[R-390] Dead horse foaling,. Monkey wrench thrown!

[Bill Hawkins]

Gotta love that subject line! Joe Foley does have a way with
words. But it's time to pour a can of Aircraft Stripper on
this to remove the gum from the works.

David Wise said:

Ok ok, for those who don't believe the math, I did an experiment.

[description of setup that uses a relative light measurement (not
exact) as a precision comparator. That is, the conditions are
adjusted to give the *same* amount of light. The same amount of
light from the same bulb filament is a pretty good indication 
that the same amount of energy is dissipated in the filament.]

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Roger replies:

[and I insert comments in brackets. These are meant to inform
and not to inflame. I'm too old and have made too many mistakes
to be thinking that I am smarter than someone else.]

Sorry Dave,

Not the same experiment.

The AC volt meter was just reading some value damped by meter
movement mass and springs. Those meter reading are very misleading.

[The meter was used on the primary of the transformer, far away
from the diode-distorted voltage that lights the bulb. If these
meters are "very misleading" why are they so popular?]

You have a dynamic resistor. It changes resistance with changes in
applied power. What happened to the current when you changed
the voltage that changed the resistance, changed the temperature
that changed the light  that changed the light meter that changed
the voltage meter.

[Actually, the experiment did not change the resistance, temperature
and light at all. Those things are all dependent on each other. If
the light is the same, so are the temperature and resistance. The
experiment measured the same light level for both cases. The light
from a lamp is related to power. In fact, small changes in power
make large changes in light, near the rated voltage of the bulb.]

When you installed the diode and read the lamp output level with the light
meter,
was the lamp output 1 /  2 or 1 /  4 or some other output of light level?

[No. The light level was the same. The voltage of the AC line
was different.]

Lamp conversion to visible light is not linear.
Filament conversion to heat is not linear.
Either one will dissipate all the energy.
Into which energy band (heat , light , X-ray) who knows.
How much of that energy does our test instrument capture and report?

[True, it's non-linear. But there's no way that light can dissipate
all of the energy in a heated filament lamp.]

Even here in your test you expect a uniform average of light to reach
your meter. The lamp make have dark and bright sides. Changing
the power across the lamp may or may not sow the same dark and
bright sides. (hot spots in filament and filament hangers shade the field)
What makes light output to meter reading a real average of light emitted?

[Seems like these variables can be controlled by keeping the
bulb and the light meter in exactly the same position while you
put a diode in series with the bulb on the secondary side of
the transformer. Using comparison at the same light level lets
you ignore variables that don't change during testing.]

I do not know how to make all light emitted be absorbed by the sensor
element.
So how do we ensure that light emitted to light received is linear?
As you changed the light pulse rate from 60 hertz to 30 hertz how
was the pulse filtered by the light meter circuit to hold an average
meter reading.

We think the thermal resistance of a lamp gives a flicker free
output. While the light is better than a florescent lamp, it still has
a flicker and you reduce the flicker rate from 60 to 30 and thus
exaggerate the factor in the test setup.

[Dunno about David's light meter, but I've never seen one faster
than half a second. 30 and 60 would not matter. Not sure the bulb
filament cools fast enough to follow that, either.]

Install the diode and lamp.
Adjust the power to get a lit lamp and a light meter reading.
remove the diode.
Remeasure the lamp light output.
Is the output twice, four times or some other value more?

[That's what we can't do without expensive equipment. Adjusting
for the same light level is simple, cheap and accurate.]

This still means nothing. It just lets you infer some thing about
the metering equipment. You can infer any thing.

[Well, no, I can't infer that my meter is wrong if a certified
calibration lab says it is right, and thousands of other labs
capable of measuring the performance of the meter would agree.]

But if you start on the correct premise, you can infer correct
thing about the response of the meters to real conditions.

[I think the correct premise is that power and light are related,
such that the same amount of light means the same amount of power.]

Measure, the voltage, the current and the resistance
with the diode and with out the diode.
Compute all permutations of voltage, power, current and resistance.
Show that all equations balance with out error.

[I think that there is only one permutation, unless you mean
different ways of writing E=I*R and P=E*I. I'll pass on the
"balance" stuff because so many people have already done it.]

The meter reading and math are models of the real world.
If the meters, math and models do not balance, then
its the meters, math and model that is in error, not nature.

[I think the meters, math and models do balance, and that many
other observers would agree with me. The math and the models
were accurately drawn from observing nature. Meters can be
misused, but they are used for their intended purpose in David's
experiment.]

Roger.

[Bottom line: a diode in series with a resistive load in an AC
circuit cuts the power in half. Since the resistance stays
pretty much constant in a heated filament emitting the same
amount of light, E and I have a constant ratio. The correct
equation for half power is P/2 = E/(sqrt 2) * I/(sqrt 2) or
P/2 = (E * I) / 2. The equation for quarter power is
P/4 = E/2 * I/2. But we have half power, not quarter power,
so the voltage cannot possibly be E/2. You get the same result
using P = (E * E) / R. The value of volts for half power must
be E/(sqrt 2).]