LSRS Ratings Aggregation Process

The initial LSRS calculations generate ratings for each component of a polygon. These components will now be aggregated or consolidated to produce a final rating symbol for the polygon.

The aggregation process produces one to four final categories with possible associated ratings (one special category and three normal categories). These final categories are:

Category Description Notes
WaterBody Those components which have standing water on them (i.e., water bodies). This is noted as "WB" in the final rating and is a special category: it has an area but no rating or subclasses.
WaterDrainage Those components which have an issue with drainage. This is a normal category, with area, rating, and possible subclasses.
DominantSimilar Those components which are similar in rating and landform to the DominantComponent. This is a normal category, with area, rating, and possible subclasses.
SignificantDissimilar Everything else which doesn't fit into the above. This is a normal category, with area, rating, and possible subclasses.

NOTE: At this point in time we do not take into consideration the Climate factor for the normal categories. It is common to all and its value is not affected by Component area. It will be re-introduced in Step 4 when determining the final polygon rating.



STEP 1: Assign Components to Categories

To populate the categories we first inspect and classify the components using various criteria.

STEP 1.1 (Populate the WaterBody category)
Inspect the components and extract all those which are water bodies into the special WaterBody category.
STEP 1.2 (Populate the WaterDrainage category)
Inspect the remaining components and classify all those which have a W subclass rating > 60 as belonging in the WaterDrainage category.
STEP 1.3 (Determine the dominant component)
Inspect the remaining components (of course it is possible that there will be none) and determine which component is the dominant component (i.e. the component with the largest area). If two or more components have the same area, then select the component with the highest soil rating as the dominant component. In the unlikely circumstance that two or more components have the same area and the same soil rating, then the highest landscape rating is used to determine the dominant component (of course it's technically possible for all the ratings and areas to be of the same value - in this case any one will do).
STEP 1.4 (Populate the DominantSimilar category)
Inspect the remaining components and place in the DominantSimilar category all those which meet the following criteria (i.e., evaluates as TRUE):

(soil rating of component being evaluated > DominantComponent.SoilRating - 15)

AND

(DominantComponent.LandscapeRating - 30 < landscape rating of component being evaluated < DominantComponent.LandscapeRating + 30)

STEP 1.5 (Populate the SignificantDissimilar category)
All remaining components (if any) are placed in the SignificantDissimilar category.

NOTE: Some categories may not have any components. This does not affect subsequent processing.



STEP 2: Summarize the special WaterBody category

For the WaterBody category only:

STEP 2.1 (Determine the WaterBody area)
Add up all of the individual component water body areas to produce a final area for the entire WaterBody category.



STEP 3: Summarize the normal categories:

Aggregate the primary ratings and areas for each of the three normal categories produced in Step 1 above. For each category:

STEP 3.1 (Summarize the soil ratings)
Combine the soil ratings for each component using a weighted average based on the area. This gives a new soil rating for the entire category.
STEP 3.2 (Summarize the landscape ratings)
Combine the landscape ratings for each component using a weighted average based on the area. This gives a new landscape rating for the entire category
STEP 3.3 (Sum the component areas)
Add up all of the individual component areas to produce a new final area for the entire category

E.g., for the normal category information shown below:

ComponentAreaSoil Factor
Rating
Landscape Factor
Rating
1206655
2206777
3105592

We get the following summary:

AreaSoil Factor
Rating
Landscape Factor
Rating
506471



STEP 4: Determine the class number

For each of the normal categories:

STEP 4.1 (Determine the class number)
Identify the class number based on the most limiting (i.e. the smallest number) of the Soil, Landscape, and Climate factor ratings, according to the following scale:
RatingClass number
80 ≤ value ≤ 1001
60 ≤ value < 802
45 ≤ value < 603
30 ≤ value < 454
20 ≤ value < 305
10 ≤ value < 206
0 ≤ value < 107

E.g., for the normal category information shown below:

AreaSoil Factor
Rating
Landscape Factor
Rating
Climate Factor
Rating
50647185

The soil factor rating is the lowest, so the class would be based on a rating of 64, i.e. class 2.



STEP 5: Aggregate the subclass values

Now aggregate the subclass values for each of the normal categories.

STEP 5.1 (Calculate weighted average)
Subclasses are combined using a weighted average (based on area) similar to the primary soil rating and primary landscape rating calculations shown above.

For example, if we extend the example from Step 3.3 above with the following subclass information:

Component Area Soil
Rating
Soil
Subclass
1
Soil
Subclass
2
Soil
Subclass
3
Soil
Rating
Soil
Subclass
1
Soil
Subclass
2
Soil
Subclass
3
1 20 66 M = 21 D = 21 - 55 T = 28 P = 22 -
2 20 67 M = 22 W = 21 - 77 P = 21 T = 10 -
3 10 55 W = 23 D = 22 Y = 21 92 - - -

The following calculations would apply:

SubClass
M((20 x 21) + (20 x 22) + (10 x 0)) / 50 = 17
D((20 x 21) + (20 x 0) + (10 x 22)) / 50 = 13
W((20 x 0) + (20 x 21) + (10 x 23)) / 50 = 13
Y((20 x 0) + (20 x 0) + (10 x 21)) / 50 = 4
T((20 x 10) + (20 x 28) + (10 x 0)) / 50 = 15
P((20 x 21) + (20 x 22) + (10 x 0)) / 50 = 17

STEP 5.2 (Sort in order of importance)
Sort the subclasses in descending order. Use the following importance table to sort subclasses with the same weighted average value.
SubClassFactor TypeImportance Value *
Aclimate0720
Mmineral soil0710
Morganic soil0709
Hclimate0620
Zorganic soil0610
Tlandscape0550
Wmineral soil0496
Worganic soil0495
Dmineral soil0485
Fmineral soil0480
Emineral soil0478
Nmineral soil0476
Norganic soil0475
Borganic soil0465
Vmineral soil0456
Vorganic soil0455
Gorganic soil0445
Plandscape0435
Jlandscape0425
Klandscape0422
Ilandscape0415
Omineral soil0405
Ymineral soil0400
*higher numbers indicate more importantance

In our processing we will determine which subclasses to use and not use based on their rating value. But it may be that subclasses will have the same value. In this case we must have some logic to determine which to keep and which to discard and in what order they should be presented. This will ensure that subclasses which happen to have the same value will be handled in a consistent manner. If the 3rd and 4th subclass both have the same value, then there will be a rational for which to keep and which to discard. The table above gives the priorities which have been assigned to each subclass.

For processing purposes every subclass is given a unique priority -- this is the "importance value" -- highest (most important) to lowest (least important) -- this is an arbitrary number from higher to lower. While a subclass may appear twice in the table (e.g. once from mineral component and once from the organic component), it will only occur once in reality. The basic premise for assigning importance is that those factors which can be easily modified should carry less weight (be less important).

For the example in Step 5.1 above, the following sort order would apply:

SubClassImportance ValueRating Value
M071017
P043517
T055015
W049613
D048513
Y04004


STEP 6: Drop the less important subclasses

At this point in time we have not dropped any subclasses (even those with zero value). We now enter a process to drop subclasses which are not needed. For each of the normal categories:

STEP 6.1 (Drop low value subclasses)
Drop all subclasses which have a value <= 20.
STEP 6.2 (Drop A and H)
If Climate is not the most limiting factor, then drop subclasses A and/or H.
STEP 6.3 (Drop A or M)
If both A or M subclasses are present, drop one based on the following logic:

if A > 20 and M >= A+5, then keep M.
if A > 20 and M < A+5, then keep A.

if A <= 20 and M > 20 and then keep M.

While A and M subclass symbols come from different factors (Climate and Mineral Soils), they effectively present the same information, thus we do not want to list them both if they are both present at this stage.

STEP 6.4.1 (Drop subclasses within factors)
Within each of the factors of Climate, Soil, and Landscape we drop subclasses if:

(Subclass_Value) < (MostLimiting_Subclass_Value * 0.40)

For example, if within the Soil factor we have a deduction value of 60 for W, and W is the most limiting subclass within the Soil factor, then we would drop all other subclasses within the Soil factor which have a value less than 24 (i.e. 60 * 0.40).

STEP 6.4.2 (Drop subclasses across factors)
Determine the deduction factors for Climate, Soil, and Landscape, by subtracting the factor rating from 100. Identify the most limiting factor deduction amount (the highest), and drop all subclasses for the other two factors if the following condition is TRUE:

(factor deduction amount) < (most limiting factor deduction amount * 0.33)

For example, if Climate has a rating of 34, then its total deduction amount is 66 (100 - 34). If climate was most limiting, then all Landscape subclasses would be dropped if the Total Landscape Deduction < 21.8 and all Soil subclasses would be dropped if the Total Soil Deduction < 21.8.

STEP 6.5 (Determine primary subclass)
Based on the most limiting factor, set the first subclass. We use the first subclass (most limiting/highest value) which occurs with the most limiting of Climate, Soil, or Landscape as the first subclass of our initial final rating (none of the above processes will have removed this limitation).
For the remaining subclasses, we select from highest to lowest from all remaining subclasses.


STEP 7: Create category codes

For each normal category:

STEP 7.1 (Determine final class and subclasses)
Produce a standard classification code from the final rating and subclasses. This consists of the Class (e.g., 3), the Primary subclass (if one exists), and one or two additional subclasses (if they exist), sorted highest to lowest. We will have three subclasses at most.

If we use the example with Climate not the most limiting factor:

Rating Subclass 1 Subclass 2 Subclass 3
Climate Factor 66 H = 35 - -
Soil Factor 64 M = 22 - -
Landscape Factor 71 P = 21 - -

We get:

Rating Subclass 1 Subclass 2 Subclass 3
Climate Factor 64 M P -

And our interim final rating is: 2MP. "M" is first since Soil was most limiting.

If the Climate factor was the most limiting, such as:

Rating Subclass 1 Subclass 2 Subclass 3
Climate Factor 55 H = 35 - -
Soil Factor 64 M = 30 - -
Landscape Factor 71 P = 25 - -

We get:

Rating Subclass 1 Subclass 2 Subclass 3
Climate Factor 55 H M P

And our interim final rating is: 3HMP. "H" is first, since Climate was most limiting.



STEP 8: Create a composite code

At this point in time, for each of the potential normal categories of WaterDrainage, DominantSimilar, and SignificantDissimilar, we have determined a rating code which consists of a class and its associated subclasses. This final step produces a composite rating code for the entire soil polygon.

STEP 8.1 (Create composite rating)
Combine the rating code for each of the categories into a single string, according to the following rules:
  1. List the normal category rating codes in order of area to which they apply, from largest to smallest, with area proportion as a decile value identifed in brackets and appended to each rating. Percent area proportions are rounded to deciles with the error distributed and calculations constrained such that total deciles do not add up to more than 10. Separate the individual ratings with a dash ("-").

    For example:
    2HMP(7) - 5P(3)

  2. List the WaterBody category first if it has the largest area, or last if does not have the largest area. This category is represented with "WB" followed by its area proportion as a decile value in brackets.

    For example:
    WB(6) - 2HMP(3) - 5P(1)

  3. If two or more of the final categories have the same rating but different areas, then the ratings are combined.

    For example: 2H(4) and 2H(5) and 3W(1) should be presented:
    2H(9) - 3W(1)

  4. If two of the final categories have the same subclass values and the same proportions but are in different overall classes, the best class is presented first.

    For example:
    3HMP(5) - 5HMP(5)

  5. If two categories have the same class values and the same proportions but different subclasses, the rating with the least number of subclasses (i.e. the fewest limitations) is presented first.

    For example: 4HMT(4) and 4H(4) and 5W(2) should be presented:
    4H(4) - 4HMP(4) - 5W(2)

  6. In the case that the class values and the proportions are the same and there are the same number of subclasses for each, the ratings are presented in alphabetical order.

    For example: 4HMT(4) and 4HDT(4) and 5W(2) should be presented:
    4HDT(4) - 4HMP(4) - 5W(2)

If we have:

Category Area (decile) Class Subclass 1 Subclass 2 Subclass 3
WaterDrainage 0 - - - -
DominantSimilar 6 2 H M P
SignificantDissimilar 4 3 H

Then our final polygon rating is: 2HMP(6) - 3H(4).


Special Data Cases for AGRASID: In the Agrasid database, there are "soil polygons" which are treated as special cases. These are Disturbed Lands. They do not have a rating and when encountered are given a final designation of "DL".