This cache is located on land owned by the Triangle Land Conservancy, https://www.triangleland.org/
I adopted this cache in 2015. Here is the original description by jcdecker:
This was originally designed as multi to honor a fellow cacher but it never seemed to get off the ground without stages being broken so I have archived the old one and started this new one. Same cache in the same spot for those that have already found it. I'm not sure about the logging rules in this situation. This cache is placed in honor of the local cacher wsgaskins reaching his 1000 find. If you have ever gone hiking with this cacher you know all too well that he believes the shortest distance to a cache is a straight line no matter what is in the way. In the Euclidean plane R^2, the curve that minimizes the distance between two points is clearly a straight line segment. This can be shown mathematically as follows using calculus of variations and the so-called Euler-Lagrange differential equation. The line element in R^2 is given by ds==sqrt(dx^2+dy^2), so the arc length between two points (x_1,y_1) and (x_2,y_2) is L==intds==int_(x_1)^(x_2)sqrt(1+y^'^2)dx, where y^'=dy/dx and the quantity we are minimizing is f==sqrt(1+y^'^2). Finding the derivatives gives (partialf)/(partialy)=0 d/(dx)(partialf)/(partialy^')=d/(dx)[(1+y^'^2)^(-1/2)y^'], so the Euler-Lagrange differential equation becomes (partialf)/(partialy)-d/(dx)(partialf)/(partialy^')==d/(dx)((y^')/(sqrt(1+y^'^2)))==0. Integrating and rearranging, (y^')/(sqrt(1+y^'^2))==c y^('2)==c^2(1+y^'^2) y^('2)(1-c^2)==c^2 y^'==c/(sqrt(1-c^2))=a. The solution is therefore y=ax+b, which is a straight line. Good luck and stay on the straight line.