This page provides a detailed explanation for how stages are established for the aggregated stage-damage function.

There are three sections of the function for each impact area, damage category, and asset category for which the software generates stages: a bottom, middle, and top portion. The stages will be the same for all functions of a given impact area, scenario, and analysis year, such as base-year without-project condition in impact area 1. The bottom portion includes stages extrapolated from the stage at the most frequent AEP event provided in the hydraulic model (typically .5 AEP) to the stage at the .9999 AEP event for graphical frequency functions. The middle portion includes stages interpolated between the stages at the AEP events provided in the hydraulic model. The top portion includes stages extrapolated from the stage at the least frequent AEP event provided in the hydraulic model (typically .002 AEP) to the stage at the .0001 AEP event. For each portion of the functions, a minimum and maximum stage are identified and then a number of equally spaced aggregation stages is calculated. The top and bottom extrapolated portions of the aggregated stages are derived differently from the middle interpolated portion of the aggregated stages as described below.

Description

Bottom Portion

Bottom Portion - Minimum Stage

The minimum stage is found by extrapolating out to the stage at the .9999 event. This is the stage that is the minimum for the bottom portion and for the entire stage-damage function. This stage represents the lowest possible stage in the river.

Bottom Portion - Maximum Stage

The maximum stage is determined by first identifying the lowest AEP event provided in the hydraulic model, typically the .5 AEP event. Then identify the input flow that corresponds to that AEP event in the flow frequency function. Finally, the maximum stage is the most likely stage in the rating curve for that flow value.

Bottom Portion - Number of Equally Spaced Aggregation Stages

The number of aggregation stages between those stages must be at least four. Otherwise, the number of aggregation stages is equal to four times the difference between the minimum stage and maximum stage of the bottom portion. This results in a minimum of four aggregations stages per foot of the function domain, and assumes the unit of measurement is in feet.

$//<![CDATA[ \begin{array}{l}\mathrm{(Stage_m_a_x\,- Stage_m_i_n) \cdot 4}\end{array} //]]>$

OR 

4; Whichever is greater

Middle Portion

Middle Portion - Minimum Stage

The maximum stage is determined by first identifying the lowest AEP event provided in the hydraulic model, typically the .5 AEP event. Then identify the input flow that corresponds to that AEP event in the flow frequency function. Finally, the minimum stage is the most likely stage in the rating curve for that flow value.

Middle Portion - Maximum Stage

The maximum stage is determined by first identifying the least frequent AEP event provided in the hydraulic model, typically the .002 AEP event. Then identify the input flow that corresponds to that AEP event in the flow frequency function. Finally, the maximum stage is the most likely stage in the rating curve for that flow value.

Middle Portion - Number of Equally Spaced Aggregation Stages

The middle portion includes the stages at all the other AEP events provided in the hydraulic model found the same way as the minimum and maximum stage; by mapping the most likely stage on the rating curve to the flow and AEP event in the flow-frequency function.

Additionally, there are an equal number of equally spaced aggregation stages between those stages, which must be at least two. Otherwise, the number of additional points is equal to four times the difference between the minimum stage and maximum stage of the middle portion, divided by the number of hydraulic profiles minus one. Again, this results in a minimum of four aggregations stages per foot of the function domain, and assumes the unit of measurement is in feet.

$//<![CDATA[ \begin{array}{l}\mathrm{[(Stage_m_a_x\,- Stage_m_i_n) \cdot 4]/(Profiles - 1)}\end{array} //]]>$

OR

2; Whichever is greater

Top Portion

Top Portion - Minimum Stage

The minimum stage is determined by first identifying the least frequent AEP event provided in the hydraulic model, typically the .002 AEP event. Then identify the input flow that corresponds to that AEP event in the flow frequency function. Finally, the minimum stage is the most likely stage in the rating curve for that flow value.

Top Portion - Maximum Stage

The minimum stage is found by extrapolating out to the stage at the .0001 event. This is the stage that is the maximum for the top portion and for the entire stage-damage function.

Top Portion - Number of Equally Spaced Aggregation Stages

The number of aggregation stages between those stages must be at least four. Otherwise, the number of aggregation stages is equal to four times the difference between the minimum stage and maximum stage of the top portion.

$//<![CDATA[ \begin{array}{l}\mathrm{(Stage_m_a_x\,- Stage_m_i_n) \cdot 4}\end{array} //]]>$

OR 

4; Whichever is greater

Example

The below example demonstrates how stages are chosen given the input information.

Input Data

Flow-Frequency Function

The flow-frequency function is entered as a Log-Pearson Type III distribution with the parameters as shown below:

The resulting tabular relationship between AEP and discharge is generated:

Stage-Discharge Function

A rating curve is entered as a triangular distribution of stage around flow with the below inputs:

Generation of Stages

Given the above inputs, HEC-FDA 2.0 generates stages as shown below.

Bottom Portion

Minimum Stage

The most frequent input AEP is .5. From the flow-frequency function above, the Input Discharge at .5 AEP (X value) is 6408.1357. From the stage-discharge function, a flow (X Value) of 6408.1357 shows a Most Likely stage of 80.02.

Extrapolating from the .5 AEP to the .9999 AEP results in 79.3808.

Maximum Stage

The most frequent input AEP is .5. From the flow-frequency function above, the Input Discharge at .5 AEP (X value) is 6408.1357. From the stage-discharge function, a flow (X Value) of 6408.1357 shows a Most Likely stage of 80.02. So, the maximum stage for the bottom portion of the stage-damage function is 80.02.

Number of Equally Spaced Aggregation Stages

The minimum number of aggregation stages between the minimum stage and maximum stage for the bottom portion of the stage-damage function is four. If the below formula results in a number greater than four, that will be the number of equally spaced aggregation stages.

Range - Maximum Stage minus Minimum Stage: 80.02 - 79.3808 = .6393

Range times four: .6393 x 4 = 2.5572

Result is less than four: 2.5572 < 4

Therefore, the number of equally spaced interpolated points will be four.

Bottom Portion Stages

From 80.0201, there will be four equally spaced stages extrapolated below, the last of which will be 79.3808. Taking the range between these two numbers (found above, .6393) and dividing by four will give the increment between each stage.

Increment: .6393 ÷ 4 = .15983

Middle Portion

Minimum Stage

The most frequent input AEP is .5. From the flow-frequency function above, the Input Discharge at .5 AEP (X value) is 6408.1357. From the stage-discharge function, a flow (X Value) of 6408.1357 shows a Most Likely stage of 80.02. So, the minimum stage of the middle portion of the stage-damage function is 80.02.

Maximum Stage

The least frequent input AEP is .002. From the flow-frequency function above, the Input Discharge at .002 AEP (X value) is 20,147.2. From the stage-discharge function, a flow (X Value) of 20,147.2 shows a Most Likely stage of 83.8269. So, the maximum stage of the middle portion of the stage-damage function is 83.8268.

Number of Equally Spaced Aggregation Stages

The minimum number of aggregation stages between the stages at the input AEPs for the middle portion of the stage-damage function is two. If the below formula results in a number greater than two, that will be the number of equally spaced aggregation stages. 

Range - Maximum Stage minus Minimum Stage: 83.8269 - 80.02 = 3.8069

Range times four: 3.8069 x 4 = 15.2275

Divided by number of input AEPs minus one: 15.2275 ÷ (8-1) = 2.1754

Result rounded is equal to two: 2.1754 ~ 2

Therefore, the number of equally spaced interpolated points will be two.

Middle Portion Stages

Between each of the stages associated with the eight input AEP events there will be two equally spaced stages. Taking the range between the stages at each input AEP and dividing by three will give the increment for the stages between those input AEPs.

Calculating the Increment

Establishing Stages from Increment

Top Portion

Minimum Stage

The least frequent input AEP is .002. From the flow-frequency function above, the Input Discharge at .002 AEP (X value) is 20,147.2. From the stage-discharge function, a flow (X Value) of 20,147.2 shows a Most Likely stage of 83.8269. So, the minimum stage of the top portion of the stage-damage function is 83.8268.

Maximum Stage

The least frequent input AEP is .002. From the flow-frequency function above, the Input Discharge at .002 AEP (X value) is 20,147.2. From the stage-discharge function, a flow (X Value) of 20,147.2 shows a Most Likely stage of 83.8269.

Extrapolating from the .002 AEP to the .9999 AEP results in 84.5025.

Number of Equally Spaced Aggregation Stages

The minimum number of aggregation stages between the minimum stage and maximum stage for the top portion of the stage-damage function is four. If the below formula results in a number greater than four, that will be the number of equally spaced aggregation stages.

Range - Maximum Stage minus Minimum Stage: 84.5025 - 83.8268 = .6757

Range times four: .6757 x 4 = 2.7028

Result is less than four: 2.7028 < 4

Therefore, the number of equally spaced interpolated points will be four.

Top Portion Stages

From 83.8269, there will be four equally spaced stages extrapolated above, the last of which will be 84.5025. Taking the range between these two numbers (found above, .6757) and dividing by four will give the increment between each stage.

Increment: .6757 ÷ 4 = .16893

Final Result

From the above bottom, middle, and top portions, the stages are shown below.