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luzulaFor years I have longed to try using a sextant--I blame Patrick O'Brian and my teenage interest in astronomy and celestial mechanics for this. Now my dreams have finally been realized! Last fall I held a guest lecture on an evening course in navigation held by an astronomer; my lecture was about the mathematical aspects of navigation (I'm a mathematician, if you don't know). I had to keep it pretty basic since there were no math requirements for the course. I talked about basic trigonometry applications to navigation and also explained the principle of old-fashioned logarithm tables. Anyway, the astronomer lent me one of her sextants, and I have now tried it out.
I got my dad to take me out in his sailing boat, out beyond the islands in the archipelago on the west coast of Sweden so that I had a free horizon to the south. I brought a GPS for independent confirmation of the coordinates. I also checked the time of true noon at our longitude (you can do that here). My aim was just to see how accurately I could determine the latitude.
An explanation of how it works, if you don't already know: very simply stated, the angle between the horizon and the sun at noon = 90° minus the latitude. For example, if you are at the equator (at latitude 0°), the sun is in the zenith (at 90°). This is only true at the time of the equinoxes, though, so if it's not the equinox, you have to compensate for that. Other things to compensate for: the time at which you take your reading (if it's not noon), the sun's radius, your height above the ocean surface, the refraction of the Earth's atmosphere, known errors in the sextant. For the moon, which is a lot closer than the sun, you also have to correct for the parallax, that is, for taking the measurements at the Earth's surface and not its center.
It was harder than I expected to get the sun in view in the sextant. The sun is small and the sky is big! Also the boat rocks and you have to avoid being blinded by the sun. Finally I got the sun in view, and could get a reading. There are sun filters, so you don't actually look straight at the sun. Also, what the sextant actually does is to reflect the image of the sun down to the horizon, and then you read off what angle the mirror is at.
Actual position (from GPS): 57° 36' 30"N, 11° 43' 41"E
Date: 20th of September 2015 (Er, I didn't get around to writing this up until now. Also, it was actually just one day from the fall equinox.)
Time of true noon: 1:06:26 PM local time
Time of my reading: 1:04:40 PM local time (pretty close--there was unfortunately a cloud in front of the sun at true noon)
My reading: 33° 21'
Taken entirely without corrections (since it was close to both the equinox and true noon), this would give a latitude of 90° - 33° 21' = 56° 39'. This is 57.5 minutes from the actual latitude, so in distance it is the same number of nautical miles, that is, 57.5 * 1.852 = 106.5 kilometers or 66.2 US miles. (A nautical mile is what you get if you divide the circumference of the Earth by the number of minutes of arc that make up 360°.) This is not that near. But we also need to correct the reading for various other things:
Correction for the height of my eyes above sea level: The higher I am, the farther away the horizon appears, and the lower it "dips" below the standard 90°-below-the-zenith level. This "dip" is 1.927 * sqrt(h) minutes, where h is my height above sea level in meters. Incidentally, this number is the same as the distance to the horizon in nautical miles. (If I wasn't too lazy to draw and scan pictures I would derive the formula for you--it involves some nice and easy plane trigonometry). Anyway, I was about 1 meter above sea level, so the dip is 2'.
Corrected reading: 33° 21' – 2' = 33° 19'
Correction for refraction: The light is refracted in the Earth's atmosphere, so that the sun appears higher than it is. This refraction gets worse the lower the angle of the sun, so that at sunset we still see the sun when it has "actually" already set. The correction also depends on air temperature and air pressure, but when the sun's angle is more than 15° you can use the formula is R = 0.96/tan(sun's angle) minutes. So for me it would be 0.96/tan(33.3) = 1.5'.
Corrected reading: 33° 19' – 1.5' = 33° 17.5'
Correction for the sun's radius: What I am actually measuring is the angle of the lower edge of the sun to the horizon, but I really want to measure from the center of the sun. So I need to add the radius of the sun in minutes of arc. This varies over the year, from 16.3' in January when the sun is closest, to 15.7' in June when it is farthest away. September is in between, so 16'.
Corrected reading: 33° 17.5' + 16' = 33° 33.5'
Now we have the angle of the sun to the horizon at our time and date. Our goal is now to find the latitude.
Correction for the date and time: If the Earth's axis had an angle of 90° to its orbit, the sun would always be on the celestial equator and we would be done now (if we had measured at true noon). But this is not the case. The sun is farthest away from the celestial equator on the solstices and right on it at the equinoxes. We need the declination of the sun on our date, that is, its height in degrees above the celestial equator. Obviously the sun's height above the horizon also depends on what time it is, since the Earth turns on its axis, so we also need the sun's so-called hour angle.
We can look these things up in the nautical almanac. Put in the date at the bottom and it will give you a table. For 11 AM GMT (=1 PM my time) we read:
declination of sun: + 1° 7.4'
GHA (Greenwich hour angle) of sun: 346° 36.8'
But the hour angle changes quickly (though the declination does not), so we need to correct it for the time in minutes and seconds. In the Table of Increments and Corrections we find that we should add 1° 10'. So:
declination of sun: + 1° 7.4'
GHA of sun: 347° 46.8'
To proceed and determine the latitude, I have to cheat and use my knowledge of the longitude. /o\ If I don't, I will get a circle on the Earth's surface, and I could be anywhere on that circle (the principle is the same as if you measure the height of a tower as seen from a distance--in the absence of any other information, you could be anywhere on a circle with a certain radius from that tower). There is of course a procedure for determining that circle and it's often used in navigation, but I will omit that (this is already tl;dr). If you do it for two celestial bodies, or for the same body at two different times, you get two circles that intersect and you can determine your position.
Let's first pretend that I managed to measure the height at true noon. The reason that Jack Aubrey and other 19th century people (or anyone who can, really) wait until noon to take their readings is that then they don't need an accurate clock nor do they need to know the longitude. We won't need to use the value of GHA from the tables, only the declination. You can tell that it's noon by taking successive readings until you notice the sun sinking again. (In fact the easiest way of determining the longitude is to have an accurate clock set to GMT and then observe the time of noon in that clock. Your noontime in GMT is then your longitude if you convert it from time to degrees, basically.)
Anyway. Height of sun at noon - declination of sun = 90° - latitude. So:
latitude = 90° - height of sun + declination of sun = 90° - 33° 33.5' + 1° 7.4' = 57° 33.9'.
The true latitude is 57° 36.5'. I'm only 2.6' off, that is, 1.852 * 2.6 = 4.8 kilometers. That's pretty good!
Let's see if it gets better when I correct for it not being true noon (now is when I have to cheat and use the known longitude). Time for some spherical trigonometry! We assume that the sphere of the sky has diameter one, so that a distance between two points on the sphere is the same as the angle between those two points seen from the center (all angles are now in radians, though I'm still stating them as ordinary degrees).
Draw a triangle on the sphere of the sky, with the north pole (A), the zenith (B) and the sun's position (C) as corners. Let A, B, C also denote the angles at those points, and a, b, c denote the corresponding opposite sides. Our goal is to solve for side c (the angle between the zenith and the north pole in the sky); the latitude is then 90° - c°.
The following are known:
LHA (local hour angle) = GHA + longitude = 347° 46.8' + 11° 43.7' = 359° 30.5' (this is very near 360° because the sun is very close to its highest point)
A = 360° - LHA = 360° - 359° 30.5' = 29.5'
a = 90° - height of sun = 90° - 33° 33.5' = 56° 26.5'
b = 90° - declination of sun = 90° - 1° 7.4' = 88° 52.6'
The law of sines on a sphere (different from the law of sines on a plane surface) gives us
sin(A)/sin(a) = sin(B)/sin(b).
Solving for the unknown sin(B), we get B = 0.4101° or B = 179.5899°; obviously the second one should be chosen considering the shape of the triangle. Now we can find c by applying the law of cosines on a sphere, which says either of the following:
cos(a) = cos(b)cos(c) + sin(b)sin(c)cos(A)
cos(b) = cos(a)cos(c) + sin(a)sin(c)cos(B)
cos(c) = cos(b)cos(a) + sin(b)sin(a)cos(C).
It would be nice to use the third one since then we'd get c directly, but we don't know C so we can't. (We can't use A and B to figure out C since the angle sum is not 180° like in a plane triangle.) Instead, we use both of the other ones and get a system of equations with cos(c) and sin(c) as unknowns. Solving that system, we get c = 32.4338°, which gives a latitude of 90° - 32.4338° = 57.5662° = 57° 33.97'. The true latitude was 57° 36.5', and the one I calculated assuming the time to be true noon was 57° 33.9'. Ha ha. I suppose the lessons here are 1) it does not make much difference if you are two minutes from true noon, 2) the number of decimals in my calculations are kind of ridiculous considering the accuracy of my reading, 3) who cares, we got to do some fun spherical trigonometry.
I am kind of proud of myself for getting a latitude that was only 2.6' or 4.8 kilometers off, though! \o/
If you read the whole of this post, I salute you. The post is not beta-read (and in fact is hardly alpha-read, since I had to rush off), so tell me if you find errors. Also, ask if you have questions. : )
I got my dad to take me out in his sailing boat, out beyond the islands in the archipelago on the west coast of Sweden so that I had a free horizon to the south. I brought a GPS for independent confirmation of the coordinates. I also checked the time of true noon at our longitude (you can do that here). My aim was just to see how accurately I could determine the latitude.
An explanation of how it works, if you don't already know: very simply stated, the angle between the horizon and the sun at noon = 90° minus the latitude. For example, if you are at the equator (at latitude 0°), the sun is in the zenith (at 90°). This is only true at the time of the equinoxes, though, so if it's not the equinox, you have to compensate for that. Other things to compensate for: the time at which you take your reading (if it's not noon), the sun's radius, your height above the ocean surface, the refraction of the Earth's atmosphere, known errors in the sextant. For the moon, which is a lot closer than the sun, you also have to correct for the parallax, that is, for taking the measurements at the Earth's surface and not its center.
It was harder than I expected to get the sun in view in the sextant. The sun is small and the sky is big! Also the boat rocks and you have to avoid being blinded by the sun. Finally I got the sun in view, and could get a reading. There are sun filters, so you don't actually look straight at the sun. Also, what the sextant actually does is to reflect the image of the sun down to the horizon, and then you read off what angle the mirror is at.
Actual position (from GPS): 57° 36' 30"N, 11° 43' 41"E
Date: 20th of September 2015 (Er, I didn't get around to writing this up until now. Also, it was actually just one day from the fall equinox.)
Time of true noon: 1:06:26 PM local time
Time of my reading: 1:04:40 PM local time (pretty close--there was unfortunately a cloud in front of the sun at true noon)
My reading: 33° 21'
Taken entirely without corrections (since it was close to both the equinox and true noon), this would give a latitude of 90° - 33° 21' = 56° 39'. This is 57.5 minutes from the actual latitude, so in distance it is the same number of nautical miles, that is, 57.5 * 1.852 = 106.5 kilometers or 66.2 US miles. (A nautical mile is what you get if you divide the circumference of the Earth by the number of minutes of arc that make up 360°.) This is not that near. But we also need to correct the reading for various other things:
Correction for the height of my eyes above sea level: The higher I am, the farther away the horizon appears, and the lower it "dips" below the standard 90°-below-the-zenith level. This "dip" is 1.927 * sqrt(h) minutes, where h is my height above sea level in meters. Incidentally, this number is the same as the distance to the horizon in nautical miles. (If I wasn't too lazy to draw and scan pictures I would derive the formula for you--it involves some nice and easy plane trigonometry). Anyway, I was about 1 meter above sea level, so the dip is 2'.
Corrected reading: 33° 21' – 2' = 33° 19'
Correction for refraction: The light is refracted in the Earth's atmosphere, so that the sun appears higher than it is. This refraction gets worse the lower the angle of the sun, so that at sunset we still see the sun when it has "actually" already set. The correction also depends on air temperature and air pressure, but when the sun's angle is more than 15° you can use the formula is R = 0.96/tan(sun's angle) minutes. So for me it would be 0.96/tan(33.3) = 1.5'.
Corrected reading: 33° 19' – 1.5' = 33° 17.5'
Correction for the sun's radius: What I am actually measuring is the angle of the lower edge of the sun to the horizon, but I really want to measure from the center of the sun. So I need to add the radius of the sun in minutes of arc. This varies over the year, from 16.3' in January when the sun is closest, to 15.7' in June when it is farthest away. September is in between, so 16'.
Corrected reading: 33° 17.5' + 16' = 33° 33.5'
Now we have the angle of the sun to the horizon at our time and date. Our goal is now to find the latitude.
Correction for the date and time: If the Earth's axis had an angle of 90° to its orbit, the sun would always be on the celestial equator and we would be done now (if we had measured at true noon). But this is not the case. The sun is farthest away from the celestial equator on the solstices and right on it at the equinoxes. We need the declination of the sun on our date, that is, its height in degrees above the celestial equator. Obviously the sun's height above the horizon also depends on what time it is, since the Earth turns on its axis, so we also need the sun's so-called hour angle.
We can look these things up in the nautical almanac. Put in the date at the bottom and it will give you a table. For 11 AM GMT (=1 PM my time) we read:
declination of sun: + 1° 7.4'
GHA (Greenwich hour angle) of sun: 346° 36.8'
But the hour angle changes quickly (though the declination does not), so we need to correct it for the time in minutes and seconds. In the Table of Increments and Corrections we find that we should add 1° 10'. So:
declination of sun: + 1° 7.4'
GHA of sun: 347° 46.8'
To proceed and determine the latitude, I have to cheat and use my knowledge of the longitude. /o\ If I don't, I will get a circle on the Earth's surface, and I could be anywhere on that circle (the principle is the same as if you measure the height of a tower as seen from a distance--in the absence of any other information, you could be anywhere on a circle with a certain radius from that tower). There is of course a procedure for determining that circle and it's often used in navigation, but I will omit that (this is already tl;dr). If you do it for two celestial bodies, or for the same body at two different times, you get two circles that intersect and you can determine your position.
Let's first pretend that I managed to measure the height at true noon. The reason that Jack Aubrey and other 19th century people (or anyone who can, really) wait until noon to take their readings is that then they don't need an accurate clock nor do they need to know the longitude. We won't need to use the value of GHA from the tables, only the declination. You can tell that it's noon by taking successive readings until you notice the sun sinking again. (In fact the easiest way of determining the longitude is to have an accurate clock set to GMT and then observe the time of noon in that clock. Your noontime in GMT is then your longitude if you convert it from time to degrees, basically.)
Anyway. Height of sun at noon - declination of sun = 90° - latitude. So:
latitude = 90° - height of sun + declination of sun = 90° - 33° 33.5' + 1° 7.4' = 57° 33.9'.
The true latitude is 57° 36.5'. I'm only 2.6' off, that is, 1.852 * 2.6 = 4.8 kilometers. That's pretty good!
Let's see if it gets better when I correct for it not being true noon (now is when I have to cheat and use the known longitude). Time for some spherical trigonometry! We assume that the sphere of the sky has diameter one, so that a distance between two points on the sphere is the same as the angle between those two points seen from the center (all angles are now in radians, though I'm still stating them as ordinary degrees).
Draw a triangle on the sphere of the sky, with the north pole (A), the zenith (B) and the sun's position (C) as corners. Let A, B, C also denote the angles at those points, and a, b, c denote the corresponding opposite sides. Our goal is to solve for side c (the angle between the zenith and the north pole in the sky); the latitude is then 90° - c°.
The following are known:
LHA (local hour angle) = GHA + longitude = 347° 46.8' + 11° 43.7' = 359° 30.5' (this is very near 360° because the sun is very close to its highest point)
A = 360° - LHA = 360° - 359° 30.5' = 29.5'
a = 90° - height of sun = 90° - 33° 33.5' = 56° 26.5'
b = 90° - declination of sun = 90° - 1° 7.4' = 88° 52.6'
The law of sines on a sphere (different from the law of sines on a plane surface) gives us
sin(A)/sin(a) = sin(B)/sin(b).
Solving for the unknown sin(B), we get B = 0.4101° or B = 179.5899°; obviously the second one should be chosen considering the shape of the triangle. Now we can find c by applying the law of cosines on a sphere, which says either of the following:
cos(a) = cos(b)cos(c) + sin(b)sin(c)cos(A)
cos(b) = cos(a)cos(c) + sin(a)sin(c)cos(B)
cos(c) = cos(b)cos(a) + sin(b)sin(a)cos(C).
It would be nice to use the third one since then we'd get c directly, but we don't know C so we can't. (We can't use A and B to figure out C since the angle sum is not 180° like in a plane triangle.) Instead, we use both of the other ones and get a system of equations with cos(c) and sin(c) as unknowns. Solving that system, we get c = 32.4338°, which gives a latitude of 90° - 32.4338° = 57.5662° = 57° 33.97'. The true latitude was 57° 36.5', and the one I calculated assuming the time to be true noon was 57° 33.9'. Ha ha. I suppose the lessons here are 1) it does not make much difference if you are two minutes from true noon, 2) the number of decimals in my calculations are kind of ridiculous considering the accuracy of my reading, 3) who cares, we got to do some fun spherical trigonometry.
I am kind of proud of myself for getting a latitude that was only 2.6' or 4.8 kilometers off, though! \o/
If you read the whole of this post, I salute you. The post is not beta-read (and in fact is hardly alpha-read, since I had to rush off), so tell me if you find errors. Also, ask if you have questions. : )
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(no subject)
Date: 2015-12-22 05:31 pm (UTC)*tries really hard*
I think I sort of mostly sort of understand the general idea? Until you got to the bit with the spherical trigonometry, which needs actual maths, of which I have none. Aw hell, this is why captains were allowed to take sailing masters with them to do the calculations, right? :)
And kudos to you for being only a couple of miles off target - two miles out of a whole earth's worth. The bit that puzzles me is how you get an accurate horizon on a wobbly ocean, though! I've seen it done on land with a bowl of water as a fake horizon, but on a boat...?
(no subject)
Date: 2015-12-22 05:41 pm (UTC)As a mathematician you should know better than to use spurious levels of decimal precision! :-)
(no subject)
Date: 2015-12-22 06:34 pm (UTC)(no subject)
Date: 2015-12-22 11:27 pm (UTC)Last, but not least, now I finally know what a nautical mile is based on. This makes my life feel more complete - and I am not even kidding about that.
(no subject)
Date: 2015-12-23 02:35 am (UTC)(no subject)
Date: 2015-12-23 10:37 am (UTC)(no subject)
Date: 2015-12-23 07:58 pm (UTC)The horizon is flat even if there are waves? I mean, the waves on the horizon are so far away that they make no difference. The difficulty is holding the sextant still when the boat is rocking underneath you.
(no subject)
Date: 2015-12-23 08:02 pm (UTC)As a mathematician you should know better than to use spurious levels of decimal precision! :-)
Better to use too many decimals in the calculations than too few, though! But yes, I suppose my final answer should've been 57° 34'. : )
(no subject)
Date: 2015-12-23 09:17 pm (UTC)(no subject)
Date: 2015-12-23 09:18 pm (UTC)(no subject)
Date: 2015-12-23 09:21 pm (UTC)(no subject)
Date: 2015-12-23 09:21 pm (UTC)(no subject)
Date: 2015-12-22 08:23 pm (UTC)This post is fascinating, not least because I fan those books like mad. I'm pretty sure I'd be sailing around in circles if navigation depended on me.
(no subject)
Date: 2015-12-23 04:46 pm (UTC)(no subject)
Date: 2015-12-23 07:21 pm (UTC)I'm glad the post was interesting! It was fun to write, as well. : )
(no subject)
Date: 2015-12-23 07:49 pm (UTC)(no subject)
Date: 2020-08-11 06:43 am (UTC)This is so interesting! I didn't know any of this. And even cooler that you actually got to do it yourself.
I managed to work out where the correction for height of eyes above sealevel comes from (well, I got 1.926*sqrt(h) but it does depend a bit on what you take as the radius of the earth...) but have no clue where the correction for refraction comes from. But maybe this is just a rule of thumb based on empirical observations?
When you mentioned the local time at true noon at the start of your post, I did wonder why sailors could not just use that to look up the longitude in some table, but I guess I was hugely overestimating the accuracy of an 18th watch that was synchronised to GMT before setting out on a sea voyage six months earlier... :D
It took me a little while to figure out the paragraph in the middle starting "The following are known:", in fact I still have not really figured it out after drawing it on paper, but will take your word for it g
Kudos on only being a few kilometres out at the end! Looks like you had a lot of fun.
(no subject)
Date: 2020-08-13 05:24 pm (UTC)Anyway, I'm glad you enjoyed the post! And did the exercises on your own like a good student, heh. *g* The correction for refraction is empirically determined--I don't think you can calculate it in any reasonable way.
When you mentioned the local time at true noon at the start of your post, I did wonder why sailors could not just use that to look up the longitude in some table, but I guess I was hugely overestimating the accuracy of an 18th watch that was synchronised to GMT before setting out on a sea voyage six months earlier... :D
That is indeed how they did it, once watches got accurate enough! Check out this guy. : ) Earlier proposed methods relied on observing the moon against the background of stars, which appears slightly different depending on your longitude, or using the moons of Jupiter as a watch. But they were never really practicable on ship. It was a difficult and pressing problem at the time!
It took me a little while to figure out the paragraph in the middle starting "The following are known:", in fact I still have not really figured it out after drawing it on paper, but will take your word for it g
I will draw a picture and scan it for you, and send by email! : )