October 24th 18, 10:54 PM
posted to uk.radio.amateur,rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Jan 2014
Posts: 329
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4NEC2?
Spike wrote:
On 24/10/2018 07:35, brian wrote:
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. Next
to it, I take a brighter bulb, where I know the brightness. This
light is NOT adjustable and is always the same known brightness. Now,
I move this bulb farther away until it appears to be exactly the same
brightness as the RF powered light bulb. At this point, I know:
1. The distance between the observer (me) and the RF powered light
bulb which I'll call A.
2. The distance between the observer (me) and the reference light
bulb which I'll call B.
3. The brightness of the reference light bulb which I'll call C.
4. 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. I'll assume the reference
light produced 1000 lux. Therefore, the brightness of the RF powered
light is:
1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux
Presumably, the reference light was calibrated for brightness at a
given RF level. Let's say it's 50 watts for 1000 lux. Therefore, the
RF power of the RF powered light would be:
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 equation 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 incandescent 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.
Before moving on to the technological side of the issue, I thought it
best to get the physics right first. Having done that one could match
lamps for all sorts of things and improve the methodology, but if the
premise of the physics is wrong, there's little point in doing that...
In any case, the intensity indicating device will also bring limitations
of its own to the method, but it may be that these are second-order
issues that have only a minor effect on the system's overall accuracy,
bearing in mind the original reason for using the set-up.
So many FURIOUS backpedals.
--
STC / M0TEY /
http://twitter.com/ukradioamateur
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