On Sun, 30 Dec 2007 13:22:30 -0800 (PST),
<bradeatspoo@[EMAIL PROTECTED]
> wrote in
<news:813841bf-5512-4c47-98f9-769d92bde6e0@[EMAIL PROTECTED]
>
in alt.algebra.help:
> I am a computer program
And a very good one, too! <g>
> and not a mathematician. I use the equations below to
> calculate Gallons when I have the Radius, Depth and
> Length of an oil tank. However, sometimes I need to
> calculate Depth when I have the Radius, Length and
> Gallons. Any help in reworking these formulas would be
> greatly appreciated.
> If iDepth <= iRadius Then
> iGals = ((iRadius * iRadius * Acos((iRadius - iDepth) /
> iRadius)) - ((iRadius - iDepth) * Sqrt(2 * iRadius * iDepth - iDepth *
> iDepth))) * iLength / 231
> Else
> iGals = ((PI * iRadius * iRadius) - ((iRadius * iRadius *
> Acos((iRadius - (2 * iRadius - iDepth)) / iRadius)) - ((iRadius - (2 *
> iRadius - iDepth)) * Sqrt(2 * iRadius * (2 * iRadius - iDepth) - ((2 *
> iRadius - iDepth) * (2 * iRadius - iDepth)))))) * iLength / 231
> End If
Judging from these formulae, it's a cylindrical tank lying
on its side. Its length in inches is iLength, its radius in
inches is iRadius, and iDepth is the depth of the fluid in
inches, measured at the centre of the tank.
The problem isn't amenable to solution in closed form:
you're not going to get a formula that gives you iDepth in
terms of iGals, iRadius, and iDepth. The easiest practical
solution is probably to write a routine that gets a
sufficiently accurate value by successive approximation,
using the formulae that you already have.
For convenience I'll introduce some additional variables.
First, set iArea = 231 * iGals / iLength; this is the area
in square inches of the submerged part of a vertical
cross-section through the tank. Its maximum value is of
course the cross-sectional area of the tank,
PI * iRadius * iRadius. Assume first that
iArea <= PI * iRadius * iRadius / 2, so that
iDepth <= iRadius.
Take iD to be an initial approximation to iDepth. It
needn't be particularly good;
iD = 2 * iArea / (PI * iRadius)
will do. Let
iX = iRadius - iD
and
iY = Sqrt(iRadius * iRadius - iX * iX).
Recalculate iX according to the formula:
iX = iX / 2 + (iRadius * iRadius * Acos(iX / iRadius) -
iArea) / (2 * iY)
Update iD:
iD = iRadius - iX
This will be a better approximation to iDepth. Repeat until
you reach a satisfactory level of accuracy. You can test
the accuracy in at least two ways. First, you can use iD
for iDepth in your formula to see whether it yields a value
for iGals sufficiently close to the known value. Secondly,
you can keep track of the changes in iD and make sure that
they're small. (You may want to use both.) Convergence was
very rapid in all of the examples that I tested.
If iArea > PI * iRadius * iRadius / 2, you can proceed as
follows. First let iArea = PI * iRadius * iRadius - iArea;
this is the area of the part of the cross-section that
*isn't* submerged. Next, carry out the procedure described
above. When you've reached a satisfactory level of
accuracy, replace iD by 2 * iRadius - iD; this will be the
desired approximation to iDepth.
Brian
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