Two tangents intersecting at V with azimuths of 255⁰12’ and 281⁰24’ respectively are joined by a 4⁰ simple curve. Without changing the direction of the two tangents, it is required to shorten the curve to one-fourth of the original length of the curve from the beginning of the curve. The stationing of the vertex of the new curve is at 10+545. Determine the following:  Tangent of the new curve. Stationing of the beginning of the curve. Stationing of the end of the curve. Long chord of the new curve

Q: A simple horizontal curve has 35 degrees of curvature and subtends an angle of 110 degrees. If the…

A: Degrees of curvature, D = 35°PC is at station  = 3 + 65.00 ft

Q: A simple curve connects two tangents AB and BC with bearing N 85°30' E and S 68°30' E, respectively.…

A:

Q: A compound curve connects three tangents having an azimuths of 250°, 270 ° and 290° respectively.…

A: The given data is: The azimuths are 250°, 270 °, and 290° The length of the chord=340m

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