Sun & Moon Calculator, Latitude-Longitude vrs UTM?

[peter_korzaan]

<p>Been using the sun moon calculator by Jeff Conrad and have found it very useful. Wonderful program. Thank You.<br>

http://www.largeformatphotography.info/sunmooncalc/<br>

But have been noticing some errors, and wondered if anyone else may be experiencing the same.<br>

In use with a local Sedona Trails map (combined topo's of the area with good info on trails) put out by Emmitt Barks Cartography and so I sent him an inquiry of where I may have been going wrong. The magnetic Declination in the Sun / Moon Calc he states is set up as Latitude Longitude and his map is set up on UTM (Universal Transverse Mercator) .<br>

Sun Moon Calc states magnetic Declination of 11.3 E for Sedona AZ.<br>

His Map on UTM 14.5<br>

He gives a calculator reference where you can convert between the two, but my question is all I have to really do is adjust my compass from 11.3 to 14.5, or add 3.2 to the displayed/printed calcs of the Sun Moon Calc to get the correct azmuth. Yes?<br>

Any other clarification on this subject would be appreciated.<br>

Below is his e-mail for reference.</p>

 

It looks like the Sun/Moon Calculator uses traditional lat-long coordinates, whereas the map uses a Universal Transverse Mercator (UTM) coordinate system, which accounts for the difference in magnetic declination. If you are using the map to get coordinates to feed into the calculator, you need to convert them from UTM to lat-long first. I found this convertor in a link from the UTM article on Wikipedia:

<a href="http://www.apsalin.com/convert-universal-transverse-mercator-to-geodetic.aspx">http://www.apsalin.com/convert-universal-transverse-mercator-to-geodetic.aspx</a>

 

For the calculator, use Clarke 1866 for the Ellipsoid, and 12 for the UTM zone.

 

The top of the map is oriented to true north, magnetic north is 14.5 degrees east.

 

Hope that helps. We use the UTM coordinate system because it is also used for most USGS maps, which is where much of the map's background information comes from. Please let me know if I can be of further assistance.

 

Martin Ince<br /> <strong>Emmitt Barks Cartography </strong> <br /> PO Box 22306<br /> Flagstaff, Arizona 86002

www.emmittbarks.com<br />

 

[William Kahn]

<p>Earth's magnetic declination changes over time. Any map older than 10 years or based on the 1927 North American Datum (NAD27) is probably off by some amount. If Emmitt Barks is using maps based on later surveys (WGS84 or NAD83), the declination shown should be more accurate, and you can adjust your compass using it.</p>

<p>In any case, either declination should be sufficient for determining the approximate azimuth of sun and moon rise/set, even if you're three degrees off.</p>

[peter_korzaan]

<p>Yes, three degress doesn't seem to be much, until your standing in the middle of Oak Creek, trying to get a sunset, creek, moon rise shot, waiting for the moon to come up in the notch of Cathedral, and its further south and comes out of the top. This has happened to me in several locations around here and is the reason why I have been investiagating the problem. Doing a closer study, the same thing happened again the other night with another notch shot I was trying to take, it again came out further south than anticipated and although the shot will pobally be Ok, its not what I wanted. The calcs are very acurate when yoiu start to learn how to use the program, so its been a teaser of why I'm off, and now the difference in declination between the map and the program is becoming the main suspect. Before I thought it was the the rise of the formations and I had gotten the altutde of them wrong. Could be a combination.</p>

[jeff_conrad]

<p>Magnetic declination shown by the Sun/Moon Calculator is indeed relative to true north rather than grid north. True north and grid north coincide at the center of the UTM zone but can differ by several degrees at the edges.</p>

<p>Magnetic declination is usually given relative to true north. For example, the US NOAA National Geodetic Survey <a href="http://geomag.usgs.gov/models/models/">calculator</a> gives a value of 11.34° E for Sedona, using the IGRF-2005 magnetic model. The Sun/Moon Calculator uses a different magnetic model (WMM-2005), but I get the same value with either model (WMM-2005 isn't available for the NGS calculator, so I had to use another program).</p>

<p>I don't see any need for coordinate conversions; adjusting for the difference between true north and grid north should suffice. But I don't quite understand Emmitt's explanation. Most USGS maps show true north, grid north, and magnetic north for the center of the map; the 7.5-minute map for Sedona shows grid north 28 minutes to the west of true north and magnetic north 14.5° to the east of true north. The map is from 1969, so the declination is far out of date, but the difference between grid north and true north shouldn't change, so I would expect magnetic north to be 11.8° E of grid north.</p>

<p>Peter is absolutely correct that 3° is a huge error (about 6 Sun or Moon diameters); if you're off that much, you might as well stay home.</p>

[peter_korzaan]

<p>Hey Jeff, great program, but what should I do to cordinate the two together? Just add 3 degress, or is there another suggestion, solution?</p>

[tomwatt]

<p>I use the U.S. Naval Observatory... you can plug in day(s) and location and it will give you sun/moon data for that time/location.<br>

It's free, and easy to use. I use it a lot in trip planning... I don't want to be still up on the mountain after dark.<br>

<a href="http://aa.usno.navy.mil/data/docs/RS_OneDay.php">http://aa.usno.navy.mil/data/docs/RS_OneDay.php</a></p>

[jeff_conrad]

<p>Peter,</p>

<p>As I said, I don't see why the difference is 3° rather than 0.5°; Sedona is almost on the central meridian for UTM Zone 12, so the deviation from true north of 28 minutes seems reasonable. But without seeing your map, I can't really give you a good answer. Is true north shown anywhere on the map?</p>

[jeff_conrad]

<p>The USNO site is unquestionably the authoritative source in the U.S. But it doesn't directly give everything that's needed for photography; in particular, it doesn't give rise and set azimuths. You can get this information by calculating Sun and Moon positions (and remembering that they apply to the center of the Sun or Moon rather than the upper limb), but it's a lot of extra work.</p>

<p>An alternative is the USNO's <a href="http://aa.usno.navy.mil/software/mica/micainfo.php"> Multiyear Interactive Computer Almanac</a> (MICA), which does include rise and set azimuths; it also gives Sun and Moon positions. It uses the JPL DE405, so it's very accurate (though there's little practical advantage over the Sun/Moon Calculator at moderate latitudes). Unlike several other programs, it also runs on a Mac as well as a PC.</p>

[akajohndoe]

<p>There a a few of these calculators that run on the Windows Mobile (SmartPhone with touchscreen) platform as well as some for the Palm platform.</p>

[jeff_conrad]

<p>On re-reading the e-mail from Emmitt Barks, the comment</p>

<blockquote>The top of the map is oriented to true north, magnetic north is 14.5 degrees east.</blockquote>

<p>suggests that the map <em>is</em> oriented to true north, and that it gives the 1969 value for magnetic declination. You may want to confirm this with Emmitt, but if it's indeed the case, I'd use the value of 11.3° E that you get from the Sun/Moon Calculator.</p>

[peter_korzaan]

<p>Jeff, yes True North is on the map, as related in the original e-mail .<br>

<em>("The top of the map is oriented to true north, magnetic north is 14.5 degrees east.")</em><br>

I'm using a good compass (Silva Type 15t 'THE RANGER' ) and staying away from any funny metal. ;-)</p>

 

<p>Thomas, before I came across Jeff's Calcs, I tried using the U.S. Naval Observatory site, but as Jeff stated, the time and azmuth was off quite a bit because you didn't have the rise of the formations. What happens is its not just the rise, but how close you are to it, or them. That translates in time for the moon to climb that high. Not only that, as it's climbing, its moving south, all referenced to time. So the azmuth changes by the time you 'see' the sun or the moon. So the red rocks in the forground may not be lit with the striking dramitic setting sun as the moon climbs into the view. Now when your trying to corordinate a sunrise or sunset shot in with the moon, its becomes more difficult because of the obstruction to either with the formations, light, angle and weather, when we have it.  Combine that with a newbie photographer, shooting film, in a state of the art 1967 Nikon Photmic Tn, you have a tentitive situation at best.  ;-)</p>

[peter_korzaan]

<p>Ah, Jeff, your last post must have come in as I was writting.<br>

Using 11.3, and verfing at two sites on the same night, ( on Monday the ninth, of March, I had a friend check where the moon came up from his deck) I found that that the time from my calc inputing of 7* rise was to be @ 6:59pm @ 88*. The actual time that it cleared the formation, which was higher, and south of the lower notch, and came at 6:14pm @ 90*<br>

So both locations were off by two degrees, and by happenstance came 'up' at the same time. Now the time may well have been correct, if I had been more north by that two degrees, as the moon would have been seen coming up in the notch earlier. The trouble is I may have been inside a formation that is named the Fin. A tight fit. ;-)<br>

When I went back into the calculator, and put in a rise of 9.5* the time came came out correct but the azmuth remained the same at 88*</p>

 

<p>Does this confuse more, or help clarify?<br>

 </p>

[jeff_conrad]

<p>Peter,<br>

It seems like the main issue here is the magnetic declination, so the best way to sort that out is to eliminate other variables. Try finding a location with a known direction to a landmark, and compare that with what you measure—the difference should give the magnetic declination. If you have trouble finding such a location, try finding a location for which you know lat/lon/elevation, and a prominent feature for which lat/lon/elevation can be obtained (from a map or from a database such as the USGS GNIS). It's then possible to get the azimuth and altitude from a <a href="http://www.ngs.noaa.gov/cgi-bin/Inv_Fwd/invers3d.prl">geodetic inverse calculation</a> ; you can use the elevation above MSL for the ellipsoid height for this purpose. If that form is too tricky, give me the coordinates and I'll run the calculation with my program (it only takes a few seconds). It helps if the feature is at a moderate distance (2–3 miles) from your location; at short distances, small errors in position result in significant errors in directions.</p>

<p>That said, I have no reason to doubt the value of 11.3°; I get the same value from about five different programs, most of which derive from either the USGS or the NOAA National Geodetic Survey. There are some limitations, though; none of the programs deals with local variations, which could be enough to explain the discrepancy you're seeing.</p>

<p>It's also very tough to get accuracy better than a degree with a compass like the Silva 15T (I've had one for 30 years). Because of this, I almost always use geodetic calculations rather than compass measurements, and usually estimate the Moon's position in a scene within about 0.1°; the accuracy is limited primarily by uncertainty in my position (or by variations in atmospheric refraction if the Moon is near the horizon).</p>

<p>The accuracy you need depends on what you are doing; if you just want the Sun or Moon in the general vicinity of a feature, a compass is fine. But if you want a tight juxtaposition with a feature (as in the image titled Pyramid Scheme near the middle of <a href="../nature-photography-forum/00CfZO">this thread</a> , you need to be very accurate indeed. I could not have made that image if I had relied on a compass and clinometer rather than calculations (but perhaps I just can't use a compass ...). There is no doubt that getting started with calculations takes a lot of work, but if you're working primarily in one area, it may be worth the effort.</p>

<p>Another approach is to use mapping software (like National Geographic's TOPO! or Maptech's Terrain Navigator) that lets you measure azimuth and distance; if you know distance and elevation difference, you then can usually get the altitude to sufficient accuracy by treating the Earth as flat and using simple trigonometry. For what it's worth, I get a magnetic declination of about 11.3° E from TOPO! 4.5.</p>

<p>Incidentally, it's not that hard to handle a Sun or Moon position above the horizon using the USNO site if you find Sun or Moon positions rather than rise and set times—it's just a bit more work than with the Sun/Moon Calculator.</p>

[paul_kent2]

<p>I hope that one day Google will add simulated shadows and sunrise/moonrise features to Google Earth. Then coordinate transformation problems will be a thing of the past. Hopefully they are reading.</p>

[peter_korzaan]

<p>Ok, I've got homework.. ;-) (fascinating) Stay tune, will hopefully be able to check it out tomorrow.</p>

<p> </p>

[tomwatt]

<p>Good luck Peter.<br>

I usually go to Topozone to snag the UTM data...<br>

but you are doing some heavy-duty calculations compared to what I've messed with.</p>

[15sunrises]

<p>I'm wondering if somebody could give me an idea of where I could find some resources regarding moonrise/set sunrise/set positions and times with regards to the horizon. Specifically I become confused when:</p>

<p>1) I'm shooting from a hill, looking down</p>

<p>2) I'm shooting from a hill, towards an object on another hill of the same height</p>

<p>3) I'm shooting from ~sea level, looking up towards an object</p>

<p>I'm sure it's simple, I'd just like to get a bit more accurate...</p>

[pjmeade]

I've been using the same calculator as Thomas Watt. Used in conjunction with the google earth ruler tool, it's possible to feed in a location and get pretty good results.

[sknowles]

<p>I use several for <a href="http://www.wsrphoto.com/mtlight.html">Mt. Rainier NP</a> because they all vary a few degrees and don't accommodate local terrain, but it's get you in the ballpark where you can move to adjust your position. Don't forget they're all calculations and close to reality but not necessarily exact.</p>

[w_t1]

<p>If only Ansel had this technology, he could have nailed <em>Moonrise</em>.</p>

[peter_korzaan]

<p>Hmmm.. getting headache ;-) checking into geodetic calculations for accuracy, but my goodness, does anyone have a web address for education?<br>

As Scott said.. "Don't forget they're all calculations and close to reality but not necessarily exact."<br>

Yet to get the accuracy that Jeff got with his Pyramid Scheme in the tread he mentioned,<br>

http://www.photo.net/nature-photography-forum/00CfZO<br>

which BTW is excellent and related to this discussion, one does have to "be there". We only get nice fluffy clouds, beautiful sunrises and sunsets, moons and thrills, and catch them in the camera if everything is 'right' on and with a great helping from serindipity. But when one goes out to try, shooting the 'vison', one would like the subject to be in the right place at the right time.<br>

Sigh.</p>

 

[jeff_conrad]

<p>Scott, there's usually something seriously wrong with a calculator whose<br /> rise and set times differ from a reference (such as the USNO or JPL<br /> Horizons) by more than a minute or so, or whose positions, including rise<br /> and set azimuths, differ from a reference by more than a degree or so.<br /> There definitely can be a difference between calculated and actual rise and<br /> set times, due to differences in atmospheric refraction, but it's rarely<br /> more than a minute. Bear in mind, of course, that the times of rise and<br /> set refer to the Sun's or Moon's upper limb crossing a level horizon (i.e.,<br /> zero altitude); if the horizon includes hills, rise will be later than<br /> calculated. With a few programs, such as the Sun/Moon Calculator and JPL<br /> Horizons, you can specify an altitude greater than zero for rise and set to<br /> accommodate hills.</p>

[jeff_conrad]

<p>W T, if only Dennis di Cicco had this technology, he could have found the date of <em>Moonrise</em> in a few minutes rather than ten years ... As to the story of how <em>Moonrise</em> was made, the 1980s version seems a bit embellished—see Reece Vogel's post at the end of <a href="q-and-a-fetch-msg?msg_id=001Wam">this thread</a> for Adams's 1942 account.</p>

[jeff_conrad]

<p>Dave T,</p>

<ol>

<li> If the camera is level (your case 2), the normal times for rise and set apply.

 

</li>

<li>If you're looking up at an object, rise will be later and set will be earlier. If you know the altitude of the object, and you're using a program like the Sun/Moon Calculator or JPL Horizons, you can specify a nonzero altitude to reflect this. Alternatively, you can generate a table of Sun or Moon positions (azimuth and altitude) using many programs (such as the USNO) to get a pretty good idea of when the object will appear; you need to remember that rise and set times usually refer to the top of the body's disk, while positions are usually given for the center of the body. <br /> As Peter is finding out, though, the hardest part of all this is getting the azimuth and altitude of a feature. </li>

<li>If you're on a hill looking down (e.g., a sunrise from Colorado's Pikes Peak), it's more complicated because of additional refraction, and most calculators (including the USNO) give the wrong anwser because they don't account for the dip of the horizon. The Sun/Moon Calculator takes this into account if a nonzero Height Above Horizon is specified. A few other calculators also do this.</li>

</ol>

[bill_clark___minnetonka_mi]

<p>Here is the URL I use:</p>

<p>http://www.sunrisesunset.com/custom_srss_calendar.asp</p>