David Martin said:Matt B said:"David Martin" <[email hidden]> wrote in message
news:[email hidden]...
Quoted message said:Matt B wrote:
>>>Probably not as soon as he would have seen the pedal reflectors.
>>
>>Thus indicating a lack of knowledge on the laws of physics.. Ever heard
>>of the inverse square law?
>
>It's no to do with the intensity, but to do with the fact that the pedals
>are moving, that makes the reflected light from them more conspicuous.
>The
>effect is similar to that of flashing amber lights, like those used to
>make
>tractors and other slow vehicles more conspicuous.
>
>BTW the inverse square law also applies to the powered lights on a bike
>;-)
Very good. Now, as your follow up to show you understand how the
inverse square law applies.
Even though, As I've already mentioned, the intensity is not the only factor
influencing the conspicuity, the fact that it is moving is also important.
Quoted message said:A car light produces 100 units of light per unit area at a distance of
1 m. A bicycle rear light 1 unit.
Taht should be about the right ratio for a 100W car headlamp to a 1W
cycle lamp.
At which distances will the cycle light be brighter than the reflected
light of the headlamps assuming the reflector is 100% efficient.?
Is this a trick question?
I'm no physicist, but I'm game for a laugh... The ratio of intensities is
100(2d)**2/d**2, as the distance that the reflected light has travelled is
twice that of the cycle lamp? Which is a constant 25:1? The reflected
light will always appear to be 25 times the intensity of the cycle lamp
light?
Quoted message said:How much brighter will the cycle light be than the reflector at 50m?
0.04 times?
I can't help thinking I've made a fool of myself here, but, hey - who cares
;-)
Please enlighten (NPI) me David, you seemed so convinced I was barking up
the wrong tree :-)
You can treat both as point sources. It is the intensity of the source
that varies.
As both reflector and light are the same distance away they light
coming from them falls of at exactly the same rate, so at 1m, 100W
light is reflected (100% efficciency) by the reflector, and 1W is given
out by the light.
At 10m, the amount of light falling on the reflector is 100/d**2 or 1W
so the amount of light the reflector can 'produce' is 1W, the same as
the bike light.
You've missed out the gain of the headlamp's reflector.
Quoted message said:
At 20m the effective power of the reflector is 0.25W, the light still
produces 1W.
The reflector isn't isotropic either, so it also has gain (1)...
Quoted message said:At 50m the effective power of the reflector is 0.04W, the light still
produces 1W.
which may also have gain due to its reflector, but probably less than
the gain of the reflector at(1).
Quoted message said:
There are obvious simplifications here, but the principle is
essentially correct.
It is, but missing out the three gains makes the result meaningless.