(41) Bulk Cylindrical Tank

[Thomas Klemm]

Quote:With inches used as the unit of measurement, the volume of the slant bottom
during filling of the first half becomes:

\(V_b=A_{seq}h\ (231\ in^3 / gal) / 2\)

Besides the fact that we have to divide by \(231\ in^3 / gal\) that's not how the volume is calculated.
The slanted cylinder isn't a slanted cuboid. Top and bottom are not the same.
Thus we can't simply divide the whole volume by 2.

Instead this formula can be used:

\(V(x)=\left [ x(\pi-\cos^{-1}x)+\tfrac{1}{3}\sqrt{1-x^2}(2+x^2) \right ]Hr^2\)

Here \(2H=h_0\) and \(h=(1+x)H\). This means \(x \in [-1, 1]\) while \(h \in [0, 2H]\).

Graph of \(V(x)\):


This program for the HP-42S calculates the volume of the slanted cylinder:

Code:

00 { 55-Byte Prgm }     ;
01▸LBL "SLANTED"        ; h
02 RCL÷ "H"             ; h/H
03 1                    ; 1          h/H
04 -                    ; x = h/H - 1
05 PI                   ; π          x
06 RCL ST Y             ; x          π          x
07 ACOS                 ; α          π          x
08 -                    ; π-α        x
09 LASTX                ; α          π-α        x
10 SIN                  ; y          π-α        x
11 X<> ST Z             ; x          π-α        y
12 ×                    ; x(π-α)     y
13 2                    ; 2          x(π-α)     y
14 LASTX                ; x          2          x(π-α)     y
15 X↑2                  ; x²         2          x(π-α)     y
16 +                    ; 2+x²       x(π-α)     y          y
17 R↑                   ; y          2+x²       x(π-α)     y
18 ×                    ; y(2+x²)    x(π-α)     y          y
19 3                    ; 3          y(2+x²)    x(π-α)     y
20 ÷                    ; y(2+x²)/3  x(π-α)     y          y
21 +                    ; V = x(π-α) + y(2+x²)/3
22 RCL× "H"             ; V*H
23 RCL "r"              ; r          V*H
24 X↑2                  ; r²         V*H
25 ×                    ; V*H*r²
26 RCL÷ "in3/gal"       ; V: in³ → gal
27 END                  ;

Registers

Code:

H=                   6.0000
r=                  72.0000
in3/gal=           231.0000

Examples:

3 XEQ "SLANTED"
16.9551

6 XEQ "SLANTED"
89.7662

8 XEQ "SLANTED"
175.0894

12 XEQ "SLANTED"
423.0134

Compare this to the table 25-3:

\(
\begin{matrix}
Measurement\ \# & H\ (total\ vertical\ level) & Gallons \\
1 & 3" & 21 \\
2 & 8" & 200
\end{matrix}
\)

Kind regards
Thomas

---

Addendum for those interested in the derivation of the formula:
\(
\begin{align*}
V(x)=\int_{-1}^{x}2\sqrt{1-t^2}(x-t)dt &= \tfrac{1}{3}\sqrt{1-t^2}(-2t^2+3tx+2)+x\sin^{-1}(t)\bigg|_{t=-1}^{t=x} \\
&= \tfrac{1}{3}\sqrt{1 - x^2}(2 + x^2) + x\sin^{-1}(x)-\left (-\frac{\pi x}{2}\right ) \\
&= x\left (\frac{\pi}{2}+ \sin^{-1}(x)\right )+\tfrac{1}{3}\sqrt{1 - x^2}(2 + x^2)
\end{align*}
\)

Then I used the identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) to end up with:

\(V(x)=x\left (\pi - \cos^{-1}(x)\right )+\tfrac{1}{3}\sqrt{1 - x^2}(2 + x^2)\)

Multiplication with \(Hr^2\) is just for scaling.

These formulas can be used with Wolfram|Alpha:

Code:

Integrate[2 Sqrt[1 - t^2] (x - t), t]

(Sqrt[1 - t^2] (2 - 2 t^2 + 3 t x))/3 + x ArcSin[t]

/. t = x

(Sqrt[1 - x^2] (2 + x^2))/3 + x ArcSin[x]

/. t = -1

-(Pi x)/2

---

Edit:
Changed R to r to make it consistent with the paper.