Good morning ladies and gents,
I am studying for my FS exam, and hope to study steady through the winter and sit sometime in the spring. Long story short, was hoping somebody was interested in walking me through this here problem. I can't find any great help so far online yet.
The problem comes from Caltrans FS study guide, unit 2, #44.
Workbook TOC
It is a symmetrical curve, grade in -1.65%, out 4.10%. PVI at 6+50, El.=34.23. The curve needs to pass 3.0' over a top of pipe elev of 36.11. Pipe is located at STA 6+87. Need to find length of curve to nearest half STA, and then calculate the PVC STA and ELEV.
On a side note, I did find a video with a methodology which got me close, but not correct! It was s= the square root of elev e - elev g / elev e - elev f. E being on the curve above the pipe, f being on the tangent line at the pipe STA, and g being the tangent IN line extended down to the pipe STA.
Then L being 2d(s+1)/s-1. d=hor dist from PVI to pipe. (That was a jumble I know, but I figure if youre familiar with that method you will recognize it)
Thanks in advance if anybody is feeling up to it
That is typical engineering problem, you have to guess at a solution, do all the math, then correct your guess again and again. It is called iteration. If you have children or grandchildren learning the new math, you will know that it is proper o guess, check and then guess again better.
Because they want the curve length to the nearest half station, the cover will be 3.0' or more.
Because the total change in grade is 5.75%, a 300' curve would have just under 2% grade change per hundred feet and a 250' curve would have a grade change of just over 2% per hundred feet. Pick one and solve or try something else. I note that Caltrans gives you 4 start point options, but the answer given does not include the length of the curve which you said was required.
Paul in PA, PE, PLS
I got a 650 foot curve with the VPC Sta. 3+25, Elev. 39.59.
This page will give you a good formula to work with. http://www.wsdot.wa.gov/publications/manuals/fulltext/M22-23/Vertical.pdf
Since the answer choices give us some hints, let's try a 500' curve beginning at 4+00, elev. 38.36'.
Elev @ 6+87 = Elev @ 4+00 + G1(687 - 400) + ((G2 - G1)/1000)(687 - 400)squared
Elev @ 6+87 = 38.36 - 4.74 + (0.0000575)(82369) = 38.36' which is 2.25' over the pipe.
Let's try a 600' curve beginning at 3+50, elev 39.18'
Elev @ 6+87 = Elev @ 3+50 + G1(687 - 350) + ((G2 - G1)/1200)(687 - 350)squared
Elev @ 6+87 = 39.18 - 5.56 + (0.0000575)(113569) = 40.15' which is 4.04' over the pipe.
Let's try a 550' curve beginning at 3+75, elev 38.77'
Elev @ 6+87 = Elev @ 3+75 + G1(687 - 375) + ((G2 - G1)/1200)(687 - 375)squared
Elev @ 6+87 = 39.18 - 5.15 + (0.0000575)(97344) = 39.63' which is 3.52' over the pipe. BEST ANSWER
Let's try a 650' curve beginning at 3+25, elev 39.59'
Elev @ 6+87 = Elev @ 3+25 + G1(687 - 325) + ((G2 - G1)/1200)(687 - 325)squared
Elev @ 6+87 = 39.59 - 5.97 + (0.0000575)(131044) = 41.16' which is 5.06' over the pipe. WRONG ANSWER
Most likely they wanted 5' of cover over the pipe.
This happens a lot in exams and test exams where the test preparer tries to change the parameters in a problem and does not check his math.
Paul in PA, PE, PLS
Check out my link above Paul. The answer is clear, it is not a guess.
vern, post: 342443, member: 3436 wrote: Check out my link above Paul. The answer is clear, it is not a guess.
Choosing a vertical curve starting at 3+25 is indeed a guess. An exact solution would be a vertical curve starting elsewhere and going exactly through the chosen point.
Paul in PA
Choosing a vertical curve starting at 3+25 is indeed a guess. An exact solution would be a vertical curve starting elsewhere and going exactly through the chosen point.
Paul in PA
PS, I just learned the difference between a Comment and a Reply.
My solution above was derived from the formula provided in the link. I never looked at the OP's link. Now that I have peeked, my answer matches D exactly.
Although you need to know how to use the formulas for the test, this spreadsheet may help with checking your work. Someone posted it last year and I tweaked it a little.
https://dl.dropboxusercontent.com/u/25124076/VertCalc.xls
Vern,
Using your equation, give me the elevation at 6+87.
Please note that I say "D" is the wrong answer.
Paul in PA
39.11 (3 foot above the pipe)
Go to profsurv.com Problem Corner #8 for a similar problem.
vern, post: 342458, member: 3436 wrote: 39.11 (3 foot above the pipe)
Vern,
I blew it. I forgot to change the 2L value as my vertical curve lengths changed.
Paul in PA
vern, post: 342435, member: 3436 wrote: I got a 650 foot curve with the VPC Sta. 3+25, Elev. 39.59.
This page will give you a good formula to work with. http://www.wsdot.wa.gov/publications/manuals/fulltext/M22-23/Vertical.pdf
That's the same answer I got running it through my handy dandy vertical curve computer program, first solving for a 3.0' cover elevation at the pipe station getting a 606.87 ' curve length and then then rounding up to to use a 650 ' curve length, which gives an elevation of 39.42 at the pipe or 3.31' of cover.
Using a 600' vertcal curve would give an elevation of 39.06 at the pipe or 2.95' of cover. This probably would also work, as what is 0.05 cover going to make?
Brosef, post: 342423, member: 9302 wrote: Good morning ladies and gents,
I am studying for my FS exam, and hope to study steady through the winter and sit sometime in the spring. Long story short, was hoping somebody was interested in walking me through this here problem. I can't find any great help so far online yet.
The problem comes from Caltrans FS study guide, unit 2, #44.
Workbook TOCIt is a symmetrical curve, grade in -1.65%, out 4.10%. PVI at 6+50, El.=34.23. The curve needs to pass 3.0' over a top of pipe elev of 36.11. Pipe is located at STA 6+87. Need to find length of curve to nearest half STA, and then calculate the PVC STA and ELEV.
On a side note, I did find a video with a methodology which got me close, but not correct! It was s= the square root of elev e - elev g / elev e - elev f. E being on the curve above the pipe, f being on the tangent line at the pipe STA, and g being the tangent IN line extended down to the pipe STA.
Then L being 2d(s+1)/s-1. d=hor dist from PVI to pipe. (That was a jumble I know, but I figure if youre familiar with that method you will recognize it)Thanks in advance if anybody is feeling up to it
Route Surveying and Design by Hickerson gives you the necessary formulas to calculate a curve through a fixed elevation.
Thanks everybody, I got it now. Big help
