Re: Compulsory low-energy light-bulbs

Cynic wrote:
On Tue, 20 Mar 2007 20:58:48 +0000, Scott <blackhole@xxxxxxxxxxx>
wrote:

As an aside, the most difficult maths problem I had to solve as a
practical exercise turned out to be one that looks ridiculously
simple. I needed to make a dipstick for the water tank on a yacht to
measure what quantity of liquid was in the tank. So I needed to
calculate where to place marks on a dipstick that correspond with
given quantities of liquid (e.g. 10l, 20l, 30l etc). The tank being a
hollow cylinder (capped pipe) of known dimensions laying on its side.
Working out the quantity of liquid from a given depth in the tank is
fairly simple trigonometry. Working the depth from a given quantity
is certainly not a trivial exercise. Try it! So far nobody has been
able to come up with a formula for solving this problem, and in the
event I worked it out by successive approximation working it the easy
way around using a programmable calculator.

I answered the more general problem here:
http://tinyurl.com/229yvn

No, you answered this (quote) "The problems is to find the volume of
fluid when the fluid is at height h from the base of the cylinder "

That's the *easy* way around! What I needed was to find the height of
the liquid when given a particular volume.

I solved it by working out the formula for the easy way around (as you
did but without the domed ends), plugging that formula into a
programable calculator, and then entering various values of height
until the answer was close enough to the required volume (i.e.
successive approximation). I felt that there must however be a
formula to work it out in one go, but I've never been able to figure
one out.

To put the problem at its simplest level, draw a circle with a
horizontal chord (put it below the center for easier visualisation).
*Given* the area of the minor circle segment under the chord,
*calculate* the perpedicular distance between the center of the chord
and the lower circumference of the circle.

Doing it the other way about is trivial (i.e. finding the area given
the distance) - the angle between the center of the circle and the
ends of the chord is calculated (simple trig as the length of two
sides of a right angle triangle are known). When divided by 360 this
gives the fraction of area of the circle represented by the sector -
which is the segment plus the isosceles triangle formed by the two
radii and the chord itself. The area of that triangle is readily
calculated by simple trig and subtracted from the fraction found.

You wrote out that formula in the first part of your article (quote):

"
Area of segment of circle radius=a, height=h is:

<snip>

A =(pi.a^2)/2 - a^2.arcsin((a-h)/a) - (a-h).sqrt(a^2 - (a-h)^2) "

But now try changing that formula around so that it solves for height
rather than area!

i.e. h= ??????


There is no analytic solution, but you can write a polynomial approximation as follows:

A/a^2 - pi/2 = arcsin((a-h)/a) - (a-h)/a . sqrt(1-(a-h)/a)
Let (a-h)/a = x so the RHS is arcsin(x) - x.sqrt(1-x). This is analytic for a closed region contained with (0,1). All we need is a suitable point where the derivative is not zero and we can apply Lagrange/Burmann's theorem to express x as a function of A/a^2 - pi/2 (or A which the same thing). It is then trivial to express h as a function of A.

http://mathworld.wolfram.com/BuermannsTheorem.html
http://mathworld.wolfram.com/LagrangeExpansion.html
.

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