Problem 231 Revisited
|There are several reasons why we are looking over problem 231.
#1. We never finished giving solutions (here I was hoping for some feed back on this problem from the readers).
#2. There is a comment on problem 231 in PS, Jan 2012, Vol. 32 No. 1, pg 39 that must be addressed.
Everybody should go back and look at the problem and read my comment of Oct. 19, 2011 at 11:01:16.
My 1st solution in 1962 got the answer but WAS NOT the type of solution the author was looking for as I have stated in my post. As I was learning surveying (Artillery, USMC) in 1962 , SSGT Pye would give me problems and see if I could solve them long hand. One was related to problem 231. If one looks at the drawing of problem 231 they will see a Quadrilateral; ADEB. It is this figure that can be solved very easy by using the Army form DA 1930, Special Angle Computation (NGS had a form by the same name; the old USC&GS Form 665a) both Forms used logarithms. There is one more Form and that will give the whole problem away. It is DA 6-18 ; Two-Point resection, see FM 6-2 Artillery Survey Aug. 1961 pg 162-164. Thats all this problem is, a Two-point Resection
All forms solve for angle's DEA(which is the sought after angle in problem 231, or angle X) and angle EDB. It takes about 5 minutes to solve for both.(by solving for both you have a check on your calculations).
If one looks at the forms, say 665a on page 176 of Special Publication #247 ,Manual of Geodetic Triangulation you will see that the formulas are very simple and can be worked on a hand calculator in several minutes.
But wait we are not quite done yet. It was in 1968 that I found a Dover book that was published in 1965 that would answer my questions about the formulas used on DA 1930 and NGS 665a. The title of the book is " 100 Great Problems of Elementary Mathematics, their history and solutions, by Heinrich Dorrie. The solution to the Two-Point Problem (which is one of the 100 listed in the book) goes to Peter Andreas Hansen (1795-1874)
In the Dover book they say that others knew the solution but the credit goes to Hansen around 1841(?).
You can also derive a formula by using the Law of Sines;
tan X = sin(EAB+DBA)sinDBA(cosACB+cos2EAB)/sinEAB(cosACB+cos2DBA)+cos(EAB+DEB)sinDBA(cosACB+cos2EAB)
This is using problem 231 labeling. Looking at form 665a the solution might be shorter.
Comments on the article in PS listed above. The listed steps on problem solving are missing the most important step (in my opinion) and that would be ; check your solution. If this was done for both problem 233 and 236 the two authors would not have these very embarrassing errors in their solutions.
Just because all angles in a problem are shown as whole numbers one does not IMPLY that the answer is a whole number. One states what he is looking for in the problem or what type of solution he is looking for, ie. no trig, no calculators
etc. Remember imply is just a nice way of saying ASSUME, and surveyors never ASSUME. One shows mathematically what the answer is.
I can give a similar problem to problem 231 with whole degrees but angle X will not be a whole degree.
Monday, January 16, 2012 at 6:02:06 PM