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Discussion: Is textbook example of applying declination correct?

in: Orienteering; Training & Technique

Nov 15, 2016 2:46 PM # 
Good Morning!

I am trying to learn how to apply declination as it pertains to magnetic compasses. In the process I found a textbook on surveying which had several examples that the author worked out for the student.

I thought I had understood the concepts and could apply them reasonably well for most examples I encountered. That is, until I came across Example 5-3 on page 129 of the book, SURVEYING written by Bouchard and Moffett. 5th edition. International Textbook Company.

Below I have carefully transcribed Example 5-3 on page 129 and presented the authors solution. I feel the author or his typist made a mistake. But because I have so little experience in applying declination, I feel insecure in my contradicting solution.

Would you please be willing to review both the author’s solution as well as mine and point out where I might have made an error? I also provided a bit of a discussion after my solution so that you would have a more clear insight into what was going on inside my mind.

Example 5-3
“The magnetic bearing of a line was recorded as S 80° 15’ W in 1880 at a place which had a declination of 16° E in that year. What is the magnetic bearing if the declination is now 4 ° 30’ E.” (exact quote)

The book's solution:
Magnetic bearing in 1880 = S 80° 15’ W
Magnetic azimuth in 1880 = 260° 15’
Declination in 1880 = + 16°
True azimuth in 1880 = 276° 15’
Declination at present = - 4° 30’

Magnetic azimuth at present = 271° 45’
Magnetic bearing at present = N 88° 15’ W

My solution:
Magnetic Bearing in 1880 = S 80° 15’ W
Magnetic Azimuth in 1880 = 244° 15’ (80° 15’ + 90° + (90°- 16°))
Declination in 1880 = 16° E (my understanding of 16°E is that mag north is East of True North)
True azimuth in 1880 = 260° 15’

Declination in present = 4° 30’ E (my understanding: mag north is East of True North)
Magnetic azimuth at present = 255° 45’ (260° 15’ - 4° 30’)
Magnetic Bearing at present = S 75° 45’ W

Thus the book suggests the answer is N 88° 15”W and I suggest the answer is S 75° 45” W

The difference between our answers might suggest where the error is:
Converting Bearings into azimuths
N 88° 15’ W = 271° 45’
S 75° 45’ W = 255° 45’

Solution Difference = 16° 00’ WHICH IS THE STATED DECLINATION IN 1880

My guess (at what I believe is the error) is that while the book called out for the 1880 declination to be “E” the
author meant to state “W”. this view is supported by the fact the author used a “+” sign when he declared what he
was going to do with the declination (add it or subtract it) when he itemized his procedure solution.

Any thoughts on this matter? Am I misunderstanding the procedure? I am the kind of guy who lacks confidence
When learning new material and so when I encounter an answer different from a text book I worry I am not
grasping the subject matter. God help the poor person when I finally understand something, however.

Thank you!
Nov 15, 2016 3:38 PM # 
I get the book's answer.

I use the mnemonic:

East is least and West is best, going from True to Magnetic.

So. If you give declination a sign as above, E is negative, W is positive, you get:

True + Declination = Magnetic

1880: True = Magnetic - Declination (but D is negative), so True > Magnetic.

True is the constant between 1880 and the Present, so...

Present: Magnetic = True + Declination, but D is still negative,
so True > Magnetic is still correct.

I got 271° 45’.

If you look at Sheet 1 of Patent 526021, you can see one of the "quadrant" dials that use the N (angle) W, and S (angle) W convention. Zero is at North and South, and East and West are both 90.

Finally, I may be wrong here. I'm as confused about declination as anyone, but the book solution makes sense to me.
Nov 15, 2016 3:52 PM # 
@whburling I don't think you have calculated the magnetic azimuth in 1880 correctly. You shouldn't have the declination (16 deg) in the conversion from magnetic bearing to magnetic azimuth. It's only in the next step (conversion to true azimuth) that the declination should appear.
Nov 15, 2016 8:32 PM # 
thank you...cedarcreek and bmay...for responding. Let me think on your suggestions
Nov 15, 2016 9:12 PM # 

Allow me to offer a different point of view. By no means do I think I am "right".
I am just trying to think as well as I can.

My understanding is that Magnetic Declination can used when seeking to understand the relationship between magnetic and true north.

When a compass measurement indicates a bearing it is always using the
current declination for its reference.

now lets go through an example:
* A BEARING on your compass reading N 30 degrees East means the bearing is 30 degrees East from the declination.

* If a DECLINATION of 14 deg West is appropriate for the region of the compass magnetic bearing (and the time of the bearing), taking into account the 14W declination means the reference of the COMPASS bearing on your compass is 14 to the west of True North.

* Thus the True north bearing is 14 degrees LESS than the magnetic bearing.

Consequently a declination of anything W always results in a True North Reading which is less than the magnetic bearing. Thus the sign of a xW declination is negative.

W is negative
E is positive

This is the opposite of what I am hearing from you, cedarcreek.

Where are the misunderstandings for me or possibly for you?
Nov 16, 2016 11:30 AM # 
You are both saying the same thing, you're just talking about different directions of conversion (true to mag vs mag to true).

"East is least and West is best, going from True to Magnetic" is the same as "W is negative, E is positive, going from mag to true"
Nov 16, 2016 12:19 PM # 
It may be worth noting that in all the orienteering and mapping I've done since 1978, I can't recall ever once doing any kind of magnetic declination calculation.
Nov 16, 2016 8:01 PM # 
I'll second what @bmay said... the magnetic azimuth in 1880 is simply the magnetic bearing plus 180 degrees.

We assume the true azimuth doesn't change over time, so our procedure is to find that next and then use that and our modern day declination to find the modern day bearing. Seems like you worked all that perfectly and the 16° error just propagated through.

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