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Thread: Road Allignment

  1. #1
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    Road Allignment

    Hi guys

    Having a bit of trouble with the below with the wording possibly rather than the mathematical end. But anyway would appreciate it if anyone could work through it.

    A single carriageway road changes direction by 25º by means of a circular curve without transitions. The distance from the intersection point to the start of the circular curve is 170m, and the superelevation on the circular curve is 2.5%. Calculate the value of μ for a vehicle travelling at 80 km/h at the middle of the curve.

    Basically what needs to be found here is R the radius as it then goes into the formula v2/127R = e + μ. Obviously its just attaining R thats tricky

    Thanks a lot in advance
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  2. #2
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    Re: Road Allignment

    Quote Originally Posted by question View Post
    Hi guys

    Having a bit of trouble with the below with the wording possibly rather than the mathematical end. But anyway would appreciate it if anyone could work through it.

    A single carriageway road changes direction by 25º by means of a circular curve without transitions. The distance from the intersection point to the start of the circular curve is 170m, and the superelevation on the circular curve is 2.5%. Calculate the value of μ for a vehicle travelling at 80 km/h at the middle of the curve.

    Basically what needs to be found here is R the radius as it then goes into the formula v2/127R = e + μ. Obviously its just attaining R thats tricky

    Thanks a lot in advance
    1. Draw a sketch! (see attachment)

    2. You are dealing with 2 different right triangles. Use the 1st one to determine the necessary interior angles. (The midpoint of the curve is located on the angle bisector of the angle of 155°)

    3. Use the indicated grey right triangle to calculate the length of R:

    \frac{85\ m}{R}=\sin(12.5^\circ)

    Solve for R.

    EDIT: I've made a typo: Instead of 72.5° it should read 77.5°
    Attached Thumbnails Attached Thumbnails Road Allignment-strassenkurve.png  
    Last edited by earboth; January 4th 2012 at 05:34 AM.
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  3. #3
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    Re: Road Allignment

    Quote Originally Posted by earboth View Post
    1. Draw a sketch! (see attachment)

    2. You are dealing with 2 different right triangles. Use the 1st one to determine the necessary interior angles. (The midpoint of the curve is located on the angle bisector of the angle of 155°)

    3. Use the indicated grey right triangle to calculate the length of R:

    \frac{85\ m}{R}=\sin(12.5^\circ)

    Solve for R.

    EDIT: I've made a typo: Instead of 72.5° it should read 77.5°
    Thanks for the reply

    Was just about to respond regarding the 72.5 and 77.5.

    I had gotten to that point myself with the diagram and im sure you noticed solving for R gives a negative value of -1281.6m which is a bit dubious. The answer that should be obtained is 767m.

    Interpretation of the intersection point is crucial and i tried the 170m length as the distance from the start of the curve to where the line changes direction. This did not provide the correct answer either. I thinnk the wording of the question is deliberately troublesome.

    Really appreciate you taking the time to reply.

    Thanks
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  4. #4
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    Re: Road Allignment

    Quote Originally Posted by question View Post
    Thanks for the reply

    Was just about to respond regarding the 72.5 and 77.5.

    I had gotten to that point myself with the diagram and im sure you noticed solving for R gives a negative value of -1281.6m which is a bit dubious. The answer that should be obtained is 767m. <--- Your calculator is still in radians mode. You have to switch it into degree mode. Then the result is 392.7 m.

    Interpretation of the intersection point is crucial and i tried the 170m length as the distance from the start of the curve to where the line changes direction. This did not provide the correct answer either. I thinnk the wording of the question is deliberately troublesome.

    Really appreciate you taking the time to reply.

    Thanks
    I obviously misundertood the question.

    If you use the grey right triangle you know that

    \frac R{170\ m} = \tan(77.5^\circ)

    Solve for R. You'll get R \approx 766.8 \ m
    Attached Thumbnails Attached Thumbnails Road Allignment-streetcurve.png  
    Last edited by earboth; January 4th 2012 at 10:48 AM.
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