On 24/10/2018 08:35, brian wrote:
In message , Spike
writes
On 19/10/2018 06:15, Jeff Liebermann wrote:

Spike contended that a distant station cannot tell the difference
between a sending station that has been tuned up using a torch bulb and
one that has used an expensive VNA for that purpose. Via some other
topics, the discussion on the torch bulb vs VNA issue has now reached
this point:

Let's start with an RF powered light at some brightness level.Â* Next
to it, I take a brighter bulb, where I know the brightness.Â* This
light is NOT adjustable and is always the same known brightness.Â* Now,
I move this bulb farther away until it appears to be exactly the same
brightness as the RF powered light bulb.Â* At this point, I know:

1.Â* The distance between the observer (me) and the RF powered light
bulb which I'll call A.
2.Â* The distance between the observer (me) and the reference light
bulb which I'll call B.
3.Â* The brightness of the reference light bulb which I'll call C.
4.Â* I'll call the unknown brightness of the RF powered bulb as D.

Let's say that the observer is 2 meters away from the RF powered
light, and that the reference light is the same brightness as the RF
powered light at a distance of 5 meters.Â* I'll assume the reference
light produced 1000 lux.Â* Therefore, the brightness of the RF powered
light is:
Â*Â* 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux

Presumably, the reference light was calibrated for brightness at a
given RF level.Â* Let's say it's 50 watts for 1000 lux.Â* Therefore, the
RF power of the RF powered light would be:
Â*Â* 632 / 1000 * 50 = 32 watts

A lamp of power or light output P has a light intensity at a point at a
distance r from it that is a function of P/4pi*r^2

Two lamps of differing power or light output, P1 and P2, and spaced
apart will have a point somewhere between them where the light
intensities are equal. At this point the distance from P1 to the point
of equal intensity is given by r1, and for P2 that distance is r2 and we
thus have the equality given by:

P1/(4pi*r1^2) = P2/(4pi*r2^2)

Simplifying:

P1/r1^2 = P2/r2^2

From which it can be seen that, if the power or light output of P1 is
known then:

P2 = P1(r2^2)/(r1^2)

If P1 = 50 units of power or light output, and r1 and r2 are 5 and 2
units of distance respectively then:

P2 = 50 * 2^2/5^2 = 50 * 4/25 = 8 units of power or light output,


Now take your equationÂ* above:

1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux

Using my notation as above, this becomes

P1/((r1/r2)^0.5) = P2

Rearranging this and separating the terms gives

P1/(r1)^0.5 = P2/(r2)^0.5

You seem to have invented the Lieberman Law of Inverse Square Roots...

If you're using incandescentÂ* bulbs , the two bulbs have different
colour temperatures . The dimmer one radiates more power in the red and
infra red or "heat bands"

My good friend Mr Planck sussed this out. If you try to use your eyes to
judge brightness, the human eye spectral response causes the redder one
to look disproportionally dimmer.

I've had these sort of problems trying to simulate sunlight using
tungsten lamps. There's a further problem introduced by the area of the
lamp, which bu@@ers up the inverse square law. You might not have that
problem with a "pea" bulb.

Spike has enough rope to play with already.

Jeff, like myself, tends to give idiots just enough to hang themselves.
He (Jeff), really is very good at it.




--
Always smile when walking, you never know where there is a camera ;-)

Remarkable Coincidences:
The Stock Market Crashes of 1929 and 2008 happened on the same
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Company led to the Great Depression.