Re: Compulsory low-energy light-bulbs

Cynic wrote:
On 20 Mar 2007 05:18:47 -0700, google@xxxxxxxxxxxxx wrote:

On Mar 20, 10:43 am, Cynic <cynic_...@xxxxxxxxxxx> wrote:
On Tue, 20 Mar 2007 08:07:46 -0000, "M.I.5¾"

<no....@xxxxxxxxxxxxxxxxxxxxxx> wrote:
You will need to tell me what *all* of the terms in that equation are
before it will make sense to me. So T is temperature. d I guess
might be ASCII for delta? but what is t,x,y & z?
No 'd' is recognised mathematical operator.
dy/dx is recogniised as meaning the 'rate of change of y as x varies'.
ITYF that the letter "d" is ASCII representation for the greek letter
"delta" as I stated, which in script would be be written as a
triangle. Usually used in maths to mean "change of" "d" is also
often used to mean distance.

No. It's an operator and is written exactly as d/dx (or rather d/
d<something>)

OK - I've not studied calculus to any great extent, as I have very
seldom had a need to use it. On the few occasions where it would have
come in handy, I've got away with using successive approximation or
physical measurement.

I was aware of the process of dividing one change (delta) in magnitude
by another change in magnitude, and then decreasing the change until
it approaches zero to get a desired result e.g. "Result approaches
(value) as delta y approaches zero" as a way of overcoming the
impossibility of dividing anything by zero to find (say) an
instantaneous velocity of an accellerating body at an instant in time
(delta t = 0). I did not realise that the condition where the change
approaches zero had its own common symbol.

So what are t, x, y and z in the equation?

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

The ends in that case were capped by sectors of a (larger) diameter sphere. The fact that the ends of the cylinder were capped with truncated spheres (not hemispheres) made the resulting formula particularly hideous, as it means calculating the volume of a sphere cut by 2 arbitrary perpendicular planes.
.

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