How Steep Should a Tunnel Slope Be for Comfortable Walking?

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SUMMARY

The discussion focuses on calculating the optimal slope angle for a tunnel that allows a person to walk comfortably while descending from a vertical shaft of 200 feet to a depth of 1.5 miles. The key formula derived is θ = sin-1(D/L) = sin-1(D/VT), where D is the depth, L is the length of the tunnel, V is the velocity, and T is the time taken to reach the destination. The participants emphasize the importance of determining the walking velocity to accurately compute the slope angle required for comfortable walking.

PREREQUISITES
  • Basic understanding of trigonometry, specifically right-angled triangles.
  • Familiarity with the concepts of velocity and time in relation to distance.
  • Knowledge of the inverse sine function and its application in angle calculation.
  • Understanding of comfortable walking angles and their implications on slope design.
NEXT STEPS
  • Research the effects of slope angle on human biomechanics during walking.
  • Learn about the practical applications of trigonometry in engineering and architecture.
  • Explore methods for calculating optimal walking speeds in varying conditions.
  • Investigate safety standards and guidelines for tunnel design and construction.
USEFUL FOR

Engineers, architects, and designers involved in tunnel construction, as well as anyone interested in the mathematical principles of slope design for safe and comfortable pedestrian pathways.

captainnofun
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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!
 
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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 [itex]\theta[/itex] denote the slope angle, we have:
[tex]\theta=\sin^{-1}(\frac{D}{L})=\sin^{-1}(\frac{D}{VT})[/tex]
where [itex]\sin^{-1}()[/itex] is the inverse sine.

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