formula for tank levels

Thanks so much Ted. I appreciate your reply post and will be "playing" with the numbers/formulas to get a better understanding. Thanks again.

I agree with you completely ! My personal problem is that I never had anything more (nor needed) than Algebra I in 9th grade (which is well over 40 year ago ) but I like to understand how "stuff" works and can work formulas (providing they are not too complicated) even if I have to look up some info, or get a little outside help on the solving. I was simply not able to find anything (other than total volume) that applied to this little problem - hence the original post.

*Some* on this group consider me to be a "noob" and evidently wouldn't be able to understand any concept of a formula (how elite-ish) - that's kinda why I was looking and asking. But then again, I don't almost set the house on fire trying to melt pennies...... Thanks again, Ted, for your info and help. Ken.
Reply to
Ken Sterling
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As a matter of fact, I'm getting an A grade in my Calc class and I _never_ said it wasn't possible. You're just being an asshole. Again.

I am assuming the tank is horizontal, like a 100 gal. (?) propane tank. Even with hemispherical ends, it isn't too simple. Although now that I think of it, it would be only a double integral as you are taking horizontal slices (dy infinitessimals).

Tim

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Reply to
Tim Williams

I believe that a triple intergal is the correct general solution to the problem. The differential volume element dV = dxdydz in cartesian coordinates, no?

So perhaps the suggestion indicates a more advanced level of learning than you suspect. I personally would never have posted the question because I know where to look and would have found an acceptable solution myself. I took second semester calc about 16 years ago, so I don't remember it too well. I do remember the grade I got because it's the same grade I got on all my math and science classes - an A.

This problem is straight from second semester calculus in any

Reply to
Rich McCarty

I believe it's one of those things that, like a lathe or a welder or a forge or ..., that you don't realize how much you can use it until you have it. Then all sorts of things you could do are suddenly within reach.

Also, I think a large part of the problem is in the teaching of math. "Solve this equation because I said you should", just doesn't cut it with teens. Choose problems that the kids are interested in and you get much better results. Of course, it usually means more work for the teacher but ...

I had the following experience with almost the same problem as yours:

At a college I taught at for some years, they had what they called the Math Learning Center. This was a desk in a college library where there would be an instructor from the science faculty available from early morning 'til late evening. Any student who was having a math difficulty could come for help.

One day, on my shift, a student approached me with the following problem:

Consider an oil drum of diameter, D, that will be used for a gas tank. The drum will be placed on its side. It is desired to make a dipstick for measuring the amount of fuel remaining in the tank. The stick is to be marked for E, 1/4, 1/2, 3/4 and F. E, 1/2 and F are obvious. 1/4 and

3/4 require some math. The student couldn't figure out how to do this and came to me for help. I assumed the student was in second semester calculus as this exact problem appeared in the text and was to be expected at about this time.

I proceeded to start to show him how to set up the integral and, from the blank stare, quickly observed that he didn't have a clue what I was talking about. On a hunch, I asked him where he got the problem. Seems his Dad had bought a boat and they wanted to install a larger gas tank. When asked what math he was taking, he replied Math 110 (*first* semester calculus). This course does not start into integration until a little bit at the end and has nothing on definite integrals.

By way of explanation, I made a sketch showing the concept. I then set up and did the integral for him explaining that he would learn how to do it himself next term. He went away happy.

Afterword: I found out that this fellow had been barely scrapping along with a C- in math. By the end of the term he was a B+ and climbing. Seems nobody had ever convinced him that this stuff was anything but a required course. Amazing what a practical example will do.

Many colleges offer night school courses in remedial math should you wish to upgrade. I'm still learning and I'll be 70 next year. I doubt that a week goes by when I'm not using calculus for a shop related problem and trig is just as usuall.

Ted

Reply to
Ted Edwards

Maybe your problem is reading comprehension? The OP was pretty clear on the configuration.

Still wrong.

Perhaps it's your mind you lost.

Reply to
Ted Edwards

In the most general case where there are no symmetries, yes. In this case, the shape is of uniform cross section so dV=LdA but dA is simply expressed as dA=2x dy where s is easily found from the equation for an ellipse.

Perhaps. But one can get a decent mark by memorizing the text without ever learning how to apply the stuff. The real test is in the word problems.

Well, its been a Hell of a lot longer for me but I use it all the time - especially in the shop.

Very good! Now how about applying it to everyday problems?

Ted

Reply to
Ted Edwards

I gathered this:

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What I gathered from your post is that you think it vertical.

Hmm so water doesn't sit horizontally? It would be the most logical course of action to take slices both parallel to the surface of the water, and of the various levels it needs to be at to estimate volume fractions.

Ah, that's a familiar sound...so I'm CC:ing it too.

Tim

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Reply to
Tim Williams

On Fri, 23 Apr 2004 01:14:00 GMT, snipped-for-privacy@netzero.net (Ken Sterling) vaguely proposed a theory ......and in reply I say!: remove ns from my header address to reply via email

(1) Just keep pouring in liquid in known amounts and measure then mark the stick. Called "calibrating" .

(2) Or, if you can work out the area of the tank at given heights and "stations", then you can get a close approximation. This can actually be done without calu...cl...cal,,, turds and buggerithms, by using triangulation of sizes small enough to be accurate for you. Works well for boat floatation (a volume of weird shapes). Clacerless is after all only the theoretical limit of lost os little bits added together.

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Reply to
Old Nick

Well, this problem is easily solved with trig. For someone who is proficient in calculus, it is probably easier. But if calculus is a problem, trig will do nicely. Bob

Reply to
Bob Engelhardt

This has been bugging me ever since someone mentioned a triple integral. It's been 42 years since I graduated and probably as long since I did an integration, but I think that I still have a grasp of the basics.

An integral is the area under a curve. For y = f(x), the area is the integral of ydx over a range of x. I.e., an area is a _single_ integral. And a volume is a _double_. Symmetrical or not. Never a triple.

For the volume in the OP's case a double integral degenerates into a single because the one function is a constant whose integral is the range.

That's the way that I remember it and I'm sticking to it.

Bob

Reply to
Bob Engelhardt

Since this has been hanging around for a while and the original question still hasn't been answered I'll take a crack at it.

" but I haven't been able to find a formula to calculate how far up the stick the 1 gallon level would be, the 2 gallon level, etc., etc."

I came up with a total capacity of 19.04 gallons and the following table for each increment of capacity. Gallons Height 1 1.61 2 2.59 3 3.44 4 4.21 5 4.94 6 5.64 7 6.32 8 6.99 9 7.66 10 8.32 11 8.98 12 9.65 13 10.33 14 11.03 15 11.76 16 12.53 17 13.37 18 14.35 19 15.82 19.04 16.00

Richard Coke

Reply to
Richard Coke

You are quite right! What I did was solve the inverse problem, i.e. How many gallons at x inches while what is really wanted is how many inches at each gallon.

My solution for elliptical ends agrees with yours but, as someone pointed out, that would be an unusual tank. More common is a tank whose ends are two semi-circles joined by straight lines. So I have reproduced your table and added the "semi-circles case".

Semi-circular top and bottom Elliptical ends with straight sides Gallons Height Gallons Inches 1 1.61 1 1.39 2 2.59 2 2.25 3 3.44 3 3.00 4 4.21 4 3.70 5 4.94 5 4.37 6 5.64 6 5.03 7 6.32 7 5.69 8 6.99 8 6.35 9 7.66 9 7.01 10 8.32 10 7.67 11 8.98 11 8.33 12 9.65 12 8.99 13 10.33 13 9.65 14 11.03 14 10.31 15 11.76 15 10.97 16 12.53 16 11.63 17 13.37 17 12.31 18 14.35 18 13.01 19 15.82 19 13.76 19.04 16.00 20 14.62 20.99 16.00

Ted

Reply to
Ted Edwards

And we both know that this is a purely academic exercise based on dimensions of unknown accuracy and an ill-defined shape. The OP's best chance at getting an accurate dipstick is to calibrate it in the tank. It's just another collision of theory and utility. Not like that happens very often on this NG.

Richard Coke

Reply to
Richard Coke

That's the way I did it for my airplane. For small tanks like mine, one gallon at a time. And got 11 3/4 gallons.

Calibrate the tank and the dip stick at the same time. Although it rather academic if the gas gauge is a wire on a float.

This tank is welded aluminum wrapper on formed flanged ends (5052H32). The ends are elliptical, the flanges hand formed between plywood blocks.

My only real boo-boo was pop riveting the tank together before welding. Something on the rivets didn't want to weld...

But it doesn't leak. (so far?!)

I gotta admit the idea of building an aluminum fuel tank was a bit intimidating. But making one piece at a time, it eventually got done, and I'm pretty proud of it.

Richard Lamb

Reply to
Richard Lamb

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