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3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting
An anonymous reader writes: Medical researchers have been steadily building evidence that prolonged sitting is awful for your health. One major problem is that blood can pool in the legs of a seated person, causing arteries to start losing their ability to control the rate of blood flow. A new experimental study (abstract) has discovered it's quite easy to negate these detrimental health effects: all you need to do is take a leisurely, 5-minute walk for every hour you sit. "The researchers were able to demonstrate that during a three-hour period, the flow-mediated dilation, or the expansion of the arteries as a result of increased blood flow, of the main artery in the legs was impaired by as much as 50 percent after just one hour. The study participants who walked for five minutes for each hour of sitting saw their arterial function stay the same — it did not drop throughout the three-hour period. Thosar says it is likely that the increase in muscle activity and blood flow accounts for this."
since smokers tend to smoke literally every hour and a cigarette takes 5-6 minutes to smoke.
I mean if I get up and go outside for a quick drag once an hour, that's a five minute walk right there.
Time is money!
You should see the smiles on the faces of slashdotters as they read this news. Seriously, hack into their computers and activate their cameras.
Don't waste your vote! Vote for whoever you want, unless you live in a swing state it won't matter anyways
Cool, just saw this on the WATCH keynote - reminders showing how much you sit/walk/stand in an hour vs the recommended levels! On the topic: I wish airliners take a note of this and show a notification, buzz the arm rest every 2 sitting hours to prompt passengers to stretch/stand for a few mins.
Right. Get up out of your chair once an hour, leave the office, and take a 5 minute walk. Come back and get back into work. Total time required: 10-15 minutes.
I don't know too many bosses who would be cool with that.
Now that I've stopped sitting and walked around for 15 minutes..... what do I do to cure the problems caused by walking for 15 minutes?
Perhaps sit for 45 minutes for every 15 minutes of Walking?
allow us to waste about 10% of the day just walking around? Given a typical developer work day of ten hours, that's fifty minutes per day. Given the hard time we get about getting up for coffee or to go to the bathroom, that will never be acceptible in corporate America.
Does your boss really micromanage your time that much?
Look for a different job. Yeesh.
Here http://pomodorotechnique.com/ a simple technique for work time use hygiene.
Really helped me put sit/walk ratio under control, among other good things. Doesn't help if you are deep in programming something big or a big part of something even bigger. But if you don't require a multi-hour flow and concentration, this may be the right thing for you.
I wonder how much of this same effect can be achieved by alternating standing with sitting. I have a sit/stand desk and switch back and forth all day long. It feels much better, I know that much.
Of course, I also take occasional walks, mostly when I need to think without distraction.
Note to ACs: I usually delete AC replies without reading them. If you want to talk to me, log in.
I don't have any evidence that standing will help as much as walking, but I was thinking this is why we should have more standing desks at the office. By standing desks, I mean the ones that convert from sitting to standing easily and encourage people to change their body positions often during the work day.
It's not just a good idea, but it's probably something to keep your work population alert and productive!
Wow, you guys are mean. It sounds like this poor guy works at EA then the moderators make him as a troll. Just plain mean.
to change mandatory work breaks from like 10 minutes per 4 hours worked, to 20 minutes per 4 hours worked.
It would be more interesting to know if the 45 minute walk a day rule accomplishes the same or similar. Personally I would dislike breaking my concentration once an hour...but going on a 45 minute walk in the morning is easy.
How long do those trips to the bathroom or for coffee take? You might already be getting your hourly 5 in...
For large sets, this will be our guide even unto death, for the LORD will work for each type of data it is applied to...
v-sit. feet on ottoman, back reclined, butt low, torso-weight on back, leg weight on heels and haunches, arm weight on elbows, hand weight on the heel of the hand, proper security-guard chair, well padded, designed for long-term sitting. wrist flexed downward (by the bigger muscle), neck flexed downward (by the bigger muscle), abs flexed instead of lower back -- again, the bigger muscle works, the smaller muscle doesn't.
it's been 21 years of programming, 15 in this same exact chair. good weight, good energy, good appetite, good drive. healthy all around, no pain, no injuries (typical broken bones as a child, including a wrist), age 35.
http://www.globaltotaloffice.c...
Three short walking breaks from hours of watching TV from the couch? Too much. Cant be done.Let me know if I summon enough will power to take three short 5 minute breaks from hours of eating junk food while watching TV, will it count? Hey! Got a bright idea. Should patent it. How about combining the bathroom breaks from TV watching to get both the walking break and the snacking break? All I have to do is to remember to leave the beer by the laz-e-boy.
sed -e 's/Chuck Norris/Rajnikant/g' joke > fact
My boss is in a different country with a 9 hour time lag.
I should use this sig to advertise my book ISBN-13 : 978-1501515132.
This is why coffee is good for your health. It makes me get up and walk twice an hour. Once to get it, and once to put it back.
I wonder if 4-5 cigs per day used in a sauntering smoke break would be healthier than stoking from DVT by sitting on your keester all day.
Legalize the constitution. Think for yourself question authority.
I learned years ago to work in CD increments. As in, put on a CD, work, when the CD is over get up, pee, get coffee, and walk around a bit to get the kinks out. Repeat as needed.
Cool! I'll get up right now, hop into my exoskeleton and walk around some. It's pimped out with camouflage paint, A/C, a radio, AND a leather seat -- no problems for me! (No in-dash GPS though; they're too darn expensive.)
And what's more, I generate much more energy that I use when running on a treadmill!
Now if I were to ACTUALLY take this seriously, I'd be up and walking for the rest of my life since I could never sit down AGAIN. Hell, I'd better order a stand-up coffin now and avoid the rush.
If the universe is someone's simulation -- does that mean the stars are just stuck pixels?
but if I took a 5 minute break every hour I'd get fired...
Hi! I make Firefox Plug-ins. Check 'em out @ https://addons.mozilla.org/en-US/firefox/addon/youtube-mp3-podcaster/
I use a program called Workpace, which monitors my typing and clicking and enforces an 8 second "micropause" every four minutes of activity. It also enforces a longer 5 minute pause every 45 minutes. Using this for a few months has completely cured my RSI-induced wrist pain. I recommend anyone who works at a computer to give it a try. There's also a freeware version called Workrave, which essentially does the same thing but isn't as slick.
Nuff said. I hope.
"The greatest lesson in life is to know that even fools are right sometimes" - Winston Churchill
No need to go anywhere!
"The Most Fun Possible on 4 wheels" is at SunBuggy in Las Vegas
I wonder if there any in-place exercises, such as isometrics, that would provide at least some of the benefits of walking without having to leave the desk or workstation. Not that I'm averse to a walk once in a while, but some bosses are averse to employees not being chained to their desks...
'The Economy' is a giant Ponzi scheme whose most pitiable suckers are the youngest among us and the yet-unborn.
I drink about a glass of water an hour while working, which is healthy by itself, and having a bathroom break every 30 minutes helps negate the effects of sitting. It's like having a built-in pomodoro timer.
Since I started walking breaks every workday I'm more productive. Walking works. We aren't meant to sit around all day.
My employer bought me one.
Then again, I'm self-employed.
www.clarke.ca
I know this a bit off-topic, and for that, I apologize, profusely. Nevertheless, when I first glanced at the paragraph from which the above quote is taken, my first thought was,
"Isn't that supposed to be "loosing?"
I now know, beyond doubt, that I've spent too much time online. Thank you, and good night.
Except that Smokers tend to stand around in groups while smoking ... the article says that one should go for a 5 minute walk every hour, not just stand in the one spot.
No. We've never agreed that the unchanging power input (my "constant electrical heating power") is "82 W/m^2". I've repeatedly failed to explain that the constant electrical heating power would only be "82 W/m^2" if the chamber walls were 0K blackbodies.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
Note that the constant rate of Dr. Spencer's electric heater would equal zero if the chamber walls were also at 150F. So any calculation of this crucial constant rate would also need to be zero in the case of chamber walls at 150F.
Since Jane's "82 W/m^2" value isn't the constant electrical heating power needed to keep the source at 150F inside 0F chamber walls, it isn't held constant. Here's where Jane actually calculated the constant electrical power heating the source inside 0F chamber walls:
So Jane's source needs 55.6 W/m^2 of constant electrical heating power to stay at 150F inside 0F chamber walls. This value is held constant. After the enclosing shell is added and temperatures stabilize, conservation of energy demands that net heat transfer out equals Jane's 55.6 W/m^2. Does it?
No, I drew that boundary for both initial and final conditions to guarantee conservation of energy. In fact, I repeatedly suggested that you check your work by drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed.
Let's do that:
Jane's constant electrical power of 55.6 W/m^2 flows into that boundary. At steady-state, power in = power out. But power out through that boundary is the
Once again, the constant electric power is sufficient to heat the source to 150F when it's surrounded by chamber walls at 0F. That's the initial condition in the experiment that we agreed on. Your "82 W/m^2" value isn't the constant electrical power sufficient to heat a 150F source inside 0F chamber walls.
Again, if you want to see why your calculation doesn't yield the power input to the heat source, just ask what power input would be necessary if the chamber walls were also at 150F. In that case Dr. Spencer's electric heater wouldn't be necessary, so that power input would be zero.
Since your "82 W/m^2" calculation can't do that, it's not the electric heater power that's held constant. On the other hand, your 55.6 W/m^2 calculation would be zero if the chamber walls were at 150F. So it represents the constant electrical power in your analysis. Hold it constant as Dr. Spencer said, and you'll obtain the correct solution if you correctly apply the principle of conservation of energy.
No, I explained why you can't add heat transfer from heat source to the inside of the enclosing plate to the heat transfer from the outside of the enclosing plate to the wall to get 55.6 W/m^2 from the shell to the chamber walls. Again, that's because any heat transfer which doesn't cross the boundary can't be included because it can't change the total amount of energy inside the boundary.
No. I've assumed that the electrical power heating the source to 150F inside 0F chamber walls is constant. (Note that this constant rate would be zero if the walls were at 150F.) That's the assumption we disagree on. I never assumed the total radiant power output of the heat source was also being put out by the outside of the hollow sphere. Maybe the fact that we disagree about what's held constant (the electrical heating power to keep the source at 150F inside 0F chamber walls) is leading to yet another miscommunication?
Again, we disagree about what's held fixed. That value you keep calculating isn't the constant electrical power heating the source.
In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
In my interpretation, Dr. Spencer's challenge is basically: "Assuming an electric heater pumps energy at a constant rate to the source, does the source temperature change after a passive plate is added?"
You've repeatedly noted that there are no other factors involved in calculating your 82 W/m^2 (41886.54 W) value. So if it's held fixed, the source temperature is also held fixed.
So it seems like in your interpretation, Dr. Spencer's challenge is basically: "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
Is that right?
No. Holding constant the electrical power heating the source is very different than holding constant the source temperature. Like Jane, let's assume the source temperature is constant (rather than the electrical heating power) and use Jane's equation and notation:
Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 ) = (e)(s) * (8908858139.78) = 55.5913 W/m^2
Now add the hollow sphere and draw a boundary between the source (T1=150F) and the inside of the hollow sphere (T2). A different "electricity_final" flows in, and heat transfer p(12) flows out.
electricity_final = p(12) = (e)(s) * ( T1^4 - T2^4 )
Now draw a boundary between the outside of the hollow sphere (T3=T2) and the chamber walls (T4=0F): "electricity_final" flows in, and heat transfer p(34) flows out. Since power in = power out:
electricity_final = p(34) = (e)(s) * ( T2^4 - T4^4 )
Combine these two equations:
T1^4 - T2^4 = T2^4 - T4^4
Solve for:
T2 = T3 = 305.47K = 90.176 deg. F.
electricity_final = 27.8 W/m^2.
So if the source temperature is held constant at 150F, adding the hollow sphere reduces the necessary electrical heating power to keep the source at 150F by a factor of two, from 55.6 to 27.8 W/m^2.
Can we agree on that?
And that's why what you're calculating isn't Dr. Spencer's electrical heating power, because it should be "zero" if the chamber walls are also at 150F.
If you draw a boundary around the heated source, you have to account for the 0F chamber walls because they're radiating power in through the boundary. Otherwise you're not actually calculating Dr. Spencer's electrical heating power, or you misunderstand conservation of energy.
Jane, didn't it seem odd that you interpreted Dr. Spencer's challenge to mean "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"
How is that different than asking "Assume x = 150 forever. Will x change?"
Isn't that a silly question? Shouldn't you at least consider the possibility that you've misinterpreted "power input to the heat source"?
I just showed that Jane/Lonny Eachus solved the "correct answer" to a different question. Instead of holding the electrical heating power constant like Dr. Spencer did, Jane/Lonny held the source temperature constant. In that case, the electrical heating power required to keep the source at 150F drops by a factor of two after the enclosing shell is added. This shows that holding the electrical heating power constant like Dr. Spencer did is different than holding the source temperature constant like Jane/Lonny did.
No, in your example the electrical heating power drops by a factor of two after the enclosing shell is added. And once again, your calculation of the power sufficient to heat the heat source would be exactly the same if the chamber walls were also at 150F. But the right answer there is zero, because an electric heater wouldn't be necessary. Is this really so hard to understand, or are you deliberately spreading misinformation?
No. Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 )
So are you disputing that power in = power out through a boundary where nothing inside that boundary is changing with time? Or are you disputing that the radiation from the chamber walls passes through a boundary drawn just inside them?
And again, if you keep ignoring that "power in" half of the equation that all Sky Dragon Slayers miss, you'll have to keep wondering why your "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!
Can we agree that the required electrical heating power would be zero if the chamber walls were also at 150F?
It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.
It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.
It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.
Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.
Once again, no. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
"Power in" has to include the radiative power passing in through the boundary. Otherwise energy isn't conserved, because power in = power out through any boundary where nothing inside that boundary is
Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.
For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.
Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.
Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.
In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Right?
Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.
But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.
... or maybe we disagree about which variable to hold constant.
Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change. Solving this problem using both sets of boundary conditions shows that Jane's solution forces electrical heating power to drop by a factor of two after the shell is added.
These two sets of boundary conditions are very different, just like Neumann boundary conditions are different from Dirichlet boundary conditions. Upon hearing that a disagreement might be caused by holding different variables constant, a real skeptic might consider working the problem again while holding that other variable constant. But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant. Jane even insists he held electrical heating power constant, despite the evidence.
So Jane won't solve this problem with the electrical heating power constant. That's unfortunate, because it's critical:
"... critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside. ..."
1. Holding electrical heating power constant while adding an enclosing shell is like doubling CO2 while holding solar heating power constant, then calculating how much Earth's surface warms.
2. Holding source temperature constant while adding an enclosing shell is like doubling CO2 while holding Earth's surface temperature constant, then calculating how much solar heating power would have to drop to keep Earth's surface temperature constant.
Even if Jane doesn't want to solve that first problem, he should recognize that it's different from the second problem Jane actually solved.
To see this difference, solve a problem with Neumann boundary conditions:
"In thermodynamics, where a surface has a prescribed heat flux, such as a perfect insulator (where flux is zero) or an electrical component dissipating a known power."
... then solve the same problem with Dirichlet boundary conditions:
"In thermodynamics, where a surface is held at a fixed temperature.
Dr. Spencer's thought experiment placed Neumann boundary conditions on the source and Dirichlet boundary conditions on the chamber walls. Instead, Jane placed Dirichlet boundary conditions on the chamber walls and the source.
No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."
Jane's even stumbled across this point:
Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?
If Ta = Tb, no electrical heating power is required. But radiant power output of (a) doesn't change. So radiant power output can't be equal to electrical heating power. Using conservation of energy, can you write down an equation which yields the required electrical heating power given Ta and Tb?
So Jane claims:
electrical power per square meter = (s)*(e)*Ta^4
The actual answer is:
electrical power per square meter = (s)*(e)*(Ta^4 - Tb^4)
Since Jane refuses to include a term accounting for radiation from the chamber walls, Jane's equation is saying that no radiation at all is absorbed by the warmer source. Why?
Of course it is! Again, this is just Sky Dragon Slayer nonsense. Absorption doesn't work like Slayers imagine. It's controlled by the surface's absorptivity, which doesn't change if the source is slightly warmer or cooler than its surroundings. All that's required for the source to absorb radiation (from warmer or colder objects) is having absorptivity > 0. Since the source has absorptivity = 0.11, some radiative power from the chamber walls is absorbed by the heat source.
Jane's been regurgitating Slayer nonsense for years:
Then how do uncooled IR detectors see cooler objects? How did we detect the 2.7K cosmic microwave background radiation with warmer detectors?
No, that's a Slayer fantasy. On the atomic scale, absorption of radiation doesn't depend on temperature because individual atoms don't have temperatures. Only very large groups of atoms have temperatures. Individual photons also don't have temperatures. Very large groups of photons from a 10C warm object have slightly different average wavelength curves than a -10C cold object, but they're very similar. This means that even if temperature somehow applied at the atomic scale of absorbing individual photons, an atom couldn't tell if a photon came from the 10C warm object or the -10C cold object.
Jane's equation claims "none at all":
electrical power per square meter = (s)*(e)*Ta^4
Since Jane's equation for required electrical power doesn't even include a term for radiation from the chamber walls, Jane's equation wrongly says that no radiation at all is absorbed by the source. None. Zero.
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source. Slayer "physics" are incoherent nonsense.
Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary. But "power in" through a boundary around the heat source looks like this:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
Jane refuses to account for the chamber wall radiative "power in" which would only be true if the source didn't absorb any of that radiation. Zero.
That's ridiculous, Jane. I'm just noting that the chamber walls are hotter than 0K, so they emit radiation into a boundary around the heat source. Therefore Jane's wrong to ignore that radiation when applying the principle of conservation of energy:
It would only be valid to omit the term describing radiation from the chamber walls if the source absorbs none of that radiation at all. This would only be true if the source's absorptivity = 0. But then its emissivity = 0, so it also couldn't emit any radiation, so it couldn't be a heat source.
So the only "heat source" where we could validly ignore the radiation from the chamber walls would be a perfectly reflective "bobble" from Vernor Vinge's Marooned in Realtime. I assure you that at our current level of technology, we haven't managed to build such a device. And even if we could, it wouldn't be a heat source.
It's truly surreal to watch Jane repeatedly double-down on nonsense which Jane claims is too ridiculous even for Sky Dragon Slayers (as if that were possible!).
Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly object to including a term for radiation from the chamber walls in his calculation of required electrical power. Since Jane doesn't even include that term, Jane's assuming that warmer objects absorb no radiation from colder objects.
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body. Otherwise Jane's calculation of the required electrical power would include a term for radiation from the chamber walls. Since Jane adamantly insists that this term can't be included, Jane's calculation assumes that no radiation at all is absorbed by the source. None. Zero.
I never said Jane objected to a term for "electrical power". I said Jane repeatedly objects to including a term for radiation from the chamber walls in his calculation of required electrical power. And Jane continues to do this:
Ranting about imaginary violations of the Stefan-Boltzmann law won't help Jane understand physics. It might help Jane to draw a boundary around the heat source and think carefully about exactly why Jane keeps ignoring the heat radiated in from the chamber wells. Accounting for that radiation doesn't "violate the Stefan-Boltzmann law" but ignoring it violates conservation of energy.
Why does Jane think the second part follows from the first? It doesn't. For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls. Maybe Jane could explain why he wrote "therefore" when his reasoning fails to describe even a simple black body problem? (Keep in mind that the gray body equation has to reduce to the black body equation when emissivities = 1.)
Jane/Lonny Eachus can capitalize "NET" all he wants, but it doesn't change the fact that
My equation for electrical power is different than the equation for radiative power out, which is why it's bizarre that Jane keeps using the equation for radiative power out to determine electrical power. That's what I've been trying to tell Jane: we don't disagree about the equation for radiative power out. The equation for radiative power out is simply a part of the equation for conservation of energy: power in = power out through a boundary where nothing inside is changing. That's why we need to use a heat transfer equation to determine electrical heating power, not just an equation for radiative power out.
I've already agreed that it's not necessary to account for cooler bodies in the temperature versus power out equation. Again, we're not disputing the equation for radiative power out. We're disputing the equation describing conservation of energy around a boundary drawn around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source
Since power in = power out:
electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source
This doesn't violate the equation for radiative power out. It simply uses that equation to account for the power flowing out of the boundary, and uses that same equation for radiative power to describe radiative power flowing into the boundary.
Jane's been calculating the required electrical heating po
After this thread is locked, this conversation can continue here.
If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out". For instance:
That's exactly the equation Jane should be using to calculate electrical heating power! It has separate terms for "power in" and "power out" so it can describe power entering and exiting a boundary. If Jane would use that equation, he'd honestly be only saying there is no net "radiative power in" from cooler to hotter.
Instead, Jane insists that electrical heating power = (e * s) * (Ta^4). Jane's ridiculous equation doesn't just say there is no net "radiative power in" from cooler to hotter. Jane's wrongly saying the source absorbs no radiative power at all.
So Jane refuses to retract his absurd claim that view factors vary as the radius ratio, which violates conservation of energy. A cynic might have expected as much, given how Jane flagrantly violates conservation of energy by incoherently ignoring radiative power passing in through a boundary around the heat source.
Jane's campaign of educating ignorant, stupid physicists about physics has only just begun. Jane still needs to educate Prof. Brown and Lonny Eachus still needs to educate Dr. Joel Shore.
Then, Jane/Lonny Eachus needs to educate the "ignorant" and "stupid" American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.