My shop is heated with an old oil burner. The tank is a 38" by 60" cylinder
which I believe is 300 gallons. Because the tanks sits on a stand on its
side I need to know how to convert the inches remaining in the tank to
gallons. The input pipe is on top and currently shows 15" on the stick.
Obviously 19" would equate to 150 gallons but how do I graduate the measure
stick to measure more than half as well as less than half? Or perhaps the
better question is how do I create a chart that equates inches to gallons?
Patrick Fischer
Olalla, WA

First step is to go tot he Lee Valley web site and order the Pocket
Reference by T J Glover. Then go to page 422 and find the ratio then the
depth factor for your tank. For a 38" tank the ratio is H/D or .39, rounded
to.40 thus the depth factor is 0.3735300. depth factor x total gallons 112 gallons. This is assuming you are giving inside dimensions, not
outside.
The Pocket Reference is a very handy book I use it frequently for some
obscure formula that comes up. It also give floor loading for joists,
airport codes, flow through pipes, steam tables, and lots of other
electrical, mechanical, chemical and structural information.

No shit, that is why I chose this formula. Page 422 is titled Horizontal
Cylinder Fillage
It states: "The following equation can be used to calculate the number of
gallons remaining in a horizontal tank if the height of the liquid remaining
in the tank and the diameter of the tank are known."
You did look at the page didn't you?

remaining
Edwin, don't get huffy. No I didn't look at the page. I know enough to
know you cannot condense volume to a linear multiplication for this example,
I don't care what the book says. One of the other posts showed a non-linear
equation which it appeared correct, but I did not work through it to see if
I thought it was correct.
Bob

Then look at the book. They give a depth factor depending on the ratio.
Rather than make comments about something you have never seen, check it out
and you will find it to be correct. To understand this you really have to
LOOK AT THE BOOK. Page 422. I got mine from Lee Valley, but it is sold
elsewhere. Sorry, but your opinion is superceded by facts.

On Sat, 17 Sep 2005 09:26:55 -0700, Patrick Fischer wrote:

d, r, L in inches.
d is measured depth.
r is radius of tank.
L is length of tank.
V in gallons (231 cu in per gal)
f(d) = 1/2 (((r^2)*arcsin(1/r)) - (r-d)*sqrt(r^2 -
(r-d)^2) + r^2 * arcsin((r-d)/r^2)
d>r; V= ((f(d) + 1/2*pi*r^2)*L)/231
d<r; V= (f(d) * L) /231
d=r; V= (1/2 * pi * r^2 * L)/231
I took calculus thirty years ago. Someone a little more current should
check this... :)

--
"Keep your ass behind you"
vladimir a t mad scientist com

Every oil company I have ever used has sticks with that info printed on it.
You stick the stick in and read off where the wil is on the stick. then wipe
it off with a rag.
Justmake sure you get the right stick. Just ask for one. They like giving
them out because it encourages oil sales.

On Sat, 17 Sep 2005 16:01:03 -0400, "Lee Michaels"

I missed the OP, but solved that problem as a kid, using high school
math. The result is sinusoidal, and not simple for anyone not
familiar with the math. Basically, you draw a circle with a chord
[initially below the half-way diameter for convenience] and draw the
sector. The area of the lower segment [the liquid you want to find]
made by the chord is the area of the sector less that of the triangle
formed by the two arms of the sector [radii] and the chord. The first
is related to the sector angle in radians, and the second is the
familiar triangle area. However, tha uses th chord length, so to tie
the two together you have to use Pythagoras' theorem. You have to
express the angle as an inverse trig function [why the "sinusoidal"
reference] in terms of the tank radius R, and the liquid height, h.
If you drew the half sector, you'd have an angle,A, a radius R off to
the right to meet the chord at the circle, a liquid height h, and a
remaining height of R-h, the third sied of a triangle of side R and
c/2, half the chord length.
Volume will be the resulting area times the lenght of the container.
...you really don't want me to go on with all of this, but you did
ask. Also, it's a bit difficult to try to write it all in plain text,
so the best bet is to look it up in a web search, or some possible
boilermaker tables you might find in a used book store.

On Sat, 17 Sep 2005 18:46:44 -0700, "Patrick Fischer"

i.e., somebody else already did the calculus and generated a chart. This
could be a really interesting problem.

+--------------------------------------------------------------------------------+
If you're gonna be dumb, you better be tough
+--------------------------------------------------------------------------------+

On Sat, 17 Sep 2005 21:18:27 -0700, Mark & Juanita wrote:

Yeah. I've realized that I had a factor-of-two error in the solution
I posted earlier. Geometry, trig, and calc are useful shop skills.
Fetching the chart from a vendor is another useful skill...

--
"Keep your ass behind you"
vladimir a t mad scientist com

On Sat, 17 Sep 2005 09:26:55 -0700, "Patrick Fischer"

There are approximately 231 cubic inches per gallon. Or, 7.481
gallons per cubic foot. To calculate the volume of a cylinder
multiply the base area by the height.

On Sun, 18 Sep 2005 23:02:50 +0000, Phisherman wrote:

But OP wants to know how much oil is left. Yes, pi*r^2*h gives the total
volume. It's trickier when the cylinder is on its side and you want to
know the volume of oil in the tank, not the tank's capacity. Calculus (or
the vendor's chart) comes to the rescue.

--
"Keep your ass behind you"
vladimir a t mad scientist com

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