assumed to have a width of two feet and an overall length of eight feet overlap between successive courses is two inches table i______________________________________ taper cone apex cone slant trapezoidal inches percourse angle u03b8 height ft 0 2 degrees 2 39 width______________________________________1 1 375 u00b0 1591 36 39 0 14 u00b0 062 4 124 u00b0 529 64 39 0 43 u00b0 183 6 875 u00b0 316 80 39 0 72 u00b0 304 9 625 u00b0 225 24 39 1 02 u00b0 435 12 375 u00b0 174 09 39 1 32 u00b0 556 15 125 u00b0 141 32 39 1 62 u00b0 687 17 875 u00b0 118 44 39 1 96 u00b0 828 20 625 u00b0 101 49 39 2 26 u00b0 959 23 375 u00b0 88 37 39 2 59 u00b0 1 0910 26 125 u00b0 77 88 39 2 94 u00b0 1 2311 28 875 u00b0 69 27 39 3 31 u00b0 1 3912 31 625 u00b0 62 03 39 3 70 u00b0 1 5513 34 375 u00b0 55 84 39 4 11 u00b0 1 72shell apex 35 264 u00b0 54 02 39 4 25 u00b0 1 78______________________________________ for the first course the cone apex angle u03b8 is defined by equ1 note that for the first course the radius r 39 defining u03b8 is drawn to the center of the lowest element 34 so that u03b8 subtends one half the width w of the element 34 diminished by the overlap hence tan u03b8 is equal to 0 5 24 34 2 34 38 2 39 12 with each succeeding course u03b8 is increased by an amount corresponding to u03b4 u03b8 which was earlier shown to be given by equ2 the cone slant height column 3 is thus calculated for each course.