) Sin theta=opp/hyps,\n    Tan   "  =opp/adj,\n    Cos   "  =adj/hyps.\n\nNow usually a blind man follows his stick now you might know where you are at the moment - get it???
1101303
That all depends on where you are located.\n\nTo generate a calendar with sunrise/sunset times for your area, go to the following website:\n\nhttp://www.sunrisesunset.com/custom_srss_calendar.asp
510653
There are three times a year where Solar time and local time are the same. January, July and then January (ok twice). Determine your position of latitude. Place a short rod in the ground, pointing north at the same angle as your latitude. (pointing to the north star) Make sure the position you pick can be seen from the South by the Sun. Mark the hours off by some smaller rods avery hour of the day. If you do this in January or July you will be closer to real or local time. The shadow of the longer rod, the gnomon, will point to the time rods you have place to mark the shadow. A sundial can also be placed on a vertical wall, with the face to the south, and the "gnomon" point down at the same angle of your latitude.
319592
The formula is to simply divide the mass of the sun by the mass of whatever planet you are interested in.\n\nThe sun's mass is: 1.989 x 10^30 kg\n\nThe mass of the planets:\nMercury: 0.33 x 10^24 kg\nVenus:   4.87 x 10^24 kg\nEarth:   5.98 x 10^24 kg\nMars:    0.65 x 10^24 kg\nJupiter: 1900 x 10^24 kg\nSaturn:   570 x 10^24 kg\nUranus     87 x 10^24 kg\nNeptune:  100 x 10^24 kg\nPluto:    0.7 x 10^24 kg\n\nSo, for instance, 332,609 earths would weigh the same as the sun.
521925
You find it in the normal coordinates way; i.e.\n((x1+x2)/2, (y1+y2)/2)\n\nThe curvature of our Earth does not affect it as the coordinates we are using is 2-dimensional.