Problem 3

Hey Mike, that link doesn't really get me to a problem 3. I found a problem but there was no side "AB". maybe the link was a little jacked up?

You do know don't you, that the two answers you posted were exactly the same (for all intents and purposes.)

For what it's worth I came up with CD = 430.10 and FA = S12 31' 48"W; my interior angles are 720 0n the nose and my closing is flat on latitude and 0.01' on departure.

And I agree; for all intents and purposes they're all the same answer.

Hi Tom A so sorry try www.scsurveyjac.org/16.html.
Mike Burkes

Hi TA,did not realize there was a new prob. I will further research my sent file.
Mike Burkes

Lee, for whatever it's worth, you will always get a perfect solution on the addition of the interior angles if all you are given are bearings or azimuths. Even change one of the bearings by say, 5º, and recalculate interior angles....one of the adjoining angles will be 5º less, and the other will be 5º more, still totaling 720º

Thank you Tom.

Hi Tom, here is problem,

Line Bearing Distance
AB S71-33-54W 632.456
BC N18-26-06E 316.228
CD N54-27-44E find
DE N45-00-00W 282.843
EF S82-24-19W 756.637
FA find 460.977

Hope this helps.
Mike Burkes
626-833-1521

okay, I worked it, and got s12º31'44"w and a distance of 430.117

same answer basically. I see when you run coordinates, they "engineered" it to get pretty even (latitudes and departure)distances. I kept most of my latitude and departures in my calculator and didn't write down rounded differences. It looks like the only differences would be in exactly how you ran your calculations and rounding errors.

How did you guys work this problem?

Could it be that the creator of the question wants to know your understanding of significant figures as much or more so than your ability to calculate the bearing and distance?

That appears to be the case from where I sit. All the same answer? No, not exactly. The answers show different significant figures.

Larry P

For what it's worth here is the result run thru FreeCOGO.

> How did you guys work this problem?
I haven't worked it, but the process is going to be to determine the bearing FA by solving for the missing interior angle EFA, then fuss with the missing distance CA until you get the latitudes/departures to sum to zero.

Larry,
Did you work the problem? The significant figures seem a bit immaterial. It appears to me, in fact, that the author purposely worked out numbers so that the latitudes and departures worked out to be relative even numbers. I think they were making problems that you didn't worry so much about writing out super-long figures. There bearings were to the nearest second, and the distances were to the thousandth of a foot. As you can see, we couldn't get the same answer to the nearest thousandth of a foot nor the nearest second. I believe the author rounded his result to the nearest hundredth even though his "givens" were to the thousandth as well.

But back to what you said, I think that is what I would concentrate on if I were writing the questions. To have practical distances, and see if the guys working the problems recorded their answers to the adequate significant digit.

Norman

I worked the problem. I was mainly asking because I wanted to see if everyone that worked it used the same method I did, or if anyone had perhaps a simpler and more "elegant" way of getting the answer.

RE:
>....the process is going to be to determine the bearing FA by solving for the missing interior angle EFA, then fuss with the missing distance CA until you get the latitudes/departures to sum to zero.

Note that it is a bit trickier than that, as there are two missing interior angles based on the bearings: EFA and FAB

Norman

> Note that it is a bit trickier than that, as there are two missing interior angles based on the bearings: EFA and FAB
Yes, I see that now. I spoke too soon.

To answer your original question... I put the two legs with the missing information at the end of the traverse and then calculated the resulting triangle using the Law of Sines and Pythagorean Theorem. I rounded everything to the hundredth and arc second.

The author definitely tried to make all of the latitudes and departures even hundreds, only five rounded to 0.01 off and even those were very close to rounding the other way.

> To answer your original question... I put the two legs with the missing information at the end of the traverse and then calculated the resulting triangle using the Law of Sines and Pythagorean Theorem. I rounded everything to the hundredth and arc second.
>
> The author definitely tried to make all of the latitudes and departures even hundreds, only five rounded to 0.01 off and even those were very close to rounding the other way.

Lee, that's how I did it as well (putting the two legs at the end of the traverse). The theory being that since the latitudes and departures will always add up to 0, you can move the legs around. (Not telling you that, since you figured it out just saying it in general)

I carried the latitudes and departures out farther than the hundredth for calculating purposes, but I am guessing the author was rounding as well.

Hi Tom, as per Bouchard and Moffitt,6th ed. pg 330, omitted measurements occurring on sides that are not adjacent:latitudes and departures of equal parallel lines are equal. I believe essentially what they do is create a different traverse by placing the known sides together creating an unknown side which is part of a triangle adjoined at the new traverse's West side. My wording is not all great HTH.
Mike Burkes
626-833-1521

Mike,
That sounds right and that is what I did (if I am understanding your explanation correctly). Look @ Lee D's post below. I think he worded the same procedure fairly well.

I don't know if there is another way to figure it out, but that way seems pretty clever to me. I would be interested to hear if someone else came up with a different procedure; (but it seems like not a lot of guys worked the problem).