Can We Go A Little Deeper? - SailNet Community
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Can We Go A Little Deeper?

Flybyknight is offline

Old 02-08-2010
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OK I will take a stab at this. First I have never took a Celestial Nav course. This looks like that kind of problem.
Going back to many,many decades of my Geometry class here it goes. Dusting off the cobwebs in my brain,.
1.Equator is = BF
2.Body = yes
3.Azimuth =W
4. Declination = S
5. Lat =S
6. Nav triangle = MCD

Melissa Renee
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Catalina 445, Hull #90
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Old 02-09-2010
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Melrna: how do you get S declination and latitude from this plot? I'm guessing D is the north pole and C is the zenith and MX is the altitude (or maybe its sine). If that's the case, M and C certainly look to be "above" the equator. I'm also confused as to how you read azimuth off the diagram.

I'm also miffed because I feel like this could have gone in my thread

s/v Laelia - 1978 Pearson 365 ketch

Old 02-09-2010
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I see a bow and arrow and range finder reticules.... Alas so much to learn its daunting...........

"Go Simple...Go Large"

Relationships are everything to me..everything else in life are just tools to enhance them.

The purchase price of a boat is just the admittance fee to the dance...you still have to spend money on the girl...so court one with something going for her with pleasing and desirable character traits others desire as well... or you could find yourself in a disillusioned relationship contemplating an expensive divorce.
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Melrna,
I sent you a PM
Flybyknight is offline
Old 02-09-2010
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Quote:
Originally Posted by Flybyknight View Post
Melrna,
I sent you a PM
I didn't get the PM.. if you want send to Melrna2001 at yahoo

AdamLein - My thoughts are this is a Celestial sphere. If that is the case I am only guessing that this problem is in the southern hemisphere and with that is how I guess my answers. Since I am a pilot I somewhat understand the concept but never had any practical applications in doing this kind of problem. If I am right in any of my answer I would be pleasantly surprise.
How I figure it out in my head that it is in the southern hemisphere is the way the arc is drawn HD in relation to what I guess is the equator.
X is the center of the earth. (three lines on a circle defines the center)
M had me puzzled and I am guessing it is sighting of some sort. The line through M with no points I have no idea except to maybe show Declination.
So now that I have made a fool of myself, I would love to see the real answer. For me it was a great brain teaser.
Melissa
PS the other threads of Quizzes here in seamanship, I just now saw and read. Great job guys. Love this stuff. Rep points for all.

Melissa Renee
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Old 02-09-2010
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Ah well I am guessing that this is some sort of projection of the celestial sphere as well. I agree that M is the body being sighted but my suspicion is that D is the north celestial pole and C is the zenith, in which case MCD is the celestial triangle as you suggested.

However I get stuck because I think this diagram is relying on conventions that are unknown to me. I'm just assuming north is D and BF is the equator as you suggested, so the line parallel to BF is the declination which is closer to D, therefore north. I'm not sure what the line parallel to AE would represent.

Since D appears to be above the horizon AE, I'm guessing that the latitude is north.

The arc from D through M to H would be the object's meridian. Since the diagram is flattened and I don't know the conventions by which it was constructed, I cannot guess at the azimuth except that it is certainly not north, since then the arc should go between D and E, rather than between D and C.

Since M is above the horizon AE, M is visible.

Alternatively you could assume that AE is the equator so that the line parallel to AE is the declension. Then BF is the horizon, D is the zenith, and the arc is M's vertical circle. I don't like this interpretation because the thing you measure directly is the altitude, which would be very difficult to plot in this case.

In the first version, CX is the vertical circle orthogonally projected, so you could place M on the line at a distance equal to the sine of the altitude you measured. You could do the declension line similarly.

Flybyknight: I feel like this sort of diagram could be very useful---possibly enabling CN without a calculator---but I don't know the conventions by which it is constructed. Do you have any references that would give a complete description?

s/v Laelia - 1978 Pearson 365 ketch
Old 02-09-2010
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If my vague memory serves me right D cannot be the north pole because of the arc is going through D. The arc is the path of the star or sighted object through the sky. Based on this the north pole has to be either B or F. Because of the way the arc is drawn, my guess it its F. Thus the sighting of M is in the southern hemisphere. Which makes my original answer wrong of the equator. I than change my answer to AE. But if I do that not sure where the horizon is on the model. If AE are both the equator and horizon, does this make the sighting on the equator?

Melissa Renee
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Melissa aced the quiz.
What you are looking at is the result of combining the horizon and the equinoctal coordinate systems in order to diagram the celestial triangle on a spherical plane.
A discussion of Celestial Coordinates can be found in Dutton's Navigation and Piloting, Chapter 19 "Introduction to Celestial Navigation"

Dick

Last edited by Flybyknight; 02-09-2010 at 01:30 PM.
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Old 02-09-2010
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Quote:
Originally Posted by Melrna View Post
If my vague memory serves me right D cannot be the north pole because of the arc is going through D. The arc is the path of the star or sighted object through the sky. Based on this the north pole has to be either B or F.
If the equator is BF as you suggested, how can the north pole be B or F?

Quote:
Because of the way the arc is drawn, my guess it its F. Thus the sighting of M is in the southern hemisphere. Which makes my original answer wrong of the equator. I than change my answer to AE. But if I do that not sure where the horizon is on the model. If AE are both the equator and horizon, does this make the sighting on the equator?
I'm similarly confused and will have to get Dutton's book and figure it out.

s/v Laelia - 1978 Pearson 365 ketch

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