Hi folks, regarding problem 3 there is a slight discrepancy between my and their answers. Go to www.scsurveyjac.org/16.html. Mine CD=430.097,Azimuth=12-31-38.82
Theirs CD=430.12 Azimuth=12-31-45.
Closures are nearly equal but my interior angles are 720-59-59.98
Theirs 720-59-53.8.
Thanks much.
Mike Burkes
626-833-1521
> Hi folks, regarding problem 3 there is a slight discrepancy between my and their answers. Go to www.scsurveyjac.org/16.html. Mine CD=430.097,Azimuth=12-31-38.82
> Theirs CD=430.12 Azimuth=12-31-45.
> Closures are nearly equal but my interior angles are 720-59-59.98
> Theirs 720-59-53.8.
> Thanks much.
> Mike Burkes
> 626-833-1521
It's a great thing you guys are doing in California, somebody should bring that to NYC.
Cheers and keep up the good work
Post Problem 3, Your Link Is To Other Problems
Problem 3 in the link is a survey bearing, missing data traverse.
Paul in PA
As a quick observation, the legs of your figure in Problem 3 are considerably longer than the legs of the figures in the first two problems.
> Hi folks, regarding problem 3 there is a slight discrepancy between my and their answers. Go to www.scsurveyjac.org/16.html. Mine CD=430.097,Azimuth=12-31-38.82
> Theirs CD=430.12 Azimuth=12-31-45.
> Closures are nearly equal but my interior angles are 720-59-59.98
> Theirs 720-59-53.8.
> Thanks much.
> Mike Burkes
> 626-833-1521
I'll look at the problem when I get a chance, but I don't understand how you can have angular misclosure when all you have are bearings. any bearing you put in there, even if a little wrong, should comput 'perfectly' when you calculate interior angles off of the given angles. (I think I am referring to the same problem Paul.
Also, if you had six angles, you would have an angular closure of 720º. Maybe you meant 719º 59' 5x"...?
Anyway it looks like a hard problem and I'll try to tackle it if I get the time.
Hi folks, yes 719 thanks much for.
Mike Burkes
626-833-1521
I get CD = 430.096
Bearing FA = S 12 31 38.7 W. Note that you found AF, not FA which was asked for, and the problem asked for bearing, not azimuth.
And I agree that angular closure should be perfect to the level of rounding error if you are given bearings. In fact, for this problem there is no redundant data so distance closure should be perfect also.
I agree with your answers but...
1. there are no closures, there really is no point in calculating the 720 degrees.
2. missing bearing and missing distance when calculated correctly is "PERFECT"
Problem 3 omitted measurements on non-adjacent sides
Hi folks,as most of you know latitudes and departures of equal parallel lines are equal. Refer to Bouchard and Moffitt,6th edition,pgs 328-330,Fig 8-28. Essentially I placed all known sides together creating traverse ABCGH and which resulted in unknown line HA and unknown problem sides of the additional triangle. When line HA is solved then remaining unknown sides can be solved. Hope this helps.
Mike Burkes
626-833-1521
I finally worked the problem, and came up with close to the same answer. If you take this as an exam, they might well grade you on putting the bearing in the correct direction and using bearing instead of azimuth. You obviously know how to work the problem, to get the right numerical answers. Kudos to that. Relook at how you came up with your angular total....if you plug in your final bearing (whichever number you use), and calculate the interior angles, you should get 720º00'00" precisely. Bill said it best since there is no redundancy.
For the record, I don't know if there is more than one way to solve the problem, but I put both unknown legs at the "end" of the traverse, since the latitudes and departures should equal zero, it was the easiest way to solve it as far as I could figure. When I ran my latitudes and departures, I could see that they were using bearings and distances to make it close to even coordinates by the end. I got +200.0146 N, by -249.9958 E. I am thinking their answer rounded to 200N by -250E and that might be where the slight difference in bearings came from. I inversed from my last known piont to the beginning point and calculated the difference in bearings between my inverse and the given bearing. I then used the law of sines to get the rest.
I got 430.095 on the distance, and FA would be SW 12º31'37.5"
I did it the same way and came up with S12 31' 48"W on FA and 430.10 on CD - rounded all calcs to 2 decimals on distance and 1" on azimuths and angles.
