Conference rusure::math

It depends on what resources are available to you.  Can you drop a float
into it and measure the string ?  If not, can you thwang the tank to locate the
liquid surface ?

Or are you merely asking how to calculate the area of a circle that
has a horizontal chord chopped off (and hence the volume by multiplying
that area by 6 ft).

For this latter question, I used to know how to do it by calculus integration.
(is there a noncalculus way ?  Although maybe even just calculating that
the area of a whole circle is pi*r*r requires calculus!)

/Eric

If the radius of the tank is r, and the chord covers 2*theta radians,
then the area consists of two pieces:  The triangle with base 2*r*sin(theta)
and height r*cos(theta), and thus area r*sin*cos, plus the segment of the
circle (shaped like a pie with a piece missing), with area (pi-theta)*r^2.

    re .0
    
    Temporarily erect the tank. Measure the height of fluid. Computing the
    volume is then trivial.

    Call the fuel company to fill the tank.  Divide their bill by their
    rate.  Subtract into 300.
    
    Turn your furnace up to full.  Measure the flow.  Measure the time
    until the furnace goes out.  Multiply.
    
    Weigh a gallon of fuel.  Pour a gallon of water into the tank.  Stir. 
    Remove a gallon of liquid from the tank.  Weigh it.  Divide the
    difference between the fuel weight and the mixed weight into the
    difference between the mixed weight and weight of a gallon of water.
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

Hey, I like that .2 circle derivation.  Neat ! Thanks.

/Eric

re .0:
some more details would help, especially about the purpose and the 
constraints.

for a house sale, filling the tank is the usual and the easiest.

is someone just converted from oil to gas fuel and wants to get rid
of the old tank, that will not work very well.

just in case something got lost along the way, here are some more
complications.  most 300 gal. tanks are 275 gal.
the cross section is a top semicircle and a bottom semicircle,
connected by straight lines.  i don't know the diameter to height ratio.
also, a fill up is often terminated by the operator based on the
sound coming out of the filler hole. it is an expensive mistake to 
overfill and spill, so there is usually part of the capacity not used.
i'd guess 5 to 50 gallons, perhaps with an average of 20.

if this is just a math problem, someone will quickly add the formula
for each of the 3 cases where the surface is in the lower or upper 
semicircle or the straight part.

if this is a practical problem, tell us more.

OBTW, compute how accurate your results will be and how many filters
you will have to change before adding the water.

Take the fuel tank to a tall building.  Bribe the custodian to tell you how tall
the building is with a barometer (but first use the barometer to measure the air
pressure).  Using a hygrometer, measure the humidity, and use this plus the air
pressure to derive the air density.  Now drop the fuel tank from the top of the
building, and measure how long it takes to fall.  Then empty it and repeat the
process.  Using the difference in drop times plus the air density, derive the
fuel mass.  Divide by the density of fuel and you are done.

Using a bicycle pump, count the strokes needed to raise the pressure in
the tank to, say, 15 PSI.  Count the strokes needed to raise the pressure
in a known volume to the same pressure.  Determine the volume of the air
space in your tank with
	tank_volume = control_volume * (tank_strokes / control_strokes)
Subtract from 300 gallons.

    from .7
    
    Ask manufacturer weight of empty tank, weigh tank in current state,
    divide resulting fuel mass by density of fuel giving volume of fuel.

    
    
    This question started out as 'how much oil is in the tank' and how 
    'much fuel am I using'.  Its become how do you compute the volume  of a
    fraction of a cylinder when you can't take the cylinder stand it on end 
    and just work out the volume of the cylinder created by the fuel.
    
    A posting in the HOME_WORK notes file came up with the following
    
    > First calculate the cross sectional area (A) of the oil.  the formula I
    > generate is:
    >
    >    A = ( PI * R * H / 2.0 ) + R*(H-R)*cos(PI*(H-R)/(2*R))
    >
    > Then the volume is generated by multiplying the cross sectional area by
    > the length, L, of the tank.  That is, V = A*L
    
    I have not had a chance to check it, but it looks good to me..
    
    jb

Isn't .10 the same as .2 ?

/Eric