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» Online Classroom   »   » Public Discussion of Cel Nav   » compass deviation Amplitudes

   
Author Topic: compass deviation Amplitudes
michael gondro


 - posted November 05, 2004 04:09 PM      Profile for michael gondro           Edit/Delete Post 
Dave,This exam question #1655 has me stumped.

22 OCT.1981.DR LAT 21*51.0 S, LONG 76*24.0 E

You observe an amplitude of the sun. Sun's center is on the visible horizon and bears 256* psc. CT reads 01h 01m 25s and 01m 15s fast. Variation is 2* E.

What is the deviation of the magnetic compass?

Ans.2.3*W. The other answers were 2.0E, 0.3W, and 0.3E.

I tried to use the sight reduction tables but got very confused please HELP. With good direction I think I can get this. Mike

David Burch


 - posted November 05, 2004 07:03 PM      Profile for David Burch           Edit/Delete Post 
We discuss amplitudes in Lesson 11 of our online course. you can use your cel nav course s/n to enroll in that course and read about them. it will pay to read through that background.

in the meantime:
we can usually solve amplitude problems quickly, but they can be prone to errors on the USCG test. remember you can always do a sight reduction to get the answer... ie once the test is done, you will never need to do an amplitude problem again.

in either approach to the solution, however, you need to know the declination of the sun and in all uscg problems, the first step is to figure out what time of day they are talking about, since the test questions use CT on a 12 hr watch face without specifying am or pm -- no one in the world records times that way, but by using such an odd system the USCG helps support navigation schools, so we do not mind. They also do not tell you this on the exam. You do have Bowditch in the exam room and that has a glossary, which would explain that CT is sometimes kept on a 12h face, but then goes on to say that these times must be labeled am or pm.

hence step one, what time are they talking about? is it 0101 gmt or 1301 gmt?
the ZD = 76/15 = 5.07 = -5h, and GMT = ZT + ZD, so ZT = GMT - ZD, and in this case ZD is negative, so the two options are 0601 or 1801 ZT.

they do not tell us if it am or pm, but we know the sun bearing is 256 which is in the west, so by local time it is pm, and hence the ZT is about 1801 and the GMT is 13h 01m 25s - 01m 15s = 13h 00m 10s GMT on Oct 22, 1981.

Remember that what we are after here is a true bearing to the sun so we can do a compass check. we can get this true bearing from a simple sight reduction or by an amplitude method. the former is usually easier and quicker for precise results, but you should know both for the test.

first lets do a sight reduction: at the time above the Dec is S 11° 8.0' and the GHA = 18° 55.4'. do a regular sight reduction from the DR position using the starpilot or by hand, to get Hc = 43° 23.5' and Zn = 257.7°. all we care about is the last number.

T = 257.7
V = 2.0 E
M =
D =
C = 256.0

Solve for M = 255.7 and then solve for D = 0.3° W. Note the answer you quote above is wrong, the correct one is 0.3° W.

Now do it by amplitude:
The article quoted shows you can compute amplitude from sin(amp) = sin(dec)/cos(lat) = sin(11.1333°)/cos(21.850°) = 0.2080, and then use arcsin(0.2080) to get amp = 12.0°.

you can also look up the amp without computation from table 27 in bowditch, but will have to interpolate for lat and dec.

the amp is the bearing of the center of the sun as it crosses the horizon assuming no height of eye and no refraction. Taking refraction into account and ignoring HE, the amp computed or tabulated is actually the bearing to the sun when the LL of the sun is about 1 SD above the horizon.

to do precise compass checks using amplitude bowditch has generated a table 28 that provides the correction for how much the bearing of the sun changes as it moves from crossing the visible horizon to reaching the correct position about 1 SD above the horizon. One must be careful in applying this correction, as well as reading the question carefully as they word this question several ways, some require a correction, others do not. If they say sun center on the celestial horizon then no correction is needed. if they say sun center on the visible horizon, then apply the correction. ("celestial horizon" is just another way of saying there is no HE and no refraction).

The correction is always applied to the observed bearing, not to the computed one. the direction of the correction is such as to move the observed bearing of the sun away from the elevated pole.

In our example, the bearing to the sun as it "crossed the celestial horizon" is 270 - 12.0 = 258.0

If you go go table 28 in bowditch for lat 21 and dec 11 you get the correction is 0.3°. We are in S lat, so S pole is elevated, so we have to move our observed bearing of 256 away from S, which is to larger numbers, so C = 256.3 and we have:

T = 258
V = 2E
M =
D =
C = 256.3.

Again solve for M = 256 and solve for D = 0.3W.

From: Starpath, Seattle, WA
michael gondro


 - posted November 08, 2004 02:28 PM      Profile for michael gondro           Edit/Delete Post 
Thank you Dave, this helps me very much, I am beginning to understand. I checked my notes and also came up with .3*W for the answer. But without
a sight reduction table,{didn't know I had one in my PC at the time} I wasn't sure of the answer.
Thank you for the clue on the wording of the exam
questions also. I'll work a few more and test this new knoledge. Mike

michael gondro


 - posted November 09, 2004 10:55 AM      Profile for michael gondro           Edit/Delete Post 
Today I tried this new info for using amplitude
for finding deviation. My calculations gave me
degrees to add or subtract from due east or west
but my deviation numbers were larger than needed,
sometimes 3 times larger. When working through the
example data you gave it works out,mine do not. It
seems like my declination figure is wrong or I'm
reading the alminac wrong. Mike



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