| my account | login-logout | resources | classroom help | support | catalog | home | get webcard |

 search | help desk | commons
 » Online Classroom   » Emergency Navigation   » Public Discussion of Emergency Navigation   » *C02 — P012 Practice Question 01 — finding position from watch alone

Author Topic: *C02 — P012 Practice Question 01 — finding position from watch alone
 David Burch posted March 30, 2006 06:07 PM                   On page 12 of Emergency Navigation, there is this statement "You are wearing an accurate watch, meaning you can figure UTC from it to within a few seconds, but have no other navigational equipment. Using the methods of emergency navigation, you could, by doing little more than looking at the sky, discover where you are to within about 300 miles. You can obtain this level of accuracy no matter where you are, on any day of the year, using only the most elementary principles and very little special knowledge."---------------- Post a reply to this question giving a precise numerical example of how this might be done. You can use the Celestial Navigation Data from the USNO AA site if that might help.----------------The following times and dates are picked at random, so there is no bias to finding a good or bad example, and they will not we worked out by us till we get some answers.----------------Part A. Let's say it is July 4, 2007 and you are located about 500 miles due east of St. Johns, Newfoundland, at 47° 34' N, 40° 13' W.Part B. Assume, on the same date, you are 500 miles north of Maui, HI at 29° 20' N, 156° 25' W.----------------How would you find your longitude and latitude with specific numerical examples and discussion of the uncertainties that would get you within the ±300 miles. From: Starpath, Seattle, WA
 Dan Cline posted May 26, 2007 06:07 PM                   Part A: Actual position is 47 deg 34 min N, 40 deg 13 min W. Assuming I have the materials to construct a plumb bob sextant (probably not permitted by the assumptions of this problem, but it's not clear to me how this problem can be answered using a watch alone), latititude can be determined within plus or minus 2 degrees using a plumb bob sextant to measure the altitude of Polaris; 1 degree of uncertainty comes from the lack of altitude corrections for Polaris, the other 1 degree of uncertainty comes from the inherent inaccuracy of the plumb bob sextant. 2 degrees of uncertainty equal 120 nautical miles of uncertainty in this position. Without a plumb bob sextant, one could estimate the height of Polaris using fist widths, which would get altitude measurement within about plus or minus 5 degrees, plus a 1 degree uncertainty for altitude corrections, gives plus or minus 6 degrees, or 300 nautical miles, for uncertainty. Another option is to find latitude by observing zenith stars, but that assumes I know the declination of various stars.Longitude can be determined from LAN; LAN is determined by finding the times of sunset and sunrise, adding them, and dividing by two. In this example, at this location on this date, sunrise is at 06 hr 49 min GMT, sunset is at 22 hr 41 min GMT; LAN = (06 hr 49 min + 22 hr 41 min)/2 GMT = 29 hr 30 min/2 GMT = 14 hr 45 min GMT. To get the longitude we subtract 12 Hrs from the GMT LAN to get 2 hr 45 min. This is converted to longitude as follows: 2 hr * 15 deg/hr + 45 min * 1 hr/60 min * 15 deg/hr = 30 deg + 11.25 deg = 41.25 deg = 41 deg 15 min W longitude. Note that this is off by 41 deg 15 min – 40 deg 13 min = 1 deg 2 min = 62 min = 62 nautical miles. This difference is attributable to the lack of correction for the equation of time, which is an unknown quantity. Note this answer assumes there is zero watch error. Part B: Actual position is 29° 20' N, 156° 25' W. Same answer as A for finding latitude. For longitude, sunrise at this location and on this date is at 15 hr 31 min GMT, sunset is at 05 hr 29 min GMT the next day = 29 hr 29 min. LAN = (15 hr 31 min + 29 hr 29 min)/2 GMT = 45 hr 00 min/2 = 22 hr 30 min. Convert this to longitude as follows: 22 hr 30 min – 12 hr 00 min = 10 hr 30 min. 10 hr * 15 deg/hr + 30 min * 1 hr/60 min * 15 deg/min = 150 deg + 7.5 deg = 157.5 deg = 157 deg 30 min. This result is off by 157 deg 30 min – 156 deg 25 min = 1 deg 5 min = 65 deg = 65 nautical miles. Again, this is due to a lack of correction for the equation of time. Again, this answer assumes there is zero watch error.In general, the longitude measurement is impacted by the accuracy of the correction for watch error and the equation of time. Each second of uncorrected watch error introduces .25 nautical mile of error in the longitude calculation. The sun is 14 minutes late on Valentine’s Day and 16 minutes early on Halloween; thus the failure to correct for the equation of time gives a maximum uncertainty of 16 minutes, which equals 16 min * 1 hr / 60 min * 15 deg/hr = 4 deg = 240 nautical miles. From: Midland, MI
 David Burch posted May 26, 2007 10:09 PM                   The Lon approach is correct, but we should go ahead and figure the EqT from the prescription on this date to see how close you get.But we are cheating on lat if we use a tool at this stage since the problem said just looking at the sky. So we need to think more on this one about getting to within 300 nmi by "just looking." What specifically would we be looking at. Seems we must rely on the usno aa site to see what the sky looks like thoughout the night... the star finder or the plot sky function in StarPilot would be even easier since it is plotted already.Also... besides looking at the stars for latitude, what about putting a stick in a board and marking the length of the shadow at noon while you hold the board level to the horizon? what do you do then? and could this get to within in 300 nmi? this is kind of a tool, but not so much as a quardrant. what would you have to have or know to carry on with the marked length of a shadow? this approach is discussed more in the next edition of the book than it was in the first. From: Starpath, Seattle, WA
 David Burch posted May 26, 2007 10:18 PM                   another thought on the earlier note. we are at 47° N in the first problem, so 62' of Lon would not be so bad in nmi as quoted. it is good exercise to think through what this really would be.in other words, how many nmi to 62' of lon at 47 N? at the equator, the answer is 62 nmi. From: Starpath, Seattle, WA
 Dan Cline posted May 27, 2007 07:36 AM                   Good point. Degrees of longitude don't equate to nautical miles -- a sloppy mistake on my part. On the latitude problem, should I assume I don't have an almanac or similar item on board? From: Midland, MI
 David Burch posted May 27, 2007 09:45 AM                   yes, in this example at the very front of the book we are just making a general statement. what can you do by just looking, and what you have in your mind... or might have in your mind. and then maybe using the length of a shadow at noon to suppplement looking at the stars. From: Starpath, Seattle, WA
 Dan Cline posted May 30, 2007 05:41 PM                   I've decided to chip away at this one little by little. First, one can convert minutes of longitude into nautical miles by multiplying by the cosine of the latitude, i.e. 1' lon * cos(lat) = distance nmi.So, for Part A, my error was 62', so I get 62'*cos(47 deg 34') = 62'*0.6747 = 41.83nmiFor Part B, my error was 65', so I get 65'*cos(29 deg 20') = 65'* 0.8718 = 56.67nmiEquation of time corrections and another stab at latitude will follow this weekend. From: Midland, MI
 David Burch posted May 30, 2007 05:48 PM                   That sounds right on the lon to miles conversion. at least the formula is right. you can also solve this with a simple universal plotting sheet without any trig or calculator. just make a section at the right latitude and measure it. We also have prescriptions in the book on how to make these universal plotting sheets with just a protractor. Furthermore, once you review how the UPS are made you will see a trick way for finding the cosine of any angle graphically, which is just what the plot trick is based upon. From: Starpath, Seattle, WA
 Dan Cline posted June 03, 2007 08:57 AM                   Longitude and the Equation of Time The equation of time can be estimated from the following (memorized) information: On Valentine’s Day (Feb 14) LAN at Greenwich is 14 minutes late (i.e., Greenwich LAN is 1214) and on Halloween (Oct. 31) it is 16 minutes early (i.e., Greenwich LAN is 1144). We also know that on May 14 Greenwich LAN is at 1156 and on July 31 Greenwich LAN is at 1206 from this mnemonic:code:`14 late 3 months later goes to 4 early16 early 3 months earlier goes to 6 late`From this I can interpolate the time of LAN at Greenwich for July 4 as follows: between May 14 and July 31 (a period of 17 + 30 + 31 = 78 days) the time of LAN changes by 1206 - 1156 = 10 minutes. There are 17 + 30 + 4 = 51 days from May 14 to July 4, so the change in LAN is calculated by adding 51 days / 78 days * 10 min to 1156, giving us 1156 + 7 = 1203. I can now correct my previous answers to A and B by subtracting 12 hr 03 min from observed LAN instead of subtracting 12 hr 00 min. So, for Part A, I determined LAN to be 14 hr 45 min GMT. 14 hr 45 min – 12 hr 03 min = 2 hr 42 min.. This is converted to longitude as follows: 2 hr * 15 deg/hr + 42 min * 1 hr/60 min * 15 deg/hr = 30 deg + 10.5 deg = 40.5 deg = 40 deg 30 min W longitude. This is off by 40 deg 13 min – 40 deg 30 min = -17 min. To convert this result to nautical miles, I multiply -17 min * cos(47 deg 34') = -17'*0.6747 = -11.5 nmi, an improvement on my last result. I note that the interpolated value I obtained for the equation of time, 1203, differs by one minute from the value obtained from the USNO calculator, 1204. For Part B, I determined LAN to be 22 hr 30 min. Convert this to longitude as follows: 22 hr 30 min – 12 hr 03 min = 10 hr 27 min. 10 hr * 15 deg/hr + 27 min * 1 hr/60 min * 15 deg/min = 150 deg + 6.75 deg = 156.75 deg = 156 deg 45 min. This result is off by 156 deg 45 min – 156 deg 25 min = 20 min. Again, to convert to nautical miles, I multiply 20'*cos(29 deg 20') = 20'* 0.8718 = 17.4 nmi, once more an improvement on my previous result.Latitude Without a Plumb Bob SextantI am now sitting in my life boat without any string, but with a pencil, a can of beans, and a piece of marine plywood (the life boat seat?), a 6-inch stick that I found drifting in the ocean, and a much better memory. On one side of the plywood I construct a crude protractor by drawing a circle on the plywood using the can as a guide for my pencil, and then dividing the circle into quarters, eighths, etc. to allow me to measure angles. I also determine the location of the center of the circle from the intersection of the lines I drew to divide the circle. (Note: without the can I could draw a circle by holding my thumb at the center of the board, the pencil in my fingers, and rotating the board underneath the pencil using my thumb as the axis. The can method is easier.)From my observations of sunrise and sunset, I have already calculated LAN. Starting about 20 minutes before LAN, I flip my board over, mark a spot on the board, and then carefully hold the stick vertical on this spot and carefully hold the board level. I then use my pencil to mark the spot on the board where the tip of the stick’s shadow falls, and do this repeatedly until about 20 minutes after LAN. I take the average of the distances from the shadow tip marks to the spot where I placed the stick, and mark this distance on the side of the stick.Turning my board over, using the mark on my stick, I measure from the center of my circle to the point corresponding to the distance from the stick base to the shadow tip on a line that passes from the center of the circle through the 90 degree point. I then place my stick vertically on the measured point and draw a line from the center of my circle to the tip of the stick. The difference between the angle of this line and 90 degrees is the observed height of the sun at LAN, Ho.Since the date is July 4, I know that the declination of the Sun is north of the celestial equator, and I need to subtract this declination from Ho to determine my latitude. I calculate the declination by first determining an angle that is equal to the number of days from the summer solstice (June 21) to July 4, divided by the number of days from the summer solstice (June 21) to the autumnal equinox (September 23), and multiplying that number by 90 degrees: angle = 13 days / (13 days + 27 days + 31 days + 23 days) * 90 deg = 13 days / 94 days * 90 deg = 12 deg (rounded). I know that the declination of the sun is 23.45 deg * cos(12 deg) = 22.9 degrees. Of course, I don’t have a calculator or trig tables with me, so I have to determine cos(12 deg) by plotting the angle on my board protractor and dividing the horizontal distance from the center of the circle in some appropriate unit such as stick lengths along the 90 degree radius to its intersection with a perpendicular dropped from the 78 degree mark (78 = 90 – 12) and dividing that distance by the radius of my can circle in stick lengths.Subtracting the Sun’s declination from Ho gives my latitude, more or less. Measurement errors can be very significant with this approach. The result will depend on my ability to hold the stick vertical while measuring its shadow, and my ability to hold the board level. Since this is July, the length of the shadow will be small relative to the height of the stick and difficult to measure accurately. I suspect that measurement error becomes less significant the longer the stick is. Measurement errors will also impact the calculation of the sun’s declination using the angle plotting method.I can double-check my work by lying on my back at night and observing the zenith stars. In part A, at latitude 47 deg 34 min N, on the evening of July 4, 2007 I should be able to observe Deneb near the zenith at approximately 0430 UTC. I recall Deneb’s declination is 45 deg north, which gives me my latitude within about 2 ½ degrees (150 nmi).In part B, at latitude 29 deg 20 min N, on the evening of July 4, 2007 I should be able to observe Alphekka at approximately the zenith at 0715 UTC. My excellent memory tells me Alphekka’s declination is 26 deg 42 min N, which is within 2 deg 34 min of my actual latitude. Without a star chart or almanac, this method is highly dependent on one’s memory of star declinations, which is a major weakness. Accuracy suffers depending on how well the zenith star’s declination matches the observer’s latitude. Finally, observational errors are a problem; determining when a star is at the zenith can be challenging, especially in heavy seas. From: Midland, MI

 All times are Pacific