Problem

nope

Well that was curt, Ranky.
Looked like a good answer to me.
I didn't read the question, however.
That may affect my opinion if I ever get around to looking at it.
Seems unlikely that I will, though.
🙂
Don

It's too early in the month to discuss this. Let's let people who might be submitting answers to the organization have a while to do their own thing.

This strongly reminds me of a prior one of their problems, but I think I've thrown it away by now. Wonder how much they changed it.

Problem, question

Mine is in white:

S 70°38'36" W 309.789
S 89°21'37" E 292.433
N 00°04'26" W 105.943
Closure Error Distance> 0.00000
Total Distance> 708.165
Polyline Area: 15489.4 sq ft, 0.4 acres

Is it just local nomenclature to label an unbuildable lot as a parcel? The easement would not create a parcel, a taking would.

Problem, question

poor form to use capital letters for the sides. Angles are capitals.... Mr. Wolfe 10th grade High school geometry instructor.

Problem, question

Mrs. Martin was not so demanding for form and it's been a while since I have thought about expressing an angle as something other than a greek letter.

Problem, question

My post above was meant in good humor and not as a serious critique. I wanted to ensure that was understood, because things can be easily misconstrued in this format.

Problem, question

I do not mind at all and I understand the scars formed by an overbearing teacher. What they teach sure does stick.

> It's too early in the month to discuss this. Let's let people who might be submitting answers to the organization have a while to do their own thing.
>
> This strongly reminds me of a prior one of their problems, but I think I've thrown it away by now. Wonder how much they changed it.
Hi Bill,yes it was submitted July,2011 and I feel our answers are correct as opposed to theirs.

Mike, if you highlight your text in the Surveyor Connect text box, then choose color, you can change the text to White. Then if someone wants to see the answer, they can highlight the blank section and see your results.

Next time, I would recommend posting your response in a more traditional format with the relative bearings and distances on their own lines. Based on filling in a couple blanks, I could not get your solution to close.

That would be a cool problem to do by hand, though why bother when you have CAD?

Could you imagine Euclid with CAD? He would self combust with excitement, then like most old timers he would have to get some kid to show him how to use it...

Ok, I saw the "May problem" title on that page and guessed that they had a different variation of the old one for the current month.

I found a copy of my coordinates from 2011 (and then Mike sent them back to me, also). I did an algebraic solution with iterations required to solve some parts of it, and checked the answer by Star*Net demo mode. The numbering should be obvious if the points are plotted.

I'd be very interested if someone finds a better way.

Point N E
1 0.0000 1000.0000
2 3.7589 663.3577
3 4.9876 553.3078
4 C 39.9975 1000.4466
5 43.7564 663.8043
6 134.3580 967.3393
7 148.5121 962.3733
8 161.8154 1000.2895
9 B 145.9261 1000.3100
10 A 43.2661 707.7142

Bill
===============
ALGEBRAIC SOLUTION METHOD
I don't see any direct way to solve this, but I can write an iterative equation that should then let everything be worked out directly if laboriously.

Call the acute angle at the left A. Start at the point where line b (which distance goes to the edge of the easement) if extended meets the middle of the easement, and drop a perpendicular to the middle of the road. The right triangle formed on our left has lower side (in the road) = 40/tan A.

So the line of b extended to the center of the easement is 486.72 - 40 - 40/tan A.

Extend the 100 ft distance to the middle of the easement for a length of 115 ft. This makes a right triangle with side 115 opposite another copy of angle A. The lower side of that triangle is b extended and is115/sin A.

So now we have an equation
486.72 - 40 - 40/tan A = 115/sin A

This can be manipulated to get
sin A = (115/446.72) * [ 1+ (cos A) * 40/115]

In this form, a guess of 20 degrees for A plugged into the right side lets us solve for sin A and find a better value for A. After a few repeats it settles down to A= 19 - 58 - 25.06

With angle A known, you can work out all the details.

===============
Star*Net .dat file

# California JAC problem 2011 July

# set origin and orientation
# Coords are North, East
B 1-2 270-38-23 !
B 1-3 270-38-23 !
B 4-9 359-55-34 !
B 4-8 359-55-34 !

C 1 0 1000 ! !
C 2 3.7 663 * *
C 5 43.75 663.7 * *
C 6 134.35 966.8 * *
C 7 148.5 961.8 * *
C 8 161.8 1000 * *
C 9 146 1000 * *
C 10 43 663 * *

# Distances
D 1-3 446.717 !
D 1-4 40 !
D 4-6 100 !
D 6-7 15 !
D 4-5 336.6633 *
D 5-10 43.9126 *
D 2-3 110.0567
D 6-10 275.1418 *
D 9-10 310.0829 *
D 4-7 115 *
D 4-8 121.8180 *
D 4-9 105.9287 *
D 4-10 292.7507 *

# Angles at-from-to
A 4-5-8 89-17-11 *
A 2-3-5 90-00-00 !
A 7-4-5 90-00-00 !
A 6-9-4 90-00-00 !
A 6-4-7 180-00-00 !
A 5-8-4 19-58-25.06 *
A 10-6-4 19-58-25.06 *
A 5-4-2 90-00-00 !
A 5-3-7 180-00-00 !
A 6-10-9 180-00-00 !
A 10-5-4 180-00-00 !
A 2-1-3 180-00-00
A 7-8-4 90-00-00
A 7-8-5 180-00-00
A 3-8-1 19-58-25.06 *

> Hi folks,first time poster I worked monthly surveying problem www.scsurveyjac.org/16.html my answers:Courses:N89 21 37,N00 04 26,S70-38-23
> Distances: 292.72, 105.93 , 310.08.
> Areaa three ways;DMD:15504.1338
> bh/2= 15504.1455
> "S" formula =15504.15813 (Sides only given s=a+b+c all over 2.
> Thanks much.
> Sincerely,
> Mike Burkes,
> 626-833-1521

Cheating and doing it in cad, I get different results. I also get a misclosure in your courses and distances by 0.02E x 0.13N. Maybe I'll work it by hand later to check myself.

I have a simple, direct solution.

Problem, question

I get the same exact answers you do with a direct solution.

Hi Jered,here is my form and numbers
agree with Bill Hart;
Line Length Azimuth Lat Dep
C-A 292.7509 270-38-23 +003.2686 -292.7327
A-B 310.0829 070-39-58 +102.6599 +292.5959
B-C 105.9287 179-55-34 -105.9286 +0.1366

Sums;Lat=-0.0001,Dep=-0.0002
Thanks much.
Mike Burkes

Wrong!

Spledeus has the correct answers above.

I have verified them independently by a direct solution.

(Highlight the blank white space and they will appear.)

Wrong!

I see my mistake now. I needed to draw that lower right hand corner with exaggeration so it was clear what distance to use - the 40.501. By my method, the equation is:
(486.72 - 40.501) - 40/tan A = 115/ sin A