Geological geometry question, can anyone explain this?

Leaving Cert Student

Attached is the q I am working on. The northern side of the road is ok for the most part although towards the very end my curve differs from that of the solution.

My real concern is the southern side of the road. I don't recall ever working on a carpark while the road is rising/falling? Perhaps somebody could try this q themselves and tell me how they do or simply talk me through the solution now? I am flabbergasted and considered myself to be reasonably good at this subject!

maughantourig

Good grief that is not a nice question. What year was that?

The best way I can think of dealing with this is:

The line |CD| is level so no cut/fill cones are needed. Ignore fall of road for that section.

The true gradient from D to E is not 1 in 15 as |DE| is not parallel to |AB|. A side edge view perpendicular to |AB| would make the line |DE| appear as 1 in 15 falling.

To find the true gradient from D to E, I would draw a skew line diagram with lines |AB| and |DE|. You need to draw an auxiliary elevation with |DE| as a true length. I think that should give you the inclination of |DE| to the horizontal plane (it's gradient).

Alternatively, you could treat it as a trigonometry problem. Given a triangle with sides P,Q and R, where R=True length of |DE|, P=Horizontal displacement between D and E, Q=Vertical displacement between D and E.

The gradient will be the angle between P and R. You can physically measure P (measure |DE| on map). Q can be determined as |AB| is 1 in 15 falling (difference in height between C and E). The angle between P and Q will be 90 degrees allowing you to find R, thus producing a triangle containing the true gradient of |DE|.

Once you have the true gradient of |DE|, you can apply your cut/fill cones and work as normal.

I could be wrong so you should probably check with your DCG teacher.

Let's hope they don't give that in June...

Leaving Cert Student

Good god sounds nasty! i have the marking scheme so i can throw that up here and maybe see if you can make sense of what they did?

ill be at the computer tonight ill do it then!

Leaving Cert Student

Here is the solution, anyone shed some light on what it is they are doing here?