• Jan 3rd 2012, 06:00 PM
question
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

• Jan 3rd 2012, 11:45 PM
earboth
Quote:

Originally Posted by question
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

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.

• Jan 4th 2012, 05:48 AM
question
Quote:

Originally Posted by earboth
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.

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
• Jan 4th 2012, 10:26 AM
earboth
Quote:

Originally Posted by question

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 $R \approx 766.8 \ m$