Okay guys & gals, I need a little help. I haven't had to calculate a compound curve longhand in a very, very long time. I have been working stuff long hand for the last few weeks, and checking myself, and my brain is just about mush at this point.
I am sure this is just one of those instances where the answer is right there staring me in the face, but I am stumped on this one. Please point me in the right direction on how you would approach this problem.
I am pretty sure after the first reply or two I will slap myself in the forehead and have one of those doah! moments.
Please don't beat me up too bad!:-O
Thanks,
Jimmy
Been a long time for me too. I see a right triangle in the first curve with an adjacent side of 400' and a hypotenuse of 450'. That makes the half delta 27-15-58 or the delta 54-31-56. Subtracting the 72-31-56 from 180 to get delta 1 plus delta 2 = 107-28-04, which makes delta 2 52-56-08. Didn't have time to work the rest. Probably been 25 years since I did that much. Maybe a 50/50 chance I am right working with an unfamiliar calculator.
Delta Curve 1 = 54°31'55.5" -- Total Delta = 107°17' -- Delta Curve 2 = Total Delta - Delta Curve 1 = 52°45'04.5"
Thanks guys. The right triangle hint and the delta 1 and delta 2 hints triggered what I needed.
There are several elaborate formulas in many of the various reference books on route surveying, but with the data given for this problem, by the time that you do the necessary calculations to solve certain aspects of the problem, just using simple trig formulas and curve formulas will have everything almost solved anyway. Once the longer radius curve is solved, then the other curve more or less is solved using the same formats.
Charles,
That is exactly how I solved it, using the simpler formulas. I broke out my Meyer/Gibson book and my Hickerson book and was trying to figure it out. I totally overlooked the simple right triangle.
Thanks to both of you for the help.
Jimmy
Drawing in the radial lines makes these kind of problems much easier to envision the answers.
The big question is: Which definition of the PI station is wanted. PC Sta.+tangent? or same as mid point of the curve?
I agree 100%. I gave answers for the PC station plus 1/2 of the arc length for the PI station.
The review packet is 98 pages long. My portion is pages 1 thru 32, and 55 thru 98. I have worked all of the problems long hand, and then, when applicable, worked the problems graphically in Carlson/Cogo to verify that my calculations are correct.
I want to make sure I am working these correctly, and giving the students/test takers the best review class I can.
> I agree 100%. I gave answers for the PC station plus 1/2 of the arc length for the PI station.
>
> The review packet is 98 pages long. My portion is pages 1 thru 32, and 55 thru 98. I have worked all of the problems long hand, and then, when applicable, worked the problems graphically in Carlson/Cogo to verify that my calculations are correct.
>
> I want to make sure I am working these correctly, and giving the students/test takers the best review class I can.
The P.I. Station is the Tangent Length added to the P.C. Station, not ½ of the arc length added to the P.C. Station. In this problem, P.I.1 = 22+50 and P.I.2 = 27+42.63
Thank you Charles. I stand corrected. I will have to go back and review my notes again.
This place is great.
also:
PCC 26+31.01 & PT 28+38.25
That's also what I got for those stations.
