# HP Forums

Full Version: (11C) Prismoidal Solver
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Prismoidal formula is used in the calculation of earthwork quantities.

The "Volume" of any prismoid is equal to one-sixth its length multiplied
by the sum of the two end-areas plus four times the mid-area.

Formula Used: V = [ h(m1 + 4m2 + m3) ] ÷ 6

Where:

m1 (Base Area)
m2 (Middle Section Area)
m3 (Top Area)
h (Height)

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Procedure: User Mode

[A] Rectangular Prism
Height [ENTER] Length [ENTER] Width [A]

[B] Cylinder

[C] Pyramid
Height [ENTER] Length [ENTER] Width [C]

[D] Sphere
Height [D]

[E] Custom Volume
Height [ENTER] Middle Area [ENTER] Top Area [ENTER] Base Area [E]

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Program:
Code:
``` LBL A x STO 1 STO 2 STO 3 X<>Y STO 4 GTO 1 ---------------- LBL B X^2 pi x STO 1 STO 2 STO 3 X<>Y STO 4 GTO 1 --------------- LBL C STO 5 X<>Y STO 6 2 ÷ X<>Y 2 ÷ x STO 2 RCL 5 RCL 6 x STO 1 R↓ X<>Y STO 4 0 STO 3 GTO 1 ------------- LBL D STO 4 2 ÷ X^2 pi x STO 2 0 STO 1 STO 3 GTO 1 -------------- LBL E + X<>Y 4 x + x 6 ÷ RTN ------------- LBL 1 RCL 1 RCL 2 4 x RCL 3 + + 6 ÷ RCL 4 x RTN```

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Example: FIX 1

Volume of Pyramid:
h = 10
w = 8
l = 6

10 [ENTER] 8 [ENTER] 6 [C] display 160

----------------------------

Volume of Sphere:
r = 3 // Radius is 3 double this up become 6 (Height = 6)

6 [D] display 113.1

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Gamo
Some more examples (FIX 3)
Rectangular prism
h=6
l=12
w=4
solution---> 288,000

Cylinder
h=24
r=12
solution---> 10.857,344

Volume of Pyramid
a) rectangular base
h=10
w=8
l=6
solution---> 160,000
h=11,313708
w=8
l=8
solution---> 241,359

Volume of Sphere
r=12 (so h=r*2= 24)
solution---> 7.238,229

Custom volume (e.g. barrel)
h= 0,9000
M= 0,2749
T= 0,1178
B= 0,1178
solution---> 0,200

Pedro
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• :