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  ARTIFICIAL HORIZON MATH
Posted by: pabrides - 02-10-2021, 11:13 AM - Forum: The Sight Reduction process - Replies (4)

It's probably an old discussion, but I still dont have this clear in my head

Of course we all know that using an artificial horizon doubles the actual elevation of a body which at some point requires division by two.

The worksheets show that when taking a shot with an artificial horizon one must add or subtract index error to hs before dividing.

Im sorry but this needs explaining.  Because index error is like a constant it does not change with elevation so why add it to hs before an AH division?  The index error is then halved.   Seems to me all corrections to hs should be made after division, that means altitude correction and index error.

Stupid me just doesnt see how index error can double just because the height is doubled while using an AH.

Lets just say your index error is 20 degrees on the arc and the AH elevation is 60 degrees. So according to the work sheet you subtract 20 from 60 then divide by 2 which equals 20 degrees elevation. Dividing first then subtracting index error equals 10 degrees elevation. You can see there is a difference although index error is not likely to be this bad.

I dont currently have any ie in my sextant so it's difficult to experiment. Someone with two sextants, one way off, might have an opportunity to check out the math with observations.

Cheers

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  AH HA MOMENT
Posted by: pabrides - 02-10-2021, 04:58 AM - Forum: The Sight Reduction process - No Replies

I was debating whether to call this post ah ha moment or IM SO STUPID.

Its been raining for days around here so when the clouds took a break yesterday I quickly grabbed the sextant, a plate of vegetable oil, a watch, paper and pencil, and hit the back yard.

It only took a moment to set up and I didnt have much time; the clouds were closing in again and already dimming the sun.  it was necessary to reduce the sextant shades.

I took the shot between ripples in the reflection and was satisfied I got a reasonable single shot. I immediately recorded HS and time and went inside.

I then turned on my short wave radio and dialed in WWV.  My watch proved 2 seconds fast.  No problem, I added the notation to the sight record.  I put everything aside and finished my laundry.

Later I put the sun shot to work.  I gathered my reduction stuff and sat down to work the problem.

I use a small plasticized card I developed which very briefly outlines the reduction steps.  I dont use a proforma perse,  but apply the reductiom math to any old scratch paper or notebook. I save tons of paper that way.  My reductions use as little as half a page in a pocket size notebook.

So off I went on my merry way doing the reduction then plotting Zn and the intercept...  Hmmmmm

Something wasnt right.

I've done this a hundred times, but my lop was short about 8 miles.  I began to doubt my shot and went over it in my head.  I redid the math, rechecked gmt, dec, d correction, hc, everything. I took the reduction to the dinner table - I still couldnt figure it out. I put the reduction away and turned on the big screen.

I started watching the movie CONVOY, that great old trucking classic, which moved me into a sense of persistance.  I hit the pause button and started reworking the reduction. - then it hit me... I'm sooooo stupid.

So here is the confession -  there are two ways to get the first difference for Hc.  You can do the math, dec increments/ 60 X d correction factor, or use table 5.  So I mistakenly did two thing wrong complicating the error.  I divided 60 by the increments,  then I multiplied by table 5 result.  I slapped myself in the head - hard.  24 hours later I'm still asking myself why I did that.

It literally took hours to find the error and correctly reduce the shot, but the result was absolutely satisfying.  The lop was almost spot on which boosted my confidence in taking sun shots.

I have the script writer of CONVOY to thank for instilling in me the desire to persevere.   I conquered my own ignorance and lived to tell about it.

Cheers

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  THINK ABOUT IT
Posted by: pabrides - 02-07-2021, 02:40 PM - Forum: General Topics Here - Replies (12)

If I disoriented you beyond doubt, so that you didnt know your location on earth, and if I dropped you in some ocean;

1... could you determine your approximate location, say within a few hundred miles, given a raft full of non-electronic navigation equipment and publications?

2... If yes, how would you go about it?

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  Apparent time
Posted by: Rumata - 02-05-2021, 03:18 AM - Forum: General Topics Here - Replies (3)

Gentlemen, hope  you  can help  me in  this  little  "predicament."

At  first ,  a few  general  terms:
 
1. Apparent time is a  measure of  the  sun'  true  position.  To  get  Greenwich  apparent  time at  any  instant  we just  convert GHA  of the  sun  into time (hh/min/ec).,  We  need  apparent  time  to  find  when  the  sun  will  be  on  the  observer's  meridian. Local  apparent  noone,  when  the  observations  for  latitude  are  made. We  know  that  it is  different  for  each meridian,  and  unless two  places  are  on  the  same  meridian- apparent time  is  different.

2. On  the  Mercator  chart  every  parallel  is expanded in  the  ration of  the secant of  its latitude.
To  avoid  local  distortion every  meridian  of  a  Mercator  chart must also  be  expaned  in  the  vicinity of  each  parallel by  multiplying  it by  the secantr  of  the  latitude of  the  parallel.

My  question:

When  the  vessel  is  located  above let's say  50 deg Lat. and  we are  trying  to find  LAN the  natural  curvature  of  meridians  would  make  GHA slightly  shorter  than  Mercator  chart  projections (  straight,  parallel  lines). And  it  means  LAN  calculated  would  be  different  from  LAN  real ,  considering expansion  ratio  of scant  of  the  latitude.  And  it means  the  moment  we  are catching  the  sun  in a sextant  is not exactly  the time  the  sun  is in Zenith.  Difference  should  not  be  significant,  but  I just  want  your  expert  opinion about  it.

Please,  advice.

Thank  you

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  Curious old navy math techniques.
Posted by: Rumata - 02-03-2021, 02:48 AM - Forum: General Topics Here - Replies (11)

Greetings,  gentlemen.  I found a  couple  of interesting  old  books  which,  considering  lockdowns  and  general inability to  cross   a street  without  damn mask,  may  help  to develop  not  very  useful  skills.  But  a  lot  of  people  have  used them  for  sure. One  of them  Mathematics for  navigators  by  Delwyn  Hyatt.  Yes, 1944. Of  course  it  is  kind  of  obsolete. But please  consider  the  fact it  was taught during  WWII to officers  who would  serve  in  the  real situations,  not in a  front of  a  simulator.  In  many  cases  its  methods  are kinda "not  straight  forward" to  say least.  But  some are  really  good.  For  instance,  Adding  and  subtracting  figures  from  left  to  right,  they  way  we write  them. I  tried  it  and  after  several  attempts  found  out  that  it is much  faster  than  traditional  ones. Also,  havesine  and other  formulas  presented  in  quite  different,  again,  not  easy  for  me to  understand,  wording  and  form.  Anyway,  real  navy  folks  really  used  it.  So,  as  a  due respect tio  their  work,  let's  try  to  understand their  daily  routines  doing  it.  Not  easy,  my  opinion.

Another one, Navigation and NAutical  Astronomy,  by  Dutton,  1943.  Again, thyere  are  a lot  of  info,  which  probably  none  of  us  would ever  use,  but  there are  also a  lot  of proves  and  explanations,  which  we  take  for granted  without  even  thinking  where did  it  come  from.  Moder editions  of  Dutton   don't  have  some  of  it.  I  consider  it  as of  a  hystorical   value.

I  am  always  fascinated  by  the  old methods  of  navigation, when Energizer'  Bunny  wasn't  even  conceived  yet.  No  batteries,  cloudy  skies,  rolling  horizon  and wet  and  misery  all  around.  And  they  did  it. Years ago  we  were taught  to  use sliding  rules.  It  gives  you  anything,  up  to  3  decimal.  Recently  saw  it  in  the  antique  shop  for  $271.  And  only  then  I  realized  how  ancient am  I.

Thank  you.

Respectfully,

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  compass rose revised
Posted by: pabrides - 01-31-2021, 10:12 AM - Forum: General Topics Here - Replies (2)

Hello Shipmates,

I've been playing around trying to learn the old compass rose.  I found out a few things and developed some memory aids. Drawing my own compass helped tremendously. See attachment.  You will note the intersecting archs around the perimeter of the compass - these are drawn to divide the circle into points.  

Here are a few more memory aids.

The 4 large points contain a single letter, N, S, E, W (cardinal points)

The 4 secondary points contain two letters combining the cardinal points. NE, SE, SW, NW

The 8 smaller tertiary points have 3 letters combining the cardinal and secondary points. NNE, ENE, ESE, SSE, SSW, WSW, WNW, NNW.

The 16 smallest points all contain the word BY: North By East, NEBN, NEBE, EBN... etc.  These refer to the cardinal and secondary points only - the tertiary points are not used in reference to "by" points.  

A cardinal letter always follows the word BY - South West by West... etc.  There is no such heading as North by North West or South West by West South West.  Also I don't think there is such a heading as North by West and a quarter North... THAT point, rather, might be given as North and three quarters West.

Just playing with a quadrant at a time lessens the confusion.   For example,  just practice with the North East quadrant first and discover how it makes sense.  The others follow the same pattern. 

Draw your own rose with paper and a draftsman compass, you can't do that with a circle of 360 degress ... try drawing half of 45 and you'll get a headache; dividing a sector into 90 equal parts is no picknic. The old compass rose has a total of 32 points with the addition of quarter marks.  very easy to make.

  If the captain ordered North and a quarter East the helmsman would stear to the first tickmark to the right of North.   If North East and a half North was ordered the wheelman steared to the second mark to the left of NE... etc.  I was too lazy to draw every quarter mark.

Next time you're bored give the old compass points a try...  you'll soon realize it has some merit over the 360 degree system....  Who needs 360 degrees anyway.  You'll just wear yourself and the stearing gear out trying to keep that precise.  Might as well make the compass 720 degrees... lol.  

I own some pocket compasses that only show tick marks at 5 degree intervals - thats only 72 compass points.  Might as well have printed the old rose and made orienteering some fun.

The really interesting thing here is that after a while you won't need written point letters to tell you what point you're looking at.   As long as you know which large point is North you'll know the others, and North may be identified with color rather than a letter from the alphabet.  I bet that even before you begin studying the old compass points you could close your eyes and from the very start imagine 8 of them;  North,  North East,  East,  South East,  South,  South West,  West,  North West.  That's already a fourth of the points. Think a bit harder and imagine the points between these such as North North East... etc.  That's half the points.   

Admittedly,  the toughest points for me to learn were the BY points such as North by East, but adding a little relative bearing grease to my brain quickly put those points in order.  

The quarter marks were almost a no-brainer.  Just remember that the BY points have no quarter.

Have fun and keep at it - that's how tradition survives modernity.

Cheers



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  diy dividers
Posted by: pabrides - 01-27-2021, 08:40 AM - Forum: General Topics Here - No Replies

Just finishing making some dividers out of old brass flat stock I've had for 10 years.

I hand sawed the brass into long wedges, used a small file to get it in shape, used some old aviation hardware with a plastic washer between the wedges, and presto... damn near perfect dividers - they work great.

Generally store bought brass dividers come in 7 and 8 inch lengths.  I couldn't decide which I might want so I made mine 7.5 inches.

Just goes to show a navigator doesnt need to spend gobs of cash to have some tools.

Next - maybe a plotting triangle from old plastic.

BTW -  the photo shows where I live in relation to the last great sea battle of ww2.  The upper divider point is on my house. The lower divider length shows the southern route the Japanese navy used to attack the America forces at Leyte.  The pencil is on the Surigao straights where the battle took place.  The Surigao southern entrance is 88 nm from my house.  Japanese lost and the surviving ships went back the same way they came.
Leyte is where general MacArthur returned to the Philippines.

My island is where Magellan was killed... lots of history here.

Cheers



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  new sun sight
Posted by: pabrides - 01-16-2021, 02:07 AM - Forum: The Sight Reduction process - No Replies

GMT jan 15 2021, 06 23 56, Ho 41 33.5

My fix is exactly 10 nm SE along LOP 3. 

I took 3 AH sightings and averaged them.  Learning to average hexadecimal was a bit tricky, but I found that making the hours of time or sextant altitude equal helps.

For example: 
83 10.7  =  82 70.7 - adding 1 degree to min.
82 33.7  =  82 33.7
82 05.0  =  82 05.0

So 82 is the obvious average for degrees and what remains is to average the minutes:  109.4 ÷ 3 = 36.5

Averaging time was a bit more difficult :

06 22 30  =  06 22   30
06 24 12  =  06 22 132  - adding 2 min to seconds
06 25 26  =  06 22 206   - adding 3 min to seconds
 so,  368 ÷ 3 = 122.7.  Move 120 seconds back to minutes with a remainder of 2.7 seconds:  06 24 03. (My watch was 7 sec fast so actual time is on the first post line above)

Of course one could change each shot to a decimal value, add together, divide by 3, then change back to the hexadecimal,  but that takes a calculater and lots of button pushing, given your calculator has a button to do this.

I'd be interested to learn how you swabbies average shots.  I tried 4 or 5 different ways to average shots all resulting in different values.  The way I've shown works best for me and put my LOP right on my position.  See LOP 3 on the worksheet.

The arithmetic mean works great if the shots are spaced near equally: (a+b+c)÷3= mean average. Simple.
But if shots are delayed due to clouds or such a better solution might be the geometric mean: a×b×c =d then take the cubed root of d. If you have 5 shots then multiply and take the 5th root... etc. If a data set has flyers that skew the average -such as in the set 2.4, 2.3, 2.6, 2.1, 2.2, 2.4, 2.3, 7.5, 5.2 - You can see that the 7.5 and 5.2 will skew the average of two somethings out of proportion. The geometric method helps reduce the skew to something more reasonable.

2.4+2.3+2.6...+5.2=29.0, 29.0÷9=3.22
2.4×2.3×2.6...×5.2= 14, 274.4, 9th root of 14, 274.4 = 2.89

Without the big numbers - 2.4+2.3+2.6+2.1+2.2+2.4+2.3 = 16.3, 16.3÷7=2.33
_________________________ - 2.4×2.3×2.6×2.1×2.2×2.4×2.3 = 366, 7th root of 366 = 2.32



Cheers



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  Navigation and other Marine Applications for SRs
Posted by: pabrides - 01-04-2021, 05:27 AM - Forum: General Topics Here - Replies (3)

I'm not sure how many here remember the slide rule (SR).  if you were in high school in the 60s then you would probably have some experience.

I went to HS in the 70s and never heard of the SR.  Only 10 years ago did I discover them as a source of enjoyment.  it feels pretty cool to know how to use some of the scales, and the more I use them the more I like SRs.  Just yesterday I bought a full sized 10in SR used from a guy in manila.  My new toy will be here in 2 days.  Up until now I've been using a 5in model.  I can hardly wait.

Anyway, a couple years ago I found this PDF book about how to use SRs for marine and aviation problems; from simple time, speed, distance calculations to cargo handling and trim.  I wrote about it here at the time.  However, maybe some of you missed that post.  Just goes to show I'm still excited about the subject.

I'm thinking some of you might like to review the book, a free download, and discover how fascinating and ingenuous the SR really is.  You don't need to have the same one they use in their example; any SR with similar basic scales will do.

http://www.survivorlibrary.com/library/t...n_1974.pdf

This site is put together by a very nice man who assembled a collection of old books that keep my mind occupied.

Check it out and tell me if you don't agree that SRs are still useful... and they don't require batteries.

Cheers



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  Dead Reckoning
Posted by: pabrides - 12-29-2020, 06:32 AM - Forum: General Topics Here - Replies (3)

I was playing with some math today figuring out the best way to use my E6-B flight computer for marine applications.   Somehow my mind went to log lines and sand timers. 

I got the idea that with todays watches, all of which incorporate a second hand, working a log contrary to the old way, which took counting knots to determine speed after a certain time, typically 28 or 30 seconds, a predetermined length of log line might easily be timed to determine speed. 

The following is a list reflecting time in seconds vs speed in knots for a log line extending 60ft:

1 kt   35.5 sec
2       17.8
3       11.8
4       8.9
5       7.1
6       5.9
7       5.1
8       4.4

So you take a fishing line 60 feet long on a cuban rig fishing spool.   Attach a log device to throw into the water which pulls the line.  Throw the log into the sea and start your timer.  When the line pulls taught stop the timer.  Determine your speed over the water using the chart above.  If the spool is held any more than 4 or 5 feet off the water accuracy may suffer.  With high cockpits a longer line would be better; just refigure the knots vs time table.

The math is thus:

1knot = 6076ft ÷ 3600s = 1.6878ft/s

You could have a log line 6076 feet long and spool it off for 3600 seconds which would tell you your speed is 1 knot.  But there's a better, faster way.

60ft of line ÷ multiples of 1.6878 gives time in seconds

Ex: 
60 ÷ 1.6878 = 35.5s for 1 knot
60 ÷ (2 × 1.6878) = 17.8s for 2 knots ...  etc.

You can help me out here and let me know if I got the figures wrong.  Any length of line would do beyond a certain minimum.   I took 60 because in an emergency 10 arms spread would about equal 60ft.  You can kinda sense here that line length doesnt get critical until you reach the higher speeds.

Have fun and give it a try next time you're out on the lake and dont forget to tell us what happened.

Cheers

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