5. Example Problems
a. Example 1
A +3.00% grade intersects a 2.40% at station 46+70.00 and elevation 853.48 ft. A 400.00 ft curve will be used to connect the two grades. Compute:
(1) Station and elevation for the curve's endpoints
(2) Elevations and grades at full stations
(3) Station and elevation of the curve's highest and lowest points
First, draw a sketch:
Figure B14 Example 1 Sketch 
For our computations, we'll use grades in percent and distances in stations.
(1) Use Equations B1 and B2 to compute the BVC and EVC
(2) Set up equations B3, and B5 to B7 to compute curve elevations.
Equation B3 

Equation B5  
Equation B6  
Equation B7 
Set up the Curve Table:
Equation B6  Equation B5  Equation B7  
Station, i  Dist, d_{i} (sta)  Elev_{i} (ft)  Grade, g_{i} (%) 
48+70.00 EVC  
48+00  
47+00  
46+00  
45+00  
44+70.00 BVC 
Why is the table upside down? We'll get to that in a second.
For each station, starting at the BVC, compute each column using the equation identified at the column top. Elevations will be computed to an additional decimal place to minimize rounding error.
sta 44+70.00:
sta 45+00:
and so on to the EVC.At the EVC each computed value has a math check indicated in red.
sta 48+70.00:
The completed Curve Table:
Equation B6  Equation B5  Equation B7  
Station, i  Dist, di (sta)  Elevi (ft)  Grade, gi (%) 
48+70.00 EVC  4.0000  848.680  2.40 
48+00  3.3000  850.029  1.46 
47+00  2.3000  850.809  0.10 
46+00  1.3000  850.239  +1.24 
45+00  0.3000  848.319  +2.60 
44+70.00 BVC  0.0000  847.480  +3.00 
There are a few other math checks in addition to those at the EVC.
In the Dist column, the difference between successive d's at full stations should be 1.0000
In the Grade column, the difference between successive g's at full stations should be k=1.35
Some texts make reference to Second Differences or Double Differences as another check. When computations were all done manually, as many checks as possible were used. We'll include it here just to show the reader how it worked.
A Second Difference is the difference between the elevation differences at full stations. Difference of a difference, got it? For this problem, we compute second differences this way:
Station 
Elev_{i} (ft) 
First Difference 
Second Difference 
48+70.00 EVC 
848.680 


48+00 
850.029 
850.029850.809=0.780 
0.7800.570=1.350 
47+00 
850.809 
850.809850.239=+0.570 
+0.5701.920=1.350 
46+00 
850.239 
850.239848.319=+1.92 

45+00 
848.319 


44+70.00 BVC 
847.480 


Note that values in the Second Difference column are equal to k, that's the math check. For you calculus buffs, the second difference is the second derivative of the Curve Equation.
It looks like a lot of computing for only two checks against k (and it is). But for those two checks to be valid requires that 5 of the 7 elevations be correct. The remaining two are the BVC and EVC.
(3) High and low point
Because this is a crest curve it will have a high point somewhere along it. The lowest point of the curve is at the BVC: 847.48 ft at 44+70.00.
Examining elevations in the completed Curve Table, it looks like the curve tops out between stations 46+00 and 48+00.
Use Equation B8 to compute the distance from the BVC:
Substitute into the Curve Equation to determine its elevation:
Add the distance to the BVC station to get the point's station:
In summary
Point 
Station 
Elevation (ft) 
High 
46+92.22 
850.81 
Low 
44+70.00 BVC 
847.48 
(4) Misc
You're not restricted to computing elevations only at full stations along the curve. Using the equations in step (2) developed for this curve, you can compute the elevation and grade for any point between 0.00' and 400.00' feet from the BVC. If you exceed 400.00', the equations will give you values (after all, they're just equations), but for part of the curve beyond.
OK, so why the upside down table? It's traditional. When standing on the alignment and looking up station, the layout of the table visually matches the stations in front of the surveyor, Figure B15.
Figure B15 Upside Down Curve Table 
b. Example 2
A 3.50% grade intersects a +2.00% at station 12+17.53 and elevation 634.25 ft. A 400.00 ft curve will be used to connect the two grades. Compute:
(1) Station and elevation for the curve's endpoints
(2) Elevations and grades at half stations
(3) Station and elevation of the curves highest and lowest points
The computations are left to the reader. The answers are shown for checking your work.
Low point at sta 12+72.07 and elev 636.80 ft.