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Hi Al,http://www.iop.org/EJ/S/UNREG/aS0aOQcupwxj4CNeiM5vaQ/article/-featured=jnl/0143-0807/23/1/304/ej2104.pdf
"Demonstration of the exponential decay law using beer froth"
This all goes back to the solution of the differential equation forwhen a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
The voltage knows nothing about how it's "supposed" to behave. It just doeswhen a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
Alan Horowitz wrote:
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
Suppose you are trying to fill up a box with balls. However, for some
strange reason, you've decided that each time you throw in balls, you'll
throw in 1/2 of the balls that will fit in the remaining space.
At the first second, you have 1/2 the balls. Next second, you'll have
that plus 1/2 of the remaining space, which is 1/2 + 1/4 = 3/4. The
third second, you'll have that plus 1/2 the remaining space, ie, 1/2 +
1/4 + 1/8 = 7/8...
So, the number of balls at any time t will be:
B(t) = 1 - (1/2)^t
Thus, after 3 seconds, there will be B(3) = 1 - (1/2)^3 = 1 - 1/8 = 7/8,
just like above.
Now, apply that same reasoning, only instead of using the ratio 1/2, use
the ratio 1/e (since we are applying arbitrary rules)
Then
B(t) = 1 - (1/e)^t
After the first second, you'll have
B(1) = 1 - (1/e)^1 = 1 - 1/e = 0.632 (that is, 63%)
Strange coincidence, isn't it? It happens because when you are charging
a capacitor through a resistor, you are throwing balls, in the form of
charges, into a box (the capacitor), and the number of charges you throw
at any given time (the current) depends on how many charges are already
on the capacitor (the voltage).
Each step of the formula above is one time constant, RC. By dividing out
the RC, you can get the answer given seconds, ie
B(t) = 1 - (1/e)^(t/RC) = 1 - e^(-t/RC)
Where B is the percentage 'filled' the capacitor is (ie, what percentage
it is of the input voltage).
Why is 1/e used instead of 1/2? That has to do with the fact that we
must have a continuous solution, not a solution based on ratios of
existing values; the rate of change of the current (ie, how many balls
we throw in per unit time) is proportional to the voltage remaining,
which is continuously changing. Using 1/e instead of 1/2 allows us to
generalize to this, in the same way as the compound interest formula
allows us to compute 'continuously compounding' interest.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
Thanks Art. I enjoyed thinking about it and writing it. The exponentialNice job Robert, I really liked it
Art
"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:6gKad.230613$D%.163996@attbi_s51...
Alan Horowitz wrote:
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
Suppose you are trying to fill up a box with balls. However, for some
strange reason, you've decided that each time you throw in balls, you'll
throw in 1/2 of the balls that will fit in the remaining space.
At the first second, you have 1/2 the balls. Next second, you'll have
that plus 1/2 of the remaining space, which is 1/2 + 1/4 = 3/4. The
third second, you'll have that plus 1/2 the remaining space, ie, 1/2 +
1/4 + 1/8 = 7/8...
So, the number of balls at any time t will be:
B(t) = 1 - (1/2)^t
Thus, after 3 seconds, there will be B(3) = 1 - (1/2)^3 = 1 - 1/8 = 7/8,
just like above.
Now, apply that same reasoning, only instead of using the ratio 1/2, use
the ratio 1/e (since we are applying arbitrary rules)
Then
B(t) = 1 - (1/e)^t
After the first second, you'll have
B(1) = 1 - (1/e)^1 = 1 - 1/e = 0.632 (that is, 63%)
Strange coincidence, isn't it? It happens because when you are charging
a capacitor through a resistor, you are throwing balls, in the form of
charges, into a box (the capacitor), and the number of charges you throw
at any given time (the current) depends on how many charges are already
on the capacitor (the voltage).
Each step of the formula above is one time constant, RC. By dividing out
the RC, you can get the answer given seconds, ie
B(t) = 1 - (1/e)^(t/RC) = 1 - e^(-t/RC)
Where B is the percentage 'filled' the capacitor is (ie, what percentage
it is of the input voltage).
Why is 1/e used instead of 1/2? That has to do with the fact that we
must have a continuous solution, not a solution based on ratios of
existing values; the rate of change of the current (ie, how many balls
we throw in per unit time) is proportional to the voltage remaining,
which is continuously changing. Using 1/e instead of 1/2 allows us to
generalize to this, in the same way as the compound interest formula
allows us to compute 'continuously compounding' interest.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
First, I should make a few corrections to what you've said. Thewhen a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
On Tue, 12 Oct 2004 16:45:50 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> wrote:
Alan,
John Popelish got a good start with "e is a natural constant that has
some very sweet properties in many applications of mathematics, and
simplifying..."
Then, it looked as thought John Jardine was going to steal my thunder
with
"the voltage knows nothing about how it's "supposed" to behave. "
This could resolve to a mater of faith Alan.
Indeed, the voltage/current "knows" nothing.
After observing what happens in such circuits, "we" (those who must
understand all things) very carefully examined what was going on and
"discovered" that there were mathematical expressions or equations which
would model what happens in nature. "We" came up with theories about
what
was going on and what was causing it to happen. "We" then found ways to
make the math fit reality.
Reminds me of Galileo writing in "The Assayer," saying:
"Philosophy is written in this grand book-I mean the universe-which stands
continually open to our gaze, but it cannot be understood unless one first
learns to comprehend the language and interpret the characters in which it
is
written. It is written in the language of mathematics, and its characters
are
triangles, circles, and other geometric figures, without which it is
humanly
impossible to understand a single word of it; without these, one is
wandering
about in a dark labyrinth."
(By the way, to anyone who has NOT actually read The Assayer from
beginning to
end, I highly recommend it!)
Mathematics is a wonderful world all of its own, independent of nature,
yet
where it often turns out that insights in that world happen to happily
suggest
relationships found in this world and where proper deductions there imply
proper
deductions here. The language is sufficiently rigorous that someone two
millennia before me can describe a circle using it and I can read it
today,
knowing absolutely nothing about their lives, their fads or interests,
their
politics or style of dress, and come away with exactly the same image in
mind
with exactly the same deductive power. In short, mathematics is a
quantitative
language that speaks across culture, time, and place. And there is
nothing we
have to compare with that.
The processes of science work to achieve a relatively objective process
that
works well. It requires the use of objective language sufficient for
rigorous
quantitative deductions (by anyone adequately trained in the language) to
specific circumstances, insists that such language both explain past
results
well and (more importantly) also make accurate and repeatable predictions,
requires quantitative prediction for discernment, and requires time and
patience
for the resulting critical opinion of others skilled in the field to
arrive at a
consensus. But mathematics *is* a key part of this objective language
used in
science because of its demonstrated congruencies with nature.
In the case of time constants, we have a natural
phenomena which is very nicely described by the equations stated
elsewhere
in this thread (the 1/e thingy). It is just like the F=MA equation.
"We"
discovered that the force applied to a mass is equal to the mass times
the
acceleration. The Mass knows nothing about force, acceleration or
mathematics. We found that this math describes nature.
One thing to keep in mind is that ideas like "density," a useful
relationship
between volume and mass, are truly discovered through hard work and
through
trying to find some kind of useful discernment regarding sinking and
floating.
One doesn't just naturally _know_ about density, as our direct senses tell
us
nothing of the kind. It's discovered and then taught and learned. And
such
relationships are about parsimonious tools for prediction.
And yes, we have been fortunate that some math describes some nature.
It is exactly like a model airplane (or whatever). We make the model to
look like the real thing. The real thing knows not of the model that we
built, but if we did a good job, I or you can now look at the model and
"know" just how the real thing looks.
It can also be that the model ignores some of the unimportant details of
the
"real thing" and still be quite useful. Or that it ignores some important
details, but that so long as we keep those boundaries and limitations in
mind
the model is still quite useful for many other things.
I like your example.
The math behind all of our sciences is just like this. *WE* found math
which models reality and because we did such a good job, we can now "do
the
math" and "know" how the real thing should behave.
We can also disappear into the mathematical universe and discover brand
new
relationships there and have some expectation that where such new
territory is
true there, it will probably be found true in the real world as well. One
can
make important discoveries using mathematics and use them to suggest what
can be
searched out and found here. Surprising, at times.
To be a little more specific, in the case of the time constant. we have
theories about current flow, charge, capacitance, inductance magnetism
and
resistance which are borne out by countless experiments and then by
subsequent usage. These theories have all had mathematics fitted to
them,
and by golly everything fits. We can now plug-in values to equations
till
the cows come home and holy-cripes! The real thing does just what the
math
predicted. Based upon the properties we have observed for each type of
component, this math works out such that this 1/e thingy fits just right.
In other words, the answer is: "It just does!"
Yup. In the capacitor case, for example, I idealized it as a simple
differential equation. Real capacitors are more complex, but the ideal is
often
close enough in practice to be useful.
Enjoyed seeing your thunder!
Jon
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
It just turns out to be 63% if you add up allwhen a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
programmers will be calling for you with torches ablaze.On 11 Oct 2004 16:11:16 -0700, alanh_27@yahoo.com (Alan Horowitz)
wrote:
when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?
And whence the number 63%?
How about this:
Charge a 1 farad capacitor to 1 volt and slap a 1 ohm resistor across
it. The resistor current is 1 amp, so the cap discharges, and the
voltage is at first declining at a rate of 1 volt per second. But
1/100 of a second later, the voltage is 0.99 volts, so the current is
only 0.99 amps, so the rate of discharge is only 0.99 volts per
second.
So we write a Basic program:
v = 1 ' charge the cap
for t = 1 to 100 ' then, for 1 second at 0.01 sec steps,
v = v - 0.01 * v ' discharge the cap by 1%
next
print v ' voltage is this, 1 second later
which simulates what I was doing above, but for a full second. The
value of v at the end is 0.36603 volts. That's close to 1/e, not exact
because I took 100 discrete steps, as an approximation to
continuous-time math. With 1000 steps, simulating 1 second of
discharge in 1 millisecond steps, you get 0.367700, even closer.
'e' is just nature's answer to a natural discharge curve.
John
Egads!. You've just given away the secret of Spice Transient Analysis. The