Question 1 of 7: Horizontal Circular Curve (Arc Definition)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Geom-A1 Surveying. Closed-book; any non-communicating calculator, ruler and protractor permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Where a datum is implied, elevations are referenced to the Canadian vertical frame (CGVD2013) and azimuths to NAD83(CSRS); US-foot stationing is retained wherever the printed question uses it.
Reference texts: Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley); Hofmann-Wellenhof et al., GNSS — Global Navigation Satellite Systems (Springer, 2008).
Given. A simple circular horizontal curve defined by:
Parameter
Value
Radius $R$
$900$ m
Intersection (central) angle $I$
$14^\circ45' = 14.75^\circ$
PI station
$1{+}948.800$ m
Definition
arc (metric)
Find. (1) curve length $L$ and tangent distance $T$; (2) external distance $E$, middle ordinate $M$ and long chord $LC$; (3) the stations of the PC (point of curvature) and PT (point of tangency).
Figure 1 — Circular curve: back tangent PC→PI, forward tangent PI→PT, arc of radius $R$ subtending the deflection angle $I$.
Approach. Apply the standard arc-definition curve formulas in $R$ and $I$, then station the PC back from the PI by $T$ and the PT forward from the PC by the arc length $L$ (never by $2T$).
Curve length (arc definition). The arc definition ties length directly to the central angle in radians; with $I = 14.75^\circ = 0.257436$ rad,
$$L = R\,I_{\text{rad}} = 900(0.257436) = \boxed{231.692\ \text{m}}$$
Tangent distance. Using the half-angle $I/2 = 7^\circ22.5' = 7.375^\circ$,
$$T = R\tan\tfrac{I}{2} = 900\tan 7.375^\circ = \boxed{116.490\ \text{m}}$$
External distance, middle ordinate and long chord. All follow from $R$ and $I/2$:
$$E = R\!\left(\sec\tfrac{I}{2}-1\right) = 900(1.008342-1) = 7.508\ \text{m}$$
$$M = R\!\left(1-\cos\tfrac{I}{2}\right) = 900(1-0.991727) = 7.445\ \text{m}$$
$$LC = 2R\sin\tfrac{I}{2} = 1800\sin 7.375^\circ = 231.053\ \text{m}$$
The long chord is slightly shorter than the arc ($231.053 < 231.692$ m), as it must be.
Stationing. The PC precedes the PI by the tangent distance, and the PT follows the PC along the arc:
$$\text{PC} = \text{PI}-T = 1948.800-116.490 = \boxed{1{+}832.310\ \text{m}}$$
$$\text{PT} = \text{PC}+L = 1832.310+231.692 = \boxed{2{+}064.002\ \text{m}}$$
