 Quick Menu


Topic Review (Newest First)

01212013 01:39 PM 
MarkCK 
Re: sextant practice
I think I am heading in the right direction on this one. Now on to step 2...

01212013 11:48 AM 
AdamLein 
Re: sextant practice
Quote:
Originally Posted by MarkCK
I guess I am still confused on the minutes. It appears a lot of the ticks on the vernier scale coincide with ticks on the arc.

I agree. Let's consider the image that's at 26°andsomething (attached). I've marked two vernier ticks with red and blue marks. The redmarked tick (6') is clearly too far to the left of the nearest arc tick, and the bluemarked tick (16') is clearly too far to the right of the nearest arc tick. That means that the correct vernier tick must be someplace in between those two, and if your tolerance for "too far to the right/left" is the same in both directions, then the correct tick is probably close to halfway between the red and blue marks (11').
You could just call it 11', but I think that the 12' tick really looks exact, whereas it's harder to judge for the 10' tick. Vernier scales work by relying on the nature of human visual perception, so to some extent you need to rely on a feeling of "clicking" when you see two ticks that line up, and "not quite clicking" when you look at ticks that don't.
I'll just mention the "divergence" issue again. Around 44', the ticks are obviously way, way off (around halfway between the nearest ticks on the arc). As you look to the left, for a while the ticks get closer, but then they get farther away again, and until 16' I think are obviously not close at all. From 6' to 16', on the other, the ticks are all quite close. Because of how the scale is designed, that's always the way it is: you never need to search the entire vernier scale for the right one. The right one will never be right next to a really really wrong one.
The ticks do look like the converge again around 58', but: the 0' tick and 60' tick are always offset from the nearest arc tick by the same amount (again due to the design of the scale). So if the 0' tick is a bit too far to the left, and the ticks converge as you go to the right from 0', then so is the 60' tick.

01212013 11:18 AM 
MarkCK 
Re: sextant practice
I guess I am still confused on the minutes. It appears a lot of the ticks on the vernier scale coincide with ticks on the arc.

01212013 02:39 AM 
AdamLein 
Re: sextant practice
Quote:
Originally Posted by MarkCK
1) Where do I place the sun on the artificial horizon.

For celestial bodies that have visible discs, the Almanac lists coordinates for the center of the disc. However, it's hard to line up the center of the sun with the horizon when you do a real sight; much easier to line up with the lower limb (edge) and then add half the sun's apparent diameter (listed in the almanac).
When using an artificial horizon, there is no horizon and you line up the sun with its reflection exactly. So the centers of the images lines up and you can use the Alamanac values directly, except...
Quote:
2) Once you have your angle from the artificial horizon do I divide it by 2 to get the angle that I use for my calculation?

Almost. You never assume you sextant is perfectly accurate; after you have removed as much error as you can, there might be some left. This remaining error is called "index error", and you need to measure it before you can use your sights. To do this you usually sight the horizon, or another distant object, and line it up with itself. Then read what the sextant says: that's the index error. This value must be subtracted from the sextant reading before any other operations are performed.
E.g. if you check the horizon against itself and it says +5' (i.e. "on the arc"), then you have to subtract 5' from your sights. "If it's on, take it off".
The index error is an error "in the sextant", so it's the first correction you make.
For artificial horizon sights, you then have to divide by 2, after taking off the index error.
Quote:
3) I am having a little trouble figuring out what my exact angle is with this sextant. I think I can figure out degrees easy enough. But what about minutes on the vernier.

I won't give a detailed explanation of vernier scales here; yours is thankfully straightforward (I have a calipers with a vernier scale indicating 128ths of an inch... a bit more confusing).
So, the 0 mark on the vernier gives you the number of degrees. Remember to look to the right of the zero to get the number, if the zero falls between two ticks on the arc. Then, to get minutes, you find the tick on the vernier scale that exactly lines up with a tick on the arc; the label on the vernier scale gives the number of minutes.
I'd put those measurements as
1) blurry, can't read
2) 64°00'
3) 83°00'
4) 26°12'
5) 52°58'
One thing that I find helps with vernier scales is to look "past" the tick you think is the right one, in both directions. For me, it's easier to see the ticks "diverge" from the most exact one, and then you can sort of follow it back to the "center" or "least diverged". Often you'll find that two ticks on the vernier scale are "inside" two ticks on the arc; when you see that, your answer is halfway between the vernier ticks.

01202013 01:26 PM 
MarkCK 
Re: sextant practice
I received my Mark 3 sextant on Friday and have been doing a little practice. Most of the time I can place the sun wherever I want it. So my first question is
1) Where do I place the sun on the artificial horizon. Am I supposed to line the sun up with the sun reflection or horizon? Or do I put the bottom tip on the sun on reflection. Or do I... The instruction were kind of vague.
2) Once you have your angle from the artificial horizon do I divide it by 2 to get the angle that I use for my calculation?
3) I am having a little trouble figuring out what my exact angle is with this sextant. I think I can figure out degrees easy enough. But what about minutes on the vernier. I have attached a few photos of some random angles. I am kind of a learning by doing type person so can someone tell me what the degrees and minutes are. I am sure this seems kind of grade schoolish but I am a visual type learner. Hopefully it will seem elementary once I have a good explanation.

01162013 11:26 PM 
Flybyknight 
Re: sextant practice
Quote:
Originally Posted by blutoyz
This looks like something cool to learn. I think that this will be on my list this year but I do think that it will take a while. I really don't get how you get latitude AND longitude from a fix but I guess that will come.

You select a convenient point on a chart, like the intersection of Lat & Lon lines, solve the Navigational Triangle, then your answer will be X distance at Y degrees from your selected point.
3 solutions above will give you a tiny triangle. The center of that triangle will be your fix.
The fix point IS your Lat & Lon.
Dick

01162013 10:51 AM 
blutoyz 
Re: sextant practice
This looks like something cool to learn. I think that this will be on my list this year but I do think that it will take a while. I really don't get how you get latitude AND longitude from a fix but I guess that will come.

01152013 03:09 AM 
Flybyknight 
Re: sextant practice
DIP SHORT of the HORIZON
Sometimes it may be necessary to view a celestial object even tho we
do not have a visible horizon due to an obstruction in our way, be it
land or another vessel. Under these conditions if we know how far (in
Nautical Miles) we are from the obstruction we can calculate the dip
short of the horizon using the waterline of the obstruction as our
horizontal reference point.
Ds = 0.416d + ( 0.566 * h/d )
Ds = Dip short of the sea horizon, in minutes of arc.
d = distance to the water line of the obstruction, in Nautical Miles.
h = height of the observers eye above sea level, in feet.
This formula is an adaption of the one given in Bowditch and works
quite well as long as the height of the eye is not great or the
obstruction is not very close. Using the above formula if the height
of the observers eye was 100 feet and the obstruction was 0.1 N.M.
away the dip would work out to be 566.'0, verses 565.'8 using the
(Bowditch) long method.
EXAMPLE:
Your eye height is 24 feet and the distance to the obstruction is
0.75 N.M. What is the DIP SHORT of the HORIZON?
Ds = (0.416 * d) + ( 0.566 * h/d )
Ds = (0.416 * 0.75) + ( 0.566 * (24 / 0.75) )
Ds = 0.312 + (0.566 * 32)
Ds = 0.312 + 18.112
Ds = () 18.424
Thus the DIP SHORT of the HORIZON is () 18.'4 when correct to the
nearest tenth of a minute.
Remember the dip is always subtracted from the sextant altitude.

01152013 01:20 AM 
MarkCK 
Re: sextant practice
Quote:
Originally Posted by Flybyknight
Two methods will serve you better:
Dip Short of the horizon,
or
a pan of very dirty motor oil and sit or stand so that the body observed will be reflected on the liquid.
There is no dip correction for the artificial horizon.
Dick

I am still very new at this. What do you mean when you say "dip short of the horizon"?

01152013 01:18 AM 
MarkCK 
Re: sextant practice
Quote:
Originally Posted by chamonix
+1 on this. Liked the video's so much I thought " what the hell ", and ordered the book he was pushing on the video's. It's the best thing I did as far as learning celestial navigation. Had read two other books on celestial navigation by the time the book arrived, but the whole process was still rather murky in my mind. His book was the resource that made everything clear. This is the secound time I've posted on this. Sorry if it looks like I'm shilling for the guy but I just can't believe what a great resource it is and how unheralded it is.

I watched a couple of his video's but which book was he pushing? I am sure I heard it at some point...

Posting Rules

You may post new threads
You may post replies
You may post attachments
You may edit your posts
HTML code is On



All times are GMT 4. The time now is 04:15 PM.
