Theoretical Navigation ?

[davidej]

For nav geeks and others interested in angles without a protractor see here:

[

Just remember that one radian is approximately 57 degrees (180/pi) and making you own protractor would be a dooddle.

 

[prv]

Doesn't moss grow on the North side of trees?

I believe that this isn't the best way of indicating north (in our hemisphere).

Indeed. When I was in Cubs, they taught me that satellite dishes face south

Pete

 

[colingr]

If the sun moves across the sky from left to right you are in the N hemisphere, or if from right to left, in the S hemisphere.

This has got me confused

 

[JayBee]

This has got me confused

Hi jokerboat

Due to the direction of the Earth's rotation all heavenly bodies (except circumpolar ones, which are always above or below the horizon) are seen to rise above the eastern horizon and set below the western horizon. If you are in the northern hemisphere, facing south, the sun will rise in the east on your left hand and move to the right across the sky to set on your right hand. The reverse is true in the southern hemisphere when facing north.

For the 'left and right' thing to make sense you need to be facing the sun's path across the sky and not have your back to it. A special case arises in the tropics when the sun can be north of an observer in the northern hemisphere and vice versa.

OK?

 

[KellysEye]

If you can see the Southern Cross low in the sky you are south of south Florida. The higher it is the further south you are.

 

[Aeolus]

Silly question time.

If I know the time (UT) and date and have a sextant, can I

(a) work out my longitude by measuring the angle of the sun and detecting local midday when it is at its highest point

(b) then estimate my latitude by taking the angle and adjusting it for the time of year.

Is this remotely feasible and how accurate could it be using no tables.

 

[JayBee]

Silly question time.

If I know the time (UT) and date and have a sextant, can I

(a) work out my longitude by measuring the angle of the sun and detecting local midday when it is at its highest point


(b) then estimate my latitude by taking the angle and adjusting it for the time of year.

Is this remotely feasible and how accurate could it be using no tables.

(a) Yes, but you need to know the greenwich hour angle of the sun (equivalent to its longitude)

(b) Yes, but you need to know the declination of the sun (equivalent to its latitude)

The arithmetic is simple, but you need a nautical almanac (or other ephemeris) for the sun's position.

The accuracy of (a) depends on the accuracy of the time
The accuracy of both (a) and (b) also depends on the accuracy of your sextant angle measurements and it helps to be stationary.

 

[bedouin]

Silly question time.

If I know the time (UT) and date and have a sextant, can I

(a) work out my longitude by measuring the angle of the sun and detecting local midday when it is at its highest point

(b) then estimate my latitude by taking the angle and adjusting it for the time of year.

Is this remotely feasible and how accurate could it be using no tables.

To get any reasonable accuracy you would need to take account of the "Equation of Time". I think my sextant has a table for this stuck to the inside of the lid so if you have the sextant and it's case you can do pretty well without the tables.

 

[AntarcticPilot]

Silly question time.

If I know the time (UT) and date and have a sextant, can I

(a) work out my longitude by measuring the angle of the sun and detecting local midday when it is at its highest point

(b) then estimate my latitude by taking the angle and adjusting it for the time of year.

Is this remotely feasible and how accurate could it be using no tables.

The first bit is resonably accurate; it is the traditional way of computing longitude using a noon site. Problem is in determining the exact moment of maximum altitude of the Sun. Easier to do the meridian passage by equal altitudes (note the time when you take an altitude of the sun an hour or so before midday, then the time when it reaches the same altitude in the afternoon, and then the meridian passage is midway between the two)

Second bit is less accurate without tables; I suppose you could estimate the Sun's declination by knowing it is nominally 0 at the equinoxes and approximately +/- 23.5 at the solstices, and guessing at refraction corrections. I suppose you'd get it to a degree. But the elevation of Polaris is probably more accurate in the absence of tables, especially if you took a series of elevations during the night and plotted them out to estimate the centre; Polaris (or the Southern Cross) isn't exactly at the celestial pole.

Incidentally, if you DID have tables, you could get longitude by a very ancient and exact method - use the Galilean Moons of Jupiter, which should be just about visible in the star telescope of a sextant, or decent binoculars. Some VERY sharp eyed people can see them with the naked eye, but that is rare. It doesn't work very well on a mobile platform because it is difficult to observe the position of the moons. But on a nice, stable island, it ought to work pretty well!

 

[JayBee]

To get any reasonable accuracy you would need to take account of the "Equation of Time". I think my sextant has a table for this stuck to the inside of the lid so if you have the sextant and it's case you can do pretty well without the tables.

If you are using a table inside the lid then surely you are not doing without the table(s). Sorry to be pedantic.

 

[bedouin]

The first bit is resonably accurate; it is the traditional way of computing longitude using a noon site. Problem is in determining the exact moment of maximum altitude of the Sun. Easier to do the meridian passage by equal altitudes (note the time when you take an altitude of the sun an hour or so before midday, then the time when it reaches the same altitude in the afternoon, and then the meridian passage is midway between the two)

But without taking account of the "Equation of Time" you could be 20 mins out - or 5 degrees of longitude

 

[onesea]

Is this not something to do with a figure on the plane of a rational horizon?

If you are smart enough with a piece of paper, short length of string and pencil you can work it all out?

 

[VicS]

Imagine waking up on a palm tree lined beach gentle breeze on a small island with no previous knowledge of how you got there.
How do you find out where you are ie North or South Hemisphere, what Lat and Long approx. What continent you are on or near, what the time is. What methods would you use to acertain where you are. Come on all you No.6's do your stuff or the big bubble will come and get you.

Please show all working out !!

(was hell of an Xmas do !)

Nigel

Go to the beach kiosk

Buy a picture post card.
That'll tell you where you are.

 

[nigelmercier]

I believe that this [tree moss] isn't the best way of indicating north (in our hemisphere).

It was meant as a joke. However, it was taken from a comedy radio series, of someone in a similar situation; perhaps actually on a boat! I can't remember what it was though.

 

[Uricanejack]

You can get time from a lunar distance measurement to within about 15-30 seconds if you know the approximate date (to within a week or so). 15 second precision needs the measurement of the lunar distance to within 0'.1 and a very accurate almanac. Both of these requirements are tricky, but not impossible, to satisfy.

With approximate time you can get latitude from two (simultaneous) altitudes (this gives two solutions for latitude, but a later repeat of the process will resolve it to one)

With latitude you can get longitude from one time sight (same comment re: two solutions).

I thought the whole point was you don't have compas aliminacs sextants watches or anything else to make it easy