aussierob wrote:
> Mr Forster you are correct to the inth degree. However, I stated at the end of
> my previous posts, that there are "other
> factors" too ! However, I stayed with the basics. P=E x I
>
> Yes, I have put an accurate ohmmeter across a bulb hot and cold and yes you
> are correct the resistance does change which will effect the outcome. But, in
> the case I'm writing about the wattage of the glo plug would only be higher
> and of course, probably, mean certain death for the glowplug
>
> That is a fact in my, "simple math" case and your "even more" scientific
> approach.
>
> So, we did arrive at an anwer that is far greater than the 75 watts... yes! I
> did state that there was an inverse law involved.
> Let me state it "very" simply ...if you "halve the voltage" there are real
> big approximate chance that you will "Not Halve the light" but that you
> will get a quarter of the light. Remember, this is not NASA ...but a simple
> little tail light bulb or a
> gloplug.
There is apparently some confusion of the 'laws' involved. I just posted a
description of the 'inverse square law' In retrospect, what you are citing is
the fact that power varies as the square of the applied voltage (for constant
resistance), which is correct.
> Now, let's go back to a guy whose jeep taillights were very dim. Is it a fact
> Mr. Foster in the real world "Basic Math" that if this gentleman inadvertantly
> did put a 12volt bulb into a 6volt jeep that his rear light would be
> extremely dim . Regardless of your post, saying you can't use simple math.
> This little tail light will not attract one single sumers eve' dancing moth !
> Not half the voltage half as bright, but, about one quarter as bright.
I agree that it obviously dimmer, but because of the complexity of the problem,
a simple analysis does not yield valid results.
Here's how it's done (if anyone really cares) To get the correct answer, you
would have to know the temperature-resistance curve of the lamp and then do an
iterateive solution with the radiation equations to determine the ultmate
filament temperature. Then, knowing the filament temperature, you could compute
the radiation spectrum (black body curve). Given that spectrum, you then do a
point-by-point integration with the spectral response of the human eye, to get
the percieved brightness.
> Now, if that is a 6 watt tail light... just how much light would we get using
> my basic math and your more thorough approach?
I don't know. It's a mess to calculate it.
> Just how are these guys on the MIL-VEH going to measure the hot and cold
> resistance of a bulb filiament. It takes about a second for the white hot
> filament to cool off beyond a red glow to a normal cold look. Then, it
> further cools down to room
> temperature. With the $40 meter from Radio Shack, do you think there is enough
> accuracy for a Mil-Veh guy to get the readings in time? I don't !
See Richard Nolton's post.
> The damm needle would barely reach a stable reading in one second. A digitial
> would hop and hunt all over the scale looking for its stable reading. By that
> time it would be all over.
>
> Mr Forster, let us take your Non basic approach to P = E x I ...then you
> were a little lapse too, in your description.
> What if the filament had a "Coiled Coil" manufacture ? Would we now bring
> inductance into the formula ? Of course we would
No. Any transients with the filament inductance are far too short to have any
effect at all on such a DC circuit.
> In a purist sense, especially at the instantaneous moment power is appleid.
> There has to be something induced in each turn of a coil, right ?
The effect is less than trivial.
> But, Mr Forster, I gave a basic approach that "hopefully" was simple enough
> to grasp what the devil might happen, if and when you double or halve the
> voltage in our old vehicles from 6V to 12V and vice versa.
Why don't we just cut the issue down to it's simplest core: The tail light
could well be dim because it's a 12 volt bulb on a 6 volt system, and just leave
it there ?
> Mr Forster, it is a fact that if I was given a 12 volt electric wiper motor
> that might wipe across the windshield and back in two second on 12 volts
> ...that on 6 volts it would NOT perform the same task in 4 seconds.
> Now, I wont even attempt to entertain the losses in the gear train or the
> pressure of the blade on the glass, either.
> Of course there are many factors involved Mostly, I feel there is the "kiss"
> factor that needs to be employed.
Again, this is not a simple problem. Fundamentally, the operating speed of a
motor or engine increases until the load torque is equal to the torque produced
at that speed and equilibrium is established. When you turn up the voltage or
step on the gas, the torque produced by the engine exceeds the load torque, and
the system accelerates due to the excess torque. As the speed goes up, so does
the load torque, so there is less available to accelerate the load. Eventually
equilibrium is re-established at a higher speed.
Thus, the speed depends on the torque-speed characteristisc of both the source
and load. With DC electric motors, it depends on the type (series, shunt, or
compound). Real life loads are not linear, and the torque-speed characteristics
are a combination of a constant (break away), linear with speed, and power-law
(with the square and cube of speed). such as aero-dynamic loads..
I'm a real fan of KISS but only as long as it works. Sometimes there aren't
simple answers.
> I think there are guys on our list that basic algebra might be not their best
> subject. However, I'm sure though, they could put and engine and tranny
> together with their eyes closed.
And I instantly yield to them in their field of expertise. These are not simple
issues, except in a qualitative sense, and I have tried to convey that.
> You are correct in you thinking Mr Forster. But this formula of Power = E x
> I
>
> Reagrds in the purist sense
> Rob Pearson Upstate NY area.
> A litter of jeeps
Take care,
Some WW II radios and an old VW in Boston
-John
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