- posted on July 18, 2003, 6:43 pm

- posted on July 19, 2003, 12:20 am

Remember your high school mathematics, the tank (at least the ones I recognize as oil tanks), are simply two half cylinders with a cube between them, so figure the volume of the round parts, and then the cube part, and add them.

- posted on July 19, 2003, 1:55 am

Actually, this is a popular problem in a calculus class. It is not simple
to solve using calculus, but there are non-calculus ways to figure it out.
I seriously doubt you will have much luck finding this already tabulated for
your tank, but here is something simple you can do that will probably answer
your purpose. To begin with, you are likely not interested in the case
where the level of oil is below the bottom of the rectangular section,
because in this case you are getting pretty low on oil and need to get some
delivered. Likewise, you are probably not too interested in the case where
the oil level is up in the top circular section--in this case you have
plenty of oil. If the oil level is in the rectangular section, then the
volume of oil is half a cylinder plus the rectangular depth times the
rectangular area.

If you want to work this out you will have to make accurate measurements of the tank (correct for the thickness of the steel walls; they are about 1/16" thick) and then compute some volumes. One gallon is 231 cubic inches.

If you are really interested in getting a dipstick calibrated for all cases, I'll go into greater detail. You will need a scientific calculator.

have

the

If you want to work this out you will have to make accurate measurements of the tank (correct for the thickness of the steel walls; they are about 1/16" thick) and then compute some volumes. One gallon is 231 cubic inches.

If you are really interested in getting a dipstick calibrated for all cases, I'll go into greater detail. You will need a scientific calculator.

have

the

- posted on July 19, 2003, 4:02 am

http://pages.cthome.net/edhome

- posted on July 19, 2003, 9:07 pm

No calculus is required just simple trig. The only hard part is the half cyclinder. R = Inside radius of the cylinder (1/2 outside width of tank minus 2 x gauge of sheetmetal) L = Inside length of tank (Outside length minus 2 x lap flange depth minus 2 x gauge of sheemetal) H = Inside height of tank (Outside height minus 2 x gauge of sheetmetal) h = measured depth of oil from bottom of tank

1. Area of a circle sector = R^2*arcos(R-h/R) 2. Area of circle segment = R^2*arcos((R-h)/R) - (R-h)*sqrt(R^2-(R-h)^2) 3. Volume of oil in gallons (h <= R) = L*(R^2*arcos((R-h)/R) - (R-h)*sqrt(R^2-(R-h)^2))/231 4. Volume of oil in gallons (h > R and h <= (H - r) =L

All dimensions in inches Angles (arcos) in radians Cu.in. per Gallon = 231 Pi = 3.1416

It would be easiest to do in a spread sheet. A correction factor could be incorporated to allow for the flexing in the sides and end caps of the tank, but this should be a small percent.

Regards,

John

- posted on July 19, 2003, 4:18 am

Frank wrote:

Check out this shareware:

http://www.kwaleak.com/downloads/tank_calculator.htm

Check out this shareware:

http://www.kwaleak.com/downloads/tank_calculator.htm

- posted on July 19, 2003, 4:29 am

have

the

I have a card from an oil company that has various tank sizes listed and the capacity at every 2" of filled height. If these tanks are existing buried tanks, knowing the dimensions could be difficult. You should know the height from previous sticking and you obviously know the galon capacity. If you know the height and the capacity, you can see if the oil company charts have a tank that matches those two numbers. Some are oval, some are cylindrical, so this could be hard to calculate if you don't know the shape.

A last alternative way would be to fill the tank from various oil levels and keep records. You should be able to see the pattern.

-- Mark Kent, WA

- posted on July 20, 2003, 2:41 am

That was all very good info. Thank you all.

have

the

have

the

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