Hi folks please reference latest JAC problem,www.scsurveyjac.org/16.html. According to author Chuck, in curve A line RP-PCC is 90 degrees to line PCC-PI and PCC-RP azimuth is 00-09-54 and angle PCC-PI-PC is 90-00-22. I also found line PC-PI to be 315.21 and azimuth 36-36-51.58 and got stumped. Thanks much.
Mike Burkes
626-833-1521
> Hi folks please reference latest JAC problem, www.scsurveyjac.org/16.html. According to author Chuck, in curve A line RP-PCC is 90 degrees to line PCC-PI and PCC-RP azimuth is 00-09-54 and angle PCC-PI-PC is 90-00-22. I also found line PC-PI to be 315.21 and azimuth 36-36-51.58 and got stumped. Thanks much.
> Mike Burkes
> 626-833-1521
Delta of Curve A = (180-00-00) - (90-00-22)
Let's call the 500' curve, Curve C.
You can compute the RP (C) from the given PC (C) and then "Point on Curve" (C) by bearing-distance intersection.
Then you have two points on Curve C.
You know the distance from RP (C) to RP (B) is 600'. You do a bearing-distance intersect holding 600 feet distant from RP (C) and a 100' offset north from line PI-PCC to get RP (B).
Then compute PRC at 100 feet from RP (B) along line RP(B)-RP(C).
Then compute PCC by bearing and distance from RP(B) using AZ 00-09-54 & 100'.
Use Distance PCC-PI and compute PI-PC by bearing and distance.
Compute RP (A) by intersection.
Hi Ryan,thanks so much for input I will work it when able.
Mike Burkes
626-833-1521
looks like one of Dave Lindell's "Problem Corner" episodes.
Wouldn't be surprised if he chimes on with this.
Mike:
Here's what I came up with.
Those aren't the answers I got. edit: I just found my problem.
I looked at it again, but will have to look at it closer later. I'm relatively convinced I did something wrong, but I came up with a radius of curve A of 202.86 and a tangent distance of 202.83.
I might have to redo it from scratch to make sure I'm not making a similar error over and over.
I got the same as Charles.
Here is the basic procedure:
1. Calculate the coordinate of RP curve C from PC (N4747.20,E4813.00) assuming 90 degrees from given azimuth of 359-58-44.
2. Calculate coordinate of a point 100' northerly from PI of Curve A (N5000.21,E5001.00). Azimuth is 90 degrees from PI to PCC (AZ 270-09-54).
3. Calculate RP of Curve B by bearing-distance intersection, bearing from coordinate Step 2 is given azimuth from PI Curve A (270-09-54) and distance is 600' (500' + 100') from RP of Curve C (coordinate calculated in Step 1). Note that when using the law of SINs, the angle you get at the RP for Curve B needs to be subtracted from 180.
4. Calculate Point on Curve by bearing-distance from PI Curve A and 500' from RP Curve C. This is similar to Step 3.
5. Calculate PCC by bearing at right angles to line PCC-PI and distance of 100'.
6. Inverse PCC to PI yields the Tangent of Curve A and the rest should be easy.
> I got the same as Charles.
I did too. I reworked it.
I got different values, than Dave, but basically followed the same process.
RP Curve 'C' N = 4747.016, E = 4313.000
point 100' north of PI Curve 'A"N = 5100.210, E = 5001.288
Hi folks,wow so cool of you all to work this I will continue to peruse and for the heck of it I will set up a traverse with those points and most likely it will close. Does that sound feasible?
Believe me I am working it with you folks and not just pluggin' and chuggin'.
Thanks so much for your inputs and keep on feeding back.
Mike Burkes
626-833-1521
Done, now get out there with transit & chain & STAKE IT!
Here is my answers (see page 4):
Done, now get out there with transit & chain & STAKE IT!
Hi DK,thanks so much for your effort and I would love to stake it if required.
Mike Burkes
626-833-1521