Another mathematical genius required!

it’s quite interesting to know how to get an approximate position offshore, with the absence of anything visible to take a bearing fix on. This subject gets touched on in the RYA yacht master theory course.

I still prefer navionics.
 
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One minute of latitude is one nautical mile irrespective of latitude. However one minute of longitude is 1 nautical mile at the equator but zero distance at the poles. One might expect at 45 degrees latitude one minute of longitude to be 0.5 nautical mile
I was thinking that at first, but of course it is not so.

Two lines of latitude are not an isosceles triangle because the two sides are curved.

A formula for determining must include a squared value or Pi.

Perhaps you can find it from this:

upload_2018-12-8_12-52-56.png
 
Thanks again, Unfortunately do not have sufficient knowledge of maths to understand the above, this is why I need to ask for help on here.

However I started this thread doubting that the very simple method described in a book I am studying for creating your own plotting sheet might not produce the correct scale. I am now pretty sure it does.

I found an online right triangle solver and entered degrees of latitude in the bottom acute angle and 1 for the length along the bottom. I entered 10, 20, 30 ,40, 50, 60, 70 and 80 degrees (latitude) and always 1 along the bottom. In this case the length opposite the right angle represents the latitude scale with the longitude scale always 1. I then converted all the answers into fractions.

I then went to the Latitude/Longitude converter and entered the same latitudes and compared the answers obtained above and they were very similar.

The only thing I did realise is the Lat/Long converter takes into account the Earth is an oblate spheroid and uses the actual measurement of a nautical mile obtained by subtending 1 minute of latitude to the Earth surface, with the Earth being a bit fatter around the Equator the length of 1 minute subtended near the Equator is longer than near the poles. Instead of using the internationally accepted length of a nautical mile of 1852 metres. This probably accounts for the slight differences when I compared the answers.

Mike
 
it’s quite interesting to know how to get an approximate position offshore, with the absence of anything visible to take a bearing fix on. This subject gets touched on in the RYA yacht master theory course.

I still prefer navionics.
Hi motorbiking, When you say there is an absence of anything visible to get a fix from I am intrigued to know how this is done?.

Mike
 
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I mean anything visible on a chart.. Obviously to navigate by celestial objects, you need a clear sky.

The normal approach to getting an estimate of your position when coastal sailing, is to take a bearing from a visible object using a handheld compass, convert that to true (magnetic north vs true north etc) and draw a line on your chart from the known object on your chart. You then repeat this with another object (and ideally another) and where the lines cross is basically where you are. You would then write the time down and repeat the exercise at intervals to plot your course.

If you cannot see anything, because you are in the middle of the ocean, then you need a sextant, a watch, etc..
see how to do it here: http://www.ncsail.info/pdf/celestial/celnavsample.pdf
 
I found an online right triangle solver and entered degrees of latitude in the bottom acute angle and 1 for the length along the bottom. I entered 10, 20, 30 ,40, 50, 60, 70 and 80 degrees (latitude) and always 1 along the bottom. In this case the length opposite the right angle represents the latitude scale with the longitude scale always 1. I then converted all the answers into fractions.
Have you not got that the wrong way round?

The latitude lines are always the same distance apart (ignoring the oblateness of the Earth), that is one degree is 1/360 of the circumference of the Earth; it is the longtidude lines which vary in distance apart depending on the latitude of your postion.

Taking the circumference of the Earth at the equator as 40,075km. (40,008 around the poles) then one degree of latitude is 111.32km. This is the same as one degree of latitude at the equator but which reduces to zero at the poles.

To determine the distance apart of one degree of latitude at a given latitude you multiply that distance, 111.32km by the Cosine of the latitude angle.

This can be seen from my diagram above, where Cr is always Er x Cosine of L. (I called it C because it is a Chord of the circle)
Therefore the same applies to the circumferences of the two.
 
Google can be useful... Probably !

"The Universal Plotting Sheet" I think this is different from other suggestions.


"The Practical Navigator"
https://www.practicalnavigator.org/navigation-workbooks.html And lots more (for free - voluntary donations appreciated).

The latter part of this 'page' may be of interest :-
http://www.simon-bradshaw.com/downloads/

Navsoft downloads http://navsoft.com/downloads.html The 'L' is missing from address given by Bradshaw htm should be html, this works ok. It is said, 'some servers don't accept 4 character extensions' - relevant ? probably.

Better late than never !!

Should have stayed clear, saw these https://www.celestaire.com/product/fujinon-techno-stabi/
Know the folly of higher powered glasses - hand tremors and platform motion... Time to tip the wallet upside down !

-0-
 
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Heretic! :eek:

The earth is flat - many flat earthers also tend to be fat earthers. Perhaps there is a connection? :mrgreen:

I've never understood flat earthers saying that you can't see the curvature of the earth when you're in an aeroplane, I'm sure I can see a slight curve when a cruising altitude.
 
I've never understood flat earthers saying that you can't see the curvature of the earth when you're in an aeroplane, I'm sure I can see a slight curve when a cruising altitude.

Optical illusion, next. Lol
 
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