Sujet : Re: Suspension losses
De : funkmaster (at) *nospam* hotmail.com (Zen Cycle)
Groupes : rec.bicycles.techDate : 17. Jan 2025, 21:58:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vmegae$7c27$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
User-Agent : Mozilla Thunderbird
On 1/16/2025 5:56 PM, Radey Shouman wrote:
Zen Cycle <funkmaster@hotmail.com> writes:
On 1/16/2025 12:18 PM, Frank Krygowski wrote:
On 1/16/2025 7:14 AM, zen cycle wrote:
On 1/15/2025 6:39 PM, Frank Krygowski wrote:
>
You should look at the energy used for the controls and think
about what becomes of it. Do that in microcopic detail.
>
If you think it all ends up as heat (IR spectrum) you have a gross
misunderstanding of electronics. First off, the indicators
dissipate energy in the visible light spectrum (this is why LEDs
are more efficient lighting than incandescent, very little energy
is used in the IR spectrum).
OK, a thought experiment: Take an adiabatic container - that is, a
_perfectly_ insulated box (a physical impossibility, but useful for
our analysis). Let the box contain whatever you like - just air,
some solid objects, whatever.
Cut an LED sized hole in it and insert an LED of your choice so it
shines into the box. Turn on the LED.
What happens to the light entering the box? Obviously, you don't end
up with a box full of light, so it isn't stored; it somehow goes
away.
>
lol...no, it doesn't 'go away'. For all intents and purposes it
suffers the safe fate as Schrödinger's cat.
>
And what happens to the temperature inside the box, and why?
>
It will rise somewhat due to the residual IR energy from the
system. Visible spectrum LEDs do emit some IR, just not nearly as much
as incandescents (given the same lux)
>
Answer: The temperature of whatever's inside the box will rise. The
energy put into _all_ spectra by the LED, including the visible
light spectrum, ultimately converts to heat.
>
So _all_ forms of energy convert to heat? You should write a paper on that.
>
>
Secondly, think about your premise that it all turns into heat -
this means no energy is available to do any other form of work.
In physics or mechanical engineering, work is defined as force
moving through a distance, or torque moving though an angle of
rotation. Valid units of measurement are the same as the units for
energy: foot*pounds, Newton*meters or Joules, etc. all of which
(interestingly) can be converted to BTUs, which are normally units
measuring heat.
And in general, you're right, energy converted to heat is not
normally available to do work.
>
I used the term 'work' more generically, in this case running a
program, setting bits in memory, etc. Yes, residual heat from the
process, but energy is used to perform whatever task, Heat is the
result of losses in the system (thermal junctions from die bonds, for
example)
I don't believe that is correct. All the energy used to run a program
does eventually get rejected as heat. I suppose either 1 or 0 bits must
have a slightly higher potential energy, but the net number of each is
not likely to change much, and the energy difference must be small.
Hmmm, so you're suggesting that if we consider computers to be heaters that also perform a computing function, that using it as a heater ostensibly gets us computing functions for free as long as we're expecting the heating function to be primary?
Sounds suspiciously perpetual-motion-esque.
There is a field of study called "thermodynamic computing", which
studies the minimum energy that must be dissipated for various logic
operations. Turns out that reversible computations, ie those that do
not irretrievably lose information, can be more thermodynamically
efficient than those that do lose information. Eventually this sort of
thinking is hoped to enable more economical computation.
RAM vs NVRAM? I think that's been done before.
-- Add xx to reply