Celestial Nav. Question

[Gos2015]

I recently obtained my celestial navigation certification but there was an exam question that still puzzles me.

The question was looking for a three point fix using Venus, Vega and Polaris. We were asked to give the Zn and 'a' (intercept) for each. I had no problem with Venus or Vega but where I had issues was with Polaris.

I get that the LHA, A. Lat allows you to enter the Polaris table using the corresponding column for the Ha and then using Ao, A1, A2, -1 degree to get your Latitude.

At the bottom of the Polaris table you can determine the Zn, which varies around 359 degrees plus a few minutes. So that gives me my Zn, but what about the intercept?

Typically you use the Declination, sight reduction tables to determine Hc and then the difference between the Hc and the Ho gives you your intercept. No declination is given for Polaris but I believe that it is 90 degrees. Sight reduction tables only deal with Delinations up to 29 degrees.

So, what I have come up with is, my intercept is Zero since Ao, A1 etc. are used to place me on the line (circle) of position based on my sextant altitude reading of Polaris. In other words, there is no intercept because I am already on the line of position.

Can anyone confirm that I am right?

Thanks, Les

[estarzinger]

It sounds like you know that for Polaris, your calculated Latitude=ho - 1° + A0 + A1 + A2.

If you are using an assumed position, I would suggest your intercept is the difference between your calculated latitude and assumed latitude.

[Gos2015]

Thanks for you posting. I hadn't looked at like that - I have been too focused on trying to factor in Declination and using the tables but in a practical sense, my Lat. by Polaris is my circular line of position, Zn as determined by the Polaris tables give me my bearing and I should know by my DR Lat. or A. Lat whether I am moving Towards or Away. The distance to the point where my Zn and Lat. intersect, gives me the Intercept

I think that's it although I would appreciate comments from others if they think otherwise.

[Benz]

It's been a while since I thought about all this, but it seems to me this is one of the drawbacks you encounter in taking courses: hypothetical problems that have no parallel in the real, practical world. Whenever while navigating I had the privilege of shooting Polaris, I would immediately reduce it to my latitude and plot that in heavy, satisfied pencil. That's the treat of Polaris: you need not treat it like any other star, so it's easy money, and doing anything other with it seems kind of silly. Any LOPs derived from other star shots were subservient to that, since it's such a foolproof calc. They could cross it if they wished, otherwise they were suspect. Am I missing an obvious reason why you would want to calculate an intercept for Polaris? Let me add that I never took courses, so my celestial was always rather shaky and seat-of-the-pants, and perhaps this is not a hypothetical-only question.
Ben

[estarzinger]

I am pretty sure my answer above is what they were looking for . . . .

But it is possible to compute Polaris exactly like any other star.. Several of the celestrial calculators do this. However the paper tables are not really set up for it. It's declination is not (exactly) 90, that's what the a1, a2, and a3 corrections are all about. Right now it's about 89 19.5.

I never used Polaris much, because I prefer to use brighter easier to find stars even if their calculations are a little more difficult. For me, I always found the primary celestial challenge at sea to be in physically finding a star and bringing it down accurately, and the calculations then to be pretty straightforward.

[Benz]

^^ I didn't doubt that you'd answered him correctly; my post was just the thought, in relation to that, that came to mind. I don't think it ever occurred to me to reduce Polaris as a regular star, because of the nearness of brighter stars to shoot; if conditions were right, Polaris was an easy, bombproof latitude. I can see why it might be included in a navigation excersise, but in the real world I would guess nearly everyone would try and use it for latitude. Anyway, I couldn't have answered the OP's question. Perhaps it's time to get to sea again....

[estarzinger]

I completely agree with you. It is a bit of an odd question, and not how you would normally use Polaris in "the real world". But perhaps that's makes it an interesting/challenging test question.