# How should I aggregate quanititative data from horizons to mapunits?

Components are each separate individual soils with individual properties and are grouped together for simplicity's sake when characterizing the map unit.

We often want to map data gathered and stored at the horizon level, but given that we're making a plan-view map, we need to aggregate the data from the multiple horizons to a single component value and without knowing exactly where the component boundaries are, we must then aggregate the data from components to mapunits. When the data we need to aggregate are quantitative such as permeability, water holding capacity, or organic matter percentage then we can perform various mathematical procedures in addition to the qualitative tools like using the value from a certain depth.

Aggregating horizon data to mapunits requires two steps. First is aggregating data from the horizons to the components and then aggregating from the components to the entire mapunit. Depth-weighted calculations are obviously of interest when dealing with horizon-level data, just as area-weighted values are used for component-level data.

- Calculating a depth-weighted sum
- Calculating a depth-weighted average
- You can also use the qualitative techniques on quantitative data

A depth-weighted sum is useful when you want a total value for a column of soil. Since the soil has different characteristics at different depths, the final value for the column must reflect that. To do so, we multiply the data values either by the percentage of the column depth that horizon covers or by the raw number of inches the horizon takes up. For example:

In the RED component of the mapunit shown above, there are three horizons. The BRIGHT_RED horizon is 9 inches deep and has a maximum water capacity of of .4 inches per inch. The other components, BRICK_RED and DARK_PINK, cover 22 and 34 inches respectively and have maximum water capacities of .1 and .22 inches per inch. The depth-weighted sum calculation looks like this:

(9*.4 + 22*.1 + 34*.22) = 3.6 + 2.2 + 7.48 = 13.28

So the maximum water holding capacity for this component is 13.28 inches of water. Here's a question for you. The answer is at the bottom of the page...think about your answer and check it by scrolling down.

Once you have values for the other components, how would you aggregate them so that you had a value for the entire mapunit?

This is very similar to calculating an area-weighted average. Multiply the data values for the horizons by the number of inches the horizon covers then divide by the total number of inches for all the horizons (i.e., the total depth). Here's a problem for you to work out...the answer is at the bottom of the page.

You want to know the average amount of organic matter in the soil. In the YELLOW component above, there are four horizons: YELLOW_GREEN (9 inches deep), GREEN (15 inches deep), DARK_GREEN (16 inches deep), and OLIVE (25 inches deep) which have organic matter percentages of 55%, 45%, 18%, and 6% respectively. What is the average percentage of organic matter for this component?

You can, of course, use the top horizon or limiting property techniques when working with quantitative data as well. We'll look at a limiting property example:

When talking about permeability, we may really be interested in the minimum permeability rate since that horizon would be the bottleneck. So we would take the minimum rate and use that value for the entire component.

Always remember
that for different variables and different mapping purposes, different
techniques are appropriate. Care is needed to make sure that the best
method is being used.** The correct answer is something
like, "It depends on what question I'm trying to answer." Maybe what
you're looking for is places where you might find flooding given a set
amount of precipitation so you'd use a limiting property approach. Maybe
you want the average holding capacity for some other purpose in which
case you would use an area-weighted average.**

The calculation for average amount of organic matter looks like this:

(9*.55 + 15*.45 + 16*.18 + 25*.06) / (9 + 15 + 16 + 25) =

(4.95 + 6.75 + 2.88 + 1.5) / 65 = 16.08 / 65 = .247

So there is 24.7% organic matter throughout the volume of the component.