Prismoidal Earthwork Volume

V = \frac{L}{6} (A_{1} + 4 A_{m} + A_{2})

More accurate than average-end-area when mid-section Am is known (or parabolic). Average-end-area V=L(A1+A2)/2 overestimates a pyramid (should be L A/3).

$L$: Station interval
$A 1, A 2$: End areas
$A m$: Mid-section area

Step-by-step solved example

L=50 ft, A1=40, Am=30, A2=20 ft². Prismoidal V?

Prismoid: A1, Am, A2 along L.

1. Formula
V=50/6 (40+120+20)=50/6×180=1500 ft³.
$V = 1500 ft^{3}$
2. Compare AEA
AEA=50(40+20)/2=1500 ft³ here because Am is the average.

Answer: 1500 ft³

10 practice questions

0/10 correct

1.Pyramid (A2=0, Am=A1/4 for linear?) linear taper Am=(A1+A2)/2

2.Average end area is

3.Simpson’s 1/3 on areas is the

4.L=100 m, A1=A2=20, Am=20 m²: V=

5.A borrow-pit grid uses

6.Shrink / swell convert

7.If Am is omitted, FE usually wants

8.Units: 27 ft³ =

9.A1=10, A2=0, L=30, linear: AEA V=

10.Mass diagram uses these volumes as