User FSimardGIS - Astronomy Stack Exchange most recent 30 from astronomy.stackexchange.com 2020-01-20T23:39:30Z https://astronomy.stackexchange.com/feeds/user/19814 https://creativecommons.org/licenses/by-sa/4.0/rdf https://astronomy.stackexchange.com/questions/33719/why-do-j2000-coordinates-change-with-time-in-stellarium/33726#33726 4 Answer by FSimardGIS for Why do J2000 coordinates change with time in Stellarium? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-10-20T02:36:44Z 2019-10-20T02:36:44Z <p>Apparently, Stellarium's "J2000.0" reports the coordinates of stars in J2000 frame, but at the epoch of the date you specify, instead of reporting the coordinates of the star at epoch 2000.0, which can be misleading to say the least. So the difference in J2000 coordinates that you noticed corresponds to the proper motion of the star. And the star you picked, 61 Cygni, actually has one of the largest proper motion in all of the catalogued stars, with about 4.1"/year in RA and 3.2"/year in declination. Looking at the differences that Stellarium reports in your example, we can notice that it matches perfectly with this proper motion : <code>31.3" - 28.2" = 3.1" ± 0.1"</code> in declination, for RA: <code>(55.27s - 54.92s) x 15 x cos(declination) = 4.1" ± 0.1"</code>. </p> <p>If you pick a star with a smaller proper motion in Stellarium, like Deneb (only 2 mas/year for each axis), the J2000 coordinates change much more slowly.</p> <p>As for the difference between J2000 RA/Dec and apparent (on date) RA/Dec, this is caused by the fact that the J2000 frame's equator/equinox was fixed at the mean position of the equator/equinox in 2000 (<em>mean</em> position means that the small periodic effects of <a href="https://en.wikipedia.org/wiki/Astronomical_nutation" rel="nofollow noreferrer">nutation</a> are averaged out). So the apparent coordinates on Jan 1 2000 are slightly different because apparent coordinates take into account nutation <em>N</em> of the rotational axis <em>R</em> around the mean <em>P</em> (precession): </p> <p><a href="https://i.stack.imgur.com/RBS6C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RBS6C.png" alt="enter image description here"></a></p> https://astronomy.stackexchange.com/questions/33595/-/33600#33600 5 Answer by FSimardGIS for A vertical stick's shadow at solar noon should be straight north/south, right? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-10-06T02:41:18Z 2019-10-06T02:41:18Z <p>Your understanding is correct, the shadow of a vertical stick on a level plane always points north or south at solar noon, except in the Tropics where sometimes it will not have any shadow due to the Sun being at the zenith, and at the Poles, where solar noon is undefined. The sun can also be below the horizon at solar noon: places inside a polar circle can experience polar night, when the Sun does not rise for at least an entire day. On those days, even at solar noon, the Sun remains below the horizon.</p> <p>This erroneous passage seems to originate from <a href="https://en.wikipedia.org/w/index.php?diff=487185695&amp;oldid=475325407&amp;title=Sun_path" rel="noreferrer">edits on 13 April 2012</a>. The original text made more sense:</p> <blockquote> <p>On the Equator, the sun will be straight overhead and a vertical stick will cast no shadow at noon (solar time) on March 21 and September 23, the equinox. 23.5 degrees north of the equator on the Tropic of Cancer, a vertical stick will cast no shadow on June 21, the summer solstice for the northern hemisphere. The rest of the year, the noon shadow will point to the North pole. 23.5 degrees south of the equator on the Tropic of Capricorn, a vertical stick will cast no shadow on December 21, the summer solstice for the southern hemisphere, and the rest of the year its noon shadow will point to the South pole. North of the Tropic of Cancer, the noon shadow will always point north, and conversely, south of the Tropic of Capricorn, the noon shadow will always point south.</p> </blockquote> https://astronomy.stackexchange.com/questions/29986/-/30001#30001 32 Answer by FSimardGIS for Official degrees of earth’s rotation per day FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-03-15T05:46:37Z 2019-07-16T01:55:36Z <p>First, we need to decide which definition of "day" to employ. There are several types of days:</p> <p><strong>Apparent solar day</strong>: the time between two successive culminations of the Sun (apparent Noon) from an fixed Earth-based observer;<br> <strong>Mean solar day</strong>: a more uniform, averaged solar day without seasonal variations;<br> <strong>Stellar/Sidereal day</strong>: the time needed for the Earth to rotate once relative to the stars;<br> <strong>SI day</strong>: a unit of time containing exactly 86,400 SI seconds defined by caesium atoms.</p> <p>Since we generally refer to the traditional day/night cycle when we say "day", this means a form of solar time. To get a more averaged value, let's use the mean solar day.</p> <p>The current formula linking the Earth Rotation Angle (ERA) to the modern approximation of mean solar time, UT1 (basically the Earth's clock following the mean day/night cycle), is by definition :</p> <p><span class="math-container">$$ERA = 2π(0.7790572732640 + 1.00273781191135448 T_u) \text{ radians}$$</span></p> <p>Where <span class="math-container">$Tu$</span> is the Julian UT1 Date - 2451545.0</p> <p>So according to this formula, a (UT1) day is 1.00273781191135448 Earth rotations, which multiplied by 360° is about <strong>360.98561°</strong>. However, the Earth's rotation and revolution are not constant, and are always changing at somewhat unpredictable rates, so the angle is not perfect, but the changes are very slow. So this is more a modern approximation rather than an exact value. Rounded to one decimal place, this gives you <strong>361.0°</strong>, a figure that will likely remain true for at least several millenia. </p> <p>If you want to know the amount of Earth rotation for every SI day, you're in luck: it is possible to consult reports of the Earth's orientation (rotation and polar motion) thanks to the IERS. Values are tabulated for each 0h UTC every day in the IERS publications, allowing to derive the angle that Earth has rotated every 86,400 SI seconds, allowing to scientifically monitor variations in Earth's rotation compared to a very constant unit of time realized by atomic clocks. Nowadays, the Earth's rotation is measured and reported thanks to radio telescopes and <a href="https://en.wikipedia.org/wiki/Very-long-baseline_interferometry" rel="nofollow noreferrer">VLBI</a> observing distant objects in the universe. Official reports of its orientation, the <a href="https://www.iers.org/IERS/EN/Science/EarthRotation/EOP.html" rel="nofollow noreferrer">Earth Orientation Parameters</a> are published on the <a href="https://www.iers.org/IERS/EN/DataProducts/EarthOrientationData/eop.html" rel="nofollow noreferrer">IERS Website</a>. </p> <p>As an example, here is a graph showing the amount of Earth rotation every 86,400 seconds constructed with IERS data of the last year:</p> <p><a href="https://i.stack.imgur.com/ICTX1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ICTX1.png" alt="Earth Rotation"></a> As we can see, there are several seasonal, periodic and unpredictable variations in Earth's rotation.</p> https://astronomy.stackexchange.com/questions/29759/-/29760#29760 3 Answer by FSimardGIS for Determine Julian Date from Gregorian without formula FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-02-26T04:48:17Z 2019-02-26T22:22:47Z <p>The <a href="https://en.wikipedia.org/wiki/Year_zero" rel="nofollow noreferrer">Year 0</a> does not exist in BCE/CE and BC/AD notation. It jumps from 1 BCE to 1 CE directly. So if we look at the sequence of years in BCE/CE notation, we have:</p> <pre><code>3 BCE 2 BCE 1 BCE 1 CE 2 CE 3 CE </code></pre> <p>If you want to calculate the number of years between, say, January 1, 2 BCE and January 1, 2 CE, you can add the year numbers to get 4, but then you need to <em>subtract</em> 1 to remove year 0 from the total, giving a total of 3 years. In your particular case, this leads to <span class="math-container">$4713 - 1 + 2010 = 6722$</span> years.</p> <p>In astronomy, we often use another year numbering system called the <a href="https://en.wikipedia.org/wiki/Astronomical_year_numbering" rel="nofollow noreferrer">Astronomical year numbering</a>, which includes a year 0 and makes things more convenient arithmetically when calculating time intervals between eras. It goes (-3, -2, -1, 0, 1, 2, etc.) To convert years BCE to the Astronomical year notation, we subtract 1 and add minus(-) in front of the year. For example, the year 4713 BCE is -4712 in Astronomical year notation. Your year interval is then calculated as <span class="math-container">$2010 - (\text{-}4712) = 2010 + 4712 = 6722$</span>. </p> <p>For the leap years calculation, since -4712 is a leap year, this simplifies matters, but don't forget to take the smallest integer greater than or equal to your result (the ceiling function). So in this case <span class="math-container">$\frac{6722}{4} - 3 = 1677.5$</span>, and the ceiling is 1678, and that is the number of leap years between your years. Also, in your initial calculation, you had forgotten to subtract 3, because 6724 / 4 - 3 = 1678, not 1681. </p> <p>Finally, the time interval between January 1 and February 28 is 58 days, not 59, because we do not count the last date as part of the interval. This gives a total of <span class="math-container">$5044 * 365 + 1678 * 366 - 10 + 58 + \frac{6 * 60 * 60 + 30 * 60}{24 * 60 * 60} = 2455256.27$</span>. </p> <p>You state that the actual answer is 2455255.77, but in fact the answer really is 2455256.27. Perhaps you typed "06:30" instead of "18:30" in the converter. </p> https://astronomy.stackexchange.com/questions/29569/-/29586#29586 2 Answer by FSimardGIS for How close to circular is the Earth's equator FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-02-13T22:29:07Z 2019-02-13T23:31:57Z <p>My answer won't be complete, from lack of time and resources, but I still wanted to share some interesting aspects here that could be helpful.</p> <p>The difficulty in aswering this question revolves around the complex and irregular shapes involved here. Also, finding a "best-fit" ellipse for comparison is not as easy as it seems, because it depends on what and how you want to model. The actual shape of the equator is rather complex and irregular. Generally speaking, it is pretty much circular, but indeed topography and the geoid complicate matters. The Earth's movement in the solar system is not a perfect ellipse either, because of the gravitational interactions with other celestial bodies. Let's review the different irregularities involved here. </p> <p>The Earth is usually modeled as an ellipsoid of revolution (oblate spheroid), and the equator as a circle. A good example of that is the WGS 1984 geodetic reference system used by the Global Satellite Navigation System. Of course, the equator is not a perfect circle, it has irregularities mainly because of topography, and even sea level itself is a little irregular too. We can approximate sea level with a <a href="https://en.wikipedia.org/wiki/Geoid" rel="nofollow noreferrer">geoid</a>, for example, here is a map of the EGM2008, a geoid used with WGS 1984 to transform ellipsoidal heights to geoid heights:</p> <p><a href="https://i.stack.imgur.com/vflxU.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vflxU.gif" alt="enter image description here"></a></p> <p>Basically, this map shows the height of the geoid (the idealized sea level without the effects of tides and currents) with respect to the WGS84 reference ellipsoid of revolution (semi-major axis 6,378,137 m, semi-minor axis 6,356,752.314 m). The differences are mostly less than 100 meters, and are caused by the irregular distribution of mass inside the Earth itself. </p> <p>Now, some studies show that the Earth's shape could be slightly better modeled by a triaxial ellipsoid, and one could try to model the equator as an ellipse, and the Earth as a triaxial ellipsoid, however, even with a best-fit triaxial, we would still need geoid corrections for the irregular mean sea level, let alone topography, and geodetic computations would be more complex on a triaxial. Other funny models and names have come up over time, like a pear-shaped (because of a slight bulge in mid southern latitudes) model as a best-fit shape. But if you look at the map above, good luck visually finding the pear shape in there, or other mathematically modelizable aspects of these bumps. We are talking about very subtle differences here, that do not necessarily need to be taken into account for most purposes when describing the general shape of the Earth. </p> <p>So depending on how you consider the shape of the equator (i.e. by topography and ocean floor, or sea level) you will arrive at a shape that is mostly circular with irregular bumps along the way. There is no authoritative agreement that I know of about an <em>eccentricity</em> of the equator. For instance, <a href="https://www.arcjournals.org/pdfs/ijms/v1-i1/4.pdf" rel="nofollow noreferrer">This study</a> proposes a flattening of about 70 meters for the equator's ellipse, <a href="https://www.britannica.com/science/geoid" rel="nofollow noreferrer">This article</a> on Encyclopaedia Britannica proposes 80 meters. </p> <p>For the Earth's orbit, for the sake of this comparison, we can use an best-fit ellipse of 149.598 million km by 149.577 million km. Of course, that is only a idealized ellipse, the real movement of the Earth in the Solar System is more complex.</p> <p>Finally, say we scale down and superimpose the Earth's orbit's ellipse on the equator to compare. The eccentricity of the Earth's orbit is 0.0167, the semi-major axis is 149.598 million km and semi-minor axis is 149.577 million km. Scaled down by a factor of 23,455 to the equator's size, this corresponds to a difference of about 900 meters in both orbit axes. So I think we can agree that a best-fit ellipse of sea level along the equator is more circular than the Earth's orbit. </p> <p>However, topography-wise, there are bumps up to over 4,000 meters in the Andes, and the sea floor reaches 5000 meters below sea level in several places. So the "topographic surface" equator would, for one thing, definitely appear more bumpy than the Earth's orbit. Unfortunately, I haven't found a example or study showing what a best-fit ellipse of the equator (including topography and bathymetry) could look like, mainly because we tend to approximate sea level, not topography itself, but with more time, tools and data, it might be possible to work out the answer with a "topographic best-fit" ellipse. </p> https://astronomy.stackexchange.com/questions/29271/-/29275#29275 10 Answer by FSimardGIS for When is Earth closest to the Sun? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-22T03:42:17Z 2019-01-22T03:58:35Z <p>Earth's perihelion timings and distances are based on the distance between the center of the Earth and the center of the Sun. They are effectively the times and distances of the closest approach between the two bodies, not the barycenters. As we can see on Fred Espenak's <a href="http://www.astropixels.com/ephemeris/perap2001.html" rel="noreferrer">astropixel page</a> about perihelion and aphelion, the time intervals between two successive perihelions tend to vary between 363 and 368 days. </p> <p>The data on astropixels seems perfectly valid. With JPL Horizons, for example, the next perihelion is calculated to be on 2020 Jan 5 at 07:47:42, at a distance of 0.98324356482 AU (147,091,144 km) which matches astropixel's values. </p> <p>I analyzed the perihelions of the 21st century to see how much the actual timing can differ from an idealized "average" perihelion moment (based on the anomalistic year length). Perihelions this century remain within 1.4 days of that average. The differences from the average are mainly due to the gravitational effect of the Moon.</p> <p>Using all perihelion times in the 21st century, and the average perihelion interval (anomalistic year, 365.2596 days) I constructed the following graph showing these differences:</p> <p><a href="https://i.stack.imgur.com/f3tkt.png" rel="noreferrer"><img src="https://i.stack.imgur.com/f3tkt.png" alt="enter image description here"></a></p> https://astronomy.stackexchange.com/questions/29259/jupiter-venus-conjunction-is-vertically-aligned-im-at-49-n-how-is-this-possi/29272#29272 5 Answer by FSimardGIS for Jupiter-Venus conjunction is vertically aligned. I'm at 49° N. How is this possible? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-21T22:21:33Z 2019-01-21T22:21:33Z <p>A combination of several movements can lead two objects in the sky to appear vertically aligned. The planets are constantly moving relative to each other, they are orbiting on slightly different planes, the Earth is orbiting and rotating, so such configurations between planets can occur frequently. </p> <p>An observer does not need to be in the tropics to see two planets being vertically aligned at some point. The planets orbit close to the ecliptic, but their orbital planes are slightly inclined relative to ours. Jupiter's orbit is about 1° from the ecliptic, and Venus is about 3°. This means that both planets can move above or below the ecliptic by a few degrees. When they pass each other as seen from the Earth in a conjunction, they are rarely on top of each other, most of the time, they are separated by a few degrees, and they are not exactly on the ecliptic line. (Such occurences of planets occulting or transiting each other do happen, but are rather uncommon: the next time Venus will transit in front of Jupiter will be on 22 November 2065)</p> <p>I'll add a few pictures to help visualizing the positions, movements and alignments involved here. </p> <p>Here is a view of the sky on 20 January at 49° North latitude:</p> <p><a href="https://i.stack.imgur.com/IsYzf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/IsYzf.png" alt="enter image description here"></a></p> <p>As you can see, Jupiter and Venus appear pretty much vertically aligned in the morning sky. The yellow line is the ecliptic. Note that the ecliptic does not pass through the planets here. Jupiter is pretty close, but Venus is a few degrees north of it. As both planets pass each other in the next days, at some point, their alignment will be perpendicular to the ecliptic (called the ecliptic conjunction), as we can see from this January 22 simulated image:</p> <p><a href="https://i.stack.imgur.com/Wjo3O.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Wjo3O.png" alt="enter image description here"></a></p> <p>But they won't appear vertical anymore from your vantage point. (In fact, they would apprear vertical again later during that day, because of Earth's rotation, but the Sun will be up, so it wouldn't be observable directly)</p> <p>A few days later, on January 26, they will appear horizontally aligned!</p> <p><a href="https://i.stack.imgur.com/aD4h4.png" rel="noreferrer"><img src="https://i.stack.imgur.com/aD4h4.png" alt="enter image description here"></a></p> <p>(Images generated in <a href="https://stellarium.org/" rel="noreferrer">Stellarium</a>)</p> https://astronomy.stackexchange.com/questions/18474/-/27872#27872 2 Answer by FSimardGIS for Closest path of solar eclipse central line FSimardGIS https://astronomy.stackexchange.com/users/19814 2018-10-03T03:53:45Z 2018-10-03T19:06:06Z <p>Your methodology seems to be based on an algorithm that uses successive approximations to obtain the desired values. I was not able to arrive quite at the same values as you mentioned, but I noticed an error in the posted script, <code>zeta = sqrt(ro^2 + x^2 + y^2)</code> clearly should be <code>zeta = sqrt(ro^2 - x^2 - y^2)</code>, so perhaps slightly different values or operations were used somewhere else too, but I couldn't find out where. I also noticed that you did not take into account the difference between Ephemeris Longitude and True Longitude in your calculation, True Longitude shifts towards the East by <code>1.002738 * At (Delta t) / 240</code> degrees. Moreover, executing more iterations could possibly converge to a better result.</p> <p>However, to compute the intersection of the Moon's shadow axis from Besselian Elements, I suggest using the following closed-form solution, which is derived from solving the intersection of a parametric line (Moon's shadow axis) and the Earth ellipsoid.</p> <pre><code>f = 1 / 298.257223563 b = 1 - f sin_d = Sin(d * PI / 180) cos_d = Cos(d * PI / 180) y0_r = y * cos_d z0_r = -y * sin_d y1_r = y0_r + sin_d z1_r = z0_r + cos_d v = y1_r - y0_r w = z1_r - z0_r a_quadratic = v ^ 2 + b ^ 2 * w ^ 2 b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1) p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic) y_i = y0_r + v * p z_i = z0_r + w * p dist_axis = Sqrt(x ^ 2 + z_i ^ 2) lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240 </code></pre> <p>Here is the explanation of the formulas:</p> <p>We use the flattening value to define the WGS84 ellipsoid, and calculate the semiminor-axis:</p> <pre><code>f = 1 / 298.257223563 b = 1 - f </code></pre> <p>Convert declination angle d to radians and execute sine and cosine functions:</p> <pre><code>sin_d = Sin(d * PI / 180) cos_d = Cos(d * PI / 180) </code></pre> <p>Select 2 points (points 0 and 1 in the code) on the shadow axis, at z = 0 and z = 1, and apply the declination rotation (_r) on them:</p> <pre><code>y0_r = y * cos_d z0_r = -y * sin_d y1_r = y0_r + sin_d z1_r = z0_r + cos_d </code></pre> <p>The x value always remains the same, so we don't need to recalculate it. Next, evaluate the parametric line's vectors v,w (delta y, z) from point 0 as a base:</p> <pre><code>v = y1_r - y0_r w = z1_r - z0_r </code></pre> <p>By substituting the parametric line's equations in the ellipsoid equation, and rearranging to quadratic form and solve:</p> <pre><code>a_quadratic = v ^ 2 + b ^ 2 * w ^ 2 b_quadratic = 2 * z0_r * w * b ^ 2 + 2 * y0_r * v c_quadratic = z0_r ^ 2 * b ^ 2 + y0_r ^ 2 + b ^ 2 * (x ^ 2 - 1) p = (-b_quadratic + Sqrt(b_quadratic ^ 2 - 4 * a_quadratic * c_quadratic)) / (2 * a_quadratic) </code></pre> <p>'p' here represents the distance from point 0 along the parametric line, expressed in Earth's equatorial radius. The contents of the square root must ideally be tested beforehand for negative values to avoid exceptions if the line does not intersect the ellipsoid. With this, we find the coordinates of our points of intersection:</p> <pre><code>y_i = y0_r + v * p z_i = z0_r + w * p </code></pre> <p>Then we convert this to geographic latitude and longitude:</p> <pre><code>dist_axis = Sqrt(x ^ 2 + z_i ^ 2) lat = Atan(y_i / (dist_axis * (1 - f) ^ 2)) * 180 / PI lon = Atan2(x, z_i) * 180 / PI - u + 1.002738 * At / 240 </code></pre> <p>Direct conversion for the Latitude is possible here because our point is on the surface of the ellipsoid.</p> <p>At the end, you can use the following logic to capture out-of-bounds longitude values:</p> <pre><code>If lon &lt;= -180 Then lon += 360 ElseIf lon &gt; 180 Then lon -= 360 End If </code></pre> <p>In terms of accuracy, you can expect pretty good results with this method, but it does not take into account some factors, like local topography, geoid height, refraction, etc. Topography can lead to differences of a few kilometers, the geoid, +-100 meters, and refraction can bend the shadow by a few meters to a few hundred meters depending on the incident angle. Also, there is no correction for the moon's center of figure, that can shift the umbral center by several hundred meters as well. </p> <p>With the above formulas for UT1 = 18h and delta t = 72.3s, the computed intersection point is at 58.26828°N, 21.57331°W, which matches F. Espenak's values. </p> https://astronomy.stackexchange.com/questions/27601/-/27616#27616 3 Answer by FSimardGIS for What would stargazing be like at the edge of our galaxy? FSimardGIS https://astronomy.stackexchange.com/users/19814 2018-09-11T00:03:13Z 2018-09-11T00:03:13Z <p>Here is a render of what an observer might see near the edge of our galaxy. The "edge" itself being rather fuzzy, let's assume that we are in an area of the edge dense enough where a few occasional stars are still visible. (As far as we know, intergalactic space contains objects and stars, but so few that seeing even one naked-eye star in the sky there would be uncommon)</p> <p>Looking at the sky towards the opposite side of the Milky Way, one could see the Magellanic Clouds, and a small quantity of naked-eye stars:</p> <p><a href="https://i.stack.imgur.com/gbiXU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gbiXU.jpg" alt="enter image description here"></a></p> <p>The Milky Way side of the sky would be interesting to look at:</p> <p><a href="https://i.stack.imgur.com/KnFjf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KnFjf.jpg" alt="enter image description here"></a></p> <p>Now, as interesting as these artistic renditions may be, there is very little that we know about actual objects in these regions. Seeing individual stars there is quasi-impossible from here; we have a general idea of how dense some regions of the galaxy are, but for the rendition part here, it is mostly procedural based on some general facts and statistics. </p> <p>The software used for these renders is called Space Engine and contains all catalogued objects, and it procedurally generates objects to fill in the gaps based on what we know about space. </p> https://astronomy.stackexchange.com/questions/26033/-/26049#26049 5 Answer by FSimardGIS for Does Universal Time really track mean solar time? FSimardGIS https://astronomy.stackexchange.com/users/19814 2018-04-25T16:53:38Z 2018-04-29T22:27:41Z <p>The rotation of the Earth, UT1 and the Earth Rotation Angle are strictly measured agaist extragalactic light sources. Indeed, nowadays, UT1 is reckoned by using the definition you mentioned, a direct linear relationship with the Earth Rotation Angle between the intermediate celestial and terrestrial frames instead of the actual ficticious mean Sun, because of the impracticality of such a measurement in the context of modern accuracies that are needed by these frames. </p> <p>Here is a good reference to read about this complex topic:</p> <p>U. S. Naval Observatory, Precision time and the rotation of the Earth, Dennis D. McCarthy, IAU 2004, <a href="https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921305001377" rel="nofollow noreferrer">https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S1743921305001377</a></p> <p>EDIT - </p> <p>Here is a different approach that may also help in the understanding of this topic.</p> <p>The important thing to keep in mind is that nowadays, with the increasing accuracy needed by many modern applications, we needed reference frames and relationships between them that are as consistent and accurate as possible. For this we reference ourselves on the most stable and easily measurable objects.</p> <p>The problem with the Sun was that its location or mean motion cannot feasably be tracked with an accuracy of a few tens of microarcseconds, like quasars. Adopting a definition based on this would have introduced inaccuracies in the definition itself and complexified the relationships, thus undermining the accuracy of the whole system, because UT1-UTC is one of the links between the terrestrial and celestial frames. So the IAU decided to adopt a definition that is "disconnected" from mean solar time, while keeping it reasonably close to the mean Sun by using the adopted ratio. The ratio will be good for decades for civil purposes, and later refinements to keep it reasonably close to the mean Sun while maintaining continuity will be much easier to make anyway. </p> <p>Another way of imagining the problem is this: say you lived on a boat (Earth) and you needed very accurate measurements of objects around you on the boat and on the sea, say fish, whales, etc (satellites and planets). You see several things around you: the lights of other boats, the lights on a distant shore. Would you prefer to align yourself on another boat's blinding and dancing light (Sun), or on a much more stable and smaller light on the distant shore (quasars), to determine your boat's orientation? Using another moving boat as a reference for coordinate systems would complicate matters and introduce measurement errors while you try to position other objects relative to it. Using the distant shore, however, would make your reference much more stable and accurate, and that does not prevent you from tracking the approximate location of the nearby boat's light. </p> <p>To sum up, we now have two main reference systems: ITRS (Earth's crust) and ICRS (distant stars), and a link between them (ERA and polar motion) that is as accurate as possible. We keep UT1 close enough to the Sun for all civil purposes, while defining it with a very simple and direct formula related to ERA for accuracy, simplicity and continuity purposes.</p> https://astronomy.stackexchange.com/questions/22978/-/23662#23662 2 Answer by FSimardGIS for Eclipses with SkyField FSimardGIS https://astronomy.stackexchange.com/users/19814 2017-11-24T05:29:57Z 2017-11-25T13:09:44Z <p>I believe the differences here are caused by the different timescales that are being used by SkyField vs NASA. </p> <p>It seems that SkyField uses the proleptic Gregorian calendar for dates in the past. However, NASA used the Julian Calendar for dates before 1582, so for example, the eclipse on 0150-12-06 (Julian) falls on 0150-12-05 in Gregorian. Also, I would use <code>ts.ut1</code> instead of <code>ts.utc</code>, as ut1 is the timescale used by NASA. </p> <p>Using the geocenter can be a efficient way to check for eclipses worldwide. When observing from the geocenter, the penumbral shadow cone touches the Earth when the following condition is true: $$\mu ≤ arcsin(\frac{r_m + r_e}{d_m}) + arcsin(\frac{R_s - r_e}{D_s})$$ Where $\mu$ is the angular separation between the Sun and the Moon, $D_s$ is the distance of the Sun, $d_m$ the distance of the Moon, and the three constant radii are:</p> <p>Sun: $R_s = 695700 km$</p> <p>Earth: $r_e = 6378 km$</p> <p>Moon: $r_m = 1737.4 km$</p> <p>This indicates that an eclipse, at least partial, is occuring somewhere on the Earth. The formula uses a spherical Earth as an approximation of its shape, so in some rare cases it can detect very near-misses by 20-25 km, but overall it is a good way to narrow down the search. Also, it is unnecessary to test this when the angular separation is outside the range 1.3° - 1.7°, since values below 1.3° always produce an eclipse, and values above 1.7° never produce one.</p> <p>I would also like to point out that uncertainties, mostly in Earth's rotation speed, make it very difficult to know where eclipses occured on the surface of the Earth for dates in the far past. The uncertainty becomes very large outside the range BC2000 - AD3000. This is the reason why NASA hasn't computed eclipse paths outside that range. According to research, the uncertainty could be as large as over 4 hours (several thousands of kilometers for eclipse paths) for dates before BC4000. See <a href="https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html" rel="nofollow noreferrer">this</a> explanation on NASA's Eclipse Web Site.</p> https://astronomy.stackexchange.com/questions/32212/sorting-out-julian-day-julian-date-julian-day-number-julian-day-calendar-and?cid=59005 Comment by FSimardGIS on Sorting out Julian Day, Julian Date, Julian Day number, Julian Day Calendar, and Julian Day Table FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-06-09T17:56:56Z 2019-06-09T17:56:56Z <i><a href="https://en.wikipedia.org/wiki/Ordinal_date" rel="nofollow noreferrer">Ordinal Date</a></i> seems to be a possible term for the year and day-of-year format. (YYYY-DDD) <a href="https://www.iso.org/obp/ui#iso:std:iso:8601:-1:ed-1:v1:en" rel="nofollow noreferrer">ISO 8601</a> uses the term <i>Ordinal Date</i> in their standardized date formats. https://astronomy.stackexchange.com/questions/32142/stellarium-software-with-de431-for-eclipse-10000-yrs-ago?cid=58872 Comment by FSimardGIS on Stellarium software with DE431 for eclipse 10000 yrs ago FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-06-04T01:04:27Z 2019-06-04T01:04:27Z The accuracy for the position of the Moon indeed isn&#39;t so bad, however the biggest uncertainty for eclipse prediction (or back-prediction in your case) is the Earth&#39;s non-constant rotation. We do not know exactly how much the Earth has rotated since then, and that uncertainty is very large for such dates (more than 4 hours, or 68 degrees in 4000 BCE, even more in 8000 BCE) which means that the Earth&#39;s orientation could very well be half a rotation off the model-predicted orientation, but we do not have enough detailed records or observations before 1000 BCE to ascertain the value of delta-T. https://astronomy.stackexchange.com/questions/32142/stellarium-software-with-de431-for-eclipse-10000-yrs-ago?cid=58864 Comment by FSimardGIS on Stellarium software with DE431 for eclipse 10000 yrs ago FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-06-03T19:13:50Z 2019-06-03T19:13:50Z @sidharth They take into account long term variations, but still the uncertainty for the position of the Moon in DE431 (as stated in the paper) is 28 m/century&#178; so for 100 centuries in the past the uncertainty is +- 280 km along-track. As for the rotation of the Earth, it is even bigger than that. https://astronomy.stackexchange.com/questions/32142/stellarium-software-with-de431-for-eclipse-10000-yrs-ago?cid=58861 Comment by FSimardGIS on Stellarium software with DE431 for eclipse 10000 yrs ago FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-06-03T16:57:34Z 2019-06-03T16:57:34Z One thing to bear in mind also is the uncertainty in the value of delta-T (Earth&#39;s rotation) and the Moon&#39;s secular acceleration. These parameters aren&#39;t known with great accuracy for such dates in the far past. <a href="https://eclipse.gsfc.nasa.gov/SEcat5/uncertainty.html" rel="nofollow noreferrer">NASA</a> estimates that the error in longitude for eclipse back-predictions 6000 years in the past could be as large as 68 degrees, which means an eclipse track could be off by thousands of kilometers. https://astronomy.stackexchange.com/questions/31725/how-to-properly-observe-solar-total-eclipse-in-2019?cid=57639 Comment by FSimardGIS on How to properly observe Solar Total Eclipse in 2019? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-05-02T14:16:32Z 2019-05-02T14:16:32Z The Argentina side is also an option, and has relatively dry regions near Bella Vista / San Juan. https://astronomy.stackexchange.com/questions/31725/how-to-properly-observe-solar-total-eclipse-in-2019?cid=57638 Comment by FSimardGIS on How to properly observe Solar Total Eclipse in 2019? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-05-02T14:13:26Z 2019-05-02T14:13:26Z You can have a look at NASA&#39;s website, concerning <a href="https://eclipse2017.nasa.gov/safety" rel="nofollow noreferrer">Safety</a> when viewing eclipses. Also, they published a list of <a href="https://eclipse.aas.org/resources/solar-filters" rel="nofollow noreferrer">Reputable Vendors</a> of eclipse glasses. For the 2019 eclipse, it would indeed be visible from the beach, however, there is a higher risk of clouds and/or fog on the coast. Places in the interior, like Vicuna, have better weather prospects, but the eclipse is low so the positioning is more critical if you don&#39;t want the sun to be hidden behind mountains. https://astronomy.stackexchange.com/questions/30098/sunset-on-mountain-from-viewing-point/30099?cid=55065#30099 Comment by FSimardGIS on Sunset on [Mountain] from [Viewing Point] FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-03-25T16:33:34Z 2019-03-25T16:33:34Z I&#39;m not sure I understand what you mean, do you refer to the fact that Earth is a spheroid (squashed sphere)? I have done the calculation using the WGS84 spheroid and 2.9&#176; is the correct elevation angle for the line of sight. The bearing is good as well. I think April 7 would be a good bet, too. The biggest uncertainty will be refraction on that day, as well as the weather :-) https://astronomy.stackexchange.com/questions/30098/sunset-on-mountain-from-viewing-point/30099?cid=55051#30099 Comment by FSimardGIS on Sunset on [Mountain] from [Viewing Point] FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-03-24T17:53:59Z 2019-03-24T17:53:59Z When I checked the azimuth of Mount Fuji as seen from Enoshima, I got a value of 276.2&#176;, a bit different from yours. Which coordinates did you use for Enoshima and Mount Fuji? The elevation angle seems OK, I got 2.86&#176; https://astronomy.stackexchange.com/questions/29589/what-causes-the-duration-of-moon-phases-the-time-the-moon-is-visible-above-the?cid=53548 Comment by FSimardGIS on What causes the duration of moon phases (the time the moon is visible above the horizon) to vary for an observer in the northern hemisphere? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-02-16T00:42:14Z 2019-02-16T00:42:14Z You mean the time interval between moonrise and moonset? https://astronomy.stackexchange.com/questions/29569/how-close-to-circular-is-the-earths-equator?cid=53440 Comment by FSimardGIS on How close to circular is the Earth's equator FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-02-12T23:05:37Z 2019-02-12T23:05:37Z Oops, my previous link is wrong for the Figure of the Earth, <a href="https://en.wikipedia.org/wiki/Figure_of_the_Earth" rel="nofollow noreferrer">here it is</a>. Also, another interesting article: the <a href="https://en.wikipedia.org/wiki/Equatorial_bulge" rel="nofollow noreferrer">Equatorial bulge</a> https://astronomy.stackexchange.com/questions/29313/why-is-the-difference-from-a-perfect-sphere-the-same-for-the-earth-and-the-moon?cid=52687 Comment by FSimardGIS on Why is the difference from a perfect sphere the same for the Earth and the Moon? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-25T02:55:44Z 2019-01-25T02:55:44Z <i>Obviously, if measured to enough decimal places, differences will appear</i>. Indeed, there are differences, but 3 significant digits is not enough to see that. The values (with more decimal places) would be 0.272657; 0.272509; 0.273095. If you calculate the flattening, you can also easily see the difference: Earth&#39;s flattening is 0.00335, while the Moon&#39;s is 0.00121. If you scaled the Earth to match the Moon&#39;s equatorial radius, its polar radius would be 1732.3, quite a significant difference. https://astronomy.stackexchange.com/questions/29278/star-rising-times-in-a-different-place-given-the-latitude-and-time-of-one-place?cid=52636 Comment by FSimardGIS on Star rising times in a different place given the latitude and time of one place FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-23T15:13:19Z 2019-01-23T15:13:19Z And what is the answer they stated? Do they simply use the difference in longitude for the calculation? https://astronomy.stackexchange.com/questions/29278/star-rising-times-in-a-different-place-given-the-latitude-and-time-of-one-place?cid=52607 Comment by FSimardGIS on Star rising times in a different place given the latitude and time of one place FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-22T17:57:54Z 2019-01-22T17:57:54Z As stated, it would be hard to determine. You need to know the declination of the star to answer properly. So unless your star is on the celestial equator, or your two observing points are on the same parallel, i&#39;d say there&#39;s a missing piece of information to perform the calculation. https://astronomy.stackexchange.com/questions/29271/when-is-earth-closest-to-the-sun?cid=52585 Comment by FSimardGIS on When is Earth closest to the Sun? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-21T23:48:26Z 2019-01-21T23:48:26Z Perihelion times are typically already referring to the moment when the Earth&#39;s center and the Sun&#39;s center are closest, not the barycenter. See <a href="https://aa.usno.navy.mil/faq/docs/apsides.php" rel="nofollow noreferrer">this explanation</a> on US Naval Observatory for details.The dates of perihelion can vary by up to about 2 days from the average, mostly because of the Moon. https://astronomy.stackexchange.com/questions/29259/jupiter-venus-conjunction-is-vertically-aligned-im-at-49-n-how-is-this-possi?cid=52580 Comment by FSimardGIS on Jupiter-Venus conjunction is vertically aligned. I'm at 49° N. How is this possible? FSimardGIS https://astronomy.stackexchange.com/users/19814 2019-01-21T22:47:13Z 2019-01-21T22:47:13Z Nice drawing, showing a very good understanding of the geometry involved in your question. However, as mentioned in my answer, the planets are not exactly on the ecliptic plane, so a vertical alignment can be seen from other places too.