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+--- Thread: (11C) Prismoidal Solver (/thread-13022.html)
(11C) Prismoidal Solver - Gamo - 05-25-2019 12:30 AM
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)
----------------------------------
Procedure: User Mode
[A] Rectangular Prism
Height [ENTER] Length [ENTER] Width [A]
[B] Cylinder
Height [ENTER] Radius [B]
[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]
------------------------------------------------
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----------------------------
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
----------------------------
Gamo
RE: (11C) Prismoidal Solver - PedroLeiva - 05-25-2019 02:07 AM
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
b) cuadrangular base
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