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Author
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Topic: Sun rising angle
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John
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posted March 22, 2021 02:21 PM
I have been reading Emergency Navigation (2nd edition) and I have a question concerning the Sun rising angle table 6-1 on page 80
I understand the basics of amplitude and the method of determining it (for any date as described in fig 6.3). So if I’m using a latitude of 50 deg N then Max Amplitude is 38 as per fig 6.5 and then on 11 Nov (say) then the amplitude will be about 26 degs. That is the sun will rise at 26 deg south of east 116 deg from north
If I understand table 6-1, I enter a latitude of 50 deg (N) at the left and an amplitude of 30 deg along the top and thus get a rising angle of 34 deg
Why are there values for rising angle greater than the max amplitude for a particular latitude? That is a rising angle of 24 deg for a amplitude of 50 deg and a lat of 50 deg
Is there a formula for calculating the values in table 6.1?
From: UK
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David Burch
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posted March 22, 2021 03:08 PM
Nov 11, dec is about S17, so at 50N, Sin Amp = Sin17/Cos 50 = 0.4548, so Amp = arcsin (.4548) = 27º
https://starpath.com/cgi-bin/web_card/courses/glossary.pl?show_def=8&cat=Celestial_Navigation
Yes, that is the right way to use that table.
And yes, there is likely an equation for anything we might ask for in cel nav, but I would be pressed at the moment to go back and derive that one again. You could also discern it from sight reduction tables, looking at Hc and Zn just after rising and then and hour later.
Best bet is use the remarkable program Stellarium and you can directly measure this angle for any conditions.
From: Starpath, Seattle, WA
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John
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posted March 23, 2021 12:38 AM
Thanks
Why are there values for rising angle greater than the max amplitude for a particular latitude? That is a rising angle of 24 deg for a amplitude of 50 deg and a lat of 50 deg?
From: UK
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David Burch
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posted March 23, 2021 11:16 AM
I am not sure how "max amplitude" applies to this table, which is for amplitude of any body, not just sun.
Have you looked at this in Stellarium? Set Lat to 50, then look west (270) then look at 220, which is amp 50 and watch the angles that stars move relative to the horizon. I assume this table describes that. The table refers readers to Figure 6-5. Let us know if that is not the case. thanks.
From: Starpath, Seattle, WA
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John
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posted March 27, 2021 05:38 AM
I’m now more confused about table 6-1, but for a different reason.
From the table at lat 0, long 0 on 21 Jun the Sun’s amplitude is 23.5 (Fig 6.2) Then the rising angle is about 67 deg.
If I plug the values into Stellarium and watch sunrise it does so near vertically that is 90 deg to horizon or 90 minus lat.
If I use NOAA Solor calculator (https://www.esrl.noaa.gov/gmd/grad/solcalc/calcdetails.html) for lat 00N and long 0.0 I get sunrise @ 5:58:11 and at 06:00 I get Altitude and Azimuth values of 0.132870948 and 66.5620055 and for 07:00 values of 13.39008495 and 65.87385959
Note that Azimuth is 66.6 which is an amplitude of 23.4 (which is what I’d expect)
I can now calculate the rising angle. If I take the change in Azimuth over a time period as A and the change in Altitude as B then the rising angle C may be found as Az = r tan A and Alt = r tan B and C= Atan(Alt/Az) =Tan B/Tan A
This comes out at 87 deg which agrees with Stellarium observations, but is a long way off from table 6-1 value off 67deg.
I’m clearly doing something wrong, but can’t work out what
From: UK
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David Burch
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posted March 27, 2021 11:17 AM
Maybe the table is wrong. Please confirm a specific example from the table, ie which Latitude and Amplitude, points to a rising angle that you believe is wrong and what do you think it should be. Thanks.
From: Starpath, Seattle, WA
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John
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posted March 28, 2021 05:51 AM
My original question concerned the Sun rising angle since the table appears in the chapter “Steering by the Sun.
I have worked out a few specific cases for sun rising angle (all cases are Long 0 hence no time zone correction and no DST)
For 28/03/21 at lat 50 sun rise is 05:54, amp is 5.7 deg . Rising angle from sunrise to 07:00 I calculate is 39.6 deg which is in close agreement with table 6-1 value of 40 deg.
For 20/06/21 at lat 50 sunrise is 03:30, amp is 39.5 deg. Rising angle from sunrise to 05:00 I calculate is 35.6 deg which is a bit off table 6-1 value of 29 deg
For 20/06/21 at lat 10 sunrise is 05:00, amp is 23.9 deg. Rising angle from sunrise to 07:00 I calculate is 82.7 deg which is a more off table 6-1 value of about 65 deg
It seems that table 6-1 is accurate at low values of amplitude, but less so as amplitude increases and as latitude decreases hence my original question as to have the table is derived.
Hope this helps
From: UK
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David Burch
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posted March 28, 2021 11:36 AM
OK. I will have look at this. Please note the table is not restricted to values at the horizon, but is an average all the way up the sky at that amplitude. It is in a chapter called steering by the sun but this is an expansion of that topic as explained in Figure 6-5 that illustrates what the table 6-1 presents.
From: Starpath, Seattle, WA
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John
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posted April 13, 2021 09:03 AM
Any progress on this question?
From: UK
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David Burch
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posted April 13, 2021 04:45 PM
Looking at your last example:
WT = Times in GMT+-0
Nautical twilight/Civil twilight/Sunrise
6/20/21 : 04:49:49
6/20/21 : 05:16:52
6/20/21 : 05:40:22
Sunset/Civil twilight/Nautical twilight
6/20/21 : 18:22:52
6/20/21 : 18:46:22
6/20/21 : 19:13:25
LAN/Equation of Time/Hc/Zn rise/Zn set
12:01:37
00:-01:-37
076°33.9'
066.0°
294.0°
LHAA AM Twilight
6/20/21 : 05:03:21
344°28.3'
LHAA PM Twilight
6/20/21 : 18:59:54
194°11.0'
=======
0540 Hc = -0º 53.9’ Zn = 066.0
0740 Hc = 26º 26.9’ Zn = 068.7
Arctan = rising angle = 84.2º and this has amplitude 90-68.7 = 21º so it seems the table is off for this example.
Thanks for pointing this out, we will add an errata. I will look into that computation as soon as I can. As i recall the Pub 214 sight reduction tables have a column called dh/dt, which is effectively what we want. Or we could work from the basic equations we have in the textbook and online.
From: Starpath, Seattle, WA
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