Logged In Members Do Not See Ads

Land Surveyor Jobs Available


Not a Surveyor

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!

You need to be a member of Land Surveyors United - Global Surveying Community to add thoughts!

Join Land Surveyors United - Global Surveying Community

Votes: 0
Email me when people reply –


  •  Hi William,

    This will help you understand bearings and Azimuth and a lot more with it.

  • Hi Dave,

    There are two ways to look at this,(1.) if you have a East declination it is a (+) ,but

     you can subtract it  from the True North Azimuth of that line. I think this is how

    the  - 4.30 came about as your math shows.The same as Bernie and Jaime did the same as you.

    (2.) I took the long way to show using Mag.North's and then the angles back to this line as referencing back to True North ,this may have been confusing but Mag.North 0 with a declination of 16 E is True North N16.00-00 E  or AZ 16.00-00 just as before 276.15-00 - 16.00-00 =260.15-00 ,the same as Mag. North 0 with a declination of 4.30 E is True North N 4.30-00 E, you can subtract again 276.15-00 - 4.30-00 = 271.45-00 ,but instead I added the Angles you will come up with the same answer. But i think this is also what the Author  was doing and that is how the - 4.30 got in the equation. A declination East ,True North is always West of 0 Mag. North. If the declination is 10.00 E this means 0. degrees Mag. North is N10.00-00 E true north it is that simple. But there are a lot more things that govern Mag. Declination. There are Variations in Declination,these are Secular, Daily ,Annual and Irregular.you also need a Isogonic Chart and to be aware of recorded local Attractions.  A Surveyors compass is as obsolete as a Gunters Chain,but it is important for a modern surveyor to understand it when retracing old lines. All magnetic object can affect this compass,that include the metal on you,the metal in the ground,natural and unnatural, Electric lines and even magnetic storms around this planet from the Sun,it is a Irregular variation.Even the compass can have the glass being Electrified,never clean it a dry cloth it must be wet. It is always a good practice to use a wet cloth and wipe the compass glass before using it. But surveying with a compass and retracing these old surveys is not as simple as some would think. And maybe that needs to be talk about.

  • Student Surveyor

    I'm casting mine with the book.

    1880 mag = S80d15'W = Az 260d15'

    1880 mag dec = 16dE so

    true bearing/Az = N83d45'W or 276d15'

    current mag bearing =  true - curent dec = 276d15' - 4d30' = 271d45' or N88d15'W

    I couldn't do it without a visual.  I hope this graphic will post (and help)


    • Government Professional

      Thanks Dave;

      Your drawing nicely illustrates the point some of us were trying to make: The sighted object never moves.... it's the compass circle relative to the variations of magnetic north that rotates.

      I hope it puts this argument to rest.... Nothing erroneous about that Moffit and Bouchard example that started all this.


  • Hi Everyone ,

     I do not know, why it was printed in this way , but here is the ?

    What is the magnetic bearing if the declination is now 4 ° 30’ E.” (exact quote)

     That would not be a neg. number. True  North can not change, the only thing that can change is where Magnetic North is, and it is asking what is Magnetic Bearing of that line. In 1880 magnetic north at N16 E True North and 0 degrees mag. North at that time.  this makes the mag . bearing of that line in ? to be 260.15-00 AZ MAG. Today 0 Degrees Mag. North is at 4.30 E that is N 4.30 E that makes the Mag. Az of that line Az 271.45-00. Mag. North today is what has changed. And 0 degrees on their Azimuth is where they point as mag.north then and now.

    • Hi Everyone,

      The only reason i can think of them using a neg (-) number is because the mag. bearing of

      0 degrees Now, has a declination of that. This means it is N 4.30 E true North.

      But it is also 0 degrees Mag. North. This makes True North 0 Degrees a North West mag. bearing of N 4.30-00 W you would also have to subtract the 4.30-00 from the line in ? of the true North Azimuth 276.15-00 - 4.30-00 =271.45-00 Mag. Bearing. The same if you calculated the AZ from Mag. North  0 Degrees. ,Angle is 271.45-00. As Mr.Marocco and Mr Gallegas have already showed. They subtracted this number from 276.15-00 for their answer.

  • Hi Everyone, This is all very simple.

    0 Degrees Mag. North in 1880 is True North 16 degrees or  N 16 E

     this makes True North of that year as Mag. North  N 16.00-00 W

    this makes 0 Degrees Mag. North Now as True North 4.30 degrees or N 4.30 E

    This makes True North 0 degrees as Mag. North Now as N 4.30 W

    and Mag. North Now  of N 11.30-00 E on Mag. North 0 Degrees 1880 line,

    that is True North N16 E .

    If someone  set their compass on a transit with a Dec. of 16 E Degrees

    then the bearing on the line of 0 Degrees Mag. north in 1880 will be N16 E True North .

    Then and now .

    If I set my compass on a Transit with a Dec. of 4.30 E then

    the Bearing on Mag. North line Today of 0 Degrees it would be N 4.30-00 E True North .

    and  Mag . North line of 1880 at 0 Degrees is still N 16 E True North.

    Then and now .

    Hope this helps .

    • Land Surveyor

      well, I still don't understand how/why its worded the way that it is... I can agree that if it was done the opposite way of my thinking, then you'd get the answer William came to... it would probably take me actually using a compass on a transit and doing a problem such as this in real life to grasp why they set it up this way... 

  • Land Surveyor
    Fórmula: MgAz+dec=TrAz entonces MgAz=TrAz-dec
    decE(+), decW(-)

    En 1880: MgAz = 260/15
    dec= 16E
    Entonces TrAz= 260/15 +16/0= 276/15

    Actual: dec= 04/30
    MgAz= 276/15-04/30 =271/45
    o lo que es lo mismo:
This reply was deleted.

Answer As your Facebook Page

Latest Discussions by Category


You are not logged in


Land Surveyor Community Forum

Most land surveying related topics are evergreen - meaning the topics of discussion are many times timeless.  If you have a question regarding Land Surveying in general, this is the place to ask your questions.    

Add Discussion

Global Forums

Hey there!
It looks like you are enjoying the community content.

That's nice! Please don't forget to mention us if you use what you just copied on your website. Help strenthen the community by linking back to us!

Thank you very much! Have a nice day! :)

Ok. Don't bother me again, for this session.