Oil storage tank - how much is left?

Garetha

Our central heating oil is in a cylindrical tank (on it's side so that the ends are circular)
I measure the oil level by dipping a stick in at the top.
The tank capacity is 1200 litres.
The tank width / depth is 1 metre.
If the dip shows 50cm - then half = 600 litres remain - but it's not linear so 25cm is less than 300 litres

My question is: how much oil is left if the dip shows 10cm, 20cm, 30cm, 40cm?
Is there a formula?

Sorry if the above sounds like one of 'those' exam questions

G_M

You're right - exam question!

But impossible to answer without knowing the length of the tank, and the diameter of the cylinder. (if it is very long and thin the answer is different to a short fat tank!)

Plus you need a mathematician which I'm not!

Garetha

The cylinder is 1 metre diameter - and it's too cold to go outside just yet! -5c last night
As I see it, a percentage of the total capacity the formula need not know the length.
So that it comes down to the area of a circle,
e.g 50 cm = 50% (of 1200 litres)
I can picture it - even draw it on squared paper and count the squares - but I can't come up with a formula

albyota

I think this a question for 'ASK ELVIS'..........or............

How it works.
All units of measure must be the same, i.e., a tank with a length of 10 ft. and a diameter of 4 ft. and liquid depth of 36 inches would have the dimensions entered : Inches (checked), length = 120", diameter = 48", liquid depth = 36". This yields cubic inches which is converted to gallons by the ratio: 231 in3/gal.
The calculation for a cylinder oriented vertically is quite simple, area of the circle x length. This gives the cubic units, inches or centimeters depending on your unit of measure.
However, calculating the volume of liquid in a tank oriented horizontally is more difficult as the area of the liquid on the ends of the tank changes with depth as the shape of the area changes as illustrated in the images.

The images below hopefully will illustrate the geometry used in solving the problem.
Total view of horizontal tank. Assumptions made: tank is level, ends are flat, measurements are ID and must be in like units.

The calculation for the horizontal cylinder is based on the formula:
[(r2 acos (r-df) / r) - (r - df) * sqrt(2rdf - df2)] x L
Where r = radius , df = depth of fluid , angles are in radians and L = length
acos = inverse cosine, arccosine, or cos-1

This yields the section area. Area x Length = Volume

Below are some graphics to help explain the geometry. (pardon the crude graphics)

minus leaving bottom segment.

For fluid levels that are > r, the volume is: (area of the circle - top segment) x length.

http://www.idcomm.com/personal/kc/cylinder.html

Garetha

Thank you albyota,

I'll examine the formulae in the morning.
It appears to be the answer - I'll see if I can create an Excel spreadsheet which will generate a table.
Up to now, when ordering oil I have guessed the quantity and then asked the delivery person to check the level as it fills. My aim is to be able to order with confidence!

There will always remain a slight error due to the distortion of the cylinder to form a surface underneath to be supported. But this is hardly significant

PS 'There are 10 types of people, those who understand binary and those who don't'

G_M

No need to create spreadsheet - there's a calculator on the link Albyota provided. Thank god!

Now I'm having nightmares about school exams!