A schoolboy circles the worldDec 21, 2007
Thus began a five-year odyssey that took Robin Lee Graham around the world. During that circumnavigation Graham fell in love, got married, had to abandon Dove for a stronger boat and became a national figure through the articles he wrote for National Geographic. When he returned to California he and his wife moved to the Montana wilderness where they have since lived raising their family. He also wrote two books as a result of his experience. Of course, it could have been otherwise, as Dove was not designed for a circumnavigation. But besides his youth, Graham had other things going for him besides luck. He had cruised the coast of California aboard the family boat and he had become a good sailor and an excellent navigator. His parents thought enough of his abilities to support his dream.
The passage of Dove was fraught with near disasters. Twice the vessel was dismasted. The second time Graham had to sail more than 2,000 miles in the Indian Ocean under jury rig before he reached port. By the time Graham reached the Caribbean little Dove was a wreck. National Geographic sponsored him to a new Luders 33 that became known as Return of the Dove. It was on this vessel that Graham completed his trip — the youngest person to ever have done so.
Let’s get aboard Dove for a sun sight. At only 24 feet long, Dove was a difficult platform from which to shoot celestial sights. Graham would make the boat more stable by heaving to.
We join him on a passage from Hawaii southeast to Fanning Island; a small island in the Line Group about 1,000 miles from Hawaii. Fanning Island is only 11 miles long by 9 miles wide and Graham had to have his celestial act together in order not to miss that speck of land. I’d guess Graham wanted to get a sun shot that would give him a good idea of his longitude. He was on a true course of 208°. He knew that if he could get a shot of the sun that was 90° off his course line that the azimuth of the sun would yield an LOP that would be parallel to his course line and give him a good idea of his longitude.
We have him at a DR of 13° 30’ N by 158° 20’ W. The height of eye is 8 feet and there is no index error. Graham shoots the upper limb of the sun at 20 hours 47 minutes and 15 seconds GMT. The resulting Hs is 62° 21.1’.
A. What is Ho?
B. What is the intercept?
C. When you plot the LOP, what is the EP?
Navigation solution: Robin Lee Graham
There is no question that Robin Lee Graham was good and lucky. To set off at the age of 16 years in a 24-foot boat to circumnavigate would intimidate many more experienced sailors. The fact that he relied solely on his celestial navigation should be an inspiration to all of us who might think that celestial belongs in the basement.
Graham was smart enough to understand that his LOP is perpendicular to the azimuth of the celestial object; in this case the sun. He is on a course of 208° True bound for Fanning Island, an island small enough to miss if his navigation is off. He is interested in getting one sun shot that will give him a good idea of his longitude. Since his course is west of south when the sun bears about 90° from his heading he can get an LOP) that will get him the results that he desires. The overcast sky is a problem so he shoots when he has the opportunity, He is at a DR of 13°30’N by158°20’W. His Height of Eye is 8 feet and there is no Index Error. He shoots an Upper limb of the sun at 20 hrs 47 minutes 15 seconds, GMT. His Hs is 62°21.1’ and the day in question is September 22. We first want to find the Ho:
3rd - 16.3’
Nothing difficult about that. Just remember to go into tables for Upper Limb. An easy place to make a mistake!
Next we want his Intercept and for that we have to do the dight reduction using HO 249 Vol. 2:
GHA 20 hrs 121°49.3’
47min, 15 sec 11°48.8’
-ass long 158°38.1’
Because the GHA is smaller than the assumed longitude 360° can be added to it in order to do the math correctly. Thus:
-ass long 158°38.1’
While we are in the daily pages we also need to find the declination at the time of the sight. Declination at 20 hours is N 0°12.15’ with a d correction of 1.0’. By inspection I see that the declination is decreasing thus I subtract the d correction which in this case is .8’. I found this by going to the rear of the NA to the 47 minute correction page and finding the value for the d correction of 1.0’. This is .8’. The declination is changing so slowly that the 15 seconds of the time does not have to be accounted for.
Dec N 0°12.7’
We are ready now to enter HO 249. We will be using an assumed latitude of 13°
When we enter the tables (declination contrary) we get the following:
Hc 62°01’ d +28’ Z 116°.
We go to Table 5 for minutes of declination and in this case the minutes are rounded up to 13’. Table 5 says that the sun has moved .6’
Intercept is 5nm Away.
Remember if the Hc is larger than the Ho the LOP is away from the celestial object in the direction of the azimuth.
When this is plotted you will find that the Estimated Position is 13°34’N by 158°28’W