I hope I never look quite this intense at it...

...I just lift my lamp that others may see.

Are we there yet?

Be assured, the end is in sight!

The final answer lies in some very real complications to my rosy hypothetical lighthouse pictures. If you haven't guessed by now, the lighthouses of which we are speaking are beacons that are very tall and have no brickwork to support them--the Sun, Moon, planets, and stars. They have a huge advantage over man-made lighthouses in that they can be seen over entire hemispheres rather than just the local coastwise region. But using celestial beacons leads to certain complications:

• As viewed from the Earth, they are moving in a most complex way. With a few exceptions, stars rise, climb higher in the sky for a few hours, then descend toward the horizon and set. They appear again the next evening, but ever more West of where they were the previous night at the same time. After a season or two, the whole sky looks different as once-familiar stars get lost in the glare of day for a few months, only to re-appear after a year has passed, right back where we first noticed them. This means that even in those extraordinarily rare instances that a bright star is in our zenith, it won't stay there long.
• We can't tell how "tall" these lighthouse beacons are. Remember that our simple plane trigonometry solution depended on knowing the height of the lighthouse.
• The Earth is round, not flat; and the sky also appears to be a big round dome over our heads.

So to get back to the question at hand, let us reconsider the lighthouse. Only this time, neither of us is standing directly under it. Furthermore, neither of us knows for certain what the height of the lighthouse may be. Now the "side view" looks something like this:

Now each of us is an unknown distance from a lighthouse of unknown height. Both of us can measure an altitude for the beacon. Your altitude is greater (that is, closer to 90 degrees) than mine.

Now we are missing all the data except the measurement of altitudes. Each of us has an altitude and it seems natural for us to compare our answers. This part is where someone had a million-dollar idea.

The decision was made to define one nautical mile as that length needed to see a difference in altitude of one minute.

At last, we come to the intercept!!

Although the diagram is nowhere near to scale, please humor me and suppose my sextant reading were 38^o 32'.6 and yours 38^o 42'.8. Then you may simply take a difference between our measurements, which is 10'.2. Since one minute is one mile (and vice versa) you know immediately that you are precisely 10.2 miles closer to the lighthouse than me. That is to say, starting from my position, the intercept, denoted p, is 10'.2 toward. To make it really work, the intercept must be toward some definite azimuth, but we don't need to choose one for this illustration.

(As an aside, I freely concede that my definition of the nautical mile isn't the most technically correct, but it is the practical effect for the Navigator on deck!)

Is that all there is to it?

Yes! Now you should be able to see for yourself how this goes. You don't really need me in the picture. You could just as well say, "Let us decide a 'convenient spot' for Al to stand on and call it my assumed position. I don't really need him to take an observation for me, because once I know his position and what time it is, and I cleverly chose both, I may use mathematics to compute what altitude and azimuth he would see from there. Since a computation comes entirely from mathematics, it doesn't matter whether we can actually measure the azimuth or not. Then we'll compare it to what I myself observed on deck."

There are 3 possible outcomes for the comparison:

• The observed altitude, denoted H_o, is exactly the same as the computed altitude, denoted H_c. We conclude that we must be standing exactly upon the same altitude circle as our "convenient spot" (assumed position) for which we did the calculation. Good guesses like this are obviously rare.
• The observed altitude H_o is less than the computed altitude H_c. We conclude that we must be standing farther away from the center of the circle computed for our assumed position.
• The observed altitude H_o is more than the computed altitude H_c. We conclude that we must be standing closer toward the center of the circle computed for our assumed position.

Let me explain now that I will use a somewhat odd notation in my Web Pages. I find a lower-case letter much easier to read than a genuine subscript, so I will use the convention of writing Ho for observed altitude and Hc for computed altitude.

The Cookbook Summary of Plotting a Celestial Fix

• We plot an assumed position, which will have some convenient Latitude and Longitude, on the sheet.
• We take our computed azimuth as a radius of a circle of equal altitude for that assumed position.
• We measure along that azimuth the number of miles specified by the intercept, either toward or away from the azimuth, following the rules:
          Ho > Hc => p toward
  and  Ho < Hc => p away
• We draw a line perpendicular to the azimuth, through the point to which our intercept led us. This line is the line of position and represents a circle of equal altitude as we actually observed it.
• This process is repeated for as many bodies as the Navigator cares to use, and the intersection of the lines of position is the fix.

If you have come this far, you now know all the necessary concepts of the method. Thanks for your attention!!!

If you want to know all the rest of the details, then...

Go to the next Celestial Navigation page

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