Volumes by Sections Using Prismoidal Formulas

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July / 2013 1 Volumes by Sections Using Prismoidal Formulas: A Case Study of Different Methods by Pat Sanders In order to illustrate the different possibilities for computing volumes by sections and their associated results, let us investigate the following example: FIGURE 1. Case Study Model My distance between sections is 100m. Section 1 at line 20+00 has a perfect radius of material that is 10m. The area above the planned line can be computed using: A 1 = πR 2 /2 = 3.14159 * 10 2 / 2. = 157.1 m 2 (EQ 1) Section 2 at line 21+00 has a perfect radius of material that is 20m. A 2 = πR 2 /2 = 3.14159 * 20 2 / 2. = 628.3 m 2 (EQ 2) METHOD NO. 1: A VERAGE END AREA Most computations of volumes by cross sections across a channel use the Average End Area approach. This is pretty simple in this case. Vol 12 = (A 1 + A 2 ) * L = (157 + 628) * 100 = 39,270 m 3 (EQ 3) METHOD NO. 2: PRISMOIDAL FORMULA #1 (WITH MIDPOINT SECTION) Some text books advocate the usage of a prismoidal formula as follows: VOL 12 = (A 1 + 4*A m + A 2 ) * L / 6 (EQ 4)

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Transcript of Volumes by Sections Using Prismoidal Formulas

  • July / 2013

    Volumes by Sections Using Prismoidal

    Some text b

    VOL1

    ooks advocate the usage of a prismoidal formula as follows:

    12 = (A1 + 4*Am + A2) * L / 6 (EQ 4)Formulas: A Case Study of Different Methodsby Pat Sanders

    In order to illustrate the different possibilities for computing volumes by sections and their associated results, let us investigate the following example:

    FIGURE 1. Case Study Model

    My distance between sections is 100m. Section 1 at line 20+00 has a perfect radius of material that is 10m. The area above the planned line can be computed using:

    A1 = R2/2 = 3.14159 * 102 / 2. = 157.1 m2 (EQ 1)

    Section 2 at line 21+00 has a perfect radius of material that is 20m.

    A2 = R2/2 = 3.14159 * 202 / 2. = 628.3 m2 (EQ 2)

    METHOD NO. 1: AVERAGE END AREAMost computations of volumes by cross sections across a channel use the Average End Area approach. This is pretty simple in this case.

    Vol12 = (A1 + A2) * L = (157 + 628) * 100 = 39,270 m3 (EQ 3)

    METHOD NO. 2: PRISMOIDAL FORMULA #1 (WITH MIDPOINT SECTION)

  • In order to compute this, we will need the area across the midpoint of the section. Using a radius of 15m, this is calculated as:

    AM = R2/2 = 3.14159 * 152 / 2. = 353.4 m2 (EQ 5)

    We now have the pieces we need in order to compute the volume:

    VOL12 = (A1 + 4*Am + A2) * L / 6= (157.1 + [4 x 353.4] + 628.3) * 100 / 6 = 36,652 m3 (EQ 6)

    METHOD NO. 3: PRISMOIDAL FORMULA #2This morning, Carlos Tejada, our HYPACK liaison for Central and South America, informed me of another Prismoidal formula. A little research on the Internet showed the formula mentioned in the computation of volumes between contours in lakes! The formula used is:

    VOL12 = ( A1 + A2 + SQRT[ A1 x A2] ) / 3. (EQ 7)

    If the areas between adjacent sections are equal, then the routine will give the same result as the Average End Area. In examples where the areas are different (such as our example), it will result in less material being reported.

    VOL12 = ( 157.1 + 628.3 + SQRT[157.1 x 628.3] ) / 3. = 36,652 m3 (EQ 8)

    SUMMARY OF RESULTSSo, using the same cross sectional areas, we have 3 different results:

    The Prismoidal #2 formula gave the same answer as the Prismoidal #1 formula without having to use a middle section!

    FIGURE 2. In HYPACK TIN MODEL

    To test the HYPACK TIN MODEL, I wrote a program to generate XYZs, based on the mathematical description of my shape.

    The resulting volume in TIN MODEL was 36,656 m3. This correlates very closely to the Prismoidal #2 result of 36,652m3! TIN MODEL came within 4 m3 of the mathematical result!

    Method Computed Volume (m3)Average End Area 39,270Prismoidal # 1 (with Middle Section) 36,652Prismoidal #2 36,6522

  • In order to compute volumes in CSV, I took the data points I had generated in TIN MODEL and cut two sections through the model to coincide with my desired sections.

    FIGURE 3. In HYPACK CROSS SECTIONS AND VOLUMES (CSV)

    From my data files, CSV reports:

    A1 = 156.4 m2 (vs. mathematical result of 157.1 m2) (EQ 9)

    A1 = 627.3 m2 (vs. mathematical result of 628.3 m2) (EQ 10)

    VOL12 = 39,184 m3 (vs. mathematical results of 39,270 m3)(86 m3 less, or -0.2%) (EQ 11)

    CSV got a slightly lower result. Some of that is attributable to how TIN MODEL outputs the data across the section.

    FIGURE 4. How TIN MODEL Outputs The Data Across The Section

    TIN MODEL outputs a data point at a user-defined distance along the planned line. In Figure 4, our mathematical model in black is a perfect semi-circle. The output from the TIN MODEL and what we will be working with in CSV is shown by the red line segments. The volume of material under those segments should be slightly less than what the mathematical shape would predict. Our result shows that!

    COMMENTS Just another example why you should be using the TIN MODEL!July / 2013 3

  • Should CROSS SECTIONS AND VOLUMES be modified to allow a prismoidal computation models? Prismoidal #3 Formula: This intrigues me. We could build it into CSV as an option for

    those who want to use it. It is NOT an Average End Area computation that is required in many contracts. However, it DOES give an answer that is closer to the TIN MODEL result.

    Lets hear from the users if they want the Prismoidal #3 formula built in as an optional selection to certain methods.4