An excerpt from Citrus Processing, Quality Control and Technology, TANK MEASUREMENT by Dan Kimball with a complete & thorough description for the attendant HP 41 calculator program: enjoy!

[attachment=6423]

BEST!

SlideRule

(10-08-2018 09:13 PM)SlideRule Wrote: [ -> ]An excerpt from Citrus Processing, Quality Control and Technology, TANK MEASUREMENT by Dan Kimball with a complete & thorough description for the attendant HP 41 calculator program: enjoy!

BEST!

SlideRule

Interesting. One thing to keep in mind, it would be very interesting to have an example developed to test the program with clear instructions and units. Pedro

(10-08-2018 09:13 PM)SlideRule Wrote: [ -> ]An excerpt from Citrus Processing, Quality Control and Technology, TANK MEASUREMENT by Dan Kimball with a complete & thorough description for the attendant HP 41 calculator program: enjoy!

Thank you. But I wonder why you posted this in the General Software Library instead of the dedicated HP41 Software Section. ?!

Having read the PDF now, may I ask two questions?

The author distinguishes two cases if the liquid level is within the slanted bottom part of the tank: the measured level h may be less or greater than half of the height h

_{0} of the slanted part. So there are two formular 25-5 and 25-7. But as far as I can see they are equivalent since arcsin(x) = –arcsin(–x). At least if the arcsin function returns the principal value within ±pi/2, as usual. I just did a (metric) HP67 version of the program and it only requires one formula.

The author also says that the radius R of the dome top has to be determined graphically or by measurement. Please correct me if I'm wrong, but I'd say that R can be easily calculated from the height of the dome top (v

_{0} in the PDF) and the tank radius r. What do you think?

Dieter

Hi, Dieter

True. Slant bottom volume only need 1 formula.

Perhaps the author don't like negative angles.

Dome Top cross section is technically not circular, thus no real radius.

(where dome and cyclinder got welded, both look sloped vertical, but r > v0)

Graphical R is to find the "best" fit ... (assumed shaped like spherical cap)

It would be much easier if gallon scale were marked directly to the tank ...

(At the minimum, provide customer with a level vs gallon plot)

I believe Albert addressed the critique very well, danks!

1) a significant number of published programs are seldom optimized:

2) this program seems to confirm same.

1) dome tops are seldom circular:

2) this example seems to confirm same.

Any interest in more related programs?

BEST!

Sliderule

(10-10-2018 08:36 PM)SlideRule Wrote: [ -> ]I believe Albert addressed the critique very well, danks!

1) a significant number of published programs are seldom optimized:

2) this program seems to confirm same.

1) dome tops are seldom circular:

2) this example seems to confirm same.

Any interest in more related programs?

BEST!

Sliderule

Yes, I have seen some other containers as horizontal cylindric tanks, horizontal oval shape (transversal side in horizontal or vertical position); and all of them also in sloped position to facilitate its drainage.

Some important comments: a) oval transversal shape has two diameters (Max. and min.); to calculate a cylindric from the same formula, just consider both D the same figure. b) the most difficult part to calculate filling volume is when the tank is sloped

Pedro

(10-10-2018 07:01 PM)Albert Chan Wrote: [ -> ]Dome Top cross section is technically not circular, thus no real radius.

(where dome and cyclinder got welded, both look sloped vertical, but r > v0)

Maybe it could be a good idea then if the calculation did not assume a spherical shape but an

ellipsoid. This could match the actual tank better than assuming a spherical dome. The volume calculation is even easier, the formula is quite short.

I tried this approach in an HP67 program, and the results look correct to me. The complete program has less than 100 lines.

Now, what do you (all) think about the idea with the elliptical dome top?

Dieter

(10-10-2018 04:58 PM)Dieter Wrote: [ -> ]The author distinguishes two cases if the liquid level is within the slanted bottom part of the tank: the measured level h may be less or greater than half of the height h_{0} of the slanted part. So there are two formular 25-5 and 25-7. But as far as I can see they are equivalent since arcsin(x) = –arcsin(–x).

Dieter

I believe you're missing the fact that the sign changes for y*sqrt(r

^{2}-y

^{2}) also.

If anyone is like me, you struggled to figure out what y is in these equations. It's the radial distance from the center of the tank to the edge of the liquid. In the first side view showing liquid, it's r minus the length of the horizontal line that makes up part of the shaded area.

To derive equation 1, it was helpful (to me) to draw a top-down view of the partially filled slant section.

Thanks for posting this. Personally I love this sort of stuff because it gives me a chance to dust off my math skills.

The slant bottom equation can also rephrase as fractions of a cylinder volume (radius r, height h)

Let x = (2 h/h0 - 1), (so that x goes from -1 to 1),

Substitute x for equation 25.7, we get:

Vb / (Pi r^2 h / 231) = f(x) = 1/4 + (x*sqrt(1-x*x) + asin(x)) / (2 Pi)

To double check:

f(1) = 1/4 + (Pi/2 + 0) / (2 Pi) = 1/2 (as expected, due to symmetry)

f(0) = 1/4 + (0 + 0) / (2 Pi) = 1/4

f(-1) = 0

For rough estimate, f(x) ~ (x + 1)/4

Update: article volume formula is close, but wrong:

https://arachnoid.com/tank_slope_bottom/
http://www.hpmuseum.org/forum/thread-115...#pid105822
for the visually challenged (or inclined)

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BEST!

SlideRule

ps: if anyone has access to the Oil & Gas Journal (archives), an excellent HP-41 program is referenced in volume 90, issue 15 {04/13/1992}

Has anyone tried running the program with the values provided in the table on page 407? I can't get the right values for the last two measurements (H=195" and 198").

I am somewhat confused by the program prompts. From what I gather the values it wants are:

DIAMETER-IN? : tank diameter (2R)

SLOPE HT-IN? : height of sloped part (h0)

CYLINDER HT? : height of cylinder part (d)

HATCH HT-IN? : height of dome above cylinder (v0)??

HATCH DEPTH : height from bottom of hatch to top of dome?

JUICE LEVEL : Height from top of dome to top of juice level.

Maybe I'm misinterpreting the "Hatch ht" prompt. But this requires it to compute R and I didn't see anything in the paper on how to do that.

As someone who worked for a company that builds tanks I'm not surprised to see the example tank roof in Figure 25-5 is a spherical dome. The picture looks like an ellipsoidal roof, but equation 25-11 is the volume of a spherical cap.

While a hemisphere (R=r) could be used for a pressure vessel it's too expensive for a simple storage tank. An ellipsoidal roof (like the sketch seems to show) is also vey expensive and would again be typically limited to pressure vessels. An example of a pressure vessel with an ellipsoidal head is your BBQ propane tank.

However the roof volume will be wrong since the roof geometry specified is inconsistent.

Vo = R - sqrt(R² - r²) = 78 - sqrt(78² - 72²) = 48 inches

The paper states and uses Vo = 36", so the roof volumes will be wrong. Measurement 4 in Table 25-3 should be 13,321 gallons, and measurement 5 should be 13,521 gallons.

Hi, David Hayden

If I am not mistaken, v0 = Hatch ht, R = Hatch depth.

It can be deduce from overflow check (LBL 07 ...), which required actual tank dimensions.

R were not computed, but graphically fitted (say a compass)

(Unless the dome is actually spherical cap, no such R exist)

In Figure 25-5, if you draw an arc (radius R), spherical cap volume ~ dome top volume.

(cross section area approximately equal, thus volume too)

---

Hi, Geoff

I also think the author dome top calculation were flawed.

(Not the formula, but how graphical fitted R was performed.)

Figure 25-5 implied R is from center to the dome edge.

If true, this implied height of cap = 78 - sqrt(78² - 72²) = 48 inches.

Since v <= v0 = 36 inches, the formula can never reach full capacity.

(10-12-2018 02:46 PM)Geoff Wrote: [ -> ]However the roof volume will be wrong since the roof geometry specified is inconsistent.

Vo = R - sqrt(R² - r²) = 78 - sqrt(78² - 72²) = 48 inches

Yes, the given values are inconsistent.

I did it the other way round and calculated the radius R = (r²+Vo²) / (2·Vo) = 90 inches.

(10-12-2018 02:46 PM)Geoff Wrote: [ -> ]The paper states and uses Vo = 36", so the roof volumes will be wrong. Measurement 4 in Table 25-3 should be 13,321 gallons, and measurement 5 should be 13,521 gallons.

For the record: with R = 90" the results are 13318 and 13509 gallons.

Assuming an elliptical dome top yields 13324 and 13532 gallons.

So for this case the results are quite close. Since the exact shape of the dome top is not defined the uncertainty of the result should be in the same order as the variation in the calculated volumes.

Dieter

(10-12-2018 02:14 PM)David Hayden Wrote: [ -> ]I am somewhat confused by the program prompts. From what I gather the values it wants are:

DIAMETER-IN? : tank diameter (2R)

SLOPE HT-IN? : height of sloped part (h0)

CYLINDER HT? : height of cylinder part (d)

HATCH HT-IN? : height of dome above cylinder (v0)??

HATCH DEPTH : height from bottom of hatch to top of dome?

JUICE LEVEL : Height from top of dome to top of juice level.

The math at the beginning, the flowchart and the program seem not to be consistent. You're not the only one who is confused. ;-)

According to fig. 25-5 Vo is the height of the dome top, not the hatch height. "hatch ht" and "hatch depth" seem to define the position of the hedge (as stated in the text above the program listing). The formulas and the flowchart consider the dome radius R, but the program does not prompt for it.

(10-12-2018 02:14 PM)David Hayden Wrote: [ -> ]Maybe I'm misinterpreting the "Hatch ht" prompt. But this requires it to compute R and I didn't see anything in the paper on how to do that.

You can determine R from Vo and r, cf. my last post above. But I don't think this formula is mentioned anywhere in the PDF.

As already pointed out the given data in table 25-3 are inconsistent. You can assume a certain R and calculate Vo from this (cf. Geoff's post), or you may assume that Vo is given and calculate the corresponding R. Which is s what I chose as I think that Vo can be easily measured while determining R graphically is more cumbersome and error prone.

The HP67 program I wrote allows to override the calculated radius. For the example in table 25-3 the radius R is calculated as 90" which gives 13318 and 13509 gallons. Overriding the calculated R with R=78" gives the same results as the PDF, i.e. 13284,4 and 13444,4 gallons.

So I'd recommend you write your own program. This way at least you know what to enter and what results you get. ;-)

Dieter

I give up. I've been staring at the program trying to figure out how it computes equation 25-13 and I just can't get it. If anyone can figure it out, or even figure out how to get the program to generate the results in table 25-3, I'd appreciate it.

Dave

Mea Culpa

My posting is from the first edition; there is a second edition. I'm trying to ascertain

if there is an

erratum for 1e which might explain the discrepancies: til then …

BEST!

SlideRule

ps: the web URL for the author is

Dan Kimball Citrus Web Page which includes an email address (pm sent)

(10-12-2018 08:34 PM)Dieter Wrote: [ -> ]For the record: with R = 90" the results are 13318 and 13509 gallons.

Assuming an elliptical dome top yields 13324 and 13532 gallons.

The numbers look good ...

To get the upper limit, assume cylindrical tank.

1" slice = Pi * 72² * 1 / 231 = 70.50 gallons

195" - 6" => 189 * 70.50 ~ 13325 gallons

198" - 6" => 192 * 70.50 ~ 13536 gallons

---

Hi, David

I am guessing you have no problem with Eqn 25-13 first term.

It is just cyclindrical tank + slant bottom volume.

Volume of spherical cap = Pi h^2 (R - h/3), where h measured from top of cap.

So, volume of liquid in the cap

= volume of cap - volume of air

= Pi v0^2 (R - v0/3) - Pi h^2 (R - h/3)

We like to measure from the cap bottom. So, let v = v0 - h, we get

h = v0 - v
Factor out the Pi, add unit conversion factor 231 in^3/gal ... you get the second term.

Using above R=90" example, and do the 198" level calculation:

First term = Pi * 72² * (180 + 12/2) = 3029199 in^3

h = 12 + 180 + 36 - 198 = 30 in

Volume of air = Pi h^2 (R - h/3) = Pi * 30² * (90 - 30/3) = 226195 in^3

Volume of cap = Pi v0^2 (R - v0/3) = Pi * 36² * (90 - 36/3) = 317577 in^3

Total volume = 3029199 + 317577 - 226195 = 3120581 in^3 =

13509 gallons
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.
(10-12-2018 10:41 PM)Thomas Klemm Wrote: [ -> ]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:

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

So the formula in the paper is wrong. Thank you very much for checking and correcting this.

Edit: But I still don't understand

why it's wrong. The liquid surface is a circle with a chopped of segment. If this area is known, why is the volume

not area · h/2 ?

(10-12-2018 10:41 PM)Thomas Klemm Wrote: [ -> ]This program for the HP-42S calculates the volume of the slanted cylinder:

You should add a RAD statement to make sure that radians mode is set.

Or modify the formula so that it works in any angular mode: pi – arccos(x) = pi·(1 – arccos(x)/arccos(–1)).

(10-12-2018 10:41 PM)Thomas Klemm Wrote: [ -> ]Changed R to r to make it consistent with the paper.

Fine. Maybe you can finally post the complete formula in terms of the variables in the paper, i.e. with h and h

_{0} instead of x and H.

H = h

_{0}/2 and x = 2h/h

_{0} – 1.

Dieter