GEFS - Vertical Curves - Symmetrical Parabolic Curves PDF

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CBLAMSIS UNIVERSITY OF THE CORDILLERAS VERTICAL CURVESVERTICAL ALIGNMENTIn road construction, to avoid sudden change in direction when two tangents with slopes greater than or less than zero intersects, it is customary to introduce a vertical curve at every such point where the angle is large enough...

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GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

VERTICAL ALIGNMENT In road construction, to avoid sudden change in direction when two tangents with slopes greater than or less than zero intersects, it is customary to introduce a vertical curve at every such point where the angle is large enough to warrant it. The vertical curve that may be used to connect tangents to gradually change slope from positive to negative (summit curve) or from negative to positive (sag curve) are compound curves and parabolic curves. Generally parabolic curves are being used because parabola effects the transition rather better theoretically than the circle, but its selection for the purpose is due principally to its greater simplicity of application. Types of Parabolic Curves 1. Symmetrical Parabolic Curves 2. Unsymmetrical Parabolic Curves Elements of a parabolic curve L = length of curve, the horizontal projection of the parabolic curve from PC to PT. H = vertical tangent offset from the vertex, V, to a point below it (summit curve) or above it (sag curve) on the curve. G1 = grade (slope) of the back tangent in % g1 = grade (slope) of the back tangent in decimal G2 = grade (slope) of the forward tangent g2 = grade of the forward tangent in decimal A = change in grade from PC to PT in % Summit curves

V

PT H PT

PC

V

H

L PC

L

Sag curves L PC

L

PC

PT

V

H

H V

PT

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GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

Symmetrical Parabolic Curves Symmetrical parabolic curve are curves with their equation as the equation of a parabola either opening upward or downward. They are called symmetrical parabolic curve because the length of curve to the left and to the right of the vertex, V, are equal and equal to one-half of the total length of curve, L. The figure shown below is a profile. The vertical axis represents the elevation at any stations and horizontal axis represent the stations. For example, sta. PC is known, sta. V equals sta. PC plus L/2 and sta. PT equals sta. PC plus L. Also, if elev. PC is known, Elev. V equals elev. PC plus Y. Elevations V

Elev. V

H PT

Y Elev. PC PC L/2

L/2 L

Sta. V

Sta. PC

Sta. PT

Stations

Relationships of the elements of a symmetrical parabolic curve

C

g1(L/2)

Figure X

(g1 – g2)(L/2) V H

- g2(L/2) PT

H B

PC

L/2

L/2 L

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GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

Relationship of the vertical distance, horizontal distance, and slope of a line.

B Vertical distance A

θ

Horizontal distance

𝑇𝑎𝑛 𝜃 =

𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = (𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐿𝑖𝑛𝑒) Or “The vertical distance between two points on a line is equal to the horizontal distance between the two points multiplied by the slope of the line connecting the two points”. So in figure X, the vertical distance between point V and point C is equal to (L/2)(g 1) and the vertical distance between point V and point PT is equal to (L/2)(- g2), the negative sign means a downward slope. The total vertical distance between PT and point C equals to the sum of (L/2)(g1) and (L/2)(-g2). 𝐿 𝐺2 𝐿 𝐿 𝐿 𝐿 𝐺1  − )= (𝐺 − 𝐺2 ) 𝑃𝑇𝐶 = ( ) 𝑔1 + ( ) (−𝑔2 ) = ( ) (𝑔1 − 𝑔2 ) = ( ) ( 2 2 2 2 100 100 200 1 Also, in figure X triangles PCBV and PCPTC are similar triangles. By ratio and proportion using the base and altitude of a triangle, 𝑏𝑎𝑠𝑒 𝐵𝐴𝑆𝐸 = 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 𝐴𝐿𝑇𝐼𝑇𝑈𝐷𝐸 𝐿 2𝐻 200 (𝐺1 − 𝐺2 ) = 𝐿 𝐿 2 𝐿(𝐺1 − 𝐺2 ) 𝐻= 800

Where: (G1 – G2) = change in grade from PC to PT = A A = |𝐺1 − 𝐺2 | – for summit curves A = |𝐺2 − 𝐺1 | – for sag curves Note: The negative sign in the quantity (G1 – G2) and (G2 – G1) represent a downward slope or an operational sign. When the sign of the grades of the back and forward tangent are unlike (positive to negative or negative to positive), the negative sign represents a downward slope. When the sign of the grades of the back and forward tangents are the same (positive to positive or negative to negative), the negative sign represents an operational sign.

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GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

Squared Property of Parabola

Figure Y

C M

Grade Elevation

g1(L/2) (g1 – g2)(L/2)

Vertical tangent offset

V Y

HY

- g2(L/2)

Curve Elevation X

N

PT

X

PC

L/2

L/2 L

Y’s are the vertical tangent offsets from the tangents and X’s are the horizontal distances from the point of tangency of the curve and tangent, where the vertical tangent offset is reckoned, to the location of the point. Y1 and Y2 are tangent offsets reckoned from the back tangents, Y3 and Y4 are tangent offsets reckoned from the forward tangent. The back tangent and the parabolic curve are tangent at point PC, thus the horizontal distance of Y1 and Y2 shall be reckoned from point PC to the point location. Similarly, the forward tangent and the parabolic curve are tangent at point PT, thus the horizontal distance of Y3 and Y4 shall be reckoned from point PT to the point location. The vertical tangent offset, H, is either reckoned from the back tangent or forward tangent, and its horizontal distance either from PC or PT is L/2. The square property of parabola states that in the same tangent, the ratio of the vertical tangent offset and the square of its equivalent horizontal distance of a point is equal to the ratio of the vertical tangent offset and the square of its equivalent horizontal distance of another point. Applying squared property of parabola in Figure Y, above. Considering the back tangent.

𝐿 ( ) (𝑔1 − 𝑔2 ) 𝑌1 𝐻 𝑌2 = 2 = = (𝑋1 )2 (𝑋2 )2 𝐿2 𝐿 2 (2 ) Considering the forward tangent. 𝑌3 𝐻 𝑌4 = = 2 2 (𝑋4 ) (𝑋3 ) 𝐿 2 (2 )

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GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

Therefore, for symmetrical parabolic curve. 𝑌1 𝐻 𝑌4 𝑌3 𝑌2 = = = = 2 2 2 2 (𝑋4 ) (𝑋3 ) (𝑋2 ) (𝑋1 ) 𝐿 2 ( ) 2 Meaning the squared property of parabola can be applied to any tangent for a symmetrical parabolic curve. You can get a tangent offset from the back tangent and can be proportioned to any tangent offset from the forward tangent. Using squared property of parabola to relate the elements of the curve. Considering the vertical tangent offset through V and PT. 𝐻

𝐿 2 (2 )

𝐿 ( ) (𝑔1 − 𝑔2 ) 4𝐻 = 2 = 2 𝐿2 𝐿

𝐺1 𝐺2 𝐿(𝑔1 − 𝑔2 ) 𝐿 (100 − 100) 𝐿(𝐺1 − 𝐺2 ) 𝐴𝐿 = 𝐻= = = 8 8 800 800 Point M is on the tangent, the elevation of points on the tangents are called grade elevations. Point N is on the curve, the elevation of points on the curve are called curve elevations. Y2 is a vertical tangent offset. The relationship of grade elevation, curve elevation, and vertical tangent offset is Curve Elevation = Grade Elevation minus vertical Tangent offset (summit curves) Location of the highest or lowest point on the curve.

Y L L/2

L/2

PC Curve Elevation Vertical tangent offset Grade Elevation

N

PT

LP

Y

X

H

M V S1

S2

S1 = location of the lowest or highest point from PC S2 = location of the highest or lowest point from PT Since the lowest or highest point is a single point on the curve, the length of curve, L, shall be equal to the sum of S1 and S2. 𝑆1 = 5 |1 3

𝐺1 𝐿 𝐺1 𝐿 = (𝐺1 − 𝐺2 ) 𝐴

𝑓𝑟𝑜𝑚 𝑃𝐶

GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS 𝑆2 =

𝐺2 𝐿

𝐺2 𝐿 = 𝐴 (𝐺2 − 𝐺1 )

VERTICAL CURVES

𝑓𝑟𝑜𝑚 𝑃𝑇

𝐿 = 𝑆1 + 𝑆2 The derivation of the formula for S1 and S2 will not be shown here. It will be the responsibility of a student to research or derive. Point M is on the tangent, the elevation of points on the tangents are called grade elevations. Point N is on the curve, the elevation of points on the curve are called curve elevations. Y is a vertical tangent offset. The relationship of grade elevation, curve elevation, and vertical tangent offset is Curve Elevation = Grade Elevation plus vertical Tangent offset (sag curves)

SAMPLE PROBLEM NO. 1 A 15% ascending grade meets a descending grade at the vertex, V, at sta. 25 + 000 and at elev. 300.0 m. The symmetrical parabolic curve connecting these tangents passes through a point at sta. 25 + 015 having a curve elevation at elev. 297.902 m. The highest point on the curve is located 27.826 m from PT. Determine the length of curve, the grade of the other tangent, and the elevation and stationing of the highest point on the curve. Given:

X’ Z

Elev. Z (L/2 – S2)g1

Elev. X’ = Elev. V + 15g1 15g1

V

YX

YZ H

Elev. HP

Elev. X = 297.902 m HP

X S2 = 27.826 m

PC

15 m L/2

L/2 L

Sta. 25 + 000 Sta. 25 + 015

Required: L, G2, Elev. HP, Sta. HP

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Elev. V = 300 m

PT

GEFS (FOUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

VERTICAL CURVES

Solution: Hint: To choose the tangent where the vertical tangent offset be reckoned, choose the tangent of known grade. For point HP and X, their vertical tangent offsets are reckoned from the back tangent, YZ and YX respectively. Using squared property of parabola to solve the length of curve, L. 𝐴𝐿 𝐴 𝐻 𝑌𝑋 800 = = = 2 𝐿2 200𝐿 𝐿 2 𝐿 ( ) ( + 15) 4 2 2 𝑌𝑋 𝐴 4𝑌𝑋 = 2 = 200𝐿 (𝐿 + 30)2 𝐿 + 30 ( ) 2 Where: Solve A and YX in terms of L, Since the grade of the back tangent is given, use S1 to solve for A, 𝐺1 𝐿 = 𝐿 − 𝑆2 𝑆1 = 𝐴 𝐺1 𝐿 𝐿 − 𝑆2 Using the relationship of the curve elevation, grade elevation, and vertical tangent offset 𝐴=

Curve Elevation = Grade Elevation minus vertical Tangent offset (summit curves) Tangent offset = Grade Elevation minus curve elevation 𝑌𝑋 = 𝐸𝑙𝑒𝑣. 𝑋 ′ − 𝐸𝑙𝑒𝑣. 𝑋 𝑌𝑋 = 𝐸𝑙𝑒𝑣. 𝑉 + 15𝑔1 − 𝐸𝑙𝑒𝑣. 𝑋 𝑌𝑋 = 300 + 15(0.15) − 297.902 = 4.348 𝑚 Substitute: 4𝑌𝑋 𝐴 = 200𝐿 (𝐿 + 30)2

𝐺1 𝐿 4(4.348) 15 𝐺1 𝐿 − 𝑆2 = = = 2 200𝐿 (𝐿 + 30) 200(𝐿 − 𝑆2 ) 200(𝐿 − 27.826) (𝐿 + 30)2 =

4(4.348)(200)(𝐿 − 27.826) 15

𝐿2 + 60𝐿 + 900 = 231.893𝐿 − 6452.664 𝐿2 − 171.893𝐿 + 7352.664 = 0 𝐿=

−(−171.893) ± √(−171.893)2 − 4(1)(7352.664) 2(1) 171.893 + 11.685 (+)𝐿 = = 91.789 𝑚 2 171.893 − 11.685 (−)𝐿 = = 80.104 𝑚 2

There are two solutions of the problem: L = 91.789 m and L = 80.104 m.

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