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That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:

Via a QUANTUM EFFECT, you fucking moron. ... [Jane Q. Public, 2014-10-03]

Charming. As I just explained, IR detectors don't have to depend on quantum effects. Classical mainstream physics allows a temperature-controlled source to detect IR from the cooler chamber walls as follows:

electricity = (e*s)*(T1^4 - T4^4)

If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).

If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.

If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.

If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.

If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.

... Further, I repeat for about the 100th time that I do not deny that some radiation is absorbed; but then it's just re-emitted. Sometimes, in a non-gray body, in a slightly different form. And ALL of that is straw-man irrelevancy, since no NET radiation absorption occurs from colder bodies to warm, which was the subject under discussion. ... [Jane Q. Public, 2014-10-03]

If you don't deny that some radiation is absorbed, then it should be very easy to write down a simple equation describing the required electrical heating power (not the radiative power out) of a blackbody source.

I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]

Jane, if we can't agree on the meaning of the term "NET", why are you still capitalizing the word "NET"? Screaming the word louder and louder is unlikely to be productive.

1. Can we agree that net heat transfer through a boundary around the source = "radiative power out" minus "radiative power in"?

2. Can we agree that net heat transfer always contains terms in both directions?

3. Can we agree that just because "radiative power out" > "radiative power in", that doesn't mean "radiative power in" = 0?

If Jane answers "no" to any of those three yes/no questions... why?

I don't need to "agree" with you about anything. I've already demonstrated how TEXTBOOK PHYSICS proved you wrong. That doesn't require any kind of "agreement". I'm just wondering when you're going to stop the dishonesty and admit you were wrong. The whole world is going to see it soon anyway, so you might as well "come clean", as they say. [Jane Q. Public, 2014-10-03]

If you're so confident that you're right, why not prove it by taking a few seconds to write dow

You didn't bother to read my reference on pyrometers, did you? ... [Jane Q. Public, 2014-10-01]

That reference shows the object (i.e. chamber wall) temperature has an effect on the temperature controlled cavity (i.e. source). Which Jane denies:

... Radiation from the cooler walls has no effect on the heat source whatsoever. This is a basic requirement of thermodynamics! ... [Jane Q. Public, 2014-09-21]

No, that's Sky Dragon Slayer nonsense. If radiation from the cooler walls really had no effect on the heat source whatsoever, the IR thermometer wouldn't work because the cooler object temperature would have no effect on the temperature controlled cavity whatsoever.

When the source temperature is held constant, its required electrical heating power is an IR thermometer.

Here's one way to see that: draw a boundary around a heated aluminum source. It's heated by constant electrical power flowing in. Aluminum cold walls at some unknown temperature T4 also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out. Using the equation which neglects reflections:

electricity = (e*s)*(T1^4 - T4^4)

If the required electrical heating power is 82.1 W/m^2, then the chamber wall is at absolute zero (-459.7F).

If the required electrical heating power is 55.6 W/m^2, then the chamber wall is at 0F.

If the required electrical heating power is 27.8 W/m^2, then the chamber wall is at 90F.

If the required electrical heating power is 0.0 W/m^2, then the chamber wall is also at 150F.

If the source needs to be refrigerated to stay at 150F, the required electrical power is negative. The same equation can be used to determine the chamber wall temperature, regardless of whether it's warmer or cooler than the source.

That's why when the source temperature is held constant, its required electrical heating power is an IR thermometer. At least, it's a thermometer when using mainstream physics. But Jane's equation is:

electricity = (e*s)*T1^4 (Jane's equation)

Since Jane's equation doesn't depend on the chamber wall temperature, uncooled IR detectors can't see cooler objects in Janeland. And we couldn't possibly have detected the 2.7K cosmic microwave background radiation with warmer detectors. But we did! How? This must be inexplicable to Slayers who are brainwashed into believing that:

... all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of electrical power, therefore the power input and power output are constant. It is not a "gradual" process. ... [Jane Q. Public, 2014-09-20]

No. Again, mainstream physics shows that electrical heating power gradually decreases to zero as the chamber wall temperature increases. That's how uncooled IR detectors can see cooler objects.

Once again, Jane insists electrical heating power = (e * s) * (Ta^4). Once again, Jane's ridiculous e

Wow, Joel, I gotta say (after reading some of the replies on this thread) that this really is pointless. These folks have no conception of the FIRST law of thermodynamics, let alone the second. The argument for warming doesn't even require mentioning the SBE, it only requires the first law, the second law, and a monotonic relation between temperature difference in ANY channel and the rate of energy transfer in that channel, subject to very broad constraints.

Funny, because he's contradicting just about every argument behind the whole idea of AGW. I like how he makes these claims but isn't able to show how it actually works. He claims you can show warming via back-radiation WITHOUT the S-B equation? When it is absolutely fundamental to the very "energy transfer" he is asserting? What garbage.

Where's the math? In the comments you show in your link he also conflates backscatter with the "back radiation". But scattering and reflection are straw-men; they are completely unrelated to heat transfer via "back-radiation", and are 100% irrelevant to Spencer's experiment.

His mention of "empirical evidence" isn't science, it's an assertion of correlation without any causal link. It's a ridiculously weak argument... in fact it's not really an argument at all.

But seriously, just a waste of time. When people just make stuff up and reject the contents of ELEMENTARY textbooks on the subject because they just don't like the conclusion those contents lead to, how can you argue with them? If somebody tries to solve the light bulb problem while pretending that it doesn't primarily cool via radiation and completely ignoring radiation, what can you do?

And this is downright hilarious in context. In incorrectly "solving" Spencer's challenge, YOU ignored basic textbook methods and math to get your answer. You used an imaginary "khayman80" method of arriving at your answer, which not only contradicts everything engineering textbooks say about heat transfer, your methodology directly contradicts the Stefan-Boltzmann radiation law, even though you used it yourself in calculations. Talk about hypocrisy. I repeat: I checked your final "answer" for temperature of the heat source and it violates both the Stefan-Boltzmann law and the second law of thermodynamics.

Further, what he was referring to in the latter paragraph were the comments in the forum... not Latour's analysis.

That's odd. Just yesterday Jane had no argument with Dr. Shore. Now Jane claims that Dr. Shore "FUCKED UP" his physics.

So? I'm still not arguing with him. I'm not even arguing with you. I've already showed you to be wrong. Let's get this straight: THIS "argument" has been with YOU, and ONLY you, and ONLY about Spencer's experiment. It's over, and you lost. All this other crap you bring up is just your way of trying to hide your own failure. It isn't working.

When a body is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate and so its temperature remains constant. When a body is hotter than its surroundings, it radiates more energy than it absorbs, and so it cools..."

NONE of the bodies in Spencer's challenge are "in equlibrium" with their surroundings. None of them. Not one. Straw-man.

Maybe the Slayers could explain how uncooled IR detectors see cooler objects?

Straw-man. Our argument involved gray bodies, not detectors of specific wavelengths or electronics that take advantage of specific quantum effects. But I have an answer anyway: they measure DIFFERENCES, not absolute radiation. You might be interested in THIS, which explains how IR pyrometers work. Hint: they don't work the way you seem to think they do.

And it's a straw-man in a different way: I repeat that I have NOT been claiming that no radiation from a coole

You're trying to solve a heating problem with a lighting solution. You would be better off with a "heating" solution. Perhaps you should be looking at something like this or this or if there's an explosion hazard maybe even this.

You're trying to solve a heating problem with a lighting solution. You would be better off with a "heating" solution. Perhaps you should be looking at something like this or this or if there's an explosion hazard maybe even this.

You're trying to solve a heating problem with a lighting solution. You would be better off with a "heating" solution. Perhaps you should be looking at something like this or this or if there's an explosion hazard maybe even this.

No, this is on par for such devices. There is little to no file system mgmt for field devices or EDRs. Just write and write. The file table is always apart from the circular memory. The amount of capacity to report is a known hack performed long ago for USB memory sticks. This is the same thing in a different package. This is probably all one chip.

You can buy your own (some come with the circular write as an option) and although the capacity is probably truthful, you can put it in a hard drive case and sell it to all the fools you can find.

What, no one ever heard of vacuum lines? Or maybe pressurized lines?

what with all the planes we lose every year to short-circuiting wires.

As I process engineer, I'd like to smack anyone who suggests temperature/humidity chart recorders. They're a pain in the ass, you can't do anything with the data, and there are so many more elegant solutions. Check out http://www.omega.com/ , http://onsetcomp.com/ , or http://www.tiptemp.com/

Yes, pumps use a simple turbine attached to a rotary encoder. They could not afford to have anything more sophisticated and they must keep operating in a environment with little care or calibration.

...let's be clear about this: all you're doing is monitoring and gathering data - there is to be no feedback signal from the homebrew rig to control the valves. There's a whole field devoted to control theory, one that is best not trifled with, especially with industrial processes that can potentially cause fatalities.

If you really want meaningful data from those process streams, you're much better off installing calibratable (calibrable?) flowmeters on those lines that cover the performance range of the process fluids you're working with. If you've got the flow, you don't need the valve position, unless it's for a secondary indication to validate the valve's performance (e.g., position vs. Cv vs. measured flowrate). The flowmeters can be hooked up to provide data for remote collection, or more simply, display data for periodic local reading. Here's a mess to start with. Whomever you buy from, you'll need to develop specifications defining the operating range, operating conditions (pressure, temperature, humidity), power requirements, tolerances, calibration frequency, etc.

They are not all foreign nationals. From http://www.bis.doc.gov/dpl/thedeniallist.asp , is "Omega Engineering", http://www.omega.com/cservice/bdwo.html , a respected supplier of process control equipment located in Stamford CT http://maps.google.com/maps?f=q&hl=en&q=omega+driv e+Stamford+CT&sll=37.0625,-95.677068&sspn=59.11905 9,95.537109&layer=&ie=UTF8&z=15&ll=41.089573,-73.5 25872&spn=0.027913,0.060725&om=1 Omega is not a fly-by-night operation; I can remember getting their catalogs over 10 years ago and had a former neighbor who worked there.

(1) capacitive switches -- inexpensive, digital (on/off)
This reverses the idea of protecting a high-impedance circuit from stray capacitance introduced by a limb or finger -- turn your accidental 'people detector' into just that:

http://www.discovercircuits.com/C/capacitance-sw.h tm

(2) frustrated internal reflection -- not necessarily cheap (needs a camera), or easy (needs video analysis) but can handle multi-touch and large screens
The idea is to shine light in from the side of a class or plastic screen, and have a camera look at the backscatter introduced by finger contact, which scatters the sidways light rather than allow it to reflect at the bounary. The camera turns the touch events into a video stream which is then analyzed to compute touch events.

http://mrl.nyu.edu/~jhan/ftirsense/index.html

(3) strain gages -- not necessarily cheap ($5 - $8 for each strain gage) but can provide a very sensitive analog signal with wide dynamic range.
Put some strain gages on several centimeters of half-inch square steel tube and you can easily measure touch events, as well as strong pushes.

http://www.omega.com/literature/transactions/volum e3/strain.html

Here is a simple, inexpensive (if you make it yourself) amplifier for strain gages that I've tried, and can vouch that it works well:

http://www.staramp.com/

A couple of Thermotron http://www.thermotron.com/ ovens,
some Omega controls http://www.omega.com/
all hooked up by their industry standard ethernet interface http://www.cisco.com/univercd/cc/td/doc/cisintwk/i to_doc/ethernet.htm
to a PC http://www2.sjsu.edu/faculty/watkins/pc.htm
and you have the same thing for less money.

Tell NASA to shoot for the moon NOT the moonpie!

I have something similar (NOMAD OM-43) that I use to record indoor temperature and relative humidity. It can be programmed and downloaded via an RS-232 port. It will run for a very long time on a lithium battery. If can be useful if you have stuff that is sensitive to temperature and humidity, like musical instruments, photographic film, magnetic tape, etc. It's also interesting to have a historical record of the actual conditions, as opposed to what they are supposed to be.

At the risk of nuking my friends at Omega into the ground,

http://www.omega.com/literature/posters/

A.

I've used Sensaphones to monitor remote computer rooms. It has alarm input contacts on it so you can connect water sensors and such. It calls a list of phone numbers when triggered by an alarm or low/high temperature and gives the problem and status to you in a voice announcement.

There are other similar slef-contained alarm dialer solutions

I have one at home and it works great.

It's a thermocouple if you use it to measure a temperature difference.

It's a Thermator if you use it to produce electrical power.

While not a major breakthrough, it's good to find more ways to use waste heat and up the efficiency of a system.

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Having a colorful poster of the radio frequencies hanging in your office really makes you look like a geek.