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

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:

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

Solve for R.

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:

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

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

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

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

Solve for R. You'll get $\displaystyle R \approx 766.8 \ m$