A walking track is constructed over the surface of a pyramidal structure having a square base of side 100 feet and base angles of the inclined faces equal to 60 degrees each. The track was laid as the shortest path from one corner of the base to its opposite one.
Is it feasible for a traveller who cannot negotiate a slope stiffer than 1 in 2 to undertake and finish walking uphill along the track?
Consider the base with one corner towards us and the path beginning at that near corner, going onto the face to our right and then continuing on the hidden right-hand face.
Construct a vertical plane through the vertex of the pyramid and through the right-hand corner of the base of the pyramid. It will of course also go through the center of the base.
Construct a small sphere centered on that right-hand corner of the base. We'll use spherical trig on its surface.
The spherical triangle is defined by our constructed vertical line, the closer right face of the pyramid and the pyramid's base.
The two angles of this triangle that lie at the base measure 90 and 60 degrees, at the constructed plane and the front face respectively. The great-circle side of the triangle between these is 45 degrees as it subtends half the plane angle of the base. Find the remaining angle of this spherical triangle (say theta) using the law of cosines for angles, where the first term drops out as it contains cos(90):
cos(theta) = sin 60 cos 45 ~= .6123724356957945
The side between this angle and the 60-degree angle represents the planar angle at the base of the front face. Call this x. By the law of sines on the spherical triangle:
sin x = sin 45 / sin(theta) ~= .8944271909999159
The path along the front right face forms one leg of a right triangle with the hypotenuse being one edge of the square base. The angle x, at the right-hand corner, is the angle in this triangle that's opposite the side formed by the path, which starts at the near corner.
Construct another sphere at the near corner, at the beginning of the path. Also construct a vertical plane containing the path, as the angle of the path on this plane will determine our answer. The plane cuts the right-hand sloping edge of the pyramid at the same place that the path does, of course. This plane and the base of the pyramid and the nearby right face of the pyramid intercept the new sphere in another spherical triangle.
The vertical plane makes a 90-degree angle with the base and the face of the pyramid makes a 60-degree angle with the base. The side of the spherical triangle opposite the 90-degree angle is the complement of the x we found above, as mentioned there it's the other angle of a planar right angle on the face.
We can then use the spherical law of sines to find the other side, which the the angle we're trying to see if it can be negotiated. Call the angle y:
sin y / sin 60 = cos x / sin 90
sin y = sin 60 cos x ~= .3872983346207416
This makes y an elevation angle of 22.78649799959715 degrees, whose tangent is the required slope and is .4200840252084029 which is below the slope of .5 that's required, and so the walk is feasible for that traveller.
Posted by Charlie
on 2012-12-06 17:20:00