Question 9 of 19: Spherical and Nested Spherical Variogram Models

Question 3.2: Spherical and Nested Spherical Variogram Models (10 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.

3.2.1 — The spherical model, mathematically. The spherical model is the most widely used because it has a genuine, finite range (unlike the exponential/Gaussian models, which only approach the sill asymptotically) and a physically intuitive near-origin behaviour: $$\gamma(h)=\begin{cases}C_0+C\left[\dfrac{3h}{2a}-\dfrac{1}{2}\left(\dfrac{h}{a}\right)^{3}\right] & 0\lt h\le a\\ C_0+C & h\gt a\end{cases}$$ where $C_0$ is the nugget, $C$ is the partial sill (structured variance), and $a$ is the range. The model rises steeply and near-linearly from the nugget at small $h$, curves over, and reaches the sill $C_0+C$ tangentially exactly at $h=a$ — the sketch below (single-structure illustration) shows the labelled nugget, sill and range on this characteristic S-shaped curve.

Given (3.2.2). Nested (two-structure) spherical model: nugget $C_0=0.1$; Structure 1 — partial sill $C_1=0.5$, range $a_1=30$ m; Structure 2 — partial sill $C_2=0.4$, range $a_2=150$ m. Each structure contributes independently and additively: $\gamma(h)=C_0+C_1\cdot\text{sph}(h/a_1)+C_2\cdot\text{sph}(h/a_2)$, where $\text{sph}(x)=1.5x-0.5x^3$ for $x\le1$ and $\text{sph}(x)=1$ for $x>1$.

Find. $\gamma(h)$ at $h=$ 0, 15, 100 and 200 m, and the total sill.

Approach. Evaluate each structure's spherical term separately at each $h$ (checking whether $h$ is inside or beyond that structure's own range), then sum both structures onto the nugget.

1. $h=0$ m — the nugget effect. As $h\to0^+$ neither structure has begun to contribute ($\text{sph}(0)=0$ for both), so the variogram value at (or immediately adjacent to) the origin is the nugget itself: $$\boxed{\gamma(0) = C_0 = 0.100}$$
2. $h=15$ m — both structures active (15 m $<$ both ranges). $$\text{sph}(15/30)=1.5(0.5)-0.5(0.5)^3=0.750-0.0625=0.6875 \;\Rightarrow\; C_1\cdot\text{sph}=0.5\times0.6875=0.344$$ $$\text{sph}(15/150)=1.5(0.1)-0.5(0.1)^3=0.150-0.0005=0.1495 \;\Rightarrow\; C_2\cdot\text{sph}=0.4\times0.1495=0.060$$ $$\gamma(15)=0.1+0.344+0.060$$ $$\boxed{\gamma(15) \approx 0.504}$$
3. $h=100$ m — Structure 1 has reached its sill (100 m $>$ 30 m); Structure 2 still active (100 m $<$ 150 m). $$C_1\cdot\text{sph}(100/30\ge1) = C_1 = 0.500\ \text{(full partial sill reached)}$$ $$\text{sph}(100/150)=1.5(0.6667)-0.5(0.6667)^3=1.0000-0.1481=0.8519 \;\Rightarrow\; C_2\cdot\text{sph}=0.4\times0.8519=0.341$$ $$\gamma(100)=0.1+0.500+0.341$$ $$\boxed{\gamma(100) \approx 0.941}$$
4. $h=200$ m — both structures beyond their range (200 m $>$ 30 m and $>$ 150 m): the total sill. $$\gamma(200)=C_0+C_1+C_2 = 0.1+0.5+0.4$$ $$\boxed{\gamma(200) = 1.000}$$