Hazard Footprints and Spatial Interpolation

Why Hazard Footprints Matter

An event may affect nearby locations with different levels of hazard intensity. A catastrophe model represents that spatial pattern with a hazard footprint: a map or set of spatial values describing the intensity associated with an event at different locations. The earlier material on hazard intensity measures distinguishes the physical quantity being measured; here, the concern is how that quantity is represented across space and assigned to an exposure location.

A location’s modeled intensity may be stored directly at a grid point, represented by a grid cell, or estimated from nearby grid values. These choices can produce different results, especially near a grid edge or event boundary. The method and boundary convention therefore need to be known before an assigned intensity is interpreted.

Definition: A hazard footprint is a spatial representation of an event’s hazard intensity. It may be stored as values at grid nodes, values assigned to grid cells, or a spatial function. Spatial interpolation estimates the intensity at a location between stored values. The event footprint’s boundary describes where the event is defined to have an effect; the grid boundary describes the area covered by the stored grid. These boundaries need not be the same.

For the examples below, intensities and coordinates are hypothetical and deterministic. No probability distribution, interest rate, or mortality assumption is involved; none is needed for assigning a specified footprint intensity to a location. Coordinates and intensity values must use consistent units and a stated coordinate reference system. In practice, mixing coordinate systems or treating degrees of longitude as a uniform distance can distort distances and locations.

Grid Values and Assignment Methods

A grid may store values at nodes (the grid’s intersections) or at cell centers. The distinction matters: interpolation formulas require values at specified sample locations. A method intended for node values should not be applied to cell-center values as if they were corner-node values without an explicit model rule.

One simple approach assigns each exposure the value of the nearest grid point. This is easy to implement and preserves the stored values, but it creates abrupt changes when a location crosses a boundary between nearest-point regions. A cell-based approach instead assigns the value stored for the cell containing the location. It is appropriate when values are defined as cell values, but it also produces step changes across cell edges. These methods are not interchangeable with interpolation.

For node values on a rectangular grid, bilinear interpolation estimates an interior value as a weighted average of the four corner values. Let \(H_{00}, H_{10}, H_{01}, H_{11}\) be the values at the lower-left, lower-right, upper-left, and upper-right corners, respectively. Define \(u\) as the location’s relative position from left to right and \(v\) as its relative position from bottom to top. Both coordinates range from 0 to 1 inside the cell.

Definition: For a rectangular cell with corner values \(H_{00},H_{10},H_{01},H_{11}\), bilinear interpolation at relative coordinates \((u,v)\) is

H(u,v)=(1-u)(1-v)H_{00}+u(1-v)H_{10}+(1-u)vH_{01}+uvH_{11}.

Here, \(u=(x-x_0)/(x_1-x_0)\) and \(v=(y-y_0)/(y_1-y_0)\), where \((x_0,y_0)\) is the lower-left corner and \((x_1,y_1)\) is the upper-right corner. This formula applies to a point inside the rectangular cell when the four values are defined at its corners. It does not, by itself, specify how to handle a point outside the cell or outside the footprint.

Why the Bilinear Formula Works

The formula can be derived in two one-dimensional interpolation steps. First, at a fixed relative height \(v\), interpolate horizontally along the lower and upper edges. The lower-edge estimate is \((1-u)H_{00}+uH_{10}\); the upper-edge estimate is \((1-u)H_{01}+uH_{11}\). Then interpolate vertically between those two estimates, assigning weights \(1-v\) and \(v\).

\begin{aligned} H(u,v) &=(1-v)\big[(1-u)H_{00}+uH_{10}\big] +v\big[(1-u)H_{01}+uH_{11}\big]\\ &=(1-u)(1-v)H_{00}+u(1-v)H_{10} +(1-u)vH_{01}+uvH_{11}. \end{aligned}

This construction also explains two useful checks. The four weights are nonnegative for an interior point, and their sum is \( (1-u)(1-v)+u(1-v)+(1-u)v+uv=1 \). Thus the interpolated value is a weighted average of the corner values and must lie between the smallest and largest corner values. If it does not, a coordinate, weight, or arithmetic error has occurred, or the point was not interpolated within the cell.

Worked Example: Interpolating Within an Unequal-Width Cell

A hypothetical footprint has node values \(H_{00}=2\), \(H_{10}=6\), \(H_{01}=4\), and \(H_{11}=8\) in intensity units. The rectangular cell spans \(x=0\) to \(x=20\) and \(y=0\) to \(y=10\). Find the bilinear intensity at \((5,5)\). The stored values are deterministic node intensities; no distribution, interest, or mortality assumptions apply.

Method and coordinates. The horizontal and vertical relative positions are \(u=(5-0)/(20-0)=0.25\) and \(v=(5-0)/(10-0)=0.50\). The cell is rectangular but not square, so the two relative coordinates must be calculated using their respective side lengths.

Weights. In the order \(H_{00},H_{10},H_{01},H_{11}\), the weights are \((1-u)(1-v)=0.75(0.50)=0.375\), \(u(1-v)=0.25(0.50)=0.125\), \((1-u)v=0.75(0.50)=0.375\), and \(uv=0.25(0.50)=0.125\). They sum to \(0.375+0.125+0.375+0.125=1.000\).

Intensity. The weighted value is \(0.375(2)+0.125(6)+0.375(4)+0.125(8)=0.75+0.75+1.50+1.00=4.00\). The answer is 4.0 intensity units.

Check and common mistake. The result lies between the corner minimum of 2 and maximum of 8. A common error is to use \(u=5/10\) merely because 10 is a cell dimension, or to reverse the unequal side lengths. The correct horizontal fraction is \(5/20\), not \(5/10\).

Interpolation Checks and a Second Calculation

A reliable calculation keeps the corner order fixed throughout. Write down which value belongs to each corner, calculate the weights in that same order, and verify that the weights sum to one. This is particularly important when a grid is presented as a table, because a row or column can easily be mistaken for a different corner.

Worked Example: Verifying Weights and a Grid Estimate

In a hypothetical rectangular cell, the node values are \(H_{00}=10\), \(H_{10}=30\), \(H_{01}=20\), and \(H_{11}=40\). An exposure location has relative coordinates \(u=0.25\) and \(v=0.60\). Find the interpolated intensity and check the weights. These are deterministic node values, with no stochastic or financial assumptions.

Method. Apply the bilinear formula using the stated relative coordinates. The four weights, in corner order, are \((1-0.25)(1-0.60)=0.75(0.40)=0.30\), \(0.25(0.40)=0.10\), \(0.75(0.60)=0.45\), and \(0.25(0.60)=0.15\).

Weight check. Their sum is \(0.30+0.10+0.45+0.15=1.00\).

Intensity. Multiplying and adding gives \(0.30(10)+0.10(30)+0.45(20)+0.15(40)=3+3+9+6=21\). The interpolated value is 21.0 intensity units.

Check and common mistake. The result is between 10 and 40, as a weighted average must be. An arithmetic trap is to calculate correct weights but add a product incorrectly; for example, the products here sum to 21, not 23. Recomputing each product separately gives \(3,3,9,6\), whose total is \(21\).

Boundary Effects: Grid Extent Versus Event Footprint

A location near an edge raises two distinct questions: is it within the area represented by the grid, and is it within the event footprint as defined by the model? A grid may stop short of an event boundary, extend beyond it, or cover a rectangular area while the event footprint has an irregular shape. A point outside the interpolation cell cannot be handled by the interior bilinear formula unless an additional boundary rule is specified.

Common edge rules include clamping a coordinate to the nearest grid edge, using the nearest available node, returning no value, or applying a separately specified extrapolation method. These choices have different meanings. Clamping extends an edge value outward; nearest-node assignment selects a stored value; returning no value signals that the grid does not support the location; extrapolation extends a spatial trend beyond the data. None is universally correct. A model must specify the rule, and a practitioner should not silently substitute one for another.

An event boundary requires separate treatment. If the model defines intensity to be zero outside a stated event footprint, then a location outside that footprint receives zero under that convention. If the event boundary is not defined to imply zero, an outside location may instead be unsupported or require another rule. A point exactly on the boundary also needs a consistent inclusion convention, especially when a polygon boundary or grid-cell edge is shared.

Worked Example: Applying an Explicit Grid-Edge Rule

A hypothetical node grid covers \(0\leq x\leq20\) and \(0\leq y\leq10\). On the left edge, the values at \(y=0\) and \(y=10\) are 20 and 16; on the right edge, they are 12 and 8. A location is at \((22,5)\). The model states that this location is inside the event footprint but outside the grid’s horizontal extent, and specifies edge clamping: replace an out-of-range coordinate by the nearest grid-edge coordinate. Find the assigned intensity. Values are deterministic; no probability, interest, or mortality assumptions apply.

Apply the rule. Clamp \(x=22\) to \(x=20\); \(y=5\) is within the grid. At the right edge, the relative vertical position is \(v=(5-0)/(10-0)=0.50\). The edge intensity is therefore \((1-v)(12)+v(8)=0.50(12)+0.50(8)=6+4=10\).

Result and check. The assigned value is 10 intensity units, between the two values on the right edge. For comparison, linear extrapolation in the \(x\)-direction would first give values at \(y=5\) of 18 at \(x=0\) and 10 at \(x=20\). Extending that trend two coordinate units farther gives \(10+[(10-18)/20](2)=10-0.8=9.2\), not 10. The model’s stated rule requires the clamped result, not extrapolation.

Common mistake. Treating every location outside the grid as outside the event would assign zero here, but the problem explicitly places the location inside the event footprint. Grid coverage and event support are different. Conversely, if the location were outside a footprint whose convention assigns zero beyond its boundary, that event rule would govern rather than grid-edge clamping.

Choosing and Auditing a Spatial Assignment

The method should reflect how the footprint values are defined, not merely which calculation is easiest. Nearest-node assignment preserves the original values but can introduce abrupt spatial jumps. Bilinear interpolation gives a continuous estimate within each rectangular cell when shared node values are used consistently, but it smooths local variation and cannot recover detail absent from the grid. Cell-based values may be the appropriate representation when each stored number summarizes an area rather than a point.

Interpolation estimates intensity; it does not establish that the estimate is physically accurate. Resolution, coordinate accuracy, grid construction, and the shape of the event boundary all affect what can be inferred. In particular, interpolation should not be mistaken for extrapolation: the bilinear weights are designed for \(0\leq u\leq1\) and \(0\leq v\leq1\). Outside those ranges, weights may become negative, so the result need not lie between the four corner values.

Locate the exposure consistently.
Confirm its coordinate reference system and identify whether the footprint stores node values or cell values.

Identify the applicable spatial region.
Determine whether the point lies inside the event footprint, inside the grid, and inside a particular interpolation cell.

Apply the stated assignment method.
Use the defined cell value, nearest node, or interpolation rule. Do not apply bilinear interpolation to locations outside its cell without an explicit boundary method.

Check the result.
For interior bilinear interpolation, verify the four weights sum to one and the result falls between the minimum and maximum corner values.

For exam-style calculations, the most important discipline is to keep the spatial rules explicit. Record the cell bounds, compute relative coordinates from the correct side lengths, preserve the corner ordering, and state what happens at both grid and event boundaries. This prevents a numerically correct interpolation from being assigned to a location where the model does not support it.

Key takeaway: A footprint’s assigned intensity depends on both its stored spatial values and the rule used to map those values to an exposure. Bilinear interpolation is a weighted average inside a rectangular node cell; boundary handling is a separate modeling choice that must be stated.

Check Your Understanding

Use the stated grid representation and boundary convention in each problem. Values and scenarios are hypothetical and deterministic.

A node cell spans \(x=0\) to \(x=8\) and \(y=0\) to \(y=4\), with \(H_{00}=1\), \(H_{10}=9\), \(H_{01}=5\), and \(H_{11}=13\). Find the bilinear intensity at \((2,1)\). (Answer: 4.0 intensity units)
A node cell has \(H_{00}=4\), \(H_{10}=12\), \(H_{01}=8\), and \(H_{11}=16\). At \(u=0.50\), \(v=0.25\), calculate the four weights and the interpolated intensity. (Answer: weights 0.375, 0.375, 0.125, 0.125; intensity 9.0)
A cell has corner values \(5,15,25,35\), in the order \(H_{00},H_{10},H_{01},H_{11}\). Find the intensity at \(u=0.40,v=0.50\), and verify that it lies within the corner-value range. (Answer: 19.0 intensity units)
A node grid spans \(0\leq x\leq20\), \(0\leq y\leq10\). At the top edge \(y=10\), its values are 16 at \(x=0\) and 8 at \(x=20\). A point at \((21,10)\) is inside the event footprint, and the stated rule is to clamp to the grid edge. What intensity is assigned? (Answer: 8 intensity units)
A model defines intensity as zero outside its event polygon and includes points on the polygon boundary. A location is exactly on that boundary but outside the rectangular interpolation grid. Which boundary rule determines whether the location is inside the footprint, and can an interior bilinear calculation be performed from the available grid values alone? (Answer: The stated polygon convention includes the point in the footprint; no, the grid values alone do not specify an interior bilinear value outside the grid, so a separate grid-boundary rule or additional data are needed.)

What You'll Learn