TANK MEASUREMENTS


Member's Correspondence OLENOLLTD


METALWORKING FLUID SPECIALIST FOR THE ATTENTION OF MR. DAVID MARGARONI 18112/92


Dear Sir. I am writing to you concerning a problem we have encountered over a considerable period of time. That is. how to


calculate the volume of oil contained in a cylindrical tank. positioned on its side, by simply measuring the depth of oil in the tank and the radius of the tank.


This is, of course, a more complex situation than when the tank is positioned upright since the cross-section of the tank is


curved and therefore making a calibration stick is no longer a trivial matter, i.e ., the graduations would not be equally spaced. I therefore set a bout finding a solution mathematically and ha ppily achieved my objective. I have decided to inform you


of this since it may be of interest a nd benefit to other compa nies a nd would probably be best brought to their attention via "Lube News". Please find enclosed the necessary calcula tions .


Yours sincerely,


4 BI.ACKFIRS LANE SOMERFORD CONGLETON CHESHIRE CW12 4QQ


Tel. 0260 298276 Fax. 0260 298267


Paul Timmis.


So. shaded area= 1tr IJ - ~ f, r Sin IJ 360


Wii!.UMj!M


But we don't know ' I}' when taking a dipstick mea surement. So we need to find the relationship between ' I}' and 'd' (the dipstick measurement) and 'r' (the tank radius).


To find the volume of oil in the tank we first need to consider the shaded area of the circle (fig. 2).


Q


3rMr ~


u r h


E:i.g._A\\ [ig.._2 Shaded area= area of sector - area of triangle Area of sector. A:


ro 0 ~ rtr


A


A TCr t'l 360


Area of Irian le Al: Fig.


A1 = 1 f, base x height :. A1 = 1f2 rh But. if we look at Fig. 4 Sin 11 = opposite =


hypotenuse !: => h = r Sin t') : . Al = 1


r f, r Sin t') t} 360


Cos t~ 2


: . ~ =


i.e. the ratio of the angle at the centre to 360" is the same as the ratio of the area of the sector to the area of the circle.


:. 11 = 2 Cos 1 [ ~ J W!l(.!mt!fW opposite hypotenuse


Cos·t ~ d] l nverse cosine or arccosine


where.


r =tank radius in metres d = depth of oil in metres


Then substitute this value of ' I}' (in degrees) in equation I. Once the shaded area is found.


Volume of oil (in cubic metres)


Shaded are a of circle (in square me tres)


Wiii.UMjjW X I


(tank length in metres)


Also note. when taking dipstick readings either take two readings at equal distances from the ends of the tank and average them or take the reading at the centre. This is to compensate for the tank not being exactly level.


r - d


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