# Almost embarrasingly simple

1. Jun 23, 2006

### captainnofun

Hi, I'm new to the forum, and loathe to admit that I'm well out of high school and have the mathematical knowledge of a snowpea. Can anyone help me with the following scenario?

A person drops down a 200 ft. vertical shaft. Now underground, they will travel through a tunnel at a downward slope until they reach a depth of 1 and 1/2 miles. My question is, how steep do I need to make my slope? Since they are walking, it would need to be shallow enough that they can comfortably remain upright, but also steep enough that they will reach their target depth within 1 hour, 2 hours if it's more logical. I appreciate any help you guys can give me!

2. Jun 24, 2006

### arildno

For simplicity, let the velocity they travel with be called V, and the time to reach their destination T (in your case 1 or 2 hours).

Thus, the length L of the tunnel must be L=VT
Agreed so far?

Now, as measured from the starting point of the tunnel, the target point lies a depth D beneath it, (D equals 1 and a half miles in your case)

Now, D and L can be seen as two sides in a right-angled triangle, agreed?
(L is the hypotenuse of this triangle!)

The slope is the angle between the horizontal side H of this triangle and L.

In general, if we let $\theta$ denote the slope angle, we have:
$$\theta=\sin^{-1}(\frac{D}{L})=\sin^{-1}(\frac{D}{VT})$$
where $\sin^{-1}()$ is the inverse sine.

As you can see, you must know what the travelling velocity is in order to calculate the slope angle.

Last edited: Jun 24, 2006