Question 8 of 13: Two-Structure Spherical Variogram Model

Question 2: Two-Structure Spherical Variogram Model (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Formula. A two-structure nested spherical model adds a nugget and two independent spherical structures, each with its own sill contribution and range:

where C0 is the nugget, C1 and a1 are the sill contribution and range of the first (short-range) structure, and C2 and a2 the sill contribution and range of the second (longer-range) structure.

Why two nested spherical structures approximate almost any classical variogram shape. Each spherical structure contributes a smooth rise from 0 to its own sill C_i over its own range a_i. Superimposing a short-range structure (small a1) with a longer-range structure (larger a2) lets the combined curve mimic: a pure nugget-effect model (both C1, C2 → 0, all variance in C0); a single spherical model (C2 → 0); a Gaussian-like smooth curve near the origin (the short structure rounds off what would otherwise be a sharp linear rise); and a variogram that appears to have TWO ranges of influence – a common real feature of ore deposits that have both short-range (nugget-adjacent, grain-scale) and long-range (structurally controlled, zone-scale) continuity. Because the two structures' sills and ranges are independently adjustable, the nested model has enough free parameters to fit almost any monotonically rising, sill-bounded experimental variogram shape without resorting to more exotic model types.

Given.

Find. The total sill, and γ(h) at h = 0, 250, 700 and 1100 m.

Approach. The sill is simply the sum of the nugget and both structure contributions; each γ(h) is evaluated structure-by-structure, using the linear-cubic spherical form while h ≤ a_i and clamping that structure's contribution at C_i once h exceeds its own range.

1. Sill. The total sill is reached once both structures have flattened out: $$\text{Sill} = C_0 + C_1 + C_2 = 0.05 + 0.20 + 0.20 = \boxed{0.45}$$
2. γ(0). By definition γ(0) = 0 exactly (no separation, no variance); the nugget C0 = 0.05 is instead the value the model jumps to for any h however small – a discontinuity at the origin, not a value at it. $$\boxed{\gamma(0) = 0}$$
3. γ(250). h = 250 lies inside both ranges (250 < 500 and 250 < 900), so both structures use the full cubic form. For structure 1: x1 = 250/500 = 0.500, Sph(x1) = 1.5(0.500) − 0.5(0.500)^3 = 0.750 − 0.0625 = 0.6875. For structure 2: x2 = 250/900 = 0.2778, Sph(x2) = 1.5(0.2778) − 0.5(0.2778)^3 = 0.4167 − 0.0107 = 0.4060. Substituting: $$\gamma(250) = 0.05 + 0.20(0.6875) + 0.20(0.4060) = 0.05 + 0.1375 + 0.0812 = \boxed{0.269}$$
4. γ(700). h = 700 has passed structure 1's range (700 > 500, so Sph1 = 1, fully saturated) but is still inside structure 2's range (700 < 900). For structure 2: x2 = 700/900 = 0.7778, Sph(x2) = 1.5(0.7778) − 0.5(0.7778)^3 = 1.1667 − 0.2352 = 0.9315. Substituting: $$\gamma(700) = 0.05 + 0.20(1) + 0.20(0.9315) = 0.05 + 0.20 + 0.1863 = \boxed{0.436}$$
5. γ(1100). h = 1100 exceeds both ranges (1100 > 500 and 1100 > 900), so both structures are fully saturated at their own sill contribution and the variogram has reached the total sill: $$\gamma(1100) = 0.05 + 0.20(1) + 0.20(1) = \boxed{0.45 = \text{Sill}}$$