I was wandering, why you changed the values from calculating the azimuth of a star. Then I tried it myself and figured that it would give you a negative value of altitude :) BTW big fan of yours. You are doing great job. Keep it up!
Hi Edward, in your Celestial Navigation website, which is very well done btw, you chose as your example leg 57 from the sample problem series _Silicon Sea._ Are any other legs available somewhere? Thanks for your time, Robert
Seems like the Selected Stars volume of the Air Almanac would be infinitely easier. You would also be given your Hc and would be well on the way to having an intercept.
How does one calculate the azimuth with a longitude east? I'm trying to figure this out, but I'm lost in the B and C step of the method. Any suggestions?
Are there calculators that take trig functions using degrees and minutes? All the ones I have ever used require radians, so converting minutes to decimal degrees, adding those to the degrees, multiplying by pi, dividing by 180 is a lot of work. Especially on a small boat pitching in the waves in the North Atlantic. Don’t know how they did it back in the day. I’m sure there was vomitus on a lot of the midshipman’s workings. 😉
¿Any idea how to locate the first aries point from any place you are on earth? Is for the purpose of practicing. Maybe some DIY clinometer gadget, that help me to locate where is the vernal point, and after that, finding a star by it´s declination and AR? Or maybe a simpler method? By the way, i like your tutorials, they are very good!
@8:14 you have tan (lat) = tan (043 35.9') shouldn't that be tan (27 10.2')? You plugged in the incorrect figure there. I do like the layout. I am trying to get the bearing of a star from my location .
I couldn't make the formula at 10:13 work, so after much research elsewhere I am now using this which does work (for me!) Az=1/(C x cos(LAT)), followed by inverse tan. I'm a big fan by the way!! (also as mentioned by others, the LAT at 8:14 is shown incorrectly in the slide (had me going for a while too!)
I tried both yours and the author's method and it gave me exactly the same result. But I prefer your method because it's easier to insert in the calculator.
The old time "Navigators", way before the age of LORAN, let alone GPS, always used a hand berring compass to get as close an azimuth as possible on every body they we are going to use for a sight!
I assume this is good as an academic exercise, but why can’t you just go outside and note the azimuth of the star. It’ll change by 15 degrees an hour and be 1 degree further west each night. So you can get a very good approximation of its location throughout the night and day by day. I’m just an armchair navigator/sailor, so perhaps I’m missing the relevance of this exercise.
Just another amateur here, my guess is that the latitude is already an approximate number (if you measure it with a sextant or something), so two approximate coordinates would lead to an entire rectangular window of error. This way you would probably have only a 1 dimensional error (the altitude of the star, so the latitude) to adjust by repeating the measurement a few times. It's just a supposition though!
The GHA (Greenwich Hour Angle)of a celestial object changes by 15° each hour but it's azimuth (Zn) does not. Look at these azimuth figures that I computed for an observer at latitude 45°N and a celestial object of declination 20°N as the LHA (Local Hour Angle) changes by 15° hourly. LHA Zn Difference 00° 180.0° 15° 211.3° 31.3° 30° 234.6° 23.3° 45° 251.1° 16.5° 60° 263.7° 12.6° 75° 274.4° 10.7° 90° 284.4° 10.0° 105° 294.5° 10.1° 120° 305.2° 10.7° 135° 317.0° 11.8° 150° 330.1° 13.1° 165° 344.6° 14.5° 180° 000.0° 15.4° 195° 015.4° 15.4° 210° 029.9° 14.5° 225° 043.0° 13.1° 240° 054.8° 11.8° 255° 065.5° 10.7° 270° 075.6° 10.1° 285° 085.6° 10.0° 300° 096.3° 10.7° 315° 108.9° 12.6° 330° 125.4° 16.5° 345° 148.7° 23.3° 360° 180.0° 31.3° GHA and LHA are equatorial coordinates and the azimuth is an horizontal coordinate. This is why they don't track in lockstep.
The uploader made a mistake, he plugged the longitude value of 043°35.9' as numerator instead of the latitude value of 27°10.2' that should be there. Solving with these figures outputs A = 0.65046 which is irrelevant to the problem being solved. The answer A = 0.35061 under the formula is correct when 27°10.2' is the numerator.
Longitude from 0° to 180° east or west is the position of the observer. GHA is the position of the celestial object counted from 0° at Greenwich westward to 360° back to Greenwich.
Yes, Zulu time and UTC are the same. The ZT after 00:30:15 does not mean Zulu Time but Zone Time which gets converted to 03:30:15 UTC by adding 3 hours since the Zone is -3.
You use the increment page for simple math calculations and then you use a calculator for complex calculations which 95% of the viewers have no idea how to formulate but which are shown specifically and easily in that atlas. Poor instruction.
![ลองสั่งของจากแอพ temu ถูกจริงหรอ?? ( part : 1/2 ) [โกงไหมครับ ep.89 ] | DOM](http://i.ytimg.com/vi/RTcZSQTiDZw/mqdefault.jpg)
I'm going to keep coming back to this video until I understand it!
I was wandering, why you changed the values from calculating the azimuth of a star. Then I tried it myself and figured that it would give you a negative value of altitude :) BTW big fan of yours. You are doing great job. Keep it up!
Yeah, I ran these numbers myself and found that Vega was below the horizon (negative altitude). But I got the same azimuth, so there's that.
The animations are well done.
Hi Edward, in your Celestial Navigation website, which is very well done btw, you chose as your example leg 57 from the sample problem series _Silicon Sea._ Are any other legs available somewhere?
Thanks for your time,
Robert
Seems like the Selected Stars volume of the Air Almanac would be infinitely easier. You would also be given your Hc and would be well on the way to having an intercept.
How does one calculate the azimuth with a longitude east? I'm trying to figure this out, but I'm lost in the B and C step of the method. Any suggestions?
Are there calculators that take trig functions using degrees and minutes? All the ones I have ever used require radians, so converting minutes to decimal degrees, adding those to the degrees, multiplying by pi, dividing by 180 is a lot of work. Especially on a small boat pitching in the waves in the North Atlantic. Don’t know how they did it back in the day. I’m sure there was vomitus on a lot of the midshipman’s workings. 😉
@Bob Hunkins, this is the calculator I use.
th-cam.com/video/rvShil1yi_g/w-d-xo.html
¿Any idea how to locate the first aries point from any place you are on earth? Is for the purpose of practicing. Maybe some DIY clinometer gadget, that help me to locate where is the vernal point, and after that, finding a star by it´s declination and AR?
Or maybe a simpler method? By the way, i like your tutorials, they are very good!
@8:14 you have tan (lat) = tan (043 35.9') shouldn't that be tan (27 10.2')? You plugged in the incorrect figure there. I do like the layout.
I am trying to get the bearing of a star from my location .
Brett Jones i saw it too, badly need it for may celestial navigation activity 😢
Yes, 27° 10.2' should have been written.
I couldn't make the formula at 10:13 work, so after much research elsewhere I am now using this which does work (for me!) Az=1/(C x cos(LAT)), followed by inverse tan. I'm a big fan by the way!! (also as mentioned by others, the LAT at 8:14 is shown incorrectly in the slide (had me going for a while too!)
I tried both yours and the author's method and it gave me exactly the same result. But I prefer your method because it's easier to insert in the calculator.
The old time "Navigators", way before the age of LORAN, let alone GPS, always used a hand berring compass to get as close an azimuth as possible on every body they we are going to use for a sight!
At 9.51 you gave the formula of Azimuth as inverse of Tan of 1/C/Cos Lat. Is there an error here? Shouldn't it be 1/(C X Cos Lat)??
It gives the same result, I tested it myself.
I assume this is good as an academic exercise, but why can’t you just go outside and note the azimuth of the star. It’ll change by 15 degrees an hour and be 1 degree further west each night. So you can get a very good approximation of its location throughout the night and day by day. I’m just an armchair navigator/sailor, so perhaps I’m missing the relevance of this exercise.
Just another amateur here, my guess is that the latitude is already an approximate number (if you measure it with a sextant or something), so two approximate coordinates would lead to an entire rectangular window of error. This way you would probably have only a 1 dimensional error (the altitude of the star, so the latitude) to adjust by repeating the measurement a few times. It's just a supposition though!
I think this exercise is necessary in the process of finding s ship's compass error by star azimuth method.
The GHA (Greenwich Hour Angle)of a celestial object changes by 15° each hour but it's azimuth (Zn) does not.
Look at these azimuth figures that I computed for an observer at latitude 45°N and a celestial object of declination 20°N as the LHA (Local Hour Angle) changes by 15° hourly.
LHA Zn Difference
00° 180.0°
15° 211.3° 31.3°
30° 234.6° 23.3°
45° 251.1° 16.5°
60° 263.7° 12.6°
75° 274.4° 10.7°
90° 284.4° 10.0°
105° 294.5° 10.1°
120° 305.2° 10.7°
135° 317.0° 11.8°
150° 330.1° 13.1°
165° 344.6° 14.5°
180° 000.0° 15.4°
195° 015.4° 15.4°
210° 029.9° 14.5°
225° 043.0° 13.1°
240° 054.8° 11.8°
255° 065.5° 10.7°
270° 075.6° 10.1°
285° 085.6° 10.0°
300° 096.3° 10.7°
315° 108.9° 12.6°
330° 125.4° 16.5°
345° 148.7° 23.3°
360° 180.0° 31.3°
GHA and LHA are equatorial coordinates and the azimuth is an horizontal coordinate. This is why they don't track in lockstep.
The solution for A appears to be wrong.Did you mix degrees and radians?Im confused.Can you help me please?
@Brian Leake Hi Brian, you got 0.65046?
I got the same.
The uploader made a mistake, he plugged the longitude value of 043°35.9' as numerator instead of the latitude value of 27°10.2' that should be there. Solving with these figures outputs A = 0.65046 which is irrelevant to the problem being solved.
The answer A = 0.35061 under the formula is correct when 27°10.2' is the numerator.
Cool.
isn´t The time zone -2, or why do you set -3?
The longitude is near 045° West.
Uhhmm can someone explain whre did the longitude come from, i thought gha and longitude are the same?
Longitude from 0° to 180° east or west is the position of the observer.
GHA is the position of the celestial object counted from 0° at Greenwich westward to 360° back to Greenwich.
I thought Zulu time and UTC were the same.
Yes, Zulu time and UTC are the same.
The ZT after 00:30:15 does not mean Zulu Time but Zone Time which gets converted to 03:30:15 UTC by adding 3 hours since the Zone is -3.
I've taken the liberty of creating a playlist for this series: th-cam.com/play/PLetA5Fi4kpULSexPc4Xul4Uq-8HoqifOd.html
Much appreciated dude .
This gave me a headache.
You use the increment page for simple math calculations and then you use a calculator for complex calculations which 95% of the viewers have no idea how to formulate but which are shown specifically and easily in that atlas. Poor instruction.