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jimh posted 01-07-2013 10:46 PM ET (US)   Profile for jimh   Send Email to jimh  
This is decidedly non-electronic, but it is definitely electro-magnetic and relates to navigation; so I am including it here.

My habits lately find me working at my bench in the late afternoon facing a west window. I watch the sun set almost every afternoon. I have a hand bearing compass, so I have begun to take observations of the bearing to the sun as it sets. This is known as the Amplitude of the Sun. It provides a method of checking compass error.

To compute the Amplitude of the Sun, we need to know the Sun's declination and the latitude of the observer. From a table I find that on this date in January the declination of the sun is South 22.5-degrees. I know my latitude is North 45.5-degrees. The Amplitude is then calculated from

Sin(Amplitude) = Sin(Declination) / Cos(Latitude)

Solving for Amplitude I get -31.2-degrees. Amplitude is measured from East or West, thus the true bearing to the Sun at sunset today at my latitude should be 270 - 31.2 = 238.8-degrees.

For my location, magnetic variation is 7-degrees West, so the compass bearing should be 238.8 + 7 = 245.8-degrees.

My observation today was 243.5-degrees. This is variance of 2.3-degrees. I attribute most of this to the observation not being made with the Sun on the horizon. Some buildings block my view, so I make my observation with the Sun still above the horizon. An observation made slightly later would be more in agreement with the predicted value.

Why do this? It keeps me from going crazy all winter. It keeps my math from getting too rusty. Try it yourself--you will enjoy it.

You can get a table of the Sun's declination from

You should know your latitude within a half-degree. The geometry is not complicated. Take a bearing on the sun set tomorrow, and check your compass or your observation.

I go outside to make the observation because the compass error is very bad at my bench. It is probably distorted by the magnets in the loudspeakers in the near-field monitors playing music from iTunes.

jimh posted 01-08-2013 07:26 AM ET (US)     Profile for jimh  Send Email to jimh     
Here is the math:

Sin(Amplitude) = Sin(Declination) / Cos(Latitude)

Declination = -22.5
Latitude = 42.5

Sin(Amplitude) = Sin(-22.5) / Cos(42.5)
Sin(Amplitude) = -0.38268/0.73727
Sin(Amplitude) = -0.51904
Amplitude = ArcSin(-0.51904)
Amplitude = -31.2-degrees

Sun's heading = 270-degrees + (-31.2-degrees) = 238.8-degrees-True

True heading = 238.8-degrees
Variation = 7-degrees-West
Magnetic heading = 245.8-degrees

Jerry Townsend posted 01-08-2013 01:59 PM ET (US)     Profile for Jerry Townsend  Send Email to Jerry Townsend     
Jim - you definitely have too much time on your hands. But, thanks - as I had not thought or read of it before - and am not sure if I will ever use it - but it is interesting. I learned something new and will have to try it out - thanks. --- Jerry/Idaho
jimh posted 01-08-2013 05:39 PM ET (US)     Profile for jimh  Send Email to jimh     
ASIDE to Jerry: I was composing my reply when I noticed the sunset tonight would be good for observation. Then I got sidetracked with hooking up a different audio amplifier to listen to a Joni MItchell live concert recording. I dragged out an old Class-A Mosfet power amp, and drove it directly from the Focusrite computer interface. It sounded so good I forgot the observation. I looked up and the Sun was gone. We did have a nice violet and dark red sky. These are all elements of Winter Boating.
6992WHALER posted 01-09-2013 05:09 PM ET (US)     Profile for 6992WHALER  Send Email to 6992WHALER     
I find it interesting that navigation is so connected to small boat electronics these days. That this post that has nothing to do with small boat electronics actually appears to fit in this forum.

Is this from you JN days Jim.

jimh posted 01-10-2013 11:40 AM ET (US)     Profile for jimh  Send Email to jimh     
I think I first heard to the Amplitude of the Sun from a segment of instruction called LIFEBOAT NAVIGATION, in which techniques were taught for simplified methods of finding position or heading. I think this is presented in the United States Power Squadron classes for NAVIGATOR.

I got my NAVIGATOR qualification from USPS in c.1988. Obtaining the NAVIGATOR rating was a substantial effort. In order to take celestial observations with a sextant, I had to drive about 50-miles to Leamington, Ontario. There is a long pier there that projects southward into Lake Erie. From that pier you can observe the horizon and have a 180-degree field to select various stars for measurement. Also, I was taking the class in the winter. I found that I had to leave the sextant outside the car for a while in order to have it reach ambient temperature. If I took a warm sextant from the car to the cold air, the observations would never be very good.

For these observations of the Amplitude of the Sun I am using a very nice hand bearing compass that I bought in c.1986. It is a VION Mini Morin 2000 compass, encased in a rubber puck protector. It is made in France. This is a very good hand bearing compass and allows sightings to be made to 0.5-degree with ease.

I see the price and very high regard for this compass has not changed very much since my purchase. I think I paid $100 for mine (new) more than 25 years ago. It still works like new.

In searching for some information on the compass, I came across this article: the-direction-of-sunrise-and-sunsetthe.html

The author mentions the possible influence of metal eyeglass frames on the compass. I had not thought of that! My eyeglass frames are steel. I will have to test to see if the eyeglass frames are influential on the compass readings.

jimh posted 01-10-2013 11:45 AM ET (US)     Profile for jimh  Send Email to jimh     
I just made a test with the hand bearing compass. I am very surprised to say that my steel frame eyeglasses do have an influence on the compass! This is quite interesting. The difference in observed bearing was 1-degree. From now on, all observations with this compass will be made without the steel-frame eyeglasses being worn.
jimh posted 01-11-2013 05:09 PM ET (US)     Profile for jimh  Send Email to jimh     
Tonight's sunset was clear and cloudless. I observed the Sun at 5 p.m. EST, or 2200 UTC, as we are keeping local time as GMT -5. The sun was observed a few minutes before sunset, which is about 5:18, so that it was still above my horizon of buildings to the West. I made the observation without wearing my steel frame eyeglasses. The bearing to the Sun at 2200 UTC was just a bit over 244-degrees per my hand bearing compass.

Using the U.S. Naval Observatory predictions for my location, this date, and this time, the true bearing to the sun should be 237.3-degrees. Adding the magnetic variation of 7-degrees-West means the magnetic bearing should be 244.3-magnetic.

My compass bearing observation was in very close agreement to the predicted magnetic bearing. This is quite interesting. The conclusion is the compass error is very small or nil. That is to be expected for a high-quality hand bearing compass being used away from other magnetic disturbances.

Also, the temperature tonight was 57-degrees-F, which made for a very pleasant observation.

jimh posted 01-12-2013 11:05 AM ET (US)     Profile for jimh  Send Email to jimh     
To predict the time of sunsets, I created a customized sunset-sunrise calendar from

jimp posted 01-13-2013 09:20 PM ET (US)     Profile for jimp  Send Email to jimp     
Back in my CG days as a Deck Watch Officer aboard a sea-going buoytender we checked gyro compass error using the azimuth of the sun, EVERY underway watch when the sun was visible. We cheated though. We followed an example in the "Azimuth Log" and had all the other tables readily available. But that was a long time ago.
jimh posted 01-17-2013 05:05 PM ET (US)     Profile for jimh  Send Email to jimh     
In reply to Jim's anecdote about his days in the Coast Guard, we used to spend a lot of time navigating a sailboat, with really only a compass to guide us. We had no LORAN, and GPS was just for the Military, if it even existed at that time.

Because the compass was on a binnacle mount on the steering pedestal and there was a rather large iron block diesel engine just below and in front, that compass had significant deviation. We had a deviation table, but I was always interested in checking and refining the data. When we had the opportunity to sail on a range, I would try to get exactly on the range and note the compass heading. In that way we could verify the deviation table. The table was quite good, as it turned out, and we were usually not more than 1-degree off. That is close enough for a sailboat. If you are 1-degree off an only sailing at 5-knots, you still won't be very far off track an hour later.

I have been watching the sun set for over week, hoping to get another observation--not so much to check anything but to verify that the Sun is returning to the North. Tonight it looked like a glimmer of the Sun's rays might break through the clouds, but no luck.

jimp posted 01-17-2013 09:45 PM ET (US)     Profile for jimp  Send Email to jimp     
I taught a 5-day sailing-cruising course on 26-41 foot sloops, yawls, and cutters out of Greenport, Long Island, NY for my college summers (1973-1976). All we had was a compass, fathometer, and a Ray Jeff RDF. Great combo for running in the fog in Block Island Sound.

The school had the compasses adjusted each year and all were fairly accurate. But it seems few of my students could keep her within 5-degrees anyway and with beating upwind and currents to 1.5 knots steering straight didn't seem to matter. The RDF was... a RDF, sometimes you believed it and sometimes it you didn't.

Steering straight didn't matter except for the one night we sailed to Block (roughly 36 miles due east from Greenport) in the fog. Visibility started at 1-1.5 miles and upon arrival was about 100 yards. I really pounded into my students that steering straight and LISTENING were very important in the fog. We dead reckoned to Block (the RDF said it was somewhere ahead of us...) and at the appropriate time watching my watch and watching the fathometer (a lot of watching here!), we hove-to and listened and heard the Block Island Great Salt Pond entrance horn just off the port bow, about a mile away. Down with the sails and we motored in about 2300. Great crew!

Fixed Coast Guard ranges on rivers or harbor entrances or natural ranges are great for comparing your compass.

Ahh, the uncomplicated life without electronic navigation! That sail to Block with a GPS chartplotter would have been boring.

Just to keep it Whaler related. While doing all this my 1963 13' Boston Whaler Sport was sitting on her mooring in Great Peconic Bay (no compass, no fathometer, no tachometer, no battery, just a 33 hp Johnson rope start and two 6-gallon plastic gas tanks).


White Bear posted 01-18-2013 12:04 PM ET (US)     Profile for White Bear  Send Email to White Bear     
Having made it a point to engrave on a window sill the exact directions of the solstices and the equinox, I can with great pleasure report that the sun has indeed begun its trip northward. This observation was made across Peconic Bay from Southold, approximately 5 miles from the Greenport mentioned in a previous post.
jimh posted 01-25-2013 01:37 AM ET (US)     Profile for jimh  Send Email to jimh     
ASIDE: I mentioned in another article I was watching ICE STATION ZEBRA, and noted a few discrepancies in the plot and action which were taking place in the Arctic. First there was a problem with the radio horizon that caused me to experience some disbelief. There was also a problem with the position of the sun which caused a substantial decrease in verisimilitude.

The scene is supposed to be in the Arctic, so we assume a Latitude of at least 70-North. The scene is dimly lit, the sky is heavily overcast, yet when the men walk, they cast distinct shadows. OK, perhaps there is a little window of clear sky and the Sun is peeking through.

The next problem is the size of the shadows. The shadow of a six foot man only reaches about two feet in length. This implies a sun elevation angle of

ArcTAN(6/2) = 72-degrees

This is a big problem for a scene set at 70-North. The sun never gets that high. Even at Noon on the Summer Solstice the sun only gets to about 43-degrees elevation at 70-North. This means that a shadow should always be longer than the height of the object casting the shadow.

Back to the movie: in a short while, the shadow lengths change, as does their direction. Sometimes in very rapid succession. The screenwriter and the lighting director did not check their notes on this.

I seem to recall some dispute about the famous explorer Peary reaching one of the Poles. Critics said he stopped short. When Peary was at the pole he had a photograph taken. Decades later the photograph was analyzed to determine if it could have been taken at the Pole. They used the shadows as a method to determine Latitude. I think they confirmed it was within a few miles of the Pole, which is probably about as close as Peary could determine himself.

jimh posted 01-26-2013 09:03 AM ET (US)     Profile for jimh  Send Email to jimh     
jimp sent this interesting photograph, a multiple exposure image that shows the track of the sun in sky. It was taken on the Winter Solstice in Juneau, Alaska, in Latitude 58-degrees 18-minutes North:

Photo: multiple explosure images shows track of sun in sky at Juneau, Alaska on Winter Solstice.

I think the Sun's elevation when on Juneau's meridian on the Winter Solstice would be about 8-degrees.

jimh posted 02-06-2013 05:45 PM ET (US)     Profile for jimh  Send Email to jimh     
It has been a long time since the weather and sky have cooperated to permit me to observe the amplitude of the Sun. Tonight we finally have a clear sky at sunset, the first one in several weeks.

I have made an observation at 5:29 p.m. local time. The sun was about to disappear behind some houses in my line of sight. As it was several minutes before the actual sunset, I will compute the azimuth from the Naval Observatory web-server. To keep my actual observation from being influenced, I did not compute the bearing until after I made the observation.

According to my hand-bearing compass, the Sun was at just a fraction over 253°. Subtracting the magnetic variation of 7°W, the true bearing observed was 246°.

According to the Naval Observatory, the Sun's true bearing was 246.3-degrees. Excellent agreement. (By the way, elevation was 2° 40')

Eureka. We have evidence of the Sun's declination returning to the North. Heck, there is still daylight as I write this. Almost 6 p.m.--it is a miracle! The dark days are almost over. We have gained ten-degrees in bearing since the initial observation.

jimh posted 02-11-2013 02:19 PM ET (US)     Profile for jimh  Send Email to jimh     
Regarding observations of the sun, perhaps the most famous observation of the sun was made by Eratosthenes. He observed the sun as part of his deduction of the circumference of the Earth.

Eratosthenes observed that on the solstice, the sun was directly overhead at local noon in a particular place, Syene. At that same moment, he measured the angle to the sun in another location, Alexandria. By assuming that the earth was a sphere and that the rays of the sunlight were parallel, be could measure the angular distance between Syene and Alexandria on the sphere.

Once the angular distance between the two cities was known, the circumference of the Earth could be deduced by measuring the geographic distance between them. (They are about 500-miles apart.) The accuracy of the deduced circumference is limited by two factors:

--the accuracy of the angular measurement of the sun's elevation

--the accuracy of the measurement of the distance separating the two observation points

According to some historians, Eratosthenes is credited with an accuracy of about two percent. Not bad for c.200 B.C.

A further source for error is the necessity for the measurements to be made simultaneously. The sun is moving rather rapidly, so that any observation of the sun changes rapidly with time. How rapidly?

The accepted circumference of the Earth is 24,901-miles. The length of a Solar day is 24-hours, plus or minus about 20-seconds, depending on the date and epoch. We can calculate the apparent speed of the sun as

24,901-miles / 24-hours = 1037.5-miles-per-hour

This is also equivalent to 0.288 miles-per-second. In other words, if we take consecutive measurements of the sun, the sun will appear to have moved relative to the Earth about one mile in four seconds. This apparent speed of motion is why celestial observations for navigation must be made with very good precision of time. To make a sight reduction with one-mile accuracy requires that the time must be known to an accuracy of less than four seconds of error.

Since Eratosthenes was said to have obtained a two-percent accuracy, we could assume his calculation was within the present-day value by that same tolerance. Two-percent of 24,901 is 498-miles. If all other factors in his observations were perfect, the time of the two observations would have need to have been coordinated to

498-miles x 1-second/0.288 x 1-minute/60-seconds = 28-minutes

In c.200 B.C. it might have been difficult to arrange for two observations about 500-miles about to occur within 28-minutes. I don't think there were watches, telegraphs, or visual signals that could have been used.

What worked out for Eratosthenes was Syene and Alexandria were roughly North-South of each other, that is, they were on almost on same meridian. Therefore local noon occurred almost simultaneously in both places. The benchmark for angular measurement to the Sun was a well in Syene. At local noon on the solstice the sun was directly overhead and sunlight illuminated the well to the bottom. On that same day, local noon in Alexandria would occur almost simultaneously because it was almost on the same meridian. The observation of local noon would be indicated by the sun reaching its highest elevation. This could be detected easily by taking a series of observations and watching the trend. Eratosthenes did not need accurate time keeping for his experiment. He just had to make the observation at local nooon on the same day the well in Syene had the sun directly overhead at local. This was a bit easier to synchronize.

The offset in the longitude is about 3-degrees. This means the time of local noon was likely about 12-minutes apart. There would be little change in the declination of the sun in only 12-minutes.

jimh posted 02-11-2013 03:35 PM ET (US)     Profile for jimh  Send Email to jimh     
There is some confusion about the actual value for the circumference that Eratosthenes calculated, and it results from uncertainty about the unit of measurement for distance that was being used. This unit of measurement was how Eratosthenes would have described the distance between Syene, now modern day Aswan, and Alexandria. A lot of conversation has been made regarding the actual equivalent modern-day value for this unit. It seems a bit arbitrary to hold Eratosthenes responsible for the accuracy of the distance measurement from Alexandria to Aswan. He was likely using a value that was the accepted value for his day. It is unlikely he measured it himself. This is the same situation we have today. If a reference source tells me the accepted value for the distance between Eratosthenes' two points is 523-miles, I use that value. I have not measured the distance myself.

An error in the geographic distance between the two measurement points would not have not have affected his angular measurement. We can judge the accuracy of the angular measurement independently. It is frequently noted that Eratosthenes found the angular distance to be 1/50th of the circumference, that is,

360-degrees/50 = 7.2-degrees

It turns out that Alexandria and Aswan (modern day locations) are not on the same meridian. However, since Eratosthenes measured the angle at local noon, we can figure out the angular separation by their latitudes:

Alexandria = 31° 11.4' N
Aswan      = 24° 05.3' N
Difference in Latitude = 7° 6.1'

Converting to decimal degrees this is 7.101-degrees.

That value compares extremely well to Eratosthenes' observation of 7.2-degrees. The error is only 0.099-degree in 7.101, an error of only 1.4-percent. I would say that old Eratosthenes made a very accurate observation of the Sun.

jimh posted 02-11-2013 03:52 PM ET (US)     Profile for jimh  Send Email to jimh     
Another consideration in Eratosthenes' calculation is his assumption that the two points were located directly N-S of each other. He assumed the distance between them was simply their difference N-S. The two points are on different meridians, that is, at different longitudes. Actual over-land distance separating them is longer than the N-S distance. The difference is about 31.8-miles. An error of 31.8-miles in 523-miles is an error of six-percent. This is another source of error that could have thrown off the final calculation of the Earth circumference.

N-S-distance = 7.101-deg x 24901-miles/360-degree = 491.2 miles

Over-land distance = 523 miles (per Google Earth)

Error = 31.8/523 = 0.06 or six-percent

jimh posted 02-11-2013 05:40 PM ET (US)     Profile for jimh  Send Email to jimh     
Other estimates of the Earth circumference we made following Eratosthenes' work. One estimate by Strabo found a much smaller circumference. It is mentioned that Christopher Columbus may have ignored the larger Earth circumference calculated by Eratosthenes. Had Columbus used the more accurate value he would have seen his plan to sail to Asia by sailing West was an impossibly long route. For more on this notion see:

jimh posted 02-12-2013 10:31 AM ET (US)     Profile for jimh  Send Email to jimh     
Another article (linked below) is quite interesting in its speculation regarding Columbus and his knowledge of the circumference of the Earth. I recommend reading it for further insight. It seems to dispel the common wisdom regarding who knew the circumference of the Earth, and on what they based their opinion. 000.html

Regarding Eratosthenes' measurement of the Sun angle, the historical literature (see above linked article) suggest he measured the angle directly with some sort of device called a skaphe. It occurs to me that today one could find the angle by measuring the height of a perpendicular gnomon and the length of its shadow, and then using the trigonometric function TANGENT to deduce the angle. This made me wonder if there were trigonometric tables for functions like the TANGENT in c.200 B.C. Egypt. A cursory search seems to indicate the Egyptians had not developed trigonometry to the point of having tables of angular functions like the TANGENT. Compare at

Today it would be much easier to measure the height of gnomon (h) and the length of its shadow (s), then deduce the angle from ARCTAN(h/s), than it would be to directly measure the angle with some graduated scale of angles.

jimh posted 02-12-2013 05:36 PM ET (US)     Profile for jimh  Send Email to jimh     
I was thinking about making a gnomon using a step ladder to hold up a piece of aluminum tubing. Then I would use a tape measure to measure the shadow length. It occurred to me that since I am located at a mid-latitude, the elevation angle of the sun would be in the 40 to 50-degree range. This is actually a benefit for measuring the shadow length, because as the elevation angle approaches 45-degrees, the length of the shadow and the height of the gnomon are the same. It occurred to me that if I wanted to approach the accuracy of Eratosthenes' measurement, which I imputed to be about 1-percent (see above), then I would have to use a 8-foot high pole for the gnomon, and make the measurement of the shadow to an accuracy of about an inch. That would be an accuracy of one-part in 96-parts (12 x 8), or about one-percent.

I will try this method, if the sun ever shows up again. (It has not been visible for several days.) I think I have a length of aluminum tubing in the garage that is 8-feet long. I can look up the time of local noon in advance, and make the observation then. I can find the Sun's declination from the Naval Observatory website. It might be interesting to see if I can make a measurement to within 1-percent.

Because Eratosthenes was in latitude 31-North and made his measurement on the Summer Solstice, the gnomon and shadow method would have cast a very short shadow. This would make the precision of the shadow's measurement even more critical. The elevation was around 83-degrees. An 8-foot high gnomon would only cast a shadow about a foot long. To measure that to a precision of 1-percent you would have to measure to about an eighth of an inch accuracy.

jimh posted 02-13-2013 02:26 PM ET (US)     Profile for jimh  Send Email to jimh     
About noon today I realized it was a beautiful, sunny, clear day. I set up my step ladder and erected a tall 141.25-inch-long pole as a gnomon. I looked up the time of local noon: about 12:46 p.m. A few minutes before then I began to measure the length of the shadow. I found the shadow was going to be over 200-inches long. I also found that a shadow that long was a bit indistinct. It was hard to tell precisely where the shadow of the pole ended. The pole is only about 1.25-inch diameter. A larger, more distinct object might be better to use.

I watched the shadow grow. The longest shadow I observed seemed to be about 203-inches. From that measurement I computed my observed height of the Sun using

[Here I have corrected the value for the Sun's declination. Initially I used a slightly different value, 13.3-degrees]

Height-observed = ArcTan(141.25/203) = 34.83-degrees

I knew the Declination of the Sun for this noon, [13.1-degrees] South. Now to compute my latitude. First, my Zenith-distance (ZD)

ZD = 90 - Height-observed
ZD = 55.17-degrees

With the Sun in the South, my Latitude is then

Latitude = ZD - Declination
Latitude = 55.17 - [13.1]
Latitude = [42.07-degrees] North

Let me assume I could have known the distance to the location on Earth where the Sun was directly overhead today at local noon on my meridian. That would be the place on my meridian at Latitude 13.1-degrees South. The difference in our position is actually 42.524 (my actual latitude) + 13.1 = 55.624-degrees. This means the distance would be 3841.3-miles [using 24,860 as the circumference through the poles]. Using this as the distance between my observation and the point where the sun was overhead, I compute the Earth circumference as

3841.1 x 360/55.17 = 25,065-miles

My measurement of the Earth circumference varies from the accepted value, [24,860], by 205-miles. This is an error of one part in 122. So I have just measured the circumference of the earth to an accuracy of about 0.8-percent. And that was on my first try!

jimh posted 02-14-2013 03:25 PM ET (US)     Profile for jimh  Send Email to jimh     
An observation taken at local Noon of the Sun simplifies the sight reduction greatly because the observed body is on your meridian. This reduces the complexity of the geometry involved to two-dimensions from three-dimensions.

The diagram below shows the relationship of the various angles of the Noon Sight when the observed body has a declination opposite to your latitude. This was the case in my observation. The Sun's declination was in the South.

Diagram of Noon Sight; copyright James W. Hebert
The Noon Sight. Drawing by James W. Hebert.

K Albus posted 02-14-2013 04:03 PM ET (US)     Profile for K Albus  Send Email to K Albus     
At your longitude, a latitude of 41.87 degrees would put you in the Ohio waters of Lake Erie, about 45 miles south of your actual location. I hope you're not planning to use your gnomon as your primary navigation tool on our next trip.
6992WHALER posted 02-14-2013 05:02 PM ET (US)     Profile for 6992WHALER  Send Email to 6992WHALER     
Kevin, Jim will only use it, if it is NMEA2000 compatible.
jimh posted 02-14-2013 05:14 PM ET (US)     Profile for jimh  Send Email to jimh     
Hi Kevin--Thanks for your comment, I now know that at least one person has read this thread.

Yes, my deduced latitude from my observation was rather coarse. This just gives you an idea of the precision necessary when making a celestial observation. Let's look at the error.

Since a Noon Sight leads directly to latitude, any error in height observed causes an equal error in latitude. If we want a sight to have an accuracy of 0.25-miles, we have to make the height observation to be very precise. How precise?

If the circumference of the earth is 24,901-miles, what angular measurement represents 0.25-miles? Let's call that angle A. We know that

A x 24901-miles/360-degree = 0.25-miles

Solving for A

A = 0.25-miles x 360-degrees/24901-miles

A = 0.00361-degrees

One-second of a degree is 0.00027-degrees, so our precision would have to be to about 13-seconds of a degree. That is really taking a careful observation. And that is just for an accuracy of a quarter mile. You can see how spoiled we have become with the GNSS solution to a resolution of a few feet.

jimh posted 02-14-2013 05:16 PM ET (US)     Profile for jimh  Send Email to jimh     
ASIDE: It took me about two hours to learn how to draw a curved line with terminating arrow heads. That was a lot longer than it took me to take the Noon Sight. I figure that if nothing else comes of this discussion, I finally learned how to use Photoshop to draw curved lines on a specific path. That was a lot harder than the celestial geometry I was trying to illustrate.
jimp posted 02-14-2013 06:11 PM ET (US)     Profile for jimp  Send Email to jimp     
JimH - For sailoring, why isn't 60nm = 1 degree of latitude, 90 degrees equator to pole = 5,400 nmi. 360 degrees (circumference) = 21,600 nmi = 24,840 statute miles?
jimh posted 02-14-2013 06:21 PM ET (US)     Profile for jimh  Send Email to jimh     
I got 24901 as the circumference, probably from Wikipedia.
jimp posted 02-14-2013 07:46 PM ET (US)     Profile for jimp  Send Email to jimp     
Interesting Internet search on "Earth circumference" turned up this:
What is the circumference of the earth?
The circumference of the earth at the equator is 24,901.55 miles (40,075.16 kilometers).
But, if you measure the earth through the poles the circumference is a bit shorter - 24,859.82 miles (40,008 km). This the earth is a tad wider than it is tall, giving it a slight bulge at the equator. This shape is known as an ellipsoid or more properly, geoid (earth-like).

jimh posted 02-16-2013 09:46 AM ET (US)     Profile for jimh  Send Email to jimh     
Some illustrations:

The Gnomon; a step ladder holding a pipe

The Shadow; not quite at local noon.

The Measurement: the shadow is indistinct

jimh posted 02-16-2013 10:50 AM ET (US)     Profile for jimh  Send Email to jimh     
I have revised my calculation to account for two errors. I used the wrong value for the declination of the Sun. (13.1 actual instead of my erroneous 13.3) This threw off my calculation of the distance on my meridian to the place where the sun was overhead. I used this as my equivalent of Eratosthenes distance between the famous well and Alexandria. It did not affect my observed value of the Sun's height. That remained unchanged.

Also, I used the more correct value for the Earth circumference through the poles, as JimP suggested. This also affected my calculation of the distance from my observation to the "well." This was appropriate because I was calculating a meridional distance, not an equatorial distance.

The result of these two corrections is that my estimate of the circumference is more accurate than I first calculated. The error is less than one percent. I afforded myself several luxuries. I did not actually measure the distance from my observation to the "well"--the location where the Sun at that moment was precisely overhead. I just computed that distance using my known location, the Sun's declination, and the known circumference. The precision of this ought to be rather good. It is certainly better than the precision I could have obtained if I tried to measure it myself. And probably better than Eratosthenes knew the distance between Alexandria and the well. The precision of that distance helps to confine my error to my observed measurement of the Sun's height.

The Naval Observatory says the observed height should have been 34.35-degrees. My observation was 34.83-degrees. This is an error of 0.48-degrees, or a variance of about 1.3-percent. This seems like a reasonable tolerance.

On the next sunny day I may experiment with a larger gnomon to cast a more distinct shadow. I think the step ladder itself will be good. It will cast a shadow of about 100-inches, so if I make a measure to at least 1-inch accuracy I should be able to get similar accuracy.

jimh posted 02-22-2013 10:34 AM ET (US)     Profile for jimh  Send Email to jimh     
Careful readers will note that in my diagram of the Noon Sight I show the observed height as if it were observed from the center of the earth, not from my location on the surface. This results in a error from parallax. A section in Dutton's explains the problem:

Excerpt from Dutton's Nautical Navigation

The worst case error would be an observation of the Sun at the horizon. If the Sun were observed directly overhead, there is no parallax error. We can compute the parallax for the worst case with simple trigonometry. The distance to the Sun is 150,000,000-km. The distance from my location to the center of the Earth is about 6,371-km. The angle formed is

ARCSIN = 6371/150000000 = less than 0.002-degrees

As Dutton mentions, the correction is only needed on celestial bodies that are relatively close. For the Sun, the correction is often incorporated in tables. An approximation of the correction can be made by factoring by the Cosine of the observed height. In my case, that was about 35-degrees.

Cosine 35 = 0.8

The parallax in my observation would be approximately 0.0016-degree. Since this error is much smaller than the general error in my observation method, it can be ignored for the purpose of calculating the circumference of the Earth.

When attempting to determine your position from observation of the Sun, ignoring the parallax is not a good approach for accuracy. The influence of parallax is described in this short narrative excerpt

which speculates that perhaps Amelia Earhart's navigator Fred Noonan may have been making a mistake due to ignoring parallax.

jimh posted 03-21-2013 02:20 PM ET (US)     Profile for jimh  Send Email to jimh     
Just a reminder that tonight the Amplitude of the Sun at sunset for most of us in North America will be just about 270-degrees exactly. I calculated the oservation for my location and came up with a Zenith bearing of 270.5-degrees tonight at 2330-UTC.

As is often said, "The Sun sets in the West." Tonight it does, quite literally. And, sorry, but I should have brought this up yesterday. The Zn yesterday was even closer at sunset, 270.1-degrees.

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