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» Online Classroom   » Emergency Navigation   » Public Discussion of Emergency Navigation   » The sun's path across the sky

   
Author Topic: The sun's path across the sky
FloridaAdventuring


 - posted December 06, 2005 09:52 AM      Profile for FloridaAdventuring           Edit/Delete Post 
It has always been my general impression that the sun rose in the east or thereabouts, was more or less in the southeast at mid-morning, was exactly in the south at high noon (LAN), was more or less in the southwest at mid-afternoon, and set in the west or thereabouts. This general idea of the sun's path seems to make a fairly useful guide for emergency land navigation when only a rough bearing is required, e.g. when you want to walk 2 miles to the east-west road north of you. In this case, general directions seem to work.

However, reading Emergency Navigation (which I am enjoying immensely) I find out that in those places and times of year when the sun reaches above 45 degrees in altitude, the above schematic of general directions may not be accurate.

So, I checked it out. According to the U.S. Naval Observatory's web site, at Orlando, Florida (N 28.53 degrees) on June 21 2005, the sun's azimuth was 128.2 degrees at 1200 EST, or Zulu - 5. Then at 1300, the sun's azimuth was 237.0--a change of 108.8 degrees in one hour! That seems incredible to me. Can it be true? I always thought the sun moved across the sky at around 15 degrees per hour (180 degrees divided by 12 hours). But 108.8 in one hour! How can this be?

What do you say about this? I'd appreciate any enlightenment on the subject.

Thanks.

From: Melrose, Florida
David Burch


 - posted December 06, 2005 06:56 PM      Profile for David Burch           Edit/Delete Post 
The sun does move across the sky at 15° per hour along its path across the sky, but when you project the positions straight down to the horizon, you will see that the bearing change with time can be much faster, depending on the height of the sun, which in turn depends on your lat and the time of year.

Within the Tropics, for example, the sun can pass over your head. At the moment it does that, its bearing has changed 180° in a just seconds.

In the Emergency Nav book we have some guidelines for when you can approximate the bearing change at 15°/hr and that is provided the noon height of the sun is less than halfway up the sky, ie <45°.

I leave it as an exercise to gather some actual data from various circumstances meeting this criteria and violating this criteria to see how this works. The height and bearing to the sun at any time on any date from any place can be found online at the first cel nav link in our inet connections, under celestial. Remmeber it is not the present height of the sun that matters, but its eventual peak height at Local Apparent Noon.

From: Starpath, Seattle, WA
FloridaAdventuring


 - posted December 08, 2005 08:18 PM      Profile for FloridaAdventuring           Edit/Delete Post 
Let me see if I have this straight: The sun DOES move across the sky at 15 degrees per hour, but its azimuth when projected down to the horizon may change more--a whole lot more--than 15 degrees per hour.

I understand the 45-degree rule. When the sun's maximum altitude for that day is 45 degrees or less, one can get fairly accurate bearings from the sun in conjunction with a watch. I can see how sailors--who may need to cover long distances--would be especially concerned since each degree of error is compounded over distance.

For land navigators, however, I wonder if the 45-degree rule would be as applicable. Let's say a cross-country hiker's immediate navigation goal is to intersect an east-west trail about a mile or so north of his position. We'll assume our hiker has no compass (dog ate it) and must rely on the sun for guidance. To find the trail, our hiker needs only to travel in a northerly direction and does not need to follow a precise bearing. Could he not reasonably use the sun at any time of year(even in a place where its maximum altitude is well over 45 degrees) and the time of day to determine a course, say within 30-45 degrees of where he'd like to go? Even such an imprecise bearing would allow him to intersect with the trail. And since the trail is no more than a mile away, not much navigation error would accrue over such a short distance.

For such imprecise work, it seems to me that one could use the following very rough approximation as a working model for baseline navigation (in which you're trying to intersect a baseline, rather than navigate to a specific point) in the northern temperate zone:

Time Position of sun to within, say, 30 deg
0600 Easterly direction
0900 Southeasterly direction
1200 Southerly direction
1500 Southwesterly direction
1800 Westerly direction

Using this approximation, could not the hiker at any time of year reasonably use the sun to walk from one baseline (say a trail)cross-country to the next(say a river)? Once on his baseline, the cross-country hiker can follow it to a specific checkpoint such as an unusally sharp bend in a river or trail, and eliminate all past navigation error. To be able to do this, the hiker must aim off to one side of the checkpoint.

I know such solar navigation works here in Florida in winter when the sun isn't too high, but I'm not sure now about summer, when the sun is well above 45.

Your thoughts please.

Thanks.

From: Melrose, Florida
David Burch


 - posted December 08, 2005 11:49 PM      Profile for David Burch           Edit/Delete Post 
My first answer was maybe misleading. The bearing change can be higher than 15°/hr but from the mid latitudes it is more likely to be less than 15. Note the most special case, however, is to be on the equator on the equinox. Then the sun bears due east all morning and then due west all afternoon.

There is much info in the emergency nav book (EN) for looking at special cases and various methods to find the sun's bearing during the day, especially if you have a watch. The suggested range of "temperate zone" is likely too broad to apply the table you present, since this covers at least 30 to 60° of latitudes (and more often even say 25 to 65), and for all seasons, since the sun's declination varies 46° throughout the year. I doubt the table could be right within the 30° limits, especially since we have an hour uncertainty in the times, unless all is reckoned relative to local apparent time, which has been measured relative to LAN, and in this case we would know due south precisely.

In EN we show that the rising angle for any body including the sun when bearing due E or W is 90° - Lat. At lat 45° N, the angle is 45°, so a sun rising at 45° will have a bearing change of 15 x cos 45 which is about 15 x 0.7 = about 10°/hr. This can be drawn to scale on graph paper using a protractor for the rising angle for any lat to see what the bearing rate would be, wihtout having to use a math. Note also, that this idea forces the sun to be due east or west at some time of day, so we do not have to know any more about the date, etc.

For arbitrary times or latitudes, one can use sight reduction tables from cel nav to see this change for any lat and date and time, since they give you height and bearing of the sun every 4 min (1 deg of LHA). We give links to pdf versions of these tables in the cel nav resources.

The main question for land navigators is how quickly do they need to know the answer. The shadow tip method will give the bearing of the sun much better than ±30° in 30 min or so on any date and just about any place. And on land--in many circumstances--you may have to just find the direction once, and then note some land mark in the right direction.... unless in the desert or arctic perhaps, or in a tall forest or jungle.

The 2102-D star finder [in our accessories catalog] is another mechanical way to practice computing sun bearings throughout the day for any latitude, but one would need The Starfinder Book [books catalog] to explain how, since the standard instructions that comes with the device are very poor.

I would propose testing the table results for 1500 at 50N and at 30N for summer and winter solstice. Again, we must be careful about what you mean by a time of "1500." (as far as the sun is concerned, there is an hour differnce between two observers 15° of longitude apart, even though both might read 1500 on their watches at the same moment.) I would guess, for example, that at 50 or 55 N the sun rises roughly SE and sets roughly SW on the winter solstice.

The table idea in general is a good one. It is effectively one form of "sun compass" discussed in the book. I would guess that something like this should work just fine, providing it were more restrictive in lat and season. As outlined in the book, in any circumstance, if one has a full day to work on it, you can construct various types of sun compasses that will work throughout the day from then on and with even a much higher precision than 30°.

the old saying we grow up with that the "sun rises in the east and sets in the west" can be very misleading for the wayward traveler who must actually apply such reasoning in a critical situation.

From: Starpath, Seattle, WA
FloridaAdventuring


 - posted December 14, 2005 08:23 AM      Profile for FloridaAdventuring           Edit/Delete Post 
Thanks for your answer. The example of the sun's bearing change on the equator at the equinox (from due east to due west in an instance)led me to see how the sun's bearing can change dramatically, even though the earth rotates just once (360 degrees) in a 24-hour period (15-degrees per hour).

Regarding the sun's bearing change throughout the day, you gave the following formula:

The sun's bearing change per hour = 15 x cosine of the sun's rising angle. (I need to dust off my trigonometry skills to remember how to compute the cosine of a number.)

If I knew the sun's amplitude and thus the direction of sunrise and sunset, could I then use this formula to plot the sun's direction each hour of the day? (Presumably, if I plotted the rising angle, I'd simply draw an appropriately rising straight line on a graph, the horizonal axis of which would be degrees and the vertical axis of which would be hours up to noon. The afternoon graph would be a straight line descending at the proper angle back to the horizontal axis.)

Not being a mariner, my uses for navigation involve finding my way through wilderness areas, and horizons are often not visible through trees and mountains. Obviously, an $8 compass is a big help in wayfinding, but in the absence of that instrument, I'd like to know a method I can use to navigate quite well by the sun. Since I can easily learn the sun's maximum amplitude in my area of operations, I can guesstimate its amplitude throughout the year, giving me approximate sunrise and sunset directions. Starting with that, I assume I could then apply the above formula to figure the sun's direction all day. I'd then have an effective sun compass which should be accurate enough (say within 20 degrees or so) for the land navigator on foot who just wants to make it to the next baseline a couple of miles away (road, river, trail, etc).

Am I correct in my assumptions?

From: Melrose, Florida
David Burch


 - posted December 17, 2005 12:53 AM      Profile for David Burch           Edit/Delete Post 
I am afraid it is not quite that simple because the formula given applies only when the cel body bears due east or west.

There are several methods outlined in the Emergency Navigation book that can be used to build a sun compass for any date and latitude, and none of them require any trig. (An ebook edition is available for $9.95.) That formula was given just to show the relationship when the sun was "crossing the prime vertical," ie bearing due east or west.

Your good questions here, however, have made me realize that we have another solution that we could generate that we have not done so far.

We have another product called the Emergency Navigation Card, and on that card there is a table we created called N(x) Table, also known as the world's shortest sight reduction tables. If you know the declination and have a watch, with that table you can compute the precise height and bearing of the sun in any circumstance.

I think we can make from that table a new specialized table that is exactly what you want. Given any latitude and date, make a table of the sun's bearing each hour of the day, which is just half a day, since it is symmetric about noon.

Instead of declination used in the table, we will use date, and instead of meridian angle of the body, we will use time relative to noon. I think this could be a very short table with an average of 6 time entries and about 12 date entries for every 15° of lat from 0 to say 60.

If you give me a specific lat and date you might care about for your expeditions, we could start with that one and see what we get.... there are such things, by the way, called Azimuth of the sun tables, which are books about a quarter inch thick and about 6 x 9 in size. The pub no 216 sticks in my mind, but i would have to look that up. We have one in the school, but i am not there now.

There is also a chance that a short list of numbers from the N(x) table and a short prescription will also do the job.

You can also of course create such a table online at any time from http://aa.usno.navy.mil/data/docs/AltAz.html

and someone from UW here in Seattle has also made this table already and put one online for every 5° of latitude:
http://staff.washington.edu/gcthomas/Guide/solaraz.html

So we will concentrate on the last approach mentioned above, so you could have one short method that is more universal.

See also: http://www.jgiesen.de/astro/astroJS/sunriseJS/

if you want to get the bearing super accurate on any date or time, you can do so here:
http://www.srrb.noaa.gov/highlights/sunrise/azel.html

and if you want to program a hand-held calculator or maybe some fancy cell phone to give you sun height and bearing at any time, date, or place, you can use these equations Sun Hc and Zn equations

In any event, i am sure we can come up with a very simple method that will do the job without all this technology. we just need discussions like this to get us motivated!

From: Starpath, Seattle, WA
FloridaAdventuring


 - posted December 18, 2005 02:10 PM      Profile for FloridaAdventuring           Edit/Delete Post 
A table would certainly be useful, but in the wilds I'm just as likely to have a compass as a table. I was hoping to find a nifty little formula that I could memorize, then use when necessary to scratch out the sun's bearings during the day for any given latitude and date.

I visited the UW website, which I found exactly on point. For example, I noticed than on June 10 at 30 degrees north latitude, the sun would be at 180 degrees at LAN (of course). But at one hour before noon, it would bear 114 degrees, and one hour after noon, it would bear 246 degrees, a change of 132 degrees in a mere two hours. At first, I had a hard time seeing how that could be, but your example of an overhead sun changing from a due east to a due west bearing in an instant made me understand how it occurs.

Then I read something (else) in EN that I found fascinating. If one were to follow the sun from one hour before noon to one hour after noon (we'll say) the AM and PM navigation errors would cancel.

For example, if it were around 1100 and I wanted to walk due south, I could simply follow the sun (as much as obstacles in my path would allow). My course would be off to the east in the AM, due south at LAN, then to the west in the PM. But by 1300, I would have walked a crescent-shaped path (once again, providing the relative absence of obstacles in my path), and the end of my path would be due south from the beginning of my path, because of cancelled navigation errors.

I could, of course, guide off the sun in any direction. If I chose to keep the sun off my left shoulder (here at 30N lat), for instance, I would walk a crescent-shaped pattern that would ultimately lead me due west from my starting point, provided I started in mid-morning and finished walking that course at mid-afternoon, or some other smaller time increment equally spaced before and after noon.

This would work from mid-morning to mid-afternoon. Before those times I could guide on the sunrise and sunset, "chasing" the sun back to the horizon if necessary by figuring its rising and setting angle as 90 degrees minus my latitude. If I knew the sun's amplitude, I could follow a course fairly accurately for a while. Fascinating. And so is EN. I do want a hard copy when it comes out again.

It seems to me uch navigation would work quite well if my intention were, say, to intersect with a road perpendicular to my path a few miles distant.

I'm working on the chapter dealing with stars right now and am captivated by it. I think I'm beginning to see how one could sail around the world with only a watch.

I look forward to other discussions.

PS Have you ever thought about taking the info in EN and putting it into a video? It seems to me that such wonderful information accompanied by scenes of beautiful yachts and oceans, and using coloful graphics would sell all over the world. Put me down for a copy.

From: Melrose, Florida
David Burch


 - posted December 19, 2005 12:03 AM      Profile for David Burch           Edit/Delete Post 
We do a planetarium class here in Seattle once or twice a year on emergency techniques, and it has been my fond goal to make a flash movie out of that presentation, but we just have too many irons in the fire at the moment.

I hope you will find and learn the techniques you need from the book.

We will work on a nice sun formula for your type of application during the holidays, if at all possible. It is easy to imagine circumstances where you could detect the sun's direction but not see much of a horizon (for height measurements) and not even be able to do the shadow-tip method if deep in the forest.

Remember, if you have a watch, which we would need to assume for this formula, then you already know a lot once you know the time of LAN. The time of LAN, too, will always be required, even for a slick formula.

Again, if you have the time to measure it, ie one full day, then that will do, but if starting from scratch you need some table for equation of time or practice memorizing it with the jingle given in the book, which starts, "14 minutes late on valentine's day..."

From: Starpath, Seattle, WA
FloridaAdventuring


 - posted December 19, 2005 08:46 PM      Profile for FloridaAdventuring           Edit/Delete Post 
Thanks. I look forward to possibly a formula. In the meantime, I'm off to learn about the (other) stars.
From: Melrose, Florida


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